Question

For the following reactions,

Where,

$k_s$ and $k_e$, are, respectively, the rate constants for substitution and elimination, and $\mu =\frac{k_s}{k_e}$, the correct option is ___.

• A. ${\mu }_A>{\mu }_B$ and $k_e\left(A\right)>k_e\left(B\right)$
• B. ${\mu }_B>{\mu }_A$ and $k_e\left(A\right)>k_e\left(B\right)$
• C. ${\mu }_A>{\mu }_B$ and $k_e\left(B\right)>k_e\left(A\right)$
• D. ${\mu }_B>{\mu }_A$ and $k_e\left(B\right)>k_e\left(A\right)$

Answer 1

The correct option is C ${\mu }_A>{\mu }_B$ and $k_e\left(B\right)>k_e\left(A\right)$

The base $C{H}_{3}C{H}_2{O}^{-}$ favours substitution reaction over elimination reaction and thus $k_e\left(A\right)<k_s\left(A\right)$.

Since the bulkier base tertiary butoxide (B) favours elimination reaction over substitution reaction,
$k_e\left(B\right)$ is higher. Since $\mu =\frac{k_s}{k_e},{\mu }_A$ will be greater as it is having lower $k_e$ value.