The Hofmann elimination uses methyl iodide, silver oxide, water, and heat to convert a 1°, 2°, or 3° amine with a β-hydrogen to an alkene. First exhaustive methylation of the amine occurs with excess methyl iodide to yield the corresponding ammonium iodide salt. The iodide ion then attacks silver oxide, forming insoluble silver iodide, and a silver oxide ion, which deprotonates water to make a hydroxide ion. Upon heating, hydroxide ion abstracts a β-hydrogen, forming a 3° amine, an alkene product and water. Preferential formation of the less substituted olefin (Hofmann product) occurs in acyclic substrates.

• Reagents: Methyl Iodide, Silver Oxide, Water
• Reactant: 1°, 2°, or 3° Amine with β-Hydrogen
• Product: Alkene, Trimethylamine
• Type of Reaction: β-Elimination (E2)

Mechanism

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Related Compounds