Question

The final product [B] is:

• A. A
• B. B
• C. C
• D. D

Hint:

LiAlH$_4$ reduces amides to amines by replacing the carbonyl group with a methylene ($-CH_2-$) group.

Answer 1

The Correct Option is D

Step 1: Identify the first reaction.
The starting compound is a primary aliphatic amine.
Reaction with benzoyl chloride $(\text{C}_6\text{H}_5\text{COCl})$ in the presence of NaOH gives an amide via Schotten–Baumann reaction.
Thus, intermediate $[A]$ is: \[ \text{Cyclohexyl–CH}_2\text{–NH–CO–C}_6\text{H}_5 \]
Step 2: Effect of LiAlH$_4$.
LiAlH$_4$ reduces amides to amines, converting the $-CO-$ group into a $-CH_2-$ group.
Therefore: \[ \text{–NH–CO–C}_6\text{H}_5 \;\xrightarrow{\text{LiAlH}_4}\; \text{–NH–CH}_2\text{–C}_6\text{H}_5 \]
Step 3: Write the final product.
The final compound $[B]$ is: \[ \text{Cyclohexyl–CH}_2\text{–NH–CH}_2\text{–C}_6\text{H}_5 \] This structure corresponds to Option (D).

Final Answer: $\boxed{\text{Option (D)}}$