Explain the inversion of configuration in $S_{N}2$ reactions and racemization in $S_{N}1$ reactions with examples.

(N/A) Alkyl halides containing an asymmetric carbon atom exhibit optical activity. When optically active halides undergo nucleophilic substitution,the optical activity of the product depends on the reaction mechanism.
$(a)$ Inversion in $S_{N}2$ mechanism: In an $S_{N}2$ reaction,the configuration of the halide is inverted because the nucleophile attacks from the side opposite to the halogen atom. This leads to a product with opposite optical activity.
Example: When $(-)-2-$bromooctane reacts with sodium hydroxide $(NaOH)$ via the $S_{N}2$ mechanism,the $-OH$ group attaches opposite to the bromide position,resulting in $(+)-$octan$-2-$ol. Here,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 enantiomers. One product has 'retention' of configuration (where $-OH$ is at the same position as the halide),and the other has 'inversion' (where $-OH$ is opposite to the halide).
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).
$(i)$ Slow step of $C-Br$ bond cleavage: The reaction proceeds through a planar $sp^{2}$ carbocation intermediate.