Question

The decreasing order of rate of $S_{N}2$ is
(I) $Cl-C{H}_2-C{H}_2-Br$
(II) $N\equiv C-C{H}_2-Br$
(III) $C{H}_3-\underset{\underset{Br}{|}}{CH}-C{H}_3$
(IV) $C{H}_3-C{H}_2-Br$

• A. II > I > IV > III
• B. II > III > I > IV
• C. II > I > III > IV
• D. IV > I > II > III

Answer 1

The correct option is A II > I > IV > III

In $S_{N}2$ reaction, the nucleophile attacks from the back side and forms a transition state and then the leaving group leaves. It is a single step concerted reaction. Thus, the rate of the reaction depends on both reactant and the nucleophile.
In primary substrate, there will be less steric hindrance so the attack of nucleophile will become easier and rate of $S_{N}2$ reaction is faster.
The order of rate of substrate for $S_{N}2$ reaction is
Primary > Secondary > Tertiary

Presence of electron withdrawing group increases the reactivity of the substrate towards $S_{N}2$ reaction.

Comparing (I) and (II)
Both $Cl$ and $CN$ are -I group. But $CN$ shows more -I effect than $Cl$.
In (IV), we don't have any destabilising group but it is a primary carbocation which is less steric.

Therefore, the decreasing order of rate of $S_{N}2$ reaction is
II > I > IV > III