Decreasing order of reactivity in Williamson’s ether synthesis of the following:
$\text{ I) }{\mathrm{Me}}_3{\mathrm{CCH}}_2\mathrm{Br}$
$\text{ II) }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$
$\text{ III) }{\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{CI}$
$\text{ IV) }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{CI}$

C—Br bond is weaker than C—Cl bond, therefore, alkyl bromide (II) reacts faster than alkyl chlorides, (III) and (IV). 
Since CH₂=CH—  is electron withdrawing therefore, CH₂ has more +ve charge on III than on IV.
In other words, nucleophilic attack occurs faster on III than on IV. Further, since Williamson synthesis occurs by SN₂ mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. 

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