Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_N2$ mechanism? Explain your answer.
$(i)$ $CH_3CH_2CH_2CH_2Br$ or $CH_3CH_2CH(Br)CH_3$
$(ii)$ $CH_3CH_2CH(Br)CH_3$ or $(CH_3)_3CBr$
$(iii)$ $CH_3CH(CH_3)CH_2CH_2Br$ or $CH_3CH_2CH(CH_3)CH_2Br$

(A) $(i)$ $CH_3CH_2CH_2CH_2Br$ $(1^{\circ})$ reacts faster than $CH_3CH_2CH(Br)CH_3$ $(2^{\circ})$ because $S_N2$ reactions are highly sensitive to steric hindrance. Primary alkyl halides have less steric hindrance than secondary ones,allowing easier nucleophilic attack.
$(ii)$ $CH_3CH_2CH(Br)CH_3$ $(2^{\circ})$ reacts faster than $(CH_3)_3CBr$ $(3^{\circ})$ because the $3^{\circ}$ alkyl halide is extremely sterically hindered due to three bulky methyl groups,making the $S_N2$ pathway very slow.
$(iii)$ $CH_3CH(CH_3)CH_2CH_2Br$ reacts faster than $CH_3CH_2CH(CH_3)CH_2Br$. Although both are primary,in $CH_3CH_2CH(CH_3)CH_2Br$,the methyl group is on the $\beta$-carbon (closer to the reaction site),causing greater steric hindrance compared to $CH_3CH(CH_3)CH_2CH_2Br$ where the methyl group is further away.