Compound A $\left(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}\right)$ shows positive iodoform test. Oxidation of A with $\mathrm{KMnO}_{4} / \mathrm{KOH}$ gives acid $\mathrm{B}\left(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}\right)$. Anhydride of B is used for the preparation of phenolphthalein. Compound A is:-

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Compound A

(C_{9} H_{10} O)

shows positive iodoform test. Oxidation of A with

KMnO_{4} / KOH

gives acid

B (C_{8} H_{6} O_{4})

. Anhydride of B is used for the preparation of phenolphthalein. Compound A is:-

Answer: (A)

The positive iodoform test marks

A

as a methyl ketone carrying a

CH X_{3} CO X^{-}

group. Oxidising

A

with

KMnO X_{4} / KOH

yields the acid

B

of formula

C X_{8} H X_{6} O X_{4}

, which is phthalic acid - its anhydride is the reagent used to make phenolphthalein, which pins the two substituents in the ortho position. Since

A

C X_{9} H X_{10} O

and oxidation of both side chains leaves a ring bearing two carboxyls,

A

must be ortho-methylacetophenone, with a

COCH X_{3}

group and a

CH X_{3}

group on adjacent ring carbons.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option