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.You need to login to perform this action.
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