Arrange the following compounds in order of incraesing `S_(N)2` reacitivity and explain:
(ii) Ethyl bromide, isobutyl bromide, n-propyl bromide, neopentyl bromide.

Explanation

`underset("Ethyl bromide")(overset(beta)CH_(3)overset(alpha)(C)H_(2)Br)` `underset("Isobutyl bromide")((CH_3)_(2)overset(beta)CHoverset(alpha)CHBr)` `underset("n-propyl bromide")(CH_(3)overset(beta)CHoverset(alpha)CHBr)`
`underset("Neopentyl bromide")((CH_3)_(3)overset(beta)Coverset(alpha)CH_(2)Br)`
The `S_(N)2` attack on the halogenated carbon is progressively hindered with increase in the number of alkyl groups at the `beta`- carbon atom. There are no alkyl group at the `beta`- carbon atom of ethyl bromide while n-propyl bromide contains one, isobutyl bromide contains two and neopentyl bromide contains three alkyl groups at the `beta`-carbon atom. Therefore, the `S_(N)2` reactivity of these bromides decrease in the order : ethyl bromide > n-prppyl bromide > isopropyl bromide > neopentyl bromide.