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

The `S_(N)2` reactions reactivity depends upon steric hindrance. More the steric hindrance slower the reaction. Thus the order of reactivity will be `1^(@) gt 2^(@) gt 3^(@)`

Since in case of `1^(@)` alkyl halides steric hindrance in increases in the order) n- alkyl halides, alkyl halides with a substituent at any position other than the `beta` - position, one substituent at the `beta`-position, two substituents at the `beta-` position, therefore , the reactivity decreases in the same order .Thus, the reactivity of the given alkyl bromides decreases in the order.
1-Bromobutane `gt` l-Bromo-3-methylbutane `gt` l-Bromo-2-methyjbutane `gt` 1-Bromo-2,2-dimethyl propane.