Predict the order of reactivity of the following four isomers towards $\text{S}_\text{N}2$ reaction. (I) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$ (II) $\text{CH}_3\text{CH}_2\text{CH}(\text{Cl})\text{CH}_3$ (III) $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$ (IV) $(\text{CH}_3)_3\text{CCl}$

The reactivity of alkyl halides towards $\text{S}_\text{N}2$ reactions is primarily governed by steric hindrance around the electrophilic carbon atom. The lesser the steric hindrance, the higher the reactivity. Thus, the general order of reactivity is primary ($1^{\circ}$) > secondary ($2^{\circ}$) > tertiary ($3^{\circ}$) .

Analysing the given isomers: (I) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$ (1-chlorobutane) is a straight-chain primary ($1^{\circ}$) alkyl halide, which has the least steric hindrance. (III) $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$ (1-chloro-2-methylpropane) is also a primary ($1^{\circ}$) alkyl halide, but it has a methyl branch at the $\beta$-carbon, making it more sterically hindered than isomer (I). (II) $\text{CH}_3\text{CH}_2\text{CH}(\text{Cl})\text{CH}_3$ (2-chlorobutane) is a secondary ($2^{\circ}$) alkyl halide, which is more hindered than the primary halides. (IV) $(\text{CH}_3)_3\text{CCl}$ (2-chloro-2-methylpropane) is a tertiary ($3^{\circ}$) alkyl halide, which is highly sterically hindered and least reactive towards $\text{S}_\text{N}2$ substitution.

Therefore, the correct decreasing order of reactivity towards $\text{S}_\text{N}2$ reaction is (I) > (III) > (II) > (IV) .