A trisubstituted compound 'A', C₁₀ H₁₂ O₂ gives neutral FeCl₃ test positive. Treatment of compound 'A' with NaOH and CH₃ Br gives C₁₁ H₁₄ O₂, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B, C₁₀ H₁₂ O₂. Compound 'A' also decolorises alkaline KMnO₄. The number of π bond/s present in the compound 'A' is _____________.

Solution

Chemistry ma’am · 3 steps

Every clue points to eugenol: a phenol, a methoxy group and an allyl chain on one benzene ring.

1. Unsaturation first. For C₁₀H₁₂O₂ the degree of unsaturation is (2(10) + 2 - 12)/2 = 5. A benzene ring uses 4 of these (1 ring and 3 C=C), so 1 more double bond sits outside the ring.

   Clue	What it tells you
   Neutral FeCl₃ test positive	A free phenolic –OH on the ring
   NaOH and CH₃Br give C₁₁H₁₄O₂	One CH₂ is added, so exactly one phenolic –OH gets methylated
   HI gives CH₃I	A methoxy group, -OCH₃, is already present
   Decolorises alkaline KMnO₄	A C=C in the side chain
   Hot conc. NaOH gives isomer B	The allyl chain -CH₂CH=CH₂ shifts to propenyl -CH=CHCH₃, same formula

2. Identify A. A trisubstituted benzene with –OH, -OCH₃ and an allyl group is eugenol (4-allyl-2-methoxyphenol). Check: C₆H₃ + OH + OCH₃ + C₃H₅ gives C₁₀H₁₂O₂. B is isoeugenol.
3. So. Eugenol has 3 π bonds in the benzene ring and 1 in the allyl C=C, so A has 4 π bonds.

The common slip: 5. Reporting the degree of unsaturation as the π bond count gives 5, but one of those 5 is the benzene ring itself, not a π bond.

The idea to keep. Degree of unsaturation counts rings and π bonds together. Subtract the rings before you report the number of π bonds.

Answer from the NTA final answer key. Solution by PrepWiser.