Question

Identify \(X\) and \(Y\) of the following reaction sequence

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Aniline is first protected as acetanilide before bromination to avoid polysubstitution. Primary aromatic amines give isocyanides with \(CHCl_3/KOH\) in the carbylamine reaction.

Answer 1

The Correct Option is D

Step 1: Reduction of nitrobenzene.
The starting compound is nitrobenzene.
On treatment with \[ H_2/Pd \] nitrobenzene is reduced to aniline. \[ C_6H_5NO_2 \xrightarrow{H_2/Pd} C_6H_5NH_2 \]

Step 2: Acetylation of aniline.
Aniline reacts with acetic anhydride in pyridine to form acetanilide. \[ C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O,\ pyridine} C_6H_5NHCOCH_3 \] Therefore, \[ X=\text{acetanilide} \]

Step 3: Bromination of acetanilide.
Acetanilide undergoes bromination with \[ Br_2/CH_3COOH \] The \(-NHCOCH_3\) group is ortho-para directing.
Due to steric hindrance, the para product is major.
Thus, the major product is \[ p\text{-bromoacetanilide} \]

Step 4: Hydrolysis of bromoacetanilide.
On hydrolysis, \[ p\text{-bromoacetanilide} \] gives \[ p\text{-bromoaniline} \]

Step 5: Carbylamine reaction.
Primary amines react with chloroform and alcoholic KOH on heating to give isocyanides.
Thus, \[ p\text{-bromoaniline} \xrightarrow{CHCl_3,\ KOH,\Delta} p\text{-bromophenyl isocyanide} \] Therefore, \[ Y=p\text{-bromophenyl isocyanide} \]

Step 6: Final conclusion.
Hence, \[ \boxed{X=\text{acetanilide},\quad Y=p\text{-bromophenyl isocyanide}} \] Therefore, the correct option is (4).