Question

What is the correct decreasing order of reactivity of the following compounds for $S_{N}2$ reactions?
${\left(CH\right)}_{3}Cl,{\left(CH\right)}_3{\left(CH\right)}_{2}Cl,{\left({\left(CH\right)}_3\right)}_{2}CHCl$ and ${\left({\left(CH\right)}_3\right)}_{3}CCl$ is:

• A. ${\left(CH\right)}_{3}Cl>{\left({\left(CH\right)}_3\right)}_{2}CHCl>{\left(CH\right)}_3{\left(CH\right)}_{2}Cl>{\left({\left(CH\right)}_3\right)}_{3}CCl$
• B. ${\left(CH\right)}_{3}Cl>{\left(CH\right)}_3{\left(CH\right)}_{2}Cl>{\left({\left(CH\right)}_3\right)}_{2}CHCl>{\left({\left(CH\right)}_3\right)}_{3}CCl$
• C. ${\left(CH\right)}_3{\left(CH\right)}_{2}Cl>{\left(CH\right)}_{3}Cl>{\left({\left(CH\right)}_3\right)}_{2}CHCl>{\left({\left(CH\right)}_3\right)}_{3}CCl$
• D. ${\left({\left(CH\right)}_3\right)}_{2}CHCl>{\left(CH\right)}_3{\left(CH\right)}_{2}Cl>{\left(CH\right)}_{3}Cl>{\left({\left(CH\right)}_3\right)}_{3}CCl$

Answer 1

Answer: (B)

$S_{N}2$ involves the formation of a pentavalent intermediate, and the reactivity depends on steric hindrance. Lesser the steric hindrance more is the reactivity of ${\text{S}}_{\text{N}}2$ .
It is a one-step reaction. From the front end, the weak nucleophile is leaving and from the backside strong nucleophile is attacking the electrophilic carbon centre.
If less crowding around the central electrophilic carbon atom then it will be easy for the incoming nucleophile. Therefore, the reactivity towards the $S_{N}2$ reaction.