A) \[\text{KI}\] and \[aq\,KOH\]
B) \[{{I}_{2}}\] and \[aq\,KOH\]
C) \[{{I}_{2}}/aq\,KI\]
D) \[HI\] and \[HI{{O}_{3}}\]
Correct Answer: B
Solution :
For iodoform reaction, we need an oxidizing agent which is provided by only \[{{I}_{2}}/KOH,\] i.e., \[\text{I}{{\text{O}}^{-}}\]ion. Hypoiodide ion first oxidses \[C{{H}_{3}}C{{H}_{2}}\xrightarrow{{}}C{{H}_{3}}CHO\] and then brings about iodination of \[\text{C}{{\text{H}}_{\text{3}}}\text{CHO}\]to \[{{\text{I}}_{\text{3}}}\text{C}\text{.CHO}\text{.}\]Alkaline hydrolysis of \[\text{C}{{\text{I}}_{\text{3}}}\text{CHO}\]then gives \[\text{CH}{{\text{I}}_{3}}.\] The other three reagents do not contain any oxidising species and hence, fail to give iodoform test.
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