Nucleophilic Addition

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Your molecule has a ketone at one end and an ester at the other. You want to reduce only the ester, but the reducing agent would attack the ketone too. So, can you hide a ketone, and get it back unchanged later? Yes, you can, using today's topic, nucleophilic addition to aldehydes and ketones. It comes up often in JEE, so let's begin.

0:27The attack: a flat carbon turns tetrahedral

So, look at the left part of the figure. The carbonyl group is a carbon double bonded to oxygen, with two groups, R and R prime, at about one hundred and twenty degrees. Oxygen is more electronegative, so delta minus, a small negative charge, sits on oxygen, and delta plus, a small positive charge, on carbon. Now a nucleophile arrives, N u minus, where the raised minus is its negative charge. Does it attack the carbon, or the oxygen? It attacks the carbon, yes, exactly! The delta plus carbon is the electrophilic centre, and the straight red arrow from N u minus shows the attack. Okay? And it comes in roughly perpendicular to the plane of the group, so from above the flat group, or from below. Right? Now the curved red arrow shows the electrons of the double bond moving onto oxygen. So the carbon goes from s p two, planar, to s p three, tetrahedral; the raised numbers are part of the names, and the arrow means becomes. We get an alkoxide ion, with O minus on oxygen, and the figure marks this step slow. With me? Then the alkoxide takes up a proton, H plus, in a fast step. O minus becomes O H, and that is the addition product. Clear?

1:45Who reacts faster? Crowding and charge

Now the big question. An aldehyde has one alkyl group and a hydrogen on the carbonyl carbon, while a ketone has two alkyl groups. So which one reacts faster with a nucleophile? The aldehyde is faster, correct, for two reasons. First, steric, which means crowding: two alkyl groups block the incoming nucleophile more than one. Second, electronic: two alkyl groups reduce the delta plus on the carbon more than one. Okay? So here is the full order, with a greater than sign between each pair. Methanal, HCHO, is the most reactive, then RCHO, then the aromatic aldehyde; A r means an aryl group, a benzene ring. Then the ketones: RCOR, then A r COR, and slowest of all, A r COA r, with two aryl groups. Got it? But wait, benzaldehyde is an aldehyde too, with only one group, the ring, on its carbonyl carbon. So why does it sit below RCHO? Many students miss this point. The ring reduces the delta plus on the carbonyl carbon by resonance, so the nucleophile is attracted less. Right? Groups on the ring matter too. An electron withdrawing group like nitro increases the reactivity, so p nitrobenzaldehyde beats benzaldehyde; the italic p means para. With a methyl group instead, p tolualdehyde is the slowest of the three. Clear?

3:20HCN, bisulphite and Grignard reagents

Now the additions themselves. Hydrogen cyanide, HCN, with a trace of base, gives a cyanohydrin: the carbon now carries O H, in brackets, and C N, and the sign like a greater than sign before the C stands for its two other bonds. But why do we need that trace of base at all? Because hydrogen cyanide is a weak acid and gives very few cyanide ions. The base generates C N minus, the actual nucleophile. Okay? Next, sodium hydrogensulphite, N a HSO three, where the small three written low counts the oxygen atoms. It gives a crystalline compound, which dilute acid or alkali decomposes again, so we use it to separate and purify. Right? And it is formed by aldehydes, by aliphatic methyl ketones like propanone, and by unhindered cyclic ketones. The reagent is bulky, so crowded ketones like pentan three one, and aryl ketones like acetophenone, do not form it. With me? Now, Grignard reagents add too, giving alcohols. Methanal gives a primary alcohol, other aldehydes give secondary alcohols, and ketones give tertiary alcohols. Got it?

4:35Alcohols and water: acetals and hydrates

Now alcohols, and this answers our puzzle. An aldehyde with an alcohol and dry H C l gives a hemiacetal, and then an acetal: RCH, then O R prime in brackets with a small two outside, so two O R prime groups on one carbon. R prime is just another alkyl group. Okay? Ketones do this with ethane one two diol, which has two O H groups, and dry H C l. Both O H groups are in one molecule, so the product is a ring, a cyclic ketal. Right? Now the key property: acetals are stable to bases, to nucleophiles and to reducing agents. With me? But when the job is done, we want the carbonyl group back. So how do we undo the acetal? With aqueous acid, exactly! It hydrolyses the acetal, and the carbonyl group comes back. So acetals protect a carbonyl group: hide it, react elsewhere, then bring it back. Got it? And last, water adds to give a hydrate, a gem diol, short for geminal diol, with two O H groups on one carbon. But this matters only for methanal and for chloral, C C l three CHO, okay?

5:50Example 1: order four carbonyls by reactivity towards HCN

Time for Example one. Arrange in order of increasing reactivity towards HCN: acetophenone, benzaldehyde, ethanal and benzophenone. Pause the video and try it first. Done? Okay, let's check. Here is a quick trick: match each compound to a type in our order. Ethanal is RCHO, benzaldehyde is A r CHO, acetophenone, with a ring and a methyl group, is A r COR, and benzophenone, with two rings, is A r COA r. Okay? Now, one word in the question decides everything, increasing. So which compound do we write first? The least reactive one, benzophenone, yes! Many students write the most reactive first and lose marks. So, benzophenone, less than acetophenone, less than benzaldehyde, less than ethanal, with a less than sign between each pair, and that is the answer!

6:50Example 2: protect the ketone as a ketal, reduce the ester

Now Example two, a harder one: our puzzle from the start. How would you convert C H three, C O, C H two, C H two, COOC two H five, into C H three, C O, C H two, C H two, C H two O H? The short dashes are single bonds. Pause the video and try it first. Done? Okay, let's check. The ketone end, C H three C O, stays the same, and only the ester group becomes C H two O H, okay? So we need a reducing agent like L i A l H four, lithium aluminium hydride. Just add it and we are done, right? No, and this is exactly the trap. Lithium aluminium hydride would reduce the ketone too. So we must hide the ketone first. With me? Step one: ethane one two diol with dry H C l turns the ketone into a cyclic ketal, so the ketone is protected. Right? Step two: lithium aluminium hydride reduces the ester to the C H two O H group, and the ketal, stable to reducing agents, survives. Got it? Step three: dilute aqueous acid hydrolyses the ketal, and the ketone comes back. Protect, react, then remove the protection. That is how you hide a ketone, and our puzzle is solved!

8:20Quick revision: mechanism, reactivity order, acetals

So, quick revision. The nucleophile attacks the delta plus carbon, which goes from s p two to s p three, and then takes up a proton. Aldehydes beat ketones, with less crowding and more delta plus. And acetals protect a carbonyl group: dry H C l to make them, and aqueous acid to remove them. Now go and try the five practice problems on this page. All the best!