Arrange the compounds of each set in order of reactivity towards S_(N)2 displacement :(i) 2- Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane,1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Arrange the compounds of each set in order of reactivity towards S_(N)

1.	Arrange the compounds of each set in order of reactivity towards S_(N)2 displacement :(i) 2- Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane,1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.
Answer» Solution :The reaction follows `S_(N)2` path if the substrate offers minimum steric hinderance to the nucleophile. Thus order of reactivity of alkyl halides by `S_(N)2` path is : Methyl halide `gt 1^(@)` halide `gt 2^(@)` halide `gt gt gt 3^(@)` halide. (i) `underset((1^(@)))("1-Bromopentane")gt underset((2^(@)))("2-Bromopentane")gt underset((3^(@)))("2-Bromo-2-methylbutane")` (ii)`underset((1^(@)))("1-Bromo-3-methylbutane")gt underset((2^(@)))("2-Bromo-3-methylutane")gt underset((3^(@)))("2-bromo-2-methylbutane")` (iii) 1-bromobutane `gt` 1-Bromo-3-methylbutane `gt` 1-Bromo-2-methylbutane `gt` 1-Bromo-2, 2 - dimethyl propane In (iii), all the halides are primary `(1^(@))`. However, the presence of bulky GROUP such as `-CH_(3)` (methyl) on `BETA` - CARBON decreases the reactivity towards `S_(N)2` due to steric hindrance. Thus, 1-Bromo-2, 2-dimethyl propane is least reactive.