A) Primary alcohol
B) Secondary alcohol
C) Tertiary alcohol
D) Ketone
Correct Answer: B
Solution :
\[C{{H}_{3}}-\underset{H}{\mathop{\underset{|}{\mathop{C}}\,}}\,=O+C{{H}_{3}}MgBr\to \underset{\,\,\,\,\,\,\,\,({{2}^{o}}\,\,\text{alcohol})}{\mathop{\underset{\,\,\,\,H}{\overset{\,\,\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\underset{\,\,\,\,|}{\overset{\,\,\,|}{\mathop{C{{H}_{3}}-C\,-OH}}}\,}}}\,}}\,\]
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