A) \[CaC{{O}_{3}}\]
B) \[NaOH\]
C) \[KOH\]
D) \[MgC{{O}_{3}}\]
Correct Answer: B
Solution :
Acetone on iodination gives iodoform in the presence of NaOH \[\underset{\text{Acetone}}{\mathop{\overset{\overset{O}{\mathop{||}}\,}{\mathop{C{{H}_{3}}-C-C{{H}_{3}}}}\,}}\,+3{{I}_{2}}+4NaOH\xrightarrow{-3{{H}_{2}}O}\]\[\underset{\text{Iodoform}}{\mathop{CH{{I}_{3}}}}\,+\underset{\text{Sodium acetate}}{\mathop{C{{H}_{3}}COONa}}\,+\underset{\text{Sodium iodide}}{\mathop{3NaI}}\,\]
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