ORGANIC CHEM TEST-07 VIDEO SOLUTION FOR RE NEET-2026

Added: 2026-06-01

[00:00:00]Goal Nations [Music] Leading Institute Hello Everyone Myself Saurav Kumar Chemistry Faculty Goal Institute In This Video I Am Going To Discuss Organic Chemistry Part Of Test Series For REET 2026 Test Number Seven Ok Question Number 48 Is Taken From Organic Chemistry. It is very easy. First of all I am telling you that benzoic acid is being reacted with ammonia and heat has also been applied.

[00:00:31]So you should know that whenever you are given reactions like this, students, wherever there is any carboxylic acid, it is not necessary that it will be benzoic acid only.

[00:00:42]If you have reacted any carboxylic acid with ammonia in this manner, then first you have to remove OH- from the carboxylic acid and H+ from ammonia and the product formed after the removal of water will form an amide. The formation of primary amide will be C double bound ONH2 and H2O will come out. This product which is formed here is called benzamide. Ok, it's done. Further, you know, it can be written here as KOBr or it can be written here as Br2 + KOH, what do we call this type of reaction? Hofmann bromide degradation is called HBD reaction i.e. Hofmann bromide degradation. From here you have to remove the C double bond O from the middle and one carbon gets reduced to form a primary amine. So C double bed O here gets converted into K2CO3.

[00:01:44]Ok? The C double bond O is formed outside to become CO2 before decarboxylation occurs. Then further it reacts with KOH there and gets converted into K2CO3. Leave all that aside. Here, whatever number of carbons you have, the given amide will form a primary amine with one less carbon. Meaning, C double bed O has to be removed from the middle and whatever alkyl, phenyl, cyclo alkyl is present here, it has to be formed by adding it with NH2.

[00:02:07]Primary amine. What else do we have to do? Look at it. This X became this benzamide for us. Y became ours, this aniline became. Further it has to be diazotized. Meaning, the student has to react to this.

[00:02:22]With Nano2 + HCl, you know that deiodination takes place at 0 to 5 degree Celsius and when it gets deiodized, the product it will form will be benzene diazonium chloride, okay and this benzene diazonium chloride has to be further reacted with HBF4, with HBF4, with fluoroboric acid, so as soon as you make it react with HBF4.

[00:02:53]HCL will come out from here and what remains after HCL comes out is called benzene diazonium fluoroborate. Ok? We call it benzene diazonium fluoroborate. Please take care.

[00:03:08]So N2+ BF4- HCl is out.

[00:03:12]But what do you have to do here?

[00:03:14]Heat.

[00:03:15]Heat is also written at the end. So remember student, wherever there is BF4-, there is three covalent bonds. There are three covalent bonds and one coordinate bond with F-. This F- acts as a nucleophile here. Ok?

[00:03:30]So when you heat F N2+ B4-, then the F- in BF4- acts as a nucleophile. And the same F- replaces N2+ here and forms chlorobenzene. Together from here comes out N2 and together comes out BF3 like this. That means the answer that comes to us is 48, that third one will form fluoro benzene.

[00:03:56]So first we react benzoic acid with ammonia. So the H2O comes out and forms benzamide here. Further KOBr means from this it is called Hofmann bromamide degradation. So C double bud O hat ke banega y humko aniline. When aniline is reacted with Nano2 + HCl, benzene diazonium chloride is formed. And if you react this benzene diazonium chloride further with fluoroboric acid or HBF4, then here HCL comes out and benzene diazonium fluoroborate is formed and further if you heat it then here F- replaces N2+ and fluorobenzene is formed. End of discussion. It happened very easily.

[00:04:35]Answer third of 48. Is it clear? I will move on to the next question. Ok? All this is very easy. After that, after 48, the next question given by us to students is a direct NCERT question from the Amines chapter.

[00:04:47]Ok? He was also from Amines. This is also from amines. Look what is being said.

[00:04:51]The correct order of basic strength is until you are given medium.

[00:04:56]Whenever you are asked about basic strength or you are asked about nucleophilicity. If the medium is not mentioned then you should consider aquatic medium. Please take care.

[00:05:07]Till the medium is not given, you should take student aquatic medium and you should know that in aquatic medium whether you have methyl or ethyl, the basic strength is more, the secondary strength is 2 degrees, you must have read that when methyl is 213 2° 1° 3° and ethyl is 231 2° 3° 1° and we make the children remember in the class that ethyl has more carbon, so the higher number 231 has to be remembered 2° 3° 1°, methyl has one carbon, that means there is less carbon, so the lower number has to be remembered 213, that is, 2° 1° and 3°, if you know like this then you can see what is given here, 1° more of ethyl has been given, which is wrong, this cannot happen because it has given 2° at the last, which is not possible at all. Unless you have been given the medium, then the aquatic medium and whether it is methyl or ethyl, there is more secondary in the aquatic medium. Ok? And here too much aniline has been given. This is clearly wrong because the value of analgin will be the lowest because the lone pair is delocalized. Its answer here will be third because 2 degree ethyl is given in this option.

[00:06:10]Then the one with 1 degree and you will be surprised to know that here children often get confused that the basic strength of benzyl amine, practically on the basis of pKB value, is more than ammonia. Because the pKb value of benzyl amine is 4.70 and the pKb value of ammonia is 4.75 and as you know the basic strength is inversely proportional to the pKb, the one which has lower pKb value has higher basic strength, so the pKb value of benzyl amine is around 4.70.

[00:06:45]In that, you may have some different data written in some points, sir, or ammonia is 4.75. In this way we can find out the basic strength here.

[00:06:53]Ok? And you know about aniline, aniline has the lowest basic strength because its pKb value is 9.38.

[00:07:00]Ok? This method and the ethyl one here is around 3.25 and this one here is around 3.29. Ok? What type of PKB value is there? So keep in mind that if you have not been given a medium, that means aquatic medium and aquatic medium has more than 2°, whether it is ethyl or methyl. Ok? And I have written down the value of PKB for you. The basic strength is inversely proportional to the value of PKB.

[00:07:25]Here ammonia is saying wrong. After that Aniline is saying.

[00:07:29]That's how you could have said it wrong. Even by looking at the options, any student can understand that if the secondary is methyl or ethyl, its basic strength will be higher. Here I can accept that children get a little confused that since there is phenyl here, if -i is added then the basic strength should decrease.

[00:07:44]But in reality it does not happen here. The pKb value of benzyl amine is practically low. Is 4.70. Therefore its basic strength is higher than ammonia.

[00:07:53]Ok? Therefore, you will get 50 answer third correct. Is it clear? I will move on to the next question, student. It's 50 marks.

[00:08:00]This is telling you that the reactivity order in electrophilic substitution reactions is: You should know where the electrophile will attack fastest?

[00:08:10]Where the electron density will be higher.

[00:08:12]That is, we can say that the rate of ESR is directly proportional to +i +m and inversely proportional to -I -M.

[00:08:24]Ok? So the rate of ESR is directly proportional to +I + inversely proportional to -I - M. The higher the electron density inside the benzene ring, the faster will be the attraction of the electrophile. The rate of ESR will be higher and the intermediate formed there, the carbocation donor, also works to stabilize the carbocation intermediate. Ok? So here you know that N double bd O, which is N double bd O actually shows both +m and -m. You must have read that N double bond O also has a lone pair on nitrogen and N double bond O also has a hetero atomic multiple bond. So N double bed O i.e. the nitroso group shows both +M and -m depending upon the situation. But -m is dominant. So overall you have to take its -m which is dominant. And you should know about OH that OH has a loan payer. So if it is a loan pair then its dominant part will be +m student. Ok? And if I talk about NO2, then what happens to NO2 is also dominant -m. Well -m -i shows both but dominant is its -M.

[00:09:30]OCH3 also has +m and -i. But the dominant one is +m. So look, +m becomes two. One OH to one OCH3.

[00:09:40]Remember, student, the magnitude of +m is greater for OH than for OCH3. I have discussed this matter many times.

[00:09:47]Ok? So the magnitude of +nm of OH is more than OCH3. This is why where will the electrophile attack faster? Phenol is compared to anisole. And if I talk about -M, what happens in -M is that -m is more dominant of NO2 compared to NO. Ok? That means nitro is more than nitrosol. Now you see, the most common here will be B, followed by D. Then A's is the lowest of C's. That means, here I am telling you that it is written in increasing order, so the lowest will be C because here the -M of NO2 will appear very powerful. This is a very strong deactivator for students.

[00:10:24]After that you will get a little bit less of A here -m. After that it will be of D because +m is a little less and according to the given option here, the maximum +m will be of oh. So B will have the highest rate of ESR. So that's what he's asking. The reactivity order in electrophilic substitution reaction is i.e. the substituent number should be fourth.

[00:10:46]Is Cadbury clear? It will be the fourth of 53. I will move on to the next question. I hope you understand all this and have made it correctly. Next is saying that an alkyl bromide produces a single alkene. Already saying that only one alkene is forming. Meaning an elimination reaction would be taking place.

[00:11:07]When it reacts with sodium ethoxide, we have told you many times that sodium ethoxide is in the presence of alcoholic medium. So whenever you are given sodium ethoxide or hydroxide in this way KOH, NaOH, methoxy, ethoxy, all these act as a strong base in alcoholic medium. And there the pathway is followed, E2 pathway is followed. And then alkene on hydrogenation produces to methyl butane. So if we say 2 methyl butane, then saying 2 methyl butane means we should get this CH3 CHCH3 and CH2 and CH3, this is called 2 methyl butane, now you see, in this the first option, what is it saying, one bromo 2 di methyl propane, well it is propane only, it will not be of any use here because as soon as I told you, in its case, one bromo 22 2 diethyl propane. You think about it yourself and see. This is what happened, one bromo and two to diethyl propane. If I react this, student, this is sodium ethoxide here. In the presence of ethyl alcohol. Wherever you have sodium potassium hydroxide like this, alkoxide, these alcoholic medium pathways take place in E2 and what happens in E2?

[00:12:29]Rearrangement does not occur because the reaction occurs from the transition state. So, from the carbon just next to where Br is released, H+ is released simultaneously. But in this case, there is no H on the carbon next to where Br is coming out. So what do I say here? No reaction. There will be no reaction in this case.

[00:12:48]Because there is no H on the adjacent carbon to be released.

[00:12:50]So the E2 pathway cannot be here. Now let me talk about One Bro Butane. Just look at this. So this is what I mean by one bromo butane.

[00:12:59]CH3 CH2 CH2Br This is correct that it will form an alkene.

[00:13:06]In the first case there is no reaction because there is no H next to the bromine. Now it is saying that it produces a single alkene, so here a single alkene will be formed but what the other person is saying is that on hydrogenation it produces two methyl butane. So brother, where is the methyl here? To pay methyl is it? So when there is no two-pe methyl, then how will we get two-methyl butane here? It doesn't exist. So try it.

[00:13:30]Wherever this type of solution is given to you, sodium potassium alkoxide or hydroxide in the presence of alcoholic medium pathway E2 will be followed. E2 means that H+ will be released simultaneously from the carbon just next to where Br- is released.

[00:13:46]So in its case, the formation that will do it for us will be one butane. Now just think, if we hydrogenate this one butene in the presence of nickel catalyst, will it be made as it is being said to be made? no way.

[00:14:01]Here I am getting butane.

[00:14:05]Whereas what is he saying about making it? This is talking about making two methyl butane. So from this point of view it became clear that there is no two-pe methyl. This means there will be no first and there will be no second.

[00:14:15]Now we are left with the third and fourth options. Let's see what happens.

[00:14:19]Your concept has become clear till here.

[00:14:22]Next we go. This might not have been known. This might not have been known.

[00:14:25]Now let's come to one bromo two methyl butane. I think this will hit because CH3 is CH2. To pay a methyl is given here. And then here's one per bromine given. And I want to react it with sodium ethoxide. This means that it will act as a strong base in the presence of an alcoholic medium. So as given alcoholic medium the pathway of Student reaction will be E2 and in E2 carbocation is not formed. The reaction takes place from the transition state, so from where Br comes out, simultaneously H+ will come out from next to it and such a product will be formed that CH3CH2 with C, CH3 with double bond, CH2 and do you know who takes Br? Will take Br-.

[00:15:11]Na+ has come out from here and an H+ has come out from here next to the Br. C2H5O- will take that H+, we have no interest in it. Further, if I do hydrogenation of this, then it is written that then alkene on hydrogenation produces, so hydrogenation means that wherever there is a pi bond between two carbons, there will be addition of one H each.

[00:15:32]Addition of one H each will form here like this. It became two methyl butane. Ok? This formed two methyl butanes. I got the answer to option number three.

[00:15:43]But I want to show you why the fourth number will not be the answer after all? let's watch. So you will write two bromo two methyl two bromo two methyl like this and if it is butane then how many carbons? If the parent chain is of four carbons then the student is gone. 1 2 3 4 to bromo to methyl. If you react here with sodium ethoxide in the presence of ethyl alcohol. Think for yourself, what will be the pathway of reaction here? E2. When E2 occurs, the carbon just next to where Br- will come out, H+ can also come out from here because it is secondary and can come out from here also.

[00:16:24]That is primary. It is a different matter that the set Japs rule will be applicable here. And according to Setzep's rule, the major will be more substituted alkene. So when H+ came out from here it became CH CH3 and when H+ came out from here it became CH2 with double bottom C CH3 and CH2 and CH3 and among these two the one above became the first one and what became the second one? It became a minor. Keep this in mind, you will say Sir, 2 methyl butane can be made from here also. After the hydrogenation to methyl butane will definitely be formed. But I am telling you that an alkyl bromide produces a single alkene. Single alkene should be formed. That single alkene is just giving you a third option. In the fourth, two alkenes are formed. One you're getting 2 methyl to butane. One to methyl one butane is being formed. And if we hydrogenate both of them, we will get two methyl butane only. But as soon as I told you this, it is single alkene when it reacts with sodium ethoxide and ethanol. So that's why you don't have to say this option. Is it clear? Because the answer to the formation of single alkene will be the third option which is in front of you. Because here there is only one type of hydrogen with respect to bromine, tertiary. Because here, with respect to bromine, one side is secondary and the other side is primary. Two alkenes will be formed. Is the concept clear? I will move on to the next question.

[00:17:46]Student, look, 55 is the third answer.

[00:17:49]After that this question is very important.

[00:17:51]Number of Number of 119° 28 minutes. This is a little bit out of NCERT, you have to see what the 120° bond angle in the compound edge will be. First of all, I want to tell you that whatever answer you get for this will be third. It will be six and six. How it will happen?

[00:18:07]let's watch. If you look here, you will be given CH3. And here a CH is given. What is given here with the double bond? CH2 is given like this in your case. You know this hybridization of carbon.

[00:18:22]What is this hybridization of carbon here? It is sp3 and what is it? It is sp2 and what is this? sp2 So first of all you will see then look at the one with h sp3. So how much hc total are you getting? Total, the angle you will get here with h, how much hydrogen is there with h? Three. So HC total here you have three. The one with h is 109° 28 minutes. This is three hydrogen for you.

[00:18:54]You will get this 109° 28 minutes. After that you will get three H CC okay? So this one is HC, HC one has become three with HCH and the sp3 carbon and three has become HC here. This is done, it will become 109° 28 minutes. In this way, first you will say that this is what happened in his case. Now let us come to which one can be 120°? Look, there are two cases in the 120° one. One here will be about CH, one will be about CH2. Two will be here. So first H CC is here, HCC is here, then one HC is here also 120 HCC HC because you should know that whatever atom and group is attached to the sp2 carbon is in the same plane in which the sp2 carbon itself is found. So you know that sp2 carbon and any atom group attached to sp carbon is in the same plane in which sp and sp2 carbon atoms are found, so this is H CC and H CC and then this CC, there are three Cs, it will also be 120° in between this, this is about CH, that is and if we look here about CH2, then about CH2, then HCH, first see this and write HC HCH, this will become 120. This one has two hydrogens, the terminal one is CH2, so that is HCH.

[00:20:26]And then after that if you see then HCC of this hydrogen and HC of this hydrogen about each hydrogen here HCC 120° and HCC 120° if you see in between this then it will become 336 oh HC HC and this is it, in its case HCH this is it and then HCC and HC and CC like this 120 in this way total total here you will get six and six.

[00:20:58]You should try to remember all this directly.

[00:21:02]If all these questions are given to you in this manner with some different structure, then it will definitely trouble you. But the chances of asking all this are less. Ok? There is no need to take that much tension in this. Just remember it simple. Even if asked, only propene can ask a normal case. In which you will get the answer in six and six. Ok? Is it clear? Let's move ahead.

[00:21:18]Our question after 59 is number 60. This is easy. First of all, do you know what we have to do first, students?

[00:21:26]Which reaction is this? There are four in the middle, CH2 is the first, CH3 is the last. That means it is normal hexane. So this is the first normal hexane given to us.

[00:21:37]Reaction of normal hexane Cr2O3 Al2O3 is called aromatization reaction.

[00:21:42]So if I do aromatization, what will it form first? Benzene will be formed first. I will do bromination of benzene.

[00:21:49]Fe here the reaction will be ESR ESR will be electrophilic aromatic substitution the electrophile will be Br+ it will attack on benzene which causes the formation of bromo benzene and when bromo benzene reacts with magnesium metal then the bond between carbon and bromine will break magnesium will come in between to form phenyl magnesium bromide and when we do hydrolysis of phenyl magnesium bromide we break it by converting water into H+ OH- and you should know that wherever there is phenyl magnesium bromide or alkyl magnesium bromide we call it Grignard reagent and this Grignard reagent is a very strong base because magnesium is electro positive it will have plus on it and minus on carbon hence it is called strong base this strong base will pick up the negative charge on carbon the H+ of water and after picking up the H+ of water the formation it will do will be benzene that is you will get answer one of 60 marks. Is it clear? So this one will be its answer. 60 marks is very easy. If you first carry out the aromatization reaction, benzene will be formed and then you will do bromination. ESR electrophile Br plus will attack bromo benzene. Magnesium will come in between phenyl and bromine in the form of phenyl magnesium bromide i.e. Grignard reagent and this type of reaction has happened many times that when we react phenyl magnesium bromide with water, ethanol, methanol or with any slightly acidic compound by taking H plus of water or H plus of any acidic compound, the product that will be formed will be benzene. Let's move forward.

[00:23:30]All this is easy.

[00:23:32]Next is even simpler, the technique used for purification of stream volatile immiscible substance is steam volatile of which? Yes sorry, this is not steam volatile, you are asking about steam volatile too, okay, I said it right, steam volatile is an immiscible substance, so whatever volatility is very high, think about it yourself, its distillation will be done, which steam distillation will it be, so remember, if the volatility is very high, then what will we have to do first, first of all we have to increase the pressure on its surface.

[00:24:06]Ok? So first we increase its boiling point. Otherwise the volatility is very high. So it will become volatile and vaporize even before it boils.

[00:24:14]So to increase its boiling point, we first have to increase the pressure above the surface of that liquid.

[00:24:21]You must have read all that in practical.

[00:24:23]So steam volatility, whatever steam volatile it is or it can be said that it is a highly volatile compound, it is an immovable substance in water, then it actually undergoes steam distillation, the answer of 67 here will be clear to you, after 67 we move forward towards 69, students, this is also very easy. The correct statement among the following is, in this I am first telling you that the penta acetate of glucose does not react with hydroxyl amine to give oxyzyme, this is absolutely correct because the aldehyde group of glucose in penta acetate of glucose is not found free.

[00:25:02]Ok? So here if you ask for penta acetate of glucose, it does not react with hydroxyl amine. Yes, it is written here that if we have glucose, then glucose reacts with hydroxyl amine and glucozyme is formed there, that reaction confirms that glucose has a carbonyl group. So if I say the reaction of glucose, then the reaction of glucose takes place with hydroxyl amine. But remember, if the penta acetate of glucose has been formed, when we have reacted glucose with acetic anhydride and after the acetylation, the glucose penta acetate which has been formed does not react with hydroxyl amine. Now he is saying that cellulose is a straight chain, this is correct.

[00:25:39]Polysaccharide made up of alpha Dglucose.

[00:25:42]This is wrong. This is wrong. It is a straight chain but it is made of beta-diglucose.

[00:25:46]So cellulose is made of beta Dglucose and starch is made of alpha Dglucose.

[00:25:53]This is wrong in alpha deglucose kahaji or cellulose.

[00:25:56]Lactose is a reducing sugar. Well, that's fine. It does not give failing test. Hey brother, what is this? Whenever there is any reducing sugar, then Fehling test or Bendik test will show positive. So what this is saying is that does not give failing test. This went wrong.

[00:26:13]Why is there a reducing sugar? Because that tolerant test gives failing test positive. So this is why it went wrong. Fructose is an example of aldohexoses. This is wrong because it contains ketone on the second carbon.

[00:26:23]Ok? So fructose is a ketohexose and glucose is an aldohexose. So this is wrong. And keep in mind here that whatever reducing sugar there is, as you have read about maltose, lactose or glucose, fructose, whatever reducing sugar there is, all of them are tolerant to the test. Fehling's test gives positive. It is saying that don't give failing test. This is where it went wrong. And till here the straight chain is correct but saying alpha Dglucose is wrong. It has been cleared up to this point in a good way. So now let us move on to the next question. This is a very easy question for you on bi molecules. The questions have been set directly from the NCERT lines which will be very useful for you. Ok? It's been a while. Purification of organic compounds was done. This question comes from the most direct line.

[00:27:05]So you should prepare NCERT thoroughly. Look here, its pKb value, you have given the value of pKa, so you should know that the acidic strength is inversely proportional to the pK value.

[00:27:18]Okay student like this. Now from here let me talk about paracresol, ethanol, meta nitrophenol and orthocresol.

[00:27:25]So the acidic strength that you will get here will be highest for meta nitrophenol i.e. C.

[00:27:34]After that you know what? The acidic strength of both orthocresol and paracresol is almost the same. It is almost equal. If you look at NCERT, the PK value of both will be given as 10.2. Both are given as 10.2.

[00:27:48]But if I look at it theoretically, then theoretically paracresol is slightly more acidic compared to orthocresol. So there's a little bit more of the paracresol here.

[00:28:01]If the PK values ​​are not being queried. If you are asking theoretically then you will say paracresol.

[00:28:07]After that comes orthocresol and the least acidic is B i.e. ethanol.

[00:28:12]Because there is alcohol. Among alcohol and phenol, the more acidic one is phenol.

[00:28:17]Because the conjugate base of phenol is resonance stabilized. So who will have the lowest PK value? Of C. So what is the lowest value? 8.3 means C will be third. Is it clear? C will have third.

[00:28:29]He is giving 8.3, that will be done. So look where is C giving three here?

[00:28:34]Hey brother, there is only one option. Second option. I got the second option. A's got four.

[00:28:39]Take A's four. Both of them have 10.2.

[00:28:41]Orthocresol, paracresol of both. It is given in NCERT in Phenol chapter.

[00:28:45]And ethanol has the highest 15.9, which means it is giving second place to B. B 's sec. I have already written the third of C.

[00:28:53]And the remaining D means which of D, which of D, which one has been given, you have used four of A, you have used four of A, so you have used four, so which one is left, in its case one is left, so make one of D, okay, it is matching automatically, otherwise, four of A has been taken, 10.2 of paracresol has been taken, two of B, that is, only one option is given, that is 15.9, three of C has become 8.3, it will be the lowest because it is the most acidic, and here one of D will become 10.2, that's it, remember all this directly in NCERT, in the box you will find what is the pK value of each [sound of deep breathing] [sound of sudden loud breathing] it will be made comfortably, I will move towards the next question, student, that is given for us, 74 good Gabriel, phthalamide synthesis cannot be used to prepare primary aromatic amine, he is saying absolutely correct, attention Keep this in mind, Gabriel's thallamide synthesis produces only aliphatic primary amines. Aromatic primary ene is not formed because if we want to form it then we will have to take aryl halide. And what happens in aryl halides? Partial double bond occurs due to resonance. Therefore, here the formation of aromatic amine i.e. aniline cannot take place from this aryl halide.

[00:30:11]And the reason is saying that the aryl halide do not undergo nucleophilic substitution reaction due to partial double bond character.

[00:30:20]Correct. You should look at the mechanism once. In Gabriel's Lattic synthesis, you will find that the alkyl halide has a complete single bond. So here the reaction takes place and the alkyl comes easily on top of the ene, after the hydrolysis of which we get the primary ene.

[00:30:34]So what is formed in Gabriel thallamide synthesis is an aliphatic primary anion and not an aromatic primary anion. The reason for this is that the aryl halide which later in the reaction gets converted into n alkyl or n phenyl substituted aniline, there we say it is n alkyl or n phenyl substituted phthalamide, that cannot happen here because there is a partial double bond, so it cannot undergo nucleophilic substitution in the reaction.

[00:30:59]This means both assertion and reason are true and reason is the correct explanation of assertion. So the reason is absolutely correct that aryllite cannot go into reaction. Hey, I understand, right? You know the mechanism, right? You look at it in such a way that it seems as if nothing is coming.

[00:31:16]You know we get thylamide from here. First we get the reaction done with the base.

[00:31:21]So this is what removes the base H+ here. And when it removes H+, it forms this. In its case, it forms here. N- That's what K+ does, right? K+ sorry OH- H+ came out from here and became water. Now whatever is formed here, it reacts with alkyl halide and which pathway is it? SN2 strong nucleophile excludes X- in this and the same formation happens in its case C double bond O C double bond O and here R comes above N i.e. alkyl. And this is not possible if you take aryl halide here. If you take aryl halide here in place of RX then it has partial double bond. It is sp2 carbon. It is held so strongly that it cannot replace the halogen and further when you do its hydrolysis, you must have seen that it breaks from here and forms primary aliphatic amine and along with it sodium or potassium phthalate is also formed. This is what happens.

[00:32:20]So this is the reason here that this RX should be completely single. If the partial doubles, it won't happen. Yes, if this is done here, if X is removed from it and CH2X is brought in between, then the reaction will take place because there is no partial double bond. Well here if the halogen is directly attached to the ring then the reaction will not take place because we have to make it directly, in its case it is said aromatic amine, so aromatic amine means NH2 directly attached to the ring and to make it we need X directly attached to benzene which will not be possible and the reaction will not take place. Ok? The partial double bond will cause one to become 74.

[00:32:58]This is just very easy. The one that does not stabilize the 2° and 3° structure of proteins. So there is a disulfide linkage.

[00:33:06]Hydrogen bond occurs. Wonderwall forces happen.

[00:33:08]This peroxide bond that you have been given between O, peroxide linkage does not happen here. Ok?

[00:33:14]So this cannot be a linkage here. There will be a forest of 75. This is given to you directly from NCERT line. After that let's go to next. Statement one here is saying that ethoxy ethane. What I mean by saying ethoxy ethane is that students, you must have seen this question somewhere, I have asked it before also.

[00:33:32]This is ethoxy ethane and one is saying methoxy propane. So methoxy propane or you can say one methoxy propane.

[00:33:40]Such R metamers are saying absolutely correct because in this there are two carbons on the left, sorry three carbons on the left, two carbons on the right, one carbon on the right, so the functional group is same and the distribution of alkyl group from both sides is different. Three carbons on the left, two carbons on the left, one carbon on the right, two carbons on the right, so compare the left with the left. Compare right with right.

[00:34:06]If the number of carbon atoms in alkyl is different, it means it is a metamer.

[00:34:11]Absolutely correct. After that it is saying but 2N so this is but 2N and one is saying that but one NR positional isomers. He is saying it absolutely correct. how easy. 1 2 3 4 1 butane 1 2 3 4 to butane. So two butene and one butene are positional isomers. So both statement one and statement two are correct. So the answer for 80 marks will be one here. Any problem? The student is saying one of 80 absolutely correct.

[00:34:44]Ok? Both statements one statement to are correct. Then, see which of the following will undergo rapid hydrolysis with an aqueous NaOH to form the corresponding hydroxyl derivatives? Wherever there is halogen, remember wherever there is halogen directly attached to the benzene ring.

[00:35:05]So whenever you have a halogen directly attached to the benzene ring, there is no SN1 or SN2. In its case, the reaction with NaOH or water, whichever is left, is a nucleophilic aromatic substitution reaction. That means you will say that rate of SNAR means nucleophilic aromatic substitution. There will be no SN1 SN2 here. When halogen is directly attached to benzene then here nucleophilic aromatic substitution takes place and that is the rate of SNaR is directly proportional -I -M and inversely proportional is +I + M. The more powerful the strong electron withdrawing group is, the more the nucleophile will be attracted. Here you are looking at two minus M and this will generate the maximum electron deficiency.

[00:35:56]Where there is halogen, the fast nucleophile will attract. Here is a NO2 -m applying.

[00:36:02]Here it is NME, it is applying +m and increasing the electron density, so the repulsion with the nucleophile will increase.

[00:36:09]And there is nothing here. So who will react the fastest here? Forest. The fastest will do one, then two, then four and the lowest will do three. So remember whenever halogen is directly attached to the benzene ring. Student, in such a condition, the reaction that will take place with the nucleophile, nucleophilic aromatic substitution, will be directly proportional -i -m because the attraction of the nucleophile will increase when the electron deficiency inside the ring is more, so the deficiency is the highest in one, the answer of 81 will be one, it is clear, after that comes number 88, it is very easy, very easy, you should know that first of all what will be formed here is ethanol, you know that ethanol has become CH3CH2OH H, its reaction will take place with PBr3. So I will replace OH. Here I will form bromoethane. The reaction of bromoethane with alcoholic KOH has been told to you many times.

[00:37:03]Whenever there is KOH, NaOH or ethoxymethoxy in alcoholic medium, it will be E2 pathway.

[00:37:09]Meaning, from where Br will come out, just next to it an H+ will come out, so this will form CH2 double bed CH2. Ok? And what will happen further here? In its case, hydrolysis. So what happens after the hydrolysis here? An H+ will come here. One, water will come here. So this will become alcohol. Okay student? It will become like this. Hey brother, please take some water. You talk about acidic hydrolysis. So take H+ OH- directly from here.

[00:37:34]Bring H+ wherever you want. Wherever you feel like it, bring it there. It became ethanol. Now this ethanol has to be reacted. Sulphuric acid. Good. 140°C Student Here SN2 pathway takes place i.e. ether is formed. So what do you do? You have to immediately take two such molecules from your mind. Here if the temperature was written as 170 degree Celsius 170 degree Celsius then we would have taken E to pathway.

[00:37:59]We would make an alkene. But here it is written 140. From here you will remove OH- from one molecule and H+ from the other, i.e. water.

[00:38:08]This will form diethyl ether.

[00:38:12]Plus there will be water outside here.

[00:38:14]Ok? Ethoxy ethane will be formed i.e. diethyl ether. So the answer to 88 out of 88 will be your second. clear? It will be seconds.

[00:38:24]Diethyl diethyl ether is formed at 140.

[00:38:32]That means 400 becomes 13, right? Yes. At 13. So this much temperature is written. Hence, SN2 pathway reaction will take place in the case of primary alcohol.

[00:38:39]Is it clear? This is how it happens, the second becomes 88. After that I move on to the next one. One last question is our 89 but in HCL, you know that peroxide effect will not be felt here because HCL is given. The peroxide effect is observed only with HBr and nothing else. So this R2O2 is given just to confuse. After this, the rest of the electrophilic addition reaction will take place here. And if it is symmetrical then Marcobini Cobb's rule will not apply here. You can add it wherever you want. H+ carbon carbon double bond, I add it here, H+ CH2 is added here, Cl- this becomes two chloro two chloro butane this becomes X further I make it react with ammonia. So I get it reacted with ammonia. Break down the ammonia. Let's do it in a very easy way.

[00:39:29]You don't have to make the length that much. Now replace this with Cl-. So what will happen if you replace Cl-? Will CH3 be added here along with CH2 CH? NH2, this became our first thing, what? Y So from here on out it's saying that The Product X is optically active. That's correct. It has one cairos centre. So when the caerus center is one, it can be optically active. Ok? After that saying that the formation of product X to Y follows elimination reaction.

[00:39:59]Hey, where is the elimination happening? If y is being formed from x then it is a nucleophilic substitution reaction. This ammonia is a nucleophile and the sigma bond is broken to form a sigma bond. So when sigma breaks down to form sigma itself, it means that nucleophilic substitution reaction has taken place here. If there is no elimination then it is wrong here. This is true. The product x is optically active because one has a center.

[00:40:22]Then the product y is propane one amine. It is wrong. Why would an amine be formed?

[00:40:27]Because when halogen is at 2° then it will produce 2° major product in its case. Why will it go to 1°? So this is wrong in his case.

[00:40:34]Ok? This is also wrong. The product y is propane one amine. And in the fourth it is saying the product Z is alkyl diazonium salt.

[00:40:42]No, you should know that alkyl diazonium salt is highly unstable. It does not exist even at 0 to 5 degrees Celsius. This N2+ exists only when the TntU+ is directly attached to benzene.

[00:41:01]So it is not only unstable but highly unstable.

[00:41:04]Therefore, the water that is immediately present in this medium, the water is automatically present in this medium.

[00:41:08]Its hydrolysis happens immediately and actually the Z that will be formed will become Z in its case butane to all, here the product will be major in its case.

[00:41:20]Ok? This will be the product with Z in its case. This means that the student with the answer of 89 will become first here.

[00:41:28]Ok? If there is one coir center in it then it can be optically active. Ok?

[00:41:32]Because first it will be made in the form of a racemic mixture.

[00:41:34]Then we can resolve it there and separate it and it can become optically active. After that no one else is matching.

[00:41:40]The formation of product x to y is nucleophilic substitution. There is no elimination. The product y is propane one amine is incorrect. The product Z is alkyl dodecyl salt and never exists.

[00:41:50]Its hydrolysis occurs immediately. It gets converted into alcohol.

[00:41:53]Is it clear? So I think the entire solution to this has been completed. So we hope you have learned a lot from this and you should keep practicing and moving forward. You should still practice for 10-20 days.

[00:42:08]Read it very well. Focus on NCERT.

[00:42:11]Whatever extra questions you get from outside NCERT, you do not need to worry much about them. It is possible that the marks may fluctuate a little.

[00:42:18]Because different tests have their own quality of questions. That may depend upon the quality of the question. It gives you fluctuating 10-20 marks. There is no need to worry about that. Do n't get depressed. You have to create a concept. While clearing the concepts, you just have to practice the questions and keep moving forward.

[00:42:34]Remember the things that need to be remembered again and again. Ok?

[00:42:37]That will give you better results. Ok? Thank you.