A) Kl 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, ie, \[I{{O}^{-}}\] ion. Hypoiodide ion first oxidses\[C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{{}}C{{H}_{3}}CHO\]and then brings about iodination of \[C{{H}_{3}}CHO\] to \[{{I}_{3}}C\,.\,\,CHO\]. Alkaline hydrolysis of \[C{{I}_{3}}CHO\] then gives \[CH{{I}_{3}}\]. The other three reagents do not contain any oxidising species and hence, fail to give iodoform test.
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