Compound $A$ with molecular formula $C_4H_9Br$ is treated with aqueous $KOH$ solution. The rate of this reaction depends upon the concentration of compound $A$ only. When another optically active isomer $B$ of this compound was treated with aqueous $KOH$ solution,the rate of reaction was found to be dependent on the concentration of both the compound and $KOH$.
$(i)$ Write down the structural formula of both compounds $A$ and $B$.
$(ii)$ Out of these two compounds,which one will be converted to the product with inverted configuration?

(B) $(i)$ Compound $A$ is $2$-bromo-$2$-methylpropane,$(CH_3)_3CBr$. Its reaction with aqueous $KOH$ follows $S_N1$ mechanism,where the rate depends only on the concentration of the alkyl halide.
Compound $B$ is $2$-bromobutane,$CH_3CH(Br)CH_2CH_3$. It is optically active and its reaction with aqueous $KOH$ follows $S_N2$ mechanism,where the rate depends on the concentration of both the alkyl halide and the nucleophile $(OH^-)$.
$(ii)$ Compound $B$ will be converted to the product with inverted configuration because $S_N2$ reactions proceed with Walden inversion.