MCQ
$\begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {Ph - CH - C{H_3}\xrightarrow{{PCC}}(A)} 
\end{array}\xrightarrow{{\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {N{H_2} - NH - C - N{H_2}} 
\end{array}}}(B)$

Product $(B)$ is

  • A
    $\begin{array}{*{20}{c}}
      {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {Ph - C = N - C - NH - N{H_2}} 
    \end{array}$
  • $\begin{array}{*{20}{c}}
      {\,\,\,O} \\ 
      {\,\,\,||} \\ 
      {Ph - C = N - C - NH - N{H_2}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
      {Ph - CH = N - N - C - N{H_2}} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
      {Ph - CH = N - C - N{H_2}} 
    \end{array}$

Answer

Correct option: B.
$\begin{array}{*{20}{c}}
  {\,\,\,O} \\ 
  {\,\,\,||} \\ 
  {Ph - C = N - C - NH - N{H_2}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$
b

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