A) \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{OH}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{HC}{{\text{H}}_{\text{2}}}\text{Cl}\]
B) \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{OH}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{H}\underset{\text{OH}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,{{\text{H}}_{2}}\]
C) \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{Cl}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{H-C}{{\text{H}}_{2}}OH\]
D) \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{Cl}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{H-}\underset{\text{Cl}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,{{\text{H}}_{2}}\]
Correct Answer: C
Solution :
Since, \[CC{{l}_{3}}\]is an electron withdrawing group Thus, product P is \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{Cl}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{H-C}{{\text{H}}_{\text{2}}}\text{OH}\]You need to login to perform this action.
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