Question

Compound ${}^{\prime }{A}^{\prime }$ with molecular formula $C_4{H}_{9}Br$ is treated with $aq.KOH$ solution. The rate of this reaction depends upon the concentration of the compound ${}^{\prime }{A}^{\prime }$ only. When another optically active isomer ${}^{\prime }{B}^{\prime }$ of this compound was treated with $aq.KOH$ solution, the rate of reaction was found to be dependent on the concentration of compound and $aq.KOH$ both.

(A) Write down the structural formula of both compounds ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$.

(B) Out of these two compounds, which one will be converted to the product with inverted configuration.

Answer 1

$\text{Structure of}A\text{and}B$

Rate of reaction depends upon the concentration of the compound only. This implies that the reaction is unimolecular. Therefore, the reaction mechanism is $S_{N}1$ and $A$ is $2$-Bromo-$2$-methylpropane (tertiary bromide) which will form a stable carbocation intermediate. The structure of $A$ is as follow:


${}^{\prime }{B}^{\prime }$ is optically active and is an isomer of ${}^{\prime }{A}^{\prime }$. Therefore, ${}^{\prime }{B}^{\prime }$ must be $2$-Bromobutane. The structure of ${}^{\prime }{B}^{\prime }$ is



$\text{Product with inverted configuration}$

The rate of reaction of compound $\left(B\right)$ depends both upon the concentration compound $\left(B\right)$ and $KOH$. This implies the reaction is bimolecular.

Hence, the reaction follows $S_{N}2$ mechanism.




Compound ${}^{\prime }{B}^{\prime }$ will have an inversion of configuration and turn out to be an inverted product.