Hofmann Bromamide Reaction and Acylation Mechanism

To determine the structure of the product T, we can analyze the reaction step-by-step:

Step 1: Hofmann Bromamide Reaction

The starting compound is $4$-methylbenzamide ($p$-toluamide):
$\text{H}_3\text{C}-\text{C}_6\text{H}_4-\text{CONH}_2$

When treated with $\text{NaOH}/\text{Br}_2$ in step (1), $p$-toluamide undergoes the Hofmann Bromamide Degradation. In this reaction, the carbonyl group ($-\text{C}=\text{O}$) of the primary amide is removed as carbonate ($\text{CO}_3^{2-}$), resulting in a primary amine with one less carbon atom:

$\text{H}_3\text{C}-\text{C}_6\text{H}_4-\text{CONH}_2 \xrightarrow{\text{NaOH}/\text{Br}_2} \text{H}_3\text{C}-\text{C}_6\text{H}_4-\text{NH}_2$

The product formed at the end of step (1) is $p$-toluidine ($4$-methylaniline), where the $-\text{NH}_2$ group remains attached directly to the $p$-tolyl ring.

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Step 2: Acylation Reaction

In step (2), $p$-toluidine is treated with benzoyl chloride ($\text{C}_6\text{H}_5-\text{CO}-\text{Cl}$).

The nitrogen atom of $p$-toluidine acts as a nucleophile and attacks the carbonyl carbon of benzoyl chloride in a nucleophilic acyl substitution reaction:

$\text{H}_3\text{C}-\text{C}_6\text{H}_4-\text{NH}_2 + \text{C}_6\text{H}_5-\text{CO}-\text{Cl} \longrightarrow \text{H}_3\text{C}-\text{C}_6\text{H}_4-\text{NH}-\text{CO}-\text{C}_6\text{H}_5 + \text{HCl}$

The resulting compound is $N$-($p$-tolyl)benzamide, in which:

• The $-\text{NH}-$ group is directly linked to the $p$-tolyl ring ($\text{H}_3\text{C}-\text{C}_6\text{H}_4-$).
• The carbonyl group ($-\text{C}=\text{O}$) is directly linked to the phenyl ring ($\text{C}_6\text{H}_5-$).

This matches the structure shown in Option C.

Correct Option: C