question_answer

In       the       following       reaction\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\frac{\text{1}\text{.Mg,}\,\text{Ether}}{\text{2}\text{.}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}}\text{X,}\]the product 'X is       [AIPMT (M) 2010]

A) \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{OC}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\]

B) \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}\]

C) \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{3}}\]

D) \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\]

Correct Answer: C

Solution :

[c] \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\xrightarrow[{}]{\text{Mg,}\,\text{ether}}\underset{\text{Grignard}\,\text{reagent}}{\mathop{{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{MgBr}}}\,\]