Beckmann rearrangement: practice and mechanism
In acid the oxime rearranges to the amide: the group anti to the OH migrates from C to N. Cyclic ketoximes give lactams (ring expanded by the N).
Example
The acid protonates the oxime OH group and water leaves. At the same time the group anti to the OH moves from C to N. Water attacks the nitrilium ion and tautomerism gives an amide. Cyclohexanone oxime becomes caprolactam (the monomer of nylon-6).
Mechanism step by step
- 1
The acid protonates the oxime OH group, making it a leaving group (H₂O).
- 2
As water leaves, the group anti to the OH migrates from C to N: a nitrilium ion forms.
- 3
Water attacks the nitrilium carbon; after tautomerism (imidic acid → amide) the amide or lactam forms.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Hydrolysis back to the ketone happens with dilute acid. With concentrated H₂SO₄ and heat the oxime rearranges and the nitrogen stays in the molecule.
- Acid does not reduce. An amine would form from the oxime with H₂/catalyst or LiAlH₄; here an amide forms.
- It is the oxime nitrogen that is inserted, not an oxygen. An ester or lactone would form from the ketone with a peracid (Baeyer–Villiger).
Often confused with: Hofmann rearrangement · Oxime formation · Acidic hydrolysis of amides and nitriles
Frequently asked questions
Which group migrates?
Always the one anti (trans) to the OH group. The E/Z geometry of the oxime therefore decides the product.
Where is it used industrially?
Cyclohexanone oxime → caprolactam → nylon-6. It is one of the largest industrial organic reactions.