JEE Advanced and IIT-JEE Questions (2001–2014): Reactions

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from an old IITJEE paper. An amine has its N H two group on the end carbon of its chain, and with nitrous acid it gives an alcohol. But the O H group is not on the end carbon. It sits on the next carbon along! So how did it move? Today we solve past JEE questions on the reactions of amines. These reactions come up often, so let's begin.

0:28Idea 1: which nitrogen gets acylated?

So, idea one. An amine nitrogen has a lone pair, so it can attack. With an acid chloride or an anhydride, the hydrogen on nitrogen is replaced by an acyl group, giving an amide. Okay? But a tertiary amine has no N H bond, the short dash on screen, so it has no hydrogen on nitrogen to replace and is not acylated. Right? With ethanoic anhydride, this is acetylation: N H two becomes NHCOCH three, and ethanoic acid forms. The arrow means becomes, and the small low numbers count the atoms. With me? An amide has a nitrogen too, but its lone pair is shared with the C double bond O group, where the equals sign is a double bond. So the amide nitrogen is a far poorer nucleophile. Clear? Now JEE Advanced, two thousand and fourteen. Three aminomethyl benzamide, with a C H two N H two group and an amide group, is treated with ethanoic anhydride. Which nitrogen gets acetylated? Yes, exactly, only the C H two N H two group gets acetylated, because the amine nitrogen is the far better nucleophile. So we get three acetamidomethyl benzamide and ethanoic acid. Got it?

1:49Idea 2: chloroform and nitrous acid

Now idea two. Heat a primary amine, RNH two, with chloroform, CHC l three, and alcoholic KOH, and we get RNC, three K C l and three water, where R is alkyl or aryl. Okay? This RNC is an isocyanide, with a very offensive smell, and only primary amines give it, aliphatic or aromatic. That is the carbylamine test. Right? Next, nitrous acid, HNO two, made from sodium nitrite and hydrochloric acid. With a primary aliphatic amine, we get an alcohol, ROH, plus nitrogen gas, N two, and water. With me? First, a diazonium ion forms. From an aliphatic amine it is unstable and loses nitrogen at once, leaving a carbocation, a carbon with a positive charge. That carbocation can rearrange, so keep this in mind for our puzzle. Clear? But at two seventy three to two seventy eight kelvin, written with a capital K, a primary aromatic amine gives a stable diazonium salt, A r N two plus, C l minus, where A r is aryl and the raised signs are charges. Got it?

3:01Idea 3: aniline in strong acid (IIT-JEE 2001)

Now idea three, aniline. In the figure, structure one has its lone pair, the two dots, on nitrogen. In structures two to four, that pair has moved into the ring, putting a plus on nitrogen and a minus on an ortho or para carbon; the double headed arrows join resonance structures. So N H two activates the ring and directs ortho and para. Okay? Now a question from IITJEE, two thousand and one. Statement one: in strongly acidic solution, aniline becomes more reactive towards electrophilic reagents. True or false? Let us check it. In strong acid, the amino group is completely protonated, giving the anilinium ion, C six H five N H three plus, where the raised plus is its charge. Right? Now statement two says the lone pair holds a proton, H plus, so it is no longer available for resonance with the ring. And that is completely true. With me? So the positive nitrogen pulls electrons out of the ring, the ring is deactivated, and the group now directs meta. Clear? So statement one is false, and statement two is true. Protonation makes aniline less reactive, not more, so don't fall for the word more in statement one. Got it?

4:23Example 1: CH3NH2 with CHCl3 and KOH gives methyl isocyanide

Time for example one, an easy one from IITJEE two thousand and six. C H three N H two, plus chloroform, plus KOH, gives a nitrogen containing compound, plus K C l and water. Identify that compound. Pause the video and try it first. Done? Okay, let's check. C H three N H two, methanamine, is a primary amine, and with chloroform and KOH, this is our carbylamine test. Right? So put R equal to C H three. With three KOH, we get C H three N C, plus three K C l, plus three water, and both sides balance. Okay? Now, many students see C H three N C and write methyl cyanide. Is that the same compound? It is not the same, and this is the trap. Methyl cyanide is C H three C N, with the carbon in the middle. Here the amine nitrogen stays bonded to the methyl group, and the chloroform carbon joins that nitrogen at the end. With me? So the compound is methyl isocyanide. The three short lines are a triple bond, and the small plus over nitrogen and minus over carbon are their charges.

5:41Example 2: amine C5H13N and HNO2, hydride shift, alcohol Y

Clear? Now example two, the harder one, from IITJEE two thousand and five: our puzzle. An optically active amine X, C five H thirteen N, with aqueous sodium nitrite and hydrochloric acid, shown with a slash, gives nitrogen and a tertiary alcohol Y, along with other products. Find X and Y, whether Y is optically active, and the intermediates. Pause the video and try it first. Done? Okay, let's check. Nitrogen gas with nitrous acid tells us that X is a primary aliphatic amine. So C five H thirteen N is C five H eleven N H two. Okay? Now for Y, a tertiary alcohol with five carbons: its O H carbon is joined to three carbons. Only one fits, two methyl butan two ol, with two methyls and an ethyl on that carbon; the small two after the bracket means two C H three groups. Got it? So X has its N H two on that same carbon, which makes it two methyl butan two amine, right? That is the trap: that amine has two methyl groups on one carbon, so it is not optically active. Our answer is two methyl butan one amine, with N H two on the end carbon, and it fits every clue. Its carbon two, C two on screen, carries a hydrogen, a methyl, an ethyl and a C H two N H two group, so it is chiral, and X is optically active. With me? Now the intermediates, following the arrows. The diazonium ion, R N two plus, loses nitrogen at once, leaving a primary carbocation on the end carbon. Then a hydrogen on carbon two moves over with its electron pair, a hydride shift, giving a tertiary carbocation on carbon two. Water attacks it, giving the tertiary alcohol. Clear? Now the last part of the question: is Y optically active? It is not: carbon two of Y carries two identical methyl groups, so it is not chiral. And the O H landed on the next carbon because of the hydride shift, which solves our puzzle. One more thing. Three methyl butan two amine also fits every clue. It is chiral too, and with nitrous acid it gives a secondary carbocation on carbon two. Then a hydride shift from carbon three gives the same tertiary carbocation, and so the same alcohol, Y, again. Some answer keys give this amine, so either answer is fine. If you got this hard one, very well done!

8:24Quick revision: acylation, isocyanide, HNO2, anilinium ion

So, quick revision. In acylation, an amine nitrogen beats an amide nitrogen, and tertiary amines are not acylated. Chloroform and KOH give an isocyanide, from primary amines only. Nitrous acid with a primary aliphatic amine gives nitrogen gas and a carbocation that can rearrange. And in strong acid, the anilinium ion deactivates the ring and directs meta. Now try these questions yourself on this page. All the best!