hello everybody Welcome to the summary Series today we are going to learn very important chapter of organic chemistry that is placed in your class 12th grade hello alcane and hello aren this chapter lay foundation for the entire organic chemistry of your class 12th by the way organic chemistry in my opinion is one of the most logical subject that I have have ever studied hence the reason I have chosen organic chemistry to do to pursue doctorate so I did my PhD in organic chemistry you should uh I should tell you um so organic chemistry is the most logical subject in the sense that every single reaction is connected with another reaction and you need need not to memorize the reaction at all I personally feel that students are afraid of so this is the reaction I don't remember this is the reaction that I cannot remember what to do you need not remember the reactions at all at first place you need not remember what do you need to do is to understand right and I'm trying to I'm going to help you with all the theoretical as well as experimental aspects of every single reaction from today's lecture and these notes are going to be well sufficient for your even if your board Examination for your J Mains Examination for your neat examination anything that you might be preparing for these are going to be more than sufficient notes you need not read anything you need not prepare anything else just just go through the notes and you keep your ncrt alongside let's start with today's topic everybody hello alane and hello arines so what does hello alane basically means and what are hello arines what are their meaning that also we are going to understand in much detail very fine detail and my promise to you after this lecture ends you'll be very confident that yes today we have known to something called organic we will be discussing method of preparation of hello alkanes then we'll discuss it chemical properties then we'll discuss hello Arin method of preparations and then its chemical properties so I have divided this session into four parts one will be method of preparation the second part second half will be of per of the chemical properties so how they participate in chemical reactions so before we begin let me quickly tell you what are hello alkanes in your ncrt there is a general symbol R used for alkes and if you connect it with X part it becomes hello Aline hello alkan how is it derived let's start with methane so CH4 is methane suppose from here I subtract one of the hydrogen and add One hogen X is a general representation for hogen that becomes a ch3x so from methane I could prepare hello methane hello methane so it can be chor it can be uh let's say CH4 you can prepare ch3cl so chloromethane bromomethane fluoromethane iodomethane anything that you wish you can prepare out of all these halogens available to us now if you replace r with something called ar ar is a general symbol used for aromatic type compounds for example which are derivatives of benzene suppose this is Benzene C6 H6 will be the general formula I replace one of the hydrogen by hogen so Benzene derivative what is this hello arine compounds so we will be separately studying a chemistry of hello arines and we will be separately separately studying uh chemistry of hello alkanes so hello alkanes helloin these are the two classification of the compounds for the discussion today let's begin with the preparation of hello alkanes starting from an alcohol let's say I start with an alcohol that's ch3 ch H2 o or any alcohol for the sake how do we actually prepare hello alkan now I will be treating it with H+ now what does H+ means what will be the source of H+ let's say I start with HX what's an acid by the way acid is a thing acid is the substance which upon added to water liberates H+ right anything that furnishes H+ or produces H+ once you dissolve that in water is an acid once I dissolve HX in water it is going to produce H+ and x minus you need to track all the positively charged species all the negatively charged species so that reaction becomes extremely simple for you what is this thing it is known as an electrophile any particle any particular species which lacks electron is known as electrophile what are the identification of an electrophile there should be either a positive charge absolute positive charge or there should be a Delta positive charge or there should be deficiency of electron incomplete octet incomplete octet that is the identification of an electrophile if you see absolute positive charge or a Delta positive charge or that atom has an incomplete octet that will be known as an electrophile so hydrogen plus has no electron it's octet rather it's duplet is nil is zero here so it's terribly it's desperately looking for two electrons what about this x minus x minus such species are known as nucleophile the species which have the species which have excess of electrons how do we identify them either they should be absolute negative charge or a Delta negative charge or a lwn pair or or double bond meaning triple bond or or simply we can say there is multiple bonds present so I'm writing double bond or triple bond even Benzene can act as a nucleophile right will you be able to identify electrophile and nucleophile awesome in this reaction now this oxygen has two lawn pairs present and hydrogen plus is present x minus is present always react an electron deficient species with on Rich species so you people have to tell me out of these two and this third one which two are going to react first always an electrophile which is in general represented by e+ and a nucleophile they two come from either two independent substances independent groups independent reactants or they might be present in the same reactant but right now we are not discussing any complicated matter we are considering one of the component has electrophile present in it the other one has a nucleophile present in it they two will do Gulu and eventually you will obtain a bond you will make an Bond uh a calent bond between an electrophile and nucleophile as a result of this gulugulu there will be a bond made right everybody let's try to figure out which one out of the two is an electrophile which one out of the entire thing is an electrophile and a nucleophile so this is nucleophile so this can obviously react back on plus but if it had to react back on H+ why did it lose H+ at first place why did it liberate why did it produce H+ so therefore the reaction between the two is nonsense never ever going to happen will this x minus react with a nucleophile will nucleophile nucleophile react with each other no will an electrophile nucleophile react with each other yes so I have identify this is my electrophile and that's my nucleophile is that point clear everybody likewise you have to do likewise you have to do the identification of electrophile and nucleophile and your entire journey of organic chemistry is very very very simple my guarantee to you everybody you just have to be very careful in identifying electrophile and nucleophile right now always remember nucleophile attacks on electrophile nucleophile attacks on electrophile so let us attack once it attacks what are you going to form there will be ch3 ch2 oxygen which has donated its pair of electron should gain a positive charge as a result of three bonds correct and whenever Whenever there is a pair of electron also present whenever you see that water is present with a positive charge oxygen has formed three bonds this will always leave such kind of groups which leaves the pair of electrons are known as leaving group leaving group any Group which leaves the pair of electron this is the pair of electron present it is leaving with the pair of electron leaving behind a positive charge is known as a leing group itself will become a neutral species now once it leaves what do you expect everybody ch3 ch2 positive charge now this positive charge species this has now become an electrophile and you had one nucleophile also present in the beginning they two will attack on each other to finally form ch3 ch2 X do you understand from alcohol how do we actually make alky halide is that point clear everybody very clear everybody Crystal Clear cc is it Crystal Clear is it Crystal Clear yes awesome CH the next part from halogenation of alkanes how do we actually do alkanes you should know that alkanes are the least reactive hydrocarbons least reactive hydrocarbons then what do we do we will be trying to do a reaction let's say I have ch3 I I have ch2 I have a ch3 the carbon which is connected to one carbon from anywhere left and right will be known as having primary hydrogen it has primary hydrogen the carbon which has two carbon neighbors those hydrogens present on it are known as secondary hydrogen so this is also primary yes this is also primary can we have tertiary yes can we have a tertiary yes can we have a quinary no this is my tertiary hydrogen this is primary hydrogen these are also primary this is also primary understood so I want to have primary and secondary both of them in the same molecule let's have it let's have it let's have it this becomes my secondary this is still primary is the identification clear everybody the primary secondary and the tertiary hydrogen in a in a alkan molecule if yes then I will react it with light H new you know plank plank was a scientist he said that the light light is made up of photons and every Photon has an energy e is equals to H new so we represent light by writing just H new okay now this light could be of any particular wavelength for uh for a reactant now you will be using X2 dihalogen now X2 can be uh we don't use generally FS2 because FS2 has a very high reactivity violent reaction it does so we will be using V2 cl2 or I2 either of the three we can use for this reaction okay as a result of which you will be getting the hello alkan as the product hello alkane as the product and out of the two out of one two and these three br2 cl2 and I2 cl2 is most reactive see I'm not comparing it with FS2 most reactive most reactive br2 is moderately reactive moderately moderately reactive and I2 is least reactive I2 is least reactive so we basically do the reaction with cl2 and br2 the more reactive species they will be least specific they are going to be least specific so what does it mean specific here specific means this chlorine does not mind if this hydrogen is to be substituted or the secondary hydrogen to be substituted or this tertiary to be substituted it will substitute all of them it want wait but br2 which is moderately reactive is more selective so we should write selective instead of specific more selective more selective so they will select hydrogen they will select hydrogen which one to be substituted so hence I can say that I may result based on I may result based on ch3 ch2 ch3 if you reacted with cl2 you will get mixture of product you will get a mixture of of product plus ch2 there is a hydrogen CH hydrogen already present this becomes chcl and this is CS3 you'll get a mixture right you'll get a mixture but when you have br2 if you have a br2 then it will prefer the one which is more stable it will form ch3 CH this in very particularly this uh compound formed would be in higher amount we also know here the tertiary one reacts in this order the tertiary one will be more secondary than the primary this will be the react uh this will be the producted distribution thirsty hydrogen will give better result secondary and the primary one clear everybody okay next one from halogenation of alkenes now alkenes are let's say I I have ch2 double bond ch2 the simplest alken possible ever if you add X2 in ccl4 like non-polar solvent there will be anti- addition there will be anti addition of X and X across carbon carbon double bond you will obtain ch2 single Bond X here another X here so this is an anti- addition so one chlorine will be towards you the other chlorine will be inside or any particular hogen so you can do br2 addition you can do cl2 addition generally we deal with br2 or cl2 but if you do it the X2 addition in a polar solvent like H H2O then there will be HX addition you have to add X+ and you have to add o minus so the compound formed would be compound formed would be Hydro hello hello hydrin this represents hello this represent hydroxy so HOH hydrin compound this will be known as HOH hydrin compound you can have here HX addition right HX addition now this HX can be h o BR H CL hoi addition ho is this clear everybody Crystal Clear sure good next from the Hydro halogenation of alkenes uh did we just discuss this yes we have discussed now I will be discussing this is the third part okay alken yeah from hydrohalogenation I would like to do this also uh suppose you have ch3 uh instead of that let's go yeah hydrohalogenation we will be adding HX let's have r c h 2 1 2 3 CH double bond ch2 let's take that example and I want to add HX to this molecule if this HX is this can be HCL this can be hbr you can use hi also we generally do not study h i uh we generally don't study HF so we will be sticking sticking ourself to just one 2 and three so the product formed would be ch3 R will be R CH single Bond ch3 there will be hogen added so I have added I have added a hogen here hogen here and H here this is this you might have studied in your books also known as maronic COV addition maronic coov Maroni Co addition which is also known as a little simple M rule M rule but I don't like this blind definition for many reasons why if we change the framework of this molecule you need to decide where the carbocatine is actually form that carbocatine has to undergo rearrangement finally you have to attach that x minus where the carbon positive charge is finally present am I clear everybody are you understanding are you following my point if you use either of the three then you have to follow marono rule you need to form carbocation C+ then you have to do rearrangement you have to do rearrangement if possible what is rearrangement you can involve hydride shift then there will be methy shift then there can be aile shift then there can be any sort of alkal shift anything anything aile shift there can be ring expansion there can be ring contraction various kinds of various various kinds of carboca uh stability rearrangement there can be ring contraction also I personally love this a lot ring contraction ring contraction various possibilities is and this actually helps you in understanding every every fundamental of organic chemistry so here you have to discuss inductive effect misic effect resonance effect hyper conjugation effect what not every aromaticity every single effect comes into this understanding of rearrangement right now if you use a combination of HCL or I can instead of that I can say hbr with peroxide with a peroxide now peroxide can be H2O2 it can be benzo peroxide BPO benzo peroxide it can be any other sort of O peroxide it can be uh per it can be per benzoic acid per benzoic acid you can use metac chlorop parb benzoic acid also m c PBA metac chloro per benzoic acid if any one of it is present along with hbr along with hbr only hbr then you are going to do anti marconic of addition now what's anti maronic of and what's maronic of addition let me quickly help you with that if you have R CH such kind of un symmetrical alken this carbon having more hydrogen this has more hyd this carbon of the double bond has less hydrogen for according to marono rule if reagent AB has to be added and a has a positive charge B has a negative charge this positive charge goes to this carbon which has more number of hydrogen and negative charge component goes to the carbon which has less number of hydrogen then the product formed would be rch double bond single Bond and uh the carbon having more hydrogen receives the positive part a here and B here but according to anti maronic of rule we do the reverse of it so this is according to maronic of and according to antiarc conov rule the positive part let me use the other color the positive part will go to the carbon having less number of hydrogen the negative part goes to the carbon having more number of hydrogen this represents NT naronov rule so according 20 maronic of rule the product formed would be B here and a here I'm sure now you have understood what does marconic of and anti marconic of rule makes sense let's proceed to from The hogen Exchange reaction the very important one is known as schad's reaction what does schard reaction means everybody schad reaction means suppose you have R uh hogen and you want to particularly make our F here it is for just for the production of flu floro alkanes you can use agf you can use you can use agf or you can use sbf3 or you can use hg2 F2 or you can use Co F2 all of these are fluorinating agent the product form would be RF this is known as Schwarz reaction very important name reation schars reaction if you want to study about finklin reaction starting from RX and reacting it with sodium iodide reacting it with sodium iodide in the presence of a polar aprotic solvent this will known as Polar atic Polar atic solvent in short p a s polar a prot solvent this is this actually helps in uh increasing the rate of the reaction by stabilizing the intermediate form so it will be r i in general if you want to form iodo alkan starting from RCL or rbr then you can named this reaction as finlin reaction very important from a laboratory perspective from a laboratory point of view very very very important reaction I hope everything is clear so far right everybody now let's discuss the electrophilic substitution reaction so we need to substitute the uh compound with the help of uh electrophile which compounds aren compounds so hello alcane compounds preparation is done let's discuss hello arines now now hello arines suppose I have Benzene ring which contains double bonds as I already told you carbon carbon double bond containing species are also nucleophile so when I react it with X2 will there be any reaction in ccl4 compare it with carbon carbon carbon double bond when I was reacting it with X2 we could do anti addition right everybody we could do anti addition one of the hogen here and one of the hogen here can we do the same here also impossible that cannot be done reason it cannot be done because these double bonds are actually not at all there they are in resonance due to Resonance the double bond has lost its nucleophilic nature nucleophilic nature now what do we need to do instead of that there will be no reaction absolutely no reaction but I want to make some hogen containing arines so I will be starting with x but in the presence of f X3 what I will do I will add a Louis acid what is this Louis acid going to do they two are going to first react they two are going to first react to form X2 plus Fe X3 will form first form Fe X4 negative or you can start with f X2 also F X2 then you can form F X3 negative and X+ is formed X Plus is form you can use FX2 you can use f X3 every uh all these are experimentally proven these are experimentally proven you can use either of it f X2 clear now what did we do here why plus there is a plus what did we do here what's the use what's the fun see earlier I was saying that the nucleophilicity of benzene has decreased due to the resonance I somehow cannot increase its nucleophilicity but the X2 can I increase its electrophilicity yes by having absolute positive charge by having absolute positive charge and to create at absolute positive charge I need to take help from leis acid I take need to take help from Louis acid you can use zncl2 you can use alcl3 you can use lots of examples that are lots of Louis acids are available but the presence of Louis acid is essential LS acid is essential if you do not use Louis acid you will not obtain Hines once you do that the compound obtained here is hogen substituted that's your haloarene compound this is one way of doing that understood everybody how do we electrophilically electrophilic substitution is done this is positively charged species I have already explained to you anything that carries a positive charge is labeled as electrophile labeled as electrophile either you increase this electrophilicity of X2 I have done that or you increase the nucleophilicity of benzene that I didn't do I didn't touch bro Benzene at all I just want to increase the electrophilicity of X clear now now from anyine how do we make uh helloin compound please pay attention here starting from analine what is analine analine has nh2 with the Benzene Ben nh2 Benzene uh Amino this is nh2 group now when you react it with N2 na2 in 0 to 5° 0 to 5° CI the temperature has to be maintained with HCL there is HCl also present 0 to 5° C this combination is going to this combination is going to produce hno2 hno2 is a weak weak very weak or it's a unstable substance which produces no+ it's a source of no+ please make it make a note of it hono or hno2 nitrous acid is a weak acid it's unstable it dissociates to produce an O okay now 0 to 5° temperature is essential if you don't maintain that you will not get the result when I was studying in I this was my very first practical to be done in the laboratory exerise very first exercise of diazotization D this is known as D diazotization so the product obtained is having n triple bond with a positive charge here and cl minus stabilizes the ne positive charge this was the very first exper experiment in the organic laboratory in my I da days now what do we do to this this is known as this is known as aromatic aromatic diazonium aromatic diazonium chloride that's basically a salt that's a salt we practically use aromatic diazonium salt uh alkal diazonium salts are very unstable so practically in a laboratory you cannot use them so you start from you start from the Benzene diazonium salt if you reacted with cubr that's CU let me change the color of the pen if you use c2br R2 cus bromide if you use cu2 cl2 that's cus chloride if you use CU CN or simply CN let's say so cupress cide all these three are going to result into the substituted product but we are interested in bromination we are interested in chlorination so chloroarene is produced bromoarene is produced and if anybody ask you what is a mechanism do they follow it's radical mechanism they do follow but you need not go into the detail of the mechanism of this particular reaction but others I am explaining wherever it is required now when you reacted with Ki it forms iodobenzene if you reacted with hbf for very one of the excellent method of preparing fluorobenzene fluorobenzene this is known as balls shean reaction balls shean reaction please check the spelling balls shean reaction we should be knowing the reaction spelling if there is a here and there it's okay this reaction when you prod when you reacted with a uh water and heat it little bit just heat little bit of heating it gives you phenol anyways this is not a part of our reaction for today we just need to know that you can prepare fluorobenzene that's known as bals shiman reaction these are known as sandmere reaction this is known as sandmere reaction these two are known as sandmere reaction these two these two this has no name you can give your name or your um Miss name your Madam's name here let me name it as on uh on my name this is Aly reaction clear okay let's proceed to the next part very very very important Benzene diazonium salts various reactions you can do let's discuss some of the properties chemical properties of hello alkanes the first one is the basis of the nucleophilic substitution reaction there are two kind of substitution reaction that we can perform on hello alkanes one is known as sn1 the other is known as sn2 reaction I will try to explain first what is sn1 and sn2 reaction then we will go into the detail of it suppose I start with ch3 ch2 um CH ch3 let's say I have this compound and U I have a let's say let's say let's say let's say uh I have hogen present here or or I have anything anything present that's let's say in general leaving group LG LG represents leaving group leing group G represents group L represents leaving so this is going to leave with the pair of electron I can do two reactions both when nucleophile reacts nucleophile reacts the product obtained beautiful thing about this reaction is the product obtained looks like they are same looks like they are same but that's not guaranteed I don't don't want you to R I don't want you to just memorize that there is something called sn1 and sn2 reaction and this is how it happens no you need to understand the fundamentals of it you need to understand the theory behind it so we expect that the product obtained after the reaction here if it is a direct substitution if you just replace LG by a nucleophile it will be sn2 reaction I will explain you I will explain every single thing about it but if you go via something called carboca and then you bring in nucleophile to get carbon nucleophile Bond where it is that we have to find out then this is known as sn1 mechanism then it is sn1 mechanism we are going to discuss that in very fine detail now so let's start everybody I have seen that I have asked when I was teaching an it goti also the moment you ask what is sn1 as sn2 reation oh everybody's in now trouble everybody body in the classroom will be in trouble what to answer we do not know see this guy see this guy I have a very uh simple and very nice uh analogy to to discuss what is sn1 and sn2 reaction although I explain this in words when I when I was teaching in my previous organizations and when I was teaching in last Akash also I was I used to always use this picture to explain what is sn1 and sn2 and that fits the best this is the very very very fine explanation of SN one SN 2 Gallery behind that equation suppose this is a um bubbly this is chbi or if you want to give them different names let's say it is a uh let me change the color yeah let's say let's say this is Tiki and this is pey there are two cats one is cheeky and another is so cheeki says I want to sit on this place and uh leaves humbly leaves and creates the space free for the cheeki cheeki comes and sits cheeki comes and sits on that empty space on that vacant space but suppose now Chiki wants to occupy that place and piki doesn't want to leave what to be done now the cheeki is very very very uh adamant that I want to sit on that place somehow I want to sit on that place so chei will try to kick away PP cheeki will try to kick away so has to get out and then cheeki will take place cheeki will take that place and occupy that place correct everybody so what I'm trying to tell you if the reaction follows if the reaction follows the first kind of pattern it will be sn1 mechanism this is sn1 mechanism and this is sn2 mechanism if the nucleophile Comes This is the nucleophile now let me rename this cat is a nucleophile and this cat is leaving group if leaving group leaves happily and then nucleophile is invited by the substrate this is your substrate or your reactant it is invited by the nucleophile please come and take the place and make a bond then it becomes sn1 mechanism see this a bond has been made again between the nucleophile and the substrate but if nucleophile wants to come if the nucleophile wants to come and leaving group does not want to leave nucleophile has to kick nucleophile has to kick away that leaving group forceful forcefully it has to do so the nucleophile nucleophile should be strong enough are you following guys nucleophile cheeki should be stronger than py to kick to push her away so that it can take that place and then py sadly standing there with a sad face uh what has what has just happened I didn't realize because I she was weak p is weak here okay now let's talk about what is sn1 ration in detail suppose I start with r i start with r there is CH double bond ch2 the very first step step number one will be the formation of Step number one is the formation of carbocation and you need to you need to study where will you form a carbocation and this reaction will be happening in the presence of HX HX is H+ and x minus combination in the first step we are going to utilize H+ so H+ will be attacked by H+ will be attacked by the carbon carbon double bond will be attacked by carbon carbon double bond so I'm writing H+ this attacks here now I have two choices either I get this carboca or or I get the second one you need to make these two carboca and then decide which out of the two is more stable keep that and then rearrange and discard the less stable one so out of these two which one is more stable depending upon what is the r so s r is the same in both the cases so we can ignore its effect but three and this two so which one out of the two you think is more stable carboca this carboca has three Alpha hydrogen nearby how many Alpha hydrogen three it has how many Alpha hydrogen just two so which one out of the two is more stable carbo the top one this is less stable so let's immediately discard that once you get the more stable carbo kattin check if it is possible to rearrange to get the stable structure this in part in this given molecule I see that rearrangement is not a possibility now you have to bring in X to attack on the most stable carbocatine site to form the final compound this is your sn1 reaction do you see that everybody this is sn1 people are actually confused with this this is not sn1 reaction this is not sn1 reaction this is addition reaction this is an addition reaction I know that majority of the student think that this addition reaction is one of the uh this reaction seems to be uh nucleophilic substitute no you are adding H and X see this in the beginning you had H+ in the last you are having h x minus you have added HX this is not sn2 sn1 reac this is not nucleophilic substitution reaction this is not that this is an addition reaction so I deliberately wanted to take this example to clear clear the distinction between or the difference between an addition as well as a substitution reaction people I am sure many of you majority of you are confused with addition and substitution reaction this is addition don't be confused at all so let me now pull in what is sn1 reaction be very very very very um careful in listening to me so I will be bringing ch3 ch2 since we need to substitute since we need to sub substitute hogen we need to substitute hogen we are discussing study of we are discussing compounds of hallo alkanes or Hines reactions I will be bringing in a nucleophile I will be bringing in a nucleophile see this this reaction and this reaction how different these two are this is an addition reaction this is not an addition this is the reaction that has a background sn1 or sn2 reation nucleophilic substitution reaction so nucleophile is substit substituting this x here so nucleophile comes here and kicks it away if this happens in a single step that is sn2 but now we are discussing sn1 reaction so it actually under goes via step number one which is carbocation formation people are confused because the first step is common in the previous case that we just saw and addition reaction and here so carbocatine formation this becomes ch3 CX ch3 plus now you have to hunt for if rearrangement is possible rearrange rearrange if needed rearrange if possible once you rearrange once you rearrange now bring X now you bring nucleophile now you bring nucleophile nucleophile will attack and you will obtain a compound ch3 CX and nucleophile comes here this is your sn1 reaction step number one carbocatine formation carbocatine formation second one nucleophilic attack the first step is slow step the second step is fast step and we know from chemical kinetic perspective the slow step is the RDS rate determining step do you see any kind of nucleophile part participating in this reaction first reaction no so the rate is directly proportional to RX is directly proportional to RX clear everybody and it becomes independent of the concentration of the nucleophile order zero order one on RX it becomes independent of the concentration of the nucleophile this is extremely important point of sn1 reaction so if I increase the concentration of the nucleophile will the rate of the reaction be changed not at all if I increase the concentration of RX or if I change the nature of RX will the rate of the reaction will change yes definitely will change is this fundamental of sn1 reaction clear everybody what is sn1 you have to generate a carboca and then you have to rearrange it finally make the stereo finally make the compound now what are the stereo chemistry stereochemical outcomes of sn1 reaction let's say you start with r ch you start with r CH double bond CH we just have discovered that we will be forming uh we will be forming positive charge on this carbon CH X here once the X leaves you will be generating our CX ch3 positive charge here we we need to know what will be the structure of this carbocatine carbocatine is SP2 hybridized carboca is SP2 hybridized it has an empty p orbital present with it it has an empty p orbital present with it so it it will be written as R carbon positive there is a ch3 and there is a hydrogen present so this has an empty p orbital with it because the hybridation of carbon plus is SP2 if it is SP2 hybridized there is one free MTP orbital at atomic orbital present now the nucleophile can come from the top nucleophile can come from the top or from the bottom or from the bottom since SP2 hybridized carboca is stable it's a planer structure nucleophile can come from the top if it comes from the top the planer structure will gain the tetraeder shape once again with the help of a new Bond here or if it comes from the bottom they will be this sort of flower made this sort of tetrah hydal structure made did you follow this everybody earlier your carbocatine is a planer if nucleophile comes from the top this will bend from the this will form a pyramid on the bottom pyramid here and a new Bond here but if it is the attack happening from the bottom this is this is your carboca stable uh planer structure nucile comes from the bottom and it forms a tetral structure like that clear so you can get two type of stereochemical outcome one you will observe if if you had if you had X here and it was my uh if you had X here let's say this is chs3 this was R this is H if you have sn1 mechanism going on there will be two products formed one the nucleophile comes where the X was present this is ch3 this is R this is H or the other one could be the nucleophile on the other side nucleophile could be on the other side so both kind of products are possible if you see X on the same side nucleophile on the same side this is known as retention retention but if you see the nucleophile comes on the other side this is known as inversion this is known as inversion practically we obtain inverted product little higher in concentration than the inverse then the retention product theoretically we should get both of them 50 50 because the chances of attack by the nucleophile from the top and from the bottom they are half off they're 50/50 but here since this bond has to break first to generate a carbo cattin this itself is a slow step so the nucleophile may not wait for its production and immediately attack on the production immediately attack via sn2 type and form more of inversion product inversion product will be generally little higher in con concentration so it will be nearly 51% or 52% and retention will be close to 49% so that total makes it 100 clear everybody so you get partial rization so the overall answer to this discussion is we obtain partial rization so what is rization rization means the presence of two kind of isomers anomers two anomers one is if is RN and shumer and the other one is SN and shumer if both of them are present 50/50 half of same same amount then it is resic mixture 100% rmic mixture pure rmic mixture but if one of it is higher in an amount the other one is little less in amount than it is partial rization process so sn1 under goes partial rization not full not complete rization then these are the few points to be noted first that it is a twostep process this number one for sn1 does not represent the order it represent molecularity this represent molecularity this represent molecularity it means in the reaction RDS how many molecules are participating in the reaction it is just one so this has molecularity equals to one rate is just dependent upon the substrate concentration and it is in independent of the concentration as well as the nature of the nucleophile so carotin forms you have to take it through the rearrangement if at all is possible and rate determining step is the formation of carbocation how do we decide which reactant will undergo the fast reaction or the slow reaction based on the stability of carbocatine if carbocatine is more stable as the stability of carban goes up the rate of reaction will also go up will also go up okay the rate of reaction is directly proportional to the stability rate of reaction will also be directly proportional how facile the formation of caroa is based on its own stability and also based on how the leaving group would like to leave if it would like to leave very easily without any problem without any trouble then it's a win-win situation then it's a win-win situation also polar protic solvent PPS polar protic solvent for example water for example alcohol we generally use these two type of solvents they will favor they will favor sn1 reaction and if the nucleophile itself is the solvent or solvent itself is the nucleophile then that reaction becom solois and solois reaction also follow sn1 pathway sn1 pathway by solois reaction now Solis you can use ammonia as the solvent as well as the nucleophile or water or any sort of alcohol ro ro soles are always first order uh sorry for are always sn1 reaction so you should also know that there is a partial rization happening not full rization let's proceed to our energy profile diagram see I have told you this will be the energy potential energy and this represent reaction coordinates or reaction coordinate means progress of the reaction reaction coordinates I will be solving some questions also in the end so you need not go away you have to complete this session everybody you have to watch this up to the end so reaction coordinates and potential energy in the beginning I had RX that dissociated form one of the inter intermediate this is known as intermediate can you name the can you name what kind of intermediate is this what kind of intermediate is this that's a carbocation carbon having a positive charge is carbo now x minus comes and you get RX so this is my reaction number two this is my reaction number one this is my reaction number two this is a slow reaction and that's a fast reaction clear so this is my reactant this is my product let's say I bring a nucleophile instead of that you if you are confused with x and x minus I bring a nucleophile and then R nucleophile is formed R nucleophile okay so the reactant will have some energy of its own as it possess some kinetic uh activation energy it will come down to one of the intermediate again it will form some compound I I'll explain you once again this is my Rx this is your R bond is just about to break let's use Dash D Dash Dash D D DH X this bond is just about to break okay this is your intermediate r with a positive charge and here again R is just about to form new bond with the nucleophile so this is your r with nucleophile Bond formed this is your product this is your reactant this is your intermediate that's your transition state these two are known as transition States so how many transition states do you see how many transition states do you see via sn1 mechanism two transition States one is where RX bond is just about to break that's step number one and the second one the step number two for of our nucleophile bond clear everybody this is known as this is the energy profile diagram of sn1 reaction now we need to know what are sn2 reactions in sn2 reaction there will be direct attack suppose now I have X present over here or in uh you can say there is X is leaving group X is leaving group this is leaving group X is leaving group now I bring a nucleophile nucleophile is so strong here that it does not wait for X to come out and uh leave behind a positive charge but this nucleophile has to always attack from the back side always have to attack from the back side this is known as backside attack now this becomes a new molecule with a nucleophile bond now in the beginning if you have X present here if you have X present here okay this is the X here let's say this is my ch3 that's R let's say that's H when you reacted with the nucleophile nucleophile will always attack from The Backs side and there will be there will be inversion there will be always inversion this is known as inverted product inverted product 100% inversion inverted product if you start with a chiral alky halide if you start with a chiral alky halide the nature of the configuration will change R will become S and S will become R you start from r r will become s after sn2 reaction you start from s you'll form R and and shumer after sn2 attack this is known as R and Sr r R and Sr configurations I'm sure you know how to calculate R and S how to figure out R and S okay clockwise rotation anticlockwise Rotation by CIP rule let's move to the stereochemistry outcome of sn2 reaction see we just discussed that you start from RX and directly there is a nucleophile coming in nucleophile attacks from the back side and kicks away and it kicks away it kicks away X so you have a direct R nucleophile reaction direct single step reaction so energy profile should also be this is my reactant and it forms a product without any formation of intermediate this is my Rx and nucleophile present this nucleophile is just about to form Bond and X is just about to leave its form so this bond is going to break this bond is going to form and that's finally your r nucleophile what is this stage called this stage is now transition state is there any intermediate form during sn2 reation no this is transition state formed okay now which factors which factors are going to decide will there be any sort of um inversion or will will there be sn1 or sn2 reaction this is the energy profile we have just discussed in the stereo chemical uh in the stereo uh chemical output we have seen that we will be obtaining inverted product 100% inversion and in the case of energy profile we have seen that we have one intermediate one transition form during sn2 reaction some points to be noted guys this is single step reaction direct nucleophilic attack and X leaves it substituted by the nucleophile direct then transition state formation happens attack of the nucleophile is the RDS so the nucleophile will decide if it is strong the rate will be fast if the nucleus itself nucleophile itself is weak the rate rate of the reation will also be very poor okay now the rate will be directly proportional to the substrate substrate as well as the nucleophile concentration both of them very important you take a stronger nucleophile easily substitution will be happening molecularity is two so there are two molecules participating in RDS the rate of the reation will be directly proportional to the electron withdrawing groups now why do I need that supp suppose I have X to be substituted I want to generate high amount of positive charge on this carbon so that nucleophile can come and easily attack I have ch3 ch2 X I have a ch3 CH X and here I have NO2 my question everybody you need to answer in the chat box in the comment box which one out of the two you think will give faster sn2 reaction faster sn2 reaction this is one substrate nucleophile is common for both both the two reactions nucleophile can come and attack here ax leaves nucleophile can also come here attacks and xx leaves which out of the two will give a faster reaction number two it is why NO2 is a strong electron withdrawing group very very very high minus I effect minus I effect very high minus effect so there is misic effect no no here it is not applicable yes NO2 shows misic effect but here it is not applicable this me inductive effect will try to create a huge amount of Delta positive charge and it will invite nucleophile better than the first structure that please come and attack on me I want to substitute hogen by youu hence the electron withdrawing groups present increase the rate of the reaction and that's what I have mentioned here that the rate of the reaction will be directly proportional to electron um electron withdrawing groups and the in inversely proportional to inversely proportional to electron donating group so withdrawing groups these are electron withdrawing groups the one which withdraws the electron now it cannot just be uh Nitro you can have CYO you can have hogen various examples so let me write you can have NO2 you can have CYO you can have hogen any Group which is present will will facilitate the kick out kicking out of the existing hogen Vol inversion happens and the polar aprotic solvents will favor S2 reation what are those polar aprotic solvents example can be acetone example can be DMF what is DMF dimethy formamide dimethy formamide you can have amide formamide uh you can also have DMSO dimethy sulfoxide dthy sulfoxide this is known as DMSO DMSO dthy sulfoxide sulfoxide this is boride amide means c bond o nh2 from formic acid if you make such kind of molecule that is known as DMF dimethy formamide this amide is made up of formic acid okay these three these many solvents but generally we use these three in our uh particular discussion I have brought to you a graphics which is which is basically created uh to show you how sn1 and sn2 reactions are uh observed or if you want to pictorially uh depict on paper how should how you should do that Suppose there are four groups present there are four groups present in the substrate one of it is labeled as leaving group The something which leaves and invites a nucleophile if leaving group leaves happily leaves happily and leaves behind a positive charge that's carbon plus this is carbo cation and that is sn1 mechanism nucleophile can attack from the top because there is an orbital present atomic orbital present there are two lobs the lob can receive the electron from the top or from the bottom to form a bond now either you get this attack or you can get this attack the top side attack or from the bottom side attack so there are two products possible there are two products possible both of them would be present but the ratio will not be 50/50 inversion product will be little higher in in compared to the retention product but if you directly substitute do you see that some sort of pentavalent carbon is present actually there are not five bonds one bond is just about to form I'm writing small F this is about to form this bond is about to break leaving group is just leaving and the nucleophile is just coming and then finally a bond forms this is inversion product this is inversion and that's retention that's retention clear everybody clear see this the position of these groups relative position of these groups is not changed wherever nucleophile was wherever leaving group was nucleophile has come to the same place wherever this dot was present this element this group was present it is still present on the same side wherever these green and the yellow dots are there these groups are also present on the same place where they were originally present but here they are relatively changed here their spatial Arrangement is changed that spatial Arrangement is known as configuration okay all right now compounds which will not show sn1 and sn2 reaction very tricky question number one if you have if you have Hine will it under go SN one or sn2 reaction answer is no why because there is a double bond character present due to the delocalization of these bonds these electrons and you generate and you will generate a double bond that becomes very very challenging to break it very very challenging to break the double bond so you have double bond character here let's move this double bond here you get a negative charge double bond double bond breaking this double bond is very very very very difficult so hence number one this will not show sn1 or sn2 reaction second case similar example ch2 venile halide venile halide due to the double bond character once again they will be once again this is resonating structure this is also resonating structure negative CH negative CH X positive this is resonating structure double bond character again it's not possible the third one cyclohexane but on the position but on the position one and four position number one 2 3 4 it is having a substitution here Suppose there is a br present or hogen present here so here also you cannot do some any sort of substitution because this carbon is very hindered is known as Bridge head carbon is known as Bridge head carbon and creating carbon positive on bridg head carbon that's very difficult why because carbon plus is SP2 hybridized having 120 Bond angle but then this is sp3 hybridized carbon having a ring once you create try to create SP2 hybridize this this ring cannot flatten this ring cannot become planer molecule it's impossible to happen also this is of one two three tertiary helide nature which is not a favored compound for sn2 type of reaction so SN one is also not possible and sn2 is also not possible remember these three examples are very important don't do any sort of sn1 or sn2 reaction here okay now the points that you should note is that iodide is a good leaving group as well as a nucleophile suppose I want to do this reaction RX I want to substitute this RX and reacting it with nii to produce r i to do our to do uh or or or or yeah let's say I want to do this reaction or the other pathway is suppose I want to do this reaction first uh I want to have instead of that okay huh let's say instead of doing instead of taking nii that's fle scene reation basically I want to take our uh kcn K CN so I will be obtaining r r k CN R CN the second stage is let's say RX reacting it with Ki first to get r i h r x reacted with k i to get r i and then reacted with r CN or k CN K CN R CN is obtained you need to tell me out of the two number one and number two which one is a better method which which one is a better method let me put it as chlorine so that the comparison becomes easy comparison becomes easy tell me which one which method you you think practically is better second method is practically better because here iodide is a better nucleophile iodide is a better leaving group also so you you do the reaction in two steps you'll get a better yield of this compound rather than doing one single step reaction you'll get less yield of of the compound so iodide is a good leaving group as well as is a good nucleophile also so in the first stage here it is acting as a nucleophile and here it is acting as a leaving group The neop penti halide also will show neopen halide is this compound that's X here although it looks like that's primary alky halide should be favoring sn2 reaction but due to this hindrance due to this hindrance of R Group the sn2 reaction also becomes very very very very sluggish and also sn1 reation shows slow nature you can obtain a positive charge here because this hogen is just lost you will obtain ch2 with a positive charge then you can do this methy shift let's do this methy shift methy goes on this carbon and you will obtain ch3 with a ch3 ch2 ch3 and finally carbocatine is formed here after rearrangement once the rearrangement is done then you can bring in a nucleophile to satisfy this carbon positive forming a calent bond remember neop penti halide they show sn1 reaction but very slow the third point is ag3 alcl3 they are also POS they are the candidates which removes hogen they are all known as hogen abstractor hogen abstractor they abstract hogen okay for example I have RX react with any one of these ag3 generally we use ag3 you can also use alcl3 you can also use sbcl5 Etc they are going to produce R plus and they can form agcl precipitate or they can form Al cl4 Negative they can also form SB CL 6 negative suppose I was using uh X as hogen if I was using X as chlorine make sense okay the solois I have already discussed with you I have R ofx if I have RX if I have RX I want to do uh reaction with water or I want to do reaction with ammonia or I want to do reaction with r o they all will fall into the category of sn1 reaction they are Soliz clear everybody okay now let's do some do some comparison of sn1 and sn2 reaction substrate what kind of substrate does sn1 like it like tertiary alky halide over secondary over primary alky halide it want less crowding less steric crowding so primary alky halide is preferred over secondary is preferred over tertiary okay the new nucleophile here should be weak in nature should be weak in nature so that it gives a chance to the alky halide to become to produce carbon positive this should be strong enough should be strong enough generally we use neutral we generally use neutral um nucleophiles over here for example water for example ammonia for example alcohol I have repeated this point many times now you should not make a mistake here generally we use negatively charged like o minus we use nh2 minus we can also use R minus majority of the time we will be using Ro minus the leaving group here Al should be a good leaving group therefore R I will be giving a better uh good reaction r i compared to R BR compared to R CL and compared to RF and here also the same water the same here also the same solvent here polar proct solvent polar protic for example water for example alcohol but here polar aprotic solvent polar aprotic solvent for example DMF aceton DMSO DMF DMSO acetone these are the examples make sense everybody comparision done let's proceed now we need to this table is from your ncrt you can now easily pick that if you use NaOH or Koh what will be the nucleophile it is H minus the substitution product R nucleophile will be r o and the compound form would be an alcohol so I'm explaining the first reation how to imagine that RX is treated with NaOH or Koh o minus is the nucleophile which will attack here to obtain r o That's alcohol this is pure sn2 reaction this will be sn2 reaction this will also water sn1 reaction again sn2 reaction sn2 reaction sn1 reaction sn1 reaction sn1 reaction sn2 sn2 reaction again this will be again sn2 reaction sn2 reaction SN one reaction and all these reactions so I want to discuss a few of those reactions we can start with even reaction in the end of the session I will be bringing to you many many many examples many many I will also give you some homework don't worry about that what is even now even represent the second part of this talk is representing elimination reactions now try to understand if I have this molecule chs3 ch2 CL uh yeah if you remove hogen the carbon which contains hogen is known as Alpha carbon and the neighboring carbon is known as beta carbon I'm repeating the carbon which contains hogen is known as Alpha carbon and the neighboring Carbon on the left right upside downside wherever it is present it is known as beta remove Alpha hogen and remove hydrogen from beta once you do that a combination of HX is done so you are going to generate Alpha carbon double bond beta I'm explaining the product obtained will have a double bond between Alpha and beta if it is happening directly then it is E2 if it is happening via carbocation it is E1 mechanism so E1 and sn1 they are similar mechanisms but here in E1 we form an alken in sn1 we form substituted product so like here we will be discussing even reaction so first of all you need to remove this compound and generate and generate and generate a carboca so ch3 once you have generate a carbocatine obviously you need to what do you need to do what do you do to carbocation once it is formed we rearrange that so I have a hydrogen present I have a hydrogen present I will be trying to move it via hydride shift to obtain more stable reson more stable carbocation that's CH with a positive charge and that has got and that has got more number of alpha hydrogen than the previous one finally I will now take out beta hydrogen from either side this beta and beta Dash both the side they are same same minus beta hydrogen beta hydrogen is lost so you will obtain ch2 double bond CH ch3 this is your alken formed and the alken formed will be either of the two type one is known as setz setz alken setf alken is the one which has higher number of higher number of high alpha hydrogen around carbon carbon double bond this is your framework you need to check whether the number of alpha hydrogen are more or less that will be the product second one is known as Hoffman product hofman alken hofman alken is hofman alken is the one which has less Alpha hydrogen around carbon carbon double bond but even will result into SN uh even will result into set of Aline can you tell me in the comment box how many Alpha hydrogen do you see here in this compound in this product how many Alpha hydren everybody how many Alpha hydr are there how many count and tell me in the comment box now what is E2 I was telling if you have carbon containing let's say this is ch3 let's say this is my ch3 with a CH hogen is present here this is ch2 and then hydrogen is present here the carbon which contains hogen is known as Alpha and the carbon neighboring carbons to this Alpha is known as Beta And this is also beta since both of them are same same so beta you will be removing these two together together together together to get this compound that's ch3 CH double bond ch2 but there is only single condition that we need to understand the reaction condition will be it is a beta hydrogen elimination beta hydrogen elimination when hogen and beta hydrogen are and anti periplanar to each other now what's the meaning of anti periplanar they should be present in their opposite planes anti periplanar should be anti parlan okay should be anti periplanar let me uh show one diagram to help you in understanding what does anti periplanar mean suppose I show that this hydrogen is present in the dash and this is present in the form of veg this is okay perfect perfect uh uh I have one group present here let's say this is the hydrogen and this is this is chlorine okay very good anti periplanar now I am going to number these hydrogen as number one number two number three this is my Alpha carbon any doubt this is my Alpha I need to take out this chlorine for sure but the hydrogen from the beta carbon should be anti periplanar to it so which one out of one and two should I take out which one out of one and two should be taken out tell me in the comment box which one out of one and two should be taken out can I take one and two no can I take this one hydrogen number one no I have to take hydrogen number two because it is present in its opposite plane so the compound form would be ch3 C with a hydrogen double bond and there is a hydrogen hydrogen here this hydrogen number one is gone this hydrogen number two this hydrogen number three so why are you not showing this Dash and veg projection here it's a planer molecule ra it's a planer molecule carbon carbon double bond SP2 hybridization understood what is E2 mechanism if the elimination is happening in a single stage it is E2 stage it is E2 mechanism the reaction that involves metals also we need to know what are metals for example I react RX for example I react RX with metal RX with sodium I will take two of these I'll take two of these so that it forms RR this is known as Woods reaction this is known as Woods reaction when you take this it happens via radical mechanism radical mechanism so it's a doubling of carbon carbon the final compound obtained is an alkan of double the number of carbon atoms you had in the beginning and at a cost of hogen halogens are lost okay now we need to discuss some of the properties of hello arines what do they do what are their properties so we can do lots of lots of lots of reactions of H arines Hines basically we will be having one one very important point before we discuss there is hogen present which can partic ipate its pair of electron in the resonance and can have a double bond character so hence it is not a good good candidate for sn2 type of reaction also not for sn1 kind of reaction so what we need to do instead what kind of reactions we do so we need to understand that in the presence of electron withdrawing groups we can do that sn2 type reaction very easily sn2 type reaction so let me go back and explain to you how do we actually do that so this is the double bond which causes a problem which will cause a problem in the substitution but you can use NaOH at a very high temperature and very high pressure high temperature high pressure this is known as do process this is known as dos process and you can prepare phenol from this you can prepare phenol from this but this reaction is very challenging dolls process is the name of the reaction at it is done at proxim 623 Kelvin and 300 bar pressure very difficult reaction to do what how is it happening at this situation I am increasing the increasing the nucle filicity of this uh uh H NaOH and that could easily kick out hogen from the existing compound otherwise in the simpler in the normal condition asent reation is not very easily possible so our aim was either to increase the nucleophilicity of no or increase the electrophilicity of this compound so I chose to increase the nucleophilicity by increasing high high temperature by making high pressure but I don't want to do that what other way you are left with is to increase the electrophilicity of this hallo alkan if I in uh substitute if I substitute it with electron withdrawing groups if I substitute it with electron withdrawing groups what will happen this carbon will experience some electron dens loss of electron density and then NaOH can easily come and attack here on the center I'm repeating the electron withdrawing group will enhance the positive charge on this carbon which contains hogen the carbon which contains hogen is our objective we need to replace this hogen correct so this carbon should carry high amount of positive charge that can be done in the presence of NaOH but having ring substituted with electron withdrawing groups how it happens let's discuss that if you react if you attach electron withdrawing groups then the positive charge on that Center will enhance I'm going to show you one example rest will be rest will be uh understood to you very easily understood to you suppose I have oros sub or Paras substituted Nitro par substitute Nitro Compound on the par position with respect to hogen I have Nitro compound present so this Nitro we know it's minus M effect minus M group this will pull out the electron density from the ring this goes here this goes here so there will be Pi Pi conjugation this generates a positive charge in the center so you draw the resonating structures you'll experience that this carbon which has hogen has now got a positive charge the positive charge on this carbon okay this positive charge is now on this carbon here let me write it inside say positive charge more you practice organic by writing you'll become Champion all the students of mine who have been uh who have studied from me they all say that sir organic is the simplest subject ever we we came across simplest subject just because they have practiced they have practiced a lot they have practiced practice practi and practice makes everybody perfect not just the men practice also make women also perfect you know you should know that so this is the resonating structure now do you notice that this carbon has got a positive charge now oh minus which was wondering earlier this carbon does not have enough positive charge should I go and attack or not it says no attach an electron withdrawing group now it's very easy for me to attack because I can form a bond now I can form a bond I need not kick out X I need not kick out X I can just form a new Bond to satisfy the positive charge positive charge negative charge they are going to form a bond very easy now it if it if it happens now what is going to done what is going to happen there is this Nitro compound let me have hogen here and Nitro on the PO Nitro on the second paros orop position double bond double bond double bond now I bring o minus from here from outside attacks on the center this negative charge will be in the resonance this negative charge will go here will come here will come here this negative charge eventually will reside on this Center on this carbon so I can write one resonating structure which involves the delocalization of this double single double Bond dash dash DH DH DH DH DH this negative charge is in resonance there is a hogen present and remember there is there is this hogen present hogen is present here NO2 is present here and now you have brought in O you have brought in o o o this is known as meaner complex meaner intermediate now the moment this pair of electron which is present here on this center right this Center this negative charge when it wants to come back this negative charge present here when it would like to come back see this is the structure NO2 this is X this is O this is O when this negative charge wants to come back and tries to form Bond it has to either kick out X or it has to kick out oh minus oh minus cannot be cannot be kicked out because X was the substit X was the group we wanted to substitute X leaves and you obtain a compound o in the place of hogen this reaction becomes now pretty easy when you substitute when you substitute hello alkane with or hello sorry hello aren with electron withdrawing groups like Nitro like coo on Oro or paros remember Oro or par positions if you try to see the effect on the metap position that's never going to work because there will be no misic effect just poor inductive effect of nitro group will be in play okay I have a question CH I think I have included that on the slide this is I was explaining to you I have brought to you a graphic also you can look into this graphic um if possible I will post the uh HD file also of this graphic in the uh comment so this is my nucleophile suppose I bring in a nucleophile I can substitute that Florine here this Florine can be substituted by nucleophile which is otherwise very weak nucleophile how I how am I increasing the how am I facilitating this reaction by increasing the electrophilicity of the hello aren ring substituting it with electron withdrawing group so there is mechanism number one that is known as via addition elimination method you can substit substitute the hydro hogen of the uh aromatic ring of hello aren ring via addition elimination method addition elimination method so first you will be adding the nucleophile then you'll be eliminating the leaving group present so let me go back and explain to you see here initially I have done the addition this is the addition of O this step is known as addition step and when o minus is present and X is leaving this is known as elimination step so if you substitute Benzene having hogen with the electron withdrawing groups now this will prefer substitution SN AR this is known as SN AR nucleophilic substitution reaction on aromatic Rings SN AR via VIA some sort of sn2 nature via some sort of sn2 nature but via addition elimination pathway so can we do via elimination addition pathway also yes that can also be possible that is known as arine mechanism so before that let's discuss some of the other points that step number one is the RDS step number one is Step number one is the formation of the formation of negatively charged species the misenheimer complex so you make the negative charg compound here and then Nitro is present and X is present still X is still there X is still present I did not touch X at all so this double bond which is in the resonance this double bond which is in the resonance will help Us in explaining there's a double bond here there's a double bond here x here x brings in the double bond here inside double bond a double bond right now fine got it double bond and double bond and then you have brought in this in the X here and O here X and O both of them are present and the negative charge is in the conjugation this carbocatine stability will decide this carboca stability will decide whether uh the compound will actually form or not right if carb carbon anion sorry not carbo carbon anion is highly stable then the reaction will proceed nicely if not then reaction will not proceed that the purpose of installing the ring with the electron withdrawing group is also Justified here negative charge is basically stabilized by stabilized by having electron withdrawing group so how do we stabilize carbon anion by attaching electron withdrawing groups this is a good situation how do you stabilize carb carbon plus by having electron donating group this is also a good situation what do you have here a negative charge species and electron withdrawing group present on the ring so this becomes the most stable structure so far X has not left X has not left X has not left the system so will the strength of CX Bond will decide the rate of the reaction no hence we say that even if this x is Florine if the x is Florine here so the rate of fluorobenzene will be the highest this looks little bit chat that the rate of floro Al floro aren is the highest and you might be thinking sir how is that possible we also know that this bond is the weakest so it should give us the fastest reaction makes sense it should give us the fastest reaction but remember we are not breaking this bond in the rate determining step rate will be decided by the step which involves the breakage of the bond rate so rate depends on the stability of the carbon anion stability carbon anion not on the bond breaking stability now I see that there is this negative charge which I want to stabilize which I want to stabilize with the Nitro here do you think that Florine and O if this is one of the molecule formed from fluorobenzene this will be more stable why minus M effect of as well as minus I effect of Nitro plus minus I effect of Florine which is the best among all the hogen will also help in stabilizing carban make sense everybody is this clear is this clear everybody all right okay good so it's It also says that the rate of the reaction this is very very very important point the rate of the reaction will be directly proportional to to the number of electron withdrawing groups that you have for example here Nitro groups if you have two Nitro group rate will be faster than having one Nitro group if you have three Nitro group the rate will be fastest the rate will be fastest so I have a comparison question here compare the rate of the reaction the first one I have just the fluo just the fluorobenzene then I have you need to answer this question in the comment box number one then you have Nitro attached on the orop position number two then I have third Nitro present on The Meta position then I have fourth see I brought to you the entire entire hello alcane Hine chapter which is helpful for you not just for your board examination but for your uh for your neat and J examination also [Music] okay let's have one more Florine here and one more five six 7 this is enough okay okay this was six this was be this will be seven this is eight okay you have to answer this order compare the rate of the reaction towards towards SN nucleophilic substitution reaction on AR nucleophile is the same for all of these okay CH next part ESR we have chlorobenzene now we want to do electrophilic substitution reaction that means that means we need to introduce an electrophile for this process electrophile so one reaction that we can do I'm sure you can see that now I can bring uh C uh let's say I want to do another chlorination Fe CL F3 or F2 then I will be getting two compounds one Oro and Pariser this is major this is Major par one is always major then I can do RX in the presence of alcl3 right so this will be CL here R and I can have par product par is again the major one par par par par par major again this is known as freedel craft alkine ation I can do rcx in the presence of alcl3 this is know known as FR craft as silation I can have um I can have a uh chlorine already present and on the orthop position c bond o r Plus on the par position CL par okay what else we can do we can also do reaction with N2 we can have analine made from here what else we can do we can treat it with NaOH at a very high temperature and high pressure we can make phenol lots of reaction can be done lots of Dos process dos process right we can do sulfonation we can do nitration also suppose I do the nitration with concentrate hno3 you will obtain Ortho and Par product once again you can have NO2 plus per various reactions can be done various various you can do sulphonation concentrated h2so4 or we can use olum Oleum you can do sulfonation again on the Oro as well as par par will be major parah will be major so many reaction that we can do clear everybody let's proceed now we have me uh the mechanism that was explained to you addition elimination and then elimination addition method so we have chlorine present here let's say I have a chlorine and you remember every carbon has hydrogen also present on it hydrogen hydrogen hydrogen now when I react it with a strong base what will strong base do the base will be preferring the hydrogen present to the next carbon of the hogen containing carbon so when it leaves here it forms when it leaves here it forms benzine you might have heard of this kind of intermediate known as benzine this is known as benzine in short it is known as Arion this is one of the intermediate just like carbo carbanion radicals uh carbine nitr this also known as benzine once now N2 comes back again this will form negative charge and nh2 double bond double bond double bond now you can bring water to neutralize this to neutral clear everybody okay good this is why add elimination addition method you first have eliminated chlorine please see that you have eliminated chlorine then you have added nh2 okay now we will be doing some reaction with the metals Woods reaction I have already discussed with you suppose I use RX which is alky halide reaction with sodium another alkal halide another alkal halide another alky halide then this will be ar this is ax this will be aril halide this is known as aril halide if R and A this forms the product then it is known as Woods fitting reaction suppose I bring in ARX with sodium plus another ARX ARX what is ARX arile halide the product formed as a result of this reaction will be a AR right everybody understood good now I have brought to you some questions we will be spending some amount of time in discussing whether we have understood sn1 sn2 E1 E2 or not okay everybody study but how many of you have understood that is the question so the first question given you here is PHS minus na+ this is a nucleophile nucleophile will attack either here or it will attack here we have seen that the nucleophilic attack on aromatic H compounds are not possible so our choice is this and here the stereochemical outcome due to this pure sn2 reaction should be inversion let's check where is BR gone and you have brought PHS PHS is present in only option number one here also it's present but the stereochemistry is the same but we want inversion and inversion has happened in the option number one your comp answer should be option number one clear now second question it says a solution of plus2 chloro to two Phile Ethan two chloro two Phile Phile is represented as pH correct so this is my carbon number let's say one this is my carbon number two it says Phile ethane is produced Phile ethane means it will be ch3 and it is uh number it is number compound number Phile ethane so this will be Phile this is ch3 this is CH and there is one chlorine also present so this will be now a compound which seems to be chyal in nature yes it and it shows plus sign it's a chyal compound now I want to do a reaction on this with SB cl5 sbcl5 and we looked at sbcl5 what does it do it's a hogen abstracted it will abstract this hogen from here and leave behind carb carbon positive charge it will create a carbocation and that carbo will undergo further attack in the presence of a nucleophile so the answer should be option number one carb not option number one it should be carboon option number D is that clear everybody option number D sbcl5 is a hogen abstractor next question what is the major product obtained in the following reaction so you have ammonia dissolved in ethanol so ammonia will be doing solois and remember Solis is uh Solis happens via what kind of mechanism SN one mechanism as in one mechanism so this substitution of bromine is not possible we will be doing it here and that's all and then the option number option number um which is the answer oh option number c is the answer direct answer ch2 nh2 very good next following is the substitution reaction in in which CN replaces CL CN replaces CL so CN is about to replace this and cl is kicked out so that you get R CN to obtain the propane nitr propane nitr means one carbon second carbon and third carbon third carbon is inclusive of nitr from RCL which should be the r r structure what what is the structure of RCL see one of the carbon is coming from cide so I think I should start with these two R and X so RX means it should be chloroethane you react ch3 ch2cl plus kcn it's ionic compound so it will form carbon carbon Bond direct sn2 mechanism ch3 ch2 and then CN this will be the compound compound formed that's propane nitr so option number one is the right answer chloroethane that's the right answer good everybody now which compound underg go a nucleophilic substitution with nacn at the fastest rate at the fastest fastest rate fastest rate out of these four which will be the fastest rate so uh n a CN since it is negatively charged nucleophile it will prefer sn2 reaction and SN to prefer primary RX let's figure out primary RX is this primary RX yes this is primary alky halide is this primary no this is secondary so option number two cannot be the answer option number three is it primary yes it's primary option number four is it primary no this is also gone so out of four we have eliminated two options A and C out of the two which one has less crowding around CBR Center this has less crowding it has a little more crowding so option number number one is the answer as the steric crowding increases rate of the sn2 will also decreases okay which of the following alky halide will undergo rearrangement in sn1 kind of reaction here so here let's break this halogen let's break this hogen let's break this hogen also and then see what which one will undergo sn1 or rearrangement this will generate a carboca ch3 ch3 ch3 CH positive charge and ch3 this will move here so methy shift yes it will undergo rearrangement here there is a hydrogen present here so you generate a positive charge on the center there is hydrogen and methy we will always be picking up hydride shift first yes there is a rearrangement possible here in this case there will be hydrogen another hydrogen present here and the positive charges generated here so there will be hydride shift and then resonance yes there is a resonance possible here option number D is the answer okay whenever we whenever you people are solving this question pause the video solve your question come and match okay great the rate of sn2 will be negligible in which of these following option this is my secondary alkal helide yes it could be possible here this is also secondary alky halide yes it's a good chance and this is also secondary alky halide there's a good chance here and uh this is a tertiary alky halide tertiary alky halide and on top of it this is Bridge head carbon so I told you sn1 and sn2 reactions are not very good in such kind of bicyc compound this is cyclic compound bicy fused Rings these are the two bicyc compounds okay identify the rate of the dehydrogenation in which of the following compound when allow so identify the rate of the dehydrogenation dehydrogenation dehydrogenation to um wait a minute I think I need to do some change dehydro it should be dehydrohalogenation let's change this dehydro halogenation dehydro halogenation so that we can remove hn X in the following compound when allowed so in the first compound this is my Alpha carbon which has bromine this is my Beta this is also beta this also beta we need to remove hydrogen from the beta carbon so the compound formed from the first one will be ch2 double bond with c and there is a a ch3 and a ch3 present check please this will be the compound obtain how many Alpha hydrogen does this have three from here three from here total six Alpha hydrogen okay now here if I form an alkan from here this is my Alpha carbon this is beta carbon so carbon ch3 will have ch double bond ch2 clear everybody so here from compound number one to compound number two the number of alpha hydrogen seems to be reduced from 6 to three so first one will be going the faster reation than the second one in the third one you will just obtain no Alpha hydrogen at all zero Alpha hydrogen therefore this will undergo very slow reaction so 1 2 3 should be the order option number one is the answer clear everybody which of the following alky halide under goes even reaction with the fastest rate even reaction that means we need to form a carboca that carbo should be stable enough now in this first case RX leaving of RX leaving of RX leaving of RX all these compounds are the same exactly the same so which one will form carbine the fastest the one which has the best leaving group iodide is the best leaving group option number four is the answer okay for uh the following reaction there is a this reaction going on number one and this reaction which is going on number two this reaction number three this is a neat 2016 question I want you to identify the nature of the reaction the first one seems to be the first one seems to be elimination reaction right everybody because I have eliminated I have eliminated bromine and hydrogen from here the second one seems to be substitution reaction third one seems to be addition reaction so elimination substitution and addition elimination substitution and addition option number c is the answer okay replacement of chlorine of chlorobenzene to give phenol requires drastic condition but chlorine of two for D nitrochlorobenzene is readily replaced by so it's comparing planer simple chlorobenzene with 24 dinitro chlorobenzene it under goes better replacement of chlorine easy replacement of chlorine but this will be undergoing slower replacement of chlorine why what's the reason what's the answer NO2 will make the ring electron Rich no NO2 withdraws electron from the metap position no not not again NO2 donates electron at the metap position no NO2 withdraws electron from oron par position yes and hence it creates extra amount of positive charge and make it more reactive clear everybody okay I have given to you now one another sheet of summary you can just have a look at sn1 sn2 E1 E2 reaction during the time of closeup of uh uh finishing up this session you can just go through all these things I have written down what about mechanism One Step mechanism twostep mechanism comparison of sn1 sn2 E1 and E2 comparison of sn1 E1 comparision of E2 and sn2 can also be done remember one question from this table itself is one question is fixed in your examination they can either directly ask you to write some features of SN one SN 2 or they can ask you to compare E1 and E2 remember to write examples also okay now I have this sn1 pictorial diagram once again I have discussed completely with you you have a leaving group nucleophile comes in rate is directly proportional to the alky halide new nucleophile comes there's a reic mixture obtained which is partial this is partial rization partial rization happens leing groups comes out the step number one is the slow step that is RDS and it has a uh empty p orbital present because of the carboca kp2 hybridization second step is the nucleic addition that's fast reaction and it's of no use to us because the rate will be uh rate will be decided by how fast the ionization happens and how the stability of carbocatine this is the energy profile diagram uh step number one step number two similarly this will be your sn2 mechanism sn2 is a direct attack and then product inversion of stereochemistry if you start with r isomer your product will be S isomer if you start with s isomer your product will be R isomer this is the sing energy profile diagram only one transition forms here transition state forms here and depending upon the stability of of the car depending upon how strong the nucleophile is the rate of reaction will also be changed accordingly this these are the few questions I'm giving you for your practice at home everybody I hope you will be able to solve these questions now please answer your uh questions in the comment box so that you uh you can get your answers checked by me and also from your fellow viewers that's all for today everybody I think I have brought to you more than enough from hello alkan hello AR chapter you absolutely need not to read anything else apart from C apart from your ncrt plus this uh your lecture today ncrt is not a sufficient textbook to understand organic chemistry this is uh I can say it I can say it to every anybody because organic chemistry is not explicitly written in uh ncrt you need to understand it from a mentor uh I think I was uh able to explain almost all the reactions I brought many other reactions from from Beyond the scope of ncrt also today it will be helpful for your neat as well as J Mains examination I'll see you in the next class guys till then stay safe and goodbye this is your teacher Dr laki Katura