Question

decreasing order of reactivity in williamson ether synthesis of the following 1 Me3CCH2Br 2 CH3CH2CH2Br 3 CH2CHCH2Ci 4 CH3CH2CH2Cl

Answer 1

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.