ORGANIC CHEM - TEST-06 VIDEO SOLUTION FOR RE NEET 2026

Added: 2026-05-26

[00:00:00]Goal Nations [Music] Leading Institute Hello dia students I am Shailesh Jhamistry faculty in Goal Institute students today we have paper of reet test six we have organic chemistry part here on our screen and you can see first question on screen you have question number seven so let us see what is asked in question number seven see in question number number seven [sound of clearing throat] in assertion what is written? N-Butane N-Butane means normal butane [throat clearing sound] and isobutane are examples of structural isomers.

[00:00:42]What the student says is that normal butane and isobutane are examples of structural isomers. Yes, how? Now look, if you write butane normal butane then it becomes 1 2 3 4. Will it happen or not, student? And if you write isobutane, you will write it like this. Now look students, what do we know?

[00:01:08]What is meant by structural isomer? There must be a connectivity difference.

[00:01:10]Should it happen or not? If the connectivity of an atom or group is different then it becomes structural. And if the connectivity is same orientation is different then it will be stereo. Will it happen or not, students? So pay attention here, this is a chain of four and this is a chain of three and CH3 is connected here, so the connectivity is different. You just think straight, one, two, three carbons and then the fourth one is attached here and in this there is a chain of three and in it CH3 is attached in the middle. This means connectivity is lost.

[00:01:40]So this will be structural. This means assertion is right. Now I read Reese.

[00:01:46]See, structural isomers have the same molecular formula but different connectivity of atoms. What we were talking about was the definition. Brother, the connectivity is different [sound of clearing throat] that's why it is a structural isomer. This means that this region is correct and this is also its correct explanation. Did it happen or not, students? This means the answer will be both assertion and reason are true and reason is the correct explanation of assertion. So which option is shown? Look, both assertion and reason are correct and reason is the correct explanation of assertion. So the correct answer to question number seven will be the first option. OK students clear? Let us now move on to the next question.

[00:02:32]So look next, which one is after this after seven?

[00:02:40]So after seven now we have question number 11. Now look, in question 11 you are given two columns and asked to match them. On one side, students have given the name of some reaction which is a name reaction and on the other side, they have given the reagent.

[00:02:56]This type of question is very famous.

[00:02:58]If we talk about NEET, around 202 also a matchmaker came up with a side reaction and reagent. So this is a common question for NEET. Let's see.

[00:03:10]So what is the Itard reaction, students? We take Tolvan in this.

[00:03:15]What do you do in this? Tolvan for.

[00:03:18]Who do we get the Tollen reaction done by? CrO2Cl2 in the presence of CH2. So first an intermediate compound is formed CH OCOHCl2 its holto. And then when it is hydrolyzed, benzaldehyde is formed.

[00:03:38]What do students call this reaction? Itard reaction. So let's keep mixing it. So which one will Itard have? This CrO2Cl2 CH2 then hydrolysis.

[00:03:47]This means A becomes four. Let's write it down.

[00:03:50]We will find the answer later. Ok? Now watch the reaction of the students to the Gatterman coach.

[00:03:55]What is Gattermann Koch reaction? So take benzene and add CO+ HCl AlCl3. So the electrophile CHO+ is formed. An electrophilic substitution reaction will occur and benzaldehyde will be formed. So the reagent to be mixed is that of Getermann Koch. Look COHCl AlCl3 this means B has become one. Then there's your Stephen reaction.

[00:04:17]What is the Students Stephen Reaction? RCN: In this we give SnCl2 + HCl, this will not reduce it completely.

[00:04:27]Will reduce partially. So there will be a pie break.

[00:04:30]H will come. What will the students become? RCH double bond NH Now if you get it hydrolyzed then RCHO is formed. You can do it like this here OH then take out OH water. Did you understand or not? So RCHO let's then SnCl2 + HCl hydrolysis of the Stephen reaction will be second.

[00:04:55]You already know about Clevemanson reduction, ZnHg connexal. So if it is an aldehyde or ketone and you give ZnHg connexal to it, then what does the CO in the carbonyl get converted into? In CH2. Come on, look at which Znhg Conkel, it is given in the third. So let's add its third option now. A has four, B has one, C has two, D has three. 4 1 2 3 Let's see in which one. Look at this. 4 1 2 3 This means we have got the correct answer now.

[00:05:25]Correct answer of 11 is second option. OK students? Let's look at the next question.

[00:05:40]Ok. So now [sound of clearing throat] next question we have question number 12. What do students say in question 12? Take a look. A positive carbyl amine test is given by, see students, what is it, the primary amine, listen carefully, the primary amine, only that carbyl amine reaction or what is it called, students, isocyanide test, the isocyanide which is formed, see what is the reaction, primary amine, take it, okay, now react it with CHCl3 and KOH, any base, then RNC alkyl isocyanide or alkyl carbyl amine, it is detected by foul smell.

[00:06:17]So you have to find out which is the primary amine.

[00:06:22]Look, I will make the structure of one and tell you. I will tell you how to see it easily later. Let's look NN di methyl aniline so aniline so you know. If you add NH2 to benzene, it will become aniline but it will be written as N-diethyl. So two H's two methyl. Now what is the method to identify amine? Primary, secondary, tertiary. I will tell you simply. n if you listen carefully. n If attached to one carbon, primary amine. If it is attached to two carbons then it is a secondary amine and if it is attached to three carbons then it is a tertiary amine. Now look at n here 1 2 3 this is a tertiary amine.

[00:07:01]So, primary amine did not occur.

[00:07:03]This means it will not give the test. Understand the matter.

[00:07:06]Now pay attention students.

[00:07:09]What are you looking for? Primary amines. Now tell me, if it is a primary amine, then should it be written like this that n methyl or n di methyl, brother, when you have added two methyls to n, then how will it remain primary and when it is not primary, then from where will the carbyl amine react, okay, it was not given first, now we will make it like this only. Now see students, methyl is written on n, so brother, methyl is written on n, after that there is aniline, meaning it will be like this, this methyl on n will be one H, aniline, then where is this primary, there should not be alkyl on n, it will not give, come to third, then methyl is written on n, it will not give, now how will it give, I will tell you its structure once, see paramethyl benzyl amine, there is nothing on n in it, how will you write benzyl amine, students put NH2 on benzyl and put methyl in para. Yes. So how many carbons is n attached to? A carbon. So it is a primary amine. This will give carbyl amine reaction. This means what is the correct answer for 12, students? Fourth option. Ok. Let's see what else comes after this.

[00:08:20]Ok. Look, there's a match again [sound of clearing throat] column.

[00:08:23]We have question number 14. Ok.

[00:08:26]So let us see what is asked in question 14. So what? Names of Amines List One gives the names of some amines and also gives their PK here. Now students, what we know is that if the basic strength increases, if the basic strength of someone is more, it means the dissociation constant of that base will be more and if the dissociation constant KB of the base is more, it means the value of PKB will be less.

[00:08:57]In simple words, if the basic strength is high then the PKB value is low. Ok? Let's see now. Now look at the students.

[00:09:05]Well, remember in the case of methyl amine and ethyl amine, if some solvent is not given. Understand the matter. If solvent is not given then you will take protic. Take water. Ok? Now remember that we used to read the orders of methyl amine and ethyl amine separately.

[00:09:22]What happened to the methyl? 2 degree 1 degree 3 degree does ammonia exist or not? Ethyl has a 2 3 1 followed by ammonia. Now let's see this ethanamine means this is your C2H5NH2.

[00:09:42]Then look at N ethyl ethanamine, that is, put one ethyl on N. Meaning C2H5 NH C2H5 N became ethyl ethanamine. Then what about students? N methyl methenamine. I mean, methenamine [sound of clearing throat] is CH3NH2, right? So NH CH3 N becomes methyl methanamine. And if you look at this, this methanamine means CH3 NH2, now it is certain that the one at 2° will be more. Because whether it is the case of methyl or ethyl, 2° was more or not the highest. okay brother. Now wait, let me change the color of the pen a little.

[00:10:22]Look here. So this is 2°.

[00:10:25]C2H5 NHC2H5 right? Look at this also CH3NHCH3 this is also 2°. Now how will you differentiate between these two students? So, it is very simple among these two. The +i of C2H5 will be greater than the +i of CH3.

[00:10:39]This means the C2H5 one will be more. So which was the strongest base in this? B.

[00:10:47]Which was the strongest base? b This means b will have the lowest pkb value. Let's say the lowest pkb value in this is three. This means b will be four.

[00:10:57]This is confirmed now. Let's search.

[00:10:58]Where is the four of b given? Look, the four of b is in both of these. We will see between these two.

[00:11:03]Let's go to B's four. Till here we can write it clearly or I.

[00:11:05]B's came out four. Let us go. Now if you talk about something less basic than this. This was the strongest base. After this it will be this one.

[00:11:13]CH3NH CH3 means C, so this was the four of B.

[00:11:18]Now look a little more than this, which one is 3.27, which one is C? The forest has come, the forest of C. After that, look, the basic ones after this are C2H5NH2 and CH3NH2, so whose +i will you have more than CH3, students? of C2H5. After this, the basic one will be A, so let's see which one will be A.

[00:11:43]Which one was after 3.27? 3.29 means second and the remaining D will become 3.38, so let's add A B C D D, so which one was 2 of A? 2 4 1 3 Let's find out where it is? 2 4 1 3 [nasal sound] 2 4 1 3 This is given in the first option itself. This means what will be the correct answer for 14? First option. OK students?

[00:12:15]Let's delete this. Look next after this after 14 okay. So next question now we have question number 15. What does it say in 15? Let's see.

[00:12:27]Incorrect method of preparation of alcohols from the following is. Meaning you have given some reaction name here. The reaction is written. It says which is the incorrect method? Mind it. He is not asking correctly. Incorrect method for preparation of what?

[00:12:47]Which one is for the preparation of alcohol? Let's see what won't make alcohol. First look at the reaction of aldehyde with CH3 MgBr followed by hydrolysis. So RCHO becomes aldehyde.

[00:13:00]If you react it with CH3MGBr. Now what does RMGX provide to students? R- It attacks carbon as a nucleophile. Opened CO.

[00:13:13]What will happen from here? R CH is O- and if you do hydrolysis it will become OH and CH3 will become alcohol. Meaning, given CH3 MgBr in number one, hydrolysis has been done in number two.

[00:13:25]So brother, alcohol is being made from this. Ok? So this will not be the answer. He is asking us from what alcohol is not made? Incorrect method of preparation of alcohol. Ok? In a second, see the reaction of primary alkyl [sound of clearing throat] halide with aqueous NaOH. Look students, I have taken this primary alkyl halide, I will react it with aqueous NaOH, so nucleophilic substitution will attack R and remove X, so what will be formed, students ROH. Now remember, if it is aqueous NaOH or aqueous KOH, then it is nucleophilic substitution and if it is alcoholic, then does E2 elimination happen or not? This means it was a nucleophilic substitution. SN2 attacked R and drove away X, so the product became alcohol. Well, this means this will also not be our correct answer.

[00:14:19]Third option see student ozonolysis of alkene. Students, if you take any alkene and do its ozonolysis, then how do you make the product?

[00:14:30]Break both the double bonds. Put a double bond O on it. a double bond O on it. Now students, both of these will not become aldehyde or ketone.

[00:14:39]Where will the alcohol be made? So he is asking you the same thing, from what will alcohol not be made? Now the answer has been received. Let's still watch the third and fourth.

[00:14:49]Hydroboration Oxidation of Propene.

[00:14:51]Students, what are the methods of hydration of alkene? One is H2OH+. Now carbocation will be made, if rearrangement is possible then get it done. Then there will be a water attack.

[00:15:02]Make it like this. Then there is OMDM oxymercuration demercuration, what do you do in that simply you add OH to Markonikoff and thirdly what do you do in hydroboration oxidation students, you do hydration with antimarkonikoff OH so this will also form alcohol, which means this is also not the correct answer, so now we have already got the answer to 15 and that is the third option, okay students, so now after 15, next is question number 16.

[00:15:37]Let's see what question 16 says.

[00:15:41]Look, in this you look at the reactant [sound of clearing throat] and look at the product. First see what changes have happened. Then we will search in it which reagent is most suitable. pay attention. Look students NO2 as it is remained. Nothing changed. Look at this COR Easter as it is left. Nothing changed.

[00:16:03]Where did the change come from? In COOH.

[00:16:08]What is COOH reduced to? CH2OH. Well, this is a very easy task that the carboxylic acid has been reduced. NO2 and Easter did not reduce. So students, one reagent is B2H6. What is the function of B2H6? That it selectively reduces carboxylic acids.

[00:16:26]But if ester contains halide i.e. halogen or nitro group, it is not able to reduce all these.

[00:16:34]So which one will be suitable brother?

[00:16:36]We had to selectively reduce CO₂.

[00:16:38]So B2H6 let's go where is B2H6? Look at this in the first option. So the correct answer for 16 is the first option. Okay students, let's move on.

[00:16:54]Okay next is question number 20, let's see what it says in 20, it is a simple question, what does it say, pay attention what happens when a mixture of acetylene, acetylene, what happened, students CH triple bond CH and hydrogen is passed over heated Lindler's catalyst, so if you have H2PDC Lindler's in it, then you will have to take C also, charcoal and BaSO4 or CaC3, conchlorine, whatever you take, it does not matter much. So for H2 PDC.

[00:17:24]Now remember, when Lindler's catalyst is used in this, complete hydrogenation does not take place.

[00:17:30]Partial hydrogenation occurs.

[00:17:33]So one pie will break H and H will be applied. Will it happen or not? So what will the product be? CH2 double bond CH2 so this will result in the formation of product CH2 double bond CH2. And where will the water come from in this, student? If you break the pie and put H then where will all the water come from? Secondly, there will be no geometrical isomerism in this that we will not write C trans. Hey brother, what is on the carbon of C double bed C? If H is the same group then where will the GI come from? Let's read the options. Ethylene and water wrong.

[00:18:02]Look at the second option. Ethane and water wrong.

[00:18:06]Ethylene is form. Correct. Acetylene and ethane wrong. This means we have already got the correct answer.

[00:18:12]Third option of 20. OK students?

[00:18:18]Next 20 onwards, let's move on.

[00:18:24]Okay look there is a sequential reaction you have question 27. Now next let us solve this. So 27 has a sequential reaction. What are you saying?

[00:18:33]Let's see. So is phenol. You will give zinc in phenol. Will reduce. Will reduce.

[00:18:40]Oh, what will come out of it? H means what did X become [nasal sound]? Benzene. Oh, what's different? H then became benzene. Now move ahead.

[00:18:50]CH3COCl Anhydrous AlCl3 Friedel Crafts Acylation. What will AlCl3 do? will pull Cl-. It will come on carbon. Plus. This became an electrophile. If you attack benzene, COCH3 will be added to benzene. So Y what happened students? So you add COCH3 to benzene, okay you added it, go ahead with ferial Craft acylation, now in Y, N2H4 and base is given, then you give N2H4 KOH, then which reaction takes place, which reduction is it, so CO will be converted into CH2. This means that the Z formed here will be combined with benzene as CH2CH3. And one is look at Y given NaOCL. This is Y. Now which reaction took place, NaOH + Cl2 or NaOCl? This is the haloform reaction. Now students, how do we make the product of haloform reaction? Wait, let me change its color a little. Let me explain to you. Look, it has CO₂CH₂3 in it. Isn't it?

[00:19:54]So this part of CH3, the student creates this hello form. Will he make it or not? So this part will become CHCl3 and from here PHCON will be formed. Let's see, W + W D is given in it and W D is written as Sodium Benzoate. This means that whatever W is, whatever W is, it will become CHCl3. Now let's go, this is Cl3.

[00:20:19]Ok? Now let us combine it with the option. Let's ask us what are the students? W and Z, so look, W is CHCl3 chloroform and Z is benzene ethyl benzene. Come on, see where it is placed? CHCl Look at this CHCl3 and ethyl benzene, this means the correct answer has already been found now. Which one will be 27? Fourth option. Are you clear, students? Ok. Let's see next after this.

[00:20:49]So after 27, see which one is after this.

[00:20:54]Look at question 28.

[00:20:56]28 Again, it's a simple question. It is given by resonance. Let's see what the question says. Select the correct increasing order of stability on the basis of resonance. He says, tell me the order of stability of all these on the basis of resonance.

[00:21:12]Let's see brother.

[00:21:15]How much pi bond is in this, student? Two pi bonds. Listen carefully.

[00:21:20]Look at the second one, there are two pi bonds. Look at the third one, a pi bond. Look also at the fourth a pi bond. What do we know, students? When you look at the stability of the canonical structure, what do you see first?

[00:21:30]Number of pi bonds more. Actually let's look at the number of covalent bonds. So brother, if you look at the covalent bond, the sigma bond will remain fixed. Sigma bond does not participate in resonance. Therefore, first of all let us see that if the number of pi bonds is more, then the stability of the canonical structure is more. So A and B have two pi bonds each. So both of these will be more stable. C and D will be less stable.

[00:21:55]Will it happen or not? Now how will you see between A and B? Just look. A has two pi bonds and B also has two pi bonds.

[00:22:02]How will you see between these two? So pay attention positive sigma pi sigma pi is like this. Ok? And look at this, positive sigma pi, positive sigma pi, so in this, extended resonance is happening in the first one. How are you doing? It was extended in the first one, right? Positive Sigma Pi Sig Pi.

[00:22:21]Look at this positive sigma pi positive sigma pi.

[00:22:23]So it was not extended. This cross happened. So, if there is cross resonance in this, then which one will have more resonance? In A. So its stability will also be more. So A happened the most, followed by B. Now let's see the difference between C and D.

[00:22:37]Look at charge separation. The number of pi bond was the same. Have seen it. So look, plus minus minus plus is side by side. Look at this, this plus minus is close. This is plus minus away. So the lesser the charge separation, the closer the plus and minus are, the greater the stability. Did someone ask why? Hey brother, plus minus is closer, the attraction is more.

[00:22:59]Wherever there is attraction in the world, energy will be released and if energy is released then stability will increase. This means less charge separation in which student was? In C. Then came C. After C came D. So A B C D. Come and see where it is kept. a B C D. Meaning second option. So what is the correct answer for 28? Seconds. OK students. Let's see further.

[00:23:23]So now after 28 next question we have question 33 so look at 33 what does it say? So in 33 it says look the best method for carrying out the following reaction a reactant and product that means reaction is given and it is asking you which of these will be suitable set of reagents so first look HCOH so brother brother what will be done with HCOH of benzene reaction no nothing is gone let's look at the second one given in benzene CH3Cl AlCl3 Friedel Crafts alkylation so what will happen first CH3 will be added to benzene understand the matter what will be added to benzene CH3 now again there is nitration students CH3 is donor group will be activator if it is activator then it will be ortho para directing right means first if you add CH3 to benzene from the second option it is ortho para directing so now if we are doing nitration then nitro will be added to para do we need para or not on meta side right okay then there is no benefit of giving H+ to KMnO4 so second also gone third look first if CH3Cl AlCl3 is added then it will be Friedel Crafts alkylation. then KMnO4H+ heat. So in this [sound of clearing throat] if you heat KMnO4H+ then the entire side chain gets oxidized into COH, so what will be formed from here, students, what happened to benzene? COH okay? And then again nitration was done.

[00:24:58]So if you add HNO3H2SO4 in it then COH is the deactivator and deactivator is meta directing. So now NO2 group will come and get attached at the meta position of COOH.

[00:25:13]And this is what we had to make.

[00:25:16]So, which option did you get?

[00:25:18]From the third. Let me tell you the fourth one also. pay attention. HNO3H2SO4 so nitro is applied. Understanding is now getting Fridl Craft done. So if there is a good deactivator attached to benzene then Friedel Crafts alkylation and acylation fails. So if you get Fridle Craft done from now on, then that reaction will not happen at all. Therefore there will be no fourth one either.

[00:25:45]So we had already got the answer.

[00:25:47]So which students became 33? Third option. OK students? Okay, let's see what happens next.

[00:26:00]So next question now we have question number 36. Let's see what it says in 36?

[00:26:07]Two columns are given to match. Here the mixture is given and here the process of purification is given. Ok?

[00:26:14]What is there? Let's see. The first contains naphthalene and sodium chloride. So, your naphthalene is one of the components that is sublimable, then you can definitely do it through sublimation.

[00:26:28]So one sodium chloride was given, one naphthalene was given, so this is written correctly in sublimation.

[00:26:32]Well look son, he is asking us for an incorrect matching pair. Please pay attention to this thing.

[00:26:37]So this is correct brother. This will not be our answer. Benzoic acid and naphthalene are not sublimation. Ok? And look at what else is there. So crystallization, yes, that's correct. Why? Look what happens in this is that we take hot water. If we have to separate benzoic acid and naphthalene, what will we do? Hot water will take hot water. Now, students, if you add a mixture of benzoic acid and naphthalene in it, then the naphthalene does not dissolve at all. The benzoic acid will dissolve.

[00:27:04]So the naphthalene will get deposited.

[00:27:07]What to do after that? Filter the naphthalene with filter paper.

[00:27:09]Filter it. The remaining benzoic acid will crystallize after you cool it and you will get it in pure form. This means this is also correct.

[00:27:19]We can separate these people through crystallization.

[00:27:21]Chloroform and Aniline There is a huge difference in the boiling points of chloroform and aniline.

[00:27:28]So we will do it through simple distillation. So this is correct. Ok? It is approximately 334.

[00:27:34]It is approximately 457 Kelvin.

[00:27:36]Then we do not have to memorize this. Boiling point. I am telling you one thing. There was a huge difference in boiling points. Right now there is a difference of more than 100°. So we will use simple distillation. So this is given correctly. Water and aniline. With what will the students use the mixture of water and aniline? It is given with water, right? So distillation will not be done under reduced pressure but through steam distillation.

[00:27:58]Actually, distillation under reduced pressure is used when one component has a very high boiling point.

[00:28:06]And which decomposes before the boiling point.

[00:28:08]Then we use this.

[00:28:10]So, it will not be used right now.

[00:28:12]What is it used for now?

[00:28:14]Steam distillation for aniline water mixture. Okay, I asked you, the incorrect matching pair has been found.

[00:28:18]What is the correct answer for 36? Fourth option. OK students?

[00:28:26]After this, see which one is there?

[00:28:31]Ok. So now, next question we have question number ah 37. What does question 37 say? Just look at which of the following structures will give a positive test with Tollens' reagent?

[00:28:46]Who gives a positive test with Tollens' reagent? Student Which will test positive with Tollens' reagent? Aldehyde is a simple thing.

[00:28:53]If you react aldehyde with Tollens reagent, then Tollens reagent is a mild oxidizing agent. It will oxidize the aldehyde and itself will be reduced.

[00:29:04]What is Tollens' reagent?

[00:29:06]AgNH3 is Hall to OH, right? So in this the silver is in +1 state. So from +1 it goes to zero. The aldehyde was oxidized and itself reduced to form silver.

[00:29:21]Silver is in zero state. This deposit gets done.

[00:29:24]So it is also called silver mirror test.

[00:29:26]Meaning, the simple thing is that aldehyde will give this test. It does not give ketones. Why wo n't it give ketones? So brother, Tollens reagent is a mild oxidizing agent, it will not be able to oxidize it. Come on, which one is the aldehyde? This will not happen. This alcohol is given. Look at this COH aldehyde, this is Easter. So the correct answer has been received.

[00:29:44]Which one will become 37? Third option. OK students? Let's move ahead.

[00:29:53]Ok. So next question. Now we have question 39. Let's see what 39 is.

[00:29:57]Watching the sequential reaction. It is visible in this. Look, it's benzene. CH3 [throat clearing sound] CH2 CH2Cl AlCl3 What will AlCl3 do?

[00:30:09]will remove Cl-. What will happen from here? CH3CH2CH2 per plus. What will be built here first? CH3CH2CH2 per plus. Now get the rearrangement done in carbocation.

[00:30:20]So shift H- from the side. So what will happen if H- is shifted from here?

[00:30:27]CH3CH+ CH3 H- after shifting. Now it will behave as an electrophile for further process and will attack benzene. Now when benzene is attacked, CH CH3 CH3 isopropyl benzene or cumene is formed. So X is this isopropyl benzene or cumene. Now look further hot alkaline KMnO4 is given. If you add hot alkaline KMnO4 to it, the entire side chain gets oxidised to form COH. What will be formed from here?

[00:31:10][nasal noise] COO- is hot alkaline KMO4 so COH not CO- right? If you want to write K+, you can write it. So Y happened.

[00:31:18]Now if NaOH + CaO is given soda lime then decarboxylation will take place so this Y was already obtained, this was Y now if you give soda lime in it NaOH + CaO heat then decarboxylation will take place what will be formed benzene okay benzene oh this will go away CO- now see after this we got Z so what is this compound formed is Z now look ahead [sound of clearing throat] Y which was formed, PCL5 given in Y is CO- consider it as COH so what will happen if you give PCL5? COCl [nasal sound] So COCl is formed from benzene, then heat NH3 in it.

[00:32:05]NH3 will become CONH2.

[00:32:10]So here we add CONH2 with benzene and then Br2 + KOH to it, so if you add Br2 + KOH to it, CO will disappear. What will happen with benzene? NH2 NH2 and what is this product that has become? W Now see what he is asking?

[00:32:32][sound of clearing throat] Z&W Z&W asking. So what did Z come up with, students?

[00:32:37]Benzene. Come on Z, what has happened? Benzene.

[00:32:43]Ok? And W is asking you or W is asking D.

[00:32:46]So W is written in it. So Z is your benzene and W is benzene with CONH2, right students?

[00:32:56]Then Br2 + KOH was given in it and W' was formed. So W, so this is it.

[00:33:03]If CONH2 reacts with benzene, then see the answer, which one will it be? So what is written first? Z should be Z, so look at this, benzene and W is not a dash, you will pay attention, if W is asked then WCONH2 has come, then where has it gone, look at benzene and in benzene CONH2 is such an option. It is not there in this. There is no second. If you look at the third one, it is not there even in the third one. If you look at the fourth, this is CH2H, so right now, according to the students, no one is correct in it. Correct? If this was asked here, pay attention. If this was asked here, Z & W D would have been asked.

[00:33:42]Okay students? This is W. So Z and W dash, if asked, which would be the W dash? You will see. The W dash was made of NH2 aniline with benzene. This was made. So benzene and aniline would have been the first option in that case. As of now, no one was correct. Okay students.

[00:34:04]But if this had happened, if this had happened then what would have happened if W dash was asked, here instead of W there would have been W dash, then the correct answer would have been benzene and aniline first option. Are you cleared, students?

[00:34:19]Ok? Ok.

[00:34:24]Let's move on to after 39.

[00:34:26][Sound of throat clearing] So next question, students, now we have question number 42. Let's look at question number 42. It is a sequential reaction. See, it is nitro benzene. It gave Sn + HCl.

[00:34:46]Now Sn + HCl reduces RNO2 to RNH2.

[00:34:50]So first NH2 will be formed with benzene.

[00:34:56]Now in this Nano2 + HCl 0 to 5° then diazonium will be formed. So after this, students, if we give Nano2 + HCl from 0 to 5° then it gets converted into diazonium.

[00:35:11]N2 + Cl- followed by HBF4 and heat. So students, if you give HBF4 first then it forms N2+ BF4- benzene diazonium fluoroborate. And when you heat it, it forms fluorobenzene and N2 and BF3 are released. Did you understand? So fluorobenzene will be formed. So P is asking, so P fluoro benzene is given in the first option.

[00:35:45]So students, what is the correct answer for 42? First option. OK students?

[00:35:53]After this look after 42 ok. So the next question is 43. See what it says in 43? Look at 43. Which of the following dibromides can give to butine upon elimination? Can you give? Please pay attention.

[00:36:12]Who can give two butine after elimination? So please pay attention.

[00:36:15]If you do elimination in this then Br will be removed. If H is removed from here then Pi will come here. Then Br from here H from here not pie. A triple bond was formed in this.

[00:36:26]Look at B. This Br, this H will come here. Then this Br here H will be removed. Pi can form 1 2 3 4 butyne in it.

[00:36:33]Correct? Look at C.

[00:36:37]Remove Br. Remove H. The pie will come. Then I removed Br and removed H. Can be made. You guys say Sir, this BR, this H is double here, then this BR, this H is double here, Pi Sigma Pi, this can also be made.

[00:36:47]Yes, it can definitely be made. But this can also be made in it. Triple bond can also come. So says can form can give to butine. Yes brother, he can give this two butane. Look at D, Brh will come here. Then Brh will come here.

[00:37:02]No, two butine will not be formed. This means B and C can give. So see in which one is B and C given? In the second option.

[00:37:09]So what is the correct answer for 43?

[00:37:11]Second option. OK students?

[00:37:21]Ok. So this was the last question of the organic chemistry part. I hope you have understood the whole question and its solution well.

[00:37:28]Okay students, thank you.