The reactivity of nitrogen mustards increases with increasing nucleophilicity of the central nitrogen atom. Select the most and least reactive from each of the following groups of nitrogen mustards.

enter image description here

My Attempt

I know that II will be the most reactive since the methyl group will donate electron density to the nitrogen atom due to inductive effects, making it more nucleophilic and hence more reactive.

I also thought that III will be the least reactive since the ester group pulls more electron density away from the nitrogen atom compared to the carbonyl group. I also thought in general that esters are stronger deactivators than carbonyl groups. However the answer is I.

After searching on the internet, it turns out that esters and carbonyl groups are roughly equally strong deactivators. So how am I suppose to determine which one will be less nucleophilic?


1 Answer 1


Your analysis of "pulls more electron density away" is a bit too simplistic. Both of them are approximately equally electron-withdrawing via the inductive effect, but in terms of resonance, which is the main factor here, they are quite different.

Compound I is an amide. Amides are non-nucleophilic on nitrogen because the N lone pair is delocalised into the C=O π* orbital. In fact, amides act as nucleophiles via the carbonyl oxygen and not the nitrogen.

In compound III, there is also the same delocalisation, but a way of looking at it is that the other oxygen (on the -OMe group) also wants to donate its lone pair into the C=O π*. In a sense this means that there is some degree of competition between the nitrogen and the oxygen, and the nitrogen lone pair is no longer delocalised to such a great extent.

Formally speaking, I think the best way of putting it is that the HOMO of compound III has a greater coefficient on nitrogen than the HOMO of compound I. You can also draw resonance forms to show that there is a smaller contribution from the resonance form with positively charged nitrogen.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.