A) \[KI\]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\[{{I}_{2}}/KOH,\]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\[C{{l}_{3}}CHO\] 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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