A) acetone
B) acetamide
C) 2-methyl-2-propanol
D) acetyl iodide
Correct Answer: C
Solution :
\[\underset{Acetyl\,\,bromide}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-Br+C{{H}_{3}}MgI}}\,\xrightarrow{{}}\underset{Acetone}{\mathop{C{{H}_{3}}\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,C{{H}_{3}}}}\,\] \[\underset{Acetone}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}+C{{H}_{3}}MgI}}\,\xrightarrow{{}}C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-OMgI\xrightarrow{aq\,\,N{{H}_{4}}Cl}\underset{\begin{smallmatrix} 2-methyl \\ 2-propanol \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-OH}}\,\]
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