Question

Rank the following compounds in order of their expected reactivity toward ${\mathbf{S}}_{\mathbf{N}}\mathbf{2}$reaction:

${\mathbf{CH}}_{\mathbf{3}}\mathbf{Br}\mathbf{,}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OTos}\mathbf{,}{\left(\mathbf{CH}\mathbf{3}\right)}_{\mathbf{3}}\mathbf{CCl}\mathbf{,}{\left(\mathbf{CH}\mathbf{3}\right)}_{\mathbf{2}}\mathbf{CHCl}$

Short answer

${\mathrm{CH}}_3\mathrm{OToS}>{\mathrm{CH}}_3\mathrm{Br}>{\left(\mathrm{CH}3\right)}_2\mathrm{CHCl}>{\left(\mathrm{CH}3\right)}_3\mathrm{CCl}$

Step-by-step solution

Nucleophilic bimolecular substitution reaction SN2

The${\mathrm{S}}_{\mathrm{N}}2$ reaction is a nucleophilic substitution reaction where bond is broken and another is form simultaneously. Two reacting species are involved in the rate determining step of the reaction. The term${\mathrm{S}}_{\mathrm{N}}2$ stands for substitution nucleophilic bimolecular.

Effect of substrate and leaving group on SN2 reactions

The substrate plays the most important part in determining the rate of reaction. This is because the nucleophile attacks from the back of the substrate, thus breaking the carbon-leaving group bond. It results in the formation of the carbon-nucleophile bond.

${\mathrm{S}}_{\mathrm{N}}2$ is a bimolecular reaction and the reaction rate depends on the alkyl halide and the nucleophile effect of the leaving group. The better the leaving group, the faster the reaction, and therefore the reaction rate is also greater.

Order of Reactivity

In this question we are comparing two effects—the effect of the substrate and the effect of the leaving group. Tertiary substrates are less reactive than secondary substrates, which are less reactive than primary substrates.

Most reactive →Least reactive

${\mathrm{CH}}_3\mathrm{OToS}>{\mathrm{CH}}_3\mathrm{Br}>{\left(\mathrm{CH}3\right)}_2\mathrm{CHCl}>{\left(\mathrm{CH}3\right)}_3\mathrm{CCl}$

Excellent good secondary tertiary

leaving leavingcarboncarbon

group group