Question

The major products P and Q from the following reactions are
Reaction 1: \( \text{P} \leftarrow[\text{(ii) H}_2\text{O}]{\text{(i) LiAlH}_4} \text{C}_6\text{H}_5\text{CONH}_2 \) Reaction 2: \( \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2 / \text{NaOH}} \text{Q} \)

• A. \( \text{P} = \text{C}_6\text{H}_5\text{NH}_2 \); \( \text{Q} = \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \)
• B. \( \text{P} = \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \); \( \text{Q} = \text{C}_6\text{H}_5\text{NH}_2 \)
• C. \( \text{P} = \text{C}_6\text{H}_5\text{-CH}_2\text{-NH}_2 \); \( \text{Q} = \text{C}_6\text{H}_5\text{COONa} \)
• D. \( \text{P} = \text{C}_6\text{H}_5\text{CN} \); \( \text{Q} = \text{C}_6\text{H}_5\text{Br} \)

Hint:

\( \text{LiAlH}_4 \) keeps the carbon count same (Reduction). Hoffmann Bromamide reduces the carbon count by one (Degradation).

Answer 1

The Correct Option is B

Step 1: Reaction Analysis:

Reactant: Benzamide (\( \text{C}_6\text{H}_5\text{CONH}_2 \)).
Step 2: Reaction 1 (Reduction):

Reagent: \( \text{LiAlH}_4 \) (Lithium Aluminum Hydride). This is a strong reducing agent. It reduces amides to amines by converting the carbonyl group (\( \text{C=O} \)) to a methylene group (\( \text{CH}_2 \)). \[ \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \] Product P is Benzylamine.
Step 3: Reaction 2 (Hoffmann Bromamide Degradation):

Reagent: \( \text{Br}_2 / \text{NaOH} \). This reaction converts a primary amide to a primary amine with one carbon atom less (decarbonylation). The carbonyl group is removed. \[ \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2 / \text{NaOH}} \text{C}_6\text{H}_5\text{NH}_2 + \text{Na}_2\text{CO}_3 + \dots \] Product Q is Aniline.
Step 4: Comparison:

\( \text{P} = \text{Benzylamine} (\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2) \). \( \text{Q} = \text{Aniline} (\text{C}_6\text{H}_5\text{NH}_2) \). Matches Option (B).