question_answer

\[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[{{H}_{2}}{{O}_{2}}/O{{H}^{-}}]{NaB{{D}_{4}}}\]Product\[X\],\[X\]is

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

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

C) \[C{{H}_{3}}\underset{\begin{smallmatrix}  | \\  CO \end{smallmatrix}}{\mathop{C}}\,HC{{H}_{3}}\]               

D) None of these

Correct Answer: B

Solution :

\[\xrightarrow{{{H}_{2}}{{O}_{2}}/O{{H}^{-}}}\,C{{H}_{3}}\underset{\begin{smallmatrix}  | \\  D \end{smallmatrix}}{\mathop{C}}\,HC{{H}_{2}}OH\]