Question

Write the structures of A and B in the following reaction: \[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{NH}_3, \Delta} A \xrightarrow{\text{Br}_2/\text{NaOH}} B \]

Hint:

- The Hoffmann Bromamide Reaction is used for amide to amine conversion by removing the carbonyl group. - This reaction reduces carbon count by 1, making it useful for amine synthesis.

Answer 1

A = Benzamide (\( C_6H_5CONH_2 \))
B = Aniline (\( C_6H_5NH_2 \))
Step 1: Benzoic Acid to Benzamide Conversion
Heating benzoic acid (\( C_6H_5COOH \)) with ammonia yields benzamide (\( C_6H_5CONH_2 \)) via amide formation.
\[\text{C}_6\text{H}_5\text{COOH} + NH_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5CONH_2\]

Step 2: Hoffmann Bromamide Degradation
Reaction of benzamide (\( C_6H_5CONH_2 \)) with Br\(_2\)/NaOH removes the carbonyl group (-CO), producing aniline (\( C_6H_5NH_2 \)).
\[\text{C}_6\text{H}_5CONH_2 + Br_2 + NaOH \rightarrow \text{C}_6\text{H}_5NH_2 + Na_2CO_3\]