Q. In a set of reactions, ethyl benzene yielded a product D: C₆H₅CH₂CH₃ —(KMnO₄, KOH)→ B —(Br₂, FeCl₃)→ C —(C₂H₅OH, H⁺)→ D 'D' would be
A) Ethyl 3-bromobenzoate (m-Br–C₆H₄–COOC₂H₅)
B) C₆H₅–CH₂–CH(Br)–COOC₂H₅
C) Ethyl (2,4-dibromophenyl)acetate (2,4-Br₂C₆H₃–CH₂COOC₂H₅)
D) 3-Ethoxybenzoic acid (m-CH₃CH₂O–C₆H₄–COOH)
Explanation
[Solution from source] Alkaline KMnO₄ oxidises the whole ethyl side chain to –COOH, so B is benzoic acid. –COOH is meta-directing, so Br₂/FeCl₃ gives m-bromobenzoic acid as C. Esterification with ethanol/H⁺ then gives D, ethyl m-bromobenzoate. Products that keep a CH₂ side chain ignore the side-chain oxidation. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.