A) \[CaC{{O}_{3}}\]
B) \[NaOH\]
C) \[KOH\]
D) \[MgC{{O}_{3}}\]
Correct Answer: B
Solution :
\[\overset{O}{\mathop{\overset{||}{\mathop{C{{H}_{3}}-C-C{{H}_{3}}}}\,}}\,+3{{I}_{2}}+4NaOH\xrightarrow{{}}\] \[\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,+\underset{sod.\,\,acetate}{\mathop{C{{H}_{3}}\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-ONa}}\,+\underset{sodium\,iodide}{\mathop{3NaI+3{{H}_{2}}O}}\,\] Therefore, acetone and iodine react in presence of \[NaOH\].
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