Discuss the relationship between nucleophilic substitution reactions and optical activity.

(N/A) Alkyl halides containing an asymmetric carbon atom exhibit optical activity. When optically active halides undergo nucleophilic substitution to form products,the optical activity of the product relative to the reactant depends on the reaction mechanism.
$(a)$ Inversion in $S_{N}2$ mechanism: In an $S_{N}2$ reaction,the configuration of the halide undergoes inversion because the nucleophile attacks from the direction opposite to the leaving halogen atom. Thus,$S_{N}2$ reactions result in a product with inverted optical activity.
Example: When $(-)-2-$bromooctane reacts with sodium hydroxide $(NaOH)$ via an $S_{N}2$ mechanism,the $-OH$ group attaches to the position opposite to the bromide,resulting in $(+)-$octan$-2-$ol. In this reaction,the reactant halide shows $(-)$ rotation and the product alcohol shows $(+)$ rotation.
$(b)$ Racemization in $S_{N}1$ mechanism: In $S_{N}1$ reactions of optically active alkyl halides,racemization occurs.
The first slow step of the $S_{N}1$ reaction involves the formation of a carbocation. The positively charged carbon in the carbocation is $sp^{2}$ hybridized,making it planar.
The nucleophile can attack this planar $sp^{2}$ carbon from either side (above or below the plane),resulting in a mixture of two different stereoisomers. One product has 'retention' of configuration,where the $-OH$ group occupies the same position as the halide ion. The other product has 'inversion' of configuration,where the $-OH$ group occupies the position opposite to the halide ion.
Since the rotations of the retained and inverted products are equal but opposite,and they are formed in a $1:1$ ratio,a racemic mixture $(\pm)$ is obtained. (e.g.,hydrolysis of $2-$bromobutane yields a mixture of $(+)$ and $(-)$ butan$-2-$ol).