Question

In $S_{N}2$ reaction, the correct order of reactivity for the following compounds is:
$C{H}_{3}Cl,C{H}_{3}C{H}_{2}Cl,(C{H}_3{)}_{2}CHCl$ and $(C{H}_3{)}_{3}CCl$

• A. $C{H}_{3}Cl>(C{H}_3{)}_{2}CHCl>C{H}_{3}C{H}_{2}Cl>(C{H}_3{)}_{3}CCl$
• B. $C{H}_{3}Cl>C{H}_{3}C{H}_{2}Cl>(C{H}_3{)}_{2}CHCl>(C{H}_3{)}_{3}CCl$
• C. $C{H}_{3}C{H}_{2}Cl>C{H}_{3}Cl>(C{H}_3{)}_{2}CHCl>(C{H}_3{)}_{3}CCl$
• D. $(C{H}_3{)}_{2}CHCl>C{H}_{3}C{H}_{2}Cl>C{H}_{3}Cl>(C{H}_3{)}_{3}CCl$

Answer 1

The correct option is B $C{H}_{3}Cl>C{H}_{3}C{H}_{2}Cl>(C{H}_3{)}_{2}CHCl>(C{H}_3{)}_{3}CCl$

Steric hindrance (crowding) is the basis of $S_{N}2$ reaction.
$\text{Rate of}{S}_{N}2\propto \frac{1}{\text{Steric crowding at 'C'}\text{bearing leaving group}}$



As steric hinderance (crowding) increases, rate of $S_{N}2$ reaction decreases.
Note : The order of reactivity towards $S_{N}2$ reaction for alkyl halides is:
$\text{Methyl}>\underset{\left(1^{\circ }\right)}{\text{Primary halides}}>\underset{\left(2^{\circ }\right)}{\text{Secondary halides}}>\underset{\left(3^{\circ }\right)}{\text{Tertiary halides}}$