Question

The best reagent for converting propanamide into propanamine is:

• A. excess H$_2$
• B. Br$_2$ in aqueous NaOH
• C. iodine in the presence of red phosphorus
• D. LiAlH$_4$ in ether

Hint:

LiAlH$_4$ is a strong reducing agent that can reduce amides to amines.

Answer 1

The Correct Option is D

Explanation:

1. Objective: Convert propanamide to propanamine by reducing the amide group (-CONH₂) to an amine group (-CH₂NH₂).

2. Reagent Evaluation:

A) Excess H₂: Catalytic hydrogenation is generally ineffective for amide reduction to amines, requiring harsh conditions and potentially causing side reactions.

B) Br₂ in aqueous NaOH (Hoffmann Bromamide Degradation): This reagent reduces amides to primary amines with one fewer carbon atom. For propanamide (C₃), it would yield ethanamine (C₂), not the desired product.

C) Iodine with red phosphorus: This combination is used for reducing carboxylic acids to alkanes and is not suitable for amide-to-amine reduction.

D) LiAlH₄ in ether: Lithium aluminum hydride (LiAlH₄) is a potent reducing agent that directly converts amides to amines by replacing the carbonyl oxygen with hydrogens, making it the optimal choice.

Conclusion:

Option (D) LiAlH₄ in ether is the sole reagent capable of achieving the desired conversion without altering the carbon chain length.

Final Answer:

The correct choice is (D).