Question

Write structural formulae of the compound A and B:
\(CH_3CONH_2 \xrightarrow{NaOBr} A \xrightarrow[Base]{C_6H_5COCl} B\)

Hint:

NaOBr is a reagent for Hoffmann Bromamide reaction. It always converts an amide (\(R-CONH_2\)) to a primary amine (\(R-NH_2\)).

Answer 1

Step 1: Understanding the Concept:
The first step involves an amide degradation reaction, and the second step is a nucleophilic substitution (acylation) reaction on an amine.
Step 2: Detailed Explanation:
1. Step 1 (Hoffmann Bromamide Degradation): Ethanamide (\(CH_3CONH_2\)) reacts with sodium hypobromite (\(NaOBr\), generated from \(Br_2\) and \(NaOH\)). This reaction removes the carbonyl group, resulting in a primary amine with one less carbon atom.
Compound A is Methanamine: \(CH_3NH_2\).
2. Step 2 (Benzoylation): Methanamine reacts with benzoyl chloride (\(C_6H_5COCl\)) in the presence of a base (to neutralize the byproduct \(HCl\)). The amino group attacks the carbonyl carbon of benzoyl chloride, displacing the chloride ion.
Compound B is N-Methylbenzamide: \(C_6H_5CONHCH_3\).
Step 3: Final Answer:
Compound A: \(CH_3NH_2\)
Compound B: \(C_6H_5CONHCH_3\)