Since Williamson's ether synthesis takes place via SN2 mechanism. So, alkyl chloride with more steric hindrance will be less reactive in nature. Out of (i), (ii), (iii) and (iv), (i) will have high steric . hindrance and will be least reactive. Alkyl chloride (ii) reacts faster than (iii) and (iv). Since CH2=CH− is electron withdrawing which makes CH2 more electron deficient than (iii) and (iv). In other words, nucleophilic attack occur faster on (ii). Out of (iii) and (iv), (iii) will be more electron deficient than (iv) due to absence of one methyl group, i.e. electron-donating group. Hence, correct order of reactivity towards Williamson's ether synthesis will be (ii) > (iii) > (iv) > (i)