Question

Arrange the compounds of each set in order of reactivity towards $S_{N}2$ displacement:

A $2$-Bromo-$2$-methylbutane,
$1$-Bromopentane,
$2$-Bromopentane

B $1$-Bromo-$3$-methylbutane,
$2$-Bromo-$2$-methylbutane,
$2$-Bromo-$3$-methylbutane

C $1$-Bromobutane,
$1$-Bromo-$2$,$2$-dimethylpropane,
$1$-Bromo-$2$-methylbutane,
$1$-Bromo-$3$-methylbutane

Answer 1

$S_{N}2$ $\text{reaction}$

The $S_{N}2$ reaction is a nucleophilic substitution reaction where a bond is broken, and another bond is formed synchronously. Two reacting species are involved in the rate determining step of the reaction. The term '$S_{N}2$' stands for – Substitution Nucleophilic Bimolecular.

$C{H}_3-Br+O{H}^{-}\to C{H}_{3}OH+B{r}^{-}$

Mechanism:

Factors affecting reactivity

When the substrate is sterically hindered, then the reactivity of $S_{N}2$ substitution reaction is least.

The rate of $S_{N}2$ reaction in $2$-Bromo-$2$-methylbutane which is a $3^{\circ }$ alkyl halide is least due to steric hindrance and rate of $S_{N}2$ reaction is fastest in case of $1$-Bromopentane ($1^{\circ }$ alkyl halide) because steric hindrance is least among these three given compounds.
$2-\text{Bromo}-2-\text{methylbutane}$ $1-\text{Bromopentane}$

$2-\text{Bromopentane}$

Hence, the increasing order of reactivity towards $S_{N}2$ substitution will be:

$2$-Bromo-$2$-methylbutane < $2$-Bromopentane < $1$-Bromopentane


$S_{N}2$ reaction

The $S_{N}2$ reaction is a nucleophilic substitution reaction where a bond is broken, and another bond is formed synchronously. Two reacting species are involved in the rate determining step of the reaction. The term '$S_{N}2$' stands for
– Substitution Nucleophilic Bimolecular.

$C{H}_3-Br+O{H}^{-}\to C{H}_{3}OH+B{r}^{-}$

Mechanism:
Factors affecting reactivity

When the substrate is sterically hindered, then the reactivity of $S_{N}2$ substitution reaction is least.
The rate of $S_{N}2$ reaction in $2$-Bromo-$2$-methylbutane which is a $3^{\circ }$ alkyl halide is least due to steric hindrance and rate of $S_{N}2$ reaction is fastest in case of $1$-Bromo-$3$-methylbutane ($1^{\circ }$ alkyl halide) because steric hindrance is least among these three given compounds.
$1-\text{Bromo}-3-\text{methylbutane}$

$2-\text{Bromo}-2-\text{methylbutane}$

$2-\text{Bromo}-3-\text{methylbutane}$

Hence, the increasing order of reactivity towards $S_{N}2$ substitution will be:

$2$-Bromo-$2$-methylbutane < $2$-Bromo-$3$-methylbutane < $1$-Bromo-$3$-methylbutane


$S_{N}2$ $\text{reaction}$

The $S_{N}2$ reaction is a nucleophilic substitution reaction where a bond is broken, and another bond is formed synchronously. Two reacting species are involved in the rate determining step of the reaction. The term '$S_{N}2$' stands for – Substitution Nucleophilic Bimolecular.

$C{H}_3-Br+O{H}^{-}\to C{H}_{3}OH+B{r}^{-}$

Mechanism:

$\text{Factors affecting reactivity}$

The steric hindrance to the incoming nucleophile in the $S_{N}2$ reaction increases with the increase in the number of substituents in the substrate surrounding the carbon atom site attached to the leaving group.

In the given question, all compounds are $1^{\circ }$ alkyl halide but steric hinderance is maximum in case of $1$−Bromo−$2$,$2$−dimethylpropane because carbon site attached to the leaving group $(-Br)$ is most hindered due to the presence of bulky tert butyl group and steric hinderance is minimum in case of $1$−Bromobutane.

$1-\text{Bromobutane}$

$1-\text{Bromo}-2,2-\text{dimethylpropane}$

$1-\text{Bromo}-2-\text{methylbutane}$

$1-\text{Bromo}-3-\text{methylbutane}$

Hence, the increasing order of reactivity towards $S_{N}2$ substitution will be:

$1$-Bromo-$2$,$2$-dimethylpropane < $1$-Bromo-$2$-methylbutane < $1$-Bromo-$3$-methylbutane < $1$-Bromobutane.