question_answer

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

A) \[\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\]

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

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

D) \[C{{H}_{3}}-\underset{Cl}{\mathop{\overset{Cl}{\mathop{\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{3}}\]

Correct Answer: D

Solution :

[d]