question_answer

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

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

B)  \[CC{{l}_{3}}\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{C}}\,H\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{C}}\,{{H}_{2}}\]

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

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

Correct Answer: C

Solution :

Since, \[CC{{l}_{3}}\] is an electron withdrawing group Thus, product P is \[CC{{l}_{3}}\,\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}OH\]