JEE Advanced Questions (2010–2016): Reactions

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from a JEE Advanced paper. We all learn the rule: aldehydes give a silver mirror with Tollens' reagent, and ketones do not. Yet one ketone, benzoin, gives the silver mirror. So how can a ketone do that? Let us find out with two JEE Advanced questions. So let's begin.

0:24Tollens' test: who makes the silver mirror?

So, first our tool, in the Tollens' box on the right of the course map. Tollens' reagent is ammoniacal silver nitrate. On screen is its silver ion: silver, A g, with N H three in round brackets and a small two after them, for two ammonia molecules, all inside square brackets; the plus written high is its charge. Okay? Warm it with an aldehyde, RCHO, where R is the rest of the molecule. The aldehyde is oxidised to RCOO minus, the carboxylate ion; the arrow means becomes. And silver goes from plus one to zero, coating the glass as a shining mirror. Right? Now let me ask you this. Benzaldehyde is an aromatic aldehyde, with its CHO group on a benzene ring. Does it give the silver mirror too? Yes, it gives the mirror too, very good! Tollens' test is given by all aldehydes, aliphatic and aromatic. With me? But do not mix it up with Fehling's solution, in the same box. It gives a red brown precipitate of copper one oxide, C u two O; the small two written low is a subscript. And aromatic aldehydes do not respond to it. Got it? Now ketones. Ketones resist mild oxidising agents, so they give no mirror. But can any ketone give the silver mirror? Yes, some ketones can, and this is the twist! Alpha hydroxy ketones, written with the Greek letter alpha, have an O H group on the alpha carbon, the carbon next to the carbonyl group, and they reduce Tollens' reagent. Methanoic acid, in round brackets, does it too. This line is the key to our puzzle. Clear? But simple ketones, with no such O H group, give no silver mirror. Right?

2:16Cyanohydrins, and what hot acid does next

Now our second tool. In the figure, the carbonyl carbon carries a partial positive charge, delta plus, so N u minus, a nucleophile, attacks it along the red arrow. The carbon turns from flat s p two to tetrahedral s p three, giving an alkoxide ion, which then takes up a proton, H plus. Here our nucleophile is the cyanide ion, C N minus. Okay? The product is a cyanohydrin. The angle sign before the C stands for the two other bonds of that carbon, and the round brackets show it carries both O H and C N on that carbon. With me? Now we hydrolyse the cyanohydrin. Tell me, what does the C N group become? It becomes a COOH group, a carboxylic acid, exactly! The O H sits on the carbon next to it, the alpha carbon, so we get an alpha hydroxy acid. Clear? One more tool, from our chapter on alcohols. Heated with concentrated sulphuric acid, H two S O four, an alcohol loses water to give an alkene. Got it? If two different alkenes can form, the major one is the more substituted alkene, with more groups on its double bond carbons. That is Saytzeff's rule, in round brackets. Right?

3:35Example 1: which compounds give a positive Tollens' test

Alright, example one, from JEE Advanced twenty sixteen. Which of these give a positive Tollens' test? Propenal; benzaldehyde; benzoin, whose formula on screen reads C six H five, C H with O H in round brackets, C O, C six H five, the dashes being single bonds; and one three diphenyl prop two en one one. Pause the video and try it first. Done? Okay, let's check. Propenal is C H two, double bond, C H, then CHO; the equals sign is the double bond. It has a CHO group, so it is an aldehyde and gives the mirror: positive, okay? Next, benzaldehyde. It is an aromatic aldehyde, and aromatic aldehydes do not respond, so it is negative, right? No, no, careful, that is the trap! Aromatic aldehydes fail Fehling's test, not Tollens' test. Tollens' reagent is reduced by all aldehydes, so benzaldehyde, C six H five CHO, is positive. Got it? Now benzoin, our puzzle. The C O in the middle makes it a ketone. So does it give the silver mirror, or not? It does, and here is why. The carbon right next to that C O group carries an O H group. So benzoin is an alpha hydroxy ketone, and those reduce Tollens' reagent. It is positive, and our opening puzzle is solved! Clear? Last, one three diphenyl prop two en one one, whose structure is C six H five, C O, C H double bond C H, C six H five. It has no CHO group and no O H on the alpha carbon, so it is a simple unsaturated ketone: no mirror, negative. With me? So the answer is propenal, benzaldehyde and benzoin, three out of four. That's it!

5:43Example 2: butan-2-one to cyanohydrin, then hot H2SO4

Now the harder one, from IITJEE twenty twelve. Butan two one is treated with cyanide ion, C N minus, and acid, to give G; then G is heated with ninety five percent sulphuric acid to give H, okay? Identify both, and pause the video to try it first. Done? Okay, let's check. Butan two one is C H three, C O, C H two, C H three, with the ketone group on carbon two and dashes for single bonds. Right? Now the cyanide ion arrives. Where does it attack, and what is G, our first product? On the carbonyl carbon, yes, exactly! C N minus adds there, and the oxygen picks up a proton from the acid. So G is C H three, then a carbon carrying O H and C N in round brackets, then C H two, C H three: a cyanohydrin. Numbering from the C N carbon, it is two hydroxy two methyl butane nitrile. Okay? Now G meets hot ninety five percent sulphuric acid. The acid first hydrolyses the C N group into a COOH group, giving the alpha hydroxy acid, just as our tool said. With me? So H is just this alpha hydroxy acid, and we are done, right? No, no, wait, that is the trap! The carbon holding that O H group carries three carbon groups, so it is a tertiary alcohol, and tertiary alcohols lose water most easily of all. Hot concentrated acid removes it as water, and we get an alkene. Clear? But which hydrogen leaves with the O H group? Two neighbouring carbons carry hydrogens: the methyl group, C H three, and the C H two of the ethyl group. Which one gives it up? The C H two, very good! Then the double bond carries three groups, while the other choice would carry only two. By Saytzeff's rule, the more substituted alkene is the major product. Got it? So H is C H three, C H, double bond, then a carbon with C H three in round brackets, then COOH, the acid group. G is two hydroxy two methyl butane nitrile, and H is two methyl but two enoic acid. That's it!

8:09Quick revision: Tollens', Fehling's and cyanohydrins

So, quick revision. Tollens' reagent gives the silver mirror with all aldehydes and alpha hydroxy ketones, but not with simple ketones. Fehling's solution does not respond to aromatic aldehydes. And butan two one plus cyanide ion gives a cyanohydrin; hot acid turns C N into COOH, and then the alkene forms. Now try the other JEE Advanced questions on this page. All the best!