MCQ
$S{N^1}$ reaction is faster in
  • A
    $C{{H}_{3}}C{{H}_{2}}Cl$
  • B

  • $\begin{array}{*{20}{c}}
      {\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
      {\,\,\,\,|} \\ 
      {C{H_3} - C - Cl} \\ 
      {\,\,\,\,|} \\ 
      {\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • D
    $C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{\underset{|}{\mathop{C}}\,{{H}_{2}}}{\mathop{\underset{|}{\mathop{C}}\,H\ }}\,}}\,-Cl$

Answer

Correct option: C.
$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,|} \\ 
  {C{H_3} - C - Cl} \\ 
  {\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$
c
(c) $S{N^1}$ mechanism proceeds through carbocation intermediate.

$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,|} \\ 
  {C{H_3} - C - Cl} \\ 
  {\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$

provides stable carbocation intermediate because it has tertiary carbon.

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