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