Hofmann elimination

The least stable alkene (the one with the fewest substituents on the carbons of the double bond), called the Hofmann product, is formed.

[1][2] The reaction starts with the formation of a quaternary ammonium iodide salt by treatment of the amine with excess methyl iodide (exhaustive methylation), followed by treatment with silver oxide and water to form a quaternary ammonium hydroxide.

When this salt is decomposed by heat, the Hofmann product is preferentially formed due to the steric bulk of the leaving group causing the hydroxide to abstract the more easily accessible hydrogen.

In the Hofmann elimination, the least substituted alkene is typically favored due to intramolecular steric interactions.

[4] The trans isomer is selectively trapped as a complex with silver nitrate (in this diagram the trans form looks like a cis form, but see the trans-cyclooctene article for better images): In a related chemical test, known as the Herzig–Meyer alkimide group determination, a tertiary amine with at least one methyl group and lacking a beta-proton is allowed to react with hydrogen iodide to the quaternary ammonium salt which when heated degrades to methyl iodide and the secondary amine.