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

Question

Assertion (A): The pK$_a$ of O$_2$N–CH$_2$–COOH is lower than that of CH$_3$–COOH. Reason (R): The –NO$_2$ group shows an electron-withdrawing effect, which increases the acidic character of O$_2$N–CH$_2$–COOH.

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 $).
\[\text{C}_6\text{H}_5CONH_2 + Br_2 + NaOH \rightarrow \text{C}_6\text{H}_5NH_2 + Na_2CO_3\]

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

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Solution and Explanation

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