JEE Advanced Questions (2013–2016): Phenols

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from a JEE Advanced question. One compound gives a dark violet colour with ferric chloride. We attach one benzyl group to it, and the violet is gone, yet a test for its aldehyde group still works. So where did the violet go? And the whole story starts from just benzene and propene. Questions like these come up often, so let's begin.

0:27The cumene process: from benzene to phenol

So, first our tools. Beside us is the course map of phenol: ways to make it on the left, its reactions on the right. The top route on the left uses cumene. Benzene plus propene gives cumene, also called isopropylbenzene, in round brackets; the arrow means gives. Okay? Next, air oxidises cumene; it is the oxygen in air, O two, that reacts, and the small two written low is a subscript. The product is cumene hydroperoxide. Right? Now dilute acid acts on the hydroperoxide, and it splits into two products. Can you find both of them on the map? Follow the arrow from that top box. Phenol, in the centre box, and acetone, written in brackets in the top box, yes, exactly! Keep this chain in mind, because both our examples sit on it.

1:18Reimer–Tiemann: putting CHO on the ring

Clear? Now the second tool, on the right of the map: the Reimer-Tiemann reaction, which puts an aldehyde group, CHO, on the ring. Phenol is treated with chloroform, CHC l three, and aqueous sodium hydroxide, N a O H, at three hundred and forty kelvin, written with a capital K, okay? The semicolon means then: acid comes after that. With me? The electrophile is dichlorocarbene, C C l two. The two dots in front of it are a lone pair on the carbon. Okay? Now let me ask you this: where on the ring does dichlorocarbene mainly attack, ortho, meta or para to the O H group? Ortho, yes, exactly! The intermediate is the ortho dichloromethyl phenoxide ion; the small italic o means ortho, and the round brackets hold the dichloromethyl group. Got it? After hydrolysis and acid, that dichloromethyl group becomes the aldehyde group. So the product is salicylaldehyde, two hydroxybenzaldehyde, with the O H and the CHO side by side. Clear?

2:22Reading the options: three tests and one ether

Advanced questions often end with statements to judge, so we need our tests. Neutral ferric chloride, F e C l three, gives a violet colour: a test for phenols, and, in round brackets, enols. Okay? Next, two four DNP, short for two four dinitrophenylhydrazine. It gives a yellow, orange or red precipitate with aldehydes and ketones. Right? Now recall ortho nitrophenol and para nitrophenol from an earlier lesson. Which one is carried away by steam? The ortho one, very good! Its O H and nitro groups form an intramolecular hydrogen bond, inside one molecule; H bond on screen means hydrogen bond. So it is steam volatile, and so is salicylaldehyde. With me? Their para isomers, written with a small italic p, have the groups opposite each other, so they form intermolecular hydrogen bonds, between molecules, and are not steam volatile. Got it? And last, the Williamson synthesis: sodium phenoxide plus an alkyl halide gives an ether. Clear?

3:30Example 1: identify T and U from benzene, propene and O2

Alright, example one, from JEE Advanced twenty fifteen. Benzene reacts with propene and an acid catalyst to give T, and T with oxygen, in the presence of a radical initiator, gives U, okay? Identify both, and pause the video to try it first. Done? Okay, let's check. Many students reason like this: propene has three carbons in a chain, so the ring simply picks up a propyl group, and T is propylbenzene. Is that right? Oh no, careful, that is the trap! The acid first adds a proton to propene, and the positive charge lands on the middle carbon: C H three, C H, C H three, with the small plus over the middle carbon, and dashes for single bonds. That is the secondary, isopropyl cation, more stable than a primary one, as in Markovnikov's rule. Okay? So T equals cumene, isopropylbenzene: C six H five, C H, then C H three in brackets with a small two, meaning two methyl groups on that carbon. It is not propylbenzene. Right? Now oxygen, with a radical initiator. Cumene has many C H bonds, on the ring and on the side chain. Which one does oxygen attack? The C dash H bond on the central carbon of the side chain, yes, exactly! That carbon is tertiary and benzylic, right next to the ring, so breaking this bond gives the most stable radical. Clear? So U is cumene hydroperoxide: C six H five, C, C H three in brackets with a small two, then dash O, dash O H, okay? That carbon now carries the hydroperoxide group. Got it? And one more trap: some students write phenol for U, but no acid has been added in this step. Phenol comes only later, with dilute acid. With me?

5:21Example 2: Reimer-Tiemann, Williamson, FeCl3 vs 2,4-DNP

Now the harder one, from JEE Advanced twenty sixteen. Cumene with oxygen, then H three O plus, gives P; the plus written high is a charge. P with chloroform and sodium hydroxide, the slash meaning together, gives Q, major, and R, minor. Q with sodium hydroxide and P h C H two B r, where P h is phenyl, gives S, okay? Read the four statements, A to D in brackets, about steam, one percent aqueous ferric chloride and two four dinitrophenylhydrazine. Which are correct? Pause the video and try it first. Done? Okay, let's check. This is our road map: oxygen turns cumene into the hydroperoxide, our U from example one, and the acid splits it. So P is simply phenol. Clear? Next, chloroform and sodium hydroxide on phenol: the Reimer-Tiemann reaction. Ortho attack gives Q, salicylaldehyde, the major product, and some para attack gives R, para hydroxybenzaldehyde, the minor one. Right? Now Q meets sodium hydroxide and benzyl bromide. The sodium hydroxide turns its phenolic O H into a phenoxide ion. So what happens next? The Williamson synthesis, very good! The phenoxide attacks benzyl bromide, and the O H becomes a benzyloxy group, an ether. So S is two benzyloxy benzaldehyde: the CHO is kept, but no free phenolic O H is left. Clear? Now the statements. Statement A says R is steam volatile. R has an O H and a CHO group, just like Q, so it should be steam volatile too, right? No, no, wait, that is the trap! In R the two groups are para, opposite each other, so no intramolecular hydrogen bond is possible. R forms intermolecular hydrogen bonds and is not steam volatile; it is Q, the ortho one, that goes with steam. So statement A is false. Got it? Now statements B and D, both about ferric chloride. Q gives the dark violet colour, and we made S from Q, okay? So does S give the violet colour too? It does not, and this answers our opening puzzle! Q has a free phenolic O H, so it gives the violet colour, and statement B is true. But in S that O H has become an ether, so no violet colour: statement D is false. With me? Last, statement C: S still has its aldehyde group, so with two four DNP it gives a yellow precipitate, and the statement is true. So the correct statements are B and C: the aldehyde test still works, but the violet colour is gone. That's it!

8:06Quick revision: cumene route, Reimer-Tiemann, FeCl3, 2,4-DNP

So, quick revision. Benzene and propene with acid give cumene, not propylbenzene; oxygen, then acid, gives phenol. The Reimer-Tiemann reaction, chloroform with sodium hydroxide, puts the CHO group mainly ortho: salicylaldehyde. And the violet colour with ferric chloride needs a free phenolic O H, while two four DNP finds aldehydes and ketones. Now try the other JEE Advanced questions on this page. All the best!