# 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?

Feb 27, 2015

The product is ${\text{CH"_3"-NH}}_{2}$.

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.