Chemistry Q66 – JEE-MAIN 2026

Home Chemistry Basic Organic ChemistryQuestion #101

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}

]

Clue 1 of 3: Determine the molecular formula of compound X

Use the given molar ratio of C, O, and H, and the molar mass to find the empirical and then molecular formula.

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Step 1: Determine the molecular formula and structure of X✦ Active

The molar ratio of C, O, and H in compound X is equal. This means the empirical formula is $C_{x} H_{x} O_{x}$ . The empirical formula mass is $12 x + 16 x + 1 x = 29 x$ . Given that the molar mass of X is 58 g mol $^{- 1}$ , we can set up the equation $29 x = 58$ , which gives $x = 2$ . Therefore, the molecular formula of X is $C_{2} H_{2} O_{2}$ . The most likely stable organic compound with this formula that undergoes the described reaction is glyoxal ( $O H C - C H O$ ). Glyoxal has two aldehyde groups and no alpha-hydrogens.

Step 2: Identify the reaction type and product Y○ Expand

Glyoxal ( $O H C - C H O$ ) is an aldehyde with no alpha-hydrogens. Treatment with 50% KOH under reflux followed by acidification is characteristic of a Cannizzaro reaction. In the Cannizzaro reaction, aldehydes without alpha-hydrogens undergo disproportionation. For glyoxal, this is an intramolecular reaction where one aldehyde group is oxidized to a carboxylate, and the other is reduced to an alcohol.

OHC-CHO \overset{50 % K O H}{\to} {HOCH}_{2} - {COO}^{-} K^{+} (Potassium glycolate)

Subsequent acidification yields the carboxylic acid product Y:

{HOCH}_{2} - {COO}^{-} K^{+} \overset{H^{+}}{\to} {HOCH}_{2} - COOH (Glycolic acid)

💡 Teacher's Secret Hint

Remember that Cannizzaro reaction involves both oxidation and reduction of aldehyde groups.

Step 3: Match the product Y with the given options○ Expand

The product Y is glycolic acid, with the structure $HOOC - {CH}_{2} - OH$ . Comparing this with the given options:

Option 1: ${CH}_{2} = CH - COOH$ (Acrylic acid)

Option 2: ${CH}_{3} - CH = CH - CHO$ (Crotonaldehyde)

Option 3: $HOOC - {CH}_{2} - OH$ (Glycolic acid)

Option 4: ${CH}_{3} - COOH$ (Acetic acid)

Option 3 matches the structure of glycolic acid.

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Chemistry Question 66 – JEE-MAIN 2026

Clue 1 of 3: Determine the molecular formula of compound X

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