MCQ
$C{H_3} - Br + \begin{array}{*{20}{c}}
  {\,\,\,\,\,C{H_3}} \\ 
  | \\ 
  {NaO - C - C{H_3}} \\ 
  | \\ 
  {\,\,\,\,\,\,C{H_3}} 
\end{array} \to ?$
  • $\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
      {\,\,\,\,\,\,\,\,\,|} \\ 
      {C{H_3} - O - C - C{H_3}} \\ 
      {\,\,\,\,\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • B
    $\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
      {C{H_3} - C} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • C
    $CH_2=CH_2$
  • D
    $\begin{array}{*{20}{c}}
      {C{H_3} - O - CH - C{H_3}} \\ 
      {\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$

Answer

Correct option: A.
$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,|} \\ 
  {C{H_3} - O - C - C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$
a

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