What is produced when #CH_3CONH_2# reacts with NaOBr? Is NaOBr a product of (or equivalent to) #Br_2 + NaOH# which are used in the Hoffmann bromamide degradation reaction?

1 Answer

The product is #"CH"_3"-NH"_2#.

This is an example of the Hofmann Degradation of primary amides to amines with one less carbon atom.

The reaction actually uses Br₂ + NaOH, but this is equivalent to using NaOBr, because the reaction produces NaOBr in situ.

The reaction involves several steps.


  1. The base abstracts an N-H proton, forming the conjugate base of the amide.
  2. The anion reacts displaces Br⁻ from a Br₂ molecule, forming an N-bromoamide.
  3. The base abstracts a proton from the N-bromoamide, forming another anion.
  4. The bromide ion leaves, and the alkyl group migrates to the N atom, forming an isocyanate.
  5. The base-catalyzed addition of water forms a carbamate.
  6. The carbamate acid spontaneously loses CO₂, forming an amine with one less carbon atom.