4-Methylacetophenone, CH3-C6H4-CO-CH3 (para), is treated with I2/NaOH and then H3O+. The acid obtained is then heated with alkaline KMnO4 followed by H3O+. The final product is

benzene-1,4-dicarboxylic acid (terephthalic acid)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ro141

4-Methylacetophenone, $CH X_{3} - C X_{6} H X_{4} - CO - CH X_{3}$ (para), is treated with $I X_{2}$ / $NaOH$ and then $H X_{3} O X^{+}$ . The acid obtained is then heated with alkaline $KMnO X_{4}$ followed by $H X_{3} O X^{+}$ . The final product is

The haloform step converts the

COCH X_{3}

group into

COOH

and leaves the ring methyl alone, giving 4-methylbenzoic acid. Hot alkaline

KMnO X_{4}

then oxidises the ring

CH X_{3}

(which has benzylic hydrogens) to

COOH

. The final product is benzene-1,4-dicarboxylic acid.