Question:medium

The main product of the reaction of CH\(_3\)CONH\(_2\) with Br\(_2\) in aqueous potassium hydroxide medium is

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Hoffmann degradation gives amine with one less carbon.
Updated On: Jun 16, 2026
  • CH\(_3\)-CH\(_2\)-NH\(_2\)
  • CH\(_3\)Br
  • CH\(_3\)CONHBr
  • CH\(_3\)NH\(_2\)
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The Correct Option is D

Solution and Explanation

The reaction in question involves treating acetamide \((\text{CH}_3\text{CONH}_2)\) with bromine \((\text{Br}_2)\) in an aqueous potassium hydroxide \((\text{KOH})\) medium. This is a classic example of the Hofmann Bromamide Reaction, also known as the Hofmann Degradation of Amides.

Concept: The Hofmann Bromamide Reaction is used to degrade amides to primary amines with one less carbon atom (loss of one carbon). Here, a carboxamide group is converted into an amine with the help of bromine and a strong base.

Reaction Mechanism: The general reaction mechanism involves the formation of an N-bromoamide intermediate which then undergoes rearrangement through the deprotonation and opening of the carbon-nitrogen bond. 

Application to Given Reaction: Here, \(\text{CH}_3\text{CONH}_2\) is treated with bromine in an aqueous potassium hydroxide medium. The reaction proceeds as follows:

  • \(\text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{CH}_3\text{NH}_2 + 2\text{KBr} + 2\text{H}_2\text{O} + \text{K}_2\text{CO}_3\)

Product Analysis: The final product of this reaction is \(\text{CH}_3\text{NH}_2\) (methylamine), as the amide is degraded to the primary amine form.

Conclusion: The main product of the reaction of acetamide \((\text{CH}_3\text{CONH}_2)\) with bromine in aqueous potassium hydroxide medium is indeed methylamine \((\text{CH}_3\text{NH}_2)\). Hence, the correct option is \(\text{CH}_3\text{NH}_2\).

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