Elimination — regiochemistry contrast
Hofmann elimination of a quaternary ammonium salt
How the Hofmann elimination of a quaternary ammonium salt works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.
In short. A 3-step mechanism. Typical reagents: (CH₃)₃N⁺–R I⁻ → Ag₂O/H₂O (gives HO⁻). Conditions: Heat 100–200 °C. Bonds formed: C=C π (less substituted); H₂O; bonds broken: β C–H; C–N⁺. Stereochemistry: Anti-periplanar H and N⁺; Hofmann (less substituted) alkene. Rate and selectivity: E2: rate = k[R–NMe₃⁺][HO⁻].
At a glance
Step 1: Base meets the least hindered β-H
The bulky trimethylammonium leaving group crowds the substrate so much that even a small, unhindered base (like hydroxide, heated) can only reach the most accessible, least hindered β-hydrogen — sterics dominate over the electronic preferences seen in E2 with small groups.
Step 2: Concerted E2
Concerted E2: the base removes that hydrogen while the C–N bond breaks and a new π bond forms, releasing neutral trimethylamine.
Step 3: Hofmann alkene + trimethylamine
The result is the less-substituted alkene (the Hofmann product) — the opposite regiochemical outcome from the Zaitsev-favored elimination you'd expect with a small halide leaving group.
Abbreviations and terms
- R
- R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
- H2
- Hydrogen gas. Reductant for catalytic hydrogenation.
- E2
- Bimolecular elimination. Concerted base removal of β-H and loss of the leaving group (anti-periplanar).
- Zaitsev
- Zaitsev’s rule. Elimination usually gives the more substituted (more stable) alkene.
- Hofmann
- Hofmann product/rule. Elimination giving the less substituted alkene (bulky base or leaving group).
How it compares with related reactions
| Hofmann elimination of a quaternary ammonium salt | E2 | Aldol condensation | |
|---|---|---|---|
| Reagents | (CH₃)₃N⁺–R I⁻ → Ag₂O/H₂O (gives HO⁻) | NaOEt / EtOH (strong, unhindered base) | NaOH or KOH (cat.) |
| Conditions | Heat 100–200 °C | Heat, ethanol | H₂O/EtOH, heat 50–100 °C |
| Stereochemistry | Anti-periplanar H and N⁺; Hofmann (less substituted) alkene | Anti-periplanar H and Br required | Mostly E-enal (thermodynamic) |
| Rate / selectivity | E2: rate = k[R–NMe₃⁺][HO⁻] | Rate = k[RBr][base]; Zaitsev product usually major | E1cB; loss of HO⁻ from the enolate is rate-determining |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the Hofmann elimination of a quaternary ammonium salt?
(CH₃)₃N⁺–R I⁻ → Ag₂O/H₂O (gives HO⁻). Conditions: Heat 100–200 °C.
Which bonds are formed and which are broken?
Formed: C=C π (less substituted); H₂O. Broken: β C–H; C–N⁺.
What is the stereochemical outcome?
Anti-periplanar H and N⁺; Hofmann (less substituted) alkene
What controls the rate or the selectivity?
E2: rate = k[R–NMe₃⁺][HO⁻]
Study guides that help with this reaction
- SN1, SN2, E1 or E2? How to decide which mechanism wins
- Nucleophiles, electrophiles and leaving groups: what makes a reaction go