Question

In $80$% ethanol at ${55}^{o}C$ isopropyl bromide reacts with hydeoxide ion according to the following kinetics:
$-\frac{d\left[RX\right]}{dt}=(4.7\times {10}^{-5}{M}^{-1}{s}^{-1})\left[RX\right]\left[{OH}^{-}\right]+2.4\times {10}^{-6}{s}^{-1}\left[RX\right]$
What percentage of isopropyl bromide reacts by the $S_{N}2$ mechanism when $\left[{OH}^{-}\right]=0.01M$?

• A. $16.36$%
• B. $83.62$%
• C. $95.14$%
• D. $4.86$%

Answer 1

The correct option is A $16.36$%

$S_{N}2$ mechanism is Biomolecular.

The only biomolecular part is

${\vartheta }_{S_{N}2}=(4.7\ast {10}^{-5})\left[RX\right]\left[O{H}^{-}\right]\to$ eq 1$[\frac{A^{\left(2\right)}}{A^{\left(1\right)}}=\frac{k_2}{k_1}\ast \frac{N_2}{N_1}]$

$S_{N}1$ is unimolecular:

${\vartheta }_{S_{N}1}=(2.4\ast {10}^{-6})\left[RX\right]\to$ eq 2

so eq 1 wrt [RX] is ${\vartheta }_{S_{N}2}=(4.7\ast {10}^{-7})\left[RX\right]$

$\therefore \frac{[RX{]}_{S_{N}2}}{[RX{]}_{S_{N}1}}=\frac{4.7\ast {10}^{-7}}{2.4\ast {10}^{-6}}=16.8$%