MCQ
${{\left( C{{H}_{3}} \right)}_{3}}CCl\xrightarrow{NaCN}A\xrightarrow{dil\,{{H}_{2}}S{{O}_{4}}}B$ Compound $B$ is
  • A
    $(CH_3)_3CCOOH$
  • $(CH_3)_3COH$
  • C
    $(CH_3)_3COC(CH_3)_3$
  • D
    All the three

Answer

Correct option: B.
$(CH_3)_3COH$
b
$CN^-$ is a strong base and since the substrate is a tert-halide, it mainly undergoes elimination reaction forming alkene $(A)$. In presence of dil. $H_2SO_4$, alkenes undergo hydration in Markovnikov’s way.

$\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,|} 
\end{array}} \\ 
  {C{H_3} - C - Cl} \\ 
  {\,\,\,\,|\,} \\ 
  {\,\,\,\,\,\,\,\,\,C{H_3}\,} 
\end{array}$  $\xrightarrow[{( - \,HCl)}]{{C{N^ - }}}$  $\mathop {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,C{H_2}} \\ 
  {\,\,\,\,\,||} 
\end{array}} \\ 
  {C{H_3} - C} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}}\limits_A $  $\xrightarrow[{(hydration)}]{{dil.{H_2}S{O_4}}}$ $\mathop {\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,C{H_3}} \\ 
  | 
\end{array}} \\ 
  {C{H_3} - C - OH} \\ 
  | \\ 
  {\,\,\,\,\,\,C{H_3}} 
\end{array}}\limits_B $

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