A) 2-methyl-2-propanol
B) Acetamide
C) Acetone
D) Acetyl iodide
Correct Answer: A
Solution :
\[\underset{\text{Acetyl bromide}}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-Br}}\,\underset{\text{(ii) Saturated }N{{H}_{4}}Cl}{\mathop{\xrightarrow{\text{(i) Excess-}C{{H}_{3}}MgI}}}\,\underset{\text{2-methyl 2-propanol}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}\ \ \ }{\overset{C{{H}_{3}}\ \ \ }{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,-OH}}}\,}}\,\]
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