MCQ
When acetamide is treated with $NaOBr,$ the product formed is
- A$C{H_3}CN$
- B$C{H_3}C{H_2}N{H_2}$
- ✓$C{H_3}N{H_2}$
- DNone of the above
✓
Answer
Correct option: C.
$C{H_3}N{H_2}$
c
(c)$C{H_3}CON{H_2}\mathop {\xrightarrow[{{\text{Hofmann}}\,{\text{bromamide}}\,{\text{reaction}}}]{}}\limits^{NaOBr} C{H_3}N{H_2}$
(c)$C{H_3}CON{H_2}\mathop {\xrightarrow[{{\text{Hofmann}}\,{\text{bromamide}}\,{\text{reaction}}}]{}}\limits^{NaOBr} C{H_3}N{H_2}$
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