ORGANIC CHEM TEST-05 VIDEO SOLUTION FOR RE NEET-2026

追加: 2026-05-26

[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 Rennet Test Number Five Question Number 93 Very simple you are simply being asked what will be the number of sigma bonds and pi bonds? Out of this, structure is most important for you.

[00:00:27]If you are making the structure then you can easily count the sigma bond and pi bond. You know, all the single bonds will be sigma. One sigma in a double bond would be one pi. A triple bond would have one sigma two pi. So first of all we learn to make the structure.

[00:00:42]What are you being told? Hex one n four ain. So saying hex one n4 ain means you have to draw the first six carbons. 1 2 3 4 5 1 6 done. Then you have Hex One In. One in means one two will have double bond and 1 2 3 4 four will have triple bond. Ok? So there will be a triple bond between four and five.

[00:01:05]So there will be four ions and one two per double bond. This will be called 1.

[00:01:10]Now I will balance it. Here there is a double bond, so to balance it two hydrogen atoms are required. So there will be four bonds.

[00:01:17]Here there is a one to three bond, so in this case one hydrogen atom is required. There is a two bond here, so here we need two hydrogen atoms, so it will become a four bond. There are already 1 2 3 four bonds here. So there will not be a single H atom here. Here also 1 2 3 1 4 is already bonded. H There won't be even one. There is one bond here. This means that to balance the tetravalency carbon, three hydrogen atoms are needed to show it. Balance is done. So you've seen Hex One in Four Ain. Now count the number of sigma bonds and see. This single is Sigma. What would this sigma and double bond have a sigma one? Pi then here is single sigma is single sigma sigma sigma sigma and what will happen to the triple bond? There will be one sigma and two pi.

[00:02:00]After that there will be a sigma here.

[00:02:02]A Sigma Sigma Sigma here. You can count and see how much sigma it is. 1 2 3 4 5 6 7 8 9 10 11 12 13 Okay? Total sigma becomes 13.

[00:02:16]All the hydrogens with single bonds with carbon will be sigma and if there is single bond between carbon and carbon then sigma will be double so one sigma one pi will be triple so one sigma two pi matter ends i.e. 13 and three which such option is coming see here it means here the answer of question number 93 will be four you will get 13 and three see there will be no problem in it at all so it is easy sir ji go ahead I will go ahead it is clear here the question which is being given after 93 might have troubled you a bit here because whatever you have studied hydroboration oxidation you have studied alkene what have you done in it alkyne has been given no problem you know when there used to be an alkene then how do you people make it like alkene is given to us it is given carbon carbon double bond and if here in the first we write B2H6 or BH3 and in the second here we write H2O2 in the presence of alkaline medium. So here we say that hydroboration is oxidation and it seems that in this case anti-Mercury Ops rule applies. Ok? And according to anti-Marcobian ops rule OH i.e. negative part goes where the number of H is more. So where the number of H is more, the negative part goes there.

[00:03:28]This becomes alcohol. According to antimicrobial opossum. One pi bond will be broken and the anti-Markovnik option is placed between the two carbons between which the pi bond breaks and the negative part OH- goes to the place where the number of H is more. The positive part went where the number of H is less. You will say Sir, we know this. You tell me further what he is saying here. That's exactly the concept I'm coming to. Please pay attention. Now what to do with this concept? The same concept has to be applied here.

[00:03:53]What has it given you?

[00:03:54]Alecaine. So when given an alkyne you have to break only one of the first two pi bonds. In the case of antimonocarbon capsules, one of the two pi bonds has to be broken and placed here.

[00:04:06]Ok? So if you apply antimonocarbon capsule then CH3 and where the number of H is more than before, there is one here and OH will go there where it is not there, that is, the negative part and if you add the positive part here then the positive part H will remain here and the negative part will come here, you will break only one pi bond.

[00:04:25]Ok? In this way according to antimarkobnikops rule one pi bond will break and OH- will be added here where the number of H is more. H+ will be added where the number of H is less. It is not there here. According to Antimarconixal. You know there's a triple bond here. So this one pi bond is removed and one pi bond is still left.

[00:04:45]And because of the OH directly attached to the one remaining pi bond, we call it the enol form.

[00:04:53]Ok? So this enol form is unstable. Immediately there will be tautomerism. Ok? So in this case what I had to do was I had to do turtermerism. And after doing tautomerism, H+ will go here. If the pi bond goes here, CH- will pick up this H+. So this will become CH3 and if H+ comes here then this will become CH2 and further here it will be C double bound O with H, that is, whatever formation the student does, he first forms propanal. Ok? And look at this reaction, in the previous test which was done, test number four, this question had also come [nasal sound] this is dilute sulphuric acid, so here dilute means water will be in this medium, water has to be broken, wherever you are asked to react alkyne, dilute sulphuric acid in the presence of HSO4, here it seems Markobnik OP's rule, okay, so according to Markobnik's rule, here the negative part OH will be added in the middle where the number of H is less, it is not there here, now the positive part will go where the number of H is more. Such a formation will do.

[00:05:57]This formation has also been done by the students, this is the formation of the in-law form. And wherever the inal form forms in this manner, it becomes unstable. So here you will have to do Turturism. Ok?

[00:06:10]So what can you do from here?

[00:06:12]Tertmerism.

[00:06:13]And see, if you get turmerism done then propanal has already formed on top. And here H+ will give two electrons. It will become C double bond O and the electron of this pi bond will be transferred here.

[00:06:22]So H+ went over CH2-. The one with OH on top, this formation was done CH3C double bed with OO, this formation of CH3 became propanone. So you can see in the first one, according to the antimarcanonical opusal OH is added here. The enol formed there was tertemized to form aldehyde and here it seems that according to the mercantile OH was placed in the middle and propanoal was formed by tertemization. So both propanal and propanone have the same formula. You can see three carbon is here also.

[00:06:53]Three carbon is here as well. One Oxygen One Oxygen. 36 hydrogens here too.

[00:06:57]Here also 325 1 hydrogen. Meaning both of them are functional isomers of each other. Because the molecular formula is the same. The structure is different. And the structure has one aldehyde and one ketone. That means the answer of both these 96 will become third here.

[00:07:11]Because propanal and propanone are called functional isomers of each other.

[00:07:17]How to clear the concept in a good way?

[00:07:19]If it were an alkene, it would have become an alcohol directly.

[00:07:21]But when hydroboration oxidation of alkyne takes place, first enol is formed according to antimarconic obsol. Now further tautomerism occurs and aldehyde is formed there. Is it clear? I will move on to the next question.

[00:07:33]What is Next telling us? After 96, our question number is 98.

[00:07:38]Which alkene is X which we react with HBr in the presence of peroxide and then we have to react the Y formed with KCN to make butane nitrile, so what do we call butane nitrile, calling it butane nitrile means CH3 CH2 CH2 and C triple bond N, you can see how many carbons are there here 1 2 3 4 ok so it is a four carbon parent, in its case you will call it butane nitrile, so you will call it including the carbon atom of CN, this is the parent chain of four carbons here. Ok? Because there is a functional group here with carbon as a point of attachment. So the numbering will be from the same carbon. The name will be butane nitrile. You see how much carbon is there here? 1 2 3 4 is carbon. Are you watching? There is already four carbon here. Here you also get the reaction done with HBr.

[00:08:26]You know first HBr is in the presence of peroxide. This R2O2 means alkyl peroxide. And it becomes alkyl peroxide, here as soon as you will have the mechanism of free radical addition reaction, it will be antibromine, but here it is symmetrical, wherever you want, you can add bromine, add bromine here and make it react with KCN, then it will be SN2 pathway and if it is SN2 pathway then CN will be added here, if you do its nomenclature then it will become 1 2 3 4, the name will be student 2 methyl butane nitrile, which is not saying this, this will be the answer, no, further here here you know, further make it react with HBr, but this R2O2 means alkyl peroxide. Ok? So anti-Markobnik ops rule will be applicable in this case. So according to the anti-Marcobian ops rule, the negative part of Hbr will go to the place where the number of H is higher. So what do you do? Br has to be added here. Where the number of H is more. Hey, there are two Hs here, it will go there and here where there is one H, where the number of Hs is less, H will go there.

[00:09:31]This is what happened with 1 2 3 carbon one became bromo propane. Further, let me react it with KCN.

[00:09:37]So what will happen if I make it react with KCN? The reaction takes place via the SN2 pathway.

[00:09:43]CN- is a strong nucleophile, so it will attack here. will exclude Br-.

[00:09:48]Due to which it will come here C triple bd N. So see, if you add this, how many carbons will it become? 1 2 3 4 Done Na Butane Nitrile. This will be our answer because here there are 1 2 3 carbons and one carbon of cyanide will come, so it will become four carbon butane nitrile. If you think about it, if we react it with HBr in the presence of peroxide, then it will take anticarbonic capsule but in its case, you see, there are already four carbons. So it becomes 1 2 3 4.

[00:10:20]Here if I react one bromo butane and one bromo butane with KCN, just think about it. So as soon as I get the reaction done here, what will happen in this case? Like if I get the reaction done here, it's 1 2 3 4, right? So 1 2 3 4 and a cyanide comes here and gets added. So 1 2 3 4 will become five carbon, right? What will the name be? Pentane nitrile. Because here the carbon of cyanide is also counted in the nomenclature. That also counts in the numbering.

[00:10:47]So there's already 1 2 3 4 carbons here. So one of carbon and cyanide came. So five carbons become pentane nitrile. This will not be the answer. And here you understand that this is CH2. It is CH2. There will be no benefit here. Because you will add HBR. You can add Br wherever you want. You can add H wherever you feel like. So the first formation that we will do is Br will become bromo ethane. Now you will make it react further with KCN.

[00:11:10]This means that the formation that will take place here, C triple bond N 1 2 3, will be named propane nitrile.

[00:11:18]Ok? So this can only form propane nitrile because there are only two carbons, so one carbon will increase to form propane nitrile. There's four carbons here. 1 2 3 4 A carbon cyanide will increase. So this will be pentane nitrile. There is already one two three carbon here. So one carbon will be added when reacted with KCN. So here it will be butane nitrile. That's what butane nitrile is telling you. This means that the answer given by the student here is 98 and it will be second. Is it clear? I hope it's clear. First you apply peroxide effect according to anti-microbial opsol. The negative part will go where its number is more. After that react it with KCN via SN2 pathway. There one carbon will increase in cyanide due to which you have to take such alkyl halide in which the alkyl halide of one less carbon is taken than the number of carbons that are being talked about to make nitrile. Is it clear? Move forward in a good way. This is 98. Then comes 1001 in which pair the second ion is more stable than the first.

[00:12:13]This is a very easy question.

[00:12:14][nasal sound] You know, this is 2 pi electrons.

[00:12:17]How did this compound happen? Aromatic. And this 2 to 4, you know, this negative charge means two electrons. So this is two 2 4 pi electrons. I mean, how did this compound happen? This student has become anti-aromatic. That means the first peacock became stable.

[00:12:29]This is useless because the first one here is aromatic and the second one is anti-aromatic. Four pi electrons, this is of no use. Look here, what is the first one in this? This first one is our aromatic compound. And this one, sorry, is anti-aromatic, isn't it? The first one. Because there are 2 to 4 pi electrons here and the positive charge does not have any electrons.

[00:12:48]So here in the B one, this is our anti aromatic, student. And if you look at this, 2 to 4 and two electrons of negative charge become six pi electrons.

[00:12:56]So if it has six pi electrons then it becomes aromatic. This is what is saying that the second ion is more stable than the first. This will be B. It will definitely happen because the second is aromatic and the first is anti-aromatic.

[00:13:07]Come into this. Here it is 2 4 2 6 to 8.

[00:13:10]Meaning this became eight pi electrons.

[00:13:12]What will we call this? It will be called anti-aromatic.

[00:13:13]So the work went on and it became 2 4 2 6 here.

[00:13:16]Well, this also happened because the second one is saying that aromatic compound is more stable than anti aromatic.

[00:13:22]So here 2 4 26 2 8 becomes anti aromatic. And here 2 24 2 6 becomes aromatic. Meaning the second one will definitely be more stable than the first one. Correct? And here you will see what this OH looks like?

[00:13:33]+m. Lone pair single positive. So lone pair single positive applies this +m. Whereas here you know that in its case +i will be applied as well as hyper congestion will be applied. Hyper congestion will also occur because what is alpha h here it is 325 to 7 alpha. So here we will say +i edge well edge hyper congestion seems to be there.

[00:13:52]But you know that the mesomeric effect, that is, the resonance effect, is dominant over the hyper conjugation end +i. That means whatever is there in this, the first one will become more stable with respect to the second one. Because +m of OH is acting on this carbocation. There seems to be a resonance effect.

[00:14:08]Whereas its methyl is showing only +I and hyper conjugation is taking place. So that's why we will say that here it will be B and C.

[00:14:14]So the end B given in the first is wrong. Here are B & C in seconds.

[00:14:20]No, this one is absolutely correct. The second option will be B & C. In the third it is saying C & D. There will be no C&D.

[00:14:27]Because in D, first is more stable.

[00:14:29]And in Forth also it is saying C & D.

[00:14:31]Well, what have you done? In the fourth, in the third, there is CD. In the fourth also, there is CD, the one with D is less stable, because he is saying that which is that pair in which the second is more stable, first same second ion is more stable than the first, so here the first is less stable, first is more stable because it is aromatic, in this the first is less stable, second is more stable because it is aromatic, here also the second is more stable because it is aromatic, the first one is anti aromatic and in this the first one is more stable with respect to the second, it must have become clear, that is, its answer will be the second option is exactly in B & C, second ion is more stable with respect to the first ion. clear? I will move on to the next question. It's 10001.

[00:15:09]After that the student says next, this question has already come before. And this is a JEE question. In this you are being asked that for the given reaction select the correct set of reagents. Look wherever you have to go from aniline to orthobromo aniline. So what do you need to do first?

[00:15:24]Mercury will have to be blocked. Otherwise the one with mercury will become major. So what do we do first to block this mercury?

[00:15:31]This aniline is sulfonated. So, when you react aniline with sulfuric acid, after the first acid-base reaction, anilinium hydrogen sulfate is formed. Ok? This is sulphuric acid, a strong acid and it gives off H+ easily. And what happens is a base picks up H+.

[00:15:53]forms anilinium hydrogen sulfate. And when you heat this Anilinium Hydrogen Sulphate, H+ comes out from here and NH2 remains here and the mercury position gets blocked.

[00:16:07]This is what we call sulphanelic acid, you might have read about it. So para amino benzene sulfonic acid means we call it sulfonic acid. Mercury was blocked.

[00:16:16][nasal sound] Further H2O means to react with acetic anhydride.

[00:16:20]So how will it happen if you react it with acetic anhydride? The mechanism that occurs with acetic anhydride is a nucleophilic addition and elimination reaction. So what will I do here? I will break this NH2 down a bit first. So let me do NH2 like this.

[00:16:38]NH- and H+. Ok? As such.

[00:16:42]Now here this will be the addition of the nucleophile and the elimination of the living group.

[00:16:47]So we will see a condition like this. Mercury has to be blocked in advance. Ok?

[00:16:52]What did we do after that? If it is acetylated then it becomes NHC double bed O and CH3 and further who is out here? CH3C double bud with OOO- here the H+ of this aniline is gone. So it's out here CH3C double bd OOOH look carefully. Nothing is there. is very easy. We have practiced such reactions in the classroom.

[00:17:16]Further, sulfonation has already taken place here to block mercury. Because I want to become an orthopedic major. If the mercury remains empty then the mercury person will make you major. We have to block the mercury, then the ortho one will be major.

[00:17:28]Further, this process is done here. is it done.

[00:17:30]Now bromination has to be done here.

[00:17:32]So as you do bromination, you will find the mercury position blocked here. So now this cannot happen on mercury because SO3H is already present here, now the acetylation that we have done here will control the reaction and it will not happen on both the ortho.

[00:17:49]Bromination will take place on only one ortho. Ok? And after doing bromination on one ortho, further heat has to be applied here.

[00:17:54]So if we heat it then it will desulphonate here.

[00:17:57]It also undergoes desulfonation. You know the desulfonation will happen here.

[00:18:01]So if desulphonation occurs then SO3 gas will come out. And the H that is with SO3 will be added to the place from where SO3 has come.

[00:18:09]So this will form NH C double bed O and CH3 ortho, bromine will remain and the H here will remain as H.

[00:18:19]SO3 gas was released. This is called desulphonation. What else should I do?

[00:18:24]Reaction has to be done with NaOH. So NaOH is a strong nucleophile. You know. And this is an amide. C Double bonded ONH is called amide. So what will happen here? This will be the addition of the nucleophile.

[00:18:36]This will break CO+ and this will break NH- I will break this. So like what happened when this happened, student here? H is left.

[00:18:45]Bromine remained here. Who broke down here? NH – It’s gone. And this OH- who will he go with?

[00:18:53]With CO+. So when OH- goes with CO+, CH3C doubled with OH becomes acetic acid. This NH- will act as a base and will pick up an H+ from here.

[00:19:04]And H+ will pick up [nasal sound].

[00:19:07]So now I will go here and make the formation that I want to make here.

[00:19:11]What is banana? I want to make ortho bromo aniline. So you can see that NH- H+ has been picked up and there is bromine on the ortho. In this way, the major product that can be formed here is ortho bromo aniline. So first of all keep in mind that if you want to make ortho bromo aniline from aniline, then first you will have to block the mercury by sulfonation. Then its activating power will have to be reduced by acetylation so that bromination does not occur on both the ortho sides.

[00:19:35]Otherwise, if NH2 is present, there will be a risk of bromination on both the ortho sides.

[00:19:39]So further here by doing acetylation the activating power of NH2 is reduced. If we further do bromination there, then it will not happen on both the ortho sides.

[00:19:47]An ortho bromination occurred. If heat is written then SO3 gets released due to desulphonation. The mercury became empty. Now after that further reaction will take place with NaOH. Addition of nucleophile Elimination of NH-. The same NH- from here when OH- is added then CH3C double bed OOH is formed, this is an acid. Acid will give H+. NH- will pick it up.

[00:20:06]So this will become NH2 and Br name is ortho bromo aniline. That means your answer to 104 should be second. Ok?

[00:20:15]This has come up in the test many times. So in this way you must have remembered it completely.

[00:20:19]Ok? I will move forward.

[00:20:22]What is the next saying in 107? Match column one with column two. Look at it.

[00:20:28]What do we have to do here? Aldehyde has to be brought from CH3CN.

[00:20:30]You know, if you want to get aldehyde from CH3CN, then we can do it in two ways. You can either, Stephen. Stephen, what would you take if you did? You can take SnCl2 + HCl. Or what can you take? Take either of the double H diisobutyl aluminum hydride.

[00:20:49]So SnCl2 + HCl occurs in Stefan reaction. And here I have taken double H, you can also take this. So either one can remain.

[00:20:57]So what does it make first? Imin. CH3 One H is added to carbon.

[00:21:03]With one H nitrogen we call it imine. And when imine is hydrolyzed, then after hydrolysis of imine, NH comes out and acetaldehyde is formed, that is, aldehyde will be formed from ethane nitrile, so see if it is SnCl2 + HCl, it means it will become double H, then it will become diisobutyl aluminium hydride, student, this means first of all let me tell you, the A here will be four, now what you have to do here is this toluene, what to make from parafluorotoluene, parafluorochlorobenzaldehyde, oh get itard reaction done, that is, chromyl chloride CrO2Cl2 will first convert that toluene into chromium complex, now further when hydrolysis will happen, benzaldehyde will be formed, so wherever we have to convert toluene into benzaldehyde, we can take itard reaction here, that is, chromyl chloride in the presence of water. That means whatever answer you get for B will be second. You should not say that this is all I can take.

[00:22:02]You must have read that there we can take Cro3 in the presence of acetic anhydride also. They can take that too. That option is not there. So if CrO3 in the presence of acetic anhydride had formed by hydrolysis, that would also have been the answer. He also makes toluene from benzaldehyde. But here you have taken the Itad reaction. Chromyl chloride prepared by hydrolysis. So this can also happen here. Where benzaldehyde is formed from toluene.

[00:22:26]What's further here? This is alcohol given to you. This alcohol has to be converted into aldehyde without affecting the carbon-carbon double bond.

[00:22:33]Ok? What is this?

[00:22:35]is aldehyde. So if we have to convert it into aldehyde then it will be better for us. Let's take PCC. PCC is a mild oxidizing agent that converts primary alcohols into aldehydes without affecting the carbon-carbon double bond and converts secondary alcohols into ketones. So here you will say that without affecting the carbon carbon double bond primary alcohol has become aldehyde. The number of carbons in an alcohol is equal to the number of carbons in an aldehyde.

[00:22:59]The decrease is not big. This means that the answer given by student C here will be one. Now let me talk about D. So in D you have reduced carbon. You can see here this is one to carbon. Whereas here it is of 1 2 3 4 carbons. This means ozonlysis has occurred.

[00:23:16]So if I do ozolysis then CH3C double broad OH CH3 C double broad OH i.e. the answer of D will be third. It will be third.

[00:23:27]So see, A is four, B is two, C is one, D is three, you will get the answer of 107 as third.

[00:23:35]Third clear concept of 107?

[00:23:38]No problem. is very easy. All these are direct NCERT based questions. If you practice a little, you will become all these.

[00:23:44]Okay, look, this question is very important. What mistake have some children made here? Here the lone pair of OH is conjugated. That was also removed. Hey brother, when it is phenol then the reaction of phenol cannot happen with Hi HBr HCl. This lone pair is conjugated and whichever lone pair of OH is conjugated with the carbon carbon double bond, how will it react with Hi HBr Cl? There will be a partial double bond. So here the reaction will actually be carbocation resonance stabilized so it will be SN1 pathway and in its case here I remove water and after removing water I will first write whatever is left, what will be X? X This OH will not go away at all. The resonance is stabilized and will not move.

[00:24:24]Here the carbocation will be resonance stabilized.

[00:24:26]This will become CH2Cl and it will come out as H2O. And here comes chlorine. Further, its reaction has to be done.

[00:24:33]Finn Calstein reaction ho gaya na ye.

[00:24:35]So further here you have to do Finn Calstein reaction i.e. reaction with sodium iodide. So remove NaCl from here and as soon as you remove NaCl, this OH will remain in its place. Nothing will go wrong with it.

[00:24:46]And it will come here CH2I, see if there is any such option?

[00:24:52]Look [nasal sound] CH2IOH, the first option is coming. So please be very careful that you can never react OH i.e. phenol with HCL HBr HI. The CH2 that comes in the middle here, then OH is not in lone pair conjugation. There is no partial double bond. But there is a partial double bond here. Due to resonance it does not break here.

[00:25:12]In alpha unsaturation, SN1 SN2 reaction does not occur.

[00:25:14]He will not move. There will be only one answer to this.

[00:25:16]After this, iodine will not be removed from the work. Only chlorine has been left here. While asking for product y is asking us this, y is not asking for product x. This is our x.

[00:25:27]Now after that it cannot happen here also because if there is a reaction with NaI here finally iodine will be formed.

[00:25:32]clear? So the answer to 119 will be one.

[00:25:36]Next I move towards the question. 113 Which of the following reactions will not form acetaldehyde? Hey brother, it is very easy.

[00:25:43][nasal sound] Do you know what reagent this is here? This is called John's reagent. Cro3 plus sulfuric acid. This is a very powerful strong oxidizing agent. Ok? So what will this make? It is a strong oxidizing agent. So this will actually create acetic acid.

[00:26:02][nasal sound] Acetaldehyde won't form here. We got our answer.

[00:26:04]Acetic acid will be formed here.

[00:26:07]Acetaldehyde will not be formed. This is the answer.

[00:26:08]This is the Walker process. Remember this. It is written PD2+ CO2+ hydrolysis, right?

[00:26:14]So here in the Walker process, acetaldehyde is formed from ethene.

[00:26:16]Remember this directly.

[00:26:18]All this is called industrial method.

[00:26:20]So you make ethene react with O2 in the presence of PdCl2 or CuCl2, it is written like this. Sorry it is written CuCl. Or you will see something like PD2+ Cu2+ written.

[00:26:32]So in such a condition you will see CuCl2 PDCl2 or PD2+ Cu2+ written in the presence of water. Air will definitely be written.

[00:26:41]So acetaldehyde is formed from ethene.

[00:26:43]Correct? You know here too. I have just told you immediately. Double H will undergo reduction first and then imine will be formed. Hydrolysis of the imine will occur.

[00:26:49]Aldehyde will be formed here.

[00:26:53]As much carbon as cyanide.

[00:26:55]And you already know this. When oxidation occurs with copper, aldehyde is formed from primary alcohol. Secondary produces ketone and tertiary produces alkene.

[00:27:04]Oxidation does not occur. Dehydration occurs in the tertiary period.

[00:27:06]So this is also not an aldehyde. This sorry it's making aldehyde.

[00:27:10]This is forming aldehyde. This is forming aldehyde. What are you asking? Not form Not form acetaldehyde. So this is John's reagent. It is a strong oxidizing agent. That will form carboxylic acid.

[00:27:22]Carboxylic acid will be formed directly from primary alcohol. So the answer to 113 will be one. Is it clear in a good way or not? This John's reagent is a strong oxidizing agent. Primary alcohol will form an acid. This is very easy.

[00:27:33]You must have seen this question many times.

[00:27:35]So in this it is saying that the incorrect matching pair is elements, it is given on one side, ppt colour is given on one side, in the lysis test, nitrogen gives phosphorous blue, sulphur gives violet colour, oh chlorine gives white ppt, white ppt is given here, blue is given here, it is wrong, phosphorus gives yellow ppt, it is absolutely correct, so nitrogen gives blue, sulphur gives violet or black also and phosphorus gives yellow, chlorine does not give blue here, it gives white colour here, AgCl gives ppt. Ok? So the answer of 116 will be third for you. It became third. Chlorine does not turn blue.

[00:28:13]After that see which of the following compound gives orange ppt with two four DNP but do not give silver mirror with tolerance reagent. Hey brother, listen, the test of two four DNP i.e. two four di nitro phenyl hydrogen is both aldehyde and ketone.

[00:28:33]Both aldehyde and ketone give. Two four DNP test gives aldehyde as well as ketone.

[00:28:38]Ok? So if any compound is reacting with two four DNP, it means that it is confirmed that there is carbonyl there. It could be an aldehyde or a ketone.

[00:28:47]But who gives the Tolent test? Gives aldehyde. Does not produce ketones.

[00:28:52]So here's the aldehyde. Talent will give the test. Here is aldehyde, the Tolent test will give. Fourth also has aldehyde.

[00:28:58]Talent will give the test. While saying Do Not Give Silver Mirror with Tollens Reagent. This will become ketones. Ketone does not give a tolerable test. Whereas two four DNP gives both aldehyde ketone. So that's what he's saying. That means the answer of 118 will get you third. The matter is clear. I'm going [nasal sound] to the next question.

[00:29:19]123 is saying Which of the following has the highest boiling point? Look what is being said in this. The first one is mercury nitrophenol. Intermolecular hydrogen bond will be formed in it.

[00:29:29]Then there is meta nitrophenol.

[00:29:34]Intermolecular hydrogen bond will be formed here also.

[00:29:37]This is ortho nitrophenol. This will be the least. Intra molecular hydrogen bond will be formed here. It will be very weak. There's phenol here. This will also form intermolecular hydrogen bonds. Ok?

[00:29:51]Here, the maximum molecular association occurs through intermolecular hydrogen bonds, when there is nitro at the para position, so the molecular association of paranitrophenol is very high.

[00:30:04]Due to which the highest boiling point here will be of para nitrophenol. Keep in mind that the molecular association of para nitrophenol will be more through intermolecular H bonds.

[00:30:13]So the highest boiling point here will be of para nitrophenol. Is it clear? I will move on to the next question. Now he is saying the major product is why in the following reaction is, first tell me what will be its first reaction in its case?

[00:30:29]Tell me quickly what will happen here? The first thing you should know [nasal sound].

[00:30:35]How many degrees of carbon is on this side of the oxygen in the middle of the ether? 1 degree. How many degrees is there here also? 1° The first carbon is to be seen. So on both sides of the ether [nasal sound] whether the first carbon is primary or secondary then the reaction pathway is SN2. And when SN2 pathway occurs then nucleophile I-in will attack where crowding is less. That means the nucleophile I minus will attack where the crowding is less, when both the sides are primary or secondary alkyl, then that means the smaller part will form alkyl halide and the bigger part CH3 CH2 CH with CH3 and this will form CH2 the bigger part will form OH, oh this will go over O- H+ and this part which is the bigger part sorry it is the smaller part will go where the crowding is less I- because we should know.

[00:31:25]SN2 reaction which is inversely proportional steric hindrance. So I- will go where the hindrance is less. The nucleophile will go where there 's less crowding [nasal sound] and the H+ will go to the larger part. Now I have to do a dehydration reaction with this X alcohol that has been formed. So dehydration when I do, remove OH. Remove OH and make carbocation and after making carbocation I know you will say it is primary sir. If E2 occurs then alkene will be formed, H+ will go away, carbocation will be formed.

[00:31:58]Who can make it that long and wide? I won't do that much. I know that if there is a primary then there is E2.

[00:32:02]Carbocation is not formed. This That is that story. I don't want to read into that. I make it directly. Whenever there are these types of alcohols you can make carbocations where there is OH.

[00:32:10]Now make a carbocation and see if rearrangement is possible. Absolutely.

[00:32:15]Because this is primary. Let me shift one to hydride here. I'll shift one to hydride. Let me bring H- from here.

[00:32:22]And after bringing H- here, CH3CH2 and CH3 with C will become CH3 and carbocation will be formed here. Is done? And when H+ is given, sulfuric acid will be left with HSO4-. The same HSO4- acts as a base.

[00:32:39]And according to Setzeps' rule. H+ will leave from where after leaving more substituted alkene will be formed. This is how the reaction happens here. So look at 1 2 3 4 1 2 3 4 to pay methyl should come. Who is like that? This is of no use. This will not happen. And this will be the second option because 1 2 3 4 2 methyl to methyl to butane.

[00:33:06]This is not going to work with one butane.

[00:33:08]That would be a minor. It will be made but it will be minor.

[00:33:10]According to Sedjeps rule we need Major. So according to the set of Zepp's rule, where the number of carbons directly attached to carbon carbon double bonds is greater, that will be major. This will never happen. Here you will get the answer of 133 in seconds. Do you have any problem? This is how to do it first SN2 pathway because both sides are primary. Both sides of the ether if primary or secondary alkyl will be SN2 pathway. I-I will not go there. I-I will not go where there is less crowding. H plus will go where it's bulky. After that, you will heat the alcohol formed with sulfuric acid.

[00:33:42]Dehydration will occur. where OH is form a carbocation. Then make it from less stable carboan by rearranging it. More stable carbocations simply remove H+ from adjacent carbons in their respective positions. The carbon with the highest number of carbons directly attached to the carbon double bond will be major according to the set pockets rule. Is it clear properly? So this is the answer to 133 seconds. clear? Now there is no question after 133 from here. The entire solution is complete, student. Ok? I hope you have learned a lot from this.

[00:34:15]You will be clearing your concepts and whatever concepts you are able to learn from all these tests, you will revise them again and again. The more you revise, the better it will be and you have to focus only on NCERT because you must have seen that this time also, in your three- match exam, most of the questions are from NCERT. Be it Biology, Physics or Chemistry, NCERT is the best.

[00:34:36]Ok? Therefore, there is no need to move around too much. If you keep a hold on NCERT then the result will be yours and that too with a good rank. The one or two questions that you get from outside will be in your copy.

[00:34:46]Whatever is there in the class notes, the teacher must have taught it. It is being given to you in the test you are taking.

[00:34:51]So you have to create the concept. The concept has to be grasped.

[00:34:55]With this all the questions will be made and you will qualify with a very good rank. Ok? Thank you.