question_answer

Consider the following reaction, \[\text{CC}{{\text{l}}_{\text{3}}}\text{CH=C}{{\text{H}}_{\text{2}}}\xrightarrow[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\text{/HOCl}]{\text{NaB}{{\text{O}}_{\text{4}}}}\]Product (P) P is

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 \[\text{CC}{{\text{l}}_{\text{3}}}\underset{\text{Cl}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}\,}}\,\text{H-C}{{\text{H}}_{\text{2}}}\text{OH}\]