What is the correct decreasing order of reactivity of the following...

Williamson ether synthesis proceeds via an $S_N2$ mechanism.

The rate of an $S_N2$ reaction is inversely proportional to the steric hindrance around the electrophilic carbon and is enhanced by the stabilization of the transition state.

Compound III ($\mathrm{C_6H_5-CH_2-Br}$) is a benzylic halide. Its $S_N2$ transition state is highly stabilized by the adjacent phenyl ring through conjugation, making it the most reactive.

Compound IV ($\mathrm{CH_3-CH_2-CH_2-CH_2-Br}$) is an unbranched primary alkyl halide with minimal steric hindrance.

Compound II ($\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-Br}$) is a primary alkyl halide but has a methyl group at the $\beta$-position. This branching increases steric hindrance, making it less reactive than IV.

Compound I ($\mathrm{CH_3-CH_2-CH_2-CH(Br)-CH_3}$) is a secondary alkyl halide, which is more sterically hindered at the $\alpha$-carbon than primary alkyl halides, making it the least reactive.

Therefore, the decreasing order of reactivity is III $>$ IV $>$ II $>$ I.

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