MCQ
$\begin{array}{*{20}{c}}
  {{{(C{H_3})}_2}CH - C - N{H_2}\,\xrightarrow[{B{r_2}}]{{NaOH}}A\xrightarrow{{COC{l_2}}}B} \\ 
  {||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$ $B$ is
  • $(CH_3)_2CH-N=C=O$
  • B
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - C{H_3}} \\ 
      {|\,\,\,} \\ 
      {\,\,\,N{H_2}} 
    \end{array}$
  • C
    $(CH_3)_2CH-NH-COCH_3$
  • D
    $CH_3-CH_2-NHCOCH_3$

Answer

Correct option: A.
$(CH_3)_2CH-N=C=O$
a
$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {{{(C{H_3})}_2}CH - C - N{H_2}} 
\end{array}$ $\xrightarrow[{B{r_2}}]{{NaOH}}$ $\mathop {C{H_3}CH - N{H_2}}\limits_{(A)} $ $\xrightarrow{{CoC{l_2}}}$ $\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} 
\end{array}} \\ 
  {{{(C{H_3})}_2}CH - NH - C} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl} 
\end{array}$ $\xrightarrow[{ - HCl}]{}$ ${(C{H_3})_2}CH - N = C = O$

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