Decreasing order of reactivity in Williamson synthesis of the following: I. Me3CCH2Br II. CH3CH2CH2Br III. CH2CHCH2Cl IV. CH3CH2CH2Cl

Question

Decreasing order of reactivity in Williamson synthesis of the following: I. Me3CCH2Br II. CH3CH2CH2Br III. CH2CHCH2Cl IV. CH3CH2CH2Cl (A) III > II > IV > I (B) I > II > IV > III (C) II > III > IV > I (D) I > III > II > IV

Tardigrade Admin · Accepted Answer

C—Br bond is weaker than C—Cl bond, therefore, alkyl bromide (II) reacts faster than alkyl chlorides, (III) and (IV). Since

C H_{2} = C H —

is electron withdrawing therefore,

C H_{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 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.

Q. Decreasing order of reactivity in Williamson synthesis of the following : I. Me3​CCH2​Br II. CH3​CH2​CH2​Br III. CH2​CHCH2​Cl IV. CH3​CH2​CH2​Cl

III > II > IV > I

I > II > IV > III

II > III > IV > I

I > III > II > IV

Q. Decreasing order of reactivity in Williamson synthesis of the following :
I. $M e_{3} CC H_{2} B r$
II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} C H C H_{2} Cl$
IV. $C H_{3} C H_{2} C H_{2} Cl$