Question

Which will undergo SN2 reaction fastest among the following halogen compounds?
A.$$C{H}_{3}C{H}_{2}F$$
B.$$C{H}_{3}C{H}_{2}Cl$$
C.$$C{H}_{3}C{H}_{2}Br$$
D.$$C{H}_{3}C{H}_{2}I$$
E.$${(C{H}_3)}_{2}CH-Cl$$

Answer 1

Hint:
The term SN2 stands for substitution nucleophilic bimolecular. It is the type of substitution reaction in which the rate is dependent on the nucleophile and the substrate.

Complete step by step answer:
Let us understand the concept of SN2 reaction. A substitution nucleophilic bimolecular reaction occurs in the following conditions –
Strong nucleophile
Polar aprotic solvent
Substrate with minimum hindrance.
The rate of SN2 reaction in alkyl halide takes place in the order:
Primary > secondary > tertiary
Now, let us look at the substrates. The one with the minimum hindrance will be the most likely to undergo SN2 reaction.
Option (e) $${(C{H}_3)}_{2}CH-Cl$$
This compound is a secondary alkyl halide. Therefore, it will not undergo reaction as fast.
Now, let us look at the rest of the options.
Option (a) – (d)
All of them are primary alkyl halides. Therefore, we will see the tendency of the leaving group.
The tendency of leaving groups is: I > Br > Cl >F.
Since, Iodine is the biggest in size, it will be able to accommodate its electrons easily and will be more stable when compared to the rest.
Therefore, the tendency to undergo SN2 can be summarized as –
$$C{H}_{3}C{H}_{2}I$$ > $$C{H}_{3}C{H}_{2}Br$$> $$C{H}_{3}C{H}_{2}Cl$$> $$C{H}_{3}C{H}_{2}F$$>$${(C{H}_3)}_{2}CH-Cl$$
Therefore, the answer is – option (d).

Note:
In SN2 reaction, no carbocation is formed. This reaction takes place in a single step. The nucleophile reaches from the opposite side, which leads to a phenomenon known as ‘Walden inversion’.