An organic compound " x " where molar ratio of C , O and H are equal, on treatment with $50 \% \mathrm{KOH}$ under reflux followed by acidification produced " y ". The most likely structure of " y " is: [Molar mass of ' x ' is $58 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

(B) $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}$

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

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}

]

(A)(see structure)
(B) $CH_{3} - CH = CH - CH = O$
(C)(see structure)
✓ Correct
(D)(see structure)

Answer: (C)

Solution

Equal molar amounts of C, O and H give the repeating unit CHO of formula mass

29

, so a molar mass of

58

corresponds to

C X_{2} H X_{2} O X_{2}

, which is glyoxal,

OHC - CHO

. Lacking any

α

-hydrogen, glyoxal undergoes an intramolecular Cannizzaro reaction with concentrated KOH, one aldehyde being oxidised while the other is reduced; acidification then frees the acid. The product y is glycolic acid,

HOCH X_{2} COOH

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

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