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