Acetamide reacts with $Br_{2}$ and aqueous KOH to form :
Show Hint
- Ethanamine
- Ammonia
- Methanamine
- Ethanenitrile
The Correct Option is C
Solution and Explanation
The reaction of a primary amide with bromine ($Br_2$) in the presence of a strong base (like aqueous KOH or NaOH) is known as the Hoffmann bromamide degradation reaction.
Step 2: Meaning
This specific reaction is used to convert primary amides into primary amines. Crucially, the resulting amine has one carbon atom less than the starting amide.
Step 3: Analysis
The reactant is acetamide ($CH_3CONH_2$), which contains two carbon atoms. During the reaction, the carbonyl carbon is lost in the form of a carbonate ion ($CO_3^{2-}$).
Step 4: Conclusion
Removing the carbonyl group connects the remaining methyl group ($CH_3-$) directly to the amine group ($-NH_2$), yielding $CH_3NH_2$, which is methanamine.
Final Answer: (C)
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(v) Aniline does not undergo Friedel-Crafts reaction.
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