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: Conceptual Overview:
The process involves two steps: first, an amide degradation reaction, followed by a nucleophilic substitution (acylation) reaction on an amine. These reactions are commonly used in organic synthesis to modify amides and amines.
Step 2: Detailed Explanation:
1. Step 1 (Hoffmann Bromamide Degradation): Ethanamide (\( CH_3CONH_2 \)) undergoes a reaction with sodium hypobromite (\( NaOBr \)), which is generated from the combination of bromine (\( Br_2 \)) and sodium hydroxide (\( NaOH \)). In this reaction, the carbonyl group is removed, and a primary amine with one less carbon atom is formed as the product.
Thus, Compound A formed is Methanamine: \( CH_3NH_2 \).
2. Step 2 (Benzoylation): Methanamine (\( CH_3NH_2 \)) then reacts with benzoyl chloride (\( C_6H_5COCl \)) in the presence of a base to neutralize the byproduct \( HCl \). The nucleophilic amino group attacks the carbonyl carbon of the benzoyl chloride, displacing the chloride ion and forming an amide.
The product, Compound B, is N-Methylbenzamide: \( C_6H_5CONHCH_3 \).
Step 3: Final Conclusion:
Compound A: \( CH_3NH_2 \) (Methanamine)
Compound B: \( C_6H_5CONHCH_3 \) (N-Methylbenzamide)