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