Question

Which of the following species is most reactive in $S{N^2}$reaction?
(A) $C{H_3} - C{H_2} - Cl$
(B) $C{H_3} - C{H_2} - Br$
(C) $C{H_3} - C{H_2} - I$
(D) $C{H_3} - C{H_2} - F$

Answer 1

To answer the question we need to know the detailed process of $S{N^2}$ reaction. And also we need to consider the reactivity, stability and electro-negativity of halogens. By considering these factors we can easily conclude which one of them is most reactive and least stable.

Complete step-by-step answer: $S{N^2}$ is nucleophilic substitution reaction for bimolecular species that means two molecules are present in the transition state. In $S{N^2}$ reaction, the whole reaction occurs in a one step process in which a nucleophile attacks the substrate, and a leaving group, L, departs simultaneously. This one step in the reaction will be a rate determining step.
Always in $S{N^2}$ reaction nucleophile attacks from backside with respect to leaving group that means $180^\circ$ away from leaving group. This is the main stable stereo-electronic requirement for $S{N^2}$ reaction.
In the $S{N^2}$ reaction penta-coordinated transition state is developed where one lobe of a $p$-orbital of the $C$atom interact with the leaving group and the other lobe of the $p$-orbital interacts with the nucleophile. And also in this reaction, the central $C$atom changes its hybridization from $s{p^3}$to$s{p^2}$.
In the $S{N^2}$ reaction, both the starting material or substrate and nucleophile are involved in rate determining state. Thus, rate of the reaction depends upon both the concentration of substrate and as well as nucleophile. $Rate = [subs][N{u^-}]$ Thus, molecularity$= 2$. Hence the order of the reaction is $2nd$ order.
Now, if we talk about halogens, from $F$to $I$, the electro-negativity gradually decreases and size of the halogen increases. So, when a nucleophile attacks the reaction carbon centre from backside large sized and more electro-positive iodine will leave fastest as the leaving group. So, the decreasing order of the tendency of halogens to leave is: $- I > - Br > - Cl > - F$
So, $C{H_3} - C{H_2} - I$species will be most reactive towards the $S{N^2}$ reaction.

Hence the correct option is (C).

Note: $S{N^2}$reaction gives inversion of stereochemistry. And we also have to consider the steric factor while considering the $S{N^2}$ reaction. And while determining the rate we have to consider substrates as well as nucleophiles.