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

Here's the explanation:

1. Transformation Objective:
The reaction requires converting propanamide to propanamine, which involves reducing the amide group (-CONH₂) to an amine group (-CH₂NH₂).

2. Reagent Analysis:

A) Excess H₂:
While catalytic hydrogenation can reduce some functional groups, it's generally ineffective for directly reducing amides to amines. It requires harsh conditions and may cause unwanted side reactions.

B) Br₂ in aqueous NaOH (Hoffmann Bromamide Degradation):
This reagent degrades amides into primary amines with one fewer carbon atom (via an isocyanate intermediate). This would convert propanamide (C₃) to ethanamine (C₂), which is not the desired product.

C) Iodine in the presence of red phosphorus:
This combination is typically used for reducing carboxylic acids to alkanes and is ineffective for amide-to-amine reduction.

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

Conclusion:
Option (D) LiAlH₄ in ether is the only reagent that achieves the desired conversion without altering the carbon chain length.

Final Answer:
The correct answer is (D).