Hofmann rearrangement: practice and mechanism
Br₂ and NaOH turn the amide into an N-bromoamide. The group migrates from C to N, giving an isocyanate that hydrolyses to CO₂ and the amine. The amine has one carbon fewer.
Example
The primary amide is brominated at nitrogen and deprotonated. The group migrates from the carbonyl carbon to the nitrogen, giving an isocyanate. Water hydrolyses it and CO₂ is lost: the amine has one carbon fewer than the amide.
Mechanism step by step
- 1
Hydroxide deprotonates the amide and the anion attacks Br₂: N-bromoamide. A second deprotonation gives the bromoamide anion.
- 2
Bromide leaves while the group R moves from C to N: an isocyanate (R–N=C=O) forms.
- 3
Water adds to the isocyanate to give a carbamic acid, which loses CO₂: the primary amine R–NH₂.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The nitrile would form by dehydration (e.g. with SOCl₂ or P₂O₅). In the Hofmann rearrangement the carbonyl carbon is lost as CO₂.
- Bromine is not built into the carbon chain, only briefly onto nitrogen. In the end an NH₂ group sits where the CONH₂ group used to be.
Often confused with: Reduction with lithium aluminium hydride · Acidic hydrolysis of amides and nitriles
Frequently asked questions
Why one carbon fewer?
The amide carbonyl carbon becomes the isocyanate carbon and is finally lost as CO₂.
Is stereochemistry retained?
Yes. The migrating group keeps its configuration (retention) because it never fully detaches from the molecule.