( C H _ 3 ) _ 3 CClNaCNA dil , H _ 2 S O _ 4 B Compound B is

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 $