Match the reactions in List-I with the features of their products in List-II and choose the correct option.…

$\underset{(\mathrm{Single} \mathrm{enantiomer})}{1-\mathrm{Bromo}-2-\mathrm{ethylpentane}} \underset{{\mathrm{S}}_{\mathrm{N}} 2 \mathrm{reaction}}{\overset{\mathrm{aq}. \mathrm{NaOH}}{\to }}2-\mathrm{ethylpentanol}$

In the above reaction the chiral carbon is not involving in the reaction as ${\mathrm{S}}_{\mathrm{N}}2$ reaction takes place via formation of transition state but not through formation of intermediate.

In the above reaction, bromine attached to chiral carbon, and it is involving in ${\mathrm{S}}_{\mathrm{N}}2$ reaction. Hence, the alcohol formed in inversion configuration.

The substrate is undergoing ${\mathrm{S}}_{\mathrm{N}}1$ reaction. It takes place via formation of carbocation. Carbocation as planar, hence, the alcohol product formed is racemic mixture as nucleophile can attack from both the planes.

The substrate is undergoing ${\mathrm{S}}_{\mathrm{N}}1$  reaction. It takes place via formation of carbocation. Carbocation as planar, hence, the two alcohol products are formed as nucleophile can attack from both the planes but the substrate already having the chiral centre, the mixture formed is diastereomeric mixture.