question_answer

In the following reaction sequence : \[\underset{({{C}_{3}}{{H}_{6}}C{{l}_{2}})}{\mathop{I}}\,\,\xrightarrow{KOH\,(aq)}\,II\,\xrightarrow[(II)\,{{H}_{2}}O/{{H}^{+}}]{(I)\,C{{H}_{3}}MgBr}\,III\,\xrightarrow{Anhy.\,ZnC{{l}_{2}}+Conc.\,HCl}\]\[\,gives\,turbidity\,immediately\] The compound I is: [JEE Online 09-04-2017]

A)  \[C{{H}_{3}}-\underset{\begin{smallmatrix}  \,\,\,| \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  \,\,| \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\]              

B) \[Cl-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}\]

C)  \[\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\]             

D) \[\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-C{{H}_{2}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\]

Correct Answer: A

Solution :