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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Aldehydes and Ketones

JEE Main 2026 · 2 April · Evening Shift · Chemistry · Aldehydes and Ketones

An organic compound " x " where molar ratio of C, O and H are...

An organic compound "xx" where molar ratio of C, O and H are equal, on treatment with 50%50\% KOH under reflux followed by acidification produced "yy". The most likely structure of "yy" is:
[Molar mass of 'xx' is 5858 g mol−1^{-1}]

Options

  1. ACH2=CH−C∥O−OHCH_2 = CH - \overset{O}{\underset{\|}{C}} - OH
  2. BCH3−CH=CH−CH=OCH_3 - CH = CH - CH = O
  3. CO=C−OH, CH2−OHO = C - OH, \, CH_2 - OH (with C-C bond)
  4. DCH3−C∥O−OHCH_3 - \overset{O}{\underset{\|}{C}} - OH

Correct answer

C. O=C−OH, CH2−OHO = C - OH, \, CH_2 - OH (with C-C bond)

Step-by-step solution

Given the molar ratio of C, O, and H in compound xx is equal, the empirical formula is CHOCHO.

Empirical mass =12+1+16=29= 12 + 1 + 16 = 29 g mol−1^{-1}.

Since the molar mass of xx is 5858 g mol−1^{-1}, the molecular formula is (CHO)2(CHO)_2 or C2H2O2C_2H_2O_2.

The structure of xx is glyoxal, OHC−CHOOHC-CHO.

Treatment of glyoxal with 50%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% KOHHOCH2−COO−K+OHC-CHO \xrightarrow{50\% \text{ KOH}} HOCH_2-COO^{-}K^{+}

Subsequent acidification yields glycolic acid (yy).

HOCH2−COO−K+→H+HOCH2−COOHHOCH_2-COO^{-}K^{+} \xrightarrow{H^{+}} HOCH_2-COOH

The structure of yy is HO−CH2−COOHHO-CH_2-COOH.

Answer: O=C−OH, CH2−OHO = C - OH, \, CH_2 - OH (with C-C bond)
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