Hofmann Elimination

Chapter 22: Amines

Reaction type: Elimination

Summary

Quaternary ammonium salts undergo an E2 elimination when heated with silver oxide, Ag₂O, in water.
Amines can readily be converted into quaternary ammonium iodides by treating them with excess methyl iodide.
AgO / H₂O reacts giving the quaternary ammonium hydroxide, silver iodide precipitates.

completely methylate then convert to the ammonium hydroxide salt

When heated the hydroxide induces a base promoted 1,2- or b-elimination giving an amine and alkene.
The regioselectivity is opposite to that predicted by Zaitsevs rule in that it leads to the less highly substituted alkene.
The less highly substituted alkene may be referred to as the Hofmann product.

1,2-elimination of the ammonium salt gives mainly the less highly substituted alkene

The outcome is dictated by steric effects of the large leaving group and the alkyl chain (more details ?)
NH₂^-and NR₂^- are very poor leaving groups (both anionic), but NR₃ is much better (neutral)

Compare this with -OH and H₂O in the dehydration of alcohols.

Related reactions

MECHANISM OF THE HOFMANN ELIMINATION
The initial steps are an example of the alkylation of an amine by methyl iodide. The mechanism of the elimination step is shown. When heated, the hydroxide removes the more accessible proton, the p bond of the alkene forms and the leaving group, a neutral amine departs.

Selectivity of the Hofmann Elimination

In general E2 reactions occur most rapidly when the H-C bond and C-LG bonds at 180^o with respect to each other. This is described as an antiperiplanar conformation. This conformation positions the s bonds that are being broken in the correct alignment to become the p bond.


	The staggered, antiperiplanar alignment is preferred because it aligns the two s bonds that become the p bond.

Lets' look at the Hofmann elimination and the selectivity of the reaction:

Highlight leaving group, N(CH₃)₃

Show antiperiplanar H

Reset colors

The CHIME image to the left shows the calculated energy minima of the sec-butyltrimethylammonium compound.

Since these reactions are E2, we need to look at the bonds that are antiperiplanar to the C-N bond that is broken.

Orient the molecule so that you can see this.

The only C-H bond antiperiplanar to the C-N bond is in the C1 methyl group and so leads to the formation of 1-butene.

Highlight leaving group, N(CH₃)₃

Show antiperiplanar H

Reset colors

The CHIME image to the left shows the of the sec-butyltrimethylammonium compound in the conformation that leads to the formation of trans-2-butene.

Again look at the bonds that are antiperiplanar to the C-N bond that is broken, this time there are two. Can you see them ?

One is on the -CH₂- and leads to the formation of 2-butene, the other is in the C1 methyl group and so leads to the formation of 1-butene.

Note that this conformation puts the large N(CH₃)₃ and the C4 methyl group gauche to each other, this destabilizes the conformation compared to the one that leads to 1-butene. Using the space filling model to check.....