JEE Chemistry 2026 Question with Solution

Given: The starting compound is a primary aliphatic amine reacting first with benzoyl chloride in NaOH, then with $\text{LiAlH}_4$ followed by $\text{H}_2\text{O}$.

Find: The final product $[B]$.

Step 1: Identify the first reaction.

The starting compound is a primary aliphatic amine. Reaction with benzoyl chloride $\left(\text{C}_6\text{H}_5\text{COCl}\right)$ in the presence of NaOH gives an amide via Schotten–Baumann reaction.

Thus, intermediate $[A]$ is:

Step 2: Effect of $\text{LiAlH}_4$.

$\text{LiAlH}_4$ reduces amides to amines, converting the $-\text{CO}-$ group into a $-\text{CH}_2-$ group.

Therefore:

Step 3: Write the final product.

The final compound $[B]$ is:

This structure matches the third structure in the provided figure.

The solution states Option (D), but its worked structure corresponds to the third displayed option. Therefore, the defensible answer from the actual chemistry is C.

Given: Benzoylation of a primary amine is followed by reduction with $\text{LiAlH}_4$.

Find: Which displayed structure is product $[B]$.

Benzoyl chloride first converts the amine into an $N$-benzoyl derivative, so the nitrogen becomes part of an amide linkage.

Reduction of an amide with $\text{LiAlH}_4$ does not retain the carbonyl oxygen. Instead, the carbonyl carbon remains in the chain as a methylene group.

So the benzoyl fragment becomes a benzyl fragment, and the product is the secondary amine:

Hence, the correct displayed choice is C.