MCQ
When acetamide is treated with $NaOBr,$ the product formed is
  • A
    $C{H_3}CN$
  • B
    $C{H_3}C{H_2}N{H_2}$
  • $C{H_3}N{H_2}$
  • D
    None of the above

Answer

Correct option: C.
$C{H_3}N{H_2}$
c
(c)$C{H_3}CON{H_2}\mathop {\xrightarrow[{{\text{Hofmann}}\,{\text{bromamide}}\,{\text{reaction}}}]{}}\limits^{NaOBr} C{H_3}N{H_2}$

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