MCQ
$+\, CH_3SNa$ $\xrightarrow{{{C_2}{H_5}OH}}$ ?

The product formed is

  • $\begin{array}{*{20}{c}}
      {C{H_3}SC{H_2} - CH - C{{(C{H_3})}_3}} \\ 
      | \\ 
      {OH} 
    \end{array}$
  • B
    $\begin{array}{*{20}{c}}
      {{{(C{H_3})}_3}C - CH - C{H_2}OH} \\ 
      {|\,\,} \\ 
      {\,\,\,\,\,\,\,SC{H_3}} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {C{H_3}SC{H_2} - CH - CH{{(C{H_3})}_2}} \\ 
      | \\ 
      {OH} 
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
      {{{(C{H_3})}_3}C - C{H_2}CH - SC{H_3}} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} 
    \end{array}$

Answer

Correct option: A.
$\begin{array}{*{20}{c}}
  {C{H_3}SC{H_2} - CH - C{{(C{H_3})}_3}} \\ 
  | \\ 
  {OH} 
\end{array}$
a

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