Rearrange the compounds of each of the following sets in order of reactivity towards `S_(N_(3)^(2))` displacement :
(i) 2- Bromo-2-methylbutane,
1-Bromopentane, 2- Bromopentaure.
(ii) 1- Bromo-3-methylbutance, 2-Bromo-2-methylbutance,3-Bromo-2- methybutane.
(iii) 1- Bromobutane, 1- Bromo-2,
2-dimethylbutane
1-Bromo -2- methylbutane.

Explanation

(i) 2-Bromo-2-methybutane,
1-Bromopentane , 2Bromopentane:
The reactivity is `SN^(2)` reaction depents upon steric hindance, more the steric hindence slower the reaction.

Since due to steric reasion the order of reactivity in `SN^(2)` reaction follows the order:
`1^(@) gt2^(@) gt3^(@)`, therefore, order of reactivity of the given alklyl bromides is :
1-Bromopentane `gt 2`- Brompebtane `gt 2-` Bromo -2 Methylbutane.

Since due to steric reason, the order of reactivity of alkyl halides in `SN^(2)` reaction follows
the order `1^(@) gt 2^(@) gt 3^(@)` therefore, the order of reactivity of the given alkyl bromies is .
1-Bromo-3-Methylbutane`gt2`- Bromo -3-Methylbutane `gt 2`-Bromo -2-Methylbutane.

Since in case of `1^(@)` alkyl halides, steric hindrance in the order : n-alkyl halides,
alkyl halide substituents at the `beta`-position, therefore , the reactivity decrease in the same order- Thu, the reactivity of the given alkyl bromides decrease in the order.
1-Bromobutance `gt 1`- Bromo - 3- Methybutane `gt 1`- Bromo-2- Methylbutane `gt1`- Bromo-2-Methylbutane`gt1`- Bromo-2-dimethylpropane.