JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

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 $^{- 1}$ ]

Solution

With equal C, O and H atoms and molar mass $58$ , let the formula be C $_{n}$ H $_{n}$ O $_{n}$ . For $n = 2$ : C $_{2}$ H $_{2}$ O $_{2}$ has mass $58$ . $✓$ This is glyoxal (OHC $-$ CHO).

Glyoxal lacks $α$ -hydrogens, so under concentrated KOH it undergoes intramolecular Cannizzaro disproportionation: one CHO is oxidised to COO $^{-}$ and the other is reduced to CH $_{2}$ OH, giving the salt of glycolic acid.

Acidification yields HOCH $_{2}$ $-$ COOH (glycolic acid) — option (3).

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