question_answer

\[C{{H}_{3}}COOC{{H}_{3}}+\] excess \[PhMgBr\]\[\to \text{product}\xrightarrow{{{H}^{+}}}X\] The product \[X\] is        [Orissa JEE 2005]

A)                 1, 1-diphenylethanol

B)                 1, 1-diphenylmethanol

C)                 Methyl phenylethanol

D)                 Methyl phenylketone

Correct Answer: A

Solution :

           \[\xrightarrow{-MgBr(OC{{H}_{3}})}\overset{\,\,\,\,O}{\mathop{\overset{\,\,\,\,\,|\,|}{\mathop{C{{H}_{3}}-C-Ph}}\,}}\,\xrightarrow{PhMgBr}\underset{Ph}{\mathop{\underset{|\,\,\,\,}{\mathop{\overset{\,\,\,\,\,\,\,OMgBr}{\mathop{\overset{|\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-C-Ph\,\,\,\,\,\,\,}}\,}}\,}}\,}}\,\] \[\xrightarrow{{{H}^{+}}}\underset{\text{1,1-diphenylethanol}}{\mathop{\underset{\,\,Ph}{\mathop{\underset{|\,}{\mathop{\overset{\,\,OH}{\mathop{\overset{|\,\,}{\mathop{C{{H}_{3}}C-Ph}}\,}}\,}}\,}}\,}}\,\]