Question:

Compound A(C9H10O) shows positive iodoform test. Oxidation of A with KMnO4/KOH given acid B(C8H6O4). Anhydride of B is used for the preparation of phenolphthalein. Compound A is:

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The acid whose anhydride is used for phenolphthalein is phthalic acid (ortho-dicarboxylic acid), so A must be an ortho-disubstituted benzene with a CH3CO- group for the iodoform test.
Updated On: Nov 5, 2026
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The Correct Option is D

Solution and Explanation

Phenolphthalein is made from phthalic anhydride, so B = phthalic acid (benzene-1,2-dicarboxylic acid), C8H6O4.
KMnO4/KOH oxidises every alkyl side chain on the ring to -COOH, so A has two side chains ortho to each other, containing 3 extra carbons in total (C9 - C6).
A gives a positive iodoform test, so it has a CH3CO- (methyl ketone) group: A = 2-methylacetophenone (o-CH3-C6H4-COCH3), C9H10O.
Check of the options: (1) para isomer gives terephthalic acid (no anhydride); (2) o-tolylacetaldehyde is an aldehyde with no CH3CO- group (no iodoform); (3) meta isomer gives isophthalic acid.
Answer: (4) 2-methylacetophenone (ortho CH3 and COCH3)
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