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Question

Acetamide when treated with $$Br_{2}$$ and caustic soda, the product obtained is:


A
acetic acid
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B
bromacetic acid
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C
ethylamine
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D
methylamine
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Solution

The correct option is C ethylamine
The given reaction is Hoffman-bromamide degradation.
The product formed is an amine. In this case, the product formed is $$CH_3NH_2$$.
$$CH_3-\overset{\overset{O}{||}}{C}-NH_2\xrightarrow{Br_2/NaOH}CH_3-NH_2$$

Chemistry
NCERT
Standard XII

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