question_answer

\[CC{{l}_{3}}CH=C{{H}_{2}}\xrightarrow{C{{l}_{2}}+{{H}_{2}}O}A;\] Compound A is

A)  \[CC{{l}_{3}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HC{{H}_{2}}Cl\]

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

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

D)  \[CC{{l}_{3}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,{{H}_{2}}\]

Correct Answer: B

Solution :

\[CC{{l}_{3}}\] is an electron withdrawing group causing \[\pi \]-electron shift from \[{{C}_{1}}\] to \[{{C}_{2}},\] and addition is opposite to markownikoff?s rule.