Question

$S_{N}1$ reaction undergoes through a carbocation intermediate as follows:
$R-X\left(aq\right)\stackrel{slow}{\rightleftharpoons }{R}^{+}\left(aq\right)+X^{-}\left(aq\right)\underset{Fast}{\overset{H_{2}O}{\to }}ROH\left(aq\right)+H^{+}\left(aq\right)$

$[R=t-Bu,iso-Pr,Et,Me]$

Choose the correct statement(s):

• A. The decreasing order of rate of $S_{N}1$ reaction is: $t-BuX>iso-PrX>EtX>MeX$
• B. The decreasing order of ionization energy is: $MeX>EtX>iso-PrX>t-BuX$
• C. The decreasing order of energy of activation is: $t-BuX>iso-PrX>EtX>MeX$
• D. The decreasing order of rate of $S_{N}1$ reaction is: $MeX>EtX>iso-PrX>t-BuX$

Answer 1

Answer: (A), (B)

The decreasing order of ionization energy is: $MeX>EtX>iso-PrX>t-BuX$

$\text{Rate of reaction of}{S}_{N}1\propto \text{Stability of carbocation}$
$\text{Stability of carbocation}\propto \text{Number of}\alpha -H$
$\therefore$The decreasing order of rate of $S_{N}1$ reaction is:
$t-BuX>iso-PrX>EtX>MeX$

$\text{Ionization energy}\propto \frac{1}{\text{Stability of carbocation}}$

$\therefore$The decreasing order of ionization energy is:
$MeX>EtX>iso-PrX>t-BuX$

$\text{Activation energy}\propto \frac{1}{\text{Stability of carbocation}}$

$\therefore$ The decreasing order of activation energy is:
$MeX>EtX>iso-PrX>t-BuX$

Hence, statements (a) and (b) are correct.