Question
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 ]
JEE Main
Chemistry
Aldehydes and Ketones
2026
JEE Main 2026 (02 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. CH_2 = CH - O \| C - OH
- B. CH_3 - CH = CH - CH = O
- C. O = C - OH, \, CH_2 - OH (with C-C bond)
- D. CH_3 - O \| C - OH
Answer
C. O = C - OH, \, CH_2 - OH (with C-C bond)
Step-by-step solution
Given the molar ratio of C, O, and H in compound x is equal, the empirical formula is CHO. Empirical mass = 12 + 1 + 16 = 29 g mol^ -1 . Since the molar mass of x is 58 g mol^ -1 , the molecular formula is (CHO)_2 or C_2H_2O_2. The structure of x is glyoxal, OHC-CHO. Treatment of glyoxal with 50\% KOH under reflux results in an intramolecular Cannizzaro reaction. One aldehyde group is oxidized to a carboxylate ion, and the other is reduced to a primary alcohol. OHC-CHO 50\% KOH HOCH_2-COO^ - K^ + Subsequent acidification yields glycolic acid (y). HOCH_2-COO^ - K^ + H^ + HOCH_2-COOH The structure of y is HO-CH_2-COOH. Answer: O = C - OH, \, CH_2 - OH (with C-C bond)
Related: Chemistry — Aldehydes and Ketones · All PYQ Banks