question_answer

Decreasing order of reactivity in Williamsons synthesis of the following:

I.\[M{{e}_{3}}CC{{H}_{2}}Br\]

II.\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\]

III.\[C{{H}_{2}}=CHC{{H}_{2}}Cl\]

IV.\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\]

A) III > II > IV > I 

B) I > II > IV > III

C) II > III > IV > I 

D)  I > III > II > IV

Correct Answer: C

Solution :

[c] : C-Br bond is weaker than C-Cl bond, therefore, alkyl bromide (II) reacts faster than alkyl chloride (III) and (IV). Since 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 Williamsons synthesis occurs by \[{{S}_{N}}2\]mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. Thus, the decreasing order of reactivity is II > III > IV > I.