Hofmann Degradation

Theory
Problems and exercises
Solutions
Where I am
Molecular panel (opens in a new tab)

Degradation is one of the most important reactions of amines. It leads to olefins.

Hofmann degradation: the quaternary ammonium salt with Ag₂O and water gives the ammonium hydroxide, which on heating eliminates a tertiary amine to form the alkene — General scheme of the Hofmann degradation

August Wilhelm von Hofmann (8 April 1818 – 5 May 1892) was a German chemist.

Look at the following example of a "double" degradation:

Double Hofmann degradation of 1,2-dimethylpiperidine: methylation with MeI, Ag₂O and heat open the ring; a second sequence gives 1,5-hexadiene

What is the mechanism of the Hofmann degradation? Let's see whether the following experimental fact gives you a clue!!!

Hofmann degradation of the 4-tert-butylcyclohexyltrimethylammonium hydroxides: the trans isomer, with the equatorial group, does not react; the cis isomer, with the axial group, gives the alkene

The tert-butyl group locks the conformational equilibrium. In the trans isomer the trimethylammonium group is equatorial and has no hydrogens anti to the nitrogen leaving group; only the cis isomer, with the axial group, has them. The E2 mechanism has this requirement. Therefore, the Hofmann elimination takes place by an E2 mechanism in which the base is hydroxide, which always attacks the least hindered position.

The reaction is kinetically controlled leading to the monosubstituted olefin which is less stable. The more substituted the olefin, the more stable.

Energy diagram of the Hofmann elimination of the 2-butyltrimethylammonium ion: hydroxide attacks the hydrogen on the less hindered carbon (more favourable TS) and 1-butene is formed, although 2-butene is the more stable olefin