ORGANIC CHEM TEST-12 VIDEO SOLUTION FOR RE NEET-2026

追加: 2026-06-06

[00:00:00]Goal Nations [Music] Leading Institute Hello Everyone Myself Saurabh Kumar Chemistry Faculty Goal Institute In this video I am going to discuss Organic Chemistry part of test series for REET 2026 Test No. 12 Question No. 111 Very easy here we have started Organic Chemistry The major product of the following reaction is you have to see first of all which is given alkyl halide.

[00:00:30]Its reaction with KOH is in which medium? Is it in aqueous medium or alcoholic medium? If the medium given here is alcoholic, then in alcoholic medium, the reaction pathway that you students will have to take here will be reaction pathway E2.

[00:00:47]Ok? This is the concept. So wherever you see strong bases written, it is written like KOH, NaOH, ethoxy, methoxy.

[00:00:54]And if you are given an alcoholic medium here, then in an alcoholic medium the reaction pathway is E2 and as you know, in E2 the reaction takes place and the intermediate is not formed from the transition state.

[00:01:07]So what do you have to do? From where the halogen will come out, the carbon just next to it will come out from the left side or the right side, H+ will come out. So you think that there is already a carbon carbon double bond here, then in this case the rule of conjugation will apply, that is, if in any compound or in any substrate a pi bond is already present, then after the elimination reaction the new pi bond which will be formed should be formed in that direction from where after the formation of the new pi bond it should be in conjugation with the earlier pi bond and should get resonance stabilized, so if you think that Br comes out from here and H+ comes out from here, between these two, a pi bond will be formed between the place from where Br comes out and H comes out.

[00:01:46]So CH2 double b CH CH2 double b CH what comes next here? Single Bond What comes after that? One H came out. There are two hydrogens here, student. I'm taking out an H.

[00:01:59]OH- will pick up H+. So, from where the halogen has come out, it will be formed with it, double bed CH. So CH2 CH CH double bed CH and what is left at the end? CH3 and so the product formed here has become resonance stabilized with the previous pi bond.

[00:02:15]This is the reason why this product will be manufactured here in major quantities. And if a student does this, he can also take out H plus from here next to where Br came from.

[00:02:25]So see what the difference will be?

[00:02:27]In this it will happen that CH2 double big CH and this middle CH2 will remain in its place as it is.

[00:02:34]And then here, from where the halogen came out, CH2 is formed with a double bond.

[00:02:39]See, will it become major? No, it will definitely become minor. It will become less because it is in double, single, single conjugation. So, pay attention that if ever there is already a pi bond in any substrate, then after the elimination reaction, the new pi bond that you have to make should be made from that direction from where it comes in conjugation with the previous pi bond and gets resonance stabilized. This means that this first option will not happen at all. It has become double, single, single, double, there is no conjugation.

[00:03:07]This will not happen at all. This pi bond has been removed and alcohol has been formed. This is wrong, this will also not happen at all.

[00:03:11]This answer would have been given if the medium here was written as aqueous. If there was water here, then in the aqueous medium, a resonance stabilized carbocation would have been formed here.

[00:03:21]Sorry, it will not be formed. It will not become resonance stabilized. If we had to do carbocation rearrangement here, then in that case, we could have done the reaction by taking SN1. If there was KOH in the aqueous medium, If it was given then aqueous medium is not given, hence here we will say that it would be E2 pathway.

[00:03:35]Given in alcoholic medium. So you will get the elimination done by doing the E2 pathway. According to the rule of conjunction, its answer of 1001 will be third for you.

[00:03:42]Is the concept clear? Just watch Medium. If there is alcoholic medium then you have to take it. If there is alcoholic medium then you have to take it in this way, E2 pathway here.

[00:03:52]And if aqueous medium is given then in case of primary and secondary you have to take SN2 pathway. And if tertiary is given somewhere, it is alkyl halide. And somewhere if it is primary or secondary but resonance stabilized carbocation is being formed then there you have to take SN1 in aqueous medium. But where resonance stabilized carbocation is not formed, in case of primary and secondary, you have to take SN2 in aqueous medium and in tertiary and where there is benzyl or allyl where resonance stabilized carbocation can be formed, you have to take SN1 pathway to get the reaction done.

[00:04:25]Okay student? Therefore, here the answer of 1001 will come third to you. Next let's move on to the question.

[00:04:31]After 1001 there is 102 in which it is telling you a very simple one 2 dibromo butane react with the access of sodamide i.e. NNH2 the major product is given to you in this, vinyl dihalide is given. So saying one to dibromo butane means CH3 CH2 and on one CH here there will be bromine and on one CH2 there will be Br. In this reaction, you have already mentioned that sodamide is available. So let me take a molecule first, NH2, and as you know about NH2, it is a strong base.

[00:05:07]So as you have a strong base, this will be a dehydrohalogenation reaction. Meaning there will be one loss of halogen and one loss of hydrogen. We call this dehydrohalogenation reaction. So first you have to get a reaction with NH2. So a Br will be formed along with a Na+ and an H+ will be formed from the carbon just next to it where there is already a halogen, student. Ok? So with one Na+ a Br- came out so NaBr and with one NH2- a H+ came out from the side i.e. ammonia came out. So what's left?

[00:05:38]In its case, CH3 is left and then CH2 is left here and Br is gone from here, so a pi bond is formed between the remaining CH. So you do not have to say the answer double bound CH NNBr because in the question you have been mentioned access of NNH2 so one to dibromo butane we wrote first reaction with one molecule here the pathway is E2 because NNH2 is a strong base so the pathway of reaction will be E2 here.

[00:06:05]What will happen now? Again, there is still one halogen left.

[00:06:08]So there is one more halogen, so along with Na+, a Br- will come out here, take it out like this and in the case of NH2-, where it has come out from, in the case of Br-, an H+ will come out from the carbon just next to it.

[00:06:22]NH2- will pick up that H+, that is, it will form CH3CH2 C triple bond CH and once again NaBr will be formed and ammonia molecule will be formed. See, is there any such option given? Two Bromo One Butin So Wrong.

[00:06:38]One bromo one butene, this halogen will not remain one. He will have to be removed. It would not be right to say one three butine also.

[00:06:46]If you have been given this access sodamide then one butine will be the answer in its case.

[00:06:50]Ok? So alkyne is prepared.

[00:06:52]Or it could have been given in such a way that it would have first reacted with alcoholic KOH.

[00:06:56]Then if it reacts with sodamide, the answer will still be one butyne. Or if you are given sodamide directly then double dehydrohalogenation will happen directly.

[00:07:05]Dehydrohalogenation will happen twice and each halogen will come out from the two carbons between which the halogen is present and one H will come out from the adjacent carbon, meaning two halogens will break and two H will come out from the adjacent carbon, so four sigma bonds will break and two pi bonds will be formed, student, here the answer of 102 will become third for you.

[00:07:23]Is it clear? I will move on to the next question.

[00:07:25]Next I am telling you that the reaction of SOCl2 i.e. thionyl chloride on alkanol i.e. alcohol to form alkyl chloride gives better yield because of oh brother this is a very easy question.

[00:07:39]Like you have been given alcohol. You know SOCl2 SOCl2, right? So I get it reacted with SOCl2. So what will the structure look like? Let me put a chlorine here. I put chlorine with S double bed O.

[00:07:51]So from here one will go out RCL one will go out in this case it is HCL and in the middle S double bed O and O together will come out SO2 gas, that is when we react alcohol with thionyl chloride then first one alkyl chloride will be formed, along with this SO2 gas comes out from here and along with this HCL gas comes out. This SO2 gas and HCL gas are easily skipped out from here. So this answer which says that the alkyl chlorides are immiscible with SOCl2 is not the correct reason.

[00:08:22]The Outer Product: To say outer product means that the by-product formed apart from alkyl chloride is also called outer product here. Of the reaction are gaseous and escape out. This is correct. Here it automatically skips out.

[00:08:36]We do not need to purify it.

[00:08:38]Otherwise, in everything else, like you take PCL3, H3PO3 is formed there. Or if you take PCL5, POCL3 is formed there along with it. It will have to be purified because it does not escape on its own. Here SO2 gas and HCL gas automatically skip out due to which it gives us pure alkyl chloride. That is why we say that the Doerges method is the best method for making alkyl chloride.

[00:09:01]Ok? This is the Darges Method.

[00:09:03]And alcohol and SOCl2 are soluble in water. It is useless to say all this. The reaction does not occur via carbocation intermediate reaction. This is true but this is not the correct reason. Here the demand of the question is saying that the reaction of SOCl2 on alkanol to form alkyl chloride gives better yield. Why is it giving better yield? The reason for this is that we get pure alkyl chloride.

[00:09:23]Because both SO2 gas and HCL gas easily escape out of the solution. We get pure alkyl chloride. Absolutely correct. This is given to you through the direct line of NCERT.

[00:09:33]So the answer to 104 will be seconds for you.

[00:09:37]Next, you're being told, student, this is a given acetaldehyde and a given propionaldehyde. So what is the actual reaction between ethanol and propanal? It is aldol condensation. See how to do it.

[00:09:47]One is acetaldehyde, one is propanal. So this can happen. Suppose we do this and react two acetaldehydes with each other.

[00:09:56]So what will happen? I will remove oxygen from a molecule. I'm going to take out two alpha H from the second one. Learn how it happens.

[00:10:03]So what do I do first? I make him react with his own self. So self aldol condensation will form four products here. So what do I do in this? OH- I'll take it. So if I add OH- and heat it, aldol condensation will take place here. Ok? So both the molecules have just been taken as identical. So oxygen was removed from one molecule and two hydrogens were removed from the other. So you write down whatever is left. What's left? CH3 CH CH3CH double bond CH and C double bond O with H and student water is out.

[00:10:33]What did it become called? We read this, but to in al, it is formed like this, 1 2 3 4 but to inal, is any option given, it is definitely given, meaning it can be formed, it is telling us that it cannot be formed, so but inal can be formed, after that we make it react with itself, that is, propenal, so if you make propenal react with itself, then you will remove oxygen from one molecule, you will remove alpha hydrogen from the other, which is alpha, this one is methyl beta, alpha H has to be removed, so methyl will remain here like this and I will make this C double broad OH. Pay attention and remove the alpha. Often students make the mistake of removing methyl and then they will get wrong answer.

[00:11:14]Oxygen is removed from one molecule and two alpha H are removed from the other so here formation will take place CH3 CH2 and CH double bed C and here it is CH3 and C double bed O with H and in its case water is formed. This will happen, right? Look brother, here this CH3 CH2 CH has to be removed from one molecule, oxygen from the other two alpha H is removed, so what will be the name of CH double bottom C with CH3C double bottom OH 1 2 3 4 and five, it means we will write its name as two methyl and after that if there is a parent chain of five carbons then it will be called pent and after that it will become two enal, is there any such option, two methyl pent two enal, it is absolutely there, so but two enal is being formed, two methyl pentuinal is being formed. This is not the answer. Now let's look at this. I will make it and cross it out. It became self-contained.

[00:12:06]Once I reacted to it with myself. Everyone reacted to it in their own way.

[00:12:11]Now I cross. Suppose I remove oxygen from acetaldehyde.

[00:12:14]Okay student? I remove oxygen from acetaldehyde and I remove two alpha H from the propanal that you have been given. Ok? So we're going to do a cross reaction where you remove the O from acetaldehyde.

[00:12:30]Secondly, we remove two alphaH from propanal here.

[00:12:32]Like what if this happened? CH3 CH double bond with C and C is done CH3 see CH3 CH and this double bond with C and with this remains methyl and then student C double bond OH is out is water. What will be the name? 1 2 3 4 It means its name will be student 2 methyl and there is a parent chain of four carbons, so you will write but and after that you will write student to enal 2 methyl but to enal, well this is also made, it means the answer is there, 2 methyl but enal one is made, one more question has to be made here, look at the cross I have got done, in the cross I have got done, I removed O from acetaldehyde and two alphas from propanal. I want to do a cross that removes the O from propanal. I work in reverse. No, now I remove O from propanal. And I remove two alpha H from the acetaldehyde. Then this will be a two- type cross, right? In total, you know that four products will be made. If both have Alpha H. If both have alpha H then four products will be formed. So oxygen from here, two alpha H, let me take this out. Like what will happen if this happens? If you heat this OH- then it will become student CH3 CH2 CH double bed CH and C double bed H with O, it will be like this, see it carefully CH3 CH2 CH double bed CH and C double bed OH what is the name? 1 1 2 3 4 and 5 means its name will be student pent to enal, its name will be pent to enal whereas what has been said in it, it has been said hex three enal, this is what went wrong, so but to enal is formed when two acetaldehydes will do aldol condensation with each other, then it becomes but to enal, after that [nasal sound] when will two methyl pent to enal be formed when two propanals will do aldol condensation with each other and the one which is being formed here, two methyl butanal, how will it be formed? When O will be released from acetaldehyde, now two alpha H will be released from propanal, then this will become the fourth option and when we do this here that we release oxygen from propanal and two alpha H from acetaldehyde, then there will be formation of penta two enal CH3 CH2 CH double broad CH C BOOH, this is how it will be formed whereas here hex three enal is given, so this is where you got it wrong. Ok? He is the one who is asking who will not become one. So in 105, the third option you have is hex three inal, this will not be made in its case, student, this is a very good question, I will move on to the next, the correct acidic strength order of the following is, many students got confused from here that Sir, chlorine has -i and here methyl has +i, so this should be less acidic. Hey listen, listen, listen, the one whose name has acid at the end is a real acid and the acidic strength of a real acid is more.

[00:15:34]What is this? It is a carboxylic acid and the conjugate base formed of the carboxylic acid is more resonance stabilized by forming the equivalent canonical structure.

[00:15:45]So it is a carboxylic acid. So remember what this is you see in carboxylic acids and phenols? It is a phenol.

[00:15:52]Which is phenol? Mercury is chloro. Mercury is chlorophenol. And what is this? It is phenol.

[00:15:58]So, one might ask, between phenol and carboxylic acid, which is more acidic? So among phenol and carboxylic acid, carboxylic acid is more acidic. All this is a real acid. So remember, the one whose name ends with acid, it is all a trick. The compound whose name ends with 'acid' has higher acidic strength compared to the compound whose name does not end with 'acid'. So the only picric acid in the phenol family is two four six tri nitrophenol in which we say that the acidic strength is more than carboxylic acid. Later, the acidic strength of phenol is less than that of carboxylic acid.

[00:16:31]So even though you can see methyl here, but in its case the conjugate base that will be formed at the top is more stable due to resonance with this C double bond O.

[00:16:41]Understand? So carboxylic acid is more acidic than phenol. Both of these will be less acidic. Now if we look at both of these, you do not find any +i etc. here. And what do you feel here? +i as well as seems to be hyper conjugation. That is, acidic strength, first of all we will say that carboxylic acid is more acidic than phenol except 24 6 try nitrophenol i.e. picric acid. Picric acid is not present here.

[00:17:04]So directly we will say that both the carboxylic acids will be more acidic.

[00:17:07]Why not put a donor here? There will be C here.

[00:17:09]After that there will be D. Ok? Because there is +i there. Now both of these have phenols. Then here we will say that chlorine adds -i.

[00:17:17]So adding -i will increase the acidic strength. So keep in mind children, if you compare carboxylic acid with phenol, then the conjugate base of carboxylic acid is more stable, the -M of C double bed O is more and at the same time it forms equivalent canonical structure between C double bed O and O-, hence the conjugate base is more stable and carboxylic acid is more acidic than phenol. Even if a donor is included in it. Ok? But among phenols, picric acid is the only exceptional case which is more acidic than carboxylic acid. That means after C there will be D and then here there will be B and the one which is least acidic will be A. The matter is over. So is there any such option as CDBA? Only one option is visible, CDBA. C D B A You people will say this in its case, keep the concept clear that carboxylic acids are more acidic than phenol except pic acid.

[00:18:10]After 106, let's come next 104X, will it be such that if we chlorinate it, it will become Benzo tri chloride, hey brother, what is called Benzo tri chloride, saying Benzo tri chloride means there will be three chlorines here and what will be formed when we hydrolyze it, so it is a simple thing, how much chlorine is there, it is called Benzo tri chloride, how many chlorines are there in it, three, so three chlorine, so three water molecules will be used here, so as many chlorines as there are bromine, that many water molecules will be used.

[00:18:42]So what will happen? Get the scope out of here.

[00:18:44]Eliminate Cl-. If you remove Cl-, how many OH will come in its place on one carbon? Three. So 3 OH on one carbon will be unstable and will be out here water. You know, right? So more than one OH on a carbon is unstable, student.

[00:19:00]The water will be outside here. So react it with as many water molecules as the chlorine given.

[00:19:04]So OH will come where the chlorine is. Now when water comes out, OH- will come out and from here an H+ will come out. So as this happens, what the formation does is C double bottom O and OH, what do we call this? It is called benzoic acid. What else? So wherever it is said in this way that benzo trichloride is given, what will we get if we hydrolyse it? So first you will get benzoic acid. So y we've got benzoic acid. So this will not be the first option. There will be no second because it contains benzaldehyde. Now let's see what x will be in benzoic acid? Whose reaction will we make with Cl2 to form benzo trichloride. Brother, where will you get benzene from? The answer to this is clearly visible that it is going to be Tallinn.

[00:19:48]Because in Tollan itself there is three hydrogen here. So there are three hydrogens.

[00:19:53]So what will we do with this?

[00:19:54]3Cl2 will react with. Will heat. The name of the mechanism will be free radical substitution reaction. And I will throw you out of here, student. If I take out 3HCL, then each chlorine will come here replacing H. Three chlorines on one carbon, this is what we call benzo chloride or benzo tri chloride. Ok? So this is what we call benzo chloride. And when you hydrolyze this, it forms benzoic acid. So there will be no benzene.

[00:20:22]Toluene will be here but it will not be benzaldehyde OI. So its answer will be third because toluene will first undergo chlorination and will form benzo trichloride. Now when hydrolysis of benzo trichloride takes place, here the student has three chlorines on one carbon, so three OH will come, which is unstable. Water comes out and C double bed OH is formed. That means Y will form benzoic acid. Is it clear? It got cleared well.

[00:20:45]Learning and maintaining all these concepts will be useful in the exam.

[00:20:47]Further telling the set of reaction which gives benzene. This is a very easy question. You know it works as a reducing agent. Ok? So it works as a reducing agent. So here H3PO2 will reduce.

[00:21:01]Will form benzene. Along with this, N2 etc. goes out from here and H3PO2 gets converted into H3PO3. So benzene is being formed here using exactly this method. So this is done.

[00:21:11]Now this will not happen here because you know that when nitrobenzene reacts with SNHCl, aniline is formed here, student. Aniline is formed. There is no formation of benzene here.

[00:21:23]Here you know that there will be reduction here also. After reduction, ZnO is removed and benzene is formed completely. So A is done and C is done. There is also a good D. So D has acetylene over here. So with iron, you know, you do polymerization. So they also definitely do the formation of benzene.

[00:21:41]So these are three reactions where benzene will be formed. So ACD is such an option only in the first and no other option is matching.

[00:21:49]In B, aniline will be formed. There will be no students here. So A C D will form benzene there. The answer of 116 will be first. Is it clear? It's easy. If you all read the questions carefully then each question will become a student's answer. No, there is going to be a problem. Ok? The problem arises when we start ticking something without understanding it. Now match column one with column two. How will you solve this? What is this? is decarboxylation. Look, this is soda lime.

[00:22:14]Ok? What do we call the mixture of CaO + NaOH? It is called soda lime. This will be decarboxylation.

[00:22:22]So what happens in decarboxylation?

[00:22:24]I have told you many times. Ok?

[00:22:26]Wherever you have to do decarboxylation.

[00:22:28]So RCON, this NaOH is what actually goes into the reaction. And the first thing that happens is the addition of the nucleophile. Ok? And the addition of the nucleophile will take place. When it opens, it will form R CO-, along with it one Na+ will go and here one ONa is already there, OH is added.

[00:22:50]Addition of the nucleophile occurs first. Then in the second step this R is eliminated. So whoever is here in place of R. The C double bed which is directly attached with O is the same one which comes out here, taking electrons to R-, then one O- will give electrons to it and Na+ will come with R- and C double bed O is formed, on top of that one O Na is already there which is given by the one above and one OH which has been added is there.

[00:23:15]Finally, this R- that you can see, this R- works as a base and the same base picks up H+ from here. So by picking up H+ it becomes RH and Na2CO3 is formed, one ONa is already there and this Na+ will come here with O- i.e. Na2CO3 is out. There is nothing for students to do.

[00:23:35]Wherever this [ __ ] is attached, you make a carbonion on that carbon.

[00:23:41]Here the intermediate carbo anion is formed on the carbon which is directly attached to the C double bond O.

[00:23:48]Carbanion is formed here. And take out one H+ from NaOH and this same H+ will get added.

[00:23:54]That means it will form PH CH3 and PA CH3 in its case, what do we call students? Tolin says. So see, you will get the answer of A here in seconds.

[00:24:04]As soon as A's second happened, I got it. I got all the answers. Now you see what this is? This is also decarboxylation. But it had CH2 in the middle. There is no CH2 here. So what happens if there is no CH2? A negative charge will be formed on the phenyl.

[00:24:14]What does phenyl mean? C6H5 and in this C6H5, take out one H+ from here and this H+ will go and get added to where the carbo anion is. So what will we get from C6H5? C6H6 i.e. benzene will be formed. So A's became two and B's became one. Benzene will be formed because here H plus will be added with phenyl.

[00:24:37]What kind of reaction is this, children? This is chlorobenzene for you. And the reaction of chlorobenzene is with sodium metal. So here the formation will take place of diaryl i.e. diphenyl. Ok?

[00:24:49]This is called Na biphenyl. So this is what happened, in its case it will form biphenyl.

[00:24:54]This reaction is called phytic reaction. Phytic reaction. You have to take two here and in between these you have to place two sodiums here and then place them here.

[00:25:03]Another such chlorobenzene. So I'll take out 2 NaCl. So here two benzenes will join together. This is called biphenyl and it is the ozolysis of benzene. So you know that when benzene is ozolyzed, there is an alternate double bond in benzene.

[00:25:18]So here if you are doing O3 zinc in the presence of water, it means this is reductive ozolysis. So reductive ozolysis is complete break it down.

[00:25:28]You will put one oxygen on each side just like a rasgulla. So now the formation that is done from here is the formation of glaaol by the student. Whenever you do reductive ozoloysis of benzene, there is one hydrogen on each carbon. So this will be divided into three parts. HC double bed O H C double bed O C double bed OH H C double bed O C double bed OH H C double bed O and C double bed OH i.e. three molecules H C double bed O C double bed O are formed with H. This is called student glyol, look at it. So glyol will be formed.

[00:26:05]A became two, B became one, C became four, D became three i.e. glyole. So you will get the answer of 11 2017, student. is very easy.

[00:26:16]You people might not be facing any problem.

[00:26:18]You guys must have made it. This is a good learning question for you guys.

[00:26:22]Next comes after 11 2017.

[00:26:26]This is 118. This is also a very good question.

[00:26:29]This is also a very good question, student. What has been given here before?

[00:26:32]KMO4 KOH So KMnO4 KOH i.e. strong oxidizing agent is given. So as soon as you are given a strong oxidizing agent and there is toluene, meaning alpha H, then as soon as alpha H is formed, that means formation will take place, student first look at benzoic acid, where is benzoic acid? So what is the first option asking? P and Q are asking, right?

[00:26:52]So P and Q are respectfully. So the P that is here will be benzoic acid first because it is a strong oxidizing agent, so there is alpha H3 in toluene, it will oxidize the entire methyl of toluene to form carboxylic acid, it will be the first option, not the fourth option, because there is alcohol in the fourth, there will be second and third because in the second you have benzoic acid, in the third you have benzoic acid, but here the answer will be second, it will not be third also, because here alcohol has been used, but you know that in this reaction CrO3 in the presence of acetic anhydride follows by hyde hydrolysis, here benzaldehyde is formed.

[00:27:26]Student, where the Itard reaction has been given to you, it is the Itard reaction where the reaction has been done by taking chromyl chloride or by side chain chlorination, there you have been given a reaction in the preparation of aldehyde ketone, when we do Tollen's reaction with CO3 in the presence of acetic anhydride, then first an intermediate is formed where the whole of CHOCH3 is formed and when we do hydrolysis of the same, then benzaldehyde is formed, so here first there is a strong oxidizing agent, so benzoic acid will be formed and further here if I take Cro3 in the presence of acetic anhydride, then by hydrolysis benzaldehyde is formed. In this way you will get the answer of 118 in seconds.

[00:28:04]Ok? There will be no problem.

[00:28:06]This question is given directly to you in the book, where there is itar reaction, this question is given next to it.

[00:28:11]Is the concept clear?

[00:28:13]You don't have any problem till here. So firstly a strong oxidizing agent will form here.

[00:28:16]Benzoic acid and Q which is given here will form benzaldehyde.

[00:28:22]This will be the second answer of 118. After that I move on to the next question, student.

[00:28:26]121 Which of the following is correct? Hyper conjugation is not possible in alkyl arene. It is wrong. Alkyl arene may have possible.

[00:28:36]If there is an alkyl arene that is directly attached to benzene. Let's say there is alpha. So if alpha is three, two, one, anything. It will be possible here.

[00:28:46]This is saying that hyper conjugation is not possible in alkylamine. He is saying wrong. He is saying wrong. Correct is asking us.

[00:28:52]In hyperconjugation the sigma electrons of C-H bond of the alkyl group enter into partial conjugation with unshared p orbital. Absolutely correct. Ok? The vacant p orbital, the sigma bond electron of CH with the vacant p orbital is saying here that there is partial conjugation. The complete thing does not come here in broken form.

[00:29:12]Ok? You guys know hyperconjugation which shows the canonical structure hypothetically. So let us show complete breaking of CH bond.

[00:29:18]That is hypothetical, imaginary. What happens in reality? In reality, the hyperconjugation that the C-H bond does is only partially elongated.

[00:29:27]Partial single bond is formed. You might have noticed that the CH bond which goes into hyper conjugation becomes a little longer.

[00:29:34]Due to hyper conjugation its bond length increases. Compare to a C-H bond that does not undergo hyperconjugation. So here we just show partial dot dot dot dot. The C-H bond does not occur in complete conjugation. It is partial. So it is said that in hyper conjugation the sigma electrons of C-H bond of the alkyl group enter into partial conjugation with unshared p orbital. Correct. When inductive and electromeric effects operate in opposite directions the inductive effect predominates. This is wrong. You know that inductance involves sigma bond. And what is given here is electromeric, pi bond is involved in it. That too in the presence of reagents. So the electromeric effect is dominant over the inductive because the sigma bond is stronger than pi pi, so when the sigma bond is strong and the sigma bond is involved in the inductive, then the electromeric effect is dominant over the inductive.

[00:30:25]Inductive is not dominant. This is wrong. After that saying the polarization of sigma bond caused by the polarization of adjacent sigma bond is called electromeric effect. It is wrong. This definition that you have written is, The polarization of sigma bond caused by the polarization of adjacent sigma bond. This is called inductive effect. Okay, this is the definition of inductive effect. In this electromeric effect, there is complete transfer of pi electrons.

[00:30:48]One atom to another atom in the presence of attacking reagent. This means that pi electron transfer occurs in the presence of the attacking agent.

[00:30:56]That is called electromeric effect. This definition given to you here is wrong. This is being said about the inductive effect. In this you will get the correct answer of 121, student. Ok? Next I go after 121. 124 Which of the following acids is a vitamin? Hey brother, this is easy. You know that there is no adipic acid. Vitamin aspartic acid is not. There is no sacchar acid.

[00:31:18]What do we call ascorbic acid here? It is called Vitamin C. The name of Vitamin C is ascorbic acid. That means this is it, your answer to Vitamin 124 will become third.

[00:31:27]Vitamin C is called ascorbic acid.

[00:31:29]Then which of the following will show geometrical isomerism. Now see where geometrical isomerism will occur? This will not happen in this because here two identical hydrogens have been given on the same carbon of the carbon carbon double bond. This will not happen here and about single bond about single bond there should be two different atom groups on any two carbons.

[00:31:46]So one H here is one bromine.

[00:31:48]But if there are no two different atomic groups about the single bond on any other carbon then this will not happen. There is no single bond in the open chain. There is no geometrical isomerism in open chains about single bonds.

[00:31:58]He went to work first. This will not happen here either because there are two methyls on the same carbon. So when there are two methyl or two H on the same carbon of carbon carbon double bond, it means there are two similar identical atom groups. There is no geometrical isomerism. This will be your answer.

[00:32:14]Why would it happen? Because there's a hydrogen here. very good. And here's a bromine. So first of all you have shown that there are two different atom groups on this carbon of carbon carbon double bond.

[00:32:25]After this come here. So you go down, you go up, you get CH2.

[00:32:30]What do you get if you come down? Please watch CH2 carefully. This is a very easy question.

[00:32:34]Both carbons of a carbon-carbon double bond must have different atomic groups. So above, H and Br got separated. But below is CH2. But what's next here after CH2? Next you have H and Br and what do you get next to that? Next to this we find CH2 here. So, you got something different, right?

[00:32:49]Regarding this carbon carbon double bond, here we get CH2 CHBr and here we get CH2 CH2 CH2, so here it is something else and here it is something else, so it is different, this means the answer you will get here, student, will be number four, it is clearly visible here that H and Br are two different things, regarding this double bond, here it is CH2 CHBr, here CH2 CH2 CH2 is different, so this is something else, this is something else and H and Br are something else, that is, there can be geometrical isomerism here regarding the double bond. It is definitely possible here. The answer to 125 will be fourth here. Is it clear? I will move ahead.

[00:33:21]126 This is also a very easy question. It is important. See what happens when you get dehydrated? First of all sulphuric acid will give H+ and when it gives H+ then it will pick up H+ from here. OH HSO4- and this lone pair here will pick up H+.

[00:33:39]So as soon as it picks up H+, this will happen here and in its case it will become H2O plus like this and then H2O+ will come out from here and will come out as water and as soon as it comes out as water, in its case these two methyls remained here, positive charge was formed here, okay and as soon as this is done here, now I will do one to methyl shift, okay and after doing one to methyl shift, this methyl will come here and after this methyl comes here, it gets formed in such a way where this carbocation will be formed here. You know this carbocation is secondary, less stable, this is tertiary, more stable. Now with respect to this more stable carbocation intermediate, HSO4- which is acting as a base will remove H+ from the adjacent carbon. It will remove H+ from here. The job of the base is to pick up H+.

[00:34:30]Now if H+ comes out then methyl will remain here.

[00:34:32]In between this, a pi bond will be formed and HSO4 - will come out taking H+. That means the catalyst used, H2SO4, is gone.

[00:34:39]This option will not be major and this will also not be major. This too will not be major. The major will be according to Setzeps rule. So according to setzeps rule will be.

[00:34:49]First make H2O and take it out. Make carbocation.

[00:34:51]Do the methyl shift and then remove H+ from the more stable carbocation next to it with respect to the carbocation formed and the more substituted alkene will be the major product according to set Jeppes rule.

[00:35:02]You will get the answer of 126 here, Student Fourth.

[00:35:05]Is it clear?

[00:35:07]Next let's go 127 The Correct Decreasing Order of Boiling Point is What would be the order of boiling point here? Dye is bromethane, bromomethane, bromoform, chloromethane. Well, I got the answer. The boiling point will be highest here. Because boiling points are directly proportional to molecular mass. Whose molecular mass is higher, meaning the Van der Waals force of attraction is higher and inversely proportional to the branching. Because as branching increases, the surface area decreases. The contact area decreases. Wonderwall F of Attraction decreases. The boiling point decreases. So here the boiling point which will be highest, student, will be of CHBr3. This is what we call bromoform. This is what it is called.

[00:35:48]After that will come dibromomethane.

[00:35:52]CH2 Br2 is called dibromoethane. This is called dibromomethane. After that will come bromomethane. So this is called bromoethane. The lowest will be cro chromethane, it is written that it should be chloromethane, so chloromethane will have the lowest. Ok?

[00:36:08]Because it has the lowest molecular mass. So look, this will not happen because more dibromoethane has been given than bromethane whereas less dibromoform, sorry, has been given in bromomethane.

[00:36:19]Excessive dibromoethane has been given. So there will be more of the bromo form, right? This is given for bromoethane. It went wrong.

[00:36:25]Here more chloromethane has been given than bromethane. This also went wrong. The answer will be third because most of the bromoform because CHBr3 will become dibromoethane. There are two bromines. Bromomethane has one and chloromethane has one chlorine so will have the lowest mass. So you can say that boiling point is directly proportional to molecular mass. And when the mass becomes the same in everyone.

[00:36:45]When the mass becomes the same in all, then you will say inversely proportional branching. And it has given more of bromomethane. Then you're given a reduction of dibromoethane bromopine.

[00:36:54]This is also wrong, student. That means the answer of 127 will get you third. Is it clear? I will move on to the next question. This is very easy. What will be the relative stability of the contributing structure? The first thing to look at is the number of pi bonds, i.e., the octet rule. It has two pi bonds. Isn't it? One to one is the pi bond in this. There is also one pie bond in this. This means X will have the highest stability. In which the number of pi bonds is more. Meaning complete octave. The peacock will be stable. Now this and this have one pi bond each. Both are charged. So here the negative charge is on O.

[00:37:25]That is, negative charge present on more electronegative atom is more stable and positive charge present on less electronegative atom is more stable. So after X, there will be Y here.

[00:37:36]Who will be the least stable? Z because here the negative charge is more electronegative. And the positive charge is less electronegative. He turned it upside down.

[00:37:41]That is not possible at all. If you look at X Y Z here, is there any option matching it? Z Y is X. X Y XZ.

[00:37:49]So X YZ came in the third option. This is directly given to you by NCERT in Some Basic Principles and Techniques of Organic Chemistry. So the answer of 128 will make the student third. The next question given to you is Match the Column Match the nitrogenous bases present in column one with their structures presented in column two. See what happens here? Hey brother, you have given it in a very different manner. The first one is thiamine. So look, this is Adenine's. This is from Adenine.

[00:38:21]Adenine has two rings. Ok? So adenine has two rings. And Ad Amin once I taught you a trick.

[00:38:27]Add amine on top. So this is clearly visible. Add amine on top means add amine on top of the big ring.

[00:38:33]This is adenine. This has become clear. B will have one. Ok? And [sound of clearing throat] what's in thiamine? 1/3 has a C double bound O and one here will be methylated next to the C double bound O.

[00:38:44]So here I think you will get the second of A. Sorry yes A will be second. So A of A you do two seconds. Ok? This will happen.

[00:38:58]Now make B into one. This has become clear.

[00:39:02]In adenine so add amen on top. That is two rings. That is clear. Now as far as cytosine is concerned, if you look at cytosine here, I think it will be the fourth one because what is there in cytosine here, student, there is an NH2 in it. A C is a doubled O. One is NH2. So this should happen.

[00:39:23]This is uracil. This is uracil. When there are two C double bed O.

[00:39:27]NH will remain in the meantime. There will be two C double bed O.

[00:39:30]We call it uracil. So this is uracil. It is clearly visible. That means in its case, you will get cytosine i.e. C, whatever will be there here, it will become four. It will become four. And it is clear that D will be three here. This will be Neurasyl. Please look at the structure a little, Babu. Ok? Explain by molecule. The structure would have been given to you there. Purine Pymidine contains all the structures you need.

[00:39:52]So keep in mind that adenine and guarine have two rings. One six member, one five member. Later, when there is one ring, then in its case you will have uracil when two C double buds are given on O3, after that in its case when one C double buds O one NH2 is given, then we call it cytosine and in adenine it will be on top, so here your answer for this will be second.

[00:40:16]Ok? This will be the second answer.

[00:40:19]You can see the direct structure of this molecule in this. I will move on to the next question. It became 130. 133 Statement one is saying that in this CH3OCH2Cl will show nucleophilic substitution by SN1 mechanism in protic medium. So in protic medium the reaction takes place through SN1 mechanism because the carbocation intermediate formed here will be resonance stabilized. Ok? So keep in mind that where the carbocation intermediate is resonance stabilized, then despite being primary, there will be SN1 pathway.

[00:40:53]He is also giving you protic medium here and the carbocation which is being formed here is also understood to be with lone pair, the carbocation will be highly stable here with lone pair, so if after the formation of carbocation, it is going into resonance, then you can have SN1 mechanism there.

[00:41:08]Absolutely statement one is correct. It is absolutely correct. To me I am saying that this one given to you neopentyl chloride will not undergo nucleophilic substitution via SN2 mechanism easily. That's correct. It ca n't be Easley at all. It may be 1° but you can see the crowding next to it, it is very crowded. So, as you know, what happens in the SN2 pathway? The living group leaves from the front side and attacks the nucleophile simultaneously from the back side.

[00:41:37]But there is too much crowding here on the back side.

[00:41:39]So due to more crowding at the back side, the nucleophile cannot approach the carbon on which the living group is present simultaneously from the back side.

[00:41:50]Meaning this statement is also correct that it does not undergo nucleophilic substitution by SN2 mechanism easily. That means statement one and statement two both are correct. Both are correct students. Ok? After the rest one and one and statement to both are incorrect. The student is wrong. One is correct. This is absolutely wrong. These statements one and two both are correct. Wherever the resonance is stabilized, there will definitely be an SN1 pathway. The number of students has become 133. After that next you will get 135 Which of the following is an electrophilic aromatic substitution reaction. Which one is the ESR here? This is a nucleophilic addition. Please take care.

[00:42:24]Wherever you have aldehyde ketone this C double broad OH. So wherever you have the reaction of aldehyde ketone with Grignard reagent. Student, here nucleophilic addition takes place and this is given to you, aldehyde will be formed, secondary alcohol is formed from formaldehyde, primary alcohol is formed from ketone, tertiary alcohol is formed from ketone, here you will find other aldehydes except formaldehyde, so nucleophilic addition reaction takes place when aldehyde ketone reacts with Grignard reagent, then the name of the mechanism is nucleophilic addition reaction, remember here it is toluene means there is sp3 carbon and reaction with Cl2 takes place in the presence of heat, so wherever there is at least one sp3 carbon, there is at least one hydrogen atom. So if reaction occurs with Cl2 Br2 in the presence of heat sunlight ultraviolet rays H the name of the new mechanism is free radical substitution reaction. This was also not answered. He is telling us electrophilic aromatic substitution. This is wrong. First, nucleophilic addition will take place. The free radical substitution reaction will take place in seconds. This is where the ESR mechanism will be. This is electrophilic aromatic substitution and FeCl3 will remove a Cl-. Cl+ will be formed.

[00:43:31]And the product formed here is chlorobenzene. So this reaction is electrophilic aromatic substitution.

[00:43:37]Now the rest here you know that this is a nucleophilic aromatic substitution reaction. This is OH-. From here OH- will remove this Cl- at high temperature and the product that gets formed here. So there will be a reaction here but it is called SNAR. SNAR means nucleophilic aromatic substitution reaction. Ok? That means a student with an answer of 135 will come third here.

[00:44:02]Where electrophilic aromatic substitution reaction takes place. So where does electrophilic aromatic substitution occur? When you have nitration of benzene and benzene based compounds, sulfonation, halogenation, Friedel Crafts alkylation, acylation. Now it seems that I have solved all the questions here, students. You just have to revise it well.

[00:44:22]Today's question has been very important. You must have learned a lot.

[00:44:27]Keep practicing like this and stay confident.

[00:44:29]Keep yourself healthy. The heat is very intense. Keep drinking water. You have to keep yourself hydrated. You should not go into too much depression or overthinking.

[00:44:40]You have to be completely confident about what you have studied and practice it. The more you practice, the more confidence you will gain and you will perform well in NEET and qualify with a very good rank.

[00:44:50]Ok? Thank you.