Question

What are X and Y respectively in the following reactions? 

• A.
• B.
• C.
• D.

Hint:

In the Hofmann rearrangement, an amide is converted to a primary amine by the action of bromine and sodium hydroxide. The carbonyl group is eliminated, resulting in a loss of one carbon atom.

Answer 1

The Correct Option is B

In this problem, we have two reactions for the given compound, which is likely an amide (since the functional group is \( {CONH}_2 \)).

1. Reaction with LiAlH4 (Lithium aluminium hydride): Lithium aluminium hydride is a strong reducing agent that reduces amides to amines. The reaction of amide with \( {LiAlH}_4 \) reduces the \( {CONH}_2 \) group to the amine group \( {NH}_2 \), thus converting the compound into aniline (\( {C}_6{H}_5{NH}_2 \)).
2. Reaction with Br2/NaOH: The second reaction, the Hofmann rearrangement, is a method of reducing amides to amines. The reaction with \( {Br}_2 \) and \( {NaOH} \) leads to the loss of the carbonyl group and results in the formation of an amine with one fewer carbon atom. This gives aniline (\( {C}_6{H}_5{NH}_2 \)).

Thus, both X and Y are aniline (\( {C}_6{H}_5{NH}_2 \)).

Thus, the correct answer is (B).