Question

In the reaction given below :

The Product ‘X’ is:

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is B

Step 1: Analyzing the given reaction. 
The given reaction involves the reduction of a molecule containing a carbonyl group (C=O) and an amide group (-NH\(_2\)) using lithium aluminum hydride (LiAlH\(_4\)), followed by hydrolysis with H\(_3\)O\(^+\). 
1. Step 1: Reduction with LiAlH\(_4\)
Lithium aluminum hydride (LiAlH\(_4\)) is a strong reducing agent. When it reacts with a carbonyl compound (such as a ketone or aldehyde), it reduces the carbonyl group to a primary or secondary alcohol. \item In this case, the carbonyl group (C=O) of the given compound will be reduced to a hydroxyl group (-OH), resulting in an intermediate amide being reduced to a primary amine group (-NH\(_2\)). The structure of the intermediate product will be:  \[ \text{H}_2\text{NC} - \text{CH}_2 - \text{CH}_2\text{OH} \] where the ketone group is reduced to an alcohol group. 
2. Step 2: Hydrolysis with H\(_3\)O\(^+\)
After reduction, the product is treated with an acidic solution (H\(_3\)O\(^+\)), which will hydrolyze the intermediate and result in a final product where the amide group has been converted to an amine group (-NH\(_2\)) attached to a hydroxyl group (-OH). This confirms the product as:  \[ \text{H}_2\text{N} - \text{CH}_2\text{OH} \] This is the final product, and it is a primary amine with a hydroxyl group attached to the adjacent carbon. Thus, the product 'X' is \( \text{H}_2\text{N} - \text{CH}_2\text{OH} \).