A trisubstituted compound 'A' C10H12O2 gives neutral FeCl3 test positive. Treatment of compound 'A' with NaOH and CH3Br gives C11H14O2, with hydroiodic acid gives methyl iodide and with hot conc. NaOH

Question

A trisubstituted compound 'A' C10H12O2 gives neutral FeCl3 test positive. Treatment of compound 'A' with NaOH and CH3Br gives C11H14O2, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B, C10H12O2, Compound 'A' also decolorises alkaline KMnO4. The number of bond/s present in the compound 'A’ is _______ .

Prashant Kotian · Accepted Answer

Correct answer: .. C10H12O2, DU = 5 (1-phenyl ring present with unsaturation in side alkyl chain) Given information: neutral FeCl3 test positive  Phenol present Compound ‘A’ also decolorises alkaline KMnO4  unsaturation in side alkyl chain Number of -bond/s present in the compound 'A’ is 4. | JEE(Main) 2023 | DATE : 30-01-2023 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 8

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