- ASN1 mechanism
- BSN2 mechanism
- CAny of (a) and (b)
- DNone of these
Explanation:
The SN1 reaction proceeds stepwise. The leaving group first leaves, whereupon a carbocation forms that is attacked by the nucleophile.
The SN2 occurs in one step, and both the nucleophile and substrate are involved in the rate determining step. Therefore the rate is dependent on both the concentration of substrate and that of the nucleophile.
For tertiary butyl chloride, it form tertiary carbocation which is highly stable. So tertiary butyl chloride undergo hydrolysis by SN1 mechanism.
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$\text{HO}−\text{CH}_2−\text{CH}−\text{CH}_2−\text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ∣ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
$[X]$ $\xrightarrow{{KMn{O_4}/\mathop O\limits^\Theta H/\Delta }}$ $\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{HOOC - {{(C{H_2})}_3} - CH - COOH}
\end{array}}
\end{array}}
\end{array}$
$[X]$ will be: