1. Tardigrade
  2. Question
  3. Chemistry
  4. Decreasing order of reactivity in Williamson synthesis of the following: I. Me3CCH2Br II. CH3CH2CH2Br III. CH2CHCH2Cl IV. CH3CH2CH2Cl

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Q. Decreasing order of reactivity in Williamson synthesis of the following :
I. $Me_3CCH_2Br$
II. $CH_3CH_2CH_2Br$
III. $CH_2CHCH_2Cl$
IV. $CH_3CH_2CH_2Cl$

AIIMSAIIMS 2017Alcohols Phenols and Ethers

Solution:

C—Br bond is weaker than C—Cl bond, therefore, alkyl bromide (II) reacts faster than alkyl chlorides, (III) and (IV). Since $CH_2 = CH—$ is electron withdrawing therefore, $CH_2$ 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.

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