Question

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

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

Answer 1

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

In $S_{N}2$ mechanism transition state is pentavalent. Smaller alkyl group will favour the $S_{N}2$ mechanism. Bulky alkyl group will have sterical hindrance.