Question

In the above reaction the attack of a nucleophile would be from which side:

$\stackrel{3}{C}{H}_3-\stackrel{2}{\underset{C{H}_3}{\underset{|}{C}H}}-\stackrel{1}{C}{H}_{2}Cl\stackrel{aq.KOH}{\to }C{H}_3-\underset{C{H}_3}{\underset{|}{CH}}-C{H}_2-OH$

• A. On the front side of $Cl$ and $C_1$ carbon
• B. On $C_2$
• C. On the rear side of Cl and $C_1$ carbon
• D. On $C_3$ carbon

Answer 1

The correct option is C On the rear side of Cl and $C_1$ carbon

This is a $S_{N}2$ reaction. Since, steric hindrance is not there, so, as soon as $C{l}^{-}$ leaves the $C_1$ carbon from the front side, $O{H}^{-}$ attacks the carbocation from the rear side of $Cl$ and $C_1$ carbon.