ORGANIC CHEM TEST-11-VIDEO SOLUTION FOR RE NEET-2026

追加: 2026-06-05

[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 number 11 starting from question number second you will be given question of Organic Chemistry in which the first question you will be asked is the reactivity order of bimolecular nucleophilic substitution reaction so first of all we should know that bimolecular nucleophilic substitution means SN2 ok so as soon as you are told here that bimolecular nucleophilic substitution reaction it means you have to understand that you are being asked the path of SN2 ok and wherever you are asked the standard rate of SN2 then you should know that no intermediate is formed in SN2 reaction.

[00:01:01]Here the reaction takes place from the transition state and that is why we say here that the rates of SN2 are inversely proportional to the steric hindrance. Ok? So wherever these types of questions are asked to you.

[00:01:17]SN2 is asked by writing. Whether you are asked about bimolecular nucleophilic substitution.

[00:01:21]Here you have to look overall to see where the crowding is less. Where attraction between substrate and nucleophile will be more and repulsion will be less.

[00:01:31]So you see where the repulsion will be the least.

[00:01:34]It has two methyls next to one bromo.

[00:01:36]Went for work. This will never happen.

[00:01:38]Next to one bromo is three per methyl. This will also not happen. There is crowding here too. There is one bromo butane here. There is no branch of any kind in it. There is no substitute. So you are given only one bromo butane. So there is no crowding here. Then you have one bromo 3 methyl. Methyl is a little further away.

[00:01:55]One bromo to two is methyl butane and one bromo to two is dimethyl.

[00:02:00]Students are crowding here the most.

[00:02:02]Ok? So this seems to be the answer I should have. I will show you the structure. This will not happen at all because two methyls have been given next to one bromo. One bromo two two dimethyl. Do you understand? This will never happen.

[00:02:11]So first you look at this. The rate of SN2 will be inversely proportional to the steric hindrance. After that I will create a structure and explain why this can happen.

[00:02:19]Like the first one given to you is one bromo butane. So one bromo butane has the least crowding. One is 1° and the adjacent carbon is 2°. There is no branching of any kind.

[00:02:32]After that you can see one bromo 3 methyl.

[00:02:36]So saying one bromo 3 methyl butane means there is bromine on one and after this the student has given methyl on three and a four carbon parent chain has been formed, this means it has become butane, so one bromo 3 methyl is a little far away, can you see what is coming after that, one bromo two methyl in the third, so there is bromine on one but now the distance of methyl has been reduced, two parent chain has been given and here also there is a four carbon parent chain. Is it clear? You can see that next to 1° there is 2° and next to that also there is 2° but here next to 1° there is 2°, what happened after that? 3° Crowding increased.

[00:03:11]And here, right next to 1°, 3° was given, and the crowding increased even more. And what is in the last? 1 bromo 2 diethyl. So I bromine one. After this one bromo 2 dimethyl. Two to diethyl and three carbon will be the parent chain student.

[00:03:27]What happened here? Propane. Just look at it. 1 2 3 So one pe is bromine. Two pe two-two methyl happened. How much has the crowding become? So all the alkyl halides are primary. But there is a lot of crowding here.

[00:03:39]One, let me tell you here in the last one where there is one bromo two two di methyl propane. There, SN2 reaction does not happen there easily.

[00:03:48]Because the crowding has become too much next door. The nucleophile will have difficulty reaching from the back side.

[00:03:53]You know that in the case of SN2, the living group comes out from the front side. The nucleophile attacks from the back side. There is a lot of crowding here in the back side. Hence the nucleophile will not approach. So the most you have to see first is 1°, you can see this 1°. It became common to all. If you look next to 1°, what is in its case? What is next to 2° and 2° as well? 2° while what is next to its 2°?

[00:04:16]3° Crowding increased, right? Here, 3° is just next to 1°. Crowding increased further. And what's next to 1° here? 4° Crowding increased further. So the least reactive student will be the last one here and one bromo butane where there is no crowding, where there is no branching, will be the most reactive. That means you will get the answer of second here, third. Is it clear here? Whenever the rate of SN2 is asked, it is said that inversely proportional steric hindrance.

[00:04:42]Now you will get the third number, the decreasing order of acidic strength of the following is. Look, this is a very good question. In this, what is being said in the first option CH2 doubled CHCOH, this will not happen because then acetic acid has been given, formic acid has been given, from here it is clear that which is more acidic between acetic acid and formic acid, formic acid is being said to be wrong, okay and if you see in this also, first acetic acid has been said, can it ever happen that acetic acid becomes more acidic than formic, this is wrong because acidic strength is directly proportional -i-M and inversely proportional +i + M and here it has been shown to be the most acidic. It is wrong. And formic acid has been given in it but it has been given wrongly. CH2Cl is given, right? When CH2Cl remains chlorine attached, its acidic strength is more than formic acid. So look, the student is saved here, the third option.

[00:05:35]What will actually happen? The most acidic acid here will be chloro acetic acid. After that there will be formic acid here. Ok? Formic acid. After that I will tell you what will happen here now.

[00:05:46]This is given to you, I am telling you CH2 double bed CH C double bed O with OH. I am telling you, why wait now? Now acetic acid will come last here. So this will come last. Firstly, you know the concept in this case is that acidic strength is inversely proportional to PK. Or we also say that acidic strength is directly proportional -i-M and inversely proportional, students +i +m, you all know all this, formic acid and this propinoic acid given to you, why is formic acid more acidic because you know that it must have been written in NCERT when you read the topic of carboxylic acid, where the acidic strength of carboxylic acid is explained to you completely by looking at the arrow, it is written there that phenyl and phenyl vinyl and phenyl. We think that the acidic strength there needs to be reduced a little. But in respect of the sp3 one, it becomes more acidic because with whom is it attached?

[00:06:46]with sp2. Whereas here it is attached with whom?

[00:06:48]With sp3. So this is attached to acetic acid with sp3. Attached here is with sp2. The electronegativity of sp2 is higher than that of sp3.

[00:06:56]So if the carboxylic acid is attached to phenyl somewhere or is attached to phenyl. That is, carbon-carbon double bond is attached with sp2 carbon, its acidic strength increases as compared to here which is attached with sp3.

[00:07:11]So our expectation is that the acidic strength should decrease here because what does this carboxylic acid show? -m So what will this show do in respect of -m? + m means it works as a donor.

[00:07:21]So +m is showing. Whereas here you will see something like this in its case, if you look at formic acid, then there is no +m showing in formic acid.

[00:07:28]Whereas here there is a group doing +m show. So by showing +m it reduces the acidic strength a little compared to formic acid. Ok? But here if we compare these two then electronegativity is seen first.

[00:07:42]And remember, if carboxylic acid is attached to sp carbon i.e. triple bond, then it will be more acidic. Then it will be highest with sp2 i.e. with double bond and lowest when it is attached with single bond i.e. with sp3 carbon. So whenever you have sp2 carbon attached to carboxylic acid or sp3 attached to carbon, the sp2 one is more acidic here. In its case, we say this because here it is more electronegative. You will find out about sp2 carbon compared to sp3 in NCERT.

[00:08:08]Wherever the acidic strength of carboxylic acids is compared by arrows, it is given there.

[00:08:13]In this formic and chloroethanoic acid, you can also read the NCERT book where it is given with an arrow, then due to the minus i of chloroethanoic acid, its acidic strength is slightly more. So remember, if the position of chlorine is second then its acidic strength is more than formic acid.

[00:08:33]But if the position of the same chlorine becomes third then it becomes less acidic than formic acid.

[00:08:38]So this is the second position, so it will be more acidic. Let me tell you its exact here, its pK value is approximately 2.85.

[00:08:45]Ok? And what happens to formic acid, if I talk about its pK value here, then it becomes 2.85 and then it becomes 3.75 and then it becomes 4.25 and finally it becomes 4.75. I am telling you this practical data.

[00:09:08]So its PK value is the lowest.

[00:09:10]Due to the -i being used here, it is more acidic than formic acid. But remember, if this chlorine position is second, one two, then it will be more acidic than formic acid. But if you do this thirdly, the distance increases a little and its acidic strength decreases a little compared to formic acid.

[00:09:24]And if we have to compare both of these, then in the second and third, you will say here that it is showing +nm. But formic acid is not showing +n. Therefore the acidic strength decreased slightly. And if you compare these two, then in the last one it is attached with sp2 here.

[00:09:38]Attached here with sp3. So sp2 has higher electronegativity than sp3. Therefore it will be more acidic. Ok?

[00:09:44]You should go to NCERT a little bit about this where it is explained with arrows. So acidic strength inversely proportional pk or acidic strength directly proportional -i -m inversely proportional +i +m. You can see a little bit of this. This practical data is given to you in NCERT. Next let's go after number three.

[00:10:04]This became number two. It's number three.

[00:10:07]After that comes number seven. The ease of solidification of the following carboxylic acids withanol. Do you know what the mechanism of stratification is? The mechanism of sterification is nucleophilic acyl substitution via the tetrahedral intermediate and nucleophilic addition elimination.

[00:10:27]Remember, nucleophilic acyl substitution occurs through the tetrahedral intermediate and the nucleophilic addition elimination mechanism is called sterification.

[00:10:35]So whenever students ask you about the rate of striation. So keep in mind what happens first in the rate of striation?

[00:10:48]Addition of nucleophile. Then the elimination of the living group. That is, the rate of sterification is inversely proportional to the steric hindrance. Ok? are inversely proportional to steric hindrance.

[00:11:02]End of discussion. Wherever the alcohol is variable, the carboxylic acid is variable.

[00:11:07]Sterification will always be inversely proportional to steric hindrance. Because the name of the mechanism is first addition of nucleophile and then elimination of water. So the addition of a nucleophile is what you read as a nucleophilic addition reaction with inversely proportional steric hindrance.

[00:11:21]So who is the youngest? Formic acid.

[00:11:23]Who is after him? Acetic acid.

[00:11:25]Who is after him? Propanoic acid.

[00:11:27]Where is the most crowding? Two methyl. Propanoic acid, that is, into isobutyric acid. So the most reactive will be X. Then it will happen.

[00:11:34]Then it will be Z. The least reactive will be W. Just remember the concept. Rate of stiffification of inversely proportional static hindrance. Because the name of the mechanism here is nucleophilic addition elimination reaction. Whenever sterification occurs, OH- is released from the carboxylic acid and H+ is released from the alcohol and this is an acid catalyzed reaction. It is reversible. The mechanism is called nucleophilic addition elimination. In this way remove OH- from carboxylic acid and H+ from alcohol.

[00:12:06]And this formation that you do is a product of alcohol. Ok? In this reaction, you can see that this alcohol acts as a nucleophile.

[00:12:14]When will he perform this fast attack here? When the size of R is small. R here is less crowded. Only then can the nucleophile attack here easily. Is it clear? So x y z will become one of w seven. clear? Move ahead.

[00:12:28]Look carefully. It must have become absolutely clear.

[00:12:33]13 come after consumption. Which of the following reactions gives isopropyl benzene as the major product? See who can provide isopropyl benzene? The first one can definitely give it.

[00:12:42]Sir, please tell me how can I give it in this case?

[00:12:47]How can the first one give isopropyl benzene i.e. cumene here? I can definitely give it. Because you know that AlCl3 will first remove Cl- and here it will remove Cl- so this becomes CH3 and this becomes CH2 and what is this? CH2 Plus. Cl- was eliminated to form AlCl4 along with AlCl3.

[00:13:05]Now the carbocation formed here will undergo hydride shifting. As soon as you do hydride shifting, it will form isopropyl carbocation. And this isopropyl carbocation will attack benzene, student. And as soon as this electrophile attacks benzene, it will form isopropyl benzene.

[00:13:26]So what do we call this?

[00:13:27]Cumin. That means A will definitely make it. Let's see what happens with B? This pi bond of alkene will give two of its electrons to H+ and will give two electrons to H+ and receive it, then what will happen? CH3 because here it is CH2, one H+ will be added. CH3 is formed and this carbocation will again attack benzene and it will attack benzene.

[00:13:49]We can show students like this. This is what we call isopropyl benzene, that is, in its case we can call it cumin. Now sulfuric acid will definitely be here too. The sulfuric acid will break first.

[00:13:59]Ah, after sulfuric acid breaks down, HS HSO4- H+ will break down here. This is the lone pair of alcohol, alcohol will pick up H+. Whom will he pick? To H+. Now the alcohol will pick up H+ so from here it becomes H2O+. But heat is not written. Dehydration you will not do.

[00:14:20]Whose formation will you make by simply removing water here? This will form isopropyl carbocation here. Do n't make it an alkene. Now the isopropyl carbocation formed here will also attack benzene and the formation it will do by attacking benzene will once again lead to the formation of isopropyl benzene i.e. cumene. Ok?

[00:14:39]You already know what AlCl3 will remove here? As soon as Cl- i.e. Cl- comes out, it will form isopropyl carbocation. And this isopropyl carbocation will attack benzene. So after attacking benzene, the formation it will do here will be isopropyl benzene, that is, all of this will be its answer. Ok? So which of the following reactions give isopropyl benzene as the major product? You will see, it will happen in everyone. The answer student will be all of these here.

[00:15:08]All of this will be the answer.

[00:15:11]First, you will remove Cl- here and form a primary carbocation. By shifting the hydride you will form a secondary carbocation. Its name is isopropyl carbanion. It will attack benzene and form isopropyl benzene. Whenever the reaction of propene is given to you, HF and H+ come from there. Generally H plus comes from HF. H+ will go where H is in higher numbers. Carbocation will be formed in the middle.

[00:15:32]where H is less in number. This isopropyl carbohydrazine will attack benzene to form isopropyl benzene. Alcohol will be provided.

[00:15:38]Now, in reaction with benzene, sulfuric acid and phosphoric acid will give H+. There, first, by taking H+, water will be formed and OH will come out and it will form isopropyl carbocation. Then it will attack and form isopropyl benzene. Similarly, here it is clear that AlCl3 will remove Cl- and form isopropyl carbocation and the same formation will take place. The answer to 13 will be fourth, that is, all of these are clear. Let me move on to the next question.

[00:16:06]13 is done, students. It is very easy. You all must have made it. Let's move on to the next question. Come on brother, let's move on to the seventh.

[00:16:17]13 is done. After 13, why doesn't ours go further? Well, 14 is this double H, diisobutyl aluminum hydride.

[00:16:27]Diisobutyl aluminum hydride can be used to carry out which of the following conversions, ester to aldehyde, can definitely happen.

[00:16:34]Students, it is possible.

[00:16:35]Carboxylic acid to ester no carbo If we want to make ester from carboxylic acid then we make it react with alcohol and there we keep acid as catalyst, generally sulphuric acid.

[00:16:46]So here we will say that to make ester from carboxylic acid, we take alcohol here. Be it dry HCL or sulphuric acid, we use catalysts in all of these. So esters cannot be made from carboxylic acid.

[00:16:58]Now we will make ketone from secondary alcohol, so brother, we take oxidizing agent here. Oxidizing agent.

[00:17:04]While what is double H? is a reducing agent.

[00:17:07]So here we have to reduce ester to make aldehyde. Ok? So here this secondary alcohol will form a ketone with an oxidizing agent. So how will this double H be used? It is wrong.

[00:17:18]How is carboxylic acid formed from aldehyde? It is formed by oxidation. So this is not an oxidizing agent. So if you take double H here then you take Easter, you can take Easter if you want. If you want, you can take nitriles. You can also take nitriles.

[00:17:34]Now, what will the name of the substance which has hydride at the end, i.e., diisobutyl aluminium hydride, actually give? H- will give. The same H- is added here and R'O- means the alcoholic part. Where does this part come from? From alcohol. And where does this part come from? From acid. So in the same acidic part, C will go into double bold O and add H- and R' O-, then it will come out like this. Then we will do its acidic hydrolysis and it will become alcohol. So if you want to make aldehyde from ester or from nitrile. If you want to make aldehyde from nitrile, then if you put double H in nitrile, first imine will be formed and when you do hydrolysis of imine, aldehyde will be formed and or you can do reduction of ester with double H, then it will form you aldehyde, so the answer of 14 here will be one, after the rest there will be no more, in this case, okay, I am going ahead, I am giving the next question 15, the correct reactivity order of nucleophilic addition is very easy, you should know first of all that the rate of nucleophilic addition reaction, the first thing we will say is that all these are inversely proportional to steric hindrance.

[00:18:40]Ok? So, steric hindrance will be called inversely proportional and when steric hindrance is same in all then it will be called directly proportional -i -m are inversely proportional. What does +I + M -i - M do? The electrophilicity of the carbonyl carbon increases. So the more electrophilicity the carbonyl carbon will have, that is, the carbon of C double bond O of aldehyde ketone is called carbonyl carbon.

[00:19:05]So the more the acceptor group, the deficiency will increase i.e. electrophilicity will increase i.e. attraction with the nucleophile will increase. So the first thing you'll say is what is an acetophenone? Ketones.

[00:19:16]What is acetophenone? Ketones. And the rest is aldehyde. So the rate of nucleophilic addition reaction of ketone is less than that of aldehyde, student.

[00:19:25]Because ketone has carbon on both sides of the C double bond O.

[00:19:27]So if there is carbon on both sides then there is more crowding. So if there is more crowding in ketone then the rate of nucleophilic addition reaction will be less.

[00:19:35]And when it is aldehyde, then on one side there will be alkyl and on the other side there will be hydrogen.

[00:19:40]And on the other hand, if there is hydrogen then the steric hindrance will be less. And if the steric hindrance is less then the rate of nucleophilic addition reaction of aldehyde is more than ketone.

[00:19:48]Remember that nucleophilic addition reactions are generally more likely to occur with aldehydes than with ketones.

[00:19:53]Now that this is clear, the first option will not be there because more acetophenone has been given than mercury nitro benzaldehyde. It is clearly wrong. Now acetophenone has been given the most here. This is also wrong in the first option. And here also mercury tolaldehyde has been given. Because here the meaning of tolaldehyde has been given as methyl which shows plus eye. That is also wrong. Answer: Its second came into the picture. Now I will tell you how. So saying mercury nitro benzaldehyde means that in respect of aldehyde there will be a nitro on mercury. Then after that, student here is benzaldehyde.

[00:20:24]So you are already understanding benzaldehyde. Then there 's Paraldehyde. So to call it mercury totaldehyde means that a methyl has to be added to the mercury position of this benzaldehyde.

[00:20:34]And you already understand acetophenone. Acetophenone means C double bond OO has phenyl on one side.

[00:20:42]One side is methyl. So if students from here ask us what will be the order of rate of nucleophilic addition reaction? So brother, first of all, the asteric hindrance is less in aldehyde. High in ketones. The lowest will be ketones. That means it will be acetophenone.

[00:20:59]Now if it adds -m to this, then -m will increase the electrophilicity over the carbonyl carbon. If the deficiency increases, the attraction with the nucleophile will increase.

[00:21:09]So the most reactive will be mercury nitro benzaldehyde. Then there would be benzaldehyde. Then there will be mercury toluidine and the least reactive will be acetophenone. So you can check it here. So wherever you ask about nucleophilic addition reaction, you have to apply the same concept that the rate of nucleophilic addition reaction will be inversely proportional to steric hindrance.

[00:21:29]clear? So the answer to 15 here will be seconds. Next let's move on to the question. It's 15 now.

[00:21:41]Next is 19 here. The Correct Order PK value is asking you. The pK value for the given compound is. Look, even before this we have seen that the relation between acidic strength and PK, the relation between acidic strength and PK is inversely. There is an inverse relation.

[00:22:00]So first of all you have to see the order of acidic strength in this, the most acidic one will have the lowest pK value.

[00:22:07]Ok? So which ones are given?

[00:22:09]One here is alcohol i.e. ethanol. One you have is mercury nitrophenol, one is mercury cresol, one is orthocresol. So the first thing you have to keep in mind is that the acidic strength of the alcohol will be less than that of phenol. So the highest pK value you will get is for ethanol.

[00:22:26]So, they have given the lowest amount of ethanol.

[00:22:28]This is wrong. This is clearly wrong because ethanol will have the highest pK value. It is giving it to you in the second option and fourth option. Remove this third option also because here also the highest pK value of ethanol is not given.

[00:22:43]Alcohol should have the highest pK value because among alcohol and phenol, phenol will be more acidic. Now come to phenol, which has nitro given at the mercury position, students, here we will call it para nitrophenol. Ok? So nitro puts -m on the mercury position and the -m works to increase the acidic strength.

[00:23:02]After that you will see that mercury will become cresol because the methyl which is at mercury position with respect to OH will put minus i a little less, the distance has increased and if methyl is given at ortho position with respect to this OH. So if you are asked what effect does the methyl group show at the ortho and para positions?

[00:23:25]So first of all you will say that methyl forms hyper conjugation as well as +i at the para position and methyl also forms hyper conjugation as well as +i at the ortho position and the way mesomeric effect happens is distance independent, similarly hyper conjugation also happens student, here it is distance independent but the inductive effect which happens whether +i or -i is a distance dependent effect, okay so here the +i of methyl will be less at the para position and if the distance from the ortho position is less then the magnitude of its +i will be more, it will be more +i so its acidic strength will be less but you will say sir in the previous test as you had told in the solution, in that the pK value of both orthocresol and paracresol is same, so if you go to NCERT from practical data then practically the pK value of both orthocresol and paracresol is 10.2.

[00:24:17]That is right. But if you go by theoretical data, then here you will say that the pK value of paracresol is slightly lower because it is more acidic than orthocresol. Whatever you have been asked this time is based on theoretical data only.

[00:24:30]So if we find the value of PK here, then the PK which is given to you, what we have written is the order of acidic strength. This and the least acidic here will be ethanol because it is alcohol. Its conjugate base will not be resonance stabilized. So if you just reverse the pK value of this, then here if I consider it as first, this becomes second, this becomes third, this becomes fourth, then the pK value of first becomes the lowest, after that of second, then of third and the highest will be of fourth, that is, of ethanol, see here the second option will be 19, para nitrophenol, paracresol, orthocresol, you are seeing this, the sign of greater than equal to two is given and the highest pK value will be of ethanol, keep this in mind, there are many questions, all this can happen again, he will repeat it, he can give all these questions, let's come to the next question in NEET, look at the 20 number, first of all, it is saying here that benzene diazonium chloride is a colored gas, this is wrong, this color is colorless, this is correct, it is not colored, so this is a colorless solid, a colored gas has been given, this is telling you wrong, so the assertion is wrong because benzene diazonium chloride What it is is a colorless solid, this colored gas is not present here. Ok? So keep in mind that it is colorless. And then in the region it is saying that direct nitration of aniline leads to oxidation product.

[00:25:57]Terry means Unusual Unwanted Temporary Product. Ok? So whatever its colour is saying, it will be in gaseous form. Can also occur in gaseous form. Can also be in solids. But in its case, at least the gas is saying the right thing.

[00:26:11]But what this colored person is saying is wrong. It is not colored. It is colorless. Please take care.

[00:26:15]And after this, direct nitration of aniline is done.

[00:26:18]Where we keep direct HNO3 and sulfuric acid. There you must have seen that a mixture of para nitro aniline, meta nitro aniline and ortho nitro aniline is formed.

[00:26:28]Where 51% is formed mercury nitroaniline.

[00:26:31]After that 47% is formed here meta nitroline and only 2% ortho one is formed. But the mixture of para nitroline, meta nitroline and ortho nitroline formed by this is a little poor in yield. Ok? What happens here is that the aniline oxidizes and forms an unusual, unwanted, temporary product.

[00:26:52]That unusual, unwanted, and temporary product is called a terioxidation product. This region is absolutely right. So assertion is true but reason is false. Here your answer of 20 marks will become your third. Ok? Please keep this in mind. Assertion Sorry assertion is false here.

[00:27:10]Its assertion was false. The reason for this became true because what is being said there is colored gas, which is wrong. It is colorless.

[00:27:17]clear? Let's move ahead, next we are going to number 20 and then 26. Now see what it is telling you?

[00:27:26]OMe om three is given. And with this access hi is given. How will you react to this? We have seen its mechanism many times. Wherever you are given OCH3 in this manner. Me here means methyl. OCH3 And do it here also OCH3 okay? And if access HI is written then you have to take total 3 HI. Now if you want to take 3HI then you have to break H+ I- here. And it remains to be seen where this bond will break.

[00:27:54]Look student here loan payer is its conjugated. So, on the side where the lone pair of oxygen is conjugated, there will be a partial double bond due to resonance. Ok? So due to the presence of partial double bond here, the bond will not break from here. This bond will break from here. Ok? So the bond will break from here because there will be a complete single bond with methyl. So here's the complete single with methyl. And what happens with oxygen and phenyl is a partial double bond due to resonance. So the bond between O and phenyl will not break. The bond between O and methyl will break and I- will go over the methyl and one H+ will go over each oxygen here and when H+ goes there will be one OH here. Then there will be an OH here. There will be one OH here and three molecules of methyl iodide will be out here. See if you see any such option? It will not be first because iodine has been applied on all three, which is wrong. The second option can definitely be there because two are on OH12 and one is on fourth. Which is the position of OME.

[00:28:53]Benzene will not be formed in this. Ok? How will benzene be formed? Where did this Ome go? And methanol will not be produced.

[00:28:59]If methyl iodide is added to it, it may be possible.

[00:29:01]Methyl iodide will be formed along with it.

[00:29:04]But in its case you will not get methyl alcohol.

[00:29:06]These three will remain OH. It will remain at one to fourth position. Ok? While solving this question, just keep in mind that you should not break the direction in which you are having resonance, i.e., from where the partial double bond is present.

[00:29:16]Wherever resonance is not happening with methyl, there will be complete single bond, there bond will break and I- will attack on it, H plus will come on O, OH position will remain at 1 2 4.

[00:29:27]That means the answer to 26 will be seconds.

[00:29:29]Next let's move towards the question.

[00:29:32]28 The correct order of rate of alkaline hydrolysis of following compound is In today's paper, mostly you have been given the concept of nucleophilic addition because what has been given to you is ester and wherever ester amide, acid halide and acid anhydride these four come, its mechanism is nucleophilic acyl substitution through the tetrahedral intermediate and nucleophilic addition elimination takes place. So here you have this Easter. C double bond OOO CH3 C double bond OOCH3 Wherever hydrolysis of ester occurs in this manner, then you will say that the rate of hydrolysis of ester will be inversely proportional to steric hindrance which is common in all.

[00:30:24]When steric hindrance is common to all.

[00:30:27]COOCH3 is common in all. Then if you look at the variable here, the variable is the atom group attached at the mercury position, in its case you will say it is directly proportional -i - inversely proportional +i +m, okay, this means the more electrophilicity it will have, rc double bound oor, isn't it, when we do hydrolysis that is a different thing, alkaline medium is given here, leave it, you will break the water directly here, okay, and wherever it comes, in this way you will say the name of ester amide acid halide acid anhydride mechanism, student nucleophilic addition elimination reaction will happen and this nucleophile will be added here, now you think using your brain, where will the nucleophile attract faster because nucleophile means electron rich species, where will it attract where there is more electron deficiency, so the deficiency will be more here, where NO2 is attached, so its electrophilicity will increase, delta plus will increase, so the attraction of negative charge of OH- will increase the more positive charge It will increase. That means this -m shows here.

[00:31:34]After that do you know what methyl mercury shows? +i edge well edge shows hyper conjugation. While chlorine shows what on mercury position? -i shows.

[00:31:44]So the highest rate of hydrolysis here will be of A.

[00:31:50]After A, D will be done and then C will be done.

[00:31:53]B will have the lowest. That means one of 28 will be the answer A D C B. So you will see that wherever hydrolysis of ester is asked or rate of sterification is asked, reaction of carboxylic acid alcohol, in both the cases there is inversely proportional steric hindrance and directly proportional -i -m is inversely proportional +i +m. That means overall the concept of nucleophilic addition becomes applicable here.

[00:32:17]clear? I will move on to the next question.

[00:32:21]So the next question is 34. And this is a reaction that you see that is similar to Friedel-Crafts acylation.

[00:32:29]Students, first you have to think from where the bond will break in this. Here this chlorine is directly attached to the carbon carbon double bond. Okay, so the carbon carbon double bond of the carbon carbon double bond when directly attached here is the halogen. If there is a partial double bond then you will say no problem sir ji resonance is there here also, C double bond is there with O also. But do you know when Cl- comes out from here, where will the positive charge be formed when Cl- comes out? Direct carbon will form above the carbon double bond. He will not go into resonance. That will be unstable. But if a positive charge is formed somewhere on the carbon carbon double bond then it does not go into resonance. But if chlorine is released here then C double bed O plus is formed and in this case resonance occurs here through back bonding. So through back bonding, resonance takes place here and C triplet bound O plus is formed. So here the positive charge is stable. Whereas it is not stable here. Ok? Therefore, you should keep in mind that whenever halogen is attached with carbon carbon double bond or carbon carbon triple bond, halogen will be attached with carbon carbon double bond or carbon carbon triple bond, in this case you will not have Friedel Craft alkylation reaction, but in this case you can see that on one side there is carbon carbon double bond with C double bond and on the other side there is C double bond with O, student, reaction will take place there. So this chlorine will not break here. But this C double bed OCL will remove AlCl3 Cl-. Because the Cl+ formed here will be resonance stabilized through back bonding. So you let this chlorine remain in this place and this chlorine will remain here, C double bed O+ will be formed and you will take out student AlCl4-, okay? This is how it will happen. Here the sufficiently stable electrophile is used.

[00:34:14]Now make it attack the benzene that you have given on the benzene ring.

[00:34:17]So all the carbons of benzene are identical. You can attack anywhere. So I make this electrophile attack this benzene.

[00:34:26]So what will happen? C double bed O plus attacked and then here at its mercury position, the chlorine which is there will remain safe and sound. If you look, will this option be there here? Wrong. This will not happen. This means that the attack was caused by removing chlorine, which is wrong. How did this get on the meta? This is also wrong. How did chlorine come to Meta? This position will not change. This position will be exactly on the mercury and the C double bed O plus will attack any carbon of the benzene which has happened here and remove the chlorine from it.

[00:34:54]So chlorine should be here.

[00:34:56]Where did the chlorine go? This was also a wrong option. That means your answer of 34 will be third.

[00:35:01]Absolutely because asking for product A is so this Cl will not come out which is directly attached to the double bond carbon carbon double bond triple bond.

[00:35:10]Is the concept clear? Be careful, such questions confuse children.

[00:35:13]Whom to remove, whom to leave. Ok? You have to keep in mind that it will be attached directly to the ring i.e. the double bond. It will not come out. C will be attached to the double bed O. It will come out because the positive charge formed there will be sufficiently stable. Ok? Now let us come to which of the following compounds gives three methyl to butanol upon reduction with lithium aluminium hydride? No problem, we will reduce it all and see.

[00:35:36]So when you reduce lithium aluminum hydride, what will be formed, student, how many carbons are there here? 1 2 3 4 5 right? That means 1 2 3 4 5 pentane to all.

[00:35:51]Whose name are you talking about making? Three methyl two butanol. Isn't it wrong? This went wrong.

[00:35:54]Because it became two panthenol.

[00:35:58]If I add lithium aluminum hydride here.

[00:36:00]When ketones are reduced, secondary alcohols are formed. So this will remain in its place and alcohol will be produced here.

[00:36:06]So 1 2 3 4 this is three methyl two butanol. The answer has definitely come. It will be seconds. Okay, let's take a look at this.

[00:36:14]What happens? I add lithium aluminum hydride. What is this?

[00:36:17]Aldehyde. When aldehyde is reduced, primary alcohol is formed.

[00:36:23]CH2OH will be formed here and then this will remain here and this will remain.

[00:36:28]So 1 2 3 4 this becomes 2 methyl butane one all, it is 1 2 3 4, right? So, he is not talking about making 2 methyl butane one all, this is an ester, if you add lithium aluminium hydride, in case of ester it will get cleaved from here and after cleavage, the formation it will make, student, is this primary alcohol which is given, 1 2 3 4, so 1 2 3 4 and later this methyl alcohol which you can see will come out, in its case it will become 2 methyl butane one all, whereas let me tell you, he is talking about making 3 methyl two butanol, which is being formed only in your second option. You must have made this. I am sure that all this comes in easy questions. Next Now After 36 39 Which of the following compounds does not give Friedel Craft reaction? Look, Friedel Crafts, whether it is alkylation or acylation, does not happen in two situations. Whether you talk about alkylation or acylation.

[00:37:25]Overall Friedel Crafts alkylation wherever you have an electron withdrawing group. Strong electron withdrawing group must be attached. Like if it is installed somewhere, nitro is installed.

[00:37:36]Wherever aldehyde is attached with benzene. If benzene is accompanied by nitro, aldehyde, carboxylic acid, acid halide, amide or you have ester, sulphonic acid, cyanide, CF3, CCL3, CBr3, Ci3 or OR2+.

[00:38:02]NR3+ is engaged in this manner. That means NH3+ is present here. In this method, if the strong electron withdrawing group is attached to benzene in the presence of de-activator, then neither Fidel Crafts alkylation nor acylation will occur.

[00:38:18]So it will not happen here at all in B. It doesn't happen in A either. Who is such an activator? So in activator you get primary amine, secondary amine, tertiary amine. It does not react because it has been taught to you, I have told you many times in the classroom, even in the test solution, this lone pair of NH2 reacts with LCL3 and that means after the acid base reaction, it forms a salt.

[00:38:48]Here comes the minus. Here comes the plus. And as soon as you know nitrogen oxygen this will be pay plus charge. So here it shows strong minus I. This becomes a strong deactivator.

[00:38:58]This deactivates the ring for further reactions.

[00:39:00]That's why the reaction cannot take place here.

[00:39:03]So Friedel Crafts alkylation aniline whether the aniline is primary, secondary, tertiary reaction does not take place in aniline. Phenol is present but is slightly poorer yield. It definitely happens in Anisole. Phenol is present but is slightly poorer yield. Please take care.

[00:39:18]Too much is not good.

[00:39:20]So where is C&D given? Has anyone given C&D? Hey brother, which one did you give me? Nitro is not present in benzene. C & D A & D is given. B & C is given. Given A and B.

[00:39:34]A & C is given. Look here, in its case, there is no option which says do not give. Well, he is saying do not give. That is why we are getting confused. Here's saying Do Not Give Friedl Craft Reaction. We thought that if he was talking about giving it then it would be in C&D. So here A and B will not give. That's what he is asking. Do not give means that will be the third answer. Did you understand or not? So no matter how strong the deactivator is, it will not be there and the active activator includes primary aniline, secondary aniline, tertiary aniline. If all this is directly attached to benzene then no reaction. Is it clear? So it will not happen in A & B.

[00:40:13]Phenol also has very poor yield. Please take care. It is not very good.

[00:40:17]Poor yield occurs. Now let's come to 41. Fluoroform is less acidic than chloroform. Brother, you are absolutely right. If I talk about acidic strength, then fluoroform is less acidic than chloroform. Absolutely correct. Often the child makes a mistake by using -i here.

[00:40:34]Look what its conjugate base you will see student stability order of conjugate base. So its conjugate base will be CF3-.

[00:40:43]And its conjugate base will be CCL3 minus. Do you know what all this is about in his case? Presence of vacant d orbital because it belongs to the third period, do back bonding here, in its case, here in the case of fluorine, only -i is being applied.

[00:41:07]Whereas here you will say lone pair of negative charge. Lone pair of negative charge delocalized in vacant d orbital of chlorine. The negative charge of the carbanion is delocalized here, which means the lone pair of the carbanion is delocalized in the vacant d orbital of chlorine. Due to which we will say that conjugate base of chloroform is more stable than conjugate base of chloroform.

[00:41:35]It is resonating in the dd orbital.

[00:41:38]So saying that CCL3- is more stable than CF3- due to Pi d Pi resonance. P Pi is the carbon one, D Pi is the chlorine one. He is absolutely right. So both assertion and reason are true and reason is the correct explanation of assertion. So here the answer of 41 will be third. Both assertion and reason are true and reason is the correct explanation of assertion. So fluoroform is less acidic than chloroform. This is absolutely correct and the region is also right because here due to resonance the conjugate base of chloroform is more stable.

[00:42:10]Is it clear? I will move ahead well. Look here now after that 43 the pair of compound which gives positive test with Tollens reagent is which one brother? I have written sucrose for you here. It does provide glucose. Glucose Tolerant Test gives both the failing test. It even gives a bending test.

[00:42:30]But sucrose is non-reducing.

[00:42:32]Fructose will give tolerant test, feling test, Bendick's test because it is a reducing sugar but sucrose will not. This caused a mess. What happened to acetophenone here? Acetophenone is a ketone. I have seen the structure today, just now. Acetophenone is a ketone. It will give hexinal but not acetophenone. The answer will be fourth. Glucose and fructose. You know that all the monosaccharides, whether it is aldose or ketose, all the monosaccharides, whether aldehyde or ketone, all of them give positive tolerance test, failing test, bending test, this is the reason that glucose, fructose are reducing sugars and sucrose is non- reducing sugar. You have been taught all this. So its fourth answer will be 43.

[00:43:16]Ok? Let's move ahead. It has happened till 43.

[00:43:19]After that I feel like I have no questions. It has reached 43. It reached 43. The entire solution to the question was till here, student. Then you will take good care of it. A very good question has been given to you from here.

[00:43:31]To learn a lot of things which you have not yet come across in the test, new things have been discovered and given to you.

[00:43:37]Questions have been taken from good topics. Especially I have seen two-three questions. Regarding the absorption region, where one is called terioxidation product, a very beautiful question has been given.

[00:43:44]Often students make the mistake here that tertiary oxidation product means the unusual unwanted temporary product which is formed along with meta ortho and the one given with mercury. I gave you a diya which has colored gas. He is wrong. It is colorless. It is colorless. It is not colored.

[00:44:00]So, the same line that has been given to you there, a very unique question has been asked. Keep practicing like this.

[00:44:05]Stay confident. Your result will be with a very good rank. And have faith in the one above. Have confidence in yourself. Trust what you have studied. You have to trust yourself the most. You have to pay for your content. You have to do it on your confidence.

[00:44:18]You have worked so hard all year. That hard work will give you very good results. Ok? Thank you.

[00:44:23]Hmm.