Question

Which of the following is the correct order of decreasing $S_{N}2$ reactivity?

• A. $RC{H}_{2}X>R_{3}CX>R_{2}CHX$
• B. $RC{H}_{2}X>R_{2}CHX>R_{3}CX$
• C. $R_{3}CX>R_{2}CHX>RC{H}_{2}X$
• D. $R_{2}CHX>R_{3}CX>RC{H}_{2}X$

Answer 1

The correct option is B $RC{H}_{2}X>R_{2}CHX>R_{3}CX$

The reactivity of primary alkyl halides is maximum and the reactivity of tertiary alkyl halides is minimum. This is due to steric and electronic effects. As the number of bulky alkyl groups increases, the approach of nucleophile becomes more hindered. This decreases the reactivity.

Also, alkyl groups show $+I$ effect and increase electron density. Due to this the electrophilic character of $C$ atom attached to halogen atom decreases. This also decreases the reactivity towards $S_{N}2$ reaction.

Hence, the correct order of decreasing $S_{N}2$ reactivity is: $RC{H}_{2}X>R_{2}CHX>R_{3}CX$