AIEEE Questions: Reactions

Transcript

0:00Introduction

Hello everyone! Let us start with a chemist's problem. Your molecule has a ketone group, an O H group, and a carbon carbon double bond. You want to turn only the ketone's C double bond O into a plain C H two, and keep the rest unchanged. The famous method uses zinc amalgam with hot concentrated hydrochloric acid. But will the O H group and the double bond survive that acid? Interesting, right? This is a real question from the A I triple E paper of two thousand and twelve. Today we solve it, along with one more, from two thousand and six, on how fast aldehydes and ketones add hydrogen cyanide. Questions like these come up often in JEE, so let's begin.

0:46Adding HCN: who reacts faster, and why

So, the first idea. On the left of the figure is the carbonyl group, carbon double bonded to oxygen, with two groups, R and R prime, the R with a small tick. The carbon is flat, with bonds at one hundred and twenty degrees, the little circle meaning degrees. Oxygen is more electronegative, so it carries delta minus, a small negative charge, and the carbon carries delta plus, a small positive charge. So a nucleophile, N u minus, with the raised minus for its charge, attacks the delta plus carbon, the straight red arrow just after N u minus. Okay? Now, hydrogen cyanide, HCN, adds in exactly this way. But why does this reaction need a trace of base? Because HCN is a weak acid, and gives very few cyanide ions. The base supplies C N minus, the actual nucleophile, and the product is a cyanohydrin, with O H and C N on the same carbon. On the right of the figure you see how: the attack gives a tetrahedral alkoxide ion, O minus, which then takes up a proton, H plus, to give the O H group. Right? Now the big question. An aldehyde has one alkyl group and one hydrogen on its carbonyl carbon, and a ketone has two alkyl groups. So which one adds faster? The aldehyde, yes, exactly, for two reasons. Steric, which means crowding: two groups block the incoming nucleophile more than one. And electronic: two alkyl groups reduce the delta plus on the carbon more than one. The greater than sign on screen means faster here. With me? Now, a benzene ring on the carbonyl carbon is an aryl group, written A r. The ring reduces the delta plus even more, by resonance, so the nucleophile is attracted less. Clear? Putting it together, here is the course's order, most reactive first. Methanal, HCHO, then RCHO, where R is an alkyl group, then the aromatic aldehyde, A r CHO, and then the ketones: RCOR, then A r COR, and slowest of all, A r COA r. Keep this order handy, because Example one is built on it. Got it?

3:09Down to CH₂: Clemmensen or Wolff–Kishner

Now our second idea. Sometimes we want the carbonyl group to become a plain C H two group, where the small two written low counts the hydrogens. On screen, the equals sign between C and O is the double bond, the little angle before each C stands for the carbon's two other bonds, and the arrow means becomes. Okay? Now look at the fourth box on the right of the map. N a B H four, sodium borohydride, or L i A l H four, lithium aluminium hydride; the small fours written low count hydrogen atoms. Can these take the carbonyl group all the way down to C H two? No, they cannot, just as the box says. They reduce it only as far as the alcohol. Right? For C H two, look at the fifth box. Method one is the Clemmensen reduction: zinc amalgam, written Z n, dash, H g, for zinc and mercury, with concentrated hydrochloric acid, H C l. So it works in strong acid. With me? Method two is the Wolff-Kishner reduction. First hydrazine, N H two N H two, then KOH, potassium hydroxide, in ethylene glycol, written glycol on screen, with heating. So this one works in hot base. Clear? Now think about this. Your molecule also carries a group that strong acid would damage. Which of these two methods should you choose? Choose Wolff-Kishner, exactly, because it works in base, so acid-sensitive groups are safe. Got it? And the other way round. Hot base can eliminate or replace a halogen, so for base-sensitive groups, we choose Clemmensen. Remember this pair, because it decides Example two. Okay?

5:01Example 1: HCN addition rate, methanal to benzophenone

Time for Example one, from A I triple E two thousand and six. Arrange in order of increasing rate of addition of HCN: methanal, labelled A, propanone, labelled B, acetophenone, labelled C, and benzophenone, labelled D, okay? Increasing means the slowest comes first. Pause the video and try it first. Done? Okay, let's check. Methanal, HCHO, has no alkyl group at all, only two hydrogens on its carbonyl carbon. No crowding, and the largest delta plus, so it is at the top of the order, the fastest. Right? Now the ketones. Propanone has two methyl groups, so it is an RCOR ketone. Acetophenone has one methyl group and one benzene ring, so it is A r COR, one step lower in our order. With me? And benzophenone has two benzene rings on its carbonyl carbon, so it is A r COA r. So tell me, which of the four adds HCN the slowest? Benzophenone, yes, exactly, the very bottom of the course's order. So, slowest first, we get D, less than C, less than B, less than A, where each less than sign means slower. Here is a small trick: write the course's order from the top, then read it backwards. And watch the trap: if you wrote A first, you gave the decreasing order instead. Clear?

6:33Example 2: C=O to CH2 with an OH and a C=C present

Now Example two, our opening puzzle, from A I triple E two thousand and twelve. So, a compound has a ketone group, a hydroxyl group, O H, and a carbon carbon double bond. Which reagent is most appropriate to turn its C double bond O into C H two? The choices are zinc amalgam with H C l, where the slash means with; hydrazine, N H two N H two, with hydroxide ion, O H minus; sodium, N a, in liquid ammonia, N H three; or N a B H four. Pause the video and try it first. Done? Okay, let's check, by ruling out options one by one. N a B H four reduces a carbonyl group only to the alcohol, never to C H two, so it is out. Okay? Sodium in liquid ammonia is not one of the two methods we just learnt for this change, so that is out too. Clear? So, two are left. Zinc amalgam with H C l is the Clemmensen reduction, the famous one. So that must be the answer, right? That is exactly the trap in this question! Hot concentrated H C l would attack the O H group, and the double bond too; on screen that is the C equals C, the equals sign for the double bond. Clemmensen is not for acid-sensitive groups, remember? So it is out as well. With me? So the answer is hydrazine with hydroxide ion, the Wolff-Kishner reduction. It works in base, so the O H group and the double bond are spared, and only the ketone is reduced. And that solves our opening puzzle. Got it?

8:16Quick revision: HCN order and which reduction to pick

So, quick revision. For adding HCN, less crowding and more delta plus mean faster, so methanal is the fastest, and A r COA r is the slowest. Okay? Sodium borohydride and lithium aluminium hydride stop at the alcohol, while Clemmensen or Wolff-Kishner take the carbonyl group down to C H two. Right? And choose by the rest of the molecule: for acid-sensitive groups, Wolff-Kishner, and for base-sensitive groups, Clemmensen. Now go and try the practice problems on this page. All the best!