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

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Alcohols Phenols and Ethers

A trisubstituted compound 'A',

C_{10} H_{12} O_{2}

gives neutral

FeCl_{3}

test positive. Treatment of compound 'A' with NaOH and

CH_{3} Br

gives

C_{11} H_{14} O_{2}

, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B,

C_{10} H_{12} O_{2}

. Compound 'A' also decolorises alkaline

KMnO_{4}

. The number of

π

bond/s present in the compound 'A' is ______.

Answer: 4

Solution

The formula

C_{10} H_{12} O_{2}

carries five degrees of unsaturation. The positive neutral ferric chloride test shows a phenolic hydroxyl; methylation by sodium hydroxide with methyl bromide, adding one

C H_{2}

to give

C_{11} H_{14} O_{2}

, confirms that free phenol, while the release of methyl iodide with hydroiodic acid shows a methoxy group already present. Decolourisation of alkaline potassium permanganate marks a carbon-carbon double bond in the side chain. The benzene ring accounts for four degrees of unsaturation, one ring and three

π

bonds, and the side-chain double bond supplies the fifth. Counting only the

π

bonds, three lie in the aromatic ring and one in the side chain, giving four

π

bonds in compound A.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Alcohols Phenols and Ethers

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