In the following sequence of reactions, \(\mathrm{C_3H_6 \xrightarrow{H^+/H_2O} A \xrightarrow[\text{dil KOH}]{KIO} B + C}\)
The compounds B and C respectively are :-
The compounds B and C respectively are :-
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- \(\mathrm{CI_3COOK}\), \(\mathrm{HCOOH}\)
- \(\mathrm{CI_3COOK}\), \(\mathrm{CH_3I}\)
- \(\mathrm{CH_3I}\), \(\mathrm{HCOOK}\)
- \(\mathrm{CHI_3}\), \(\mathrm{CH_3COOK}\)
The Correct Option is D
Solution and Explanation
KIO (hypoiodite, from I\(_2\)/KOH) first oxidises the secondary alcohol to acetone \(\mathrm{CH_3COCH_3}\).
Acetone has a \(\mathrm{CH_3CO}\) group, so it undergoes the iodoform reaction: \(\mathrm{CH_3COCH_3 + 3KIO \to CI_3COCH_3 \to CHI_3 + CH_3COOK}\).
Hence B = \(\mathrm{CHI_3}\) (iodoform) and C = \(\mathrm{CH_3COOK}\).
Answer: (4) \(\mathrm{CHI_3}\), \(\mathrm{CH_3COOK}\)
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