An organic compound “x” where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced “y”. The most likely structure of “y” is: [Molar mass of ‘x’ is 58 g mol-1]
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- CH3-CH=CH-CH=O
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Answer
(3) [figure]
Solution
C : H : O = 1 : 1 : 1 ⇒ empirical formula CHO (formula mass 29).
n = 5829 = 2 ⇒ molecular formula C2H2O2 = OHC–CHO (glyoxal).
Glyoxal has no α-hydrogen ⇒ with concentrated (50%) KOH it undergoes Cannizzaro reaction; one CHO is oxidised and the other reduced in the same molecule (intramolecular):
OHC-CHO 50\% KOH, HO-CH2-COO K H3O HO-CH2-COOH
∴ y = glycolic acid (2-hydroxyethanoic acid).
Answer: (3)