Transcript for:
General Organic Chemistry (G.O.C) Energetics Session Notes 📘

[Music] hey guys good evening and welcome back again to your an academy Nate English Channel I hope all of you are doing amazing all of you doing great so my dear students quickly let me know in the chats if all of you can hear me if I'm perfectly Audible and visible to everyone of you let me know in the chats quickly with some fire emojis yes good evening people good evening good evening and welcome back good evening and welcome back so quickly give me some fire emojis in the chats so that I'll get to know I'm perfectly audible visible to you quickly people everyone everyone means everyone everyone means everyone yes perfect guys perfect perfect perfect yes I think I'm perfectly audible vises as the chats are running quite fast yeah so guys are you all ready for the session are you all ready for this amazing session are you all ready for this amazing session people are you all ready for this amazing session yes perfect perfect perfect wonderful so my dear students as you all must be doing today it's going to be the day 13 of our game of need Series right so till now we have covered 12 chapters yes and today it's going to be the 13th chapter which we are going to cover in detail that to from the basics from the scratch right and I'm pretty much sure you guys know it what exactly we are going to do today yes and my dear students uh there's a mistake over here yeah so you need to change it it's not day 12 it's date 13 right so people today we shall be doing the general organic chemistry go right one of the high weighted chapters of your Chemistry perfect and this particular chapter is involved almost I won't say almost I would say this particular chapter it's involved everywhere when it comes to your organic chemistry right so if this particular chapter is stronger that means your entire organic chemistry is automatically going to be strong if there are doubts in this particular chapter that means there will be doubts in the entire organic chemistry so so I've decided to start exactly from the basics from the scratch as if you have never studied this particular chapter before okay we'll be starting from the scratch from the basics and slowly slowly slowly will be scaling up the things as well it just I would want you guys to call up everyone everyone whoever is preparing for n 2024 or even 2025 I would want you guys to call up everyone right let everyone taste go by Wasim sir this time yeah all ready all ready people all ready yes guys I want the chats to run continuously yes need students J students everyone everyone is welcome right this is for everyone this is for everyone this is for everyone I don't want the chats to stop at all I do not want the chats to stop at all and a lot of people are asking how long will it take to complete the general organic chemistry my dear students have no idea it can take 8 hours it can take 10 hours it can take 12 hours till the chapter is over I'll be with you and you will have to be with me as well so are you ready to be with me till the end are you ready to be with me till the end quickly quickly then only I'll be starting are you ready to be with me till the end perfect guys perfect this is something which I wanted from you yeah this is something which I wanted from you so people before starting this particular chapter before starting this particular chapter the ones who have not liked the session yet I can see not everyone has liked the session yet I would want everyone of you to like the session right now if you have not shared this particular video with everyone I would want you to share this with everyone whosoever is preparing for me 2024 2025 or a j share this particular video with everyone and the ones who have not subscribed the channel yet I would want you guys to subscribe to the channel as well right perfect okay so before starting this chapter few announcements I need to make over here and then we shall be starting yeah so dear students as you all must be knowing there is one mega event Mega offline event which is going to happen in Chennai on 23rd of December that means coming Saturday right so this mega event is going to start exactly at 5:00 p.m. right the venue of this particular event you can see over here right I have mentioned the exact address where where exactly you guys are supposed to come you guys can come with your friends you can come with your family members your parents whosoever you want to be with everyone is most welcome in this particular event and in this event there will be I mean we need english team complete need english team will be there in the event we'll be addressing you in person right there'll be a lot of NE Toppers as well who have qualified this particular need examination with amazing scores and they'll be addressing you at the same time so if you are preparing for n 2024 or even 2025 right you guys are most welcome in this particular event right the chenai event which is going to happen on 23rd of December sharp at 5:00 p.m. okay and and and and my dear students in order to attend this particular event you just have to fill one Google form right you just have to fill one Google form that Google form you can find in the description of this particular video or you can see the chats basically right you can see the chats in the chats there's a link which is pinned you just have to click on that link and after that right after that you just have to fill the Google form if you are really willing to join this particular event okay and it's not going to take you more than 5 to 10 seconds to fill that form as well yeah so do that right now number one number two number two my dear students so today we are completing the chapter General organic chemistry go okay and as you all must be familiar with the fact after completing a chapter here on YouTube what do we do we do extra problem practice we do extra problem practice and extra problem practice how exactly you going to attend that see this is the video click on more after clicking on more here you'll find a link go practice questions go practice questions there's a link you just have to click on this particular link once you click on this particular link my dear students here you have to click on notify me and tomorrow at sharp 11: a.m. tomorrow at sharp 11: a.m. we are going to have extra problem practice for the chapter General organic chemistry are you ready for it the ones who have not who have not clicked on the link yet I would want you guys to do it right now right now with me right and click on this notify me so that you remain you will remain updated about this particular extra problem practice session which is going to happen tomorrow at 11:00 a.m. sharp on the unacademy platform yes are you all done with this are you all done with this people quickly are you all done with this are you all done with this yes so should we start now should we start now I think a lot of people have gone to attend this particular session yeah my dear students everyone I would want everyone to click on this notify me here so that you can remain updated about the extra problem practice session which we are going to do tomorrow at 11:00 a.m. sh I think we need to start the session now just give me a second just give me a second we are going to start just give me a second people okay so let's get going then let's get started with the chapter General organic chemistry let me exactly show you the session flow first of all session flow from where exactly we are going to start and where exactly we are going to end the session so my dear students we are going to start from the basics of the organic chemistry since you know in your organic chemistry lot of Concepts from chemical bonding are involved right so I'll be covering those Concepts perfect which are from the chapter chemical bonding perfect I'll be covering those Concepts first so that you'll be fit to attend this particular session so that you'll be fit to cover the entire organic chemistry with us Point number one right after this we'll go to inductive effect and its applications right then we'll be talking about the resonance types of conjugation cross and extended conjugation steric inhabit inhabitance of resonance resonance energy okay momic effect and its applications then we shall be covering the topic aromaticity and at the end of the session we shall be covering the hyper conjugation and its applications as well now a lot of students must be thinking what about IUPAC nomenclature what about IUPAC my dear students if if if if if I teach iupsc also in this particular session then it's going to exceed 14 hours for sure right and one YouTube stream can I mean one YouTube stream can be there for maximum 12 hours perfect so related to your IUPAC IUPAC tomorrow in the afternoon tomorrow in the afternoon you'll be getting one recording I'll be providing One recording in the afternoon tomorrow right tomorrow in the afternoon I'll be providing you one record ing in which you shall be covering your complete iacc as well and that recording will be maximum for 2 and a half hours that is a 2 and a half hour recording which you'll be getting tomorrow right on IUPAC Perfect all isomerism does not include here isomerism needs another 10 to 12 hours right that's again a v topic perfect isomerism we'll do we'll do isomerism as well but not in this particular session that also needs some like 8 8 to 10 hours minimum so let me know once in the chats if you all are ready with your pen and paper let me know in the chats quickly if you all are ready with your pen and paper let me know people quickly if you all are ready with your pen and paper everyone everyone everyone let me have some water first because the session is going to be long quickly guys everything will be done relax everything will be done just just remember and follow whatever I'm saying that's more than sufficient for you after attending today's session and after attending tomorrow's 11: a.m extra problem practice session as well whatever module whatever book you want to solve on gooc you can do that you can do that and you'll be solving those questions like this I'm telling you that yeah perfect guys purification will be also done purification will be also done I told you already I'm not going to skip a single thing session duration I have no idea okay I'm starting the session and let's see uh till the chapter ends I'll be here with you and you are going to be with me too perfect guys so let's get going then let's get started I hope you all are ready with your pen and paper and please and please whatever points I'll be noting on the screen whatever points I'll be writing on the screen I would want you all to take a note of every single point because every single point is going to be equally important my dear students and at the same time at the end of the session I'll be providing you the handwritten PDF of this particular session also in my telegram group I I'm pretty much sure you already are there in the telegram group yeah if not then do join it right now okay perfect guys perfect yes J students can also watch this session no issues you can watch this session so people let's get going then so I'm starting starting this session from a term I'm starting this session from a term that is charged specie this particular session I'm starting from the term that is charged specie first of all if you ask me what is a charged specie what is a charged speci my de students any atom any atom or group of atom any atom or group of atoms having charge on it any atom or group of atoms having charge on it having charge on it is something which you call as the charged speci whenever you see an atom containing any type of charge be it positive or negative or whenever you see a group of atoms carrying any sort of charge be it positive or negative you'll be calling that as the charged speci right so charged species is nothing it is the atom or group of atom atom or group of atoms having any sort of charge on it is something which you'll be calling as charged SPAC right now my dear students for the charged species for the charged species we Define a term for the charged species we Define a term what is that term that term is volume charge density so the term which I'm going to Define volume charge density vcd first of all this particular term we always Define for a Charged speci right we always Define for a charged speci and how exactly you guys Define it volume charge density is nothing it is the ratio of charge present on the spey to that of size of the spey so this particular term vcd volume charge density it is defined only for the charged species number one and this vcd it is nothing it is the ratio of charge present on the speci to that of size of the SP right people now first of all try to understand one one one simple thing which I'm going to tell you right now volume charge density it is directly proportional to charge on specie it is inversely proportional to size of the SP correct now first of all imagine that imagine I'm taking two species over here for example this is one species carrying positive charge this is one more species which is bigger in size containing the same amount of positive charge so how many species have we taken over here we have taken two species one species carrying positive another specie carrying equal amount of positive as well right so first of all if I'm telling you that the charge present on both the species is same the magnitude of charge on both the species is same where the difference is lying the difference lies in the size the difference lies in the size this particular specie has got more size if this specie has got more size that means its volume charge density will be less because vcd is inversely proportional to size right so more the size of the specie lesser is going to be the vcd Lesser is going to be the volume charge density right and over here since these species are carrying the positive charge so if I ask you which volume charge density you are defining for these species you will say it is the positive volume charge density which I'm defining for these two species right so I would say positive volume charge density of this particular species will be less than that of positive volume charge density of this particular speci as simple as that similarly my dear students let's say this is one more species which carries the negative charge right I've got one more species which is bigger in size carrying the same amount of negative charge first of all if I ask you if I ask you if I ask you which volume charge density should I Define for these two species you'll exactly say it is the negative volume charge density which you are going to Define for these two species as simple as that now my dear students imagine that the magnitude of charge present here is same as that of magnitude of charge present here imagine the magnitude of charge on both the species the same where is the difference the difference lies in the size more the size more the size lesser is going to be the volume charge density so among these two species which particular species has lesser negative volume charge density I would say this particular species it'll be having lesser volume charge density so Point number one point number one volume charge density is defined for a Charged speci first of all for any charged speci we Define the volume charge density which is the ratio of the charge present on spey divided by size of the spey this volume charge density is of how many types it's of two types positive volume charge density negative volume charge density positive volume charge density is defined for those species which carry positive charge negative volume charge density is defined for those species which carry the negative charge right right and remember from now onwards more the charge on the species more the volume charge density right lesser the size of the species more the volume charge density be it positive or negative I hope this is clear to everyone now people why am I giving you this term this volume charge density because because because this volume charge density is related to stability this volume charge density it is related to stability let me tell you please and please take a note of every single thing let me tell you stability stability is inversely proportional to volume charge density stability is inversely proportional to volume charge density what is meant by this particular Point how can we get different types of questions from this particular point that is something which you'll get to know in some time first of all remember it directly from now onwards stability of the charged specie is inversely proportional to vcd more the V CD more the volume charge density lesser is going to be the stability lesser the volume charge density more is going to be the more is going to be the stability right perfect basically what is the logic for it why they're inversely related my dear students it's pretty much simple if charge density if volume charge density on the speci is more that means its potential energy is more and if potential energy is more stability automatically will be less right as simple as that more positive more more volume charge density means more potential energy and more potential energy means less stability that's it so just remember this particular statement from now onwards stability is inversely proportional to volume charge density now where do I utilize these terms where do I utilize these terms before utilizing these terms before showing you the application of these terms over here there are few more things which you need to know first of all right and what are those things have a look I believe every one of you would have come across the term electro negativity right I believe every one of you would have come across the term electro negativity how do you exactly Define the term electro negativity my dear students look at the definition the tendency of an element the tendency of an element to attract the bonded pair of electron towards itself the tendency of an element to attract the bonded pair of electron towards itself that defines what that defines electr negativity of an element for example for example my dear students have a look let's say let's say there's a calent bond between A and B perfect I'm assuming that the electro negativity of B is greater than that of electro negativity of a this is something which I'm assuming let's say there's a calent bond between a and b and I'm assuming electr negativity of B is more than that of a now what does this ter mean what does it mean that electr negativity of B is more it simply tells you that B has got more tendency B has got more tendency to attract this bonded pair of electron towards itself right now you tell me if B has got more tendency to attract the bonded pair towards itself will the electron density towards B increase or decrease electron density towards B will increase and that increase in the electron density towards B is represented by Delta negative and here you'll be writing Delta positive as simple as that right as simple as that perfect my dear students one more thing just tell me one thing just tell me one thing in your periodic table which element is considered to be maximum electr negative it is Florine remember this followed by oxygen followed by nitrogen I believe all of you would be knowing this particular thing right in your periodic table chlorine is the most electronegative element followed by oxygen followed by nitrogen Point number one point number two in your periodic table if you move from left to right if you move from left to right left to right what happens to electro negativity generally if you ask me I'll tell you one simple thing I'll tell you electro negativity electro negativity is inversely proportional to size electro negativity it's inversely proportional to size now as you move from left to right as you move from left to right generally size decreases generally size decreases if size decreases that means electro negativity will automatically increase so remember one simple thing from now onwards when you move from left to right electro negativity it increases it increases generally right similarly my dear students you tell me you tell me when you move from top to the bottom when you move from top to the bottom in the periodic table when you move from top to the bottom in the periodic table what happens to size size increases generally if size increases if size increases electro negativity decreases electr negativity decreases perfect this is one simple thing which I want all of you to remember from now on right right people perfect one basic thing which I told you which I would want all of you to remember from now on wordss right and my dear students remember this particular statement as well electro negativity it's inversely proportional to size because I'll be using this particular statement many of times I'll be using this particular statement many of times now people there are few more things which we can conclude from the term electro negativity there is there is a lot of other information which we can conclude from the term electr negativity and what is that information try to understand try to understand what kind of other information you get from the term electr negativity I mentioned over here few points I mentioned over here few points just remember them from now onwards the tendency of an element the tendency of an element to hold its valence electrons electro negativity gives you the idea about the tendency of an element electro negativity gives you the idea about the tendency of an element to hold its valence electrons to hold its valence electrons my dear student that element that element which is more electr netive in nature that will be having more tendency to hold its valence electrons that will be having more tendency to hold its valence electrons for example let's say I've got carbon and Florine let's say I've got carbon and Florine which one is more Electro negative Florine is more Electro negative if Florine is more electronegative that means Florine will be having more tendency to hold its valence electrons to hold its valence electrons so electro negativity gives you the idea about the tendency of an element to hold its valence electron more Electro negative element can better hold its valence electrons as simple as that as simple as that right so Florine has got more tendency to hold its valence electrons Point number one point number two electr negativity gives us the idea about the negative charge holding capacity of an element electro negativity it gives the idea about the negative charge holding capacity of an element what does it mean it's pretty much simple it's pretty much simple more Electro negative element it can sustain it can hold negative charge more Electro negative element can sustain negative charge it can hold negative charge for example let's say you have got oxygen carrying negative charge you have got oxygen carrying negative charge let's say you got carbon carrying negative charge you got oxygen carrying negative charge you got carbon carrying negative charge just tell me one thing in the chats which one is more Electro negative oxygen if oxygen is more electr negative that means that means means oxygen can hold this negative charge in a better way than that of than that of carbon it means that oxygen can hold this negative charge in a better way than that of carbon right so in short one simple thing which you have to remember from now onwards what is that simple thing electro negativity it gives us the idea about the negative charge holding capacity of an element more electronegative element means more electronegative element means more will be its tendency to hold the negative charge right more will be its tendency to hold the negative charge as simple as that do remember this particular statement as well so I would say the negative charge holding capacity of oxygen will be more than that of negative charge holding capacity of carbon now if I ask you one simple thing over here where do you think is the negative charge more stable let's talk in terms of stability where do you think is the negative charge more stable the one which can hold this negative charge the one which can hold this negative charge properly I would say there only this negative charge will be stable you tell me in the charts where do you think negative charge is more stable absolutely it's going to be oxygen absolutely it's going to be oxygen right absolutely it's going to be oxygen negative charge on oxygen will be more stable than that of negative charge on carbon and my dear students remember one important thing because this thing which I'm going to write over here I'll be using it many of times from now onwards what is that do remember the negative charge the negative charge holding capacity take a note of it the negative charge holding capacity of oxygen the negative charge holding capacity of oxygen is greater than is greater than the negative charge holding capacity is greater than the negative charge holding capacity holding capacity of 10 carbon atoms 10 carbon atoms now from this particular statement you can get the idea of how stable of how stable this negative charge on oxygen will be right take a note of the statement the negative charge holding capacity of oxygen is greater than the negative charge holding capacity of of 10 carbon atoms now you can get the idea of how much this negative charge is stable on oxygen right this is one veryimportant important point I'll be using it frequently here perfect now guys this is something which I've already discussed electro negativity of an element it is inversely proportional to size right it is inversely proportional to size at the same time there are few more there are few more Logics which you need to remember have a look my dear students when you talk about the orbitals when you talk about the orbitals you have got s orbital right you have got p orbital you have got D orbital you have got F orbital right if I ask you which orbital is closer towards the nucleus you'll say s orbital is closer towards the nucleus s orbital is closer towards the nucleus right now you understand it like this you understand it like this for example if the electron is present in this s orbital for example similarly there's one more electron present in P one more electron present in D one more electron present in F can you let me know which electron will be experiencing maximum attraction which electron will be experiencing maximum attraction I'll say that electron which will be closer towards the nucleus that will be experiencing maximum attraction now my dear students can you relate this attraction with electro negativity can you relate this attraction with electro negativity you should be able to relate this attraction with electro negativity right tell me tell me I'm writing a statement electron negativity of orbitals is also defined which orbital has got more electro negativity tell me that tell me that in the charts s followed by P followed by D followed by F I'm relating attraction with electr negativity right Perfect People perfect remember this particular Point as well now now now then there is there are few more basic things which I want to share with you try to understand something important my dear students in the entire organic chemistry the there is one element which is pretty much common that is carbon there is one element which is pretty much common which element is that that is carbon right can you let me know what kind of hybridization carbon shows I'll say carbon shows SPI hybridization carbon always shows SPI hybridization carbon always shows SPI hybridization where I value can be either one it can be two or it can be three right what does that mean that means if you particularly talk about carbon if you particularly talk about carbon how many different types of hybridizations carbon can show carbon can show three types of hybridization carbon can show three types of hybridization right one first one is going to be SP then it's going to be SP2 then it's going to be sp3 carbon can show exactly three types of hybrid ations right SP hybridization it can show SP2 hybridization carbon can even show sp3 hybridization right now my dear students you tell me one thing you tell me one thing I hope in your earlier classes you would have at least studied hybridization term once what happens in hybridization basically hybridization which is a complete theoretical concept which which does not have any experimental background hybridization it's a complete theoretical concept it does not have any experimental background one important Point number two number two try to understand what exactly I'm going to say hybridization if I talk about hybridization what is hybridization all about if you remember in hybridization intermixing of atomic orbitals takes place intermixing of atomic orbitals of an atom takes place right which leads to the formation of hybrid orbitals which leads leads to the formation of hybrid orbitals and you should be knowing number of atomic orbitals participating in the intermixing is equal to the number of hybrid orbitals formed number of atomic orbitals participating in the intermixing is always equal to the number of hybrid orbitals formed for example I'm talking about carbon what is the outermost configuration of carbon it is 2s2 and 2 P2 correct so this is your 2s orbital these are your three p orbitals right perfect this is your s this is your P this is p this is p one is PX p y pz now my dear students if if if I would say it like this if the S orbital of the carbon and p orbital of the carbon I'm assuming that these two orbitals are participating in the intermixing if these two orbitals participate in the intermixing that means if two orbitals are participating I'll be getting two hybrid orbitals number of atomic orbitals intermixing intermixed is equal to the number of hybrid orbitals formed and this particular hybrid orbital I'll be calling as SP hybrid orbital this particular hybrid orbital I'll be calling as SP hybrid orbital correct similarly imagine that let's say this s and 2p this s and these two P orbitals have participated in the intermixing so three Atomic orbitals if participating in the intermixing it will lead to the formation of three hybrid orbitals like this right and each hybrid orbital you'll be calling as SP2 hybrid orbital right similarly if all these four Atomic orbitals will participate in the intermixing it leads to the formation of four hybrid orbitals and I'll be calling each hybrid orbital here as sp3 hybrid orbital correct something which you all must be knowing so carbon shows how many types of hybridization SP SP2 or sp3 perfect now people one thing one thing in case of SP hybridization in case of SP hybridization how many hybrid orbitals do we get we get two hybrid orbitals this is your first hybrid orbital and similarly my dear students I can say this is your second hybrid orbital perfect these are your two SP hybrid orbitals this is sp even this is sp hybrid orbital perfect okay if I ask you these hybrid orbitals if you remember they are they are placed in such a way that there should be minimum repulsions between them right so that means they are placed at an angle of what right here they are placed at an angle of 180° perfect s SP2 SP2 hybridization how many hybrid orbitals do we get here we get three hybrid orbitals this is your first hybrid orbital this is your second hybrid orbital right this is your third hybrid orbital and they are placed in such a way that there are minimum repulsions between them so at what angle they'll be they'll be placed at an angle of 120° perfect similarly if I talk about SP three hybridization in case of sp3 hybridization how many hybrid orbitals do we get we get four hybrid orbitals we get four hybrid orbitals and they are placed in such a way that there should be minimum repulsions and maximum stability between them perfect and they are placed at an angle of what 109° 28 minutes I believe this is something which you all already know I believe this is something which you all already know perfect if not then remember it from now onwards now guys tell me one thing tell me one thing see if I ask you how this SP hybrid orbital was made basically when one of the S orbital and one of the p orbital of carbon when they intermixed they resulted in the formation of two hybrid orbitals sp sp can you let me know will there be any s character present in this hybrid orbital absolutely there'll be some s character present in Sp hybrid orbital as well how do you get the percentage of s character in any hybrid orbital it's pretty much simple try to understand in your SP hybrid orbital in your SP hybrid orbital which is formed when S and P intermixed so there is some s character here in this hybrid orbital there is some P character as well how do you get the percentage of s character tell me how many s orbitals participated One S orbital participated divided by how many total orbitals participated two total orbitals participated in the intermixing multiply the 100 the value comes out be 50% what is meant by this 50% it means this particular hybrid orbital it contains 50% of s character it contains 50% of s character what about this hybrid orbital what about this hybrid orbital if I ask you about the percentage of s character in SP2 hybrid orbital so One S orbital participated and in total three orbitals participated multiplied 100 the value comes out be 33% so I would say in this SP2 hybrid orbital in this SP2 hybrid orbital there will be 33% of s character similarly what is going to be the S character here in sp3 hybrid orbital it's going to be 25% it's going to be 25% it's going to be 25% now people you tell me one thing you tell me one thing SP SP2 sp3 SP Bond angle is 180 here it's 120 here it is 109 As you move from left to right here is the bond angle decreasing or increasing 108 sorry 180 120 n on moving from left to right is the bond angle increasing or decreasing Bond angle is decreasing on moving from left to right Bond angle is decreasing B angle is decreasing at the same time what is happening to the percentage of s character 50 33 25 on going from left to right percentage of s character is also decreasing when you move from left to right here Bond angle is decreasing as well as percentage of s character is decreasing so can I say Bond angle is directly proportional to to percentage of s character Bond angle is directly proportional to percentage of s character right percentage of s character is it clear did you get the idea of this particular thing did you idea did you get the idea of this particular statement quickly let me know in the chats everyone so so I would say more the bond angle more the S character and if you remember s character is related to electro negativity if you remember s character is in related to electro negativity s character is related to electro negativity like this more than S character more than electro negativity more the S character more the electro negative so from this particular slide what you have to remember you have to remember something like this Bond angle it is directly proportional to percentage of s character right percentage of s character but is directly proportional to electro negativity right so my dear students this is the conclusion of the entire thing which I was teaching you till now I hope this is clear to everyone I hope this is clear to everyone I hope this is clear to everyone yes quickly let me know in the chats with some fire emojis if you got this particular concept because this is important this is important quickly people quickly everyone everyone in the chats this is something which is important I'll be utilizing this concept everywhere yeah in the chats everyone I want the chats to go up continuously done understood done understood perfect guys if this is done understood if this is done done understood then then let me tell you one more thing let me tell you one more thing can I write a simple statement like this can I say electro negativity of hybrid orbitals is also defined you are going to answer this question electro negativity of hybrid orbitals is also defined if I have SP hybrid orbital if I have got SP2 hybrid orbital if I have got sp3 hybrid orbital can you tell me which hybrid orbital has got more s character it is sp s character is directly proportional to electro negativity so can I say electro negativity of SP hybrid orbital will be greater than that of electro negativity of SP2 which will be greater than that of electro negativity of sp3 so guys do remember this particular statement as well electro negativity of SP hybrid orbital will be more than that of SP2 than that of what than that of sp3 because of the percentage of s character yes is it clear is it clear in the charts quickly end the charts quickly end the charts quickly end the charts quickly everyone so I believe whatever statements are mentioned here I believe these statements will be absolutely clear to you number one electro negativity is directly proportional to personage of s character electro negativity of SP hybrid orbital is greater than that of SP2 is greater than that of sp3 you know the logic you know the reason Point number three point number three this is just one example which is going to clear you one simple thing which is going to clear you one simple thing see guys let me make the molecule again like this this is H this is C then you have got triple bonded carbon right then you have got CH and double bond ch2 double bond ch2 now people tell me just one simple thing tell me about this particular carbon how many Sigma bonds it's forming how many Sigma bonds it's forming it's forming one and two it's forming two Sigma bonds it's forming two Sigma bonds it's forming two Sigma bonds try to understand one more thing I hope you know the term called a static number I hope you know the term called a static number what is static number static number means Sigma plus lone pair Sigma plus lone pair if the steric number comes out be two the hybridization is sp if it is three the hybridization is SP2 if it is four the hybridization is sp3 okay now if I talk about this particular carbon how many valence electrons carbon has how many valence electrons carbon has four four valence electrons how many Sigma bonds it's forming it's forming two Sigma bonds it's forming two Sigma bonds carbon has four valence electrons and how many Sigma bonds it's forming it's forming two Sigma bonds now tell me how many total bonds this carbon is forming 1 2 3 4 Carbon is forming four bonds right and in order to form four bonds can you say carbon would have utilized all its four valence electrons absolutely carbon would have utilized all its four electrons or its four valence electrons carbon had four valence electrons and it has utilized all those four valence electrons for the bonding so how many lone pairs this carbon will be having it'll be having zero lone pair if I ask you static number of this carbon Sigma plus lone pair 2 plus 0 that is two if staic number is two that means hybrid is sp so this particular carbon has got the SP hybridation here right yes right people in the similar way look at this particular carbon look at this particular carbon how many siges forming one and two so again I'll say this carbon is sp hybridized how many how many Sigma this is forming one Sigma 2 Sigma and one more with hydrogen right so three siges forming so it's SP2 this particular carbon how many siges forming one and two with hydrogen so this again SP2 right so I believe from the structure you can easily let me know you can easily let me know the hybridation of different atoms yes hybridation of different atoms I believe this is clear just tell me one thing just tell me one thing do you know how to calculate loone pairs do you know how to calculate loan pairs just tell me that yes or no yes or no do you know how to calculate loone pairs quickly guys everyone in the chats what do you think yes or no honestly there is no need to be shy or something I'm here to teach you everything that's why I told I'll be starting exactly from Basics one example I'll take by means of which you'll understand how to calculate the lone pairs okay one example I'll take one example I'll take then I'll come to this particular case which I have drawn over here I'm keeping this case on halt one simple example I'll take which will make you clear every single thing see for example if you have got the molecule xf4 if you have got the molecule xf4 if I particularly talk about Xenon Xenon is a noble gas right if it's a noble gas right in its outermost shell how many electrons will be there eight valence electrons so Zenon has got eight valence electrons tell me how many surrounding atoms it has it has four surrounding atoms the number of surrounding atoms is equal to the number of Sigma bonds the number of surrounding atoms is equal to the number of Sigma bonds now you tell me Zenon had eight valence electrons and it's forming how many bonds four bonds so out of eight valence electrons can I say Zenon has utilized four of its valence electrons to form four bonds out of eight valence electrons Zenon has utilized four of its valence electrons to form four bonds so how many valence electrons will be still there there'll be four valence electrons left on Zenon as such in its outermost shell so these valence electrons they have not participated in the bonding and you'll be calling these as the lone pairs so Zenon has how many lone pairs Zenon has two lone pairs four electrons means two lone pairs simple right for example take the case of water H2O take the case of water if I talk about oxygen oxygen it's a group 16 element how many valence electrons oxygen has how many valence electrons oxygen has six there are six valence electrons with oxygen how many surrounding atoms are there two surrounding atoms the number of surrounding atoms is equal to number of Sigma bonds right so out of six Val electrons it has used two valence electrons to form two Sigma bonds so how many electrons will be left on oxygen as such I'll say four four electrons means how many lone pairs I'll say two lone pairs this is one lone pair this is one more lone pair correct simple and basic simple and basic for example you have got nh4 positive this is the last example which I'm showing here nh4 positive look at this nitrogen how many valence electron it has five but there's a positive charge if there's a positive subtract this subtract this positive so the value comes out to be four how many surrounding atoms are there four surrounding atoms the number of surrounding atoms is equal to the number of Sigma it had four valence electrons and it has used all these four valence electrons to form four bonds so how many lone pairs will be there no there'll be no lone pair on nitrogen perfect and I believe if you got to know how to calculate Sigma how to calculate loone pairs you can calculate the steric number which is Sigma plus lone pair Sigma plus lone pair Sigma plus lone pair and once you get the static number you can easily talk about the hybridization if the static number is five hybridization is sp3d if the static number is six hybridation is sp3d2 right and so on perfect I believe this basic thing is clear right now people try to understand try to understand one thing I gave you one particular molecule over here if you look at this particular carbon it is sp hybridized this particular carbon is SP2 hybridized let me just show you these these two carbon atoms which I'm going to take into consideration these are the two carbon atoms which I've taken into consideration one is sp one is SP2 if this is sp hybridized can I say it s character will be more if it s character is more that means it's electr negativity will be more if its electro negativity is more that means it has got more tendency to attract the bonded pair towards itself so this particular bonded pair will be attracted towards this carbon due to which it gets Delta negative and this particular carbon gets Delta posi to right right people so what happened due to the difference in hybridization due to the difference in the hybridization can I say a partial polar character got developed here right can I say due to the difference in hybridization a polar character got generated over here that is the point which I have mentioned over here the molecule can become partially polar the molecule can become partially polar because of what because of the difference in the hybridization is this statements clear is this statement clear is this statement clear people say yes or no in the chats everyone everyone everyone everyone someone is saying sir you are looking so sad today why would I look so sad happy with the work with I do yeah that's the reason why you guys are here quickly people is it clear I think this particular slide will be absolutely clear to you electro negativity is directly proportional to S character s character is directly proportional to bond angle remember that SP hybrid orbital its electr negativity is more than that of SP2 than that of sp3 one more Point number three due to the difference in hybridization a partial poke a partial po character gets developed in a molecule yeah I believe this is clear now guys now comes one more important thing now comes one more important thing have you heard about something called as angle strain these are some points Guys these are some points which we are frequently going to use these are some points which we are frequently going to use try to understand there's a term which I mentioned over here angle strain angle strain in cyclic compounds angle strain in cyclic compounds what is meant by it and why do I need it try to understand if you look at this particular carbon if you look at this particular carbon right there'll be some two hydrogens which will be associated with it perfect right carbon has to form four bonds 1 2 3 4 perfect if I ask you what is the hybridation of this carbon you'll directly say it's sp3 hybridized four Sigma bonds four Sigma bonds no lone pair that me means thetic number four sp3 this sp3 hybridized if this carbon is sp3 hybridized so what is the expected Bond angle here what is the expected Bond angle here expected Bond angle is 109° 28 minutes this expected Bond angle but my dear students if I ask you if I ask you look at the angle between these two bonds is it is it exactly 10928 you'll say no it is just 60 it is just 60 if I talk about the actual Bond angle if I talk about the actual Bond angle actual Bond angle here is 60 actual B angle here is 60 look at this particular molecule look at the other molecule look at this particular carbon it's forming these are two hydrogen's associated with it now what about it hybridization it's again sp3 perfect what is the expected Bond angle expected Bond angle is 19° 28 but what is the expected Bond angle here expected Bond sorry what is the actual Bond angle here actual Bond angle is 90 actual Bond angle is 90 right actual Bond angle is 90 now my dear students try to understand one simple thing and remember it from now onwards remember it from now onwards whenever in a molecule whenever in a molecule the actual Bond angle is not equal to the expected Bond angle whenever in a molecule the actual Bond angle is not equal to the expected Bond angle we say that molecule is under angular strain we say that molecule is under angular strain so this particular molecule is under strain can I say even this particular molecule is under strain absolutely this particular molecule is under strain as well this particular molecule will be also under strain this particular molecule will be also under strain so can I say all these molecules here they are under strain absolutely they are under strain now what is the consequence of this strain what is the consequence of this strain try to to understand what is the consequence of this strain what will happen due to this strain that is important have a look I'm going to make the first molecule over here I'm going to make the first molecule over here again perfect this is your molecule which I took in the beginning right and these are the two hydrogen atoms these are the two hydrogen atoms now my dear students if I ask you this angle should be how much I'm using the term should be this angle should have been this angle should have been 109 this angle should have been 109 so this particular angle it should have been 109 it should have been 1009 but how much it is it is less than that if this angle has decreased what about the outer angle can I say outer angle would have increased this angle it should have been 109 but it's 60 so this angle has decreased so outer angle would have increased yes so if outer angle has increased before that let me tell you these bonds over here these are called as exocyclic bonds these are called as exocyclic bonds and these particular bonds these particular bonds these are called as endocyclic bonds these are called as endocyclic bonds love you tell me this particular Bond angle should have been 109 but it's it's 60 it's less than that that means the outer Bond angle would have increased and more the bond angle more the electro negativity more the bond angle more the elect negativity right more the bond angle more the electro negativity so my dear students just compare one thing just compare one thing tell me about this particular carbon and this particular carbon tell me tell me which particular molecule among these two is under more strain here the difference is more here the difference is comparatively less so this molecule is under more strain if this is under more strain so can I say this particular Bond angle here will be more as compared to this particular Bond angle Bond angle is directly proportional to electr negativity so if I ask you what about the electr negativity of this particular carbon one and two can you tell me which carbon will be more electr negative among these two is it going to be one or two which carbon one or two which one will be more electronegative you tell me that only which carbon will be more electr negative quickly in the charts absolutely this carbon will be more Electro negative than this then this then this is it clear so in short what is the conclusion here what is the conclusion here conclusion is very much simple more the angle strain more the electro negativity more the angle strain more the electro negativity I believe this particular statement is clear more the angle strain more the electro negativity so I can I can write a simple statement do remember this electr negativity is directly proportional to what it is directly proportional to angle strain more the angle strain more is going to be the electro negativity I believe this particular point is also clear to you yes right people is it clear okay if it is clear if it is clear over here I've taken two rings this is a three membered ring there's a four membered ring and I have joined them I have joined them right my dear students if you look at this particular carbon which I'm calling as carbon one for example I'm calling this carbon as Carbon 2 for example okay now these two rings can you let me know which ring is under more strain PR membered ring it is under more strain than that of four membered ring four membered ring is under more strain than that of five member than that of six membered so can I say can I say this particular ring is under more angular strain if this ring is under more angular strain more The Strain more the electr negativity so can I say electro negativity of this particular carbon electro negativity of this particular carbon will be more than that of this okay if electr negativity of this carbon is more so it has got more tendency to attract the bonded pair towards itself so I'll say if it has got more tendency to attract the bonded pair towards itself that is the reason why this carbon will get Delta negative this carbon will get Delta positive agreed now tell me one thing can I say due to the difference in the angular strains due to the difference in the angular strains again partial polar character got developed in the molecule yes can I say due to the difference in the angular strains again I would say partial polar character got developed in this molecule so do remember one thing due to the difference in the angular strains partial polar character can get developed in the molecule yes or no in the chats everyone yes or no in the chats everyone yes or no in the chats everyone quickly yes or no everyone in the chats quickly everyone in the chats guys this is complete Basics which I'm teaching you and you have to remember each and every Point agreed perfect wonderful wonderful guys now your screen will be blinking a bit because electricity is over so maybe we might have to shift to dark mode soon and I hope everyone of you is fine with that yeah we might have to shift to the dark mode soon and the people who who watch my sessions regularly they would know exactly what this dark mode is all right so people the things the basics which I was teaching you till now way do we use this Basics where do we use this Basics where do we we use this Basics where do we use this Basics before utilizing all these Concepts in the questions there is one more thing which I would want to share with you my dear students there's a term called as stability of the charges there is a term called as stability of the charge for example for example let's say we have got the carbon carrying negative charge we have got nitrogen carrying negative charge carbon carrying negative nitrogen carrying negative tell me one thing whether these elements whether they belong to same period or same group what do you think carbon nitrogen they belong to same period they belong to same period my dear students whenever you have to compare whenever you have to compare the stability of the charges whenever you have to compare stability of the charges on the elements which belong to same period you will be giving your answer as per electro negativity you'll be giving your answer as per electro negativity and as per electr negativity is concerned negative charge is more stable on more electr negative negative charge is more stable on more Electro negative element right whenever you have to whenever you have to compare the stability of the charges on the elements which belong to same period at that time electr negativity plays a role negative charge is more stable on more Electro negative element similarly opposite of this opposite of this positive charge will be more stable on positive charge will be more stable on less Electro negative element less Electro negative element less electronegative element now look at the case for example carbon carrying negative nitrogen carrying negative oxygen carrying negative these elements they belong to same period they belong to same period and whenever you have to compare the stability of the charges on the elements which belong to same period which concept electro negativity and negative charge is more stable on more Electro negative right which one is more elative oxygen so negative charge be maximum stable here followed by this followed by this correct this is the stability order of the negative charges this is the stability order of the negative charges for example you have got carbon carrying positive nitrogen carrying positive oxygen carrying positive again have a look these elements they belong to same period in the period we prioritize electro negativity and you know positive charge is more stable on less Electro negative element positive charge is more stable on less Electro negative so which one is less Electro negative among these carbon so positive charge will be more stable here followed by this followed by this this is the stability order this is the stability order right let me know if this particular slide is clear let me know if this particular slide is clear whenever you have to compare the stability of the charges on the elements which belong to same period you'll be talking in terms of electro negativity negative charge is more stable on more Electro negative positive charge is more stable on less electr negative as simple as that as simple as that clear Point number one point number one point number one my dear students since I told you since I told you positive charge is more stable on less Electro negative since I told you positive charge is more stable on less Electro negative on less Electro negative now the question is can positive charge be more stable on less Electro negative anytime can positive charge be more stable on less electr negative anytime see I think I'm telling you the reverse way since we know positive charge is more stable on less Electro negative now the point is can positive charge become more stable on more Electro negative is there any case like that as well yes yes my dear students try to understand this is the carbon carrying positive this is the oxygen carrying positive carbon carrying positive oxygen carrying positive carbon and oxygen do they belong to same period or same group they belong to same period right and in the period we look after electro negativity we know normally positive charge is more stable on less Electro negative so where should be the positive charge more stable positive charge should be more St on carbon because that is carbon is less Electro negative than oxygen but here the case is reversed positive charge is more stable on oxygen here why is that because look at the octed of oxygen 2 4 6 and 8 the octed is complete look at the octed of carbon 2 4 6 this is incomplete octed this is incomplete octed here you have got complete octed here you have got complete octed let me tell you positive charge can become more stable on more electronegative as well if the octed is complete if the octed is complete that the statement which I have written over here positive charge can become more stable on more Electro negative element when it's octed is complete is this point clear is this point clear normally what is the trend carbon oxygen they belong to same period same period electro negativity plays a role positive charge is normally more stable on less Electro negative so this positive charge on carbon should be more stable but here oxygen carrying positive is more stable positive on oxygen is more stable why is that because here the oct is complete here the oct is incomplete so do remember positive charge can become more stable on more Electro negative element if the oct is getting complete yes or no in the charts everyone everyone everyone quickly in the charts my dear students remember these points remember these points remember these points now comes one more point now comes one more Point whenever whenever you have to compare the stability of the charges on the elements which belong to same group whenever you need to to compare the stability of the charges on the elements which belong to same group at that point of time you are not going to consider electr negativity you are going to prioritize volume charge density you are going to prioritize volume charge density along the period always talk in terms of electro negativity group talk in terms of vcd talk in terms of vcd and I hope you know your volume charge density is equal charge present on the specie divide by size of the speci and you must be knowing already have discussed stability is inversely proportional to volume charge density right these are the two things which you have to remember nothing else now we try to understand try to understand try to understand chlorine carrying negative chlorine carrying negative bromine carrying negative iodine carrying negative do these elements do they belong to same period or same group you tell me the answer do they belong to same period or same group these elements belong to same group and in the group BCD plays a role in the group vcd plays the role the charge on every species is same but the size is different this particular specie has got more size more size means less vcd less volume charge density means more stability now tell me where is the negative charge more stable I would say this is the stability order of negative charge yes I hope you're getting this I hope you're getting this again chlorine carrying positive chlorine carrying positive bromine carrying positive iodine carrying positive now you have to answer me you have to answer me whether these elements whether they belong to same period or same group whether these elements belong to same period or same group they belong to same group in the group vcd plays the role volume charge density plays the role now you tell me the charge everywhere is same the charge everywhere is same but but the difference lies in size iodine it has got more size more size less volume charge density less the volume charge density more the stability so so where is the positive charge more stable this is going to be the stability order say yes or no in the chats if it's clear say yes or no in the chats if it's clear everyone quickly people everyone in the chats guys I want the Jo to be high all the time it's a big Marathon it is a big Marathon okay so you have to keep me motivating as well with your chats I want the chats to run continuously and I can see not everyone has liked the session yet so do like the session right now right now means right now or else you are over you are gone yeah everyone people everyone everyone in the chats yeah perfect whatever Concepts we have discussed till now I'm going to utilize all the concepts now in the questions but before showing you the questions just a quick revision of everything just a quick revision of everything whenever we need to compare the stability of the charges on the elements which belong to same period we have to talk in terms of electro negativity negative charge is more stable on more Electro negative positive charge is more stable on less Electro negative number one number two whenever we need to compare the stability of the charges on the elements which belong to same group volume charge density plays a role volume charge density plays the role stability is inversely proportional to vcd that's all that's all now people try to solve the questions which I'm giving you almost almost I have got more than 500 questions today almost almost more than 500 questions today all right arrange the following in the stability order can you let me know where is the positive charge maximum stable I would want to see your answers first I would want to see your answers first I would want to see your answers first where is the POS charge more stable among all these four five species quickly everyone quickly you should be in a position to answer this question you should be in a position to answer this question you should be in a position to answer this question everyone everyone means everyone are you sure are you sure what kind of logic did you use here see guys carbon positive nitrogen positive oxygen positive Florine positive these elements belong to same period in the period electron negativity plays a role right positive charge is more stable or less Electro negative positive charge is more stable or less Electro negative look at the second question carbon carrying negative nitrogen negative oxygen negative Florine negative carbon nitrogen oxygen Florine again they belong to same period electro negativity plays a role electro negativity plays a role negative charge is more stable on more Electro negative negative charge is more stable on more Electro negative so this is the stability order sulfur carrying negative chlorine carrying negative where is the negative charge more stable sulfur and chlorine same period electro negativity electro negativity plays a role negative charge is more stable or more electr negative so which one is more electr negative among these chlorine so negative charge is more stable here oxygen negative this is oxygen this is sulfur this is SE this is T they belong to same group they belong to same group we have to consider vcd here we have to consider vcd here now charge everywhere is same look at the size down the group size increases size is maximum here size is maximum ha and you know BCD it inversely proportional to size it's inversely proportional size more the size less the vcd less the vcd more the stability this is the order of stability of these negative charges yes perfectly done perfectly done perfectly done yes or no look at the more questions more questions on your screen have a look nitrogen carrying negative phosphorus negative arenic negative antim negative do these elements belong to same period or same group everyone same period or same group they belong to same group vcd plays a role more the size less the vcd lesser the vcd more the stability right this is the stability order of the negative charges right tell me here oxygen carrying negative sulfur carrying negative oxygen sulfur oxgen sulfur same group same group vcd more size less vcd more stability is this is this the order quickly let me know here let me know here if you look properly this is single bonded carbon this is double bonded carbon this is triple bonded carbon where is the S character more where is the S character more single bonded double bond or triple bonded where is s character more here the S character is more so I would say electr negativity of this one is more right but we know positive charge is more stable on less Electro negative right we know positive charge is more stable on less Electro negative this one is less electr negative so this is the stability order done simple trick to remember that's it okay look at the next one carbon carrying negative double bonded carbon carrying negative triple bonded carbon carrying negative where is the S character more where is the S character more s character is more hair right s character is more ha SP hybridized if s character is more here that means its electro negativity is more and you know negative charge is more stable on more Electro negative negative charge is more stable or more electr negative so is this going to be the stability order absolutely this is going to be the stability order absolutely this is going to be the stability order these questions tell me their answers tell me their answers people negative charge on three membered ring negative charge on four membered ring five membered six membered which one has got maximum angle strain which one has got maximum angle strain have a look people here have a look here which one is under maximum angle strain I'll say first one is under maximum angle strain first one is under maximum angle strain right more the angle strain more the electr it so electr negativity of this carbon will be more than that of this one right and negative charge is more stable on more electr negative this is the stability order positive charge is more stable on less Electro negative this is the stability order now tell me this particular scenario carbon carrying negative and nitrogen carrying negative where is the negative charge more stable you need to answer this you need to answer this everyone you need to answer this you need to answer this people quickly quickly where is the negative charge more stable on carbon or nitrogen so normally what what should you think normally you'll think like this carbon and nitrogen they belong to same period in the period electr negatively dominates right and you know negative charge is more stable on more Electro negative so this has to be the stability order but if I ask you is this the actual stability order this is not the actual stability order this is not the actual stability order remember this remember this people remember this this is a special case it is a special case here you have got SP hybridized carbon the carbon which you have here it's SP hybridized nitrogen it is sp3 hybridized nitrogen it is sp3 hybridized nitrogen do remember electro negativity of electro negativity of SP hybridized carbon is more than that of electr negativity of SP C3 hybridized nitrogen if you want to remember the values the value here is 3.25 here the value is three electr negativity values I'm giving right now which one is more Electro negative I would say this particular speci is more Electro negative if it is more Electro negative then it's evident negative charge is more stable and more Electro negative so this is the order this is the order so this was the first type of question which we did here arrange the charges on the base of their stability right right people let me know everyone with the in the chats with the fire symbol if every single thing is clear then I'll show you one more type of question which we can do from the same concept which I taught you tell me first of all in the chats with the fire emojis guys I don't want the chats to be slow at all I don't want the chats to be slow at all I don't want the chats to be slow at all people I don't want the chats to be slow at all let me first of all see how many people are watching us and what is happening in the chats because I'm unable to see your chats from there is it live I think I need to refresh it once okay this is the a refresh screen but I'm unable to see the chats again wow nice good okay where are your live chats here are your live chats guys I want the chats to run fast what is this what is happening they are stagnant right the charts are stagnant if every single thing is clear till here I want the charts to light up with the fire fire Emoji everyone if every single thing is clear to and I believe I've taught an entire death guys yes all clear all clear so we have got Zan we have got Lakshmi we have got a lot of people well I'm unable to see now we have got suja we have got killer signs Skiller science is a very old student by the way since we it's been almost 7even months we have started this Channel and I have seen the Skiller Science Guy from day one itself on this channel wonderful good oh we have got y manour s as well in the chats right we have got y manzur in the chats as well wonderful we got SVA Kumar as well SVA is also a old student larika is a old student nice nice nice good good good good we have got Nikki we have got namata we have got Priya all are regular students here good job guys I want you to be regular Till the End by the way tomorrow we have got one more amazing session guys in the evening at 700 I'll be revealing one more important one more amazing surprise with you right tomorrow at 7:00 p.m. on this particular Channel yeah and I can see the likes are pretty much less why is it don't I deserve a single like from you right don't I deserve a single like guys okay so should we move on so first step of the question is done and dusted let's move on now let's move on now to to to to to one more type of question acidic strength acidic strength try to understand what exactly I'm going to say my dear students for example I'm going to take an acid over here this is acid H Aquas it is an acid it's an acid right I introduced the acid in water I've introduced the acid in water now what this acid is going to do in water this acid in water is going to donate H positive you know it right you know it so this acid when you introduce prod this acid in water it's going to donate what it's going to donate H posit after donating H positive it is going to get converted into a negative after donating H positive it's going to get converted into a negative this is something which you were calling as acid this is something which you were calling as acid and after the liberation of H positive after the liberation of H positive this acid it got converted into a negative right this a negative here I will be calling as the conjugate base this is the conjugate base of the acid which I've taken over here right this is acid acid after donating H positive it is getting converted to a Nega this a negative I'll be calling as conjugate base here this is something which I'll be calling as conjugate base here now do remember do remember strength of the acid strength of the acid strength of the acid people strength of the acid strength of the acid do remember it is directly proportional to Ka value I believe you would have studied K somewhere ionization constant of the acid which I'm going to teach you in the upcoming chapter equilibrium right if you do not remember it yet remember it from now onward strength of acid is directly proportional to Ka which is inversely proportional to P of K which is inversely propor P of K so basically there will be a set of questions which will be asked on strength of acid there'll be a set of questions which will be asked on Ka there'll be a set of questions which will be asked on PK now do remember strength of acid is directly proportional to the stability of the conjugate base this is the only statement which I would want you guys to remember here strength of the acid it is directly proportional to the stability of the conjugate base strength of this acid is directly proportional to the stability of its conjugate base more stable the conjugate base stronger the acid more its Ka value less its PKA value simple as that right do remember this strength of acid is directly proportional to stability of conjugate base now I'm going to do many equs on this particular one statement strength of the acid is directly proportional to stability of its conjugate base right just remember it and have a look see how I'll be dealing with the questions my dear students among these two you have to identify which AET is stronger r o r sh which one is stronger among the two which one is stronger among the two the concept which I gave you right now you're going to utilize that concept itself r o rsh which one is more acidic which one is more acidic how exactly you're going to do it try to understand first of all if I talk about Ro if I talk about Ro my dear students after after the liberation of H positive after the liberation of H positive the conjugate base which we get over here that is R negative R negative is the conjugate base of this acid similarly we have got RS after the liberation of H positive after the liberation of H positive it gets converted to RS negative so I made their conjugate bases I made their conjugate bases now look at the stability of these conjugate bases look at the St of these conjugate bases tell me oxygen carrying negative sulfur carrying negative oxygen carrying negative sulfur carrying negative oxygen and sulfur same group vcd plays a role vcd plays a role more the more the size less the vcd more the stability so I would say this conjugate base is more stable and that acid which gives more stable conjugate base strength of that AET will be more right so which acid is stronger our sh is stronger than Ro simple Yes simple the kind of Concepts which I gave you I'm utilizing those Concepts only I'm not tell something new look here make their conjugate bases make their conjugate bases first of all take H posit out from here you'll you'll be getting a negative on carbon take H POS out from here you'll be getting negative on nitrogen take H posit out from here you'll be getting negative on oxygen take H POS out from here neg on Florine NE on Florine now you tell me carbon with Nega nitrogen neg2 oxygen neg2 Florine negative all these elements they belong to same period in the period we talk in terms of electro negativity negative charge is more stable on more Electro negative negative charge is more stable on more Electro negative so negative charge will be more stable on Florine so the conjugate base which you get from HF that is more stable if the conjugate base is more stable if the conjugate base is more stable I would say the acidic strength of these acids will follow the same order that inded absolutely H2O H2S take H positive out there'll be a negative charge on oxygen take H positive out negative charge on sulfur right if on taking H positive out negative charge on oxygen on taking H positive out negative charge on sulfur now look at oxygen and sulfur oxygen sulfur same group in the group vcd more the size more the size less vcd more stability so the conjugate base which I get from H2S that is more stable if that is more stable if that is more stable that means this has to be the strength of these acids right done look at the next one look at the next one see take H posit out from here carbon will carry negative take H positive out from here this carbon will carry negative take H positive out from this carbon will carry negative so you have made you have made the conjugate bases now see single bonded carbon double bonded carbon triple bonded carbon triple bonded carbon has got more s character more s character means more electro negativity and you know negative charge is more stable on more Electro negative so the conjugate base which you generate from here that is maximum stable and more stable the conjugate base better is the acid more is the strength of the acid so this is the acidic strength order yes okay tell me about this one tell me about this one quickly this is Edge tell me about this one among these two which one is more acidic you have to tell me the answer among these two which one is more acidic you have to tell me the answer Doctor Strange you need to watch the session from the beginning then you'll get to know what is s corrector which one is more AIC guys this concept also I've given you this concept also I've given you this concept I've given you too which one is more acidic you have to answer me this and the answer has to be correct it has to be appropriate if you give the wrong answer here it'll be a heartbreak for me it is it is actually going to be a heartbreak for me if you give give the wrong answer I can see majority has given the correct answer wonderful good job see guys what you'll be doing make their conjugate bases take HED POS out from here you'll be getting something like this H C triple bond carbon carrying negative take H positive out from here it is going to be nh2 carrying negative right now these are the conjugate bases these are the conjugate bases look at their stability this is sp hybridized carbon this is sp3 hybridized nitrogen electro negativity of SP hybridized carbon is more and you know negative charge is more St on more electr negative so this is the stability order of these conjugate bases and whatever is the stability order of the conjugate bases same is the order of their acidic strengths done and does it right yes is it clear I believe every single thing whatever I taught you till now is clear so the basics which was required to learn the general organic chemistry that is clear and by learning the basics you could solve two types of questions by learning the basics you could solve two types of questions first question on the basis of first question was stability of charges on different species second question was the AIC strength remember acidic strength is directly proportional to stability of conjugate base nothing else you have to remember that's it is it clear are you ready for the inductive effect now are you ready for the inductive effect are you ready for the inductive effect I want the chats to move fast they're pretty slow I want the chats to run fast they should run they should not walk they should run they should run yes basic strength questions will be given too but wait for it we have to do certain Concepts first and the ones again who have not liked the session yet like the session right now I want the likes to be more than 1K quickly I can see not everyone has done that guys it does not I mean you need not to pay for the likes it's free yeah it is free but when you enter into your Medical College then I'll make you pay okay just remember that then you enter into your Medical College I'll make you pay or whatever I've done on YouTube whatever the team has done on YouTube yeah and I believe at that time you should and you will and I'll make sure you do okay perfect hello guys let's move on to something called as inductive effect this again one important concept and simple at the same time simple at the same time have a look how do you exactly Define the inductive effect my students I'll be writing a simple statement over here then I'll make you understand what that statement exactly is all about induct to effect whenever whenever an atom or whenever an atom or whenever an atom or a group a group is attached is attached to a carbon chain whenever an atom or a group is attached to a carbon chain is attached to a carbon chain the chain the chain becomes partially polarized IED the chain becomes partially polarized due to the transmission of Sigma electrons due to due to the transmission due to the transmission of Sigma electrons or let me write du to the transmission of Sigma electron density this effect is something which we call as inductive effect this effect is something which you call as inductive effect now what does it mean how can we understand this how can we understand this have a look people for example I'm taking a carbon chain over here let's assume you have got this particular carbon chain right this is the carbon chain which we have okay this is the carbon chain which we have and this particular carbon imagine it is attached with hydrogen imagine it's attached with hydrogen right my dear students in this particular chain can I say there are only Sigma bonds present yes in this chain there are only Sigma bonds present right so can I say in this chain there is only Sigma electron density absolutely in this chain right now there is only Sigma electron density now what exactly I'll be doing I'll be replacing this hydrogen I'm going to replace this hydrogen by an atom or a group of atoms I'm replacing I'm going to replace this hydrogen by an atom or group of atoms for example these were 1 2 3 4 5 these were five carbon atoms in the chain one 2 3 this is four and this was five right now I'm replacing the hydrogen by an atom X by an atom X right by an atom X and similarly my dear students similarly I'm again going to replace this hydrogen by an atom y so these are your 1 2 3 4 5 five carbon atoms this was your carbon chain now instead of hydrogen here I'm attaching what here I'm attaching an atom or group y an atom or group y so what I exactly did I took a carbon chain which includes only the sigma electron enity now instead of hydrogen I placed X and here I placed y my dear students let's look at this particular scenario first of all I'm assuming I'm assuming over here I'm assuming I'm assuming electron negativity of X is more than that of electro negativity of carbon now you tell me if electro negativity of X is more than that of electro negativity of carbon what's going to happen can I say this x will pull the sigma electrons towards itself absolutely if x is more electronegative than carbon that means X has got more tendency to attract the bonded pair towards itself so this particular bonded pair can I say it will shift towards X due to which this x gets Delta negative and this particular carbon gets Delta positive now this particular carbon which got Delta positive can I say it is going to attract this bonded pair as well towards itself right it's going to attract this bonded pair towards itself as well due to which this will also get Delta positive now this will attract this this will also get Delta positive perfect if you try to understand when I attached an atom or a group to the carbon chain to the carbon chain can I say the carbon chain got partially polarized have a look on the charges the carbon chain got partially polarized and the carbon chain got partially polarized due to what due to the transmission of Sigma electrons whenever you see this kind of the effect in which upon the introduction of an atom on a or a group the carbon chain becomes partially polarized due to the transmission of Sigma electrons due to the transmission of Sigma electrons you call this particular effect as the inductive effect so I would say this group this group or atom it is showing which effect right here it is showing inductive effect it is showing inductive effect Point number one point number one point number two my dear students imagine that electro negativity of carbon is greater than that of electro negativity of Y if electro negativity of carbon is more than that of Y that means carbon has got more tendency to attract this bonded pair towards itself right carbon has got more tendency to attract this bonded pair towards itself due to which this carbon gets Delta negative and this y gets Delta positive right now my dear students this carbon carrying Delta negative this Delta negative can I say it's going to Ripple this electron pair if it ripples the electron pair if it repuls the electron pair I'll say this carbon will get Delta Nega now similarly to the repulsion this carbon will get Delta negative so what happened over here what happened over here when you introduced an atom or a group The Chain got partially polarized due to the transmission of Sigma electrons so again I would say this y over here is showing which type of effect it is showing inductive effect it is showing inductive effect now my dear students try to understand one very simple thing one very simple thing tell me what do you see exactly is this x is this x increasing the electron density in the chain or decreasing this x is pulling the sigma electrons towards itself so this x it is decreasing the sigma electron IND so can I call this X as electron donating or electon withdrawing here I'll be calling it as electron withdrawing group this x will be called as electron withdrawing group here this y this y should I be calling this as electron donating or electron withdrawing this Y is donating right it's donating it is pushing Sigma electron in in the chain right so I would call it as electron donating group my dear students that particular group that particular group which withdraws Sigma electron density from the chain that particular group Group which decreases the sigma electron density in the chain we say that group shows minus I effect so I would say the group or atom X which I attached over here it is showing minus inductive effect from now onwards remember minus means withdrawing plus means donating minus means withdrawing plus means donating right over here this electron donating if it is electron donating I would say this particular group is going to show plus sign so do remember this statement as well that atom or group which decreases the sigma electron density in the chain due to the transmission of Sigma electrons that group is electron withdrawing and we say that group shows minus I effect and here I would say the group is showing plus side effect it's electron donating donating is plus withdrawing is minus donating is plus withdrawing is minus that's all is this clear I hope you got the idea of what this what this inductive effect is all about about and my dear students this inductive effect is a weak effect it's a weak effect what is meant by this particular term it's a weak effect what is meant by this particular term it's a weak effect my dear students let me tell you let me tell you we do not consider the inductive effect after three carbon atoms if you would have analyzed this particular statement here if you analyzed this particular case over here this carbon I showed positive this positive this positive after that I did not show any charge this carbon I showed negative this carbon I showed negative this carbon I showed negative after that I did not show any charge indictive effect we are only going to take up to three carbon NS the charges du to inductive effect we are only going to take up to three carbon atoms after that inductive effect vanishes after that inductive effect vanishes after that inductive effect vanishes I believe this is clear okay I believe this is clear these two statements I already told you look at these statements and understand whether these are clear to you or not that atom or group which decreases the sigma electron density in the chain we say that group shows minus a effect that group or atom which increases the sigma electron in in the chain we say that atom or group shows plus I effect right this is understood now this is understood now this is understood now now my dear students there is one more term Plus plus I and minus I we say they are relative to each other we say they are relative to each other what does that mean what does that mean we say they are related to each other try to understand one thing look at this particular molecule here you have got three member rink here you have got four member ring I would say this particular carbon is more Electro negative this particular carbon is comparatively less electr negative we know due to angle strains this is more electr negative so it will attract this bonded pair towards itself due to which it gets Delta negative this gets Delta positive if you particularly look at this particular group if you particularly look at this group what do you think is this group is this group pulling the electron density pulling the sigma electron density towards itself I would say yes this group is pulling the sigma electron density towards itself and that group which pulls the sigma electron inity towards itself we say that group shows minus I effect now people look at this particular group look at this particular group is this group pulling the sigma electron and you're pushing it is pushing the sigma electron in towards the other one right so I'll say this particular group is showing plus I so basically this plus I and minus I they are related to each other I would say this group is showing minus I with respect to this one and this is showing plus I with respect to this one right this group it's showing minus I with respect to this and this is showing plus I with respect to this so plus I and minus I they are going to be related to each other they related terms they related terms one is withdrawing with respect to another another is donating with respect to another yeah similarly look at this particular case here the carbon here the carbon is sp hybridized right here the carbon is SP2 hybridized perfect right so this carbon will be more Electro negative if it is more electr negative it will attract the sigma electron towards itself it will attract the bonded pair towards itself due to which it gets Delta negative this gets Delta positive now people just try to understand one simple thing look at this particular group look at this particular group this particular group is it pushing the electron dens or pulling the electron it is pulling the electron inate if it is pulling the electron inate that means it is going to show my side this particular group is it pulling the electron density or pushing the electron density it is pushing the electron density right so this group is showing plus I this group is showing plus I I I'll say this particular group is showing minus I with respect to this one and this shows plus I with respect to this one so again I can categorically say this plus I and minus I they are relative terms they are relative terms right they are related to each other I believe this particular statement is also clear to you now guys in order to do the questions on inductive effect we have to remember certain things in order to do the questions we have to remember certain things Point number one all the C ionic groups all the cic groups they show minus I effect all the cic groups they show minus I effect right all the kic groups kic groups means group carrying positive group carrying positive group carrying positive any Group which carries the positive charge remember it's always going to show minus I it's going to withdraw the sigma electron in state it's going to be electron withdrawing it's going to be electron withdrawing Point number one point number one all the kic groups they show minus I effect they are withdrawing in nature right minus I power of kic group minus I power of kic group will be always greater than that of minus I power of neutral group minus I power of kic group will be always greater than that of minus I power of neutral one more thing guys I will have to use all these things in the questions that's why I'm keep I keep on telling you all the time to remember all these things all the kic groups they show Min power mini power of Kat group will be greater than that of mini power of neutral group all your halogens chlorine chlorine bromine iodine all your halogens they show they are withdrawing in nature they show minus effect minus ey effect all your halogens they are withdrawing in nature they electron withdrawing in nature they show minus ey effect right groups like o nh2 NO2 CN they are also minus I they also show minus I now the point is why now the point is why understand my dear students if this o if this o is attached to a carbon chain if this o is attached to a carbon chain oxygen is more electronegative than carbon so it will attract the bonded pair towards itself right it's behaving like electron withdrawing if it is electron withdrawing so minus I right look at this one if this nitrogen is attached to the carbon chain look at the first atom nitrogen if this is attached to the carbon chain nitrogen is more electron than carbon it will attract the sigma electron towards itself right again Min nitrogen attach this with carbon chain again minus I this is CN this is C triple bond n attach this with carbon chain attach this with carbon chain what you'll observe what you'll observe look at the hybridation of this particular carbon it's SP look at the hybridation of this particular carbon it's sp3 right this one is more Electro negative if this one is more Electro negative it will attract the sigma electrons towards itself right attracting so it is behaving like electron withd Drawing Group so this is going to show what this is going to show minus I so CN also shows minus I I'm showing you exactly I'm showing you exactly how to identify whether the group is showing plus I or minus I right I'm exactly identifying how to I'm exactly showing you how to identify whether the group is showing plus I or minus I people try to look at this particular case this is c dou bond o imagine this particular group is attached with a carbon chain imagine this particular group is attached with a carbon chain what you will observe carbon oxygen which one is more Electro negative oxygen if oxygen is more Electro negative so this will carry Delta negative this will carry Delta positive now this Delta positive on carbon will attract these Sigma electrons to itself right perfect so is this group behaving like electron withdrawing or electron donating this is electron withdrawing so I'll say it will directly show mini right it will directly show mini look at this particular case this will also Min side because this minus Delta this is plus Delta this is minus Delta this is Plus plus Delta this minus Delta this plus Delta so all these groups whatever are mentioned on the screen they are going to show minus I effect they're going to show minus I effect all these things are clear say yes or no ones in the chats I will have to utilize all these things in the questions people I'll have to utilize all these Concepts in the questions everyone quickly quickly everyone so from this particular slide what we have to remember minus all the c iic groups they show I they show minus I they are withdrawing right minus I power of kic group will be greater than minus I power of neutral group second thing and third thing all these halogens all the halogens they are electron withdrawing they show minus sign and all these groups all these groups whatever are mentioned on the screen they show minus I right when attached to the carbon chain yes perfect people I believe this is clear now similarly similarly similarly what are Plus showing groups what are plus I showing groups my dear students just opposite of what I told you just opposite of what I told you plus I means electron donating plus I means electron donating all the anic groups all those groups which carry the negative charge all those groups which carry the negative charge all the an groups they are electron donating they show plus I they show plus I they show plus I number one number two plus I power plus I power of anionic groups plus I power of an groups will be greater than that of plus I power of neutral groups so anic groups they are more electron donating than that of your neutral groups than that of your neutral groups all your alkaline all all your alkal groups alky alkal group alkal group ch3 right this is methy this is methy right C2 H5 this is eile perfect all your alkal groups alkal groups they always show plus I that means they are electron donating or withdrawing they electron donating why they are electron donating have a look on this one as well imagine this is ch3 what imagine this is ch3 right this is ch3 here it is attached to a carbon chain it is attached to a carbon chain now among carbon and hydrogen which one is more Electro negative you tell me among carbon and hydrogen carbon is more electr negative so this carbon will be definitely getting Delta negative right because carbon is more electr negative if carbon gets Delta negative this is a bond pair here this is a bond pair here there will be repulsions there'll be repulsions this negative charge will repel this Bond pair and will push the sigma electron density in the chain if it is pushing the sigma electron density in the chain so I'll say this particular group is it donating or withdrawing this is donating and your donating here are called as plus I all your alkal groups be it methy ethy propile butle whatever is right they're always going to be donating right as for plus I yes let me know in the chats quickly let me know in the chats quickly people let me know in the chats quickly everyone everyone means everyone everyone means everyone quickly I have to tell you one more important statement there are two three few things which I need to tell you about inductive effect then we'll see its application we'll see the types of questions which are asked my dear students one more important thing plus I whenever you hear the term Plus plus means electron donating and right now I'm talking about plus I power plus I power it is directly proportional to number of branching at the free valent carbon plus I power it is directly proportional to number of branching at free valent carbon what does it mean imagine this is a group imagine this is a group imagine you are attaching this group with a carbon chain imagine you're attaching this group with a carbon chain this is a carbon chain with which you attach the group that carbon of the group that carbon of the group which is directly attached with the chain you'll be calling that as free valent carbon this is your free valent carbon this is your free valent carbon as per this particular statement plus I power is directly proportional number of branches if I ask you how many branches this free valent carbon has this is one of the branch which it has this one more Branch so this free valent carbon it has got two branches here it has got two branches here right look at the other case if this is the group and you're attaching this with the carbon chain you're attaching this with the carbon chain this carbon is directly attached with the carbon chain so this is your free valent carbon this is your free valent carbon and 1 2 and three there are three branches which are associated with the free valent carbon do remember one thing more branches means more plus I more branches means more plus I more branches means more plus I now you tell me you tell me one simple thing you tell me one simple thing if I talk about this particular group this is group one and this is group two which group is maximum branched at the free valent carbon the second one right so it's plus I power be more so this group will have more plus I power as simple as that because branching is more yes simple simple now now now if the branching is same if the branching is same if the branching is same if the branching is same then plus I power is directly proportional to number of carbon atoms if the branching is same then plus I power is directly proportional to what number of carbon atoms what does that mean uh invite code for tomorrow's 11:00 a.m. extra problem session this is guys it might ask you the code V 1 is the code C10 is the code for tomorrow's for tomorrow's 11:00 a.m. extra problem session of go okay perfect all right look at this particular thing guys what what I was talking about if the branching is same for example this is the group which is attached to the carbon chain this is the carbon chain for example this is the free valent carbon if this is a free valent carbon how many branches just one branch it has right one branch it has similarly if this is a group if it is attached with a carbon chain this is a free valent carbon here is a prevalent carbon how many branches it has it has two branches right this is the prevalent carbon here it also has got one branch only why am I writing two Branch it has one branch right this particular carbon how many Branch it has it also has one branch perfect so whatever free valent carbons we had in these three cases all the free valent carbons are having one one branch perfect one one branch now if the branching is same how do you relate plus I power plus I power power is directly proportional number of carbon atoms tell me in which branch carbon atom is more carbon atoms are more here so this particular group will be showing more plusi this will be showing more plus I followed by this particular one followed by this particular one so do remember if the branching is different more the branches more more the branches at prevalent carbon more the plusi if the branching is same then then what do you have to do then you have to look at the number of carbon atoms in the branch more carbon atoms in the branch means more plus I I believe this is clear yeah perfect similarly if you talk about hydrogen do you know how many isotopes of hydrogen do we have how many Isotopes hydrogen has you should know it how many Isotopes hydrogen has quickly three protium duum titium proteum duum titium and all these isotopes of hydrogen they show plus I power basically they show plus I power and do remember plus I power of Trum is maximum then that of duum then that of what then that of proteum on this particular thing again there will be few questions that will be asked perfect are you ready to solve the questions now are you ready to solve the questions people are you ready to solve the questions are you ready to solve the questions are you ready to solve the questions you can only solve the questions if you remember all these things this was the first thing which you have to remember some common minus a showing groups remember these particular points common plus I showing groups right similarly branching right how it affects how it affects a particular set of questions I'm going to give you you have to answer you have to answer a particular set of question there is a lot of application as for this inductive effect right a particular application arrange the following according to the mind Min I power arrange the following according to their minus I power you tell me arrange the following as per their minus I power quickly quickly guys chlorine chlorine bromine iodine which one will be having more Min side which one will be having more M side you know it chlorine is more withdrawing than chlorine then bromine then iodine right chlorine attached with carbon chain oxygen attached with chain nitrogen with chain which one is more withdrawing this will be more withdrawing simple this is a cic group this is a neutral group all the cic groups they have got more plus sorry more minus so this more withdrawing this more withdrawing look at the next one oxygen attached with the chain nitrogen attached with the chain oxygen is more withdrawing more mini more mini right more Min side this is your SP hybridized carbon because it's forming two Sigma bonds this is your SP2 this is your sp3 this is your sp3 where is the S character more s character is more here electron negativity is more here so it will be more withdrawing this will be more withdrawing this is how you arrange these groups on the base of their mini power simple simple this is how you are going to arrange them on the base of their mini power now tell me about these tell me about these among these three one two or three which one will show more minus I power all the concepts I gave you all those Concepts only you have to utilize all those Concepts I'll give you break once indc effect is completed indc to effect and its application part is completed then I'll give you break and after the break we'll start the resonance quickly it's very evident guys if this group is attached with a carbon chain this group will carry Delta positive because oxygen carries Delta negative electro negativity difference oxygen is more elect negative so Delta negative Delta positive even this there is electro negativity difference here so this Delta positive this Delta negative right perfect now you tell me just one thing carbon oxygen electr negativity difference is more than that of carbon nitrogen if the electr negativity difference here is more that means the magnitude of positive charge here is more right so magnit magnitude of positive charge hair is more than that of magnitude of positive charge hair right if the magnitude of positive charge here is more can I say it will it will pull Sigma electrons more towards itself right so what has to be the electron withdrawing nature mini power this is the Min power simple and basic right nf3 NH3 both are cic groups nf3 NH3 a nitrogen is attached with Florine a nitrogen is attached with what it's attached with hydrogen chlorine is more electron chlorine is more electr negative than nitrogen so chlorine will increase the intensity of positive charge on nitrogen chlorine will in increase the intensity of positive charge on nitrogen due to which it becomes more withdrawing right NH3 ph3 ash3 one simple thing nitrogen phosphorus arsenic antimony do they belong to same period or same group they belong to same group in the group vcd plays a role in the group vcd plays a role right in the group vcd plays a ro this one has got lesser size lesser size means more volume charge density more volume charge density means more positive volume charge density if over here you have got more positive volume charge density so it's going to pull the sigma electrons more towards itself so it will show more minus I simple and basic O2 positive sh2 positive same reason oxygen carrying positive sulfur SE oxygen sulfur SE they belong to same group again vcd lesser size more positive volume charge density more positive charge volume density means more withdrawing so this is more withdrawing more Min side right now nr3 positive NH3 positive nr3 positive NH3 positive can you let me know in the chats what do you think which one will show more minus sign quickly which one will show more M sign I'm asking you this is the question I'm asking you quickly which one will show more minus sign quickly guys everyone quickly normally it seems like this should have more mini it seems like because nitrogen here is attached with hydrogen right here nitrogen is attached with alky groups alkal groups are donating so normally a common man thinks that alkal groups are donating alky groups are going to decrease the intensity of positive charge here alkal groups are going to decrease the intensity of positive charge here this has to be more minus I normal people think like that but we are not going to think like that we are not going to think like that see we have got nr3 positive this is R this is R this is R nitrogen carrying positive this is n this is hydrogen this is hydrogen there's one more hydrogen carrying positive right perfect let me first of all tell you the first one has got more minus I there are lot of reasons for it one reason is solvation reason right or in simpler words just to remember it just to remember it what you should remember exactly so that you can remember it have a look r r r alky alkal are these bulky groups you can remember it like this are these bulky groups these are are bulky groups if they are bulky groups will there be repulsion yes there'll be repulsion if there is repulsion Bond angle increase or decreases Bond angle increases if Bond angle increases electro negativity automatically yes right perfect electro negativity automatically increases so which nitrogen do you see more Electro negative here the first one right if first one is more Electro negative it's going to draw more electron density from the carbon chain so it will show more minus there's one more logic that is solvation I'm not going into the details right I believe this particular statement is clear I believe these sort of questions you can easily solve now guys this is the minus I series since I have showed you exactly how to check which one has got more minus I which one has got more plus I on the base of all those points this is the series which you have to remember my nf3 positive will show maximum minus I power followed by nr3 positive followed by NH3 positive all these three are cattiny groups then comes your neutral groups right then comes these you have to remember this particular series you can create a nimonic based on your own thing yes yes Danish you can watch it perfect this is the minus I series similarly there's a plus I series as well there's a plus I series as well first of all look at all these These are anic groups and you know all the anic groups they do what all the an groups they they have more plus I power then the neutral groups so that's why an groups are the first then comes your neutral and all you have to remember this particular series as well you have to remember this particular series as well one is for one is for minus I series and another series is plus I series guys okay plus I series one thing which I told you already one thing which I told you already I believe I believe you remember this both plus I and minus I are considered only up to three carbon atoms both plus I and minus I they are considered on only up to three carbon atoms because inductive effect it's not a strong effect it's a weak effect okay so this was all the theory which was needed in the inductive effect now is the time for questions now is the time for questions but before doing the questions I need to tell you certain things which will help you which will help you which will help you in order to solve the equations my dear students the first point which I would want you guys to remember that is stability of carbocations stability of carbocations as per inductive effect as per inductive effect remember stability of carbocation carbocatine is stabilized by electron donating groups carbo caes are destabilized by electron withdrawing groups carbo catins are stabilized by electron donating groups they are destabilized by electron W drawing groups that means in short since till now we have studied only inductive effect so I'm not going to take any other effect into consideration I'm just going to solve the equations as for what I'm just going to solve the equations as per indictive effect electron donating as per indictor means plus I electron withdrawing as per indictive means minus I so the first type of question will be done from this particular equation which I gave you stability of carboca right electron donating group it stabilizes carboca electron withdrawing group it destabilize carbocation perfect so stability of carbocatine is directly proportional to plus I inversely proportional to minus I point number one point number two stability of caronian stability of caronian stability of caronan caronan is stabilized caronan carbon carrying negative caronan is stabilized by electron withdrawing they are destabilized by electron donating groups they destabilized by electron donating groups they are destabilized by electron donating groups and my dear students electron withdrawing as per inductive means minus I electron donating as per IND to means plus I so this is one more question which we are going to do there's one more question which we are going to do right there's one more question which we are going to do the third one the third one acidic strength acidic strength strength of acid as I told you already strength of acid strength of acid is directly proportional to Ka is inversely proportional to PKA and let me tell you strength of acid it is directly proportional to what you know it already it is directly proportional to stability of conjugate base stability of conjugate base and let me tell you this conjugate base it is stabilized it is stabilized by electron withdrawing groups conjugate base is stabilized by electron withdrawing it is destabilized by electron donating groups if I only talk inductive effect if I only talk in terms of inductive effect electron withdrawing means m s electron donating means plus I so few more questions you can get from this particular thing few more questions you can get from this particular statement few more questions you can get from this particular statement since I till now taught you only inductive effect we are going to do the questions only based on inductive effect now yeah now similarly the fourth question the fourth question that is strength of a base strength of a base strength of a base my students if I ask you what a base is as per Lewis concept as per leis concept what how do we Define a base does the base accept a pair of electron or lose a pair of electron tell me that does a base accept a pair of electron or lose a pair of electron a base loses a pair of electron donates a pair of electron a base donates a pair of electron right you know that it donates a pair of electron correct does the base accept H positive or donate H positive tell me that does a base accept H positive or donate H positive does the base accept HED positive or donate H positive it accepts HED positive base accept H positive can I say h positive can attack on the base H positive can easily attack on such a base where the electron Den will be more H positive will except will H positive will attack easily on such a base wherein the electron density at some sight will be more agreed right so tell me strength of base base for the h positive to attack easily electron density has to be more or less it has to be more how electron density is going to be more can I say electron donating groups electron donating groups electron donating groups if attached to a base will increase the strength of the base and electron withdrawing groups electron withdrawing groups if attached with the base are going to decrease the strength of the base simple right now electron donating means over here in terms of inductive effect it's plus I electron withdrawing in terms of inductive effect it's minus I so can you solve all these questions can you solve all these questions by considering only inductive effect till now we have studied only indictive effect now anything can be asked from this particular slide anything anything means anything anything means anything now now now let's try to solve certain questions so that it'll be absolutely clear to every one of you absolutely clear to every one of you okay just a second people just a second just a second the first and the basic question you are going to answer me in the chats now I want the chats to move fast among these two among these carbocation which which one is more stable among these two which one is more stable quickly who is this model bro please stop spamming please stop spamming let people study you are writing East or West alak sir is the best of course he is the best we appreciate him right we love him every teacher in the country loves him because of the contribution which he has made right but don't write all these things let people study right let people study okay right so look at this one now look at this one tell me among these two which is which is more stable quickly guys you are not supposed to take this much of time you're not supposed to take this much of time quickly one or two one or two one or two quickly a wait wait wait I'll do this at the end I'll do this at the end tell me among these three tell me among these three carboca stabilized by plus I see this is plus I showing plus I showing plus I showing so 3 plus I showing since I only taught you inductive effect yet so I'm going to answer the questions only on the Bas of inductive effect so three plus I sh groups how many two plus I showing groups how many only one plus I showing group right so more plus I showing groups more the stability 1 2 3 done question number one question number one look at this one look at this one all these are carbocation right carbon carrying positive carbon positive carbon positive here you have got ch3 attached CD3 attached NO2 attached carboca are stabilized with plus I carboca are stabilized there electron donating groups now ch3 ch3 shows plus I of course right CD3 it also shows plus I right NO2 it shows minus I right so so minus i showing group it destabilizes the carbocation so stability of the C at the end right now among one and two which one is showing which one does show more plus I which one shows more plus I is it ch3 or CD3 I hope you remember 3um shows more if you remember in that that part 3um more followed by D followed by H similarly CD3 CD3 will show more plusi than that of ch3 right more plus I means more stability so B is the order simple look at the next one look at the next one no group over here you have got NO2 attached over here you have got NO2 attached NO2 is it withdrawing or donating it's withdrawing right it's withdrawing it shows M sign inductive effect is it distance dependent yes it's distance dependent so can I say NO2 will be more withdrawing in the third case NO2 will be more withdrawing in the third case and if in the third case NO2 is more withdrawing it will destabilize it will D stabilize this particular speci to the maximum so this the stability of C is the least right a will be maximum followed by B simple ABC is the order ABC is the order ABC is the order nothing to do tell me about answer of this particular question a b or c again these are carboca again these are carboca CD3 electron donating plus I CD3 plus I CD3 plus I right now inductive effect is distance dependent where is the distance less here the distance is less so more promting plus sign nature more promting plus sign right so what has to be the stability order plus I increase the stability of carboca so c will be maximum stable right followed by B followed by a right look at the next one look at the next one look at the next x one this your alkal group plus I this again your alkal group plus I this again your alkal group plus I right now which alkal group shows more plus I where the branching is more where is the branching more b b so this will be maximum carbocatine followed by What followed by C followed by a BCA is the order yes clear guys the chart should run continuously the charts should run continuously you're supposed to answer all the questions all the equs look at this one a b or c which one is more stable all these are carbocation carboca are stabilized with plus I see this is alky showing plus I right NO2 showing minus I right CN showing minus I plus I groups they stabilize the carboca plus I groups they stabilize the carboca so a will be maximum stable right now among B and C which one is less electron withd drawing CN is comparatively less electron withd drawing than NO2 so c will be more stable followed by B if you remember the series series I gave you right NO2 is more with shows more minight than that of CN now look at this one look at this one positive charge is here and positive charge is here right look at this double bond this is SP2 SP this is SP2 this is also SP2 but here all are sp3 SP2 means more withdrawing withdrawing they D stabilize the carboca right so stability of B is more than that of a yes right going ahead going ahead look at this one CD3 ch3 ct3 plus I showing even this is plus I showing even this is plus I showing now which one is more plus I showing ct3 is more plus I showing than CD3 then ch3 and plus I showing group they stabilize the the carboca so stability of C will be maximum followed by What followed by a followed by B this the stability order oh yes welcome to the night mode welcome to the night mode welcome to the night mode okay look at the next one look at the next one people look at the next one carbon carrying positive here carbon is carrying positive carbon carrying positive right over here Nitro group is ATT attached here Florine is attached here no such group no withdrawing group is attached as such right perfect so this shows minus I this shows minus I and minus I groups they destabilize the carboca there you do not have any electron withdrawing group so stability of a is maximum now this is more withdrawing if this is more withdrawing it will destabilize the carboca to the maximum so B at the end here is going to be C so ACB is order ACB is order yes ACB is is the order look at the next one look at the next one look at the next one I think this is the question which we already solved C carrying positive here positive so one plus I showing group here over here two plus I showing groups here you have got three plus I showing groups right so stability order will be like this because plus I showing groups they stabilize the carboca right now since since I did not teach you hyper conjugation or anything yet I'm solving the equ directly on the Bas of indect effect now this particular carbon carrying it's getting electrons from one side this particular carbon carrying positive it's getting electrons from three side this particular it's getting electrons from two sides right perfect right so where will be the positive charge compensated the maximum positive charge is compensated the maximum here right if the positive charge intensity is decreased here the maximum the maximum decrease in the intensity of positive charge is here right so it'll be more stable so two will be more stable than that of three and at the end it's going to be one now ch3 ch2d duum chd2 CD3 positive right more than D more the D more is the stability right Nitro CN F Nitro CNF or are all are electron withd drawing so minus I minus I minus I the one which shows lesser minus among the three the one which shows lesser minus I among the three which one shows lesser minus I so third shows less minus I so it stability will be more followed by two follow followed by one right perfect I believe you are you are able to solve these questions easily now now now one more one more among these two species which one is more stable can you let me know among these two species which one is more stable carefully look suur carrying negative oxygen carrying negative rest everything is same this part is same even this part is same sulfur carrying negative oxygen carrying negative you tell me the answer sulfur and oxygen they belong to same group same group means vcd volume charge density right volume charge density sulfur more size less vcd less vcd means more stability so this is the stability order right perfect this question we have already solved this is your SP hybridized carbon carrying negative this is your this is your sp3 hybridized nitrogen right this is your sp3 hybridized nitrogen perfect now you tell me which one is more Electro negative which one is more Electro negative SP hybridized carbon is more Electro negative negative charge on an electr negative element is considered to be stable so this is the stability order done perfect perfect tell me the order of these questions I think yeah okay now these are not your caroa these are not your caroa these are your carbon carbon carrying negative carbon carrying negative carbon carrying negative caronian are stabilized by electron withdrawing minus i showing ganians are stabilized by minus i showing groups yeah now ch3 it's plus I CN minus I NO2 minus I minus i showing groups they stabilize the caronian now which one shows more minus I Nitro Nitro shows more minus sign so C then B then a this is the order right guys keep on telling me the answers in the chats I'm looking at your chats look at the next one look at the next one all these are carbanions this shows minus I right even this Florine shows minus I perfect CD3 it shows plus I and since these are caronian caronian are stabilized by electron withdrawing so there is electron withdrawing this electron withdrawing now which one is more electron withdrawing nf3 positive is more electron with drawing it shows more minus sign right so a is more stable followed by B followed by C right done and dusted done and DED now look at other cases look at the other cases just a second just a second people just a second just give me a second okay look at this particular case again these are caronian carbon carrying negative negative negative tell me where is the stability maximum where the stability is maximum I need the answer I need the answer where is the stability maximum what are the concepts I gave you use those Concepts only nothing else I'm going to give you the break after this inductive effect is done quickly quickly what do you think see this is oxygen it will be withdrawing electrons here the oxygen is comparatively close right here there is no oxygen wherever oxygen is close wherever withdrawing group is close there only Carboni and stability is maximum so B will be maximum stable followed by a followed by C right this question I believe you should you should easily be able to solve suf carrying negative oxygen carrying negative right sulfur carrying negative oxygen carrying negative sulfur and oxygen they belong to same group vcd plays a role this has got less vcd less vcd means more stability so negative charge is more stable in this so a is more stable than b a is more stable than b now compare B and C compare B and C A is more stable than b compare B and C now oxygen carrying negative nitrogen carrying negative oxygen nitrogen they belong to same period in same same period electro negativity plays a role negative charge is more on more elect negative negative charge is more stable on more Electro negative because oxygen is more Electro negative so B is more stable than c b is more stable than so order has to be a b and then C when you compared these two you took vcd into consideration when you compared these two you took electro negativity into consideration right since they belong to the same period electro negativity plays a role negative charge is more Electro negative and more I mean negative charge is more stable on more electr negative and at the end it'll be carbon carting negative right so this has to be the a AB B CD right guys AB CD has to be the Order Perfect now guys look at this particular question what do you think what do you think quickly okay okay okay okay sorry sorry sorry sorry quickly what is the answer of this question what is the answer of this question that was my previous organization by the way where I used to work okay uh quickly tell me the answer of this particular question acidic strength acidic strength acidic strength acidic strength if you remember guys if you remember just a second acid strength I hope you remember that table which I gave you I hope you remember the table which I gave you acidic strength acidic strength strength of acid is directly proportional to stability of conjugate base and conjugate base is stabilized by minus I so in short I would say minus I groups they increase the strength of acid minus I groups they increase the strength of the acid right plus I groups they decrease the strength of the acid now now you going to give me the answer since you know it already you are going to give me the answer now quickly guys just a second it was slide number 60 I believe this is slide number 60 yes okay so first of all identify the site from where hydrogen is going to donate hydrogen is going to go away as H positive this is the site from where hydrogen is going to leave as H positive this is the site from where hydrogen is going to leave as H positive this is the site from where hydrogen is going to leave as H positive here you have got electron withdrawing right here you have got electron donating electron donating electron withdrawing groups they increase the strength of the acid so I would say a strength will be maximum followed by B followed by C right now majority of the students will do a mistake I believe here in this question tell me which one is more acidic I think a lot of students will do a mistake here quickly which one is which one is more acidic here guys after looking at this question do not straight away C plus I minus I plus M minus I do not do that do not do that why because this is carboxilic acid this is carboxilic acid this is phenol this is phenol and generally you know it carboxilic acids are stronger acids than phenols carboxilic acids are stronger acids than phenols so definitely A and B is is going to be more acidic than C and D okay A and B is going to be more acidic than C and D now among A and B among A and B this is minus i showing this is plus I showing withdrawing electron withdrawing group it increase the strength of acid right so a will be maximum AIC followed by B now among C and D this is withdrawing there is no withdrawing group so I'll say C then D I believe this particular question is done again right agreed right guys okay what about this one what about this one first of all look at the site from where hydrogen is going to go away as H positive I identified the site right over here one Florine two withdrawing three withdrawing more the withdrawing groups more the acidic strength more the withdrawing groups more than acidic strengths done and dusted look at the next one this is nf3 positive okay okay first of all this is your phenol basically this is phenol but this is carboxilic acid carboxilic acids are stronger acids than that of phenols so c will be maximum acidic now among A and B strength of acid is directly proportional to minus I minus I this shows more minus I this is more withdrawing this more withdrawing so after C I'll prioritize a then I'll prioritize B right I believe this is also clear I believe this is also clear okay okay what about this particular equation one or two one or two see from here hydrogen is going to go away as H positive from here hydrogen is going to go away is H positive perfect Now 1 2 at Carbon number two there's withd drawing 1 2 3 at Carbon number three there is with drawing so where will the Florine show more prominent minus I where the distance is less where the distance is less so in the first case more with drawing more with drawing means more acidic strength right okay now now first of all this is the site from where hydrogen is going to go away his H Falls to perfect now you know strength of acid is directly proportional to stability of conjugate base perfect stability of conjugate base right or directly tell me one thing which which ring over here will be more withdrawing angle strain concept angle strain three membered four membered five membered six member three M ring more angle strain more electro negativity more withdrawing more withdrawing if it is more withdrawing that means it will stabilize the conjugate B the maximum right so this is the stability order of the conjugate base as well as the stability as well as the order of the strength of the acids right okay here you might do a mistake here you might do a mistake you might do a mistake in one and two which one is more acid acidic in one and two which one is more acidic tell me that in one and two which one is more acidic tell me that among one and two which one is more acidic quickly are you sure are you sure among one and two which one is more acidic have a look guys first of all make their conjugate bases take hydrogen out as H pos2 this will be cf3 NE right take hydrogen out as H pos2 this will be ccl3 Nega these are the conjugate bases see these are your acids and these are their conjugate bases now check the stability of these conjugate bases check the stability of these conjugate bases have a look guys try to understand carbon carrying Nega carbon carrying negative right but this particular carbon is attached to three Florin atoms right this is carbon and you have got three Florin atoms perfect look at this particular carbon it is attached to three chlorine ATS three chlorine items these are the conjugate bases basically which we have now tell me one thing in the Florine do you have any vac orbitals in Florine are there any vacant orbitals no right in chlorine if you remember chlorine in chlorine there are vacant D orbitals chlorine configuration is 3s2 3p5 but there is 3D as well 3D Z there are vacant D orbitals so if there are vacant D orbitals in chlorine can I say this if there's a vacant D orbital in the chlorine can I say it can show the back bonding it can show the back bonding if it shows back bonding back bonding stabilizes the carb carbonium or I would say back bonding it stabilizes your conjugate base so which conjugate base is more stable this conjugate base is more stable more stable the conjugate base stronger the acid more conjugate more stable the conjugate B stronger the acid look at the next one take hydrogen out as H positive negative charge hydrogen out is H positive negative charge right perfect now this is sulfur sulfur out of most configuration 3 S2 3p4 and there is 3D as well containing zero oxygen oxygen is simple 2 S2 2p4 right there is no D there's no vacant orbital here you got vacant orbital if there's vacant orbital there is a chance of back bonding and back bonding stabilizes the conjugate base so wherever the conjugate base is more stable that acid is going to be stronger I believe this is clear to everyone yes so with this your inductive effect is clear now is the time for what resonance guys so do you want a break do you want a dinner break right now yes do you want a dinner break right now then we'll start the resonance yes but guys come back are you are you going to come back let me first of all see what you are saying let me open your chats once are you planning to come back tell me that if you are not planning then I'll end the session it's okay are you planning to come back I need that promise from you okay tell me one more thing whatever I've taught you till now is every single thing clear honestly can you do the questions based on the concepts which I gave you till now honestly you are going to let me know honestly you guys are going to let me know I want everyone to let me know quickly whatever I've taught you till now is every single thing clear so I would want every one of you to join back again on time right so what time will be meeting it's 7:45 right now 7:45 right now so maximum it will take you 30 to 40 minutes for dinner right even I'll have dinner so so we'll be back at 8 8:25 we'll be back okay so session resumes at 8:25 perfect have dinner come back because it's a long Marathon going to happen okay it's a long Marathon you have to be with me till the end okay right but be back on time guys be back on time on time on time perfect I'll see you after the break till then you carry on till then you carry on guys is everyone back I guys done with your dinner yes are you all done with your dinner let me know quickly in the chats yes yes I'm done I'm done I'm done guys I'm done what about you are you all done oh people are still having it uh guys tell me one thing like whatever Concepts we discuss till now is every single thing clear whatever Concepts we discussed till now is every single thing clear here let me know in the chats quickly yes suan yes yes quickly in the chats whatever we discussed till now is every single thing clear so let me let me know quickly guys all the things are clear to them all right all right so the last topic which we discussed what was that that was indictive effect and its applications right we saw different Ty typ of questions based on inductive effect in which you were supposed to talk about the stability of carbocation stability of caronan acidic strength right strength of acids strength of bases all those things were discussed in detail so now my dear students we are going to move into one more important topic of this particular chapter you can call it as the most crucial part of the chapter and that is going to be resonance just let me know once in the chats how many of you have studied resonance before someone is asking sir when is the break again will complete resonance its applications and a lot of things related resonance right and after two two and a half hours I'll give you one more break for 15 minutes okay all right so let's move on then let's have a look on what resonance exactly is all about my dear students before letting you know the exact meaning of resonance we should know first of all which kind of molecules do exhibit this property which of the molecules can show resonance which of the molecules can not show resonance and after that I'll let you know what this resonance exactly is so my dear students I'm going to write one simple statement over over here try to understand the statement is like this whenever whenever a true structure whenever a true structure of a molecule cannot be drawn whenever true structure of a molecule cannot be drawn or or if any structure or if any structure cannot explain cannot explain all all its properties all its properties we say we say the molecule exhibits resonance we say a molecule exhibits resonance this is one very very very important point my dear students whenever you will be unable to draw true structure of the molecule whenever you'll be unable to draw the actual structure of the molecule or if any structure of the molecule will not be able to explain all its properties if any structure of the molecule will not be able to explain all its properties at that point of time we say that the molecule exhibits resonance now what is meant by it let's try to understand this a bit more in detail try to understand guys first of all I'm sure everyone of you would have come across the term Benzene right Benzene every one of you would be knowing now my dear students experimentally it has been observed experimentally it has been observed that in Benzene all the bond lengths are equal experimentally it has been observed in case of benzene all the TC Bond length are equal all the carbon carbon Bond length are equal this has been observed experimentally Point number one and at the same time it has been observed that Benzene it does not behave like a normal alen it has been observed that I mean it has been decked my dear students that in case of benzene all the carbon carbon Bond lengths are equal experimentally it has been observed and apart from that it has been observed that in case of benzene I mean Benzene does not behave like a normal Al correct if I ask you guys what is the structure of benzene some of you will come up with this particular structure some of you will come up with this particular structure right and some of you will come up with this particular structure some of you will come up with this particular structure if I ask you what is the structure of benzene some of you will say this is the structure of benzene and some of you will say this is the structure of benzene now my dear students let me call this as carbon number one let me call this as carbon number two this is three this is four this is five this is six right do you know one thing that bond order bond order is directly proportional to bond strength right which is inversely proportional to bond length I believe everyone of you would be knowing this bond order is directly proportional to bond strength which is inversely proportional to bond length right now guys try to understand if you look at this particular structure this is the structure which you gave me about Benzene right some of you said this is the structure of benzene and some of you said this is the structure of benzene if you look at the first structure in the first structure there's a double bond between this carbon and this carbon there's a double bond between this carbon and this carbon so between carbon number one and carbon number two there's a double bond as for this particular structure right between carbon number two and carbon number three there's a single bond between carbon number two and carbon number three there's a single Bond now if I ask you as for this particular structure there's a double bond between these two there's a single bond between these two as for this structure will this particular Bond length and this particular Bond length will these Bond l be same or different as per this structure I'll say C1 C2 C1 C2 Bond length won't be equal to C2 C3 Bond length correct very simple see if you look at this particular structure as for this structure there's a double bond between one and two there's a single bond between two and three if there's a double bond and a single Bond definitely their bond lengths won't be equal right but my dear students students my dear students already we know something about Benzene we know in case of benzene all the CC Bond lengths are equal but as for this structure CC Bond lengths are not coming out to be equal this Bond length as per the structure will be less and this Bond length as for the structure will be more right so as for this structure the CC Bond lengths are not equal so if I ask you is this structure able to explain all the properties of enzine you'll directly say no this structure is not able to explain all the properties of benzene because we know in case of benzene all the CC Bond lenss are equal but as for this structure CC Bond lengths are coming out to be different because somewhere there's a double bond somewhere there's a single Bond right similarly if you look at this particular structure if you look at this particular structure let me call this as one let me call this as two this is three this is four this is five this is six as for this structure between one and two between one and two there's a single Bond right and between two and three there's a double bond perfect so if I ask you as per this particular structure of Ben Benzene will C1 C2 Bond length will C1 C2 Bond length be same as that of C2 C3 you'll say no right they won't be equal as for this particular structure they won't be equal perfect because as for this structure there's a single Bond here there's a double bond here perfect right so if I ask you are these two structures able to explain all the properties of benzene these structures are not able to explain all the properties of benzene this particular structure is not unable to explain all the properties even this particular structure is unable to explain all the properties of benzene whenever you Cann not draw the true structure of the molecule or if any structure will be unable to explain all the properties of the molecule at that point of time we say the molecule exhibits resonance so can I say Ben Benzene will be exhibiting resonance absolutely Benzene will be exhibiting resonance absolutely Benzene will be exhibiting resonance right because none of the structures of benzene are able to explain all its properties and any such molecule whose structure cannot explain all its properties we say that molecule exhibits resonance okay clear guys so Benzene is the molecule which exhibits resonance perfect now what this resonance exactly is all about before talking about resonance there are few points which you need to know before understanding what this resonance is there are few things which you need to know my dear students have a look exactly what those things exactly are have a look exactly what those things exactly are have a look people just a second let me create a space over here let me create a space over here okay my dear students in order to understand the term resonance you should know first of all what is a localized electron pair you should know first of all what is a localized electron pair and how do we Define the localized electron pair when an electron pair is shared only between two at if an electron pair is just shared between two atoms we call that particular electron pair as a localized electron pair for example for example this electron pair it is just shared between A and B this particular electron pair it is just shared between A and B right any such electron pair which will be just shared between two atoms you call that particular electron pair as the localized electron pair you'll be calling that as the localized electron pair when the electron pair is shared by just two atoms you'll be calling that as the localized electron pair now what will be the meaning of delocalized electron pair then what will be the meaning of delocalized electron pair my dear students any electron pair any electron pair which will be shared any electron pair which will be shared between more than two atoms any electron pair which will be shared between more than two atoms that is what you will be calling as delocalized electron pair for example have a look let's say I've got a molecule this is the molecule let's say this is one more electron pair which I'm keeping between A and B this is one molecule right this is one electron pair which I kept between A and B correct now my dear students for example if I draw the structure of this molecule again if I draw the structure of this molecule again and by chance if the same electron pair which was earlier between A and B now if the same electron pair is between B and C earlier this electron pair which I'm representing by cross it was between A and B now in one more structure of the same molecule in one more structure of the same molecule the same electron pair is between B and C so the first structure is telling you that this particular electron pair is shared between A and B the second structure is telling you that the same electron pair is shared between B and C now tell me whether this electron pair is shared by two atoms or three atoms I'll say this particular electron pair is shared among three atoms any such electron pair which will be shared which will be shared by more than two atoms you call that particular electron pair as delocalized electron pair so this particular electron pair which we have it is not shared only between two atoms it is shared between three atoms and any such electron pair which will be shared by more than two atoms you call that electron pair as the delocalized electron pair correct right now my dear students the point is which electron pair will be called as delocalized electron pair the point is which electron pair will be called as delocalized electron pair a simple and basic statement a high energy electron a high energy electron pair can be considered as delocalized like your Pi electrons now what is meant by this statement since I told you delocalized electron pair is that electron pair which is shared between more than two atoms now which electron pair can be de delocalized my dear students over here is mentioned high energy electron pair can be delocalized now what is high energy electron pair you must be thinking what is high energy electron pair let me tell you first of all what is high energy electron pair try to understand exactly what is high energy electron pair have a look for example imagine this is an S orbital of one atom imagine this is s orbital of one more atom right now my dear students when these Atomic orbitals will approach each other when these Atomic orbitals will approach each other what will happen will there be actual overlapping will there be headon overlapping absolutely there will be head-on overlapping absolutely there will be head-on overlapping right and due to headon overlapping which bond is formed it is the sigma bond which gets formed due to headon overlapping so this particular bond which got formed here by the actual overlapping this particular Bond I'm calling as Sigma Bond correct that's something which I call a sigma Bond now you tell me one thing the sigma bond which gets formed over here right this particular common electron density can I say this particular common electron density it is under the influence of both these nuclear directly this particular electron density is under the influence of both these nuclei directly right now now try to understand one more thing let's say this is your p orbital this is one more p orbital I'm keeping their X's parallel if I keep their XS parallel tell me will there be actual overlapping or parallel overlapping there will be parall overlapping and due to parall overlapping which bond gets formed Pi bond gets formed so what will happen these lobes they'll Bend towards each other and they'll be parall overlapping right so this is your p orbital and this is your one more p orbital and this particular electron density right this common electron density it results in the formation of what it results in the formation of Pi Bond perfect now how do I make it at the end try to understand this is the nucleus even here you have got the nucleus perfect so let me make it like this let me make it like this this is the common electron density at the top this is the electron inity here right this is the nucleus this is the nucleus so in between I would say over here there'll be a plane over here there'll be a plane on which electron density won't exist and you call this particular plane as noal plane I hope you know that right now my dear students just tell me just compare these two just compare these two this is a plane on which no electron density lies electron density lies on the top and below the plane electron density lies above the plane and below the plane correct now tell me whether this electron density which you call us Pi electron density right whether this electron density whether this electron density is under the direct influence of nucleus the electron density is not under the direct influence of the nucleus if the electron density was under the direct influence of the nucleus now tell me which electron density among these two Sigma electron density or Pi Elon Pi electron density which electron density will be experiencing more force of attraction by the nuclear can I say this electron density will be experiencing more force of attraction can I say this particular electron density will be experiencing more force of attraction by the nuclear because it is under the direct influence of these nuclear right can I say this electron density this Pi electron density it is under less force of attraction it is subjugated to less force of attraction that electron density which is subjugated to less force of attraction that electron density is something which you call as high right that electrons those electrons which are under less force of attraction by the nucleus which are subjected to less force of attraction by the nuclear that is something which you call us high energy electron density that is something which you call as high energy electron density so what is the high energy electron density why did I tell you all this because the point was very simple initially I told you first of all what is a delocalized pair delocalized electron pair is that electron pair which is shared by more than two atoms now the point was which electron pair can be delocalized I told you high energy electron pair can be delocalized like your Pi electrons right like your Pi electrons is this clear high energy electron pair that electron pair which will be subjugated to less force of attraction by the nuclear that is something which you call as high energy electron pair and that high energy electron pair that high energy electron pair can be delocalized yes is this point clear is this point clear guys try to understand carefully what I'm saying first I made you clear what is delocalized electron pair it is that electron pair which is which is shared by more than two atoms now the point was which electron pair can be delocalized it is the high energy electron pair like your P electron pair which can be delocalized and delocalization delocalization of high energy electron pairs delocalization of high energy electron pairs is something which is what you call us resonance did you get the idea of what this resonance is all about did you get the idea of what resonance is all about so resonance is just defined as the delocalization of high energy electron pairs like your P electron right let me know once in the chats if it is clear let me know once in the chats if it is clear the delocalization of high energy electron pairs like your Pi electrons right that is something which is what you call us resonance perfect now guys before making you understand before showing you how do we draw the resonating structures etc etc there is one approach which you need to learn right there is one approach which is which you need to learn that is something something which you call as orbital approach of resonance but butd again before making you understand the orbital approach of resonance there is one more thing which I want to let you know your Pi electrons which are your high energy electrons right they can be delocalized these P electrons are exactly of two types one you call as real P electrons another one you call as differential P electrons if I talk about your P electrons P electrons are of two types basically right P electrons are the ones which are subjugated to less force of attraction by the nuclear right these are high energy electron pairs right and these high energy electron pairs get delocalized basically now these P electron these are of two types one is real Pi one is differential Pi what are real P electrons real Pi electrons are the ones which are present either in double bond or triple bond real Pi electrons are the ones which are present in either double bond or triple bond whereas your differential Pi lone pair you call as differential Pi or negative charge with an unshared pair of electron lone pair and negative charge with unshared pair of electrons these are what you call as differential Pi electron these are what you call as differential P electron these are what you call as differential Pi electrons and my dear students there is a particular energy order of these real pi and differential P electrons which you need to remember there is one energy order which you need to remember regarding these real P electrons and differential P electrons and what is that energy order same negative charge with unshared pair of electron this will be always having more energy followed by the lone pair followed by the pi electron pair of this double bond followed by these two Pi electron pairs of these triple bonds just remember this directly first of all what are these electron pairs what are these electron pairs what the real Pi is all about what this differential Pi is all about these are the types of Pi electrons Pi electrons are high energy electrons and only high energy electrons they get delocalized I mean your high energy electrons they get delocalized they show delocalization right they show de delocalization now people now let me make you understand the actual concept of resonance now let me make you understand the actual concept of resonance I don't understand first of all if I ask you what is the structure of benzene some of you will come up with this structure some of you will come up with this structure right okay now if you exactly have a look on how this structure is formed actually how the structure is formed actually if you have a look on how this particular structure is formed can you do that can you make the orbital diagram of this particular benzene can you make the orbital diagram of this particular Benzene or or I'll do one thing I'll make it comparatively simple let me make it like this let me make it like this right I asked you what is the structure of benzene some of you told me this is the structure of benzene now I'm going to explain how the structure was formed basically how the structure was formed first of all my dear students if you look at this particular carbon can you let me know it's hybridization this particular carbon it's hybridization see one Sigma 2 Sigma and one more Sigma with hydrogen three sigma so it's SP2 hydr even this is SP2 this is SP2 this is SP2 every carbon here is SP2 now my dear students what is the outermost configuration of carbon it is 2s2 2p2 you know that right this is your 2s orbital and this is your 2p these are your three 2p orbitals now every carbon is SP2 hybridized every carbon is SP2 hybridized so can I say every carbon will contain one pure p orbital as well absolutely every carbon will contain one pure p orbital as well so basically basically the hybridation involved here is SP2 right the hybridation involved here is SP2 and I would say there'll be one pure p orbital present at every carbon and my dear students let me tell you in case of SP2 in case of SP2 hybridization what is the geometry as for SP2 hybridization it's triagonal planer it is trigonal planer right you must be knowing this this is trigonal planer what is meant by trigonal planer geometry what is meant by trional planer geometry see if I talk about this particular carbon it is attached with hydrogen as well on this side right first of all the carbon has undergone SP2 hybridization right which leads to the formation of three hybrid orbitals so there'll be three SP2 hybrid orbitals and apart from that there'll be one pure p orbital as well perfect now my de students these SP2 hybrid orbitals what they have done one of the SP2 hybrid orbital has done the overlap form the sigma Bond one more has formed one more Sigma Bond one more has formed one more Sigma Bond correct right perfect and this particular carbon it has got p orbital as well it has got p orbital as well and that p orbital will be perpendicular to the plane containing these bonds see first of all this bond this bond this Bond they're present on the same plane they're present on board now per perpendicular to the board there'll be a p orbital perpendicular to the board there'll be p orbital so this is a p orbital here even this has got its p orbital this has got its p orbital this has got its p orbital this has got its p orbital even this has got its p orbital you know it now right right guys so if I talk about Benzene all the carbon atoms are SP2 hybridized and every carbon has got one pure p orbital as well perfect now guys try to understand try to understand if I exactly want to make the orbital diagram how exactly I'm going to make it try to understand it carefully try to understand it carefully let me make a structure like this let me make a structure like this I told you every carbon has got a pure p orbital every carbon has got a pure p orbital right every carbon has got a pure p orbital as per this structure there's a pi bond between these two carbon atoms there's a pi bond between these two carbon atom how a pi bond gets formed Pi bond gets formed by the parall overlapping right can you say as for this particular structure as for this particular structure these two orbitals have par overlapped they have par overlapped and formed a pi Bond here similarly p orbital of this and p orbital of this they have par overlapped and formed a pi Bond right similarly these two have par overlapped and formed a pi Bond right as for this structure as for this structure there's a pie Bond here there's a pie Bond here that means the p orbital of these two carbon atoms would have parall overlapped and formed a p Bond here a pyone here a pyone correct now my dear students one more thing one more thing one more thing if I ask you what is the structure of benzene a lot of you told me this is the structure of benzene but there will be few students who will tell me that this is not the structure of benzene this one is the structure of benzene lot among you will tell me this is the structure of benzene now can you tell me how this structure would have got formed if you look at its orbital approach see every carbon will contain every carbon is SP2 hybridized right and there's a pure p orbital with every carbon here right so if you make its structure or let me make its structure over here let me make its structure over here let me make its orbital diagram over here every carbon has got a pure p orbital every carbon has got a pure p orbital now people look at this particular structure there's a pi bond between these two what does that mean that means the p orbital of this and p orbital of this they would have par overlapped and formed a pi Bond so so I would say as for this particular structure this orbital and this orbital they have par overlapped right similarly p orbital of these two they would have par overlapped that means these two would have par overlapped right similarly these two would have par overlapped perfect these two would have par overlapped correct so I give you two structures of benzene one structure of benzene is this one structure of benzene is this right this particular structure of benzene is formed when these orbitals overlap and this particular structure of benzene is formed when these orbitals overlap right perfect but is this structure able to explain all the properties of benzene no is this particular structure able to explain all the properties of benzene no but I would say but I would say some of the properties of benzene some of the properties of benzene the structure will be explaining that means this structure will be explaining and some of the properties of benzene this particular structure will be explaining right some of the properties of benzene this structure will be explaining and some of the properties of benzene this particular structure will be explaining so basically related to Benzene some of the information you are getting from this particular structure and some of the information you're getting from this particular structure if you combine the information if you combine the information what do you get see guys this particular structure is telling you this particular structure is telling you that these two orbitals have par overla but this particular structure is telling you these two have not overlapped these two have overlapped right so in the actual structure of benzene what would have happen happened in the actual structure of benzene what would have happened in the actual structure of benzene what would have happened I told you guys I told you some of the information this is giving us some of the information this is giving us we have to compile both the informations we have to compile both the informations we have to compile both the informations right we have to compile both the informations now try to understand try to understand as for this particular structure overlap is between these two overlap is between these two but as for this structure overlap is between this and this right as for this structure overlap is between this and this but as for this structure overlap is between this and this right as for this particular structure overlap is between this and this par overlap as for this particular structure overlap is between this and this so we compiled all the informations from these two structures these are the proposed structures of benzene these are the proposed structures of benzene which we gave and after giving the proposed structures of benzene we compiled all the information we are merging the information and we got something like this we got something like this right we got something like this so this particular structure try to understand in this particular particular type of structure if you see if you see this particular type if you see this particular structure this particular structure is telling you that in case of benzene there is the simultaneous overlap there is a simultaneous parall overlap of these P orbitals right this particular structure is telling you there is a simultaneous parallel overlap there's a simultaneous parallel overlap of P orbitals perfect and my dear students let me tell you let me tell you one simple thing let me tell you one simple thing the overlap the overlap of orbitals the parallel overlap of orbitals orbitals can be P or D is something which is called as resonance basically the parallel overlap of your as in case of benzene the parallel overlap of your P orbitals right in such a way in such a way that the electron density is uniform throughout that the electron density is uniform throughout is something which you call as Benzene sorry which is is something which you call as a resonance perfect so did you get the idea of what I said did you actually get the idea of what I said did you actually get the idea of what I said now we will try to understand this is the structure which we proposed about Benzene this the structure which we propose about Benzene we compiled the information we got the actual structure of benzene in the actual structure of benzene these P orbitals are simultaneously and partially overlapping these are simultaneously and partially overlapping perfect now let me tell you this particular structure how you are going to denote it how you are going to denote it how you are going to denote it how you are going to denote it if I ask you will you be able to do that will you be able to do that tell me one thing as for this structure there should be double bond between these two atoms as for this structure there should be single bond between these two atoms so merge the information merge the information merge the information this structure is telling you there should be a double bond between these two this structure is telling there should be a single bond between these two but in the actual structure it will be neither pure single Bond nor it will be pure double bond it will be partial double bond here it will be a partial double bond here similarly in the actual structure it will be partial double bond here in the actual structure it will be partial double bond everywhere it'll be partial double bond everywhere that is the reason why in Benzene all the bond lengths are equal that's the reason why in Benzene all the bond lengths are equal got it my dear students this particular structure is what you call as the actual structure of benzene or you can call it as the resonance hybrid as well this actual structure of benzene is something which you call as a resonance hybrid resonance hybrid and these particular structures these are the proposed structures which we gave award Benzene right this is something which we call as canonical structures this this is one of the canonical structure of benzene or you can call it as the thetical structure or you can call it as the resonating structure perfect similarly this one this either you'll be calling as canonical structure or resonating structure or theoretical structure now you tell me one thing which is very important here one thing which is very important my dear students in this canonical form if I ask you if I ask you whether the bonding is localized or delocalized in this particular structure you can categorically say there's a pi bond between these two you can say there's a pi bond between these two there's a pi bond between these two right but if I ask you where is the pi Bond here it can be here it can be here it can be here it can be here so where is the delocalization actually present is the delocalization present in the canonical form or is the delocalization present in the hybrid delocalization is present in the hybrid right delocalization is the present in the hybrid got it delocalization is present in hybrid you cannot say with 100% probability that your Pi electron pair is here or here or here or here it is delocalized it is delocalized but here in this proposed structure in this canonical form you can categorically say your Pi bond is here Pi bond is there Pi bond is here similarly here Pi bond is here Pi bond is there Pi bond is here so those proposed structures those proposed structures which give you which give you the idea about the localized bonding those proposed structures or what do we call us canonical structures but this particular structure this particular structure what does it contain it contains delocalization of Pi electrons and this is something which we call as resonance hybrid is this clear is this clear people say yes or no in the chats if you got to know this so so from now onwards if I ask you is delocalization present in the hybrid or delocalization is present in the canonical structure or delocalization is present in both what do you think I would say delocalization is present in the hybrid delocalization is not present in the canonical form right so my dear students the partial overlap of these P orbitals the partial overlap of these P orbitals is something which we call as a resonance now what happens due to Resonance if I ask you what happens what is the consequence of resonance my dear students due to Resonance what is the resonance delocalization due to delocalization will the pi electron density be subjugated to only two nuclei or more than two nuclei I'll say due to delocalization Pi electron density it'll be subjected to it'll be subjected under the influence of more than nuclei if the pi electron density is subjected is subjected to more than 2 nuclei what will happen force of attraction on the electron deny will increase if force of attraction increases if force of attraction increases if force of attraction increases what will happen will stability increase or decrease force of attraction increases means energy decreases energy decreases means stability increases do remember this yeah perfect guys so in short what happens due to delocalization due to delocalization of these high energy electrons due to delocalization of these Pi electrons due to the delocalization of this P electron density I would say this P electron density it it comes under the influence of of more than two nuclear due to which force of attraction increases if force of attraction increases energy decreases and stability increases right perfect that is the reason if delocalization is in this particular hybrid if delocalization is here that means its energy will be the least and its stability will be the maximum right so do remember energy of the resonance hybrid is the least and its stability will be maximum because over here resonance is present over here delocalization is present among these two resonance is not present delocalization is not present I hope this is clear I hope this is clear so so guys in nutshell in nutshell have a look that structure which contains localized bonding that structure which contains localized bonding wherein you can clearly say that Pi electron is between these two carbons P electron is between these two carbons so that structure which contains localized bonding but cannot explain all the properties of the molecule that's what you call as theoretical structure or resonating structure or canonical form or canonical form okay or canonical form perfect right okay one more thing look at this particular statement none of the canonical structures is the true structure of the molecule you saw it none of the canonical structure is the actual structure of the molecule perfect it is just it is just from the canonical structures we get some some information we compile that information and make the actual structure of the molecule which is something we call as a resonance hybrid and in that resonance hybrid basically your delocalization is present your resonance is present and due to delocalization due to delocalization as I told you due to delocalization what happens due to delocalization what happens the high energy electrons they come under the influence of more than two nuclear due to which force of attraction increases energy decreases stability increases as simple as that right so the structure which contains delocalized bonding but can explain the structure which contains delocalized bonding but can explain the properties of the molecule is something which you call as resonance hybrid so this is basically your resonance hybrid or Benzene this is the hybrid of benzene this is the hybrid of benzene I believe this is clear to everyone I believe this is clear to everyone I believe this is clear to everyone right perfect and due to delocalization what happens due to delocation what happens can I say partial Pi bonds are formed due to Delo what what did you see what did you see due to delocalization what happened partial P bonds are formed right you saw that you saw that okay partial P bonds are formed now my dear students this is something which is which all of you must be knowing see over here there's a single Bond it's a pure single Bond it is a pure double bond and it's a partial double bond it's a partial double bond I'll say bond order here will be one I'll say bond order here will be two and bond order here will be greater than one and less than two correct yes and you know your bond order it is inversely proportional to bond length right perfect so Bond length in case of single Bond will be maximum followed by partial double bond followed by pure double bond correct yes and you know your bond order it is directly proportional to bond strength wherever is the bond order maximum bond order is maximum here two pure two it bond strength will be maximum followed by partial double bond followed by your single Bond right all these things will be utilized all these things will be used in the equations but but before doing the questions there are few more things which I need to tell you related to Resonance if you ask me what are the conditions what are the conditions for resonance what are the conditions for resonance this is again one more important Point what are the conditions for resonance my dear students for resonance there should be presence of parallel P orbitals with high energy electrons to planer to each other coer to each other Co ler to each other for example this is ch2 this is double bond this is CH this is double single bond this is ch2 and this is positive okay this is the molecule which I made over here my dear students try to understand if you ask me what is the hybridation of this particular carbon this carbon it's forming how many Sigma bonds two with hydrogen one with carbon two with hydrogen one with carbon three sigma bonds so it is SP2 hybridized if it is SP2 hybridized there'll be a p orbital there'll be a p orbital this one it is also forming three sigma bonds this carbon again there'll be p orbital right perfect now look at this particular carbon it's again forming what it's again forming three sigma bonds there'll be a p orbital here there'll be a p orbital here and as for this particular structure as for this particular structure which two parall P orbitals have overlapped these two parallel P orbitals have overlapped and they have formed a Pi Bond they have formed a pi Bond right can you draw it one more structure in it see this is one of the canonical forms it's one more canonical form how will it look like in one more canonical form I would say these two P orbitals would have overlapped so the pi Bond would have been shifted over here Pi Bond would have shifted over here correct over here in as for this particular canonical form these two orbitals have parap but in one more canonical form there would have been one more possibility that these two would have overlapped these two would have overlapped correct so what is the condition basically condition is there should be presence of parall P orbitals with high energy electrons condition is there should be condition is there should be presence of what there should be presence of parall P orbitals with high energy electrons like you saw here these two orbitals have par overlapped and formed a pi bond in one more canonical form I would say these two would have these two would have this is ch2 this is CH and this is your ch2 for example try to understand what I'm going to say these are the pure P orbitals so in one more canonical form these two would have overlapped par right these two would have overla par perfect and your Pi Bond should have been here as for this particular canonical form perfect now if I ask you is this the true structure is this the hybrid this is not the true structure this is not the actual structure of the molecule even this is not the actual structure of the molecule this is giving me some of the information this is giving me some of the information if I compile the information if I compile the information I'll get if I compile the information I'll get the actual structure of the molecule now how the actual structure of the molecule will look like try to understand actual structure will be like this this for example ch2 I'll put a single Bond everywhere first of all I'll put a single Bond everywhere now as for this structure there's a double bond between these two but as for this structure there's a single Bond but in the actual structure there will be neither pure single nor pure double so it'll be a partial double it'll be a partial double here as for this structure there's a single Bond as for this structure there's a double bond but in the actual structure neither pure single nor pure double it will be partial this structure is telling me that this carbon carries positive right perfect but this structure is telling me this carbon does not carry any positive so let me write it as Delta positive here now this particular structure is telling me this car is carrying positive this is telling me no this carbon is not carrying positive so I'll be writing it as Delta positive so this is something which I'll be calling as resonance hybrid this is your hybrid actually this is your hybrid now you tell me is resonance is delocalization present in first case second case or third case everyone in the chats is delocalization present in first case second case or the third case is delocalization present in the first case second case or third case I want every answer dation first second or third dation is present in third it is present in hybrid due to delocalization your high energy electron pair right due to delocalization your Pi electron pair it would how come across more force of attraction due to which its energy would decrease stability would increase right absolutely guys wonderful so so do remember this particular case there should be the presence of parallel P orbitals with high energy electrons which must be c-plant to each other and this is something which I already told you due to delocalization of P electrons the pair of electrons are subjugated to more force of attraction due to which potential energy decreases stability increases therefore we say that resonance increases the stability therefore we say that resonance increases the stability right now guys there are few more points related to Resonance then we'll do lot of questions based on these okay the first point resonance hybrid resonance hybrid is always more stable right because in the hybrid itself delocation is present in the canonical forms delocation is not present in the hybrid delocalization is present in the canonical forms delocalization is not present and wherever delocalization is present wherever resonance is present there only I would say stability is more right so do remember resonance hybrid is always more stable because resonance is present in the hybrid not in the canonical form right told you already told you already okay second important Point all the canonical forms should have almost same energy for example you have got the molecule you got the molecule n it's one of the canonical form is C1 it's one of the canonical form is C2 it's one more canonical form is C3 it has got three canonical forms for example it has got three canonical forms now I will be calling them canonical forms of this m only if their energy is almost same these three I'll be call calling the canonical forms of this molecule only if the energy is almost same for example what are canonical forms first of all these are proposed structures by us so I can give one more canonical form C4 for example its energy is far far far greater than this if energy of this canonical form is far far far greater than these three then I won't be considering this as the canonical form all the canonical forms first of all canonical forms are proposed by us right and all the canonical forms they should have what they should have similar energy almost same energy right they should have almost same energy okay almost same energy now my dear students one more thing out of all the canonical forms that structure which is the most stable gives the better representation of the true molecule so basically if you have got the molecule and you are proposing it structures canonical forms C1 C2 C3 these are the canonical forms of this molecule M perfect that canonical form which will be most stable among all these three that canonical form which will be most stable among all these three your true molecule will look more like that your true molecule will look more like that I hope you trying you getting what I'm trying to say you got the molecule you got the molecule you proposing its canonical forms now let's say among these three C1 is the most stable canonical form if C1 is the most stable canonical form that means your actual molecule will look more like C1 it will look more like C1 that means C1 will give more information about the actual molecule about the true structure of the molecule same will give more information about it your actual molecule will resemble more like what will resemble more like C1 yeah is this point clear is this point clear guys so that canonical form which will be the most stable your actual molecule will look more like that that canonical form which is highest stable let me know ones in the chats if this point is also clear if this point is also clear all right have a look on this particular thing as I told you look at this particular Point all the canonical forms should have same this is one more important Point all the canonical forms they should have same total number of shared and unshared electron pairs for example see if you need identify whether these two are canonical forms are not so there's a condition both these canonical forms should have same number of shared plus unshared electron pairs now see how many do we have this one electron pair 2 3 4 and five so in total it has got five electron pairs now this should also have five electron pairs then only I would call them as the canonical forms 1 2 3 4 5 this also has five electron pairs in total so all the canonical forms do remember they should have same number of electron pairs shared plus unshared shared plus unshared right this was one more important thing now guys we are actually going to we I hope you got the exact definition of resonance let me quickly summarize it let me quickly summarize it guys in order to understand resonance resonance is considered as the most difficult topic if it is taught actually right okay if it is taught in real sense basically otherwise it's easy you know the conjugation do this that sh the bonds that is easy but the actual meaning of resonance is basically right it is difficult to understand and I hope it is clear to everybody now yeah can you quickly let me know if it is clear so quickly summarize it whenever a true structure of the molecule cannot be formed or if any structure structure of the molecule cannot explain all the properties of it you will say that the molecule will exhibit resonance right you'll say that the molecule will exhibit resonance perfect what are the conditions what is the meaning of first of all localized and delocalized electron pair localized electron pair is that electron pair which is shared between two atoms delocalized shared by more than two atoms right now which electron pairs can be delocalized high energy electron pairs can be delocalized like your Pi electrons which are subjugated to less force of attraction they can be delocalized and the delocalize the delocalization of high energy electrons like Pi electrons is something which you call as resonance if you look at the orbital approach as for orbital approach what is resonance it the partial overlap of P or D orbitals which are parall and co-planar to each other right perfect now one more thing one more thing your what how do you define the how do you define the canonical structures canonical structures are the ones in which bonding is localized in which bonding is localized delocalization is not present in the canonical form delocalization is present in the hybrid if delocalization is present in the hybrid that is the reason why in the hybrid your Pi electron density comes under the influence of more than two nuclei due to which force of attraction increases right energy potential energy decreases stability increases that's the reason why your hybrid is more stable that is the reason why the hybrid is more stable yeah that's the reason why hybrid is more stable perfect and at the same time at the same time condition for resonance there should be the presence of parall P orbitals which should be co-planar to each other Perfect Right your true structure will look more like like that canonical form which is highest stable among all the canonical forms this is one more important point which we have discussed okay in all the canonical forms the total number of electron pairs should be equal one more important point right one more important Point perfect and all the canonical forms should have similar energy all the canonical forms should have similar energy now now now now we are going to move into conjugation now you will easily do the resonance etc etc now this is the easy part right now this is the easy part see guys basically wherever there's conjugation wherever there is conjugation resonance will be there conjugation can be of many types Pi Sigma vacant orbital conjugation something which we have to discuss in detail Pi Sigma free radical conjugation Pi Sigma negative charge with unshared pair of electron right Pi Sigma Pi conjugation wherever you see Pi Sigma I mean these sort of conjugations do remember in that molecule resonance will be there in that molecule resonance will be there right in that molecule resonance will be there for example for example for example for example this is ch2 this is double bond this is CH this is single bond this is ch2 positive for example just an example just an example just an example first of all SP2 this is also SP2 this is also SP2 everywhere or let me just make it like let me make it like this see guys over here there's a pi Bond as you can see it is pi and after Pi you have got Sigma after Pi you have got Sigma after Pi you have got Sigma so first you have Pi then you have Sigma if you look at this particular carbon if you look at this particular carbon if you look particularly at this carbon what about the outermost configuration of carbon it's 2s2 2 P2 right 2 S2 2 P2 correct 2 S2 2 P2 so this is 2s containing two electrons 2p containing two electrons perfect now after excitation is done 2s will be containing one electron and 2p will be containing three electrons 2p will be containing three electrons now there is a positive charge here positive charge here means take one electron out POS charge means take one electron out let's say one electron is here let's say that one electron is taken out from here now this particular carbon how many Sigma it's forming three sigma three sigma means SP2 hybridization this SP2 and there is one vacant orbital one vacant orbital one vacant p orbital so this carbon it has got a vacant p orbital over here right so in this molecule do you see Pi Sigma and vacant orbital present absolutely here we have Pi Sigma and vacant orbital wherever in a molecule you see Pi Sigma vant orbital understand resonance will happen in that molecule understand resonance will happen in that molecule similarly Pi Sigma free radical resonance will happen Pi Sigma negative charge with unshared pair of electron resonance will happen Pi Sigma Pi conjugation resonance will happen okay perfect now people one by one we are going to discuss this conjugation the first type of conjugation is your Pi Sigma vacant orbital conjugation your Pi Sigma vacant orbital conjugation your Pi Sigma vacant orbital conjugation now guys firstly I'll make you understand this particular thing after that I'll show you how to draw the structures exactly see first of all this is your C2 okay if you look at this structure this is one of the canonical forms which is given to me this is one of the canonical forms which is given to me now after looking at this canonical form I will have to make more canonical forms after looking at this particular canonical form I will have to make more canonical forms but before that have a look exactly whether any type of conjugation is present here or not see I can see this is a pi Bond I can see here you have got Sigma now it has has to be Pi Sigma vacant orbital then only I'll say there is conjugation and then only I'll show I I I'll say that this molecule will undergo resonance now how do you check whether you have got a vacant orbital here or not the same approach which I showed you few minutes back same approach which I showed you a few minutes back right tell me outermost configuration of carbon is 2s2 2p2 is 2s2 2p2 right this is 2s containing two electrons and 2p containing two electrons one here one here after the citation is done one electron will be here and three electrons will be here 1 2 and three now there's a positive charge here so take one of the electron out if you take one of the electron out this is the structure now this is the structure now okay now what is the hybridation of this carbon it is forming three sigma right so it's SP2 hybridization three sigma means SP2 so this SP2 hybridization perfect and there's a presence of a vacant p orbital here so this particular carbon it has the vacant p orbital it has the vacant p orbital even this is spt SP2 this is SP2 right if you exactly have a look on the orbital approach if this is SP2 this is SP2 that means it will be having one p orbital and here one p orbital and this structure is telling you that these two P orbitals have overlapped parall now other structure will tell you that these two can also overlap parall right perfect now I'm not going into the details of that since over here I got to know that we have got the Pi Sigma Pi Sigma we have got the pi Sigma vant orbital conjugation so I'll directly say that resonance will happen here and how do we make its other structures other structures in this structure these two P orbitals have overlapped in the other structure these two will overlap so what does that mean in short shift shift this P Bond here when you shift this P Bond here it will become ch2 positive it will become ch2 positive there was a double bond here there was a double Bond here now there will be a single bond this CH you writing as such initially there was a single Bond here now it's going to be a double bond now it's going to be a double bond initially this was ch2 positive but due to the double bond it is not that positive will be compensated nullified this will be ch2 now so this is one more resonating structure one more canonical form and this canonical form is telling you this canonical form is telling you the p orbital of this and the p orbital of this they have overlapped now some information about the actual molecule this is given some information about the actual molecule this is given and if you compile the information you'll get the resonance hybrid you'll get the resonance hybrid so make the hybrid guys make the hybrid make the hybrid here make the hybrid here try to understand the hybrid looks like this first of all I'll write ch2 I'll put single Bond everywhere I'll put single Bond everywhere correct now as for this structure there should be double bond between these two but this structure is saying single Bond so in the actual molecule it's neither going to be pureing single nor pure double it will be partial double similarly this is single this is say double so in the actual one it is partial double right this structure is telling you that this carbon should have positive so make Delta positive over here this structure is telling you no this carbon has to be positive so keep Delta positive over here so this is something which I'll be calling as a hybrid this is the resonance hybrid people I hope you got to know how to make it right I hope you got to know how to make it perfect I hope you got to know how to make it now look at this one look at this one look at this one again you can see this is pi this is pi Sigma and over here you will find a vacant orbital you'll find a vacant orbital this is SP2 hybridized this SP2 hybridized even this is SP2 hybridized right every carbon here is SP2 hybridized so every carbon will be having a p orbital p orbital and those P orbitals will be parall to each other right and the partial overlap of the parall P orbitals is something which which you call as a resonance right now guys try to understand this is pi this is Sigma and here you got the vacant orbital there you got the V so there's a conjugation if there's a conjugation so you'll be making its resonant structures so shift this P Bond here first right when you shift this P Bond here this was ch2 positive in the beginning now it's going to be ch2 positive charge compensated now there is a there'll be a double bond here right there'll be a double bond here since you have shifted this P Bond towards this right so I would say this particular carbon will be getting positive right it'll be getting positive rest everything is same now again if you see this is pi Sigma V orbital Pi Sigma positive Pi Sig V orbital so again resonance so this Pi Bond will shift here when this Pi Bond shifts here the other resoning structure will be like this guys see exactly how do we make the resonating structures see there was a positive here now Pi bond is Shifting right from which carbon Pi bond is Shifting in this structure in this structure your Pi Bond was was between these two in this structure your Pi bond is between these two so basically this is the carbon at which Pi bond is present but Pi Pi bond is shifting from this carbon that carbon from which Pi bond is Shifting that will get the positive charge just everything is same again if you see Pi Sigma positive Pi Sigma positive that means shift it in this direction when you shift it in this direction so what is one more conjugating structure you will get this is ch2 and others this is this right and this will get the positive now again Pi Sigma positive pi Sigma positive means shift this in in this direction so what do we get at the end this is ch2 this is double bond sorry not the double bond this is ch2 this is a single Bond and rest rest have a look there'll be a double bond here there'll be a double bond here there'll be a double bond here perfect right guys now if you category look at if you categorically look at one more important thing here one more important thing here guys look at the first canonical form and the last canonical form if I ask you are these same or different canonical forms are these same or different first one and last one are these same or different are these same or different are these same or different quickly first and the last are these same canonical forms are different see canonical forms they contain localized bonding this canonical form is telling you there's a pi bond between these two but the last one is telling you there's a single bond between these two right so these two canonical forms they are different right no doubt they're different but but but first and last no doubt they're different canonical forms but let me tell you their energy and stability will be the same look at both of them look at both of them look at both of them sorry there's a charge here look at both of them right so they are basically different canonical forms but they're but but their energy and stability will be the same okay so first and last let me write it here first and the last they are actually different canonical forms they are actually different canonical forms but their energy and stability is same but their energy and stability is the same is the same right so finally if I want you guys to make the hybrid you can easily make the hybrid let's make the hybrid people let's make the hybrid see every Bond here is getting the partial double bond character this is getting partial double bond why because here it was supposed to be single but here it is double so it'll be partial double right every Bond there'll be partial double bond character there'll be partial double bond character number one as for the first structure this should have positive as for this structure this should have positive as for this structure this should have positive and as for this structure this should have positive perfect so this is something which you will be calling as a hybrid right this is something which is what you'll be calling as the hybrid perfect is it clear people is it people is it clear people quickly have a look you have got again Pi Sigma and here you got positive Pi Sigma vant orbital basically right pi Sigma vant orbital if you see every carbon here is SP2 hybridized every carbon here is SP2 hbd so it's again Pi Sigma vacant orbital if it is pi Sigma vant orbital right th structures so it's one of the structure will be like this so Pi Bond got shifted and now this gets the positive right now again you got Pi Sigma positive Pi Sigma positive that mean shift it here when you shift it here there'll be a structure like this now there's a there's a p Bond here here and this particular carbon will will be getting the positive perfect now do you see first one and last one first one and last one are they same or different they're different but their energy and stability will be same tell me one thing how do we make the hybrid here how do we make the hybrid every bond is getting partial double bond right every bond is getting partial double bond as for the first structure Delta positive a as for the second structure Delta POs to here as for the third structure Delta POs to here so this what you'll be calling as hybrid and hybrid will be maximum stable as for all these structures here okay so this was your this was your this was your Pi Sigma vant orbital conjugation right now tell me one thing will the molecule show resonance or not what do you think will the molecule show resonance or not will will there be resonance in this molecule you are going to tell me will there be resonance in this molecule what do you think what do you think I'm asking you it's a simple question yes or no see what are you seeing here you're seeing Pi Sigma positive Pi Sigma V orbital right so there is conjugation right so majority is saying there is conjugation nice good if there is conjugation that means it should undergo resonance but people just try to understand if it under goes resonance this Bond will shift here right if this Bond shifts here what do we get we got RC this is oxygen since a bond is getting shifted from oxygen so oxygen carries positive now and there'll be a double bond here this will be ch2 if resonance happens here this has to be it canonical form but if you look at this form oxygen was not carrying any charge oxygen is carrying positive charge if you remember positive charge on an electron negative element makes it highly unstable positive charge on an electr negative element makes it highly unstable so I would say if this I'm calling as canonical form one this I'm calling as canonical form two I would say energy of canonical form 2 will be will be higher than that of canonical form one but we know as for the property of canonical forms is concerned canonical forms they show have same energy similar energy similar energy but its energy is very very very high so this cannot be its canonical form so this molecule won't undergo resonance understood understood tell me that is this point clear guys all these are very very very important points I'm not sure whether you would have studied all this before or not tell me in the chats quickly quickly yes I want everyone to say it in the chats quickly perfect perfect guys see normally normally if you look at these two molecules what you'll quickly say you'll say there's a conjugation I Sigma positive Pi Sigma positive you'll directly say these two will undergo resonance right if the concept of resonance is not clear if you would have superficially discussed resonance you'll Clearly say there's a conjugation Pi Sigma positive Pi Sigma positive so it will undergo resonance but I'll say this will not undergo resonance why what is the reason what is the reason my dear students conjugation is basically like this this is pi then there should be Sigma then there should be a vacant orbital right but look at this particular Nitro how many bonds it's forming it's forming four bonds if it is forming four four bonds it octed is complete right it octed is complete if it octed is complete will there be any vacant p orbital will there be any vacant p orbital with this nitrogen it is forming four bonds this nitrogen is forming four bonds so it's octed is complete so will there be see guys it octed is complete here if it octed is complete that means now its configuration is going to be 2s2 2p6 right so do you see any vacant orbital with nitrogen do you see any vacant orbital with nitrogen nitrogen does not have the vacant p orbital here right because all its P orbitals are carrying the electrons now so it is not going to show resonance look at this one Pi Sigma vacant orbital should be present here it is forming three sigma bonds three three sigma bonds so 3 2 are six and two electrons here 6 + 2 8 its octed is also complete if its octed is complete it is not going to have any vant p orbital if there is no vacant p orbital if there is no vacant if there is no vacant p orbital it's not going to show any resonance here got this point so these two not going to show any resonance because there is no vacant p orbital here there's no vacant p orbital here the oct of this nitrogen and oxygen is complete if the oct of nitrogen oxygen is complete there'll be no W be or with them yeah all right tell me tell me whether this molecule will show resonance or not this is pi Sigma positive Pi Sigma positive whether they'll show resonance or not see phosphorus is forming 3 + 1 four bonds so 4 to its octed is complete right sulfur its octed is complete nitrogen its octed is complete right will this show resonance this will show resonance this will show resonance in phosphorus see nitrogen phosphorus nitrogen phosphorus nitrogen does not have any D orbital but phosphorus has has vacant D orbitals phosphorus has vacant D orbitals right phosphorus has vacant D orbitals so I would say phosphorus over here it has vacant D orbital it has vacant D orbital so the conjugation will be Pi Sigma vacant orbital Pi Sigma vacant orbital right I had categorically mentioned in the first case it has to be Pi Sigma vacant orbital that vacant orbital can be P or D that vacant orbital can be P or d right no doubt it looks like it's forming four bonds its octed should be complete but it has got vacant D orbital guys right if there is a vacant D orbital that means it will show resonance look at this one sulfur sulfur four bonds is forming octed complete now tell me sulfur does it have vacant de orbitals sulfur has de orbitals right there is a vacant de orbital with sulfur what is the electronic configuration of sulfur outermost it is 3s2 it is 3p4 there is 3D as well which contains zero electrons so it has vacant orbitals if it has vant D orbital it will show resonance but nitrogen nitrogen nitrogen if you look at the configuration of nitrogen 2 S2 2 P3 is there any D there is no D in nitrogen there's no D in nitrogen right there's no D in nitrogen if there is no D nitrogen this will not show resonance is it clear is it clear is it clear guys I hope you can analyze from now onwards which one can show resonance which one canot okay among these three among these three quickly among these three is the first one going to show resonance yes or no everyone is the first one going to show resonance this is pi Sigma and it will have this phosphorus will have a and d orbital it will show resonance right what about the second one I sigma look at this Boron it's forming how many bonds three bonds it's forming three bonds right it's forming three bonds quickly tell me is this particular molecule going to show resonance or not quickly tell me is it going to show resonance or not if you look at Boron guys what is the outermost configuration of boron it is 2s2 2p1 correct this is 2s carrying two electrons and 2p carrying one electron 2p carrying one electron now after the excitation is done after the excitation is done one electron here one electron here right okay how many bonds it's forming it's forming three sigma bonds three sigma bonds means SP2 this SP2 is there the vacant p orbital yes this boron has vacant p orbital so Pi Sigma vacant p orbital absolutely it will show resonance right look at this one look at this one is it going to show resonance Pi Sigma again the similar scenario this will show resonance so all these three will show resonance right all these three will show resonance I hope this is again clear to everyone now there is one more conjugation so till now I showed you one conjugation what is that what is that that is your Pi Sigma vacant orbital conjugation the second one is pi Sigma free radical conjugation Pi Sigma free radical conjugation right have a look Pi Sigma radical Pi Sigma radical conjugation is there so it will undergo resonance right so shift the pi Bond on this side when you shift the pi Bond on this side this ch2 will get a radical right and this was a double bond here now it's going to be a single bond this is CH now now it's going to be a double bond so it's going to be ch2 like this so this is one more canonical form of it now if you want to make it hybrid so make it simple ch2 everywhere keep a single bond for CH this is ch2 right now there's a double bond here but this structure is telling single Bond so neither single nor double partial double single but this is saying double neither single nor double partial double now look at this particular resonance hybrid see as for this particular structure is concerned this having a radical so have a radical here as for this structure this having a radical so have a radical here Delta radical right so this is the hybrid this is the hybrid people clear this is the hybrid make its resonating structures how will you make its resonating structures see Pi Sigma radical so shift this Pi bond in this direction when you shift the pi bond in the direction this is ch2 now it's going to be a double bond right that carbon from which the bond has been shifted from this carbon bond has been shifted I'll put a radical here rest everything is same now it's again Pi Sigma radical if it's Pi Sigma radical shift in this direction when you shift in this direction what do we get try to understand what do we get exactly this is your Pi here you'll be getting the radical and here is a double bond perfect now again this is pi Sigma radical right so shift in this particular direction so what do we get again this is ch2 this is double bond and rest I believe everything is similar only right now this particular carbon gets the radical because the bond has been shifted from this particular carbon now have a look this Pi Sigma radical Pi Sigma radical shift this Bond here when you shift the bond it is going to be at the end ch2 radical it is a single Bond rest all the things this is like this now tell me one thing thing first and the last are the same or different resonating structures first and last same or different same or different quickly first and last same or different first and loss same or different they are different resonating structures but their stability and energy that is that's equal yeah so I hope this particular conjugation is also clear now guys there is one more conjugation what is that first one was Pi Sigma viant orbital second one is pi Sigma free radical third one is pi Sigma negative charge with unshared pair of electron negative charge with unshared pair of electron have a look this is your Pi this is Sigma this is negative charge with unshared pair of electron so shift it just shift it after shifting what do we get this will be your ch2 negative this will be your ch2 negative there'll be a single Bond here this is ch right there'll be a double bond here and this will be your ch2 perfect simple now make the hybrid how will you make the hybrid first put a single Bond everywhere first put a single Bond everywhere after putting the single Bond everywhere now this is double here this is single so hybrid will contain partial double single this is double hybrid will contain partial double now as for this particular structure this one carries negative charge right so this is Delta Nega as for this one the other carbon carries a negative charge so Delta negative so this is your hybrid this is your hybrid I believe this also clear I believe this also clear right now similarly guys there is one more conjugation there is pi Sigma lone pair conjugation okay one more example we have here one more example we have here right one more example we have here have a look guys look at this one this is pi this is Sigma this is negative right with a unshared pair of electron so again conjugation shift it right when you shift it what do we get at the end this is ch2 there's a single Bond here this is CH there's a double bond here and this is O now this will carry the negative because this bond has been shifted now if I ask you what is the hybrid how the hybrid looks like this is ch2 single Bond ch right this is O perfect now as for this particular structure there's a double bond here as for this structure there's a single Bond so in the hybrid it will be partial double in the hybrid will be partial double right this is going to be Delta negative and this is also going to be Delta negative so this is your hybrid over here yes I hope you can easily do all these sort of things right I hope you can easily do all these sort of things okay can you make the the resonating structures of these two you should be in a position to make the resonating structures of these two have a look this is pi Sigma negative charge with the unshared pair right Pi Sigma negative charge with unshared pair what will happen shift this negative charge here shift this P Bond here so what do we get this is O so what do we get this is O now there is a double bond here perfect now see what all shift would have happened now since this is a double bond right this this this Pi bond has shifted in this direction so now this carries the negative charge rest everything keep it as such now this is pi Sigma negative so shift it here shift this one here what do I get this is oxygen this is double bond and all the things all the things rest this is double this is negative perfect in this particular direction now again this Pi Sigma shift this in this direction shift it in this direction what do I get I get oxygen this is double bond right and see all the things this is now netive right there's a double bond here there's a double bond here perfect now again this Pi Sigma negative shift it and shift it so what do I get at the end I got something like this perfect so these are all the resoning structures which we have here now people tell me first last are they same or different they are different but their energy stability both is same yeah perfect now I think this one you can easily do on your own this one you can easily do on your own now here is a chuck for you this is your Chuck mate will the following molecules show resonance or not I've got four molecules here among these four molecules which one will show resonance which one will show resonance among these four among these four which one will show resonance among these four will first one show resonance this is pi Sigma negative but it has to be Pi Sigma negative with unshared pair this has to be the actual scenario I'm just mentioning Pi Sigma negative but it has to be Pi Sigma negative with unshared negative with unshared so is this going to show resonance no way look at the second option look at the second option look at the second option tell me one thing what is resonance the the partial overlap of P orbitals right the partial overlap of orbitals which can be P or D correct now guys try to understand try to understand if you look at this particular carbon if you look at this particular carbon what is the if you talk about carbon it is 2 S2 2p2 right so this is s and these are 3 p orbitals of carbon okay first of all first of all there's a pi Bond here if there's a pi bond that means p orbital of this and p orbital of this they have par overlapped so one of the p orbital is consumed it is consumed right one of the p i mean this p orbital and this p orbital they have partially overlapped right so this p orbital is gone now how many orbitals are left SN 2p so this will undergo SP2 hybridization SP2 hybridization means how many hybrid orbitals three hybrid orbitals one of the hybrid orbital would have formed this Sigma Bond one of the hybrid orbital would have formed this Sigma Bond and one hybrid orbital will be carrying this positive charge one hybrid orbital will be carrying this positive charge right now for resonance for resonance for resonance now you tell me will this molecule show resonance or not will it show resonance what is the resonance will it resonance yes or no quickly quickly so that positive charge is basically present in what SP2 orbital it should have been in s it should have been pure p orbital right so it is not going to show resonance look at this one again here it is present in Sp SP2 this radical is also resonant SP2 so none of the molecule will show resonance among these four none of the molecule will show resonance among these four okay none of the molecules will show resonance among these four these are some typical cases guys which I'm showing you okay this is one important statement lone pair has to be present in the pure p orbital when you talk about Pi Sigma lone pair conjugation when you talk about Pi Sigma lone pair conjugation the lone pair has to be present in pure p orbital there is one important thing which I'm going to tell you now first of all have a look have a look guys this is pi this is Sigma this is a lone pair is this lone pair over here conjugated yes this is conjugated so basically oen has two loone pairs oxgen has two loone pairs out of the two loone pairs one loone pair will participate in the resonance that loone pair which is participated in the resonance that is not considered while calculating the hybridization that loone pair which participates in the resonance that is not considered for hybridization so one of the loone pair you don't have to consider for for hybridization now if I ask you what is the hybridation of this oxygen quickly see it's forming one Sigma 2 Sigma so two Sigma and one lone pair you have to consider you don't have to consider two you have to consider one so two Sigma one lone pair static number three if static number is three so it is SP2 it is SP2 it is SP2 that means this this loan pair it is present pure p orbital that's what I mentioned over here lone pair has to be present in the pure p orbital lone pair has to be present in the pure p orbital perfect right so make its resonating structures so shift this lone pair in this direction shift this bond in this direction what do you get you get ch2 neg2 you get ch2 neg2 there is a single bond this is CH this is double bond this is O carrying only one lone pair and there'll be a positive charge here perfect so this is its one more canonical form look at this particular nitrogen this nitrogen it's forming three sigma bonds it's forming three sigma bonds and this lone pair is conjugated this is pi Sigma lone pair it is conjugated perfect now this nitrogen if I ask you it's hybridation it's SP2 if it is SP2 there'll be a p orbital and that p orbital will be carrying this lone pair right so shift it when you shift it it's going to be ch2 negative H2 negative there's a single bond this is CH there's a double bond this is going to be your n H2 positive here perfect so this is going to be its canonical form similarly can you make the canonical forms for this one can you make canonical forms for this one this Pi Sigma lone pair Pi Sigma lone pair this oxygen it has got two loone pairs basically it has got two lone pairs basically but out of the two lone pairs one lone pair will participate in resonance so you do not have to count that loone pair for hybridization now tell me the hybridation of this oxygen it's forming two Sigma and one lone pair two Sigma one lone pair that means steric number three so it's SP2 that means this particular lone pair it is present in pure p orbital and that is the criteria lone pair has to be present in the pure p orbital correct now you can do all the Wonders right shift it in this direction shift it here there'll be a negative charge when there is a negative charge here shift it in this direction this will shift in this direction there'll be a negative charge then shift this in this direction and shift it here I mean you can easily make these structures you can easily make these resonating structures now okay is it clear now my dear students once you are familiar with how to draw the resonating structures how to draw the resonating structures now it is the time to compare it is the time to compare the stability of the resonating structures it is the time to compare the stability of resonating structures this is the question which is frequently asked how do you check the stability of the resonating structures these are the three resonating structures which are given to us we need to compare their stabilities right de students the first and the most important Point first and the most important point if you need to compare the stability of these resonating structures what you should do what you should do first of all try to calculate the number of covalent bonds see first of all this is ch3 here ch3 here ch3 here right this is NH this NH here this NH here this is ch3 here this is ch3 this is ch3 so all these things are same everywhere all these things are same everywhere so what I'll be doing I won't be counting the exterior bonds I'll be counting the interior bonds I'll be counting Counting the interior calent bonds tell me how many interior calent bonds do we have here 1 2 3 4 5 and six six interior coent bonds right look at this one 1 2 3 4 5 6 six interior coent bonds here absolutely how many here 1 2 3 4 5 6 six interior coent bonds Here My Dear students whenever you need to compare the stability of the resonating structures the first point is you have to count the covalent bonds interior calent bonds that structure wherein calent bonds are maximum that will be maximum stable that will be maximum stable this is the first point which you'll be checking all the time calculate the maximum calculate the interior covalent bonds that resonating structure that resonating structure which will be containing maximum interior covalent bonds will be maximum stable but here in these three the interior calent bonds are same so you cannot decide on the base of interior coent bonds you cannot decide on the basis of interior calent Bond right but your priority is always coent bonds number two number two look at that resonating structure look at that find that resonating structure in which atoms do not contain any charge is there any resonating structure like that in which atoms do not contain any charge is there any resonating structure like that in which atoms do not contain any charge I can directly say first one right I can say first one that resonating structure that resonating structure in which atoms do not contain any charge atoms do not contain any charge that is going to be stable that's going to be simple first you are going to check what first you are going to check number of bonds then you are going to see that resonating structure in which the atoms do not contain any charge there is the structure in which atoms do not contain any charge so first one is maximum stable first one is maximum stable now among two and three among two and three have a look properly what exactly we are going to say have a look my dear students what do you think 1 2 3 4 5 6 okay what do you think among two and three which one among two and three which one quickly quickly among two and three which one which one is more stable what do you think what do you think see if you look at these two structures carefully this oxygen carrying negative oxygen carrying negative same right same perfect look at this particular oxygen look look at this particular oxygen oxygen is carrying positive here oxygen is carrying positive here oxygen is carrying positive positive charge on an electr negative element makes it highly unstable right so among two and three among two and three which one is more stable I'll say three is more stable followed by two so is this the order this is going to be the order here right this is going to be the order here let's have a look on one more question have a look on one more question check the order of the stability of these resonating structures check the order of the stability of these resonating structures have a look guys first of all calculate the total number of internal coent bonds 1 2 3 4 5 6 7 8 eight coent bonds 1 2 3 4 5 6 7 seven coent bonds 1 2 3 4 5 6 7 8 eight covalent bonds wherever bonds are minimum that is least stable so 1 2 3 two is least stable two is least stable now I'm one and three this is the resonating structure in which atoms do not contain any charge so one is maximum stable followed by three followed by two all of you got this order 132 all of you got this order 132 perfectly done perfectly done wonderful okay tell me the answer of this question among these two which one is more stable first calculate the bonds 1 2 3 4 1 2 3 4 bonds are same bonds are same now this one does not carry any charge on atoms so this one will be more stable 1 2 3 calculate the bonds calculate the bonds 1 these are 6 + 1 7 8 9 10 10 internal bonds 2 2 + 6 is 8 8 + 2 is 10 10 2 2 + 6 is 8 8 + 2 is 10 so bonds are same bonds are same so I'm not going to talk in terms of bonds now what is the next thing this particular resonating structure its atoms do not contain any charge is maximum stable now among two and three among two and three see oxygen carrying positive here also oxygen is carrying positive carbon carrying negative here also carbon is carrying negative so these are the two resonating structures wherein same atoms carry similar charge oxygen positive oxygen positive carbon negative carbon negative and when same atoms carry similar charge at that point of time what do we say at that point of Time how do we how do we distinguish more stable less stable these are like charges or unlike charges these are unlike charges positive negative unlike positive negative unlike more the distance between positive and negative can I say more will be the potential energy of the system as the distance increases between unlik charges potential energy of the system increases and if potential energy of the system in increases stability decreases so which molecule will have more potential energy this resonating structure will have more potential energy less stability so three is least stable and here is two I was talking about unlike charges right more the distance between unlike charges more is the potential energy of the system lesser is the stability right so 1 2 3 is the order of the stability here all right among this and this okay these are three BAS basically in the in given in the beginning two four four Sigma bonds I mean four calent bonds four coent bonds three coent bonds so third one is least stable leave that third one is least stable now among one and two among one and two check it out see this is the nitrogen carrying positive nitrogen carrying positive perfect this nitrogen is carrying negative and here carbon is carrying negative nitrogen carrying negative carbon carrying negative nitrogen carbon they belong to same period same period Electro negative electro negativity plays a role negative charge is more on more Electro negative so this particular first one will be maximum stable followed by two followed by three do you agree with all this yes 1 2 3 is the order okay tell me the next one I'm giving you one minute of time one minute of time I'm giving you tell me the this is calling this as resonating structure one calling this as two calling this as three and calling this as four what will be the stability order of these resonating structures I'm asking you you should be in a position to answer this quickly quickly guys quickly everyone you should be in a position to answer this particular question as well so one of the option I got from sentil Kumar 4321 43 2 1 this is one of the option I'll be taking four options from you SED is saying the same same thing 43 2 1 everyone is I think saying 4312 I got one more option 4312 okay 3 412 3 412 now any any other are you doing some permutation combination here okay let's have a look let's see what is going to be the actual answer first of all count the total number of internal coal bonds 2 3 4 5 right five 1 2 3 4 5 6 this is six right 1 2 3 4 5 5 perfect 1 2 3 4 5 6 this is six right so you calculated what did you calculate you calculated the total number of internal coent bonds now you tell me more the bonds more the stability so it is evident that three and four will be more stable than 1 and two it is evident that three and four will be more stable than 1 and two now among three and four among three and four which one is more stable among three and four which one is more stable see carbon carrying negative here here also carbon is carrying negative nitrogen carrying positive nitrogen carrying positive so in these two resonating structures I'll say same atoms carry similar charge same at ATS carry similar charge right same atoms in these two resoning structures carry similar charge are the charges like or unlike they're unlike they're unlike more the distance between unlike charges more the potential energy of the system less the stability so among three and four which one is more stable four then three now in two and one in two and one in two and one in two and one in two and one what do you say in two and one what do you think in two and one n 2 and one what do you say negative 1 carbon positive one carbon right negative on carbon positive on carbon perfect now here will you use the distance the distance rule between unlike no after this carbon there is nitrogen nitrogen has got the lone pair right after this carbon there's a nitrogen nitrogen lone pair so lone pair and negative they'll Ripple Ripple means more will the energy of the system less is the stability so two will be more stable than that of one did you get this guys you should be able to solve these questions I'm giving you the typical questions these are not easy ones I'm giving you the typical ones wherein you can do a mistake wherein you can do a mistake tell me among these two which one will be more stable this easily you can this easily you can 1 2 3 4 5 1 2 3 4 5 6 right two is more stable one yeah directly count the bonds from the bonds itself you got to know everything right from the bonds itself you got to know everything every possible thing I hope this is clear to everyone now let's move on to one more topic that is cross and extended conjugation have have you ever come across this particular topic cross an extend conjugation have you ever come across this particular term cross and extended conjugation guys I think the J is over is the J over do you want me to cancel the class from here do you want me to cancel the class from here is J over is the Jo over is that it is that it are you done you done I want everyone to say it is the Jo still still high is the Joe still high no right now I'm not giving any break we'll complete this topic then maybe we'll see is fire still left or it's over is fire still left or it's over are you all active guys don't do this don't do this okay don't sleep before me I'm not going to be the one who's going to write the exam okay and I'm not sure whether you would have solved these sort of questions before or not I don't think these are typical examples which I'm giving you typical examples which I'm giving you h okay yes sharp magnet J can also watch it hello let's have a look on one more topic this is again important cross and extended conjugation right cross and extended conjugation now what is cross and extended conjugation all about see guys see guys just a second look at this particular slide once from this particular slide I'll make you understand what is cross and extended conjugation all about if you read the first point whenever the first point says whenever a particular electron pair whenever a particular electron pair competes for delocalization whenever a particular electron pair competes for delocalization with two Pi electron pairs with two Pi electron pairs which are not conjugated it is termed as cross conjugation it is termed as cross conjugation what is meant by this you'll get the idea in some time if the pi electron if the pi electron pair gets delocalized in the same direction it's called as extended conjugation try to understand what it means my de students if you look at this particular molecule look at this particular molecule carefully carefully look at this particular molecule here you have got a pi Bond right here you have got a pi Bond here you have got a pi Bond look at this Pi Sigma Pi Pi Sigma Pi there is conjugation over this particular side Pi Sigma Pi right similarly Pi Sigma Pi again there is conjugation again there is conjugation on both the sides there is conjugation correct can I say this particular Pi electron pair this particular Pi Bond it is simultaneously in conjugation with this one as well as this one this particular electron pair this particular Pi electron pair it is simultaneously in conjugation with this Pi Bond as well as this Pi Bond whereas if I consider these two Pi bonds are these two Pi bonds conjugated this is pi Sigma Sigma Pi they are not conjugated with each other they are not conjugated I'll say this particular Pi bond is a conjugation with this one as well as this one whenever you see this kind of the scenario in which a particular electron pair simultaneously will be in conjugation with will be in conjugation with two electron pairs will be in conjugation with two Pi electron pairs this is something which I call as cross conjugation whenever a particular electron pair whenever a particular electron pair whenever a particular high energy electron pair is simultaneously in conjugation with is simultaneously conjugation with two other high energy electron pairs but those two high energy electron pairs are not conjugated this particular system is what you call as cross conjugation whereas if you look here this is pi Sigma Pi Sigma Pi the conjugation is in a single direction right whenever conjugation is in single Direction this is something which you'll be calling as extended conjugation you can call it as linear conjugation you can call it as linear conjugation let me tell you people extent of resonance extent of resonance in Cross conjugation is always less than that of extent of resonance in extended conjugation extent of resonance in Cross conjugation is always less than that of extent of resonance in in extended conjugation and do remember extent of resonance decides the stability extent of resonance deci decides the stability more the extent of resonance more the stability more the extent of resonance more the stability more the extent of resonance more the stability right so tell me one thing over here a pi electron pair is simultaneously in conjugation with these two electron pairs but these two electron pairs themselves they are not conjugated this kind of the system is something which we call as cross conjugation But Here There is a conjugation in One Direction you call it as linear conjugation or extended conjugation extent of resonance in Cross conjugation is less than that of extent of resonance in extended conjugation right wherever extent of resonance is more stability is more just if you remember two things you are done if you just remember these two things you are done tell me the answer which one among these two is more stable which one among these two is more stable first or second more stable first of all my dear students if you have a look this is pi Sigma Pi Sigma lone pair so conjugation is throughout right over here this is this is pi Sigma lone pair correct this is pi Sigma lone pair even this is pi Sigma lone pair so can I say this lone pair is in conjugation in this direction as well as this direction so can I call this as the example of cross conjugation it is example of cross conjugation extent of resonance in Cross is less if the extent of resonance in Cross is less stability is less so which one is more stable I'll say first one is more stable right and my dear students this concept you are going to use at the time when both these molecules will have same P electrons have a look 2 4 6 2 4 6 so same same same number of high energy electrons are there both these molecules are having same number of high energy electrons so whenever same number of high energy electrons are there whenever same number of high energy electrons are there you at that point of time you are going to use the concept of cross and extend conjugation otherwise not otherwise not tell me in this case see this is pi Sigma Pi the conjugation now Pi Sigma Pi conjugation so I'll say this high energy electron pair and say this high energy electron pair this high energy electron pair is simultaneously in conjugation with this one as well as this one so this is the example of cross look at this one Pi Sigma Pi Sigma Pi so conjugation is throughout conjugation is throughout right so this is cross conjugation extent of resonance and cross is less extent of resonance in Cross conjugation is less so stability is less right look at this one this negative pi Sigma negative pi Sigma Nega so this high energy electron pair is conjugate on this side as well as this side right okay but have a look on this one this is pi Sigma Pi Sigma air conjugation is Crow basically right this is your cross this is your extended right extent of resonance in Cross is less if extent of resonance in Cross is less stability is less so this is the stability order this is the stability order is it clear is it clear people quickly is it clear I believe this is clear all right tell me one more example just a second okay one more one more thing from which question comes that is negative charge holding capacity negative charge holding capacity of an atom negative charge holding capacity of an atom my dear students as I told you in the starting of the lecture the negative charge holding capacity of one oxygen atom the negative charge holding capacity of one oxygen atom it is greater than the negative charge holding capacity of 10 carbon atom the negative charge holding capacity of one oxygen atom is even greater than the negative charge holding capacity of 10 carbon atoms for example oxygen is carting negative carbon is carting negative right oxygen carbon they belong to same period same period electro negativity negative charge is more more Electro negative right so negative charge of course will be stable on oxygen it'll be more stable on oxygen but if you want to see how much stable this negative charge on oxygen will be you'll directly say negative charge holding capacity of oxygen is greater than the negative charge holding capacity of almost 10 carbon atoms from that you'll get the idea how stable negative charge will be on oxy oxygen as of carbon as of carbon now you tell me one thing now you just tell me one thing look at these two molecules this is O Negative this o Nega oxygen carrying negative but over here over here is there any Oxygen there's no Oxygen but I would say resonance will be there this will come here this will go here and there'll be a negative here then negative comes here this will go on this side negative comes on here then this will Shi sh here and this will shift here negative comes here perfect that's something which will happen so in this particular molecule where where all negative charge is getting generated it was there now the negative charge is getting generated at this position at this position at this position perfect no doubt there is dispersal of negative charge there is dispersal of negative charge here negative charge is coming on 1 2 3 4 four carbon atoms but here they negative charge than oxygen so can you compare the stability of negative charge no doubt here negative charge is getting dispersed on four oxygen atems but in this case in the first case in the first case in the first case negative charge is on oxygen so tell me where is the negative charge more stable where is the negative charge more stable I'm asking you first or second first or second quickly where is the negative charge more stable first or second guys you have to tell me the answers in the chats quickly look at this statement I told you negative charge holding capacity of oxygen is greater than negative charge holding capacity of 10 carbon atoms a negative charge is spread on four carbons air negative charge is alone on oxygen but you know negative charge carrying capacity of oxygen is greater than the negative charge carrying capacity of 10 carbon atoms right so where is the negative charge more stable in the first case in the first case right okay tell me among these two among these two which one is more stable among these two which one is more stable which one is more stable where is the negative charge more stable so first of all no more Electro negative element or everywhere there is carbon carbon right so resonance can happen and due to Resonance this negative charge can come here similarly due to Resonance this negative charge can first come here then the same negative charge will come here then the same negative charge will come here right so where is the dispersal of negative charge more in the second case right in this since everywhere there's carbon again everywhere there's carbon no electronegative element nothing right so here you'll be simply seeing the dispersal of charge wherever dispersal of charge is more stability is more right so where is the negative charge more stable okay look at this one look at this one here you have got oxygen no oxygen here no oxygen it's evident if this under goes resonance negative charge will automatically come on oxygen if this under goes resonance negative charge can come here then then it can go downwards then it can come this side right negative charge spread on 1 2 3 4 four carbon atoms but here one oxygen atom alone is enough right one oxygen atom alone is enough yeah one oxygen atom alone is enough to sustain the negative charge right correct people correct correct correct okay few more examples few more examples tell me among these two among these two which one is where is the negative charge more stable where is the negative charge more stable quickly you have to answer this question where is the negative charge more stable among these two I'm asking you first or second first or second first or second quickly quickly quickly see this can undergo resonance and this Bond will shift in this direction due to which oxygen will get negative so negative charge is on one carbon and one oxygen but here already the negative charge is on oxygen now when it under goes resonance this negative charge can come here or the same negative charge can come here or the same negative charge can come here here the negative charge was on oxygen apart from that due to Resonance it is coming on three carbon atoms so leave you cannot decide on the base of oxygen oxygen oxygen in both the cases you have got one oxygen atom but here the dispersal is more here the dispersal is more more the dispersal more is going to be what more is going to be stability so second is more stable yes second is more stable you can you can check it like this as well over here you can find two resonating structures here you can find more than two resonating structures if if the negative charge bearing capacity is same more the resonating structures more the stability in short more the resonating structures more the stability in short because more the resonating structures means more the dispersal of charge more the dispersal of charge means more is the stability look at the next one look at the next one first or second first or second quickly first or second first or second I'm asking you quickly quick quickly quickly guys see if there is resonance this negative charge can come on this oxygen as well so negative charge here is coming on two oxygen atoms negative charge here is coming on two oxygen atoms here negative charge is on one oxygen and due to Resonance it can maximum come on these carbons it can come on these carbons right it can come on these carbons but here the negative charge du to Resonance is coming on two oxygen atoms and you know negative charge carrying capacity of one oxygen is is greater than that of 10 carbons over here you got two so which one is where is the negative charge more stable of course one of course one right some is seeing still you're sing two why two no here negative charge is coming on two oxygen atoms due to Resonance here you have got only one oxygen atom and apart from that there are carbon atoms right two oxygen atoms carrying I mean negative charges spread between two oxygen atoms see how stable it is going to be yeah I hope this is clear okay if this is clear tell me the answer of this question tell me the answer of this question stability order among these three where is the negative charge more stable first second or third arrange the following on the base of stability order quickly quickly now this question everyone has to answer properly every everyone has to answer properly people are saying ACB 1 32 people are saying 1 3 and two are you sure are you 100% sure see this will go come here this will come here so the negative will come here so negative is coming on to oxygen atoms right there is no oxygen so nothing to do with that one oxygen is already there and apart from that negative charge can come here it can come here it can come here perfect so two oxin will be playing a greater role so a Then followed by C followed by B absolutely you all are right good job that means you getting things wonderful wonderful okay tell me the answer of this question a b c d e f arrange them on the base of stability arrange them on the base of stability till till then I can have some water arrange them on the Bas of stability arrange them on the Bas of stability arrange them on the base of stability let me see who is going to answer this question correctly Guys these are six molecules these are not only four there is EF also in the bottom there is EF in the bottom as well people are saying D EB CFA is that is it is it DB CFA let's have a look first of all look at all these structures and see where do we have maximum oxygen atoms no oxygen one here you have got nitrogen as electr negative one oxygen one oxygen okay here you have got two oxygen atoms if here you have got two oxygen atoms that means due to Resonance this negative charge can come here as well so negative charge is spread between two oxygen atoms right so D is going to be maximum stable leave D aside done and dusted right now look at those molecules wherein there is one oxin atom here you have got one here you have got one but apart from that no doubt this will undergo resonance and oxygen will get the negative so among these two there is one oxygen but at the same time this will carry more resonating structures more charge dispersal more stability so e will be more stable than that of B so these two are gone okay now a a C and F A C and F in a and F there is no Electro negative element but here you have got electr negative element nitrogen so it will undergo resonance negative charge will come on nitrogen negative charge on Electro negative element makes it highly stable so c will be more stable so leave cide now among a and F there's no oxygen no electronegative element so decide directly on the basis of resonating structures there are two resonating structures here will be many resonating structures right there you'll get many resonating structures right there you'll get many resonating structures more the resonating structures more the charge dispersal more the charge dispersal means more the stability so this is the stability order how many of you gave this order how many of you gave this order DB CFA Deb CFA someone is saying 4 Five Guys I'm I'm unable to relate what is 453 to 61 is it same as it of d DB CFA so honestly the ones who give the correct answer just say yes in the chat honestly who gave the correct answer just say yes in the chats quickly just say yes in the chats once yeah perfect okay if this is perfect then then then just a second tell me where is the negative charge more stable among these two where is the negative charge more stable now it is simple okay this is your SP hybridized carbon this is your SP2 hybridized carbon s character more more than S character more the electr negativity negative charge is more stable on more Electro negative right these sort of questions are pretty much simple now guys there is one more important thing okay before talking about that important thing tell me the stability order here one or two one or two quickly one or two see this is your SP2 hybridized even this is your SP2 hybridized if hybridation is same if hybridization is same now see whether resonance is happening in anyone there you can see it can undergo resonance right but this cannot undergo resonance right this cannot undergo resonance so resonance is possible in a so wherever there is resonance more stable so a is greater than b yeah okay first one see hybridation in both the cases same so I did not decide on the everywhere everywhere whenever these sort of question comes first point is you will have to decide on the base of hybridization if you are unable to decide on the Bas of hybridization then go for resonance okay then go for resonance I hope this is clear now guys guys now what is resonance basically the partial overlap of pp orbitals the partial overlap of orbitals right if you remember how did we Define resonance it is the partial overlap of parallel P orbitals correct overlap if it is the overlap if it is the overlap can I say 2p 2p overlap extent of overlap in 2p 2p is more than of 2 P 3p is more than the 2p 4p okay so again I'm making you understand one simple thing see how did we Define resonance the partial overlap of parallel P orbitals right so resonance is basically overlap resonance is basically overlap so extent of resonance extent of resonance depends on the strength of overlap extent of resonance depends on the strength of overlap and strength of overlap in 2 P2P is maximum followed by 2p 3p followed by 2p 4p right and do remember extent of resonance is basically directly proportional to what it is basically directly proportional to stability right and extent of overlap it is directly proportional to strength of overlap as well strength of overlap as well from this particular statement also questions are asked now the point is what kind of questions now the point is what kind of questions are asked from this particular statement see guys understand understand one thing can you let me know where will be the extent of overlap more extent of overlap where will be the resonance more where will be the extent of resonance more where will be extend of resonance more where will be stability more tell me that tell me that tell me that quickly my dear students if you look at this particular carbon it's SP2 hybridized right it SP2 hybridized so this is the vacant 2p orbital here this particular loone pair this oxygen has got two loone pairs among which if I talk about one loone pair one loone pair is in 2p orbital as well so here it will be 2 P2P overlap but here it's going to be 2p 3p overlap right now which overlap is effective 2p 2p over overlap is effective if 2p 2p overlap is effective extend of resonance in the first one will be more stability of the first one will be more simple right is this clear is this clear or if you want to have a look on the first case again just have a look just have a look just have a look properly first of all this carbon this carbon it has got 2 S 2 2p2 right this is 2s carrying one orbital 2p getting three orbitals now this particular carbon it is forming how many Sigma bonds it is forming three sigma bonds right one here one here one with hydrogen three sigma bonds perfect it's forming three sigma bonds perfect it's hybridation is SP2 and there will will be one pure p orbital so this is SP2 hybridized and there is one pure p orbital similarly there is one pure p orbital here and these two orbitals have overlapped par and formed a pi bond this understood this is understood now if you look at this particular oxygen it carries two lone pairs out of these two lone pairs is any lone pair conjugated Pi Sigma lone pair one of the lone pair is conjugated let's assume this lone pair is conjugated that lone pair which is conjugated is not considered for hybridization now tell me how many Sigma bonds 1 2 two Sigma bonds and one lone pair so two Sigma bonds one lone pair 2 + 1 3 2 + 1 3 means this SP2 if this is SP2 that means this lone pair it is present in this lone pair it is present in pure p orbital one of the loone pair is present in pure p orbital right it's present in pure p orbital now which orbital of oxygen which orbital of oxygen It 2 p orbital of oxygen so the overlap here is going to be 2p 2p overlap but here it's going to be 2p 3p and you know strength of overlap 2p 2p overlap is more effective than 2p 3p so wherever overlap is more you'll see res extent of resonance is more correct even guys if you remember I had told you in the beginning I had told you the order of the order of high energy electrons I had told you the order of energy of high high energy electrons basically I told you negative charge with unshared pair its energy is more if you remember right then comes your lone pair then comes your this particular electron pair of Pi Bond then comes your the electron pair in the triple bond the pi Bonds in triple bond this particular stuff I've told you already if you remember this particular stuff I've told you already this particular stuff I've told you already I hope you remember I gave you the energy order I gave you the energy order right I gave you the energy order already what is resonance if I ask you what is resonance it is the delocalization of high energy electron pairs resonance is the delocalization of high energy electron pairs now that electron pair which will be having highest energy that will get delocalized the maximum the one which gets delocalized the maximum there only I will say extent of resonance will be more right are you trying to get all these things whatever I'm saying is it clear what is resonance resonance is the delocalization of high energy electron pairs now that electron pair which will be which will be having highest energy will be delocalized the maximum and whever delocalization is maximum I'll say extent of resonance is maximum and wherever extent of resonance is maximum I'll say stability is maximum I'll say stability is maximum right so this is the energy order this is the energy order why am I giving you this energy order there's a logic behind that now tell me among these two okay answer is already written among these two where is the extent of resonance more look at this this is negative with unshared pair of electron this is the lone pair here right Pi Sigma lone pair resonance Pi Sigma negative charge with unshared pair of electron now which is is this electron I mean are these electrons or these electrons among these two which electron pair is high energy this is the higher energy electron pair the one which is high energy electron pair will do more delocalization more delocation means uh extent of resonance will be more extent of resonance will be more means stability will be more yes right okay what about this one order of extent of resonance my God answer is already given here what is this what is this anyways this is is pi this is Sigma vacant orbital I Sigma vacant orbital right this is pi this is Sigma and this vacant orbital correct where is the extent of resonance more where is the extent of resonance more which concept you you guys are going to use here what is resonance you'll you'll simply remember the definition of resonance delocalization of high energy electrons right higher the energy of higher the energy of the electron pairs more is a delocalization more de delocalization means more resonance the answer is done now you tell me the answer of this question this is the pi Bond of the double bond this is the pi electron pair of the double bond this is the pi electron pair of the triple bond which one is having more energy Pi electron pair of double bond has is has got more energy than that of Pi electron pair of triple bond so I would say the first one in the first one extent of resonance will be more simple right clear clear guys all right can you answer this question 1 2 3 which one is where which resonating structure is more stable give the order give the order over here 1 2 3 three bonds 1 2 3 three bonds one two two bonds right I think you can easily do this we have done these sort of resonating structure equations I'm not going to waste my time on this now let's move on steric inhibition of resonance steric inhibition of resonance this is something guys which will eat your brain this topic will eat your brain but before discussing this topic I need a break for 10 15 minutes you also take a break for 10 15 minutes but be back this is something from which you you'll be getting the toughest questions this is something very very very important you are not going to understand this on your own no way and the questions which I come up with this right in this particular topic the questions which I come up with by solving those questions I mean nothing is left you can easily solve all all the types of questions asked okay so should we take a break so what what time you guys will be back I think 11:10 so is it a promise you'll be back at 11:10 is it a promise that you guys will be back at 11:10 uh 11:10 okay guys be back okay be back be back or if I after coming back if the students would be less maybe I will have to cancel the session then yeah so I'll expect all of you all the students for watching me right now I'll expect you guys to be back at 11:10 so that we can start hysteric inhibitions of resonance right see you see you in some time what's up what's up guys all back with the new energy yeah let me know let me know in the chats is everyone back is everyone back okay so let me know quickly if all the things are clear till here yes or no in the chats with the fire emojis I don't want you guys to sleep so keep on typing in the chats let me know let me know quickly are all the things perfectly clear till here are all the things perfectly clear till here now guys we have to discuss something really really important I want some fire emojis in the chats first I want some fire emojis in the chats quickly everyone everyone I want some fire emoies in the chats then I'm going to start I need some fire emojis in the chats then I'm going to start people quickly okay so let me know once in the chats have you studied this particular topic before steric inhibition of resonance s iir have you studied this particular topic before let me know with yes or no in the chats let me know with yes or no in the chats so majority is saying no then let's get going let's get started and let's see what this steric inhibition of resonance is all about my dear students try to understand try to understand very carefully because this is something which is which might confuse you a bit okay this might confuse you at a bit all right so let's try to understand what this sir steric inhibition of resonance is all about my dear students for example over here I'm taking one Benzene ring right this is one Benzene ring at the top of the Benzene ring I'm am attaching a group let's say this is nitrogen this is r and this is R this R is the alkal group okay first of all if I ask you how many electrons are in the valence shell of nitrogen you'll say five out of five electrons nitrogen have used how many electrons for bonding 1 2 3 so can you say there'll be a lone pair on this nitrogen yes there's a lone pair on nitrogen absolutely there's a lone pair on nitrogen okay number one similarly over here what exactly am I going to do I'm going to take one more Benzene ring this is a Benzene ring for example at the top of it I'm attaching this is n this is R this is R right and and there's a lone pair on nitrogen and apart from that I'm attaching one more group R over here perfect these are the two molecules which I taken into consideration now first of all do you see this lone pair in conjugation with this ring Pi Sigma lone pair absolutely it's in conjugation right similarly Pi Sigma lone pair conjugation is there perfect conjugation is there my dear students can I say this lone pair will participate in resonance absolutely this loan pair will participate in resonance so basically this loan pair will shift here due to which right here you'll get a double bond so this loan pair is going to participate in the resonance similarly here also the lone pair is going to participate in the resonance what is resonance resonance is partial overlap of P orbitals so basically if you look at this particular nitrogen it is forming three sigma bonds three sigma bonds since those this loone pair it is participating in resonance so I'm not going to count this for hybridization so if I ask you what is the hybridation of nitrogen here leave this lone paraside 1 2 3 so it is SP2 this SP2 hybridized right if I talk about this particular carbon even this SP2 hybridized perfect correct now my dear students this nitrogen is SP2 hybridized right nitrogen's outermost configuration 2s2 2p3 right this is 2s and these are 3 p orbitals of nren perfect nitrogen has undergone SP2 hybridization SP2 hybridization that means there is a p orbital which is carrying this lone pair right so can I say on this nitrogen since it's SP2 hybridized perfect SP2 hybridized means trigonal planer trional planer means all these bonds this bond this bond this Bond they are lying on the same plane imagine these three bonds they are lying on the board yes they are lying on the board board in the similar way this carbon is also SP2 hybridized trional planer so this bond this bond this Bond like all these bonds are basically lying on the board on the board Perfect Since this nitrogen is SP2 so it will have be having one pure p orbital which will be perpendicular to the plane for example like this similarly this carbon it is also SP2 so it will be also having one p orbital so if I show it to you like this if I show it to you like this all these bonds all these bonds all these bonds they're present on the board and this nitrogen has got the p orbital this carbon has got the p orbital so these P orbitals can overlap right what is resonance partial overlap right so these P orbitals can overlap these are two parallel P orbitals which can overlap right now my dear students try to understand one simple thing due to the partial overlap can I say a double bond character gets generated over here between Nitro n and carbon so between nitrogen and carbon I would say between nitrogen and carbon due to Resonance a partial double bond character gets developed due to nitrogen and this carbon partial double bond character gets developed look at this particular scenario look at this particular scenario this nitrogen has got the p orbital even this has got the p orbital now my dear students you have attached a bulky group over here this is a bulky group this is a bulky group over here this is a bulky group can I say between two bulky groups there will be repulsions the hindrance there will be repulsions due to repulsions these P orbitals which were supposed to be parallel due to repulsions the orientation of this particular group will change what does that mean that means the P orbitals which were supposed to be parall like this right parall co-planar orientation of this particular group for example will change like this it was supposed to be like this it will change like this if the orientation of this this particular group chains that means this p orbital which was at the top its orientation will change it will go like this it will go like this due to the change in the orientation of p orbital will there be now effective overlap or ineffective overlap between these two you tell me in the charts will there be effective overlap or ineffective overlap quickly will there be effective or ineffective overlap now effective or ineffective quickly there will be ineffective overlap right yes so these two P orbitals they were supposed to be parallel and co-planar now due to the repulsions the orientation of the above group has changed and this particular white colored p orbital right it will change it will its orientation will change if its orientation Chang what will happen extent of overlap extent of overlap will it increase or decrease can you let me know where is the extent of overlap more among these two where is the extent of overlap more I say extent of overlap will be more here so basically when you attached a bulky group here extent of overlap decreased here or I would say extent of resonance that decreased drastically right I'll say upon the introduction upon the introduction of the bulky group understand it carefully upon the introduction of the bulky group what happened what happened I'll say extent of resonance extent of resonance decreased drastically decreased drastically so due to Sir due to Sir due to steric this effect is what you call a steric inhibition of resonance right due to S the extent of resonance that drastically decreases no doubt resonance will be there resonance will be there but but the extent of resonance will be less extent of resonance will be less and due to Resonance there'll be partial double bond character here right absolutely there will be partial double bond character there will be partial double bond character here also between nitrogen and carbon but you know upon the introduction of this bulky group extent of resonance has decreased now you tell me one thing due to Resonance this Bond it was supposed to be it was a pure single Bond first it was a single Bond right it was a single Bond and due to Resonance can I say a double bond character got established here yes the double bond character got established similarly here also it was a it was a single Bond but due to Resonance I would say double bond character got established here double bond character got established here right now you tell me one thing where is the extent of resonance more in the first one extent of resonance is more if the extent of resonance is more in the first one can I say this particular Bond would have got more double bond character comparatively this particular Bond would have got lesser double bond character can I say something like that just to make you understand just to make you understand for example for example now the bond order of this particular bond is let's say 1.5 imaginary numbers here it will be bond order will be 1.3 correct yes yes people is this clear these are the imaginary numbers which I'm giving you just to make you understand so basically what I'm trying to convey over here due to steric inhibition of resonance extent of resonance that extent of resonance decreases right due to Resonance this particular single Bond was getting double bond character even this got the double bond character but wherever extent of resonance will be more there only double bond Bond character will be more so double bond character in this particular case double bond character in the first case will be more and double bond character in the second case will be comparatively less right now you know bond order is directly proportional to bond strength and which is inversely proportional to bond length so if I have to compare this particular Bond length let me call this Bond length as one let me call this as Bond length as two can you give me the order of bond length can you give me the order of bond lengths here the double bond character is more double bond character more means less Bond length so one will be less than two 1 will be less than two is this clear is this clear people is this clear people see due to Resonance double bond character got established due to Resonance double bond character got established but where is the extent of resonance more here the extent of resonance is more right if the extent of resonance is more that means her double bond character will be more in this particular Bond if double bond character is more here that means this particular Bond link will be comparatively less and this particular Bond link will be comparatively more I believe it's clear I believe it's clear now right why all this happened because this group and this group they showed some repulsions due to which orientation of this group changed right its p orbital slightly got tilted due to which ineffective overlap happened extent of resonance decreased drastically right extent of resonance decreased drastically I hope this particular point is clear I hope this particular point is clear so do remember one thing theic inhibition of resonance s s it is directly proportional to repulsions basically more repulsions more repulsions more repulsions means more s more s means less extent of resonance right so can I say something like this s iir s i it's directly proportional what it's directly proportional to repulsions between these bulky groups right and if repulsions are more extent of resonance will be if repulsions are more this p orbital will go more away from it extent of resonance will be less right so s is inversely proportional to what it is inversely proportional to extent of resonance and repulsions repulsions are directly proportional to what size of these groups so it's directly proportional to size as well perfect right I hope this is clear if this is clear just remember one thing just remember one thing then I'll make you understand some beautiful questions I'm giving you just just remember remember first of all this particular slide these are the groups which basically show sir right R CH co co nh2 NO2 Co CCL all these groups all these groups they contribute towards what they contribute towards s they show sir you have to remember that you have to remember them now where do I use this particular concept guys please and please have your eyes properly on the screen now have your eyes properly on the screen now see where do I use all this s concept see what kind of questions are asked in s try to understand first of all this particular Bond length is represented by X this particular Bond length is represented by y I have to compare these Bond lengths how to show which Bond length is more which Bond length is less first of all you know bond order is directly proportional to bond strength and bond strength is inversely proportional to bond length right this is something which you all know this is something which you all know my dear students is do you see any sort of conjugation here do you see any sort of conjugation here quickly what do you think do you see any sort of conjugation here yes you can see this is chlorine chlorine it has got three lone pairs right because in its outermost shell there are seven electrons so 2 4 6 1 7 right if you look at this particular this is pi this is Sigma this is lone pair this lone pair can come here this can go here perfect so can I say this bond which was supposed to be pure single due to Resonance this bond is getting partial double bond character this Bond got the partial double bond character but over here do you see any conjugation there's no conjugation it's a pure single Bond it's a pure single Bond here here you have got partial double bond character here it is pure single Bond where is the bond order more bond order is more here more the bond order lesser the bond length right so X will be less than y agreed is it clear say yes or no in the chats if it's clear everyone say yes or no in the chats if this point is clear this was a basic question this was a basic question yeah guys everyone in the chat say yes or no in the quickly don't take this much of time okay if this is clear let's move on to one more let's move on to one more okay now I'm going to use all the concepts which we have discussed till now properly see guys this carbon carbon Bond length is represented by X this is represented by y this is represented by Zed I have to arrange these Bond lengths how to arrange these Bond Lanes okay try to understand there's no conjugation it's a pure double bond it is a pure double bond it's a pure double bond here if you look at nitrogen nitrogen has got the lone pair it can show resonance it can show resonance now try to understand try to understand this Bond it was it was double it was a double bond now due to Resonance this double bond it is getting partial single Bond character it was double in the beginning right due to Resonance it became partial double it was mentioned as purely double but due to Resonance we know it's going to be partially double yes it is going to be parti double look at this particular case look at this particular case oxygen carries a lone pair here perfect so one of the lone pair will participate in the conjugation this double it will also get partial single Bond character right perfect so this is purely double this is partial double this is partial double now try to understand one thing try to understand one thing nitrogen and oxygen compare their electro negativity can I say oxygen is more Electro negative than nitrogen yes oxygen is more Electro negative than oxygen sorry oxygen is more electr negative than nitrogen if oxygen is more electr negative than nitrogen can I say this oxygen will be holding its its electron pair firmly can I say this oxygen will be holding its electron pair firmly as compared to this nitrogen if this oxygen is holding its electron pair ly tell me whether the extent of resonance in this particular case be more or less than that of second one among two and three where will be the extent of resonance more here comparatively this loan pair is easily available this is not easily available right wherever loan pair is easily available I can say there only there only extent of resonance will be more right extent extent of resonance will be more right extent of resonance will be more so extent of resonance is more hair extent of resonance is more hair you got to know extent of resonance is more hair Now understand one thing this was a double bond due to Resonance it's getting single Bond character due to Resonance it bond order is decreasing understand it like this it's bond order is decreasing for example due to Resonance if extent of resonance is more here can I say bond order will be decreased more can I say bond order will be decreased more here bond order will be decreased more here as compared to this one right bond order will be decreased more so here you'll be having lesser Bond bond order less bond order as compared to this one less bond order means more Bond length less bond order means more Bond length right less bond order means more Bond length so among these two I'll say y Bond length will be greater than Zed will be greater than Zed and since it's a pure double bond its Bond length will be least did you get this concept did you get this concept say yes or no in the chats did you get this concept see guys here it's a pure double bond so Bond length least Bond length least pure double bond Bond L least now due to Resonance due to Resonance bond order is decreasing here it was pure double bond now it's partial double pure double partial double right okay this dou double bond is getting single Bond character to Resonance this double bond is also getting a single Bond character there right now extent of resonance is more here wherever extent of resonance is more see bond order is decreasing technically bond order is decreasing technically hair also bond order is decreasing now where extent of resonance is more bond order would have decreased more so bond order would have decreased more hair lesser the bond order more the bond length so y greater than Zed greater than x yes yes let me take few more examples let me take few more examples so that it will be properly clear see first of all this is a pure double bond pure double bond here you have got oxygen right it has got the lone pair so it will participate in the resonance it will participate in the resonance due to Resonance this won't be pure double it'll be partial double so this double is getting single Bond character basically right this double is getting single Bond character now here as you can see this again oxygen right this will go in this way perfect right or I can write it like this I can write this molecule like this have a look have a look properly this is O then you have got methy this is your CH this is double bond this is CH this is n this is double bond and this is O this is the structure basically correct now what's going to happen this oxygen carries the lone pair perfect it it has got the lone pair this will come this way right perfect this will shift in this way and this will go here for example so this particular double bond this particular double bond due to Resonance due to Resonance it is getting partial double bond correct I mean this double bond is getting single Bond character due to Resonance du to Resonance is getting single Bond character now you tell me one thing it's a pure double bond here pure double bond this is partial double bond this is partial double bond pure double bond will have least Bond length will have least Bond length correct we'll have least Bond L now among these two double bond is getting converted double bond is getting single Bond character double bond is getting single Bond character that means bond order is decreasing here also bond order is decreasing here also now if I ask you among these two where is the extent of resonance more my dear students have a look it is starting from here it is ending here see over here extend of resonance is starting here resonance starting here ending here right so where among these two Where is the extend of resonance more I'll say extend of resonance is more here right it is extended conjugation extent of resonance is more here more the extent of resonance due to Resonance this double bond its bond is decreasing more the resonance more the decrease in bond order so this bond order will be less so this Bond length will be more right so Zed will be more than y y will be more than x can you understand this can you understand this quickly so it's 3: 1 is it clear is it clear now is it clear people right I hope you got it so that's why I'm telling you be attentive here be attentive here be attentive now compare these Bond lines here it's it's a pure single Bond pure single Bond pure single Bond pure double bond over here nitrogen has got lone pair Pi Sigma lone pair resonance right Pi Sigma Pi Sigma lone pair resonance so this will get partial double bond this will get partial double bond this is pure single bond this is pure double bond pure single bond has got maximum Bond length and pure double bond has got minimum Bond length these two we already know I'm telling you the bond length order right we just have to compare the these two we just have to compare these two understand carefully due to Resonance this it was a single Bond the single bond is getting double bond character this single Bond due to Resonance is getting double bond character now where is the extent of resonance more this spy Sigma loone pair this Pi Sigma Pi Sigma L pair extent of resonance is more here more the extent of resonance due to Resonance it was single it getting partial double so it's bond order is increasing here also bond order is increasing it was single now due to Resonance it getting partial double bond character so bond order is increasing basically right so wherever extent of resonance is more bond order will increase more there only so here bond order will be more this particular bond order will be more comparatively bond order more means less Bond length bond order more means less Bond length means less Bond length so four will have less Bond length than that of three did you get this did you get this did you get this yes manula you're right did you get this concept I hope you got it okay one more one more one more have a look this particular Bond length and this particular Bond L one or two give me the relation between one and two see guys understand this is pi Sigma Pi Sigma lone pair right this is pi Sigma Pi Sigma lone pair it was a it was a single Bond here also single Bond now due to Resonance it will get double bond character due to Resonance this Bond will get double bond character this NC bond this will also get double bond character so due to Resonance bond order in both the cases is increasing due to Resonance bond order is increasing in both the cases now people what kind of resonance it will show this will go here right this will go here this will come here and this will go here per perfect so negative charge at the end on this particular carbon this will go here perfect this will go here and this will go here negative charge on oxygen where is negative charge more stable negative charge on carbon negative charge on oxygen oxygen has got more negative charge carrying capacity so where is the extent of resonance more first or second extent of resonance is more in second more the extent of resonance see due to Resonance bond order is increasing more resonance more increase in bond order so this bond order will be more more B order lesser Bond length so two will be less than one right yes perfect I hope this is clear now now now compare the bond length compare the bond length this particular Bond length is represented by one this is represented by two okay now first of all first of all what will happen due to Resonance due to Resonance you'll get a double bond character here due to Resonance you'll get a double bond character here right so single bond is getting converted into partial double bond this single Bond due to Resonance is getting converted into partial double bond so in both the cases bond order is increasing bond order is increasing from single to partial double single to partial double bond order is increasing in both the cases now try to understand try to understand this is O2 on the side of it you have you have got BR can I say this BR is going to show s on this side can I say it will experience s on this side right similarly similarly you have got NO2 on this side as well I mean you have got BR on this side as well so on this side also can I say s will operate so NO2 from two sides s is operating from two sides s is operating here only one side s is operating only one side s is operating due to Sir extent of resonance will be less extent of resonance will be less due to Sir where is the S more here sir is more extent of resonance will be lesser right and due to Resonance what was happening bond order was increasing bond order was increasing but extent of resonance is less here that means extent of resonance is more here more extent of resonance means more increase in bond order more increase in bond order right so bond order here will be more more bond order means lesser Bond length right one less than two be careful guys be careful yes either you can say one less than two or you can say two greater than one perfect I believe this is clear okay one more one more one more have a look have a look this Bond length I'm calling one this Bond length I'm calling two compare the bond lengths quickly compare the bond lengths quickly try to understand carefully carefully carefully due to Resonance it first of all this was a single Bond due to Resonance it will get partial double bond character due to Resonance this will get partial Double B character right it will get partial Double B character right so basically it was single now it's getting partial double bond so its bond order is increasing or decreasing its bond order is increasing basically similarly due to Resonance its bond order is also increasing it's getting partial double bond character its bond order is increasing but here we have got s here we have got s repulsions s due to S extent of resonance decreases ha extent of resonance will be less ha extent of resonance will be more more the extent of resonance due to Resonance Bond bond order is increasing ha if extent of resonance is more bond order will increase more hair if bond order hair is more that means this Bond length will be less so two will be less than one or one will be greater than two you can say like that yeah you're killing it nice good right perfect one more question one more question compare the bond compare the bond lens this is one this is three and this is two this is single due to Resonance it will get partial double bond character it was single due to Resonance it will get partial double bond character one is getting partial double bond character this two also is getting partial double bond character due to Resonance three is also getting partial double bond character due to Resonance but if you look carefully this will operate s from here this will operate s from here right over here there won't be any s due to S extent of resonance is less so among all these three where will the extent of resonance more ex of resonance is more here right due to Resonance bond order is increasing this was one initially now it's partial double bond right due to Resonance bond order is increasing so extent of resonance will be more here more the extent of resonance more will be the increase in bond order more will be the increase in bond order so bond order here will be maximum so Bond length will be minimum three minimum right now one and two they'll be equal one and two they'll be equal right one and two will be equal yes one and two will be equal because here on one side s on here also on one side you're getting s perfect I believe this is also clear perfect and you should be able to solve these sort of questions now guys you should be able to solve okay one more which might confuse you a bit one more which might confuse you a bit sorry guys don't look at time try to understand things try to learn things don't look at time it's going to take still long time it will take a lot of time it will take 3 4 hours more so do not look at time look at the slides try to understand it try to understand it properly okay so tell me about this one see due to Resonance this will get double bond character this is get dou character this will get double bond character right initially it was a single Bond it was a single Bond everywhere everywhere it was a single Bond now they're getting double bond character so all the three bonds are getting double bond character right all the three bonds are getting double bond character so bond order of all these three bonds is increasing due to Resonance your resonance bond out of all these three bonds are increasing now s from this side s from this side s from this side right this will operate s from from this side perfect now guys three and two in three and two sir is operating only on one side but here Sr is operating on two sides right Sr is operating on two sides if Sr is operating on two sides extent of resonance will be least here extent of resonance will be least here means means what does that mean extent of resonance is least here that means extent of resonance in these two will be comparatively more if extent of resonance in these two is more that means bond order would have increased more in these two right bond order more means Bond length less so two equal to 3 which will be less than one did you give this answer 2 equal to 3 less than one right I believe it's clear okay one more type of question one more type of question compare the acidic strength of hydrogen which hydrogen is more acidic which hydrogen is less acidic try to understand compare the acidic strengths of hydrogen these are the hydrogens which I have given right you have to compare their acidic strengths how do you compare their acidic strength my dear students if you take this hydrogen out as H positive there'll be negative charge on this oxygen if you take this hydrogen out as H positive negative charge here take it out negative charge here take it out negative charge here perfect now that hydrogen which after leaving gives more stable conjugate base that hydrogen which after leaving gives more stable conjugate base that hydrogen is going to be more acidic that hydrogen is going to give that hydrogen will be more acidic which will give more stable conjugate base now we are taken H positive out oxygen carrying negative now it can participate in resonance if it participates in resonance can I say negative charges shared by two oxygen atoms due to Resonance it is on two oxygen atoms and you know how stable negative charge on oxygen perfect here negative charge on sulfur here negative charge on one oxygen perfect so I'll say this conjugate when hydrogen leaves positive whatever is left here that is first of all resonance stabilized and and negative charge comes onto oxygen atom so this particular hydrogen will be maximum AIC now look at let's compare this and this oxygen carrying negative sulfur carrying negative oxgen and sulfur do they belong to same period or same group same group same group volume charge deny plays a role more size less volume charge density more size more size less volume charge density more stability so s negative is more stable than O Negative right so third will be stronger than second now compare second and first compare second and first oxygen carrying negative nitrogen carrying Negative they belong to same period electro negativity negative charge is more more negative so I would say two will be more acidic than one so this is the order of it which you were supposed to shck is this clear say yes or no in the charts if it is clear yes say yes or no in the charts if it is clear if it is clear if it is clear this hydrogen or this hydrogen which one is more acidic you need to tell me first second which hydrogen is more acidic do not go directly do not jump directly into resonance first check with the help of hybridization check with the help of hybridation take this H positive out negative here take this H positive out negative here this negative is at SP2 carbon this negative is at SP carbon negative charge is more stable on more Electro negative this is more Electro negative right negative charge is more stable so I'll say acidic strength of second hydrogen will be greater than that of first so first you have to see hybridization then go for resonance look at this one look at this one look at this one when you take HED positive out this is negative when you take HED positive out this is negative perfect right this SP2 this SP2 both are SP2 you are unable to decide as per hybridization now what is left resonance is left now now check this negative I mean this getting resonance stabilized if if this hydrogen leaves positive if this Hy Le is positive the conjugate base which we get that's resonance stabilized so I'll say one will be more acidic than two here right one will be more acidic than two I hope you can easily solve these sort of questions from now onwards right now guys we are going to jump into something called as misic effect we are going to jump into something called as misic effect this is again one very important thing first of all let me know in the chats if every single thing is clear till here or not not everyone in the chats everyone in the chats after momic effect we have to do aromaticity then hyper conjugation Okay so 3 3 and a half hours more that's it but before that tell me in the chats if every single thing is clear tell her or not mesomeric effect mesomeric effect misic effect perfect guys so what is Mom let's have a look at its definition then I'll make you understand what it exactly means look at its definition once whenever any group whenever any group creates a charge separation between itself and the conjugated system through resonance whenever any group creates a charge separation between it self and the conjugate system through resonance we say the group is showing misic effect now what it means exactly try to understand this guys for example over here I'm taking a perfectly conjugate system this is a conjugate system which I have right this is a conjugate system which I have first of all tell me is this a conjugate system Pi Sigma Pi Sigma Pi Sigma it is one conjugated system it is a conjugated system here I've taken a conjugate system now my dear students towards this conjugate system let's say I'm attaching a group I'm attaching a group is it not writing I'm attaching a group for example X right let's assume that on x there's a lone pair let's assume that on X there's a lone pair so I have taken a conjugate system and towards this conjugate system I have attached a group this group carries a lone pair first of all do you see one thing can I say this group is in conjugation with this conjugated system have a look Pi Sigma lone pair yes Pi Sigma lone pair the group is also in conjugation with this particular conjugate system can you say between this group and this system there will be resonance absolutely there will be resonance so I would say this negative will come here this will go here this will go here right and so on right can I see something like this can I see something like this so this group will show resonance with this system right this loan pair will come here right and my dear students when the loan pair comes here me say this x will get Delta positive right and this this conjugate system will get Delta negative in short this conjugate system will get Delta negative so do you see charge separation getting created between conjugate system and the group between the group and the conjugate system charge separation got created and this charge separation got created due to Resonance between the group and the conjugate system whenever you attach a group towards a conjugate system that creates the charge separation between the group and the system through resonance we say that group shows misic effect so this particular group over here is showing what is showing what it's showing misic effect it is showing misic effect now you tell me one thing you tell me one thing is this group increasing the electron density in the conjugate system or decreasing the electron density in the conjugated system through resonance it got Delta so it is increasing the electron density in the conjugated system through resonance that group which increases the electron density in the conjugate system through resonance we say that group shows plus M effect plus misic effect plus misic effect right similarly my dear students one more example one more example understand this is C double bond C single Bond C double bond C right this is example of a conjugate system now towards this conjugate system I'm attaching a group the group is like this x double bond Y and I'm assuming electron negativity of Y is greater than that of X I'm assuming electro negativity of Y is greater than that of X towards this conjugate this is the example of perfectly conjugated system which I've taken and towards this conjugated system I've attached a group X dou Bond y y is more electr negative than x what's going to happen can I say this group is in conjugation with this system have a look Pi Sigma Pi this group is in conjugation with the system what will happen resonance will happen between the group and the conjugated system resonance will happen between these two and if resonance happens between the two if resonance happens between the two what will happen I'll directly say this Bond will shift in this direction this will shift in this direction this will shift in this direction and so on right so due to which y gets Delta Nega and this side it gets this conjugate system gets Delta positive so tell me again one simple thing when I attached a group towards a conjugated system charge separation got created between the group and the conjugated system through resonance can I say this particular group is operating misic effect here yes it is showing isic effect now is this group increasing the electron density in the system or decreasing Delta positive it is decreasing the electron density in the system and my dear students that group which decreases the electron density in the conjugated system through resonance we say that group shows we say that group shows minus M effect minus M effect this is increasing electron density this is decreasing if this is increasing electron density so I'll be calling this as Elon donating group through misic effect this I'll be calling as electron withdrawing group through misic effect I believe these two terms are clear these two points are clear right as I told you in the beginning minus always means withdrawing plus always means donating minus always means withdrawing plus always means donating that's all that's all I hope this is clear and if this is clear let's have a look on few more things I believe this is something which I told you already when the group increases the electron density in the conjugated system the group exhibits plus M otherwise mind s okay otherwise mind s let's take few examples let's take few examples for example I'm taking phenol into consideration guys try to understand carefully what I'm going to say let's say this is Benzene towards Benzene I'm attaching O oxygen there are two lone pairs let me show just one okay so is Benzene the example of a conjugated system yes the Benzene is the example of conjugated system and towards this Benzene what I attached I attached a group now tell me oxygen carries a lone pair Pi Sigma lone pair can I say this group is in conjugation with this system yes the group is in conjug conjugation with this system due to which due to which what's going to happen resonance is going to happen between the two and if resonance happens I'll say this particular electron pair will come here this Pi Bond will break in this direction so what do I get I got something like this there's a point which I have to mention over here there's a point which I have to tell you over here this becomes o o posit right it becomes a double bond right here comes a negative charge rest everything is same now Pi Sigma negative this negative will come here this Bond will break in this direction so what do I get have a look what do I exactly got what did I exactly get try to understand try to understand this is double right this is a negative this is a negative this bond is same and this is same O positive perfect now this is pi Sigma negative this negative will come here this Bond will break in this direction what do I get here what do I get here this is your Pi this is your pi and this is negative and on the top we have got O positive perfect again this is pi Sigma Nega Pi Sigma negative right this will come here this will go here what do I get at the end what do I get at the end this is the final thing this is O Okay this is oh now guys try to understand one simple thing can I say just tell me one thing just tell me one thing when you attached o towards a conjugate system is the is the electron density in the conjugated system increasing or decreasing through resonance just tell me that is the electron density increasing or decreasing in the conjugated system due to Resonance have a look negative negative negative in the conjugate system electron density is increasing through resonance through resonance right so can I say this o Group which group edes is it going to be plus M or minus M it is showing plus M it is showing plus M number one number one number two do you see electron density getting increased negative charge coming only at Oro and parap positions do you see this negative charge is only coming at Oro and Par positions over here you checked exactly this negative charge when you attached plus M showing group towards the conjugate system you saw the negative charge getting developed only on Ortho and parap positions perfect do remember this particular Point as well when you attach a plus M showing group towards a conjugate system like Benzene the negative charge will be developed on Ortho par position what does that mean that means this position is called as meta that means that means misic effect misic effect can I say it does not operate at metap position it does not operate at meta position right what does that mean that means the charges are not getting generated established on the positions yes so is this o orthop parad directing here I hope this is clear so remember misic effect will be operated only at or and Par it won't be operated at matter right this is one important thing perfect similarly guys one more example I'm giving you one more example I'm giving you one more example I'm giving you this is your conjugate system your Benzene if this is Benzene I'm attaching c bond o h here CH group CH group now people you tell me one thing I attached the group towards this perfectly conjugated system okay let's see what happens this is if you look carefully look at this group this is X double bond y x double bond y where Y is electr negative than x where Y is more electr negative than x already few minutes back only I showed you X double bond y y more electr negative than x it's going to show minus I sorry minus M it's going to show minus M so this is first of all we got to know categorically this is minus M this is this group shows minus M if this group shows M what is going to happen this Pi Bond it will shift in this direction and this Pi Bond will shift in this direction what do we get at the end what do we get try to understand this is your C now the oxygen carries a ne2 this is hydrogen attached with oxygen there's a double bond here right and from this particular carbon double bond got shifted so this becomes positive this is rest everything is same now look here Pi Sigma positive Pi Sigma positive Pi Sigma vacant orbital resonance will happen conjugation right this will shift here and when it shifts here what do I get this is your C H this is your o with negative right this is your double bond and my dear students look at this carefully look at this carefully this is your double bond and over here you got the positive right again Pi Sigma positive Pi Sigma positive means this will shift here and when it shifts here what do I get this is negative this is your H check the other things check the other things this double bond will remain as such this will remain as such and your positive got developed here right now Pi Sigma positive Pi Sigma orbital right so this will shift here and this will shift here what do I get at the end this is CH this is double bond o there's a single Bond here now it's a single Bond here rest everything is going to remain like this right now guys just try to understand one simple thing when you attach this particular group towards a perfectly conjugated system can I say charge separation got created this is positive negative positive negative positive negative charge separation got created between the conjugate system and the group so I'll say this group is operating it is showing misic effect now is this group increasing the electron density or decreasing I will say this group is decreasing the electron density in the system see positive got developed positive got developed positive got developed right this group when attached to the conjugated system it is decreasing the electron density through resonance in the conjugated system right so I'll be saying this group is showing it's electron with drawing it is showing minus it is showing minus number one number two where at what position the positive charge is getting generated Oro para Oro is the positive charge getting generated at meta no that is something which I already told you right through momic effect through momic effect positive as well as negative charges in the Benzene ring they are generated only at Ortho and Par positions so in short I'll again say plus M as well as minus M it does not it does not operate at meta positions right positive as well as negative through either plus M or minus M they get only generated at Ortho and Par positions okay clear now guys there are some things which you have to remember what are plus M showing groups what are plus M showing groups whenever you see a conjugate system right and if the group is attached to a conjugate system if the first atom of the group carries a lone pair if the first atom of the group carries a lone pair this group will show plus M if the first atom of the group carries a lone pair it will show plus M if the first atom of the group carries negative charge with a loone pair it will show plus M it will increase the electron density in the conjugate system it will increase the electron density in the conjugate system so do you remember do you remember this particular statement plus M showing groups if the first atom of the group carries the lone pair or the first atom of the group carries the negative charge with the unshared pair right it is going to show plus M it's going to increase the electron density in the conjugate system through resonance through resonance perfect number one number one minus M showing groups minus m showing groups minus M showing groups if the first atom of the group if the first atom of the group if the first atom of the group if it contains vacant orbital if it contains vacant orbital this vacant orbital can be either P or D it can be either P or D what will happen what will happen see this is basically your if you look at it carefully this is basically your SP2 hybridized carbon so there is a p orbital right so what is what is resonance resonance is going to happen between them resonance is basically the partial over overlap of these P orbitals right so partial overlap will happen so resonance will happen basically perfect and what is going to exactly going to happen I'll say this will shift in this direction this will shift in this direction and so on so if this atom it carries the vacant orbital it carries the vacant orbital perfect if it carries the backend orbital see now it is not necessary that that this atom should have positive charge it can be without positive charge as well look at this bh2 look at Boron look at Boron Boron is forming how many bonds three bonds boron's configuration is 2s2 2p1 boron's configuration is 2s2 2p1 right this is 2s caring two electrons this is 2p carrying one electron now after excitation what's going to happen this is one and this is one this is one perfect now it is forming three bonds so SP2 hybridize this is SP2 there will be a pure p orbital here which is vacant right so this atom it can have positive charge or it cannot have positive charge condition is there should be a vacant orbital present there should be a vacant orbital present right there should be a vacant orbital present and that group will show minus M effect that group will show minus M effect now now people if this is clear I'm giving you certain groups you have to identify which group will show plus n which group will show minus N I think you're sleeping are you sleeping are you sleeping quickly let me know quickly don't sleep you are not I'm not letting you sleep I'm not going to let you sleep I need some fire emojis in the chats first I need some fire emojis in the chats I need some fire guys I don't know how my energy is getting in increased slowly do you realize that do you realize that so your energy should increase drastically right it should increase drastically first I would want fire emojis from everyone let the chats be fast then only it will give me the idea that you guys are in the session perfect go ahead tell me whether this group shows plus him or minus M when it is attached to the conjugated system like benzene when it's attached to a conjugate system like this will it show plus M or minus M have a look it is negative with a lone pair right it will show resonance so this will be plus n right this carries negative this will be plus him right carrying lone pair carrying lone pair right if you attach this with this particular ring what is going to happen it will increase the electron density through resonance plus M what about this one again a pair it will show plus M right o r it will show plus M correct it will show plus M now what about this group O2 positive O2 positive will it show minus M I told you if the first atom of the group it can carry positive or cannot carry positive will it show plus or minus quickly quickly people are saying minus m is that is that is that I'll I'll say it is neither going to show plus M nor minus M I'll say it is neither going to show plus M nor minus M why is that see when you attach it with a conjugated system this is your conjugate system for example there's a positive charge in oxygen right there's a positive charge on oxygen no doubt no doubt but how many lone pairs oxygen has two they say one lone pair will be there only with oxygen there will be one lone pair with oxygen right check the octet of oxygen it's forming three bonds three bonds means six electrons 6 plus 2 is eight it's OCT is complete there is no vacant orbital the condition is the first atom of the group should have vacant orbital there's no vacant orbital there's no vacant p orbital here so if it's not vacant it's not going to show plus n sorry it's not going to show minus n yeah look at this one o R2 R2 positive there will be a lone pair here one lone pair oxygen has two but one will be there due to other one there is positive again 1 2 3 three bonds means six electrons 6 plus 2 8 no vacant orbital nitrogen octed complete no vacant orbital nitrogen octed complete no vacant orbital sh there are two lone pairs there are two lone pairs right so it will give away lone pair right or resonance so plus M correct look at this one look at this one attach this with conjugate system right attach this with conjugate system what is going to happen this will shift here this will shift here so it is it is pulling the electron density towards itself right through resonance minus n pr3 positive minus M or plus M quickly quickly guys is it minus M or plus M see phosphorus carrying positive right see no doubt it's forming four bonds so OCT complete but but but but phosphorus it has got vacant D orbitals vacant D there is a vacant D orbital if there's a vacant orbital that means minus n vacant orbital minus n here lone pair plus n right vacant orbital minus n clear this is how you identify which group shows plus M and which group shows minus M right this is how you identify which group shows plus M which group shows minus M I hope this is clear to to everyone right all right identify the group showing plus or minus identify the group showing plus M and minus M quickly identify the group showing plus M and minus M quickly see if it is attached to a conjugated system what will it do what will it do this is basically X double bond or triple bond y x double bond or triple bond Y and Y is more electr negative than x so this minus M if you remember if you remember remember if you remember right if you remember this oxygen it carries a lone pair it carries a lone pair if it attached to a conjugated system if it is attached to a conjugate system it can donate this lone pair through resonance right so this will show plus M right this nitrogen carries a lone pair it will donate this loone pair it will show also show plus n correct this what about this one if you're attaching this with the perfectly conjugated system what is going to happen what is going to happen this is X double bond y y is more electr negative than x this will show minus n even this will show minus n same reason right this will show minus M same reason right now you tell me bh2 what is it going to show what is it going to show Boron OCT is not complete it is forming three bonds so six electrons so it will be having a vacant p orbital it will show minus n this carbon it will show minus n H2 grou right because there's a vacant p orbital perfect now people Florine chlorine bromine iodine what do they show since they have got lone pairs all the hogen they have got three three loone pairs right they have got three three loone pairs so they can give away they can contribute these loone pairs right through resonance so they'll show plus n they'll show plus s they'll show plus M but particularly my dear students when you talk about halogens when you talk about halogens hogen they as for mic effect they should show plus but at the same time they show minus indc effect as well particularly in case of halogens if hogen show plus n they're going to be electron donating due to minus I they are electron withdrawing but let me tell you halogens they are net electron withdrawing hogen show both plus M as well as minus I but inductive effect dominates over misic effect only in case of halogens inductive effect dominates or misic effect inductive effect dominates over misic effect in case of particularly halogens so hogen will be net electron withdrawing I'm particularly talking about halogens I'm particularly talking about halogens right I'm particularly talking about halogens okay now few points to remember then we'll start the questions what are the points have have a look plus M power of anionic group this is the anionic group is greater than that of plus M power of neutral group The something which we discussed in inductive effect as well right plus I power of an groups used to be greater than that of plus I power of neutral groups similarly here also plus M power of an groups will be greater than that of plus M power of neutral groups number one minus M power of K groups minus M power of K groups will be greater than that of minus M power of neutral groups right will be greater than that of minus M power of neutral groups perfect misic effect is directly proportional to ex of overlap what does it mean have a look I've taken two examples I've taken two examples so imagine this is your conjugated system and you attaching o with it right similarly you have got one more conjugate system and you are attaching sh with it sulfur also has two lone pairs oxygen also has two lone pairs now since this is going to show resonance even this lone pair is going to participate in resonance due to which this will also show plus M and this will also show plus M now since the group and the system is showing resonance what is resonance overlap partial overlap of partial overlap of orbitals partial overlap of parallel orbitals which are co-planar right since this contains p orbital this contains p orbital so there will be this carbon it will be having 2p this will be having 2p so the overlap here is going to be 2p 2p and the overlap here is going to be 2p 3p right now extent of overlap extent of overlap effective overlap will be here 2p 2p if more effect to the overlap more is the EXT of resonance more the EXT of resonance more with the plus M power so do remember plus M power of O will be greater than that of plus M power of sh simple plus M power of o will be greater than that of plus M power of H because here it's going to be 2 P2P overlap here it's going to be 2p 3p overlap right got it got it guys all right arrange the following as per their misic power arrange the following I think these sort of questions you can do on your own arrange the following as for their misic power for example o and O this will show plus M this will show plus M now which one shows more plus M which one shows more plus M which one shows more plus M how you are going to remember them how you are going to remember it see bond is between o and H here the bond is between O and C carbon electro negativity difference here is more if electro negativity difference here is more that means oxygen will be getting more Delta negative here oxy will be getting less Delta negative more Delta negative means more plus so this will show more plus corre this is how you are going to remember it look here nitrogen hydrogen here nitrogen hydrogen as well as carbon right so negative charge will be maximum in the first case right more negative charge more plus M perfect now my plus M power of anic groups that's greater than plus M power of neutral groups so 1 and two will have more plus M than three and four now among one and two among one and two among one and two which will have which will have more plus see this oxygen more Electro negative it will hold its electron pairs towards itself so availability of this is less availability of this is less so extent of resonance is less here comparatively extent of resonance will be more so two more plus n followed by one now among these two it'll be three then two I hope this is clear I hope this is clear right now at the same time Florine if attached to a conjugate system chlorine if attached to a conjugate system see chlorine has got the lone pair so this group this Florine and the conjugate system they are going to show resonance what is the resonance partial overlap of parallel orbitals which are c-plan basically right so here the overlap has to be due to Resonance what is the resonance overlap right here the overlap will be this is p this is p now this is 2p this is 2p the overlap here is 2p 2p here the overlap will be 2p 3p here the overlap will be 2p 4p for here the overlap will be 2p 5p more the EXT of overlap more the plus M power so this is the Plus in power of these halogens as well this is the Plus in power of these halogens as well right this is the plus M power of these halogens as well right I believe this is clear now guys this is your plus M series plus M series which needs to be remembered at any cost this is your plus M series which needs to be remembered nh2 its plus M power is maximum then O Negative nh2 nhr you have to remember this particular series similarly minus M series also you have to remember then only we can do questions minus M series also we have to remember then only we can do the equs this is your minus M Series so it is my request to remember plus M series minus M series as well right and once you remember these two series you can easily solve the questions but before solving the questions there are few more points which I would want to share with you few more points which I would want to share with you then we'll do we'll go ahead with the questions I told you halogens they are electron donating as per plus M withdrawing as for minus I but for halogens inductive effect dominates or misic effect that's the reason why hen are net electron withdrawing that is the reason why hogen are net electron withdrawing so halogens for hogen minus I dominates over plus right okay now when hogen is conjugated with reference net electron withdrawing nature of Cl is more than that of F this is the scenario which I want to show you Florine is attached to the conjugate system chlorine is also attached to the conjugate system so Florine it's lone pair it will donate its lone pair it will show plus M as well as Florine shows minus I right this will show plus M as well as minus I if you talk about the plus M for the plus M to happen it's going to be 2 P2P overlap here it's going to be 2p 3p overlap right perfect so I would say plus M power of plus M power of Florine should be more plus M power of chlorine should be comparatively less minus I power of hogen of Florine is more minus I power of chlorine is less you know it right okay plus M power of chlorine is more than the plus M power of chlorine because here it's 2 P2P overlap here it's 2 p3p overlap if I talk about inductive effect minus I power of chlorine is more minus I power of chlorine is comparatively less but we know halogens what are they they net electron withd drawing if you compare the magnitude of both these effects if you compare the magnitude of both these effects eventually you'll get to know Florine is electron withdrawing this is also electron withdrawing and do remember in this particular case electron withdrawing nature electron withdrawing nature of chlorine will be more than that of electron withdrawing nature of chlorine it will be more than that of electron withdrawing nature of chlorine remember this particular Point remember this particular point right they show this shows plus M minus I plus M minus I plus m is more plus m is less minus I is more minus I is comparatively less but do remember both of them they are net electron withdrawing and net electron withdrawing nature of chlorine ha will be more than that of chlorine I hope this is clear I hope this is clear and guys one simple thing and we'll start the questions one simple thing and we'll start the questions momic effect it always dominates over inductive effect except in case of hogen for example you have got a group the group has got the tendency to show misic effect as well as inductive effect as well as inductive effect as well as hyper conjugation as I should leave hyper for example you have got a group it has got the tendency to show momic effect as well as inductive effect remember momic effect always dominates over indc to effect always dominates or indictive effect momic effect always dominates or indictive effect except in case of halogens except in case of halogens for example you got a group X right let's say it can show plus M it can show plus I as well right you'll be considering it's plus M this will be given the priority this will be given the priority perfect let's say you got a group it can show minus M it can show minus I as well but you'll be prioritizing misic effect so misic effect dominates or inductive effect except in case of halogens in halogens indc to dominates or misic and already one thing which you must be knowing momic effect does not operate at metap positions it does not operate at metap position okay now people we have got a lot of questions and in order to solve the questions let me tell you certain tricks let me tell you certain tricks first of all if you talk about the stability of the carboca if you talk about the stability of carboca carbo Catans are stabilized by electron donating groups destabilized by electron withdrawing groups electron withdrawing group right all the plus showing groups they are donating all the minus groups they are withdrawing correct so people if you want to arrange whenever you'll be given certain carboca and you have to arrange them on the basis of their stability you know momic effect dominates for everyone first you'll prioritize plus M showing group plus M showing group is going to stabilize the carbo and the maximum then plus h then uh then plus I right after plus I it's going to be minus I after minus I it's going to be minus H after minus H it's going to be minus M since H we have not discussed yet you just note it down that's it this is going to be the order which you'll be following plus M showing group it's going to stabilize the carbo and the maximum right then you'll be prioritizing plus M right sorry plus H after plus h you'll be prioritizing plus h sorry plus I right similarly whenever you'll be asked equation in which you have to give the stability order of the caronian caronian are stabilized by electron withdrawing groups and destabilized by electron donating groups destabilized by electron donating groups electron withdrawing minus and we know minus M will dominate over minus H minus H will dominate over minus I after minus I it's going to be plus I after plus I it's going to be plus h and then at the end it's going to be plus M this is the sequence which you have to follow this is the sequence which you have to follow that's all sequence which you have to follow right similarly third strength of acid strength of acid is directly proportional to K which is inversely proportional to P of K right which is directly proportional to stability of the conjugate base stability of the conjugate base of the acid right and conjugate bases are stabilized by elect electon withdrawing electron withdrawing so directly proportional to electron withdrawing inversely proportional to electron donating groups and in electron withdrawing which one to prioritize minus M to prioritize the maximum then minus H then minus I then plus I then plus h then at the end plus M right this is the order which you'll be following similarly strength of a base strength of the base is directly proportional to KB which is inversely proportional to P of KB which proportional to P of KB right perfect and strength of Base strength of Base strength of Base electron donating groups increase the strength of the base electron withdrawing groups they decrease the strength of the base right and here plus M dominates over plus h plus h dominates over plus I plus I dominates over minus I then you have got minus H then add the the end you have got minus M so my dear students this is the sequence which you have to follow and different types of questions can be asked if I mean different types of question can be solved if you remember these particular sequence if you remember these particular sequences right and I hope you guys are going to remember them from now onwards this is the final order which you have to remember and you can easily solve the question easily easily easily you can solve the questions have a look understand one thing if you properly have a idea of certain things this series and this series it is same this order and this order is same right similarly first and the fourth order is also same right first and the fourth is also same first and the fourth is also same now let's let's try to solve the questions let me see if you can solve these questions or not see guys arrange the following according to their stability order these are your carbo caes right carbo are stabilized by electron donating groups and you know plus M dominates over plus h plus h dominates over plus I plus I dominates over minus I minus I dominates or minus H minus H dominates or minus this is the order which you have to understand first of all understand this is carbon carrying positive carbon carrying positive carbon carrying positive you have checked your site this is your site common site everywhere now this is O here o at par oxygen has got the lone pair right it's going to donate this loone pair in the form of resonance so it is going to show plus M right o at meta at meta you are not going to operate momic what you'll operate here this oxygen more Electro negative you'll operate it's minus I right you'll operate minus I from here Nitro at para Nitro at para Nitro at para it's going to show minus n Nitro is minus I'm showing group right Nitro at meta minus I sh sulfur has got the lone pair it can donate so plus M add par okay your carboca are stabilized by plus M showing groups which one is plus M this is plus M this is plus M now which one shows more plus M oh shows more plus n oh shows more plus n so first one is maximum stable followed by followed by Fifth so plus M we have prioritized now is it anyone plus h we have not discussed plus h yet leave it aside now plus I which one is plus I nobody's plus I leave plus I also now minus I minus I this is minus I this is minus I right this is minus I even this is minus I minus I means electron withdrawing electron withdraw drawing groups they destabilize the carbocation right which one among the two is more withdrawing this is showing more minus I this shows more minus I more withdrawing this is going to destabilize more as that of this right I'll say two will be more stable than four and at the end you'll be prioritizing this three so is this the order what people have written 152 43 1 52 43 absolutely correct wonderful wonderful yes perfectly done okay this was our first question look at the second question look at the second question these are the caronian carbon carrying negative these are the caronian everywhere these are the caronian people these are the carians and carians are stabilized by electron withdrawing priority order minus M minus h - i+ i+ H+ n this is the priority order Now understand this nh2 nitrogen carrying lone pair it will show resonance right it will go in this way so it's plus M right at meta position nitrogen is more electr than carbon so minus I so at para will give away so it is where to show plus M this pr3 pr3 vacant D orbital phosphorus has vacant D orbital it will show minus m right now which one which one is showing minus M there's only one showing minus M that has to be prioritized which is going to be more stable so four is more stable followed by minus m is gone now minus H leave it aside minus I anyone showing minus I yes second is showing minus I after second after minus I is there anyone showing plus I no now there are two plus M showing this is plus M showing this is plus M showing now which one has got more plus M which one shows more plus n power nitrogen carrying lone pair oxygen carrying in loone pair now tell me nitrogen less Electro negative this loone pair is more available more available for resonance so it will show more plus n right it will show more plus M this is more plus M this is less plus M right more plus m means more electron donating which will destabilize it maximum destabilize it maximum so one at the end before one it's going to be three so this is going to be the order stability order agreed people plus say 4 2 3 1 4 2 3 1 yes 4 2 3 correct okay right guys perfect Mo ahead Mo ahead look at this particular question arrange the following as for the K order that means acidic strength order acidic strength order how do you compare the acidic strength what is the order acidic strength is directly proportional to stability of conjugate base right acidic strength is directly proportional to stability of conjugate base conjugate base is stabilized by electron with drawing minus M dominates or minus hus H dominates or Min I then comes your plus I then comes your plus h then comes your plus M now have a look this this is o o o everywhere you saw your reference this is NO2 from from para it will operate minus M CN from par it will operate minus n right for these two index to effect dominates index to effect dominates so I'll consider it's minus I it's minus I even though it shows plus M as well it shows plus M as well due to which both of them are electron withdrawing but chlorine comparatively is more electron withdrawing than Florine right that's something which we have discussed already now first of all you have to prioritize minus M this is minus M this is minus M which one shows more minus M Nitro shows more minus M right so one is maximum followed by two now here this is withdrawing this is withdrawing this is more withdrawing so four then three this is the order yeah is it clear you should be able to solve these sort of questions now all right look at this particular question let me see if you can solve this one or not again you have to arrange them on the Bas of acidic strength again you have to arrange them on the basis of their acidic strengths okay acidic strengths atic strengths quickly what do you think about this particular equation what do you think about this particular equation what do you think about this particular equation guys it's again very simple this is o oh o oh everywhere the reference is same now you have to arrange them on the base of Ka order step Ka acidic strength acidic strength is directly proportional to stability of conjugate base conjugate base is stab stabilized by electron withdrawing in electron withdrawing minus M dominates over minus H then you have got minus I then you have got plus I you got plus h you got plus 7 this is the order which you'll be following right now people tell me whether this particular group which if is it going to show minus M or plus M this is n positive forming four bonds there's no vacant orbital it is not going to operate minus M from here it's not going to show minus M by the way okay if you remember it does not have the vacant orbital okay so it will directly show minus I right it will directly show minus I perfect what about this one there is a vacant D orbital it will show minus n it will show minus n this will show minus I this will Al also show minus I because they are present at the matter positions now tell me the order tell me the order people tell me the order which one is showing minus M there's only one showing minus M that's two followed by minus H leave minus H aside we have not discussed that yet then minus I minus I minai see guys if this is carbon number one this is two this is three this is four so mini is operating from fourth carbon but we know inductive effect we do not consider up to three carbon atoms so you do not consider it's inductive effect don't consider it inductive effect now look at these two look at these two nitrogen carrying positive phosphorus carrying positive nitrogen phosphorus nme3 positive pme3 positive which one shows more minus I nitrogen phosphorus they belong to same group in the group vcd plays a role so here volume charge density positive volume charge density will be more positive volume charge density means more electr negative more minus I so this will show more minus I so after two it's going to be three after three it's going to be four and at the end it's going to be one did anyone give this particular order 2 3 4 1 wonderful 2 3 4 1 yes it's a correct order okay I hope this sort of a question you can also solve from now onwards all right what do you think about these two arrange the following according to the KA order I mean which which one among the two is more acidic H Co phc HC or PC HC or pH quickly what do you think HC or PC see guys if you talk about this phc pH means this is a fenile ring right this is your C double bond o and this is o right look at this particular one this is X double bond y y is electr negative than x right perfect why is so this will operate minus M what does that mean that means this operating plus M so this pH is operating plus M from here and here there's hydrogen nothing more acidic more acidic this operating plus n acidic strength is directly proportional to stability of conjugate base conjugate bases are stabilized by electron W drawing right electron withdrawing this electron donating less acidic so first one will be more acidic first one will be more acid right first one will be more acidic now guys here you can do a mistake here you can do a mistake I want you to do that mistake what should be the answer of this what should be the answer of this quickly what should be the answer of this question what should be the answer of this particular question see first of all if you look at these particular carbons which carry the negative charge if you look at these particular carbons which carry the negative charge be careful with this question guys be careful with this question see first of all if you look at this particular carbon if you look at this particular carbon first of all just let me make you understand it like this just a second just a second now it's fine perfect now if you look at this particular carbon carbon's configuration is 2s2 2p2 so this is 2s and this is 2p 2s and 2p correct this carbon it has already formed a pi Bond so one of the p orbital has already participated in the parall overlapping so this p and this P this p and this P they have overlapped now the hybridation is SP2 SP2 right so how many hybrid orbitals do we get three hybrid orbitals one hybrid orbital has formed a bond here one hybrid orbital has formed a bond here and one hybrid orbital that will be containing this negative charge one hybrid orbital will be containing this negative charge right and all these three hybrid orbitals this one this one and this one they'll be present on the same plane because it's SP2 SP2 means trigonal planer SP2 trigonal planer so this hybrid orbital this one and this one they are present on the board and there is one p one p which is already participating in the overlapping right so this Nitro this Nitro Nitro at par we know it shows minus M but this minus M this minus M tell me one thing tell me one thing whether I'll be considering minus M of nitro hair or minus I of nitro what do you think whether I'll be considering minus M of nitro or minus I of nitro see guys if this negative charge was in p orbital if this negative charge was in p orbital then only this negative charge would have spread through resonance right then only this negative charge would have spread through resonance then only stability would have increased but this negative charge it is out of plane of resonance it is not in the plane of resonance this negative charge it is not in the plane of resonance it is not not in the plane of resonance right it is not in the plane of resonance if it was in the plane of resonance if it was present in the pure p orbital then I would have taken minus M then minus M would have stabilized it then minus M would have stabilized okay but this negative it is not in the plane of resonance here right so you'll be taking since we are talking about the stability of carbon caronian you'll be considering its minus I here right even though this is carbon number 1 2 3 and four it is operating from from at Fourth carbon right and inductive effect dies after third okay here it is meta so minus I here here it's again minus I you are not going to consider it perfect and here ch3 it's going to be plus I I believe you can easily give the order here I believe you can easily give the order here my point is this negative charge here is present in SP2 hybrid orbital it is present in SP2 hybrid orbital it is not present in the plane of resonance right it is not present in the plane of resonance if it is not present in the plane of resonance then then if it is not present in the plane of resonance then there'll be no overlap no overlap means no spreading of this negative charge no spreading of negative charge means no increase in St stability through resonance right I hope this sort of equ question you can easily solve okay tell me about this one tell me about this one again guys this is positive this is positive this is positive right positive positive positive so again the positive charge is present on the Benzene ring Bic positive charge is present on the Benzene ring in the earlier case negative charge was present on the Benzene ring whenever you see such kind of scenarios positive charge present on Benz ring ring or negative charge pressure on Benzene ring momic effect does not stabilize them momic effect does not stabilize them because those positive and negative charges they are out of the plane of resonance they're out of the plane of resonance so which effect will be operating here inductive effect only right inductive effect only perfect I hope you can easily solve this one as well only effect will operate only indictive effect will operate okay one last question on momic effect okay last but one this is last but one last but one arrange the following as for the KB order KB means basic strength first of all this is the site this is the basic site here which is common everywhere this is the basic site here which is common everywhere now see nh2 it will operate plus n o it will also operate plus n this ch3 it will operate plus I this will operate minus M and uh when you talk about the basic strength your plus M groups they increase the basic strength to the maximum followed by plus h followed by plus I after plus I it is minus I after minus I it is minus H after minus H it is minus M okay now which one is showing plus M this is showing plus M this is showing plus M right plus M of nh2 is more so one is more basic than second right then plus I so 1 2 3 4 is order 1 2 3 4 is the order 1 2 3 4 is the order anyways this is something which you call as gadan right which is maximum basic this is maximum basic already this is something which you should know gine maximum basic correct corre guys now there is one question there is one question have you seen this particular sort of question before just tell me that arrange the following according to the bond angle around nitrogen Bond angle around nitrogen have you seen this sort of equation before just tell me that just tell me that once see guys what this question is all about and how do we solve this arrange the following according to according to what according to the bond angle around nitrogen around nitrogen see this is CH here right this is your CH here this is C ho o CH means this is C double bond o and this is H correct first of all if I talk about this particular nitrogen if I talk about this particular nitrogen and this particular nitrogen both these nitrogen they have got lone pairs right are these lone pairs conjugated yes these lone pairs are conjugated yes these lone pairs are conjugated right yes these lone pairs are conjugated are these localized or delocalized loone pairs these are delocalized loone pairs because this is pi Sigma loone pair delocalized loone pairs these are if they are delocalized should I consider them while calculating the hybridation of nitrogen I'm not going to consider them right I'm not going to consider them because delocalized here also it is delocalized my dear students imagine imagine if this was localized lone pair then the steric number of nitrogen would have been three bonds one loone pair 3 + 1 4 its hybridization should have been sp3 at that time if this was localized if this was localized its hybridization should have been sp3 right due to delocalization if the loone pair is now delocalized so three sigma bonds its hybridation is coming out to be SP2 right due to delocalization its hybridation is coming out to be SP2 Bond angle here was 109° 28 minutes here the bond angle should have been 120° right here it should have been 9° 28 minutes here in case of SP2 it is 120° perfect if the lone pair was localized hybrid addition of this nitrogen should have been sp3 but due to Resonance it is SP2 due to Resonance it is SP2 and due to Resonance you are getting the partial double bond character as well due to Resonance you getting the partial dou Bond character as well so tell me due to Resonance due to Resonance is the bond angle increasing or decreasing due to Resonance if there was no resonance if this loone pair was localized hybridation should have been sp3 but since it's participating in delocalization it's participating in resonance so it's hybridation is changing from sp3 to SP2 so Bond angle is increasing so due to Resonance Bond angle is increasing due to Resonance Bond angle here also is increasing B angle hair is also increasing and due to Resonance this gets double bond character even this gets double bond character if I ask you where is the extent of resonance more can I say extent of resonance is more in this case extent of resonance is more in this case so more the extent of resonance more the extent of resonance more the bond order I mean more the bond order more the double bond character more the extent of resonance more the double bond character here can I say something like that can I see something like that see guys try to understand one simple thing which I'm trying to convey due to Resonance due to Resonance Bond angle is Bond angle is increasing Bond angle is increasing so where is the extent of resonance more extent of resonance is more in the second case more the ex of resonance more the increase in bond angle so where is the bond angle more Bond angle around nitrogen is more in second than first simple tell me you got this question due to Resonance Bond angle is increasing right now wherever xtent of resonance is more increase in bond angle will be more as simple as that yeah got it perfect now now guys there's a topic called as aromaticity aromaticity tell me first of all if momic effect is clear to everyone see this is a simple topic this is not difficult or something tell me if aromat tell me if misic effect is clear to everyone tell me if misic effect is clear to everyone tell me if momic Pi this clear to everyone guys you will have to revise this lecture tomorrow by the way you will have to revise this lecture otherwise it'll be all over otherwise it'll be all over and you have to remember all the things whatever I have told you in the session you will have to revise this particular lecture again tomorrow otherwise it's going to be really difficult to remember this all right let's talk about something called as aromaticity see very simple and basic thing I'm not going into the details I'll just tell you that amount of theory required which is used to solve the problems okay now people try to understand what exactly I'm going to say let me tell you in organic chemistry in organic chemistry there were some organic compounds which were experimentally found out to be extra stable in organic chemistry let me write it over here in organic chemistry experimentally experimentally some of the compounds some of the compounds were found out to be were found out to be extra stable where found out to be extra stable than expected in organic chemistry there are some compounds which are found out to be which were experimentally found out to be extra stable than expected and those compounds which were experimentally found out to be extra stable than expected those compounds were given a name as aromatic compounds those compounds were given a name aromatic compounds similarly one more set of compounds was found experimentally which was found out to be highly unstable than expected highly unstable than expected and those compounds were termed as anti- aromat those were termed as as anti-aromatic and there were few compounds in the organic chemistry which were found out to be whose stability was found out to be between these two whose stability was found out to be between these two those were called as non-aromatic compounds so let me repeat it let me repeat it in organic chemistry a particular set of compounds were found experimentally which were having stability extra than expected those compounds were called as those were given the names aromatic compound some more compounds were there which were found out to be highly highly highly unstable those were given the name as antiaromatic some com compounds were seen experimentally whose stability was between aromatic and anti-aromatic those were called as non-aromatic now guys first of all first of all first of all if I ask you what is the stability order of these what is the stability order of these particular compounds you'll say aromatic aromatic compounds will be maximum stable followed by non-aromatic followed by anti-aromatic followed by antiaromatic this is something which has been observed experiment right this is something which has been observed experimentally okay now guys the point is what is the point here the point is in aromatic compounds in aromatic compounds and anti-aromatic compounds there were some things which were common in aromatic compounds there were some common features which were seen similarly in antiaromatic compounds some common features were present some common features were present what were those common features in aromatic compounds they were found out to be cyclic they were found out to be cyclic they were found out to be planer they were found out to be planer there was complete cyclic conjugation present there was complete cyclic conjugation present and and there were 4 n + 2 4 n + 2 pi electrons were involved in the cyclic conjugation that means these compounds were following the huckles rule they were following the huckles rule these were the few things which were common in these aromatic compounds see a particular set of compounds were found out to be experimentally in organic chemistry which were found out to be extra stable than expected those were called as aromatic compounds in aromatic compounds few things were common they were cyclic they planer complete cycli conjugation was there 4 n plus 2 pi electrons were involved in the complete Cy conation right 4 n + 2 2 pi means if n is zero that means either 2 pi electrons were involved if n is 1 4 + 2 6 Pi electrons were involved or four is n is 2 10 pi electrons were involved n is three 14 Pi electron were involved in the complete cyclic conjugation now in anti-aromatic compounds again there were some common features they were again cyclic they were planer also they were planer also it was observed that there is complete cyclic conjugation there is complete cycli conjugation there is complete cyclic conjugation but there were 4 n Pi electrons involved in the complete CYCC conjugation right if n is one that means either 4 Pi electrons were involved in the complete cyclic conjugation or 8 Pi electrons or 12 Pi electrons or 16 Pi electrons etc etc non-aromatic see they were they are planer these are also planer these were non planer these were nonplanar these were nonplanar if the molecule is neither aromatic nor anti-aromatic then what is it it is non- aromatic itic now see how I'm going to utilize this particular concept see how how I'm going to utilize this particular concept TR to understand guys rest no need to remember anything rest no need to remember anything rest no need to remember anything look at this particular molecule all the carbon atoms are SP2 they're SP2 hybridized absolutely it's a Benzene ring SP2 hybridized right SP2 means trigonal planer so first of all the molecule is cyclic the molecule is planer as well because every carbon is P2 right now is there complete CYCC conjugation Pi Sigma Pi Sigma Pi Sigma complete cyclic conjugation how many Pi electrons are participating in complete cyclic conjugation 2 4 6 6 Pi electrons that means this comes under the category of this is your aromatic this is your aromatic right this is your aromatic because six P electrons involved in the cyclic conjugation means aromatic perfect this is your aromatic look at the second one look at the second one see this is SP2 this is SP2 this is SP2 that means planer SP2 means planer cyclic as well see Pi Sigma positive Pi Sigma positive complete cyclic conjugation only two Pi electrons are involved in the complete cyclic conjugation 4 n + 2 pi that means this aromatic 2 pi electrons means aromatic following the huckles rule right look at this one cyclic right all the carbon atoms SP2 cyclic cleaner 2 four four Pi electrons involved in the complete cyclic conjugation if four electrons are there that means it's supposed to be anti-aromatic right this antiaromatic simple look at this one simple look at this particular carbon there'll be one hydrogen here one hydrogen here so 1 2 3 4 four Sigma so this sp3 sp3 planer or nonplanar means nonplanar nonplanar means non-aromatic right because aromatic as well as anti-aromatic they are planer if it is non planer it's non aromatic look at this one this particular carbon it is sp3 hybridized so nonplanar means non-aromatic simple non plar means nonaromatic where did this slide go why is not showing anything okay just a second we need to give some rest to the screen I think something is wrong here something is wrong here stopped working yeah finally now it's fine now it's fine just a second okay all right guys first of all tell me is this lone pair in conjugation this lone pair is not in conjugation is it conjugation it's not in conjugation right if it is not in conjugation you have to consider this also for hybridization 1 Sigma 2 Sigma lone pair two Sigma and one lone pair this is SP2 hybridized SP2 hybridized nitrogen right SP2 this lone pair is not involved in the conjugation okay now all the other atoms are also SP2 so planer how many Pi electrons 2 4 6 six P electrons are involved in the complete CYCC conjugation so aromatic right look at this one among these two lone pairs only one lone pair is conjugated Pi Sigma lone pair Pi Sigma lone pair only one lone pair is conjugated so leave that aside and the other one count for hybridation 1 Sigma 2 Sigma one lone pair 2 Sigma one lone pair steric number three so this SP2 so this is planer all the other carbon atoms are already SP2 planer right planer cyclic 2 4 and six six Pi electrons are involved in the complete cyclic conjugation so this aromatic again right exocyclic bonds you are not going to consider exocyclic bonds you are not going to consider now this is cyclic every carbon is SP2 and there are two Pi electrons involved in the complete cyclic conjug ation so this also aromatic this is also aromatic right this is also aromatic I believe it's clear I believe it's clear my dear students for aromaticity what is happening uh screen is not responding properly just give me a second not showing the slides h just wait guys just wait it will it will respond it will respond what do I do with the screen let's wait for 1 minute okay let's wait for 1 minute let's give it some rest I don't know let's wait for 1 minute what's up how are you all P wants break yeah P wants break okay I'll go a little slow so what's up what's happening are you all alive are you all alive are you all alive someone is saying my brain is not braining anymore good nice brain should not brain anymore I'll do one thing I'll save this part of the slide save these part of slides and then we'll upload it again uh I think the screen is stuck screen is stuck I think something is wrong something is wrong oh this happened in HSP sir's class also and what he did exactly not doing anything stuck okay I'll also leave it for some time let's leave it for some time so guys how is the preparation going tell me how is the preparation going till then let's CH Chit Chat how is the preparation going going all well on the scale of 10 on the scale of 10 on the scale of 10 are you sure 78 that's great that's nice nice great guys that's amazing good so are you following all the marathons all the marathons zean where are you studying exactly which offline you are studying that's amazing guys follow biology as well why you are not following biology is follow biology as well that's again important that's great shat that's great Sai following from day three uh Dharma will decide that we have not decided who is going to take this isomerism he the screen is stuck by the way nilam you have Amika man teaching you right ma'am consider I rate her at the top when it comes to content nobody can beat un become the content trust me on that nobody nobody m I think I'll have to restart the screen but I don't know whether I would be able to give you the hand ritten slides then wait I'll just restart the screen once I don't know what will happen it's not shutting down also wait wait wait wait guys I'm not sure about the slides now okay I'm not sure about the slides now I don't know whether the slides will be saved or not please don't blame me don't blame I'm really sorry slides are gone uh If anyone among you like have uh taken down down the notes please take the photos of that and put on the telegram okay so that everybody can have it uh mhm two more minutes guys two more minutes and yes you'll be getting the Hat in slides there is something wrong wait yeah it's sweet let's wait guys let's wait something happened good you all right yes we are back it's okay I I'll give you the slides okay so am I again audible visible to you am I again audible visible to you so perfect guys let's continue let's continue guys let's continue so basically for the molecule to be aromatic it has to be cyclic it has to be planer there should be complete cyclic conjugation and there should be 4 n + 2 pi electrons involved in the complete cyclic conjugation that means either it can be 2 pi or it can be 6 PI right or it can be 10 pi etc etc now guys try to understand first of all if I ask you whether this loone pair is conjugated whether it's localized or delocalized whether this loone pair is localized or delocalized look at this particular nitrogen right there's a double bond here there's a double bond here is it localized or delocalized tell me that it is localized loan it is not going to participate in the resonance if it is is not participating in the resonance we have to count this lone pair for hybridization so 1 Sigma 2 Sigma and one lone pair so two Sigma and one lone pair 2 + 1 is three that means this is SP2 SP2 means planer all the other carbon atoms they are already SP2 so this molecule overall is planer Point number one the molecule is planer this is cyclic as well is there complete cyclic conjugation see pi Sigma Pi Sigma Pi Sigma there is complete cyclic contribution how many electrons how many Pi electrons are participating in the complete cyclic conjugation 2 4 6 six Pi electrons if six Pi electrons are participating in the complete cyclic conjugation tell me is it following huckle's rule huckle's rule was 4 n + 2 pi electrons should participate if n value is 1 so 4 1 are 4 + 2 6 right so six comes in this category right so this molecule is going to be aromatic I hope this is clear look at this particular thing oxygen has two loone pairs out of which one is delocalized pi Sigma lone pair Pi Sigma lone pair this is conjugated one of the lone pair is conjugated so leave that one lone pair right now one Sigma 2 Sigma and one lone pair 1 2 3 only localized lone pairs are taken into consideration when it comes to hybridization localized lone pairs are taken into consideration when it comes to hybridization so 1 2 3 2 Sigma 1 lone pair St number three steric number three means SP2 right all the other atoms are already SP2 perfect so cyclic ler now 2 4 and six six P electrons are involved in the complete cycli conjugation so this is again going to be aromatic right as I told you exocyclic bonds are not taken into consideration number one number two all the atoms in the ring they are SP2 right it is cyclic there are two Pi electrons involved right 2 pi electrons means if n is zero you get 2 pi that means it is following huckle rule so it has to be aromatic as well correct it has to be aromatic I hope this is clear now guys for aromaticity we have to we have to consider only the circumferential resonance for the aromaticity only circumferential resonance has to be taken into account what is circumferential resonance say this is pi Sigma Pi Sigma Pi Sigma this is something which you call a circumferential resonance so you do not have to consider this particular Pi Bond when it comes to aromat only circumferential resonance you have to consider only those Pi electrons only those Pi electrons which are showing the circumferential resonance only they have to be taken they have to be taken into consideration perfect so see every carbon here is SP2 okay so it is cyclic as well as planer now complete cyclic conjugation complete circumferential resonance is there absolutely now how many electrons 2 4 6 8 10 12 14 so there are 14 Pi electrons involved in the complete cyclic conjugation right so is it following Huck's rule if n is 3 4 3 is are 12 12 + 2 is 14 so this is going to be aromatic it is following huckles Rule corre now tell me about this one this is Benzene 6 Pi electrons everything is there it is aromatic we know it look at this one look at this one every carbon is SP2 cyclic planer now 2 4 6 8 10 12 14 14 P electrons are involved so it is going to be aromatic absolutely right it's going to be aromatic right guys it's going to be aromatic now see all the atoms here are SP2 hybridized so it is planer it is cyclic at the same time planer as well as cyclic complete cyclic conjugation is also there Pi Sigma Pi Sigma everywhere right now 2 4 6 8 10 there are 10 P electrons involved right there are 10 P electrons involved so it is following the huckles rule because if n is 2 4 2 are 8 8 2 10 so 10 P electrons that means it is supposed to be aromatic right n Pi aromatic moving ahead moving ahead okay this is something which we already discussed correct look at this one look at this one first of all the compound overall is cyclic now let's see whether it's planer it will be planer only if all the atoms are SP2 hybridized it will be planer if all the atoms are SP2 hybridized tell me whether this is localized or delocalized this is localized you have to this for hybridization as well so one Sigma 2 Sigma and lone pair 1 2 3 3 means SP2 this is SP2 and all the other carbons are already SP2 so this is planer this is cyclic now how many electrons is there complete cyclic conjugation Pi Sigma Pi Pi Sigma Pi Sigma Pi Sigma Pi Sigma Pi Sigma Pi right there is complete cyclic conjugation now how many Pi electrons 2 4 6 8 10 there are 10 P electrons involved in the complete cyclic conjugation this lone pair it is not participating in the resonance it is not conjugated right so 10 pi means this aromatic correct now this particular loone pair have a look Pi Sigma loone pair Pi Sigma loone pair is this localized or delocalized this is delocalized this is delocalized don't consider for hybridization one Sigma 2 Sigma 3 Sigma 3 Sigma means SP2 so this is cyclic this is pler as well planer as well now complete cycle conjugation Pi Sigma lone pair Pi Sigma loan pair complete cyclic contribution how many Pi electrons are participating in complete cyclic contribution 2 4 and six six Pi electrons means following the huckles rule it's aromatic correct what about this one oxygen has two lone pairs out of which one is in conjugation one is delocalized one is delocalized right so this is one Sigma 2 Sigma and lone pair 2 Sigma one loone pair steric number three so this SP2 SP2 means since all the carbons are SP2 involving this oxygen this also SP2 so this molecule is planer and how many Pi electrons 2 4 and these two six six Pi electrons are involved in the cyclic conjugation so this is going to be aromatic right this is going to be aromatic perfect guys perfect look at this one now look at this one cyclic as well as SP2 every carbon is SP2 so planer Pi Sigma positive Pi Sigma positive complete cycli conjugation two Pi electrons are involved in the complete cyclic conjugation following huckles rule so aromatic right look at this one 2 and four four Pi electrons four Pi electrons means antiaromatic because 4 Pi comes into the category of 4 n PI right this comes into the category of 4 n + 2 pi perfect right so this is antiaromatic look at this one two plus radical 2 + 1 3 3 Pi 3 Pi does not come into any category so it has to be it is neither aromatic nor ant anti-aromatic it is nonaromatic yes look at this one first of all this Boron it's SP2 hybridized this is SP2 this is SP2 so it's pler it is cyclic right this is pi Sigma boron has got the vacant p orbital Pi Sigma vacant orbital Pi Sigma vacant orbital so complete cyclic conjugation how many electrons I'll say two Pi electrons are participating in the cyclic conjugation so it is following the huckles rues so it supposed to be aromatic now look at this particular carbon it's sp3 hybridized sp3 means non-plan non planer aromatic as well as anti-aromatic they are planer aromatic as well as anti-aromatic they are planer right this nonplanar that means it's non-aromatic nonplanar means non-aromatic right perfect look at this particular case leave EXO aide SP2 SP2 SP2 SP2 everywhere cyclic also planer also right yeah cyclic also planer also now tell me now tell me is it going to be aromatic or antiaromatic quickly aromatic or anti-aromatic aromatic or antiaromatic quickly aromatic or anti- aromatic quickly absolutely this will be aromatic what about this one cyclic planer also because every is every carbon here is SP2 cyclic conjugation also Pi Sigma Pi Sigma Pi Sigma Pi complete cyclic conjugation 2 4 6 so six Pi electrons are involved in the complete cyclic conjugation so this is aromatic right see this is sp3 if this is sp3 that means it is nonplanar nonplanar means non-aromatic yeah perfect I hope you're following this properly okay tell me what about this one what about this one see 2 4 4 Pi electrons that means anti- aromatic correct anti- aromatic look at this one this particular carbon it's sp3 hybridized sp3 means non-planar nonplanar means non-aromatic right non-aromatic this is 2 4 6 and 88 Pi electrons means antiaromatic 8 p electrons means antiaromatic right look at this one 2 4 6 6 Pi electrons means aromatic 6 Pi electrons means aromatic okay I hope you are easily following this what about this one 2 4 6 6 Pi electrons this is aromatic right 2 4 6 6 Pi electrons this aromatic because they are already planer they are cyclic as well complete cyclic conjugation is also present now see 2 4 6 six Pi electrons so they are also this also aromatic right this also aromatic now look at this particular one planer cyclic conjugation everything is there right then have a look 2 4 and six six P electrons that means aromatic right Perfect 2 4 and six six P electrons aromatic out of these two lone pairs one lone pair will participate so 2 four and six six P electrons means aromatic why are all these aromatic aromatic what is happening see guys 2 4 6 8 and 10 10 pi electrons means hule rule 4 and + 2 pi right so this again aromatic right see in the middle you have to do nothing you have to talk about the circumferential resonance right all the atoms present at the circumference they are SP2 hybridized as well as it's cyclic so 2 4 6 8 10 10 pi electrons means aromatic right uh look at this one so nothing to do with this 2 4 6 8 10 10 P electrons means aromatic perfect this is parabeno quinon this is p parenzo quinon two and four so four Pi electrons four Pi electrons means it has to be anti-aromatic it has to be anti-aromatic it should be anti-aromatic but I told told you all this aromatic non-aromatic this all experimental so as for us it should be anti-aromatic but it has been observed that it is stable at room temperature at room temperature it is stable if it is stable at room temperature it from anti-aromatic we say it is nonaromatic it is comparatively stable okay this all experimental data by the way this all experiment these are all experimental facts which I'm telling you right now as for our rules it has to be anti-aromatic but experimentally it has been seen that it is comparatively stable so from anti-aromatic it becomes non-aromatic okay you have to remember this directly okay now what about this particular case what about this particular case quickly what about this particular case quickly what about this particular case see guys remember aromatic plus aromatic Plus nonaromatic or let me write it like this aromatic plus anti-aromatic makes the overall molecule as antiaromatic aromatic plus antiaromatic makes the overal molecule as anti-aromatic right perfect see this particular ring this is aromatic 2 46 6 pi and this sp3 this non-aromatic so aromatic plus non-aromatic remember aromatic plus non-aromatic makes the overall molecule as aromatic so overall molecule over here is aromatic because this aromatic plus non-aromatic makes it aromatic okay aromatic plus anti-aromatic makes it antiaromatic aromatic plus non-aromatic makes it aromatic right for example see this is aromatic this is non-aromatic aromatic plus non-aromatic makes it aromatic okay this is aromatic this is aromatic overall molecule is aromatic right now what about this one see I can make it like this as well I can make it like this as well absolutely you can make it like this as well now what about this ring this aromatic 2 46 what about this ring this is 2 and four so this ring is aromatic this ring is anti- aromatic so aromatic plus anti-aromatic makes it anti- aromatic okay getting it so this is anti-aromatic perfect guys now among these two among these identify which one is aromatic which one is antiaromatic which is non-aromatic have a look this oxygen has got two lone pairs out of which one will participate in resonance one will be counted for for hybridization so 2 4 and six six P electrons involved in the complete cyclic conjugation this again aromatic right perfect right guys is it clear this nitrogen it has got a lone pair Pi Sigma lone pair Pi Sigma lone pair so again Pi Sigma lone pair Pi Sigma lone pair complete cyclic conjugation so how many electrons four Pi electrons are involved in the complete cyclic conjugation so it has to be antiaromatic right now this ring this is aromatic this ring it's aromatic because 6 Pi + 6 Pi 6 Pi here also 6 Pi here also so this is aromatic this is aromatic so aromatic plus aromatic makes it aromatic only right now guys understand you you have to remember these things okay this particular structure the cyclic the cyclic conjugated unsaturated hydrocarbons these are called as anolin this is cyclic this is conjugate as well Pi Sigma Pi Sigma conjugate as well unsaturated double bonds unsaturated as well cyclic unsaturated hydrocarbons these are called as anolin and this contains four Pi electrons so this is what you'll be calling as anoline 4 this is what you'll be calling as anoline 4 perfect this is what you'll be calling as anoline 8 this is what you'll be calling as anoline 8 now anoline 4 since there are four Pi electrons four Pi electrons means it's going to be anti-aromatic it's going to be anti-aromatic this is going to be anti- aromatic right but anine 8 as well as anoline 12 anoline 8 as well as anoline 12 okay anoline 8 as well as analine 12 if I talk about anoline 8 anoline 8 anoline 8 it actually has got a tub like structure it actually has got a tub like structure due to which due to which it was looking like planer but then it becomes non planer and when it becomes nonplanar automatically it becomes non-aromatic so your anoline 8 and anoline 12 it is the example of non-aromatic it is tub shaped it is tub shaped do remember this particular thing as well do remember this as well right similarly I told you anoline 12 that is nonaromatic what is this this is basically Benzene this is anoline also right I'll be calling this as anoline this is anoline 6 this is anoline 6 basically right what is it it is aromatic right six P electrons following the huckles rule aromatic Perfect all right look at this look at this 2 4 6 8 10 10 so it has to be what it has to be aromatic it has to be aromatic because it's following the huckles rule 4 n + 2 pi but but but if you look at this particular carbon how many bonds is forming 1 2 3 so there will be one hydrogen here this also forming 1 2 3 so there'll be one hydrogen here between the electron clouds of these hydrogens between the electron clouds of these hydrogens the molecule overall becomes nonplanar so if it is nonplanar it becomes non-aromatic then it becomes non-aromatic right here I think this molecule we've already done this is 2 4 6 8 10 10 P electrons 10 P electron means following huckles rule so it is going to be aromatic it is going to be aromatic right it is going to be aromatic so tell me whether all the things still here are clear or not tell me whether all the things are clear here or not so for the molecule to be aromatic it should be cyclic it should be planer complete cyclic conjugation should be there and 4 n + 2 2 pi electrons should be I mean 4 n plus 2 electrons should participate in the cycli conjugation okay 4 n + 2 pi means it can be either 2 pi or 6 Pi or 10 pi or 14 Pi etc etc right if in the ring all the atoms are SP2 then the ring is planer right then the ring is planer that's again something which we already know clear till here I believe all the things are clear here yeah for the molecule to be anti-aromatic it has to be cyclic first then it has to be planer there should be complete cyclic conjugation but there should be four n Pi electrons involved in the complete cyclic conjugation that means either four electrons or eight electrons or 12 electrons or 16 electrons Etc yeah okay now comes one more point that is resonance energy energ what is resonance energy guys see I'll write its two definitions the amount of energy which is used for delocalization the amount of energy which is used for delocalization the amount of energy which is used for delocalization that's what we call as resonance energy or or I will say the energy difference the energy difference between the energy difference between the resonance hybrid and the most stable canonical form energy difference between the resonance hybrid and the most stable canonical form is something which is what we call as a resonance energy so basically if you talk about the resonance hybrid or let me make you understand it like this just a second let me take this example ch2 dou Bond CH nh2 yeah ch2 double bond CH this is nh2 this is the molecule which I've taken right so first of all this nitrogen it has got the lone pair is this lone pair in conjugation yes it's in conjugation so this is basically one canonical form which is given to me can we make it another canonical form we can do that right this will shift here this will shift here when you make its one more canonical form it becomes ch2 negative ch2 negative it becomes single this is CH this is double bond this becomes nh2 positive this is its one more canonical form this is the second canonical form correct this is the second canonical form if I would want to make it hybrid hybrid also you can make this is ch2 put a single Bond first of all everywhere right put a single Bond then get the partial double bond character here here is the partial dou one character this is your Delta negative this is your Delta positive so this is your resonance hybrid this your hybrid right now guys if you remember if I ask you is the delocalization is the delocalization present in the canonical forms or the delocalization is present in the hybrid where is the delocalization present delocalization is present in the hybrid right delocalization is present in the hybrid what is delocalization what is Deliz ization de delocalization of what delocalization of high energy electrons my dear students that amount of energy which is used for delocalization of high energy electrons that amount of energy which is used for the delocalization of high energy electrons that amount of energy is what you call as resonance energy Point number one what is resonance energy it is the amount of energy which is used for delocalization delocalization is present in the hybrid it is present in the hybrid so in the Hy hybrid whatever amount of energy is used for delocalization that amount of energy is what you call as resonance energy okay that's called as resonance energy or or there is one more way the energy difference between resonance hybrid and most stable canonical form these are the two canonical forms now if I ask you among these two canonical forms which canonical form is more stable more stable the one which does not carry any charge right here the atoms do not carry any charge so this is more stable more stable means lesser energy so energy of canonical Form 1 will be less than that of energy of canonical form 2 so stability of canonical form one stability of canonical form one will be more than that of stability of canonical form two and resonance hybrid it is the maximum stable among all among all right so my dear students if I plot the energy diagram here let's say this energy this is energy if this resonance hybrid is the highest stable so its energy will be the least let's say this point I mean this line represents the resonance hybrid right now this canonical form it is more stable means less energy so C1 energy C1 energy will be less than that of C2 let's say this is C1 energy and this is c2's energy right the energy difference now among these two canonical forms which one is most stable T1 is the most stable canonical form right the energy difference between the energy difference between the resonance hybrid and the most stable canonical form that is something which you call as a resonance energy that is something which you call as resonance energy so this energy gap here I'll be calling as resonance energy or I can write it like this resonance energy is basically it is the energy of it is the energy of resonance hybrid minus the energy of most stable canonical form well energy of hybrid is less than that of energy of most stable canonical form so I would say the overall value will come out to be negative the overall value will come out to be negative so what does this negative sign signify here this Nega sign I'll say negative sign signifies negative sign tells us tells us how much resonance hybrid is stable how much resonance hybrid is stable as that of what as that of the most stable canonical form this negative sign it indicates that it tells you that it gives you the idea about how much resonance hybrid is stable than that that of most stable canonical form okay right right guys is it clear now as I told you the amount of energy which is used for delocalization is called as resonance energy people delocalization is present in the hybrid so in the hybrid some amount of energy is used used for delocalization if some amount of energy is used for delocalization the amount of energy left with hybrid will it be more or less if some amount of energy is used for delocalization the amount of energy left with the hybrid will be less lesser the energy more the stability that is the reason why hybrid is maximum stable because some part of energy is used for delocalization and and the energy carried by the hybrid after that is less lesser the energy more the St stability let me know once in the chats if it's clear let me know once in the chats if it's clear let me know once in the chats if it's clear quickly guys let me know once in the chats if it is clear everyone everyone means everyone everyone means everyone look at one more case let's say I've got the molecule M here I've got the molecule n I've got two molecules right okay let's say I'm making one energy diagram for this particular molecule okay this is its for example energy of resonance hybrid and this is the energy of most stable canonical form right similarly for this particular molecule let's say I'm making energy diagram this is the line which represents energy of the resonance hybrid and here somewhere it is the energy of most stable canonical form now you know whatever is the difference whatever is the difference between the hybrid and the most stable canonical form that's called as a resonance energy and over here this is called as resonance energy as well perfect now where is the resonance energy more M or n so I'll say resonance energy of n is greater than that of resonance energy of M now you tell me wherever resonance energy is more I'll say there only more energy is utilized for delocalization and wherever more energy is utilized for delocalization there only energy with the molecule remains less energy with the hybrid remains less stability is more right so more the resonance energy more is the energy used for delocalization lesser the energy hybrid has more is the state stability right so which one is stable among the two if you talk about the stability order the one which has more resonance energy will be more stable so in short you can say something like this stability is directly proportional to Resonance energy stability is directly proportional to Resonance energy is this clear stability is directly proportional to Resonance energy and you can see this resonance energy it is directly proportional to number of conjugating sites as well number of conjugating sites what is meant by the number of conjugate sites you'll get the idea in some time first tell me whether this point is clear to you or not tell me whether this particular scenario is clear or not right stability is directly proportional to Resonance energy right and Here My Dear students we can we are now fit to solve the questions among these two where is the res resonance energy more what do you think among these two where is the resonance energy more where is the resonance energy more guys have a look Pi Sigma Pi Sigma Nega here Pi Sigma Nega where are the conjugating sides more first or second quickly where are the conjugating sites more conjugating sites are more here because throughout there is conjugation more the conjugating SES more the resonance more the resonance energy more the stability resonance energy is directly proportional to stability okay more the conjugating SES more the resonance energy more the stability that's all that's all look at this one Pi Sigma Pi Sigma everywhere there's conjugation right all these are conjugating sides look here Pi Sigma Pi Sigma Pi and similarly Pi Sigma Pi so conjugating sites here are comparatively more conjugating sides here are comparatively more okay conjugating sides here are comparatively more correct correct cor right guys so I believe you should be able to give the answer easily right more the conjugating SES more the conjugating sites more the resonance energy more than stability yes correct all right look at these two look at these two look at these two look at these two where is the resonance energy more see Pi Sigma Pi Sigma lone pair all are conjugating sides but here this is the case of cross conjugation right at once at once it will be conjugated at once it will show resonance with one of the two right this lone pair is conjugated from here this lone pair is conjugated from here as well at once it will show conjugation with one of the two now tell me where is the where are the conjugating sides more here the conjugating sides are more it's a linear conjugation right more conjugating sites more resonance energy right more conjugating sites more resonance more more resonance energy perfect now guys tell me one thing can you make the canonical form of this one can you make one more canonical form of this one you should be able to make one more canonical form of this particular one which will be like this it will be like this this is one more canonical form well what about the stability of these canonical forms they are different canonical forms but their stability and energy is same right and these canonical forms whose stability and energy is same we call them as I'll be calling them as equivalent resonating structures equivalent I'll be calling these two as equivalent resonating structures right well if you make its resonating structure if you make it one more resonating structure it will come here this will go here right so its resonating structure will be like this it's resonating structure will be like this perfect now should I be calling them as the should I be calling them as equivalent they're non equivalent do remember resonance energy of equivalent resonating resonance energy of that molecule which forms equivalent resonating structures right that is always more that is always more these two are equivalent resonating structures they have got same energy same stability equivalent resonating structures these are right these are non-e equivalent that molecule which can form equivalent resonating structures its resonating energy is always more its stability is always more remember it directly remember it directly okay remember it directly tell me one thing now tell me one thing now among these two among these two what do you think among these two I think these should be your homework questions now now I think these should be your homework questions now these should be your homework questions now yeah which one one or two one or two if you look at the first case guys if you look at the first case it will undergo resonance it'll be like this O2 double bond these are equivalent resonating structures look at this these are equivalent resonating structures this is r c bond o Nega c bond negative these are equivalent resonating structures right and that molecule which forms equivalent resonating structures its resonance resonance energy will be more okay its resonance energy will be more now among these two where is the resonance energy more what do you think among these two where is the resonance energy more quickly so this is your this is your homework question you can give it try afterwards okay all right among these two I I hope it is visible to the screen among these two among these two which one has got more resonance energy first or second quickly where are the conjugating sides more six more than six two Benzene Rings fused right perfect correct so more the more the conjugating sites more is the resonance energy correct now look at this one look at this one this is aromatic 2 4 six six P electrons aromatic this is non- aromatic this is open this aromatic this is non- aromatic okay for the molecule to be aromatic or anti-aromatic it has to be cyclic if it is open then it non aromatic so this is aromatic this is non aromatic it is more stable more stable means more resonance energy right you can solve questions like this as well Perfect Right see this is non-aromatic and this one 2 to 4 4 pi means anti-aromatic which one is stable non- aromatic more stable more resonance energy right look at this one look at this one look at this one see this is pi Sigma here Pi Sigma Pi Sigma more conjugating SES more conjugating SES means more resonance energy more resonance energy right now Guys these are two rings which are fused right two fused Rings similarly here you have got two fused Rings how do you check which one does have more resonance energy this is decided as per the F rule what does f rule say F rule says that more the Benzene like rings more the Benzene right Lings more the Benzene like Rings more the resonance energy now have a look this is your Benzene ring absolutely this is Benzene ring now is this your Benzene ring for the Benzene ring there should be three Pi bonds but there are only two Pi bonds so in this particular molecule how many Benzene like rings are there only one Benzene like ring one benzenoid ring here here if you look at this particular one 2 4 6 this is a Benzene ring if you look at this one 2 4 6 this is the benzing so two Benzene like rings two benzenoid rings more the benzenoid Rings more the resonance energy right look at the next one look at the next one see this is Benzene like this is not Benzene like so one benzenoid ring Benzene like Benzene like so two Benzene like like Rings more the Benzene like Rings more this resonance energy more than stability correct now here this is Benzene like 2 46 this is not this is not so one Benz like ring here it is like this this is Benzene like even this is Benzene like even this is Benzene like so three Benzene like Rings more Benzene like Rings more than resurance energy correct yes I think all these things are clear I think all these things are clear guys I think let's take a break for 10 minutes okay my legs are hurting a lot let's take a break for 10 minutes this is the last topic which we have to complete this is the last topic which we have to complete we'll just take a break for 10 minutes but everyone be back after 10 minutes okay everyone be back because this is the last topic which we have to complete it will not take more than an hour right so I'll be back just a second uh the time right now is 2:4 session resumes at 25 right or 215 let's keep it as 218 or 220 is it a or p.m a.m 220 take a break freshen up come back at this time where to go nowhere I just want to sit for by like 10 15 minutes so that I can teach you this concept and we'll be done okay so be back at 220 till then you guys see you in sometime is anyone alive is anyone alive is anyone alive is anyone alive quickly let me know in the chats is anyone alive yeah okay one last topic which is left in gooc that is hyper conjugation so guys this is again one of the important topics of the chapter right let's see what this hyper conjugation is all about and and and what kind of questions are asked from the topic hyper conjugation see I'm not going much into the details of the theory right I'll only tell you that part of the theory which is required to solve the questions of the term hyper conjugation so first of all first of all first of all let's get to know when does hyper conjugation take place in a molecule when hyper conjugation takes place in a molecule the way in the resonance I told you when resonance can happen in the molecule in the similar way let's get to know first of all when hyper conjugation can happen in a molecule see guys look at this particular statement whenever CH Sigma bond is adjacent to SP2 hybridized carbon at that time a phenomenon operates which is something what you call as hyper conjugate try to understand what exactly I'm going to talk about my dear students for example let's say you have got a molecule which has got an SP2 hybridized carbon you have got a molecule that contains SP2 hybridized carbon adjacent to SP2 hybridized carbon imagine there is one sp3 hybridized carbon adjacent to SP2 hybridized carbon imagine there is one sp3 hybridized carbon and with this sp3 hybridized carbon if there is hydrogen or hogen attached if there is hydrogen or hogen attached remember in the molecule hyper conjugation is going to operate in the molecule hyper conjugation is going to operate right so if in a molecule adjacent to SP2 Hy hybridized carbon there is sp3 hybridized and towards this sp3 hybridized carbon if there is hydrogen or hogen attached remember hyper conjugation is going to operate in such a molecule right now this SP2 hybridized carbon it can be this SP2 hybridized carbon it can be what it can be either a carbocatine like this it can be a carboca right it can be a carbon free I it can be carbon free radical this is SP2 hybridized even this SP2 hybridize it can be carbon double bond carbon right it can be it can be the carbon of this particular double bonded carbon this is also sp sp SP2 this is also SP2 or it can be a fenile ring or it can be a Phile ring right it can be the carbon of the fenile Ring right right whenever you see such kind of the scenario in the molecule wherein with the SP2 hybridized carbon there is sp3 hybridized carbon attached and towards this sp3 hybridized carbon there is hydrogen or hogen attached remember remember remember remember I would say the molecule in the molecule hyper conjugation is going to operate right remember in the molecule hyper conjugation is going to operate so first of all before making you understand what this hyper conjugation is let's exactly see whether you will be able to identify whether in a molecule hyper conjugation is there or not have a look people have a look for example this is one of the scenarios these are some molecules just a second let me just arrange them just a second people just a second uh just a second so for example this is hydrogen this is hydrogen okay just a second let me just there is a this is ch2 positive this is ch2 positive this is this okay and and here just a second this is ch2 positive here and in this case in the last case this is a ch3 which is attached here can you let me know out of all these molecules where do you exactly see the hyper conjugation getting operated can you let me know in the chats is hyper conjugation going to operate in the first molecule try to understand try to understand people I already told you the criteria for hyperon conjugation see this is if you look at this particular carbon this is your SP2 hybridized carbon right because this carbon double bond this SP2 towards SP2 this is your sp3 hybridized carbon attached and with this sp3 hybridized carbon there is hydrogen attached right so in the first molecule hyper conjugation is going to take place right yes look at the second one look at the second one this is your SP2 hybridized carbon absolutely SP2 hybridized adj s to this this is sp3 hybridized and with this sp3 hybridized hydrogen is attached so it will show hyper conjugation right it will show hyper conjugation look at the next one look at the next one look at the next one is this going to show is hyper conjugation is going to operate here what do you think will hyper conjugation operate here in this quickly quickly what do you think in this one yes or no yes or no yes or no yes or no quickly guys the answer is a big no why because this is sp3 hybridized carbon this has to be SP2 okay look at this one this is ch2 radical this is SP2 this is sp3 and hydrogen is attached with the sp3 so hyper conjugation is is going to operate right this is SP2 this has to be sp3 but it's also SP2 right in this one it's not going to operate in the next one in the next one this carbon right here is SP2 right this carbon here is SP2 and adjacent to that you have got sp3 of course and with this hydrogen is attached so hyper conjugation is going to operate here perfect so you got to know how to identify whether in a molecule hyper conjugation is going to operate or not perfect you got to know in a molecule hyper conjugation is going to operate or not well my dear students one more thing that sp3 hybridized carbon which is attached to SP2 hybridized carbon this is something which you call as Alpha carbon right and the hydrogens which are attached with the alpha carbon these are called as Alpha hydrogen these are called as Alpha hydr hen right these are called as Alpha hydrogen now let's have a look exactly since you got to know how to identify since you got to know how to identify whether hyper conjugation is going to operate in a molecule or not now it is the time to understand what this hyper conjugation exactly is all about try to understand few things look at the definition it is the sigma electron delocalization it is the sigma electron delocalization due to the overlap of p orbital of SP2 hybridized carbon with the adjacent CH Sigma Bond now what is meant by this what is meant by this statement see guys try to understand for example I'm taking the molecule ch3 ch2 positive okay ch3 ch2 positive my dear students if you look at the hybridation of this particular carbon it's SP2 and this is sp3 correct SP2 sp3 right and with this sp3 carbon there are hydrogen's attached so hyper conjugation is going to operate here okay now what is hyper conjugation basically how can we understand this see first of all if this carbon is SP2 hybridized that means that means if I talk about carbon in particular carbon has got 2s and 2p right 2s and 2p if this carbon is SP2 so this is s this is p and the this is s and these are 2 p so it has undergone SP2 hybridization and there is one pure p orbital as well right so whenever there is SP2 hybridation how many hybrid orbitals do we get we get three hybrid orbitals perfect so this particular carbon if I talk about there will be three hybrid orbitals basically for example this is one hybrid orbital this is one more hybrid orbital and this is one more hybrid orbital perfect these are the three hybrid orbitals of this particular carbon perfect and apart from the three hybrid orbitals there will be one p orbital which will be perpendicular to the plane containing these hybrid orbitals so there's one pure p orbital like this there is one pure p orbital like this this is SP2 hybridized this is SP2 hybridized orbital this is SP2 hybrid hybridized orbital this is pure p orbital right now guys as you can see two hydrogens two hydrogens one hydrogen has come from this side and made a sigma Bond one more hydrogen would have come from this side made a sigma Bond right now one one more SP2 hybridized one more SP2 hybridized of this particular carbon right see this particular carbon is sp3 if this is sp3 can I say if this is sp3 can I say sp3 means four hybrid orbitals let's say let's say this is one of the hybrid orbital of this sp3 hybridized carbon or let me make it bigger in size this is sp3 hybridized orbital of this particular carbon then there are in total four hybrid orbitals out of which one I have made and this is the sigma bond which is formed in between right direct overlap similarly this is for example one more hybrid orbital sp3 this is one more sp3 and this is for example one more sp3 right and since there are three hydrogen attached so one hydrogen would have overlapped here one hydrogen would have overlapped here and one hydrogen would have overlapped here so this is the orbital diagram of the molecule which I took this is the orbital diagram this is the orbital diagram of the molecule which we took perfect now my dear students this particular bond this particular Bond it is between what it is sp3 hybrid orbital of carbon and S orbital of hydrogen right now among hydrogen and carbon which one is more Electro negative carbon so can I say the sigma electron density will be more towards carbon the sigma electron density will be more towards carbon absolutely the sigma electron density will be more towards WS carbon the sigma electronc will be more towards carbon correct now guys try to understand see this is SP2 this is SP2 and this is SP2 so they are this is triagonal planer SP2 means triagonal planer so this particular bond this particular Bond and this particular Bond imagine these three bonds are present on board and this p orbital this p orbital it is perpendicular to the plane containing these three bonds so imagine one of the lobe of P will be towards youu and one more lobe will be behind the board one lobe is towards U and another lobe of p is behind the board right similarly this is sp3 sp3 so if this is sp3 I'll say I'll say this particular orbital this sp3 hybrid orbital one will be towards you this particular one will be behind you and this one will be also behind the board correct so can I say this p orbital which is towards you and this sp3 hybrid orbital which is also towards you can I say they are going to overlap they are going to partially overlap what kind of overlap this is right so this is the overlap this is the overlap which is happening this is the overlap which is happening so what kind of overlap this is can you say this is Sigma P overlap basically right this was your CH Sigma Bond right consider one thing this p orbital one of the lobe is towards you right similarly this sp3 hybrid orbital it is also towards you so they are parall they are parallel they'll do the sideways overlap okay perfect perfect guys so should I call this as Sigma electron delocalization or Pi electron delocalization this is something which I'll be calling as Sigma electron delocalization and this Sigma electron delocalization is something which you call as hyper conjugation this also called this is what you call as hyper conjugation or what you call it simply as you can call it as Sigma P overlap you can call it as Sigma P overlap and this Sigma P overlap this is something which we call as hyper conjugation I hope this is clear to everyone right I hope this is clear to everyone yes let me know once in the chats if this concept is clear this Sigma P overlap is something which you call as hyper conjugation basically yeah right people let me know once in the chats see basically this particular bond this Bond and this Bond right they were present on the board right this p orbital is perpendicular to the board one towards you and another another low behind the board right similarly imagine this sp3 hybrid orbital of the carbon it is towards you if this is towards you this is towards you I say these two are parallel they'll show the parallel overlapping and this is the basically delocalization of what this is the delocalization of Sigma electrons this is the delocalization of Sigma electrons right and sigma electron delocalization is what you call as hyper conjugation which is also called as Sigma P overlap right or you can call it as sp3 P overlap as well sp3 P overlap as well you can call this overlap as sp3 P overlap as well I believe this is clear this is the orbital diagram first of all this is the orbital diagram first of all now every time you are not supposed to make this orbital diagram see see how exactly you guys will be making the hyper conjugative structures this is ch2 positive first of all tell me one thing what happened here Sigma electron delocalization took place and you know due to delocalization due to delocalization whether the delocalization is Sigma or Pi due to delocalization iation what happens due to delocation what happens what happens due to delocation what happens stability increase or decreases due to delocation what happens stability increase or decreases stability increases right so this was basically your carbo so can I say due to this hyper conjugation the stability of carbo would have increased absolutely the stability of carbo would have increased right perfect so can I say hyper conjugation increases the stability of carbocation I hope this is clear I hope this is clear to everyone yes now now how do we exactly make it you are not supposed to make the orbital diagram every time just remember one simple thing see this is your SP2 hybrid carbon adjacent to it it is sp3 right that means this is your Alpha carbon these are your Alpha hydrogens so first of all let me make it like this let me make it like this this is C this is H this is H and this is H this is Sigma this is ch2 with a positive here correct now this is the perfect candidate for hyper conjugation right now how you are going to show it how you are going to show it just guys do one thing you just have to shift this Sigma Bond over here right you just have to shift the sigma Bond over here like what happened here right what happened here you just have to shift this Sigma Bond over here when you shift the sigma Bond over here what will happen this is C this is H this is H and there'll be H this H is this H is not completely lost as I PA to why is that I'll tell you the reason this is a double bond this is now ch2 now this H positive which was supposed to be completely lost it is not completely lost here it is still it is still infl influenced by this electron density this electron density will be holding this H positive comparatively so this hge positive is not completely lost right this H positive is not completely lost here this electron density will be holding this H postive comparatively towards it okay now similarly guys similarly similarly since since it is a carbon single Bond carbon and towards this carbon single Bond carbon free rotation is possible you know it so in one scenario in one scenario this particular orbital would have come towards you and done the overlap with this one in another scenario this particular orbital would have come towards you right and done the overlap with what with the p orbital perfect so it is not only this H positive which can can do the I mean which can participate here right it can be this H positive as well it can be this hydrogen as well yeah okay so one more one more hyper conjugating structure will be like this this is carbon this hydrogen will be as such now this hydrogen will lose as will be lost as H positive not completely lost though right this is H this is double bond this is ch2 right similarly in one more scenario this is carbon this is hydrogen this uh this is hydrogen and here you'll be writing H positive there's a double bond here this is ch2 perfect right these are the possible cases which can happen correct what I'm trying to say along this single Bond free rotation is possible free rotation is possible if free if if along this single Bond free rotation is possible that means in one scenario in one scenario these two will overlap in another scenario this particular one and this particular one they'll overlap in another scenario this particular one and this particular one they will overlap right so tell me how many hyper conjugate how many Alpha hydrogens it had how many Alpha hydrogen were here three Alpha hydrogen three Alpha hydrogen were here perfect three Alpha hydrogens right so can I generalize a statement here number of hyper conjug getting structures involving CH cage involving CH cage is always equal to the number of alpha hydrogen is always equal to the number of alpha hydrogen is always equal to the number of alpha hydrogen so in total this was already one hyper conjugating structure this is one more this is one more this one more in total how many how many hyper conjugating structures are there four four right so can I say total number of hyper conjugating structures total number of hyper conjugating structures is equal number of alpha hydrogens plus one number of alpha hydrogens plus one right these are the total number of hyper conjugating structures these are the number of hyper conjugating structures involving CH cavage involving CH cleavage right clear guys so can I say more the hyper conjugating structures more the delocalization of Sigma electrons and more the delocalization means more the stability right yes you can say something like that perfect you can say something like that I hope you got to know the exact idea now tell me if you want to make the hybrid if you want to make the hybrid over here how this hybrid will look like how this hybrid will look like can you tell me that see this is the carbon and this is your ch2 perfect I'm putting a single Bond here right I'm putting a single Bond here as for this structure there's a single Bond as for this structure there's a double bond so it's a partial double bond right perfect in one of the cases this hydrogen is leaving as H positive in one more case this hydrogen is leaving as h in one more case this hydrogen is leaving as H POS perfect right right guys is this clear is this clear this structure is telling you that carbon carry is positive this structure is telling you carbon is not carrying any charge so I'll write it as Delta positive over here so this is your hybrid this is your hybrid this is your hybrid over right this is your hybrid over here perfect this is your hybrid over here right so my dear students I hope this particular point is clear hyper conjugation it's also called as sp3 P overlap or you can call it as Sigma P you can call it as Sigma P overlap now now now let's have a look on the types of hyper conjugation let's have a look on types of hyper conjugation let's have a look on types of hyper conjugation let's have a look exactly on types of hyper conjugation so first of all look at this particular molecule this is SP2 hybridized carbon adjacent to it there is sp3 right this called as Alpha carbon and these are the alpha hydrogen so this will undergo hyper conjugation it will show hyper conjugation and which hyper conjugation I'm talking about Sigma Pi hyper conjugation Sigma Pi hyper conjugation so how you are going to make its hyper conjugating structures see guys first of all you just have to shift this Bond here and this Pi Bond will be shifted in this direction so what do you get you get the first structure like this this is hydrogen as H positive this is hydrogen as such hydrogen as such there's a double bond here right this is CH this is single bond this will be ch2 negative perfect this is one hyper conjuga structure in another case this hydrogen would have left as Edge positive right so another hyper conjugating structure will be something like this this C this hydrogen as such this is H positive here and this is hydrogen there is double bond here right this is your CH and this is your ch2 negative perfect in another case this hydrogen would have left as H positive so this was H this is H and this is H positive double bond CH single Bond ch2 negative perfect now tell me how many Alpha hydrogen were here this is Alpha hydrogen Alpha hydrogen Alpha hydrogen there were three alph nitrogen so if I ask you how many hyper conjugating structures will be there which involves CH cleavage this involves CH cleavage this involves CH this involves CH number of alpha hydrogens is always equal the number of hyper conjugating structures involving what involving involving what involving CH cage right how many total hyperic structures are there how many total 1 2 3 4 four right you already know total number of hyper conjugan structures is equal number of alpha hydrogen plus one okay right so what kind of hyper conjugation this was this was Sigma Pi hyper conjugation this was Sigma Pi hyper conjugation correct this was Sigma Pi hyper conjugation right now there is one more hyper conjugation which is going to be Sigma positive hyper conjugation there is one more hyper conjugation which involves Sigma positive hyper conjugation have a look this SP2 hybridized carbon with this it is sp3 that means this is the alpha carbon these are the alpha hydrogens so hyper conjugation will be there right how do you exactly show the conjugating structures over here see first of all if I ask you how many hyper conjugating structures will be there which involves CH Bond cleavage number of alpha hydrogens are three and you know number of alpha hydrogens is equal to number of hyper conjugating structures involving CH Bond cleage if I ask you how many total hyper conjugating structures will be there number of alpha hydrogen's plus 1 3 + 1 that is four right that is four toal hyper conjugating structures now make the structures shift this Bond over here shift the sigma Bond over here this will be C this will be H positive right this will be your H this will be your H this is double bond and this is ch2 correct now in one more case in one more case I would say this hydrogen would have lost right as it's positive not completely lost you know it already what is the logic right this is ch2 in one more scenario this carbon this hydrogen will be as such this hydrogen will be as such and here you'll be writing H positive and this is ch2 perfect so these are the possible four hyper conjugating structures over here right okay and you know I've already told you number of hyper conjugating structures involving CH clage involving CH clage involving CH clage it is is always equal to what it is always equal to the number of alpha hydrogens it is equal to the number of alpha hydrogens here the alpha hydrogen were three so there are three hyper conjugating structures which involve CH cleavage right if I ask you how many total hyper conjugating structures will be there that is total hyper conjugating structure total hyper conjugating structures will be number of alpha hydrogen Plus+ one so 3 + 1 comes out be four which you can see over here right I hope this is clear so this was your Sigma positive hyper conjugation now similarly there is Sigma or electron hyper conjugation there is Sigma odd electron hyper conjugation there is Sigma odd electron hyper conjugation there is Sigma or electron hyper conjugation as well Sigma or electron hyper conjugation first of all this is SP2 hybridus carbon adjacent to this sp3 right right so this Alpha carbon these are alpha hydrogens so it will also show hyper conjugation how will you make it structures how will you make it structures how will you make it structures you'll make it like this see you will make it like this so first of all first of all this odd electron will come here one of the electron from here will come here and one of the electron will go towards hydrogen right so you'll be getting C this will be H radical this will be H this will be H and in the middle it's going to be a double bond and this is going to be C H2 here right this is one of the structures hyper conjugating structures similarly one more this is C aboh hydrogen as such this will be H radical this will be H and this is double bond ch2 this is one more hyper conjugating structure similarly one more this is H this is H and this is H radical over here this is double bond this is ch2 this is one more hyper conjugating structure so structure over here perfect now you tell me one thing how to make the hybrid over here how do you make the hybrid how do you make the hybrid so I'll write first first carbon then I'll write ch2 then I'll put a single Bond right here there's a single Bond here there's a double bond so it's going to be partial double bond right partial double bond right so here it's h radical or you can say h Delta radical right H Delta radical this is also H Delta radical correct and this carbon is having radical this carbon is not having any radical so this is Delta radical this is Delta radical right so similar guys we did not make the hybrid in this particular case how the hybrid will look like here how the hybrid will look like here see this is C single Bond ch2 right as for this structure there's a single Bond but here is a double bond so it's a partial double bond here right best you know this is H this is H H this H positive so it's h Delta positive this can be written as H Delta positive this can be also written as H Delta positive this can be written as H Delta positive right and what all see as for this carbon carrying positive here it's not carrying positive so it's Delta positive here also so this is your hybrid in this particular case right this is your hybrid in this particular case did we make the hybrid in the first case first hyper conjugation was Sigma Pi second was Sigma positive third was Sigma radical right what about the hybrid in this particular case what about the hybrid in this particular case so this is carbon this is carbon right single Bond ch right I'll put single Bond everywhere first of all right now the first structure this is going to be H Delta positive this is going to be H Delta positive this is going to be H Delta positive right there's a single Bond but this structure says it's a double bond so partial double bond double bond here is a single Bond so partial double bond correct partial double bond perfect and here I'll be writing Delta negative so this is your hybrid over here this is your hybrid over here this is your hybrid over here perfect so you have got three types of hyper conjugations which we discussed one was Sigma Pi hyper conjugation another was Sigma positive hyper conjugation another one is Sigma OD electron hyper conjugation right now my dear students generally generally according to according to to M and M according to M and moan hyper conjugation is exactly of two types one is called as sacrificial hyper conjugation one is called as isovalent hyper conjugation now what is sacrificial and what is isovalent my dear students just look at the definition look at the definition the type of hyper conjugation where charge where charge separation takes place where charge separation takes place the type of hyper conjugation where charge separation takes takes place see see see guys just a second if you look at this hybrid has the charge separation takes taken place this is positive this is negative charge separation has taken place and that particular hybrid that particular hyper conjugation wherein charge separation takes place over here positive positive this negative charge separation has taken place and let me tell you let me tell you that that type of hyper conjugation in which charge separation takes place that is something which you call a sacrificial hyper conjugation and its example Sigma Pi hyper it's example is Sigma Pi hyper conjugate right right perfect similarly my dear students one more is isovalent wherein charge separation does not take place for example Sigma positive or Sigma odd see this is Sigma odd is there any charge separation there is no charge separation similarly Sigma positive everywhere there is positive there is no charge separation there is no charge separation right there is no charge separation here so you have got two types of hyper conjugations one is sacrificial and one is isovalent one is sacrificial and one is isovalent right my dear students this is something which I've already told you number of hyper conjugative structures involving CH cage involving CH cage is equal to what is equal to the number of alpha hydrogens right is equal to the number of alpha hydrogens now my dear students the point is this hyper conjugation also we break into two parts one is plus h one is minus H one is plus h and one is minus h plus h plus means donating minus means withdrawing donating withdrawing now try to understand over here this is a ch3 group this is the c ch3 Group which is attached with the conjugated system this is a ch3 group which is attached with the conjugated system if you look at the particular case this is SP2 this is sp3 so this Alpha carbon these are alpha hydrogen so hyper conjugation will operate right this Bond will shift here this Bond will shift a right perfect so let me tell you one thing I mean if I ask you whether this group whether this group is increasing the electron density in the conjugated system or not quickly whether this particular group is increasing the electron density in this conjugate system through Sigma electron uh through Sigma electron delocalization what do you think yes or no see this Bond you're shifting here right and these bonds you're shifting perfect so can I see this particular group it is increasing the electron density in the conjugated system with with the help of Sigma electron Sigma electron delocalization and that group which increases the electron density in the conjugate system so Sigma electron delocalization we say that group shows plus h effect that group shows plus h effect so ch3 ch3 group what does it show it shows plus h effect it shows plus h effect right it shows plus h effect it shows plus h effect as a told you already misic effect it is not going to misic effect that does not operate at the metap positions similarly hyper conjugation similarly hyper conjugation hyper conjugation also does not operate at metapos hyper conjugation does not operate at metapos see this is your ch3 group attached right this is this is your SP2 hybridized carbon this sp3 these are alpha hydrogens right right so plus h hyper conjugation will happen here right plus h hyper conjugation will happen here if you look at all these scenarios negative charge getting created at Oro negative charge getting created at par right perfect so can I say due to hyper conjugation the charges they also get created only at Ortho and parap positions so I would categorically say your hyper conjugation it operates at oropa it does not operate at metapos right your misic effect as well as well as hyper conjugation they do not operate at meta they do not operate at meta positions okay now guys what are the applications of this hyper conjugation what are the applications of this hyper conjugation let me tell you plus h effect plus h effect it stabilizes the caroa it stabilizes the caroa plus hyper conjugation plus h effect it stabilizes the carboca right similarly my dear students your hyper conjugation it stabilizes your alken as well it stabilizes your alen as well it stabilizes your alken as well for example have a look plus h effect it increases the stability of alkan it increase the stability of alkan try to understand this is alken here you got alken here you got alken right now this is SP2 this is SP2 this is the sp3 right so these are the alpha hydrogen so how many Alpha hydrogen are here three Alpha hydrogen SP2 SP2 this sp3 this is sp3 these are alpha hydrogen these are alpha hydrogen so how many six Alpha hydrogen over here zero Alpha hydrogen right more the alpha hydrogens more the alpha hydrogens more the alpha hydrogens means more are the hyper conjugative structures more the hyper conjugative structure means more is the delocalization of Sigma electron right more is a delocalization of Sigma electrons and more delocation means more stability right so wherever Alpha hydrogen are more directly you will say that alken is more stable I'll say third alkan is more stable followed by two followed by one just you have to count the alpha hydrogen that's it just you have to count the alpha hydrogen that's it more the alpha hydrogen more the stability right more alph hydrogen more is the stability of the alen okay does plus h effect apply on carboca this is ch3 this is first of all SP2 hybridized this is sp3 hybridized right so these are your Alpha hydrogens perfect is hyper conjugation going to happen here yes hyper conjugation will will be operated due to hyper conjugation due to hyper conjugation what will happen to the stability of carbo will it increase or decrease will it increase or decrease my dear students plus h effect see this ch3 ch3 group this ch3 group it shows plus h it shows plus h and plus h plus h plus means donating right and all the electron donating groups they increase the stability of carboca right okay so do remember plus h groups plus h showing groups they also increase the stability of the carbo catins they also increase the stability of caroa right they increase the stability of caroa now if you remember earlier I gave you the series for plus M minus M plus I minus I in the similar way I want you guys to remember this particular Series this is the plus H series this is the plus h Power Plus plus h power of ch3 is more than that of ch2d where is dyum chd2 CD3 this is something which again you have to remember this again something which you have to remember and if you remember guys I have already told you misic effect misic effect misic effect how many effects we we discussed till now misic hyper conjugation and inductive remember misic effect it always dominates or hyper conjugation and Hyper conjugation always dominates for inductor that is the reason if you remember I used to give you that series minus m- h- I right perfect if you remember I used to give you that series for solving the questions remember misic always dominates over hyper conjugation hyper conjugation always dominates or inductive always dominates or inductive right one more important thing now let's have a look on few questions and we are almost done see guys carboca plus h increases the stability of carboca right is there any Alpha hydrogen here no how many Alpha hydrogen do we have here we have three Alpha hydrogen how many Alpha hydrogens do we have here we have two alpha hydrogens right more than Alpha hydrogens more the hyper conjugative structures that means more the stability of carbocation right so 2 is greater than 3 greater than 1 this is the stability order of these carbocations right your Alpha hydrogen they stab the carbole how many Alpha hydrogen here this is your SP2 this sp3 hydrogen is attached here so only one alpha hydrogen right over here this your Alpha carbon this is your Alpha carbon two hydrogen are attached here one hydrogen is attached so three Alpha hydrogen so this carboca is stabilized by three Alpha hydrogen this carboca 1 2 3 three Alpha hydrogen 2 and two 3 to 5 to 7 this carboca is stabilized by s Alpha hydrogen more the alpha hydrogen stabilizing the carbo more the alpha hydrogen stabilize than carbo more is going to be the stability of the carboca right more Alpha hydrogen means more hyper conjugative structures more hyper conjugative structures means more stable the carboca right correct correct guys what about these two this is your SP2 this is your SP2 as well sp3 sp3 right perfect sp3 sp3 right now you tell me which one is going to be more s this is CD3 this is CD3 over here you have got ch3 plus h power of ch3 is more than that of CD3 right we have discussed it we have discussed it plus h power of ch3 is more than that of CD3 right now you tell me the answer plus h power of CD3 is more than that of sorry ch3 is more than of CD3 tell me which one is more stable quickly which one is more stable one or two you should answer this now yeah plus h of ch3 is more than that of CD3 and plus h what does it do it increase the stability of the carbocation so first one is more stable than second right look at this one do you see any Alpha hydrogen here this is C and three methy groups this is C and three methy groups C and three methy groups it's like this if I talk about this carbon this is C and this is ch3 whole thce right if I talk about this this is the carbon which I'm talking about correct C ch3 thce this is SP2 towards SP2 sp3 is attached but this sp3 carbon does not have any Alpha hydrogen it does not have any Alpha hydrogen because this carbon is attached to three other carbons so no Alpha hydrogen over here 3 3 6 39 nine Alpha hydrogen more than Alpha hydrogen more the stability of the caroa right more the stability of the carboca my dear students till now I was showing you plus h now there is a group which can show minus H as well which can show minus H as well there's a group which can show minus H as well for example you have a cx3 group cx3 where X is a hogen c H3 group it shows plus h cx3 group it shows minus H it shows minus H have a look this is your SP2 carbon this is sp3 which is your Alpha carbon and I had told you in the beginning towards this Alpha carbon if either hydrogen is attached or hogen is if either hydrogen is attached or hogen is attached hyper conjugation is going to operate right but here this hyper conjugation if you look at this hyper conjugation this this group is not going to show plus h it is going to show minus H why is that look at this one imagine this is your conjugated system imagine this your conjugated system you're shifting this bond in this direction and this bond in this direction right and you getting this kind of the structure similarly one more similarly one more so in the conjugated system in the conjugated system when you attach cx3 did the electron density in the conjugated system increase or decrease through Sigma electron delocalization it decreased it decreased that's the reason why I'll tell you I that's the reason why you'll remember that your cx3 it is going to show it is going to show minus Edge effect it is going to show - HF ch3 shows plus 3 cx3 shows sorry ch3 shows plus h see X3 shows minus one is donating another one is withdrawing C X3 is withdrawing it is withdrawing as simple as that and there look at this one this is cx3 this is cx3 this is not ch3 this cx3 attached to a conjugated system is it increasing the electron or decreasing the electron density in the conjugated system it is decreasing the electron density in the conjugated system through Sigma electron delocalization right so this is what is showing minus H here right PX3 decreases the electron density in the conjugate system therefore it shows minus H remember ch3 shows plus h cx3 shows minus H right okay already I told you already I told you ch three plus h plus h plus h operates at Oro and Oro and para that means due to plus h of the group the charges in the conjugated system are generated at Ortho and Par in the similar way in the similar way when you attach minus H group with the conjugate system again the charges will be present again the charge will generate at Ortho and para so what does that mean that means your minus H that means your minus h it also does not operate at meta position if you have a look this is your cx3 this is your cx3 group it shows minus H so when you shift this Bond here shift this Bond here a posit charge gets created a posit charge gets created here so positive charge will either get created here then it will get created here or it will get created here right so due to the hyper conjugation either plus H or minus H either plus H or minus H charges get created only at Ortho and parap positions charges do not not get created at metap position it is just the differences the difference is simple when plus Edge group is attached negative charges are generated at orthop par when minus Edge group is attached positive charges are generated at Ortho and Par positive charges are generated at Orth and Par right so your plus h minus H they are they do not operate from meta position they do not operate at meta positions now this is the series of minus H which you need to remember see I3 it shows maximum minus H cbr3 ccl3 cf3 this is the minus H series minus H series this is a minus H series guys which you have to remember as well okay so there are few questions which you need to solve and we are done and we are done and we are done arrange the following as per k k means acidic strength acidic strength if you remember when you talk about acidic strength first you give priority to you give priority to minus n then minus H then minus I then plus I then plus h then then plus n right because acidic strength is directly proportional to stability of conjugate base and conjugate base is stabilized by electron withd drawing groups right and minus M will dominate misic will dominate over hyper conjugation that will dominate over inductive now you give me the answers you give me answers first of all look at the common site this is the site which is common to everyone this is the site which is common to everyone here right this is the site which is common to everyone this is the site from where hydrogen is going to leave as H positive okay now tell me one thing this ch3 it is present at para so ch3 is going to operate plus h ch3 at meta at meta you do not consider hyper conjugation you you'll be taking only its inductive effect plus I right C X3 at perah minus H right C X3 at perah minus H right see X3 at meta minus I minus I right now which one is showing minus M nobody is showing minus M so leave minus M aside which one is showing minus H this is showing minus H this is showing minus H now which one shows more minus H cbr3 as per series shows more minus H so 1 2 3 4 four will be maximum acidic followed by three right then comes your minus sign right now one and two one and two one and two minus I is gone now plus I plus h this is plus I this is plus h so two at the end it's going to be one did you get this order let me see the charts 3 4 5 2 1 3 4 52 1 43521 right cbr3 has got more minus H power then ccl3 the series which I gave you a few minutes back the series which I gave you a few minutes back the series which I gave you a few minutes back basic strength KB order basic strength basic strength basic strength plus M will be given first priority then plus h then plus I then minus I then minus H then minus M right this is your site this is your basic site right this ch3 what is it going to operate plus h from para right C X3 minus H from par from meta plus I from meta minus I now is there anyone showing plus M no is there anyone showing plus h yes this is showing plus h so one will be maximum basic followed by H is gone is there anyone showing plus I yes so three so three is also over then minus I minus I is the fourth one after minus I minus H that is the third one that is the second one did you write this answer in the chats people are writing 13 42 1342 absolutely correct absolutely correct wonderful wonderful guys one last question of the session and I think I should keep this as the homework what do you say should I keep this as the homework or should I solve this arrange the following as as AIC strength acidic strength minus M dominates over minus h- H dominates over minus I then you have got plus I then you have got plus h then you got plus s right call this as one this is two this is three this is four this is five and this is six quickly give it a try first you give it a try then I'll solve first you are going to give it a try then I'll solve first you are going to give a reply then I'll solve and the answer has to be correct guys answer has to be correct I'm telling you answer has to be correct this is the last question of the session and with this our general organic chemistry will be done and dusted quickly quickly quickly quickly first you solve this question tell me it's answer then we can talk I want its answer guys I want it answer I already gave you the series like right quickly someone is saying our brain is not proceeding yeah see guys it's very simple first of all this is your acidic site from where hydrogen is going to leave as H positive this is your acidic site from where hydrogen is going to leave as H positive number one number two this ch3 is present at the parap position so plus h plus h add the meta position plus I right C X3 at parah so minus h o ch3 O3 this is o r basically o r shoes o shoes plus M plus M this is O ch3 O3 at meta O3 at meta shows minus I because mic does not operate right Nitro Nitro at at para it shows minus M now can you prioritize it can you prioritize it first of all is there anyone showing Minn yes I can see there is one showing Minn so six is maximum acidic six is maximum so this is gone is there anyone showing minus H minus H look at all these yes third one is showing minus H minus H is gone is there anyone showing minus I yes there is one showing minus I that's fifth right so Min I is over is there anyone showing plus I is there anyone showing plus I yes second is showing plus I so write second afterwards after I is there anyone showing plus h yes I can see there is one showing plus h so one after plus h is there anyone showing plus M yes there is fourth showing plus M so did anyone write the order like this 635 214 63 63 someone has written 63 4152 63 4152 no right this is O3 right this is plus M yeah this is the actual order of this particular question guys right and with this our general organic chemistry is done and D right General organic chemistry is done and dusted right people okay now now now one thing is there one thing is there guys please and please do watch the same session again at 2X or 1.5x tomorrow okay do watch the same session at either 1.5x or 2x tomorrow otherwise you are going to forget everything otherwise you're going to forget everything right it is 3:21 right now by the way this session needs to be watched again because the kind of questions which I've showed you I deliberately I deliberately came up with some questions which are tricky basically which I don't think you would have solved in your earlier go classes okay I want you guys to remember them I want you guys to remember them exactly from 7 to 62 live watching do you see where the competition lies this is the actual competition which I keep on telling you all the time this is the actual competition that is not the actual competition that is just fake right uh PDF I will have to see I don't know whether I'll be able to send you the PDF right I think I need to sit for like 2 minutes yeah and tomorrow at 11:00 a.m. what do we have we have got extra question practice right at 11:00 a.m. tomorrow extra question practice is there extra question practice do attend that as well so this was our day 13 day 13 I mean night 13 it is over so tomorrow tomorrow there is one surprise tomorrow M today there is one surprise in the evening do join that session as well okay do join that session someone is saying in half an hour Milkman will be there in our house is that is that yeah yes 10: a.m. amikam has kept the special class at 11 I have kept so guys are you are you like in any offline institution or you guys are doing the self study everyone is sleeping in my house too you need not to worry I'm the only one who is awake all right so majority are doing the self study perfect is anyone from you I mean is anyone among you from Chennai anyone among you from Chennai yes or no in the chats okay A lot of people are from Chennai so the ones who are from Chennai will be waiting you on 23rd of December right I hope you know the venue where to come 23rd December Saturday 5:00 p.m. yeah remember the date no it's not being jealous with Chennai people it is just that we are starting the offline Journey from Chennai after Chennai we'll come to like Andra etc etc everywhere we'll come we're starting from Chennai so you guys are most welcome there hello guys I think it's time for you and me to sleep take rest right let's catch up tomorrow in the special class okay hello guys you take care God bless you all see you tomorrow at 11: a yeah God bless you take care take care it