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QJEEM23Chace1: JEE Main 2023 Chemistry - Class 12 JEEM23Chace1 Solution: A trisubstituted compound ' ', gives neutral test positive.…

A trisubstituted compound ' A\mathrm{A} ', C10H12O2\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2} gives neutral FeCl3\mathrm{FeCl}_{3} test positive. Treatment of compound 'A' with NaOH\mathrm{NaOH} and CH3Br\mathrm{CH}_{3} \mathrm{Br} gives C11H14O2\mathrm{C}_{11} \mathrm{H}_{14} \mathrm{O}_{2} , with hydroiodic acid gives methyl iodide and with hot conc. NaOH\mathrm{NaOH} gives a compound B,C10H12O2\mathrm{B}, \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2} . Compound 'A' also decolorises alkaline KMnO4\mathrm{KMnO}_{4} . The number of π\pi bond/s present in the compound ' A\mathrm{A} ' is _____________.

Step-by-Step Verified Solution

A:C10H12O2\mathrm{A:C_{10}H_{12}O_2}

DU of A =22−122=5=\frac{22-12}{2}=5

1 DU is due to Ring (Benzene ring)

4 π\pi -bonds will be there

(3 π\pi -bonds in ring and 1 π\pi -bond outside ring) as it decolorises alkaline KMnO 4_4 .

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