Question:

Which from following compounds is obtained when propionamide is treated with \(\text{Br}_2\) and concentrated aqueous KOH solution ?

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Hofmann degradation removes one carbon atom as carbonate and gives a primary amine.
Updated On: Oct 1, 2026
  • \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)
  • \(\text{CH}_3\text{COCH}_3\)
  • \(\text{CH}_3\text{CH}_2\text{NH}_2\)
  • \(\text{CH}_3\text{CH}_2\text{Br}\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
In Hofmann bromamide degradation, an amide reacts with \(\text{Br}_2\) and aqueous KOH to give a primary amine with one carbon atom fewer.

Step 2: Reaction:
\[ \text{CH}_3\text{CH}_2\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + 2\text{KBr} + \text{K}_2\text{CO}_3 + 2\text{H}_2\text{O} \]
Propionamide has three carbons, so the amine has two: ethylamine.

Step 3: Check the Other Options:
(A) has four carbons, so it has more carbon than the amide. (B) is acetone, which has no nitrogen. (D) is ethyl bromide, which is not formed since nitrogen is retained. So (C) is correct.

Final Answer:
The product is ethylamine, option (C). \[ \boxed{\text{(C) } \text{CH}_3\text{CH}_2\text{NH}_2} \]
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