Predict the correct order of reactivity of the following four...

The rate of an $\mathrm{S}_{\mathrm{N}}2$ reaction is inversely proportional to the steric hindrance around the electrophilic carbon atom. Since all the given compounds are primary alkyl bromides, we must compare the steric hindrance caused by branching at the $\beta$ and $\gamma$ positions.

(III) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$ is a straight-chain primary bromide with no branching, offering the least steric hindrance.

(IV) $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$ has branching at the $\gamma$-carbon, which introduces slight steric hindrance compared to the unbranched chain.

(I) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{Br}$ has branching at the $\beta$-carbon (a tertiary $\beta$-carbon), which significantly increases steric hindrance to the approaching nucleophile.

(II) $(\mathrm{CH}_3)_3\mathrm{CCH}_2\mathrm{Br}$ (neopentyl bromide) has a quaternary $\beta$-carbon, providing maximum steric hindrance and making it practically unreactive towards $\mathrm{S}_{\mathrm{N}}2$ reactions.

Therefore, the steric hindrance increases in the order (III) $<$ (IV) $<$ (I) $<$ (II).

The reactivity towards $\mathrm{S}_{\mathrm{N}}2$ follows the reverse order: (III) $>$ (IV) $>$ (I) $>$ (II).

Answer: (III) $>$ (IV) $>$ (I) $>$ (II)