Products of Aniline Multistep Reaction Involving Bromination and Diazotization

To determine the major product of the given multistep reaction, let us analyze each step sequentially:

Step 1: Acetylation

Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) reacts with acetic anhydride ($(\text{CH}_3\text{CO})_2\text{O}$) in the presence of pyridine.

• Mechanism: The lone pair on the nitrogen atom of aniline attacks the carbonyl carbon of acetic anhydride, undergoing a nucleophilic acyl substitution to yield acetanilide ($\text{C}_6\text{H}_5\text{NHCOCH}_3$).
• Purpose: The bulky acetamido group ($-\text{NHCOCH}_3$) moderates the activating effect of the $-\text{NH}_2$ group (due to resonance delocalization of nitrogen's lone pair into the carbonyl group) and prevents polybromination while offering steric hindrance to the ortho-positions.

$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH}$

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Step 2: Bromination

Potassium bromate ($\text{KBrO}_3$) reacts with hydrobromic acid ($\text{HBr}$) to generate bromine ($\text{Br}_2$) in situ:

$\text{BrO}_3^- + 5\text{Br}^- + 6\text{H}^+ \longrightarrow 3\text{Br}_2 + 3\text{H}_2\text{O}$

• Reaction: Acetanilide undergoes electrophilic aromatic substitution with $\text{Br}_2$. Since the $-\text{NHCOCH}_3$ group is strongly para-directing and sterically bulky, bromination occurs predominantly at the para-position to form $4$-bromoacetanilide (p-bromoacetanilide).

$\text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{Br}_2 \longrightarrow \text{p-Br-C}_6\text{H}_4\text{NHCOCH}_3 + \text{HBr}$

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Step 3: Acid Hydrolysis

Heating $4$-bromoacetanilide with aqueous acid ($\text{H}_3\text{O}^+$) hydrolyzes the amide linkage, regenerating the primary amino group to yield $4$-bromoaniline (p-bromoaniline).

$\text{p-Br-C}_6\text{H}_4\text{NHCOCH}_3 + \text{H}_3\text{O}^+ \xrightarrow{\Delta} \text{p-Br-C}_6\text{H}_4\text{NH}_2 + \text{CH}_3\text{COOH}$

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Step 4: Diazotization

$4$-bromoaniline is treated with nitrous acid ($\text{NaNO}_2 + \text{HCl}$) at $273\text{--}278\text{ K}$ ($0\text{--}5^\circ\text{C}$).

• Reaction: The primary aromatic amine is converted into a diazonium salt, giving $4$-bromobenzenediazonium chloride.

$\text{p-Br-C}_6\text{H}_4\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273\text{--}278\text{ K}} \text{p-Br-C}_6\text{H}_4\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$

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Step 5: Gattermann Reaction

Reaction of $4$-bromobenzenediazonium chloride with copper powder and hydrobromic acid ($\text{Cu/HBr}$) replaces the diazonium group ($-\text{N}_2^+\text{Cl}^-$) with a bromine atom ($-\text{Br}$).

$\text{p-Br-C}_6\text{H}_4\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu/HBr}} \text{1,4-dibromobenzene} + \text{N}_2 + \text{CuCl}$

Thus, the final product of the reaction sequence is 1,4-dibromobenzene, corresponding to structure (B).

Correct Answer: B