A) \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}} \\ & {{H}_{3}}C-\overset{|}{\mathop{\underset{|}{\mathop{C}}\,}}\,-Cl \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,H \\ \end{align}\]
B) \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}} \\ & H-\overset{|}{\mathop{\underset{|}{\mathop{C}}\,}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}\]
C) \[\begin{align} & \,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}} \\ & Cl-\overset{|}{\mathop{\underset{|}{\mathop{C}}\,}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,H \\ \end{align}\]
D) \[\begin{align} & \,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & H-\overset{|}{\mathop{\underset{|}{\mathop{C}}\,}}\,-Cl \\ & \,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}} \\ \end{align}\]
Correct Answer: A
Solution :
[a] \[C{{H}_{3}}\text{ }CHClC{{H}_{2}}C{{H}_{3~}}\] Priority order is
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