A) \[III>II>IV>I\]
B) \[I>II>IV>III\]
C) \[II>III>IV>I\]
D) \[I>III>II>IV\]
Correct Answer: A
Solution :
Williamsons ether synthesis involves \[{{S}_{N}}2\]pathway, ie, involves formation of transition state and back side attack. Thus, less hindered alkyl halides react readily in Williamsons synthesis. The order of hindrance is Allyl halide \[<{{1}^{o}}<{{2}^{o}}<{{3}^{o}}\] As the size of halide ion increases, reactivity towards \[{{S}_{N}}2\]reaction increases. Thus, bromides are more reactive than their corresponding chlorides towards Williamsons ether synthesis. Hence, the decreasing order of reactivity towards Williamsons ether synthesis is \[C{{H}_{2}}=\underset{III}{\mathop{C}}\,HC{{H}_{2}}Cl>C\underset{II}{\mathop{{{H}_{2}}C}}\,{{H}_{2}}Br\] \[<C{{H}_{3}}\underset{IV}{\mathop{C{{H}_{2}}C{{H}_{2}}}}\,Cl>M{{e}_{3}}\underset{I}{\mathop{CC{{H}_{2}}}}\,Br\]
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