A) \[n-\] butyl alcohol
B) tertiary butyl alcohol
C) secondary butyl alcohol
D) isopropyl alcohol
Correct Answer: B
Solution :
\[C{{H}_{3}}COC{{H}_{3}}+C{{H}_{3}}MgBr\xrightarrow{{}}\] \[\underset{\text{Tertiary}\,\text{butyl}\,\text{alcohol}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{CH}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-OH}}\,+mg<_{\text{Br}}^{\text{OH}}\]
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