A) Methyl acetate
B) Benzaldehyde
C) Ethyl formate
D) None of the above
Correct Answer: A
Solution :
All esters (except\[HCOO{{C}_{2}}{{H}_{5}},\]i.e., ethyl formate) on reaction with Grignard reagent to form tertiary alcohol. Hence, methyl acetate gives a tertiary alcohol on reaction with excess of\[C{{H}_{3}}MgI\]. \[\underset{methyl\text{ }acetate}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,-OC{{H}_{3}}}}\,+C{{H}_{3}}MgI\xrightarrow{{}}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OC{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OMgI\xrightarrow{C{{H}_{3}}MgI}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OMgI\] \[\xrightarrow{{{H}_{3}}{{O}^{+}}}\underset{tertiary\text{ }alcohol}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OH}}\,\]
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