Question

Which of the following compounds is obtained when acetamide is warmed with bromine and excess KOH (aq) solution?

• A. \( \text{CH}_4 \)
• B. \( \text{CH}_3\text{CH}_2\text{NH}_2 \)
• C. \( \text{CH}_3\text{COOH} \)
• D. \( \text{CH}_3\text{NH}_2 \)

Hint:

In the Hofmann rearrangement, the reaction of an amide with bromine and KOH leads to the formation of a primary amine by removing the carbonyl group.

Answer 1

The Correct Option is D

Step 1: Understand the reaction mechanism.
The reaction described involves acetamide, which is \( \text{CH}_3\text{CONH}_2 \), reacting with bromine and excess KOH. This is a typical Hofmann rearrangement reaction, which leads to the loss of the carbonyl group and the formation of an amine.

Step 2: Reaction details.
The Hofmann rearrangement involves:
- The bromine and KOH causing the carbonyl group to be replaced by a hydroxyl group, forming a primary amine.
- The reaction conditions (excess KOH and bromine) lead to the cleavage of the carbon-nitrogen bond in acetamide. The nitrogen atom is left with a hydrogen atom, forming methylamine, \( \text{CH}_3\text{NH}_2 \).

Step 3: Apply the conditions to the options.
- Option (1) \( \text{CH}_4 \) (methane) is not formed in this reaction.
- Option (2) \( \text{CH}_3\text{CH}_2\text{NH}_2 \) (ethylamine) is not the product, as the reaction does not add extra carbon atoms.
- Option (3) \( \text{CH}_3\text{COOH} \) (acetic acid) is the starting material, so it is not formed.
- Option (4) \( \text{CH}_3\text{NH}_2 \) (methylamine) is the correct product of the Hofmann rearrangement.

Step 4: Final conclusion.
Thus, the correct product of the reaction is:
\[ \boxed{(4)\ \text{CH}_3\text{NH}_2} \]