A) a secondary alcohol
B) a tertiary alcohol
C) a primary alcohol
D) \[C{{H}_{3}}-C-C{{H}_{3}}\]
Correct Answer: A
Solution :
\[C{{H}_{3}}CHO+C{{H}_{3}}MgBr\xrightarrow{{}}\] \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-OMgBr\xrightarrow{H.\,OH/{{H}^{+}}}\]You need to login to perform this action.
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