Question

Alcohols can be prepared by haloalkanes using hydroxide ion in aqueous media through $S_{N}1$ and $S_{N}2$.
The $S_{N}1$ reaction is a two-step reaction. In the first step, the carbon-halogen bond breaks to generate a stable carbocation. In the second step, the nucleophile reacts rapidly with the carbocation. This reaction follows first order kinetics.
The $S_{N}2$ is a concerted reaction in which transition state is achieved where bond making and bond breaking occur simultaneously. This reaction follows second order kinetics.
Consider the following reaction sequence.


Which of the following statement is correct about the mechanism of this reaction?

• A. A carbocation will be formed as an intermediate in the reaction
• B. $O{H}^{–}$ will attach the substrate (i) from one side and $C{l}^{-}$ will leave it simultaneously from other side
• C. An unstable intermediate will be formed in which $O{H}^{–}$ and $C{l}^{-}$will be attached by weak bonds
• D. Reaction proceeds through $E_1$ mechanism

Answer 1

The correct option is A A carbocation will be formed as an intermediate in the reaction

Reaction proceeds in two steps. In step I, the polarised $C-Cl$ bond undergoes slow cleavage to produce a carbocation and a chloride ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.
It can be estimated that the reacrtion may proceed either through $S_{N}1$ or $S_{N}2$ type. Since in the prodcut, the spatial arrangement of bonds to the carbon centre are preserved which that the reaction is followed by $S_{N}1$ mechanism.