Question

Among the following pair of $S_{N}1$ reactions in which pair the second reaction is more reactive than the first?

• A. $\left(i\right)M{e}_{3}CCl+H_{2}O\to M{e}_{3}COH\left(ii\right)M{e}_{3}CBr+H_{2}O\to M{e}_{3}COH$
• B. $\left(i\right)M{e}_{3}CCl+C{H}_{3}OH\to M{e}_{3}C-OC{H}_3\left(ii\right)M{e}_{3}CCl+H_{2}O\to M{e}_{3}C-OH$
• C. $\left(i\right)\underset{\left(1M\right)}{M{e}_{3}CCl}+H_{2}O\to \left(ii\right)\underset{\left(2M\right)}{M{e}_{3}CCl}+H_{2}O\to$
• D. All of the above.

Answer 1

The correct option is D All of the above.

$\text{(a). Rate of}{S}_{N}1\text{reaction}\propto \text{Leaving group ability}\text{Leaving group ability:}{F}^{-}<C{l}^{-}<B{r}^{-}<I^{-}$
So in (a) second reaction has a higher rate.

$\text{(b). Rate of}{S}_{N}1\text{reaction}\propto \text{Polarity of the solvent}$
Due to the $+I$ effect of $-C{H}_3$ group, the $O$ atom of $C{H}_{3}OH$ molecule becomes less electronegative than that of $H_{2}O$ molecule. Therefore, $H_{2}O$ is more polar than $C{H}_{3}OH$.
So, in (b) second reaction has a higher rate.

$\text{(c). Rate of}{S}_{N}1\text{reaction}\propto [R-X]$
Higher the concentration, higher is the rate.
In second reaction, the concentration $[R-X]$ is greater, so second reaction has a higher rate.