Hi everyone, welcome to Vidantuj English. I'm Ricksha Koshan your chemistry master teacher and as you all requested that ma'am we want VIP triple E session. So here I am. Okay. Whatever whenever I promise something I always always make sure that I am on that point what I promised to you right. So I have selected even uh the last shift that has happened right uh according to that also if you see chemistry it is actually very scoring for vit reason being it's not very high level of chemistry they're asking you very moderate level easy predictable questions they're asking you right and here you can easily fetch full marks in your chemistry section at least right so let us start with the organic chemistry questions and then we will do some important questions of physical chemistry and inorganic chemistry as well. So I have taken most of the questions of organic chemistry. Reason being if you see my video on this channel only if you see this over here right you will see that there is already I have taken uh one live two lives in fact on VIT. You can see even all model questions of English and aptitude sir has taken. You can also go through this video here right and we have taken other vit e session as well as you can see vit e most important concepts and question that has crossed a lot of views right so you can watch that right this is for maths this is for physics this is for chemistry so in this particular part I have taken inorganic and physical questions along with the questions and the concepts in detail right so today I'm going to take your organic chemistry questions So let us start this without wasting any time. Okay. So everyone have your notebook and paper along with you. So everyone let us join and put green hearts in the comment section for VIT students. Okay. Yes. Everyone Ratita Muru and Civesh Bharini right everybody let us start this session without wasting any time. Right. Okay. First question on your screen. I will start telling you side by side what are the different different topics that we can ask you in your exam right so first question here is which is the most reactive towards nucleophilic addition reaction right so whenever we talk about nucleophilic addition reaction see these are some of the standard questions right okay so which one of the following is reactive towards nucleophilic addition reaction nucleophilic addition means on the carbon there should be electron deficiency if someone is poor Right? The nucleophile which will get attracted towards something which is electron deficient. Right? The rate of reaction will be more than that. So they are asking you most reactive here. So what you can see most reactive towards nucleophilic addition reaction. So you have to see this carbon this carbon this carbon and this carbon here which is most electron deficient. That means if someone is taking out electrons here. So here we can see it has electron withdrawing group that is taking out electrons from it right via which two things here that is minus m and minus i. What about this one? Here it is donating through plus I effect. So donation means more electron density that means this surely cannot be the answer. Here you have no effect. And in this case what is happening? In this case, it is donating through CH3 and it is also taking away the electrons like this because see simple part is whenever on the benzene ring students remember these two things that will really really help you in the exam. Whenever on the benzene ring if you have a group that has lone pair of electrons then it is going to show plus M effect then it is going to donate the electrons in the benzene ring itself. It will lose the electron. It will take away the electrons, right? Okay. Then if you see you have a group which is double bondedly or triple bondedly attached to more electronegative element. This is less electro negative. This is more electrogative. Always it is going to show minus m effect here. So what it will do? It will show you minus effect here. It will take the electron here and take the electron from the benzene ring. Right? So this is plus m and this is minus m. So as you can see this carbon here is attached to the benzene ring which is also attached to double bond with oxygen which is more electrogative. So it is going to take the electron from the benzene ring that means this is definitely not electrond deficient and here you see no effect. So answer will be D which is showing you electron withdrawing group here. So answer is going to be D is the correct uh session duration will be 1 hour or 1.5 hour. Okay. Only most most important questions I have picked here. Right. Okay. I know you have less time. So according to that only the session will be don't worry about that. Right. Okay. Now next question. Let us see that. Next is next question students try to answer the questions also. Right. Okay. The product formed in the reaction in the of cumine with O2 followed by treatment with dilute HCl. Now all the naming reaction should be on your tips. Right? all the naming reactions. If you're thinking ma'am I have no idea right now what to do just go through my naming reaction and important reagents video go through that that is just 1.5hour video that will really really help you and I promise you that you will see questions from organic chemistry from there only right okay so the products formed in the reaction of cumine with O2 followed by treatment with dilute HCl right so here we are talking about cumine process what is cumine cumine is when you have isopropile isopropile on the benzene that is CH this is CH3 this is CH3 right now whenever you add oxygen here it has a peroxide linkage that means this is converted to this is CO this is CH3 and this is CH3 this also they can ask you sometimes they also ask that what is the intermediate that is formed here right after that doing the hydrarolysis You get two products that is I can see in the options. I can see in the options second one is SC tone. So let us see how many of you have answered it. B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B BB BB very good. So answer is B is the correct answer. D process, cumin process, diotization process all are very very important for you. Right. Next is confirmers. Okay. So confirmers how to remember? Right. First is in ethane and cycllohexane in confirmers they will ask you questions from stability only. Right? Okay. So one thing to remember that in ethane and cycllohexane which one of the following is more stable confirmer. So if you talk about ethane here in ethane see firstly ma'am what is eclipse to all those students who have no idea ma'am what is eclipse what is tag we are totally dependent on you see this right. So if this is one structure one is represented the backward atoms one is dotted will represent the forward atoms. Okay. So here it is one some atom here some atom here some atom here. If you have zero angle between this let's say one one layer of atom one carbon is in front one carbon is in back. The atoms that is attached to this carbon the atoms that is attached to this carbon these two are representing backwards and front. So one is solid is representing let's say backwards and dotted is representing let's say forward or you can say more accurately is solid is representing the forward backward is represented by dotted right and one other is when you have 60° angle right so here what we have this is called eclipse is you know eclipse means when something hide something okay so like sun eclipse right like moon eclipse we have right so here this is called eclipse. Okay. And this is called staggered. Now you yourself feel that ma'am this will cause more repulsions. More repulsion always means less stability. That is why always staggered is going to be more stable as compared to eclipsed here. Next is regarding the confirmers of cylohexane. Right? So confirmers of cycllohexane it's very simple. Conformers of Py ex are let's say you have chair confirmer, you have board confirmer, you have twisted board confirmer, right? So out of this chair confirmer is the most stable confirmer here. So answer will be uh yes answer will be staggered and this one more question related to this I'll tell you that they can ask you one more question related to this. Okay, that is simple. Those who have no idea those students also can answer this. Okay. See there is one structure which is anti and one is Bosch. Okay. So usually what is anti? Let's say these are the two structures. This question also they can ask you regarding propane or especially butane this question is asked right? So if you talk about anti- structure in anti-ructure you will have things like this like this is CH3 group and this is CH3 group. This is H. This is H. This is H. This is H. Right? This is I'm talking about butane. One carbon here, one carbon in front, one carbon in backwards. And in this carbon you have one CH3 group attached. In backward atom also you have one CH3 group attached. This is what I have formed. If CH3 PH3 are anti to each other, gosh means they will be Gosh means they will be at this much repulsions. Okay? they will be at 60° angle with each other. Now in usual case what we say that because here in anti you have less repulsion. So always anti is more stable as compared to gosh right but what makes this question important is gosh effect. Okay. So students what is gosh effect? Anybody in the chat box what is gosh effect? Right. So one question they can ask you regarding this simple question or they can ask you this very very repeated question that is anti and gosh anti is more stable if there is no hydrogen bonding. Let's say you have a structure. Let's say you have a structure where you have an uh where you have hydrogen bonding, right? Let's say you have a structure where what you are saying is hydrogen bonding. Here you have let's say you have CH2O. Here you have CH2O. Right? And here also you have CH2O or directly you can also add O here. Here also O can also be there like totally your choice whatever you want to add here. It doesn't matter. What matters is if there is any hydrogen bonding involved, right? So we know that even now this is the gosh structure, right? And this is the anti- structure. But we know that here we will have hydrogen bonding. That is why gosh is more stable. And that is called gosh effect. That is called gosh effect. Very very important for your exam. Right? That is called gosh effect where you have hydrogen bonding that will be more stable in general case anti is more stable as compared to gosh. Okay understood students. Next question is next question is how many hyper conjugable hydrogen atom does this molecule have? Right? So always remember we hyper conjugation is not very very simple question very simple yet if you not gone through hyper conjugation you will find it what is this map right directly what you have to remember whatever is the main thing that main thing can be carocetine that can be a double bond that can be a free radical that can be anything right the carbon that is attached next to it the carbon that is attached next to this carbon this carbon and this carbon that is your alpha carbon first condition is alpha carbon should be sp3 hybridized okay what I'm writing first condition is alpha carbon should be sp3 hydraized and at least it should have one at least one alpha hydrogen it should have at least one alpha hydrogen it should have that will make it uh like it can produce hyper conjugating structures only then now we will apply these two things over here if you draw this structure students if you draw this structure here right this is the double bond this is CH3 and this is CH3 now if you see this right I'm telling the crux of everything right that we have discussed this is alpha this is the main thing that is the double bond right so this is alpha this is alpha this is alpha this is alpha if you see how many hydrogen can be there to complete the valency two hydrogen to complete the valency two hydrogen. So 1 2 3 4 5 6 7 8 9 10. Total you have 10 alpha hydrogen. That means you will have 10 hyper conjugating structures. Always remember number of hyper conjugating structures. number of alpha alpha hydrogen is equals to number of hyper conjugating structure and more the hyper conjugating structures more will be the stability of the compound right so here we can write C is the correct answer very good I can see that in the chat box that ma'am 10 is the correct answer very good next question is here now this is I would say from one of the most important reactions and most important reagents of your organic chemistry that is repeatedly asked in the exam. Right? So first is two brooentane is heated with potassium ethoxide. Now let us see how many of you actually identify this thing. Potassium ethoxide is what students let me know in the chat box. Are you able to identify what kind of reaction is this? Ethoxide means C2 H C2H5 C2 H5 O N A that is ethoxide right potassium means K here it will be right now this thing ma'am what kind of reagent is this what kind of Very good very good very good so here it is nothing ethanol do you know this reagent Hello. Hello. Hello. Hello. I'm so sorry. I'm so sorry. Where I was? Where I was when I was teaching you this. I am so sorry. Now I'm no audio. Uhoh. See how consumed I get when I'm teaching. I have no like h what is going on or I'm I feel bored and reaction. That's my energy when I solve organic chemistry, physical chemistry. in organic the kind of energy that you should have in your exams right no nothing I see when I am solving the questions okay so everybody now let us see this okay I'll again solve this for you again I'll do this for you right so basically here what I talking about is that here you have potassium ethoxide that is C2H5 K that is nothing but alcoholic Q now you have two kind of reagent one is aquis Q other is alcoholic Q. So if you talk about alcoholic Q and aquis Q let us draw the compound firstly here that is two bromo two broentine right. So now here now what if we are using KOH but that is aquis right. So directly it will do the substitution. So Br will be substituted by O no worries nothing right. But what if here you have KO alcoholic right? The reagent is C2 H5. Okay. This just to confuse you. They have given it like this. Right? Otherwise what you will do here. Here directly you will eliminate the dehydrohalogenation. One hydrogen will be eliminated. Now the question is ma'am here you have CH3. Here you have CH2. So two products can be formed here. If you replace hydrogen from this side this product can be formed and if you replace hydrogen from this side this product can be formed. So here which product you will choose A or B? Everybody let me know in the chat box. A or B we will choose the this product not this product because according to szip rule more alkyated the one in which double bond has more alkyle groups attached that is preferred right so that is the major product so answer will be now here again the question is whether trans or cyst will be formed right so I can see the answers in the chat box that ma'am trans is formed here reason is because trans always has less repulsions if If you see this is your trans this is one this is two this is uh one this is two one more question I'm telling you here right one more question I'm telling you here regarding GI right if you have if they ask you which of the following shows geometrical isomer this is one this is two this is one this is two right so beta in this case remember one thing let's say this is three or let's say this is four okay remember one thing 1 should not be equal to 2 3 should not be equal to four if you're talking about geometrical isomers will be present only then geometrical isomer means cis and trans if you have two same atoms on the opposite that is trans if you have two same atoms right side by side that is called sis okay so trans will have less repulsion that is why we will answer C is the correct answer here. Right? Not six. This is where students can get confused that which is the correct answer here. Right? Next question is based on the stability of the carbonion. Now students one rule I want all of you to have it in your head. That should be crystal clear in your head. That is B A R HI. Everybody spam it in the chat box. B A R HI. Ma'am what is this BI? Trust me this will make questions of stability of intermediate like in seconds you can solve the questions then trust me you will not have any confusion after that you can easily solve this sometimes we know all the effects still we are not able to understand which will be more priority right so this means back bonding this means aromaticity this means resonance this means hyper conjugation and this means inductive effect right okay this is aromaticity this is back bonding now Firstly we will look for back bonding right? Is there any back bonding in any of the compound? Back bonding ma'am means what? In back bonding like you have oxygen here and here you have lone pair and here you have carocatide. One is electron rich other is electron deficient. That is back bonding right that is called back bonding. Or you can have one boron here and one carbon ion here. Now carbon ion is electronrich but boron we know that from inorganic chemistry boron is electrond deficient that is also back bonding. Okay, just remember electronrich and electron deficient if they are on the adjacent position then back bonding will happen here. We can't see any electron deficient comp atom in place of next to the electron rich. Right? So that means back bonding is not possible here. Right? Next is aromaticity. Can you see aromaticity should follow PCC? What is PCC? Planer conjugated. from this question I'm revising the full GOC right I'm revising full GC right that you should apply here PCC means planer cyclic conjugated right so if you see here planer cyclic conjugated that is there only in this compound which is complete conjugation is present cyclic it is it is planer also electrons are 1 2 3 4 5 6 it has six pi electrons that means 4 n plus 2 pi electron it has that means it is aromatic in nature for this students always remember always remember one table I'm drawing it for you on the side part here right that is if you have important charts I'm giving you for your exam right say let's say you have PCC if PCC is followed if PCC is not followed then it is nonaromatic PCC is not followed maybe not it's not planer it's not cyclic it's not conjugated it's nonpl non-aromatic if PCC is followed And there you have four n pi electrons or 4 n + 2 pi electrons. This is anti-aromatic and this is aromatic in nature. And if you see the stability aromatic then you have non-aromatic then you have anti-aromatic is the least stable compound. If you see any compound they have given you stability of intermediate. Any compound comes out to be anti-aromatic that will be least stable among all of them. Right? Least stable compounds are anti-aromatic compounds. Understood? So here what we will say that fourth is going to maximum stable. So this is eliminated. This is eliminated. Right? Okay students have answered C is the correct answer. Let us see that. Right. After this let us see if there is any resonance here. Resonance is present over here. Right. So that means after that you will have third. Then let us see hyper conjugation. In carbon ion you don't have hyper conjugation right in carbon ion hyper conjugation is not valid right unless and until you're talking about minus h which will not be there in your service right. So last part is inductive. So here we know that this is carbon which is 2° and this is carbon which is primary. And we know that in the inductive effect which is more dominating. See simple thing is if something bit one more thing I'm telling you about your carbonion stability I'm writing it here right I'm writing it here remember that if you have if you have negative if you have let's say negative if you have something negative and you are donating the electron somebody's already under stress if you are under stress I'm giving you more stress you'll be like ma'am please go I don't want more stress right if I come here and start wasting your time. You'll be like, "Ma'am, I don't want to listen to you." Right? So here if I am teaching you something which is useful for you, right? I'm taking away stress from you. Then you are more stable here, more happy, right? So negative needs negative. That means a carbonion needs something which is taking out stress from them. Then it is becoming stable. And if you have carbon ion then if you take give the electron if someone is poor you give the money to them they will be happy right but if you take the money from them they will be unhappy here right. So here what we can see that in the carbonion here you are giving the electrons with one f with one CH3 and here you are giving the electrons from two sides definitely this will be unstable. So second will be less stable as compared to first. So answer is A is the correct answer here. Right? It needs less electron. It wants that I am already negative. Don't give me more electrons. I am already full of electrons. Right? So that is the reason B is going to be the correct answer. Is that clear to everybody? Very good. Very good. I can see your energy in the chat box. So that is B is the correct answer. Very good. Right. Okay. Super example. Yes. Now you will remember this forever that okay negative means stress positive means money okay just remember this for your exam right aa one more thing regarding what kind of question they can ask you from go this is what I told you that when I'm saying that this is going to be a important session that means not only questions but I'm going to tell you some important paper leaking questions I'm going to tell you right so this is the etching right same kind of question they can ask you based on acid rather than giving you carbonion they will give you some HA. We know that if they ask you the stability here, remember that more is the stability of this carbonion. More will be this will be a better ask right acidity depends on the stability of carbonion and stability of carbon. You have already answered that ma'am. We can easily answer that right. Okay. Next is ma'am pKa value is inversely proportional to acidity. Remember that. Okay. Next important thing is basicity. If they ask you basicity in any question, remember that basicity is directly proportional to number of or you can say directly proportional to electron availability. Electron availability. If on that species you have more electrons availability that will be more basic in nature right directly you can answer that right. Okay. Now that was about GOC how they can ask you the questions based on GOC. Now one question is based on enolines. Now tell me in the chat box how many of you have studied enolines. Enolines is a part of aromaticity, anti-aromaticity and non-aromaticity. Right? Usually what we see ma'am anolines are very very difficult. No beta they are very easy. I'll tell you how. Right? Enoline is monocyclic like you have benzene. Right? Benzene is the example of six enoline. the number of pi electrons or the number of carbon will represent the anoline number. So we call it six enoline because it has 1 2 3 4 5 six carbons. This is called four enoline because it has four carbons. Okay. So according to that we name the anoline. Now you have you will have always even number of enolines. From 4 till 14 you can have your enolines. In fact till 18 you can have the enolines. From 4 till 18 you can have the analines. But remember one thing right? What you have to remember the non-aromatic analines. Nonaromatic analines. What are these? 8 10 and 12. Just remember 8 10 12 that's it right? Don't make too much pressure of enolines in your head. Right? If they ask you very simple and direct question they will ask you eight. If you see 8 means 1 2 3 4 I think that is going to be 1 2 3 4 5 6 7 8 right? This is your aolon. It does not exist as like this. Rather it exists in a tub-like structure like this. Okay? It exists in a tub-like structure like this. That is why it is non-planer. Hence, it is anti-aromatic. So, they can ask you why 8 is sorry non-aromatic because of its tub-like structure. 10 12 are nonplaner and that is why they are non-aromatic. Reason being because of their hydrogen hydrogen repulsion or you can say angle stream ma'am what about 4 6 14 16 18 aroline you didn't decide you didn't told us about that right so now regarding other anolines what you have to remember is number of carbon atoms let's say they ask you for they are asking you for 14 anolines 14 anoline means it will have 14 pi electrons. Number of that enoline will represent the pi electrons. 14 pi electrons are fall into the category of 4 n pi 4 n pi + 2. If you put 4 n + 2 4 n + 2 if you put n is equ= to 3 3 4 12 + 2 is 40. That means it is going to be aromatic in nature. This is what I'm telling. If they ask you 18 don't learn anything that okay 18 alanine will be what ma'am we are ma'am we forgot that in the exam you don't have to learn anything that is four if you see 4 n + 2 again if you put n is equals to 4 that is 18 right again it is aromatic let's say ma'am let's talk about 16 enoline 16 enoline has 16 pi electron so it will be 4 n pi if you put n is equals to 4 here right okay that's how you will answer this. Is that clear to everybody? Is that clear to everybody or not? Right? So maximum the amount of question that they can ask you from GOC that I have covered from one question only. Right? Okay students, is that clear? Right everybody? Now next question is ring expansion, ring contraction and ring carocetine rearrangement. Okay students ma'am we can't do this ma'am. Please ma'am easy easy question you tell us I'll tell you I'll make this easy for you you give me just 2 minutes okay I'll make this easy for you right now do you know one thing that in SN1 mechanism SN12 just remember 1 2 1 1 2 1 do you mean by that if you have one SN1 mechanism that means it is a two-step mechanism in two-step mechanism you will have carocetine then nucleophile will attack right so SN1 mechanism Means there will be formation of carboation here. That is clear. Right? Now if you put this here, if I draw this and if I put I take X negative from there, I'll get this carboatine. Right? Now this is only possible if it is the SN1 mechanism. Right? Now here ma'am what is ring expansion contraction? Simple thing is if you have positive charge outside the ring, positive charge outside the ring will cause ring expansion. Just write it here. Positive charge, positive charge. Where you want to write positive charge outside ring, outside ring, outside ring, then ring expansion without thinking do ring expansion. Ma'am, how do we have to do ring expansion? Very simple. Put 1 2 3. Okay. Now what you will do? Simple thing is you will draw the same thing here. You will draw the same thing here. But now attach 1 and three. Attach one and three. Okay ma'am. Attach one and three. So I'll attach one and three. Then put positive on two. Put positive on two. That's you have your compound ring expansion. That's it. That was ring expansion. Right? So now this does not look good. So rather because pentine exists like this. So now I will add O negative because we are adding water here. So I'll add O negative here. And this will be your answer. So answer is A is the correct answer. Right? Okay. So A is the correct answer. Uh one student has answered O here. We think this should be the answer if this was SN2 mechanism because in SN2 mechanism you don't have any carboation formation in carocetine there will be ring expansion there will be ring contraction so take care of ring expansion and ring contractions okay right that I have told you where you will use now I know you are a little bit scared after seeing if you are seeing this for the first time then you are like man okay this was a good question Right. So now ma'am will I be do that in the will I be able to do that in the exam? Let's see it from this side. This is not the same question. I'll give you the hint. This is a question of ring contraction. Right? And trust me in VLE E they don't ask very high level question from ring expansion ring contraction. very high level question of ring expansion and contraction are asked in G advance. In VAT triple E these state exam or especially university exams right they ask you in VA good questions are asked that is why this concept they can ask you but from a very simple perspective right so again it is SN1 reaction that means it is carocetine formation it is there right so here what you will have this is the cyclloutane right take out x negative here you will have positive here remember whenever you have all the ring You have the positive charge. You will apply ring contraction here ma'am. How we will do that? Same copy phase that we did in ring contra expansion. This is one. This is two. This is three. What I'll do join one and three. Okay ma'am. We will join one and three. Now put positive on two. Okay ma'am. Put positive on two. Right now write it beautifully like this. Okay ma'am. Now you have to add O negative here. Very simple. O negative is added there. So it will be 1 2 this is O that is CH2 to O. That means C is going to be the correct answer. Let me see how many of you have answered it. Very good. Very good. Very good. Sura Ashish very good. Uh okay. Okay. Okay. Very good. Very good. Right. Very good. Now ah very good one student has also answered. Shish is saying dancing resonance. Very good. Now this is very stable compound due to dancing resonance right. This is very stable due to dancing resonance because you know like you dance similarly this will dance. How this dances? Those who wants to know this how this dances. Okay. So when you're dancing on your one leg sometimes you're dancing on both legs. Sometimes you have one leg up right? So how you dance? Similarly, this will dance. Okay? So how it will dance? You will have a double bond shifting here. This will be double bond. This is the positive. Right? This is the dancing resonance. Or this bond can shift here. It can also look like this double bond. And this is one to positive here. So that is your dancing resonance. Right? That is your dancing resonance. How to find anti-aromaticity? C if it follows PCC then you have to check for 4 N pi electrons. If there is four N pi electron that is going to give you anti-aratics which are the most unstable compounds. Next question see in every question you're going to learn something right. It's not like repeated questions I'm giving you right. Which of the following alkyle helide is respectively most and least electrofphilic in SN1 reaction? Most and least. Now ma'am in electrofphilic. Okay ma'am electrofilic you are talking about in electrofile that means in SN least electrophilic in SN1 reaction. Okay they're talking about SN1 in S1 you have the formation of carocetine isn't it? We have the formation of carocetine here right now if you see the formation of carocetine if I take Br here carocion will be over here this carbon will be carocatine. If I take Br here, carocation will be over here. If I take caroc Br here, carocation will be over here. Right? They're asking you which of the following is respectively most and least electrofphilic. Right? Okay. So electrphilic means some nucleophile is attacking here that is why it is electrofile. If nucleophile is attacking here so this person will be electrophile that person is a nucleophile. Right? So here that means I have to find the stability least stable first I'll tell you whenever this is a bridge head position this is the bridge head position this position is the bridge head position ma'am do we have to know everything about this no till your exam you don't have to know anything like full detail on this just remember if they give you something like SN1 reaction this kind of thing is given to you and Br is attached attached here that means positive charge will be over here which is not at all stable why because this bridge head position cannot handle sp2 hybridization it is like I cannot handle sp2 hybridization because of some angle strain it cannot handle right that you don't have to get into right just remember here carocetine or here it is not stable at all that is the reason first is the least stable here so first first answer we have that C is the correct answer. We have the answer that C. Why third is most stable? Because if you put carboatine here, it is one degree. If I put the carocation here, if I see the carocetine, that will be something like this. Now students, we know that this is a tertiary carocetine. If I put carboatine here, this is 1 2 3. But at this position, it is not very stable, right? at as compared to this position and this two position this position is more stable. Right? So answer will be this tertiary is the most stable here. So C is the correct answer. Is that clear to everybody? Right? Is that clear? Okay. One more thing regarding uh this uh carboation right that is rearrangement. Okay that is rearrangement. Students just remember one thing that whenever you have the formation of carocetine whenever you have the formation of carboetine either ring expansion or ring contraction or sometimes ring formation but ring formation will not be there in your exam right you focus on ring expansion ring contraction or carocetine rearrangement. Rearrangement means like you have for example carocetine like this carocatine is formed right now what it will do it will this will be your hydrogen over here this hydride will shift on this position and you will have the formation of this carocetine that is through hydide shift that is called rearrangement of carboation and whenever you see any SN1 reaction there will be rearrangement of carocetine okay there will be rearrangement For example, I'll give you one example here like athetic hydration. Athetic hydration if you remember whenever you add acid H positive and then you add water right. For example in alken you are adding it you form the carocation. So firstly you rearrange it only then you form the carocatine like in hydroboration oxidation you follow. I'll also give you two other things here. In hydroboration oxidation, hydroboration oxidation you follow antimarovnikov addition. Right? In demmercuration, oxy mercuration, demmercuration, OMDM, oxy mercuration, demarcuration, you follow marovnikov rule. Now anybody who does not draw marovnikov antimarovnikov rule. So marovnikov rule means marovnikov rule means negative part is attached to less number of hydrogen. Antimarovnikov means negative part is attached to more number of hydrogens. Okay, is that clear? Okay. Yes, students. Right. Uh uh. Oh my god, I'm so sorry about it. Right. I told you to give like when I told you to give 100 comments over there. Now students reminded me that 100 comments are completed. Now you take the lecture. I thought you guys are not interested. If I would have known that you guys are interested, trust me, even if takes me night like day, I have taken classes at 11:00 p.m. also, right? So it just if you are if you guys are willing to do hard work, I am always there, right? I am always there with you. Right? Okay. So now next question is okay. So if you see this here, now these are the sequence here I'm taking up questions. Till now we have done enough of go. Now we are discussing question where you have let's say zinc HCl. First you should remember that ma'am zinc HCl is what your double bond O is converted to H2 here right? Zinc HCl. One more thing this is clemenzin reduction. Zinc usually we have zinc amalgam and then you have HCL clemensen reduction. Okay, in clemenzen reduction you use zinc amalgam then ethl. But if you talk about voler reduction same thing will happen double bond O is converted to H2 but in that case what you will have here is the hydraine in W Kishner you will have NH2 NH2 I'm just writing it here WV Kishner WV Kishner you have NH2 NH2 and then you have Q that is in the basic conditions you have right so directly what you can do firstly which we think will be formed. 1 2 3. This is H2. This is 1 2 3. Now can you tell me what is this? At a very high temperature 773 Kelvin when you use CR23 that is aromatization. Okay. aromatization that means this 1 2 3 4 5 6 membered ring will be converted to benzene ring aromatization will happen and hence the answer will be A is the correct answer here okay A is going to be the correct answer here right next question is very very important question very very important reaction that is can you guess that anyone of you who can guess this right the major product firstly just guess don't give me the answer ABCD right firstly just tell me in the chat box what is the name of this reaction just give me the name of this reaction okay right name of this reaction let me see who is answering first here who is answering first here in organic we are going to take that take that don't worry right so very good that is ma'am eldol condensation right so here you have alpha hydrogen here you have alpha hydrogen right so exactly Which will happen? Can you tell me? You have two answers. Either this alpha hydrogen will be acidic or this alpha hydrogen will be acidic. Which will be more acidic here. Very good. That is L2. Now you should understand which will be more acidic students. So we know that this alpha you have and this alpha you have. So exactly what happens right? So simple thing is here you already have what you are doing it is already taking oxygen from this carbon is already taking electrons from the benzene ring. You should know the reason why this is more acidic in the exam who will tell you that this is more acidic. What if you have taken this more as more acidic? So you should know the logic behind that. Right? So this is the benzene you're donating the electron. This carbon is already very happy. This carbon is already very electronrich. That is why this hydrogen is like okay this person is acidic. This person is electronrich. So I don't need to like he's not crazy about me. This carbon is like this hydrogen is not totally fallen like mad in love with me. So this hydrogen will be like I'll go right I am ready to go. Right? So this person is not fully committed. So I'm ready to go. But this carbon this hydrogen here this is alpha carbon. This hydrogen here this carbon is electronri because this is electronri. This sorry this is electro negative. This is electropositive as compared to oxygen. This is electron deficient. Then this carbon is like this. Sorry this hydrogen is like this hydrogen here that is on the alpha carbon. P is like this is electron deficient. So it needs me. That is why it is staying here only. Right? So that is why this is more acidic in nature. Once you know which is more acidic simple thing you can do right what is that simple thing one thing that you can do is you can make the alpha you can first see this is the alpha carbon because of B's wants to abstract the proton so this will be CH2 negative over here and then what you can do this will this is the negative part this will attack over here electrons will jump here right so what you will have this is benzene C double bond O CH2 This is linked to this carbon here. CH2 this is linked to this carbon here where you have the electrons. That is why this hydrogen will be attached here. So here you will have O and this is CH3. One part is through full mechanism you can solve this. Other thing is through direct method. What is that direct method? H one is CH3 C double bond O H right. Other is you have benzene and on that benzene you have CO you have COH3 right you have CO3 now let us see how I'm going to write it this is a method of LDL condensation those who find LDL condensation difficult because of the mechanism they can do this method which will give you always even if they have given you some complicated LDL condensation you will be able to answer this what you have to do the alpha hydrogen that you are using right we using this alpha hydrogen you will mark it separately right and the carbonal carbon of other molecule you will open this separately now beta what you have to do you just have to join this carbon with this carbon right join carbonal carbon with the alpha carbon so this is CH3 this is T and this will be converted to O because this hydrogen will be shifted and this is converted to O here. So this is O. This is H. This is CH2. This is C double bond O. This is benzene. Okay, that's how directly you can answer this. Right, my benzene is little bit funny. So I'll just have to write the benzene again. Right, that is your benzene. Now remember one thing. This is LDO part, right? You also have to use condensation. So from alpha and beta position take out O and H here. So what you will have in the end this is CH3 this is CH double bond CH this is C O and this is your benzene. But sometimes in the option they don't condense the thing they don't take out water they just the options are just related to this thing. Right? So we have to see what are the options that are given to you over here. The options are related to O only. So that means you don't have to uh eliminate water here. Right? So directly answer will be C is the correct answer here. Understood everybody? Understood? Yes. Okay. Ch. Now next is now we go through wood's reaction. How many of you know wood's reaction everybody? Thank you. Thank you. Thank you for the lovely comments. Thank you uh Sah. Thank you so much. Now one bromo 3 chloro. Now students see this. We all know what is wood's reaction right ma'am sodium you add ma'am ether you add what we do ma'am unsymmetric unsymmetrical is now symmetrical as possible you have ma'am rx ma'am from here r is formed through free radical mechanism right okay now what I used free radical mechanism very important thing I'm going to tell you here right okay so they are saying one bromo one bromo cyclopetane this is one bromo if this is bromo three chloro if this is fluo right now they are saying on reaction with this it will give you what students so first is just remember it follows free radical mechanism right so free radical mechanism means homolytic cle homolytic ma'am I don't want to get through this homolytic ma'am my brain is not that much working okay your brain is not working don't worry right just put one dot one dot like this bindi you put two two dots here right so here it is one dot and this is one dot right just remember that like dictionam always has bindi on the forehead that unicorn that always is on the forehead so that is how unit dot is always on the forehead that is how you will have this dot dot over here that is free radical now students they cram that okay this is going to be what mam okay this will join no you should remember that this will follow inter or intra Right? Okay. See, always inter is preferred. Right? That means only one molecule will join together to give you the answer. So that means this C is going to be the correct answer here. Then ma'am why you told us intra you want to confuse us or not? No. If let's say your inner product is anti-aromatic. Let's say your inter product is anti-aromatic then you will use intra. Ma'am is there any example of that? One thing you should be clear on inter and intra both situations are possible. Preferred is intra. Intra means in one molecule only those two dots will join and that will give you one bond. That is preferred. But if this is becoming let's say you have one compound that is let me just think about compound. Okay, you have this Br and you have this Br. Okay, this is the compound that is given to you. They are saying that this is going through woods reaction. Okay, that means you'll be like okay ma'am this is going to form something like this one one dot here. Right now the question is ma'am if it goes through intra which is preferred it will join together and it will give you this kind of product. Now tell me is this aromatic is this anti-aromatic? Always intra is the formed but if you have with with intra if you're getting anti-aromatic compound like this anti-aromatic which is it is having four pi electrons then this is not preferred then you will have to use inter that means two molecules these will join together. How they will join together? Like this. They will join together. Like this. And that will give you the new compound here. Right? That is how the product will be formed. Okay students? Got it or not? Is that clear? Simple. It is not complicated. Right? Okay. So, it's not complicated, right? So, uh I think that's it. Now, next question is William Sansson. This is a very very repeated question. Ma'am, you have given us the answer. I'm so sorry. But here you have to predict. See, Williamson synthesis is RX and you have R O N A Roza Rona Roar, right? So, rosa RX R Na. So, this NA and X will be eliminated. You will have R O R that is ether here. Right? Okay. Now one thing to remember here if you see this here with NH it will have the formation of R2 NA this is nothing but this is RX right and here it is giving you nothing but ether so that is why it is called William sun synthesis William ether synthesis reaction so answer is I think all of you have answered it that is B is the correct answer okay B is the correct answer right okay Now next is uh I think yes we have done that right right okay now next is next is consider the reaction so this reaction right why I have picked up this question because here two three important reactions we can revise from here right first is if something like this is given to you right you should know that tolerance this is nothing but tolerance test Right? And here you are using copper at 573 Kelvin. Now students I'm giving you some important thing here very very important regarding your oxidizing nature. So if you have 1°ree alcohol you have 2°ree alcohol you have three degree alcohol right. So one is if you have 1° 2° 3° first case second case and this is the third case right? These three cases are very very important for you. Right? First case is mild oxidizing agent. Second is strong oxidizing agent. Third is copper at 300° C. Right? So these are the three cases from where they ask you very direct simple questions. Right? In between also right. So everybody I want to see that in the chat box also. How many of you remember that? If you see the first part here, if you see the first part here, that is one degree. One degree alcohol in mild is going to give you alihide. Two degree is going to give you ketone. 3° will not show any reaction. Right? With strong oxidizing agents like you have jones reagent. Jones reagent means you have CO3 plus H2SO4 H2SO4 H2S4, right? like you have KNO4 right? Okay, here you have uh CR3 simple or you have PCC or you have PDC. These are mild oxidizing agent. Tolins reagent is also mild. Pelling is also mild. Pelling is also mild. Okay. So I have written all the reagents here. Right now this is going to give you one degree is going to give you caroxyic acid. Caroxyic acid 2° will give you ketone only 3° will not give you anything. Right? Now last one is uh when you have copper copper with bond and two degree will give you mild answer that is alihide alihide and ketone. But if you have three degree this tell me students if with three degree what it will give you? With three degree it will give you which thing everybody with three degree. And how many of you have liked the session right? How many of you let us see how many of you have liked the session? See 417 are watching. 417 or 437 students are watching and likes are 320. Right? That is not good. Right? That is not good. Like should be beyond 700 or something till now. Those many students who have come here. Right? Okay. So everybody do like the session. I don't have this habit of asking this again and again but I it feels really happy right I have spilled my dinner also for you right so that is your duty to like the session right okay so now here it is going to give us very good can I see that yes very good very good very good very good that is going to give us Elken yes Raj has answered that ma'am yes Raj was the first one to answer this that ma'am elken is the correct answer here very good right Now here let us see go through this reaction again right like there is some compound X which is having one oxygen on copper at 3° you should be able to understand that there is some either tertiary secondary or primary secondary or tertiary alcohol is there right then with that it is giving tolerance test mirror is observed that means silver mirror is observed now remember that tolerance test is given by how many things it is given by aromatic or alifhatic alophatic alihides. It is given by alpha hydroxy ketone. Only one ketone that gives tolerance test. Alpha hydroxy ketone. I know all of you know aromatic aliphatic alihide but only some of you know that there is alpha hydroxy ketones that gives positive tolerance test. And yes the questions are asked directly which of the following will give you tolerance test. You know what in that case what they will give you alpha hydroxy ketone and very rare students. Alpha hydroxy means you have double bond on the alpha position you have O alpha hydroxy ketone right okay then one is hemi acetyl. Hemi acetyl also give you ketone. Last one is formic acid also give you positive tolerances. Oops. formic acid that is HO right okay now here what we will do ma'am okay this we have done positive means in this case if this many see one trick I'll tell you not even trick one logic simple like uh common sense I'll tell you here if specially they have asked you question on tolerance reagent you should remember this this and this also right if not specially they have asked you this then you can ignore that then definitely they are talking about aromatic or alifhatic alihides right so here we know that with two carbon it's not possible to have aliphatic aromatic alihide so this will be alifhatic alihide only so this is your alifhatic alihide alifhatic alihide alifhatic alihide is possible with copper at 3° that is possible in what is possible when you have primary alcohol now you know that this is primary alcohol. Ma'am, this is primary alcohol. So answer you already have this A will be ethanol, right? So A is ethanol. That is your primary alcohol. This is acetalihide. Acetalihide is CH3 CHO. That is going to give you primary that is give you going to give you positive tolerance test. That means answer you can already give answer you already have here. But ma'am is here to give you knowledge. Right? I am not solving the paper for you here. Right? So my duty is not to give you the answer. The answer you can get even from any book right these questions are there in books only you can get the answer anywhere right you can get okay B is the correct answer that is mentioned in the book what's my duty to explain you things right because I am present here right so this you know how to answer okay how to use your brain answering this right so that is clear after this what they are saying that you have a this thing here okay this O is going to give you some acid you have Okay, see here you have LDOL condensation, right? O negative on alihide that is going to give you LDOL condensation. We know that we have acid alihide. Now students tell me how LDOL condensation will be there. This is everyone draw it on your notebooks. Okay, draw it on your notebooks. I already told you how to answer this. Right? This is TH2 and there is one alpha hydrogen here. One alpha hydrogen here. Now we know that these two things will join. that is this is CH3 this is C double bond O this carbon is going to join with this hydrogen here this is CH this is your CH after that we will have elimination array array here this will be converted to O okay so this is CH this is converted to O after dehydration this is converted to CH3 CH double bond CH2 right that is alpha beta unsaturated Why I solve this? so that you can practice eld condensation right now only if you have your exam tomorrow also I don't want to waste even a single minute I want full revision here right okay so that is there right next is al we have answered no ma'am what is this kind of question ma'am what is this in your NCRT you have a table right where you have semicarbazone shift base oxy all of these are formed that table is very important go through that table if right now just write it somewhere in your uh wall or somewhere write it that if you have not gone through that go through that I'll also send you that table on the J channel vidantu j channel everybody if you have not uh like if you have not subscribed to the vidantu j channel that is totally for free of cost you can see the link in the description I I'll send you all the important pdf and the chart also in that particular uh channel only now see this thing if you have some ketone And you have NH2. Here you can draw NH2. And this is Z. Zed can vary. Zed can be hydrogen. Zed can be O. Zed can be a big benzy, benzene also anything zed can, right? So directly what will happen? Water will be eliminated. And here you will have C double bond N and this is Z. This is the compound that is formed here. Right? Now if you see in this case NH2Z this is the Z part. This is the Z part. This is the Z part. So here what you will have estate alihide is CH3 C double bond O. So here see instead of C double bond O what you will have? This is hydrogen right? This is your alihide. Instead of C double bond O you will have N zed. Zed is this whole part that is NH CO NH2 right this compound is formed. Okay understood everybody right understood that is your hydra zone is formed here right? Yes everybody next next this we have done right like this we have already done this n C N H this is what I have written right uh uh uh let me see if I have written this only right nu yes yes okay by mistake sometimes in hurry I forget one hydrogen uh nitrogen no so that's why I thought next is the heating of phenile methile ethers with HI produces now in ethers you have also one very important reaction That is if you have phenile methile. Okay, you have phenile and methile. One is phenile. This is O. This is CH3. Now whenever you add HI, what happens? That this is positive, this is negative. Hydrogen will go to some place and iodine will go to some place, right? Either it can get hydrogen attached, either it can have iodine attached, right? So what will happen here? This bond see either this bond can break or this bond can break. This bond is very difficult to break because of the double bond character because of the resonance. That is why this bond is not possible to break. That is why you will have the answer that is phenol plus CH3 X is formed. That is CH3 I is formed. Right? Okay. Understood? Right? Now remember one thing more here that in Williamson synthesis reaction remember one thing that primary helite the helide should be helide should be primary helide should be primary right you will prefer in william's reaction helide should be primary okay that is the important part right okay okay next is this also we have done this also we have done now students Next part is next part is the heating of this we have done uh in the reaction this is given to you the electrofile involved first you have to tell me the name of this reaction students the name of this reaction what is the name of this reaction everybody what is the name of this reaction yes very Good. The name of this reaction is reertimer reaction. Right? Remote teamer reaction. Remration. Remma reaction. And this is reertimer reaction. Right? But when you have CH, CL3, chloroform and NO, you have the formation of dchloroarbine that is CL2. That is your D is the correct answer. That is the electrofile here. Can you see there is one more reaction. Similarly like this there is see here we know that you have the formation of caroxilic salicylic alihide okay that is formed right. There is one more reaction that is carbilamine reaction where you use the same reagents but there you convert NH2 to NC right that is why it is also called isocoscanide reaction. This R NH2 R NH2 is converted to RNC in the carbamine reaction. The reagent is same the electrofile is same in that case as well. Right? Okay. Yes. Now students are saying ma'am there is one more reaction that is col that is similar in a different way. Colbe reaction you have phenol and you add here you add here CO2 you add Q or NU right with hydraysis it is converted to salicylic alihide it is converted to salicylic alihide one more thing ma'am can I make salicylic alihide here yes if you use if instead of this instead of this in phenol you use CL4 plus N then also you can make salicylic acid here right in rearr also you can make salicylic acid if you use CL4 this question they can ask you directly very simple direct question like in 1 second you can answer that right that if they use CL4 here then you can make CO right if they use CC CH3 then you will have CH if they use CO2 and KOH that is reaction that is CO if they use carbide I mean that is a different but that is similar to reertimer because of the reagents that are involved here right okay next is next question is the structure of intermediate a in the following reaction is if you see the structure here this is cumine right now they are asking the intermediate that I have told you in the before also that intermediate also they can ask you cumine is very like highly frequently asked question right so here what we did instead of h you will have O here peroxide linkage will be there. So answer will be A is the correct answer. Is that correct? Right. Next we have is next we have is the major product of the reaction. The major product of the reaction. Now if you see this here whenever you are doing ammonolysis students whenever you are doing this is C double bond O this is O this is C double bond O right. Okay. So this is you are adding NH2 here what will happen? This will be replaced by NH2 and NH2. Then you are heating it with strong heating. What will happen? This a wait minute this the what will happen? This hydrogen and this NH2 here that will be eliminated that is minus NH3 and can you name this compound? Can you name this compound answer? So I know you already know that what is the answer? You will be left with NH here. Here this is called fthalamide that is using gabriel fthalamide reaction. Here it is fthalamide is formed. This is using gabriel fthalamide reaction again a very important reaction right which is used for the preparation of primary amines and that to primary alifhaticamines. Anyone of you, anyone of you can tell me what is the reagent of gabrielthalamide reaction? reagents are first you add Q then you add RX right then you add I think uh firstly you add something with K we add I totally I forgot that right this is RX then you add I think NU right something NU so what will happen first this will be replaced by uh K right or Na whatever the ion is there Right? Then it will be replaced by R here. Then this bond will break after the hydraysis. Here you will have OA. Here you will have O NA. And here you will have attachment of H and H. That is R NH2 will be formed. That is why we say only primary alphetic amines are used here. Right? Okay. Next is identify the compound. The compound you have here in the sequence. Now if you see this here, this is the CH3. Right? Now whenever you are using Cl2 HU right remember one thing student this Cl2 H mu this is going to give you free radical mechanism. So it will give you alilic or alyic or benzilic substitution alyic or benzilic substitution. Right? So this is the benzene here. Aylic means what? Alic means you have C double bond C. The hydrogen here will be substituted. Benzilic means you have benzene here. And to this carbon if you have any hydrogen that will be substituted. This is benzilic. This is elyic. Okay. This is benzilic. This is alic. That means here this hydrogen that you have here 1 2 3 hydrogen will be substituted. But the question is ma'am how many hydrogen's will be substituted? two hydrogen's will be substituted here. I'll tell you why. Because you want to form CHO, right? So here two hydrogen's will be substituted CL CL and this is H. After this you add KO. After this you add water here. So this will be replaced by OH. So this will be C O and this is H. Now what will happen this O negative H will replace and you will have the formation of basically this is going to happen. This is O. This is O. So what will happen here? This will be eliminated. This will be eliminated. This bond will shift here. You have the formation of benzaldihide. Right? You have the formation of benzaldihide. Right? Okay. Understood? That is why this is a confusing question. Okay. Somewhere from you can also take CCL3. You can also take CHL3. So answer is CH. Wait a minute. C H2. Huh? This is CH CL2. Yes, A is the correct answer. Okay, CHL2 chlorines are replaced. Now understand one thing. If you have in this place, if you have Br2 HU or you can also have NBS. NBS is N brothenomide. Ma'am, do I need to learn the structure? No. Just remember if in the region they have written NBS that means they are talking about alyic or benzilic substitution, right? With bromine, right? Okay. Now next is the reaction between the reaction between benzaldihide and acetan in presence of same question right. So now this we have repeated so many times so I will not repeat it again and again right. So now here they have asked you about the cross ldole. So we know that this is benzaldihide this is hetenon. So we know that here you have alpha hydrogen. LDole condensation is whenever you have alpha hydrogen canisar is when you don't have alpha hydrogen. So if any of the reactant has alpha hydrogen in that case LDOL condensation will be there because you have two different different reactants that will be crossdol condensation right that will be crossdole condensation. The product formed by reaction of alihide with primary amine is right the product formed by reaction of alihide with primary amine.hide with primary amine. Okay, primary mean okay so simple is just the same we did this is C double bond O this is this right alihide is CH3 and this is hydrogen here right and here primary amine means you have NH2 and this is R here so what will happen you will have C double bond O this is N this is R this is CH3 and this is H here now what is this called C double bond O N anybody aromatic shifts base ketone carbon hydroxilic acid. What is this for? Yes, very good. That is called shift's base is the correct answer. Yes. Right. This is again from that table that I told you. That is very important. Right. Okay. Which of the following reactions is appropriate for converting acetamide to methanamine? Acetamide is what ma'am? CH3 C double bond O. Right. Uh NH2. Yes. This is acetamine. Then you have you convert it into methanomine right that means you have to take out this should be eliminated. The best method is Hoffman broomemide degradation right that means you add Br2 you add NaOH or any base here directly this C double bond O is vanished that is why this is called Hoffman bromide degradation because you have one carbon less is formed here right okay next question is nitration of enoline nitration of enoline in strong acidic medium also gives meta nitroenoline Now this is a very very interesting question. Let me just clear this out what we are talking about here right three four concepts will be there. Okay. So understand one thing that whenever you are doing enaline. Now firstly how many of you know that this is orthoparadirecting or metadirecting? Tell me this if this orthoparadirecting or metadirecting enoline. Okay, I'll give you once you have nitration what kind of products are formed. Right? By the time I want to see that if you're able to answer that whether this is orthoparadirecting or metadirecting I told you when on the benzene ring you have directly a group attached that is having the lone pairs right then loan pair it will donate here that means it is plus m and plus m is always going to be orthoparadirecting. Very good. Very good. One student has answered it right. Very good. Very good. Very good. S has answered it. Very good. So here what you will see what you will see one ortho product on nitration I'm talking about. Right. Here you have concentrated HNO3 HNO2 and concentrated H2SO4. Two things I'll talk about bromination also. Okay. I can see that in the comments. Right? So NO2 and this is NO. Ma'am this was orthoparadirecting but you have 2% orthoproduct you have 47% meta product 51% par what you are saying ma'am this is orthoparadirecting how can it form metadirecting product how can it form let us see that so what exactly happens here this is converted to in these acidic condition what happens in these acidic conditions what happens we know That enolene is a good base, isn't it? Enolene is a good base. How many of you agree that enoline is a good base, right? Because it has lone pair of electrons and lone pair means basicity, right? Now here can I say that this thing will take out one electron from the acidic medium. This wants to accept the proton. It will accept the proton will be converted to analium. This question they will ask you why you have the formation of why you have the formation of meta product 47% because of the enaline which is metadirecting why it is metadirecting because it is taking out electron from the benzene ring then any group takes out electron from the benzene ring that is metadirecting you don't need to prime it any group take out the electron metadirecting any group gives the electron orthoparadirecting so that is why meta product is formed here right okay ma'am okay okay ma'am got it that okay that's all but ma'am this first question they can ask you that how much nitrating mixture how much meta product is formed second question why it is formed because of the endium third question how you can make how you can make how you can make one product only that is tarot how you can make only this product here how you can make that by doing acid dilation that You can add acid and hydide or you can add acid SI acidic chloride here whatever you want to add in presence of pyodine right here what will happen you will have the formation of this kind of product. Now let us see this question is also asked okay in statement based this question is also asked. Third question they can ask you here is how you can make par product here by doing acetilation. In acetilation what happens? How can you actually prepare par product by doing just acetilation? Because these lone pairs are now busy in the elect busy in the resonance with the carbonal group. That is why what you will have here after nitration this is NH CH3 you will have NH NO2 group here. After the hydration, after the hydration you will have only the para product here that is not 51 full 100% para product you can form. Similarly with the same process you can form bromo mono bromo para product also by doing acetilation then adding Br2 here you can get the paraproduct of bromine as well. Okay, understood everybody right now let us see the question what we were asking you and nitration of enolene in strong acidic medium also give meta because of in absence of substitutant no in electrophilic substitution no in spite of substitute no right in acidic medium enoline is present as analinium ion so answer will be D is the correct answer is that clear to everybody D will be the correct answer here correct Okay. Shall I move ahead? Right. Shall I move ahead? The correct order of basic strength. Now I told you the basic strength here. Mythile substituted am means in aquis solution. Right? Now students what I told you for basic strength. See the direct proportionality here. Basic strength means more the availability of electrons. Right? Now ma'am here they are asking you in aquis medium. That is the different question. Here they're asking you this in aquis medium in gaseous medium you can say 3° then 2° then 1° this is in vapor form or in gaseous medium right because 3° has the maximum donating power electron will be maximum but in case of methile see in case of methile I'll just directly write here in case of aquis medium aquis medium if you have methile it will be or if you have ethile in case of methile it is 2 1 3 2 1 3 I'll just write it here 213 in case of ethile it is 231 direct question can be asked on methile or ethile right that is the order here right okay so answer will be uh 2 1 3 that is D is the correct answer next question on your screen is which of the following amine will give carbolamine test. Carbolamine is given by only primary amine right okay that can be aromatic that can be alifhatic so this is not primary this is not primary this is the only primary amine that you have so C is the correct answer here right okay which of the following compounds will form significant amount of meta product during monitration reaction significant amount of meta product we have seen this that meta product is formed by enolene here because of the enolinium ion. So answer is D is the correct answer here. Right? Okay. Understood? Now for all the students here, right? One important thing if you have your exam, see I have not finished yet. Here I have yet to complete physical and inorganic. Some important questions I have picked out for you. Hardly it will take half an hour because physical and inorganic students I have completed here right you can go and watch this lecture here right you can go and watch this lecture here I have completed physical and inorganic in 2 like 2 hours you can say right if you watch this at 2x speed it will be 1 hour or 1.5 hour hardly it will take you but all the concepts will be revised here right if you don't want to go through any questions here then students this is the video section here you will See one lecture for physical chemistry important formulas. Wait a minute. Let me just show you that. Uh no no no no. If you see this here then let me know. Must do this questions last minute sample paper NCERTT top questions organic conversions. Okay. Where is that? Why I'm not able to find my own bid lectures? H can you see that I think huh here it is all formulas in physical chemistry you can just type physical chemistry formulas vidant to j English dika ma'am okay so you'll get all the formulas in 53 minutes you can revise for physical chemistry very beneficial for your exam right at least one day before you must go through then for inorganic chemistry Right? If you don't want to go through this lecture, right? What you do here for inorganic chemistry and for like too many lectures we have already taken. Right? So here I just show you the important important two three important things here in the uh uh uh uh uh okay so here it is all naming reactions complete revision for organic chemistry. Complete revision is there complete organic chemistry revision. You can see this here from my handwritten notes only I have taught you all the naming reactions right. Okay. So important reagents also I have taught you some important reagents don't have any name right but they are important right. Then there is uh important exceptions that you can see here right the important exceptions right you can see here all the important exceptions important lines we have covered here you can also go through that right now I'll come back to the lecture again that is your physical chemistry some important questions in detail I have covered physical chemistry already right now I'll shall brush up some concepts here right okay this is not like a revision revision thing this is basically basically you're testing okay let us see if you are able to answer this or not right so if you have attended my lecture before then I think you will be able to answer this right at least I have that confidence on me right the bohar orbit radius for hydrogen atom is approximately this the radius of first excited state if they're talking about first excited state that means in uh orbit you have to add it right you have to 1 + 1 that is two second excited shape means 1 + 2. Third excited state shape means 1 + 3. According to that you will answer your excited shape. Hydrogen atom means z is equ= to 1. Here it should be zed is equals to 1. Now we know that the radius is r is equals to.53 n² by zmstrong because the options are inst. So I'll put the value also instrong. Put the value of n and zed here and you will get your answer for the radius. Right student in physical chemistry uh in the detailed lecture also indirect questions also sorry in the direct formula video also I have taken up all the formulas right so if you don't have enough time I would advise you to go through the formulas directly okay next is which of the following is not permissible arrangement of electrons in an atom so not permissible arrangement they are saying not permissible arrangement so not permissible arrangement means you have to see Remember two three things here. L is equals to 0 to N minus one. So that means L and N can never be equal. M is equals to - L 2 + 7. Right? So if you see this here that is not permissible arrangement. So that will be uh it is 5 3 this is not equal that is not there in any case. But we can see here L is equals to 2. So m should be min -2 to +2. So this is not possible. So that is not permissible. So b is the correct answer. Okay. Next consider the following processes. So this process is given to you that is half of a to b this is given. This is given. Now student hes law is very important. Okay. We don't have any as such formulas. I feel even the student who has no idea about what is thermodynamics they can even solve this. Here you see no chemistry is involved here. This is not chemistry. This is a question of mathematics. His law is not a question of chemistry because reactions are given to you. What you have to you have to see how you can form this reaction by using these three reactions. Now if you add these 1 2 3 here. Let's say you add here half a plus you have 3 b plus you have e plus you have a give rise to uh uh uh I think a should be multiplied by two right h this should be multiplied by two if you multiply this by two you will get a to 2 b right then you add it this is a + 2 uh plus 3 b + 3 b A + 3 B + E + A. Okay. Then here you have 2 B + 2 C + D + 2D. Right? Now let us see what do we have here. Uh we know that this 3 B and 2 B will be cancelled. So this will be B here. This uh will be A. This will be 2 A. Right? This also we have written E. It will be E. Right? Then here it will be 2 C 2 C plus A you need to cancel this right. So you need to reciprocate this o you need to reciprocate this also or this one. So let us say directly only why to use that much hu. So I think so yes they have reciprocated it. This is a + 2 b at least one thing will be reciprocated that is inverse you will take of one thing they have taken inverse of this thing. Let us solve it like that only. Here you have a + 2 b. a give rise to 2 b because you need to make this formula here. Right? And here you have 3 b give rise to 2 c + d. You need to delete a here. That means if here you have a here in this side you should have a. So it is a + e this is 2d. Now if you see that this is multiplied by 2. So this answer will be multiplied by two that is it is going to be 300. If this is reciprocated it will be minus 350. This is going to remain the same. You need to add these three things here. When you add this what you get? This is uh a a will be cancelled right and here you will have 3 b + 2 d right here you have 2 b + 2 c + e right. So here we know that 2 b this is b + 2 d this is 2 c + c right so that is going to be your answer here. So you just need to add 300 + 125 minus 350 and you will get your answer here. Right? That is based on the H's law. No chemistry is involved here. This is direct mathematics. You have to get that equation. Right? Next question is at 500 Kelvin the reaction N2 + 3 H2 give rise to 2 N3 the KP value is given to you. KC they are asking. So directly the relation between KP and KC KP is equals to KC RT delta NG delta NG we know that this is gaseous only you have to take so this is 2 -4 that is minus2 put the value here that is minus2 right so you will get KP is equals to KC by delta NG no divided by RT² KC they are asking you I feel so that will be equals to KP is equals to KC by KC they are asking you right okay so directly you can answer the value of KC here right okay that will be delta NG so KP will be this H minus 2 now so directly you can answer it KC is equals to KP * RT² you don't have to write it like this if this is given in RT sometimes they don't even give you the value of like they give the options in RT only that is even more easy right okay so now you have to put the value of R here now for R remember one thing R can have different different values see this now in physical chemistry this is a confusing part I have seen student even of J advanc they feel it very I don't know why they feel it difficult they get confused where to put the R value here right if things are given in calories or kilo calories you will have two value here If you things are given in jewles right then you will use the value 8.314 that is nothing but 25 by3 right if things are given in let's say uh liter or atmospheric we are talking about this is mostly used in solution chapters right so here we use point here we use 0821 that is nothing but 1 by 12 right so that is how you are going to put the value of r t will be given to you in the exam Right. Okay. Next is elevation of boiling point is 1 molar solution of glucose is 2k and depression. See we know that elevation is TB is depression is want factor of glucose is given. So one half factor will be one. So is KF into mality delta TB. Colative property is very important. Definitely you will see questions based on that. Right? So 1 m that means KB if you see this is also 2k uh depression the depression increasing point is also given to be 2k. So that means delta tf and deltat tb are equal that is 2k right. So directly you can say kf into marity is equals to kb into marity. Here mity of boiling point if you see uh elevation in boiling point. So here the mality is given to be one right. So here I'll put the value one right this is boiling point this is freezing for freezing the mality is given to be two okay so this is KB this is KF they're asking you KB and KF so you have to calculate KB KB is 2 KF so answer will be D is the correct answer here right okay that is not the correct answer D is the correct answer I think by mistake it is there so Huh D is the correct answer here. Right. Okay. D is the correct answer. Right. Next. D uh D sorry. Okay. Next is the halflife of a reaction inversely proportional to the square of initial concentration. See direct formulas. I'll tell you those students who are thinking of skipping chemical kinetics remember these two three formulas. You remember right? T half is directly proportional to initial raised to power 1 minus n. If n is 0 1 2 order will be given to you. You can calculate the dependency here. Next is if units are asked to you then what you can do mole per liter. Mole per liter 1 - n/s n is the order. Right? Third is reaction for first order reaction. Don't because this is even if you have never studied like at least for boards you must have studied right so you just need to know this boards so you have seen the questions of first order reaction that is K is equals to 2303 divided by T log of initial concentration by final concentration if it is given in percentages take initial concentration as 100. If it is given in fraction take initial concentration as one. Right? So these are the crux of the full kinetics chapter I have told you right. What is important and one more formula that is important K is equals to A uh sorry K is equals to A E^ minus E A by RT you know 99.99%. You will see one or two question based on any of this formulas right okay from chemical kinetics. Now let us see this here. Uh they are saying half life of a reaction is inversely proportional right to the square E not that means we know that actually the formula is a n minus one that means n -1 is equals to 2 is order is equals to 3. So order is 3. You just have to equate this a 1 - n or you can also do this thing. It is 1 by a square. So it can be a minus2 here. So you can also write like this that 1 - n = -2. So ma'am n is equ= to 3. Okay. Whatever you do answer will be three is the correct answer. Is that clear? Next. Now here what you have to remember that ma'am this question is given to you. Simple thing is remember that whenever you are given reduction potential right that is directly proportional to oxidizing power right more reduction potential it wants to get reduced it wants to oxidize something it has a very high oxidizing power right oxidizing power is high reducing power will be less reducing power will be less or which thing will be raised oxidation potential will be less. Mostly they will ask you I have seen question on this only. Mostly you will see in previous year question you must have seen question on this only. Right? So reduction potential is directly proportional to oxidizing power here. So if you see this here ma'am how to calculate reduction potential? In reduction potential in reduction always oxidation number will decrease. In oxidation always oxidation number will increase. So if oxidation number is decreasing here right? So here also it is gaining gaining gaining gaining means all of them are reduction potential in reduction potential it is gaining the electron right. So here it is what maximum reduction potential means maximum oxidizing power. So maximum reduction potential is of CO so that is maximum then you have uh PB then you have C then you have BI. So answer will be B is the correct answer. Is that clear to everybody? I can see you guys very very less answering the questions. Why why why why you not answering the questions? Right? Try to answer it right now. Next is uh given the equation given the equilibrium constant. So this reaction is given to you. Equilibrium constant is given to you. Calculate E not value. Right? So students two three formulas you must know. Right? One is the formula of delta G is equals to minus NF E very important. Next is this formula. Here they have given you relation between K equilibrium and delta G. That is very important. Sorry, K equilibrium and E not cells. That is these all formulas are there in the physical chemistry formula session that is a 53 minute session. Okay, you can watch that. E not cell is equals to 0591 by N factor log of KC. KC is given to you here. N factor how you will calculate number of electron loss and gain. So if how you will see this here we can see that Cu is getting converted to Cu2 positive Ag positive is getting converted to Ag here that means it is losing two electrons here this means it is gaining one electron but number of electron should be same so this is multiplied by two this is multiplied by two this is multiplied by two and hence n factor is going to be two here you apply this formula here you will get your answer directly right okay next is very important question right now currently this question is important for you right ma'am okay tomorrow morning my exam don't worry I'll just end it in next 15 minutes you can go and revise physical chemistry formulas naming reactions before sleeping I would advise you then don't study anything in the morning please don't study anything in the morning okay so which of the yes you can study your important notes just for one hour not more than that right uh which of the following Facts regarding ionic conductance of following ions is correct. Right? Simply remember one thing. Ionic conductivity is directly proportional to size of ion inversely to size of hydrated ion. Why? You know in mostly students they remember this by size of hydrated ion. Don't remember by size of hydrated ion because if the size of ion is small more hydrated molecules can attach then the size of that uh ion which is having smaller size the hydrated ion will have more size that is why hydrated ion and iron has opposite sizes right just remember it from the size of iron. So here we know that uh ruby. So this is say friendship. This is CR E right. So if you see here sodium, potassium and rubidium over here the size is decreasing. So ionic conductivity will also decrease. So you will have sodium minimum then you will have potassium. then you then maximum will be your so answer will be D is the correct answer okay understood next question is for an electrochemical cell this question is given to you now why I have taken up this question see n equation is very important right very important direct formula you will use here but the problem with students is formula you can see from my formula sheet also which I have given you. You can see from the revision of 53 minutes of formula also directly but the application is difficult part in this question. Application you have to see. So see how to apply this. First you should know that the formula is E is equals to E not cell minus of you have 059 you can say or here they have given you 6 only right divided by N factor log of this is product by reactant. Now this representation is given to you. Firstly pick out the product by reactant. So here we know that Sn is getting converted to Sn2 positive right PB2 positive is getting converted to PB right that means product by reactant is SN2 positive by PB so I can see that here I'll put log of SN2 positive raised to power the stoometry divided by PB2 positive raised to power the stoometry we can see that here what is happening two electron electrons it is losing or gaining. It is getting converted to SN2 positive. Oxidation number is increasing. It is losing two electrons that is minus2 electron. Here it is gaining two electrons. Here it is gaining two electrons. So n factor also you can write 06. This is also two. Now ma'am how do I answer the value of e not here right? E not. So students for that you should remember this one thing here. Whenever you are calculating the value of E not what you have to see E not cell will be equals to E not of cathode reduction potential minus E not of anode reduction potential please don't make mistake this mistake I know students who know everything they have done enough practice this is where they lag behind right they put oxidation potential here reduction potential you need to put now who will tell us that they have given us oxidation potential or not. Will that be given in the question? See, it will be given in the question. But you have to identify yourself whether it's oxidation potential or reduction potential. See here you have if ox reduction potential means reduction potential means oxidation number should decrease. That is the way to understand. Right? So here oxidation number is plus 2 to0. It is decreasing. It is also decreasing. That means both of them are reduction potential. Now ma'am how to tell which is cathode which is an O? We have justified that both are reduction potential. Who will tell you that which is cathode? Nowhere it is mentioned which is cathode which is an old. Remember in the see one thing is if you can see that here you can see loss of electron here you can see gain of electron. Loss of electron means oxidation means anode. Gain of electron means reduction means cathode. But then is there any direct thing also that you can tell us? So always remember that in this case what you have in this case what you have here this is the salt bridge always before the salt bridge this is anode after the salt bridge this is cathode always it is like that always it is like that AC like you turn on the AC that's how you remember AC right okay so that means in this cathode instead of writing cathode can I write this instead of writing cathode I will Write PB 2 positive2 PB here I'll write SN2 positive to SN you can calculate the E not value here put it here all the other things I have told you directly you can get your answer is that clear to everybody right is that clear rest is the calculation part that you can do yourself right this may seem dangerous to you but this will not seem dangerous to you understood if you know what I am teaching you if you know how to apply this easily you can solve the questions based on n equation right okay next is according to arenous equation whenever there is arinius equation one thing should come in your mind that is e k is equals to a e^ minus e a by rt right now the uh graph is between log of k and 1 by t right for graph based question what you remember convert Convert it into ln of k. This is ln of a. This is minus e a by rt. Convert it into log base 10. That will be log of k is equals to log of a minus 2.303 ea by rt. Ma'am I don't know how to do the conversion but then you have to cram this. Okay. Then it's not in my hands that you have not done enough CBST practice. So this was there in CBST also. If you have not done any like and you have not gone through any coaching, you have not done uh like CBSC or statewood any stateood exam that you are in right for example you have not done any competitive exam preparation this was there in your board exams right this you must have done right so till here now the competitive part I'll tell you right now this is y always it is y versus k okay just Remember, wait a minute. Okay. Always it is y versus x. Okay. So here it is y versus x means here this is y. y is equals to this is mx and this is t. Right? So x is given to be 1x t. So I'll take 1x t as x. So that means m is equals to slope is going to be minus 2.303 303 uh 2303 EA by RT right not only R C that is intercept is going to be log of A so slope if you see that is going to be negative that is intercepting at log of A right and this is log of K this is 1 by T ma'am what if they have asked here ln of k ln of k versus 1x t then you will Take this then this will be your y this will be c this will be x and then this will be your m that is slow right so here answer is minus e a by 230 wait a minute I should have divided it here I don't know why I multiplied it here I'm so sorry beta I should have divided it here by mistake I have multiplied it here right this is e a by 2303r R because here you need to multiply you converting now here you will multiply by 2.303 then you will divide the whole term by 2.303 so that will be 2.303 303 over here. So answer will be this is your slope that is minus E A by 2.303R that is A is the correct answer here. Right? Okay. Next question is regarding inorganic chemistry. Now let us quickly see some questions of inorganic chemistry here. Right? Inorganic chemistry students. Uh first is uh see four ma'am see inorganic is not a big deal to gain marks right. So first is you should do VBT then you should do vases. Basically ma'am do we have to do all the theories in chemical bonding do hybridization right all the previous year questions of hybridization should be on your tips same questions are repeated they are not going to form a new molecule just for you right okay then you should go MOT right is a must must do right these two from chemical bonding then from coordination uh application of VBT one molecule how to calculate magnetic moment all of these how to calculate the hybridization then CFT you must do these are the four topics that are must do after that what you can do exceptions of inorganic chemistry you can do if you have done this easily you can get full marks in inorganic chemistry it is definitely not a very big deal to get full marks in inorganic chemistry predictable they don't even change the how they can change the compound right they are not going to make new exceptions for you right then you can solve it in seconds Right. So here it is XPF6 on complete hypothesis yield X the molecular formula. So now here you should know this reaction. Right. What my idea is in bit E they will rather give you directly they will ask you the hybridization rather than giving you the reaction. But if they want to ask you some three four tough questions then this can be one of them. This is tough only because of this reaction. Otherwise directly how will you solve it? So we know that they're on ma'am how to apply this central metal atom you see in center metal atom see in what group it belongs to it belongs to group number 18 so eight is the outermost veence electrons right now here we know that oxygen always forms a double bond right so it is 1 2 3 4 six electrons are occupied in bond pair so it has two electrons that means one lone pair you have now if you calculate sigma plus loan pair. Sigma plus loan pair. Sigma is what? Sigma. How many sigma you have? 1 2 3 sigma. 3 sigma loone pair is one. That is four. This should be on your tips that I'm writing it here. If you have two, it is sp. If you have three, it is sp2. If you have four, it is sp3. Five, it is sp 3d. Six, it is sp3d2. Right? So here this is linear trional planer tetrahedral trional trional biparameal octahedral that should be on your tips. Okay. Now this is four that means sp3 geometry is tetraedle. Can you see here what they have asked you? They have asked you the shape. Shape you should know. So for shape what we will do? See simple thing uh I just have molecule here. which I'll just show you so that you don't do mistake in the exams. See these with these D like 3D structures I have explained in the uh full detailed video that I have completed important concepts you can if you're like ma'am I want to see more see these this concepts in detail then you can go through that video right okay so now here this is tetra take out one bond pair put lone pair here because in shape you don't consider the lone pair so what we will do this is nothing but this is pyramidal shape right so what we will do this is the non This is one lone pair. This is 1 2 3 4 5 6. So that is your pyramidal shape here. So answer is XO3. And the shape is pa middle shape here. Right? Don't confuse it with planer. This is the planer shape. Planer shape is in one plane. Right? You don't have any angle with my hand here. Right? But in pyramidal shape you can see the angle with my hand here. Right? So that is a pyramidal shape. Right? Okay. Next is which of the following is paramagnetic in nature. So now if you see the paramagnetic nature here paramagnetic nature students see this very important paramagnetic nature means two things you have one is diiamagnetic one is paramagnetic. Paramagnetic means you should have at least one unpaired electron ma'am. Okay, unpaired electron. Now, how we will solve this unpaired electron students? Very simple thing, right? You will apply MOT here. Whenever these kind of molecules are given to you, you will apply MOT, right? So, here unpaired electron, you have to solve. See, firstly tell me the total number of electrons. Here you have 14 electrons. Here you have 6 + 6 carbon it is 12 electron. 14 - 1 that is 13 electron. This is 16 minus 16 + 2 that is 18 electrons right firstly I'm writing one thing for bond order right because sometimes they ask you bond order right so what if number of unpaired electron and bond order is given to you bond order then what you have to remember then just remember that with 14 you have three okay 14 has three bond order So when you go below 16 17 18 or when you go above that is 13 12 uh 11 and 10. Okay.5 will be subtracted. So this will be 2.5 2 1.51. Here also it is 2.5 2 1.5 and 1. If they ask you the bond order if they have asked you unpaired electrons here right? So what you will do? You will have to make the diagram here. Right? So if you see this here you have 14 12 13 18. So I'll just write it here 12 13 and this is 18. Right? Now one by one we will solve this. Right? For 14 if you have equals to or less than 14 it is going to be 1 2 21 2 uh sorry 2 1 2 1 right? Out of 14 subtract KK star. K star is 8 electrons. See if you have more than eight electrons always subtract eight electrons here. There is no need to draw full diagram here. Right? So here it is KK star you will subtract here. So I'll just uh divide it into four batches so that we can draw everything here. So this will be 8 electrons minus 8 it will be six electrons. So 1 2 3 4 5 6. So this is definitely diamagnetic because it is having no unpaired electron. Right? C2 we know that this also is less than 14 so it will also follow 1 2 sorry 2 1 2 1 right so 12 - 8 is equals to what uh 4 electron so it will be 1 2 3 4 again it is diiamagnetic right n2 positive if you see N2 positive here N2 positive whenever you have ions don't look for the electrons in the 9 look for the electron that is there in the neutral specy. In the neutral specy of N2 you have this thing that is 14 electrons right? So here we have taken out one electron here that is 1 2 3 4 and this one electron here that means this is the unpaired electron that's why this is paramagnetic in nature right one thing you should know that O2 is paramagnetic that is the reason that is the point of this theory so this is O2 we know that that is having 16 electron that is greater than 14 that means it will follow 211 so I'll Write it here. Less or equal to 14 or equal to 14. It is having 21. 21. More than 14. It is having 1 2 21 2 1. Okay. More than 14 it is having 1 2 1. So what I'll do here? This is 1 2 2 1. Okay. So from this 18 I'll subtract 8 electrons from KK. That is 10. Now fill 10 electrons. 1 2 3 4 5 6 uh wait a minute. 1 2 3 4 5 6 7 8 10 electrons. 10 electron. 1 2 3 4 5 6 7 8 9 10. Right. So that is why this is also diagnic. Okay. We have completed whatever they can ask you. How long this lecture is? Maybe 10 minutes more. Okay. So we are proceeding towards ending the lecture only the complex having a spin only magnetic moment. So now the questions are related to spin only. Now remember one thing here which are mostly the questions that I'm teaching you right now are uh like most probable questions in your exam. And if you see this here this is number of unpaired electron this is magnetic moments right. So what we will do here? What we will do? Number of unpaired electron is 1 2 3 4 5. Okay. Then magnetic moment will be 1.7 2.7 3.8 4.9 5.9 something like this. Right? We know that the formula is mu is equals to under root n +2 where n is the number of unpaired electrons. Right? But we don't need to use that formula in competitive exams. Directly you can use this. Right? Now here you will see that this is 2.82. That means directly we can know that number of unpaired electron is two. Now you have to see which of the calling has unpaired electron is two. That's now if you see unpaired electron 2 nickel this is in zero oxidation state. Nickel in zero oxidation state. Nickel is atomic number 28. Now let me know in the comment section who have confusion regarding the electronic configuration. Right? If you have confusion regarding electronic configuration of 3D series. Ma'am please write the electronic configuration of 3D series from scandium to uh zinc. Then let me know here. I'll write it. Okay? If you already know that's good. If you if even one student says I'll write it. Right? So in nickel we know that it is 3d8 that is 4s2 3d8 here. Right? So here we know that eight electrons you have here. Right? So carbonyl if you see there is zero unpaired electron will be there. Right? Because carbon is a strong fant. What what will happen here? You will have d here it will be 1 2 3 4 5 6 7 8 9 10. This is 2 + 8 that is 10 electrons in total. Right? Next we have this thing. This is X -4 is equals to -2. So X= to +2. That is you have total nickel 2 positive. That means two electrons you will take out from 4s. That is 3d8. Right? Okay. Now here if you draw the diagram, this is the diagram. 1 2 3 4 5 D orbitals you have to fill. This florine is weak field again. So it will be 1 2 3 4 5 6 7 8. You can see that there are two unpaired electrons. So already answer will be B is the correct answer. You don't need to see the other. Right? Okay. Can I say that? Uh I can't see any any of you asking for the electronic configuration. That means I think you guys are okay with the like ma'am we know the electronic configurations right? Okay. So next is which of the following ions is colored? See DNF block and coordination chemistry. If you don't know the 3D series electronic configuration, you won't be able to pass that. Trust me on that. You won't be able to do that if you have no idea about the electronic configuration. Now if you see colored now here we have given copper. Copper is what? 4s1 3d10. Copper positive is what? Take out from 4s1 electron it is 3d10. That means no DD transition. See if you want to see colorless colorless either it will be D0 it will be D10. So it will be colorless. We want colored so it's not color. It's not color. Then we have CO2 positive. It is 3D9 that is definitely color right because here you can see 3D9 which is not under the category of 0 or 10. So that is not colorless. Right? that is color. Titanium. We know that titanium is sub T that is T means titanium it is 22 4 S2 3d2 right. So here you take out four electrons it will be 3d0 that means in that category. Venadium is 23 that means 4 s2 3d3 you take out five electrons it will be 3d0 that is under that category. So answer is B is the correct answer. Okay. Yes, everybody. Right. Okay. Next we have next we have this very important question regarding the oxidizing nature of KMO4 with KI. So what you need to remember that when you have KI in the acidic medium what will happen in the alkaline medium what will happen? One it is converted to I2 other it is converted to IO3 negative. So in the acidic medium it is converted to I2. In the alkaline medium it is converted to iodate ion that is IO3 negative. In all the other oxidizing like when you talk about SO42 negative it will do the same thing for acidic and alkaline. But this is the different part here right? Okay. Yes. So that's it. That's it. I'm again asking anybody anybody anybody let me see if anyone of you is asking students please if you write in English that will be better for me to understand right okay okay let's find unity in diversity right okay let's find unity in diversity let's not be any language barrier on this channel that's the reason this channel was created vantu jish right so that all the students right who are preparing for different different exams s they should not uh uh it it should not be like because they don't know Hindi or because they are not good in their regional language that is the reason they are not able to understand the topic so if you know English good to go right and we know that English is very important if you're preparing for engineering exam right so you'll end up in some good MNC that obviously even the placement interviews are in English right so that's why I'm saying uh please I would h I'll be happy if you write the comments in English so that even I can understand what you are saying right okay uh which question you which thing you are asking me to explain this one this question or this one uh what you are asking me Ashman what you want me to explain ma'am can I get session PDF session PDF now you want to get see tomorrow to you will get because I don't think so team will be available right now right so team will not be aail available right now. So right now you can't get the session PDF but by tomorrow it will be updated but I'll ask right okay I'll ask them to create the session PDF on priority and I'll ask if they can do it today right okay okay I'll ask the team I'll just message them now only now uh which question you want we are asking ma'am very good session thank you thank you ma'am configuration please okay uh Uh members was a super chat. Ma'am, where is the super chat? Where is the There is There was a super chat, ma'am. There was a super chat. Where was the super chat, Arun? Where was the super chat? Tell me. You can now give the super chat, Arun. Okay. Yes. Now, if you are asking me about this question here, then in the acidic medium, this I negative is getting converted to I2. In the basic medium this is getting converted to IO3 negative that is a factual question. Okay. Then here what you have if you're asking me about here always remember that whenever you have let's say whenever you have colored compounds there will be DD transition right. So if you have colorless compound the electronic configuration will be D0 or D10. We know that in CU positive it is D10. In titanium 4 positive, this is titanium 4 positive. This is benadium 5 positive. Right? That is zero or 10. Only in this case you have 3d9 that is that is why only this is called. Now for those students who are looking for the electronic configuration see this here right ma'am. Thank you so much. Configuration question please ma'am. Configuration question is nothing. You see this here. This is nu. Okay. Now here just write 4s 4s everywhere. Right? Write 4S 4S and 3D 3D everywhere. 4S 4 S 4S 4S 4S 4S 4 S 4S 4 S 4S. Write 3D 3D everywhere. 3D 3D 3D 3D 3D ma'am 3D ma'am 3D ma'am 3D ma'am 3D ma'am Now ma'am what you have to do this is from 21 22 atomic number I'm writing 25 26 27 28 29 30 Now you see how to write the electronic configuration 2 1 2 1 2 2 3 2 4 chromium Skip chromium. Skip copper. Right. Okay. Then you write manganesees 25, iron 26, cobar 27, then 28, then 29. This I told you to skip. Chromium and copper is exceptional. Okay, that you have to skip. Zinc is 13. Now, should I write 4s3 3D 0? No. Right. S can only have two electrons. Right. So 4 is 2 and what you are left with because here before this you have argan. So this will be 2 and 10 right now here this exception will be this should be 2 and 4 but because of the half filled configuration this will be 1 5 this should be 2 and 9 but because of the fully filled configuration this will be 1 and 10. Right? Okay. Understood? Right? That's how you will answer the electronic configuration. Once you know the electronic configuration, right? Then you can easily answer question like this. What you have to do whenever you take out the electron, you take out the electron from 4s orbital, not from 3d. Right? So we are taking out the electron from 4s. Right? Then from 3d you have zero. We are taking out electron from 4s in 3D then you have zero. Right? Okay. Understood ma'am. Tips tips. tips tips. So for your tips students for see depends upon when you have your uh shift right if you have your exam tomorrow then I would really really advise you it's almost 10 so I will advise you just go through the formulas please don't read anything new go through the formulas of chemistry physics and mathematics right we have taken sessions here in the live you can see our vitri e session over here Right? Wait a minute. Let me just show you that vit e session over this is my organic chemistry reactions. Right? You can see that is just for 1.5 hours. Right? This is my exceptional video. Right? Exception all exceptions in one go. You can watch that. Right? And this we have taken important concepts here of vit right. Let me just show you that important concepts of vitri e. This was the important 50 questions. I'll just ask you to go through VAT triple E important concepts here. Six days ago we have taken most important concepts of maths, physics and chemistry. You can go through that right that will really help you. And you know what we have also taken right as you can see here Shia has taken for English and aptitude as well right because that is already also there in your sy. So go through this English and aptitude that is just for ba go through that because the questions are not difficult in VTE regarding aptitude and uh English general question if you have gone through once you'll be able to do that right okay then I think that's it if you're able to do these questions that will be enough for you right okay that's it now the only thing is vit e long concept sessions were for Those students who don't have their shift tomorrow who have their shift tomorrow don't do anything just revise the formulas that's it and sleep revise the formula sleep that's it right okay for those students who have days left for their exam then don't worry go through these important sessions that I have told you you'll be able to manage right don't worry about that right and for those students who want to get themsel enrolled in the offline coaching ing students, right? So those students who are thinking about taking a job and get enrolled to the offline coaching. So students at this page you will get your Vidamu city center nearby your city. You can see Tamil Nadu, Andhra Pradesh, Telangana, Punjab, Maharashtra, Delhi, Bihar, Orisa, Rajasthar, Kundiri, Karnataka, Uttar Pradesh. Right? So depending upon which is near to you, you can visit the center and you can tell them that okay I am a student of sher sir I'm a student of nam ma'am vika ma'am you can tell them and they will surely help you just have to take our name there right and the counseling the guidance will be given to you and you can also get the scholarship depending upon your marks right so you have 40% scholarship on j courses right you can start and enroll yourself offline coaching and if you're looking for 90% % scholarship then you can go for this IV SAT program that is instant vid without scholarship admission test right students so here ma'am how can I get 90% scholarship here that is register yourself here and give the offline see in offline you have to give the test right 90% scholarship is a big deal we want to see you that yes you have done this only by yourself right no cheating nothing was there right so based on that we will give you your scholarship up to 90% if you have scored good in that exam right so I would advise you if you're looking for if you're thinking about a draw right even if you're not thinking about a drop right let's say you're not thinking about a drop right I'm sure even then give this test at least get yourself enrolled here once you have completed all your exam give this test right that would be your backup plan right that would be an amazing backup plan right Okay. And for all those students who are preparing for J advance then students uh wait a minute where is this where am I? Uh uh uh uh yes. Okay. So if you see this here in the description box students in the description box over here right you will see different different test series right? Then if you're looking for J advanc batch then J advanc rank accelerator book is there for all the students who are preparing for J advance and you know what I'm also in fact Sher Namata ma'am and your dsha ma'am we are going to take up session on this rank accelerator top 20 questions we will take up from this book right so you can see the questions from this book and you can see the value of this book we have already taken so many sessions here based on this Right? Which will give you the value of this book. Right? So here these books have this book has really really interesting amazing questions which will open up your brain for more interesting questions. Right? Okay. So that is for J advance students. And if you're someone ma'am I am totally ma'am my my brain is not braining ma'am right J advance is very near what to do ma'am I am not able to solve the question by myself. Then what you do enroll to e love batch over here. So this E club batch is basically for J advanced aspirants to pick up you from this level of J means to J advanc level. Here previous year questions, club questions, doubt sessions are there in the class itself. You can go through that batch and this will really really help you to fill that gap that you have in J means and J advance. Okay. So that is there. Everything is in the description. If you have any other question then let me know in the chat box. any other sess question and these session how much mark we gain see I have to given you all the marks here right I have given you all the important topics right it totally depends if you're able to apply whatever I have taught you in the exam you might get full marks also in chemistry right it's not like I have like only taken up some question in one question continuously I'm speaking from past how many two hours I think more than two class. So this will help you right. So don't worry about that. You have gained marks but that application part is important right to brush the concept. Please go through the physical chemistry formula video. Please go through the naming reactions. Please go through the exceptional video. This will hardly take you 2 hours if you watch that in 2x speed. That will brush up the concepts for you. Right? You can see if exceptions you don't want to do you can ignore that. would at least go through naming reactions and physical chemistry formulas because that will cover the full physical chemistry and full uh uh in sorry full physical chemistry and organic chemistry regarding inorganic chemistry I have taken theories part here rest you can see my important concept video or you can just even these questions were enough if you're brushing the concepts right okay yes okay so I'll now ask the team right I'll message the team that they provide you the link here only today itself but I can't promise you that but uh yes tomorrow morning it will be provided okay then bye-bye take your students all the best for your exam all the very best and please don't uh panic in the exam you should have your one strategy and that strategy is be relaxed okay there will be so many things that will be odd at the exam hall right you already have given so many other competitive exam you might have given J means also the same thing will be there you will have some unexpected things not only in the paper but also by the examiner also right so those things will definitely uh disturb you will definitely create disturbance in the exam hall but your main target should be you should try to keep as calm as possible and try to do those questions that you know first you do those questions right then go for the difficult ones Please don't get disheartened in the exam that okay this question I don't know what to do if you feel that okay this question I don't know just skip that question it's not like every question is going to give you the same mark right so if you even if you solve difficult question or easy question that is going to give you the same mark so I would advise you firstly solve chemistry because in the previous shifts student have told that okay ma'am that chemistry was very easy right it was predictable questions were there, right? So, firstly try to solve chemistry questions. then go for whether you like maths or physics according to that you can choose right okay then bye-bye all the best for your exam all the best and do your 100% don't be panic don't be don't have stress anything just relax right now uh please take good sleep right now don't wake up the whole night right that will ruin your exam if you have exam tomorrow take proper sleep your brain will only work when you have proper sleep that's my advice to Lots of love and this heart is for you. Uh all the best. Now this is Viksha Koshal signing off. Have dinner and sleep and take rest. Bye-bye. Take care. See you all in the next class. And while I am ending this, I want all of you to write it down in the comment section. Hashtag # # chemistry done. Okay. # chemistry done. If after watching this lecture you have confidence then write down # chemistry done chemone in the comment section. I'll wait for your comments. That will also give me energy that okay you guys have benefited out of this uh uh lecture and you have revised everything in this lecture. Okay, bye-bye. Take care. See you all. All the best.