JEE Main 2026 · 2 Apr · Shift 2 · Chemistry Q66

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ChemistrySingle correct+4 / -1Question 66 of 75

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 $^{-1}$ ]

\ce{CH2=CH-COOH}

\ce{CH3-CH=CH-CH=O}

\ce{HOOC-CH2-OH}

\ce{CH3-COOH}

Answer and worked solution

Official answer: C

Identify x. Equal moles of C, O and H give the formula $(\ce{CHO})_n$ ; $29n=58$ means $\ce{C2H2O2}$ , glyoxal $\ce{OHC-CHO}$ .

Reaction. Concentrated KOH causes an intramolecular Cannizzaro reaction: one CHO is oxidised and the other reduced, giving $\ce{HOCH2COO-}$ . Acidification gives glycolic acid, $\ce{HOCH2COOH}$ .

(C)

Topics:Named reactions Empirical and molecular formula

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JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 66