Question

Decreasing order of reactivity in Williamson synthesis of the following is

I. Me₃CCH₂Br II. CH₃CH₂CH₂Br

III. CH₂ = CHCH₂Cl IV. CH₃CH₂CH₂Cl

(a) III > II > IV > I 

(b) I > II > IV > III

(c) II > III > IV > I 

(d) I > III > II > IV

Answer 1

The correct option is: (c) II > III > IV > I 

Explanation:

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 S_N2 mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. Thus, the decreasing order of reactivity is II > III > IV > I.