An organic compound (A) has molecular formula `(C_(5)H_(10)O)`. It does not reduce Tollen's reagent but forms an orange precipitate with 2, 4-DNP reagent. It forms a carboxylic acid (B) with molecular formula `(C_(3)H_(6)O_(2))` when treated with alkaline `KMnO_(4)`, yellow precipitate on treatment with NaOH and `I_(2)` under vigorous conditions. On oxidation it gives ethanoic acid and propanoic acid. Sodium slat of (B) gave a hydrocarbon (C ) in Kolbe's Electrolytic Reduction. Identify (A), (B) and (C ) and write the reactions involved.

Explanation

The compound (A) is `CH_(3)CH_(2)COCH_(2)CH_(3)`. The reactions are explained as under:

`underset((A))(CH_(3)CH_(2)COCH_(2)CH_(3)) overset([O])to underset("Propanoic acid")(CH_(3)CH_(2)COOH)+underset("Ethanoic acid")(CH_(3)COOH)`
`underset("Sodium propionate")(CH_(3)CH_(2)COONa) underset("Electrolytic reduction")overset("Kolbe.s")to underset((C ))underset("n-Butane")(CH_(3)CH_(2)CH_(2)CH_(3))`
Thus A, B, C are
`A=CH_(3)CH_(2)COCH_(2)CH_(3)" "B=CH_(3)CH_(2)COOH" "C=CH_(3)CH_(2)CH_(2)CH_(3)`