Hofmann elimination: practice and mechanism
Excess CH₃I methylates the amine completely to the trimethylammonium salt. Ag₂O and heat cause elimination: the less substituted alkene (Hofmann product) is favoured.
Example
Exhaustive methylation turns the amine into a quaternary ammonium salt, a good leaving group. Silver oxide supplies hydroxide and heating causes an E2 elimination. Because of the bulky leaving group the less substituted alkene is favoured (Hofmann rule).
Mechanism step by step
- 1
CH₃I methylates the amine three times (SN2): a quaternary trimethylammonium iodide.
- 2
Ag₂O exchanges iodide for hydroxide (AgI precipitates).
- 3
On heating OH⁻ removes a β-H and trimethylamine leaves (E2): preferably the less substituted alkene.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The bulky NMe₃⁺ leaving group steers the base to the more accessible H. The less substituted alkene forms, not the Zaitsev product.
- Ag₂O does supply hydroxide, but heating causes elimination, not substitution. An alkene forms.
- The ammonium salt is only the intermediate after step 1. With Ag₂O and heat trimethylamine is eliminated.
Frequently asked questions
Why the Hofmann product and not Zaitsev?
The large N(CH₃)₃⁺ leaving group hinders access to the inner H atoms. The C–N bond is also hard to break, so the transition state has carbanion character, and primary carbanions are more stable.
What was the reaction used for historically?
For structure elucidation of alkaloids: repeated Hofmann eliminations removed the nitrogen from the molecule step by step.