Hofmann Elimination

by Kiley Lynch

Two key reactions of amines to talk about today. They’re very different reactions, but often seen together. 

First of all, yesterday I said that amines are good nucleophiles. Waaaay back in Org 1 you probably talked about nucleophilic substitution (remember the SN2? backside attack?) 

Amines, being nucleophiles, can certainly participate in this reaction with alkyl halides. However, you probably didn’t see many examples of amines in this reaction. Why not?

Well, amines have the same relationship with alkyl halides that I have with potato chips. It’s hard to have just one.

A primary amine (like that shown) will react with an alkyl halide like CH3-I, to make a secondary amine. Once that’s formed, however, the secondary amine is still a nucleophile. It can react with another alkyl halide, making a tertiary amine. Which can then add a final alkyl halide until we have made a nitrogen with four alkyl groups, and there aren’t any lone pairs left. I didn’t draw out the full mechanism – it’s here. 

It’s impossible to get this reaction to just stop after one alkylation. So even if you add only one equivalent, you’ll still get a mixture of product. If we add a large excess of alkyl halide, we can get one product – the “quaternary ammonium” salt. 

Quaternary ammonium salts are not very exciting chemical species. However, they do perform one interesting reaction. 

The ammonium (NR3)+ is a good leaving group. Treatment with a base will lead to  E2 reactions, where we form an alkene. This is called the Hofmann elimination. A common base for this is silver oxide (Ag2O) although many other bases will also work. 

This E2, however, has a twist. Unlike most E2 reactions, which form the most substituted double bond (the Zaitsev product) these ones end up always forming the least substituted double bond. That’s weird, because the Zaitsev is usually favored due to more substituted alkenes being more stable. So what’s going on here?

The answer lies in the transition states. If you compare the transition states leading to the two alkenes (and draw out our old friend, the Newman Projection) you’ll find that the transition state leading to the Zaitsev product has more steric hindrance. That’s because the leaving group is very bulky, and it has a steric clash with the carbons adjacent to it in the Zaitsev product. Meanwhile, the Hofmann elimination transition state doesn’t have this problem with steric hindrance. So the transition state is lower in energy, and therefore, the product is formed at a greater rate.

Bottom line: ammonium salts are bulky leaving groups, and from them we obtain the “non-Zaitsev” alkene.

Tomorrow: a nifty way around this problem helps us make primary amines. 

Thanks for reading! James