MCQ
$\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {C{H_2} - OH} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}} \\ 
  {C{H_2} - OH} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_2} - SH} 
\end{array}$ ${ + \mathop {C{H_3}MgBr}\limits_{\left( {Excess} \right)}  \to xC{H_4}}$

What is the value of $x$ in the above reaction ?

  • A
    $1$
  • B
    $2$
  • $3$
  • D
    $4$

Answer

Correct option: C.
$3$
c
$(c)$ Because $3$ acidic $H$ are present

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$\begin{array}{*{20}{c}}
  \,\,\,{{C_2}{H_5}} \\ 
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