Question

The reactive intermediate(s) in the Hoffmann bromamide formed is/are:

• A. R-N=C=O
• B. R-$\overset{O}{\underset{||}{C}}$-NH-Br
• C. R-$\overset{O}{\underset{||}{C}}$-$\overset{\mathbf{:}}{N}$
• D. R-$\overset{\mathbf{:}}{N}$

Answer 1

Answer: (A), (B)

R-N=C=O, R-$\overset{O}{\underset{||}{C}}$-NH-Br

Answer 2

The Hoffmann bromamide degradation reaction converts a primary amide (R-CO-NH2) into a primary amine (R-NH2) with one less carbon atom. The reaction proceeds through several intermediates.

1. Formation of N-bromoamide: The amide reacts with bromine in the presence of a base (e.g., NaOH) to form an N-bromoamide:

   R-CO-NH2 + Br2 + NaOH → R-CO-NH-Br + NaBr + H2O

   So, R-CO-NH-Br is an intermediate.

2. Formation of N-bromoamide anion: The N-bromoamide is deprotonated by the base to form an N-bromoamide anion:

   R-CO-NH-Br + NaOH → R-CO-N-Br + H2O + Na+

3. Rearrangement to Isocyanate: The N-bromoamide anion undergoes a concerted rearrangement. The alkyl or aryl group (R) migrates from the carbonyl carbon to the electron-deficient nitrogen atom, and simultaneously, the bromide ion departs. This step leads to the formation of an isocyanate:

   R-CO-N-Br → R-N=C=O + Br-

   So, R-N=C=O (Isocyanate) is a crucial intermediate.

4. Hydrolysis of Isocyanate: The isocyanate is then hydrolyzed by water (present in the aqueous basic solution) to form a carbamic acid:

   R-N=C=O + H2O → R-NH-COOH (Carbamic acid)

5. Decarboxylation: The carbamic acid is unstable and readily decarboxylates (loses CO2) to yield the primary amine:

   R-NH-COOH → R-NH2 + CO2

Based on the accepted mechanism of the Hoffmann bromamide degradation, the key reactive intermediates among the given options are:

• R-N=C=O: This is the isocyanate, a well-established intermediate in the reaction.
• R-CO-NH-Br: This is the N-bromoamide, which is formed in the initial steps of the reaction.

The other options are not considered stable or transient intermediates.