Determining the Degree of Unsaturation and $π$ Bonds in an Unknown Aromatic Compound

A trisubstituted compound ' A ', C_10H_12O_2 gives neutral FeCl_3 test positive. Treatment of compound 'A' with NaOH and CH_3Br gives C_11H_14O_2, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B, C_10H_12O_2. Compound 'A' also decolorises alkaline KMnO_4. The number of π bond/s present in the compound ' A ' is

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Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why is the number of pi bonds 4 if the degree of unsaturation is 5?

The degree of unsaturation (DBE) accounts for both rings and pi bonds (DBE = rings + pi bonds). The benzene ring provides 1 ring and 3 pi bonds. Together with 1 alkene pi bond, there are 4 pi bonds and 1 ring, totaling DBE = 5.

Answer

Accepted answer: 4

Compound 'A' has a degree of unsaturation of 5, comprising 1 benzene ring (3 $π$ bonds, 1 ring) and 1 alkene double bond (1 $π$ bond), giving a total of 4 $π$ bonds.

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Step-by-step solution

Given: Compound 'A' has the molecular formula $C_{10} H_{12} O_{2}$ . It gives a positive neutral ${F e C l}_{3}$ test, indicating a phenolic $- O H$ group, forms ${C H}_{3} I$ with $H I$ (indicating a methoxy group $- {O C H}_{3}$ ), and decolorises alkaline ${K M n O}_{4}$ (indicating an aliphatic double bond).
Strategise: Calculate the Degree of Unsaturation (DBE) of compound 'A' from its molecular formula $C_{10} H_{12} O_{2}$ using the formula $D B E = C + 1 - \frac{H}{2}$ . Identify the rings and $π$ bonds contributing to the DBE.
Execute: For $C_{10} H_{12} O_{2}$ : $D B E = 10 + 1 - \frac{12}{2} = 11 - 6 = 5$ The positive neutral ${F e C l}_{3}$ test indicates a benzene ring, which accounts for 1 ring and 3 $π$ bonds (total DBE = 4). Since the total DBE is 5, the remaining unsaturation is $D B E - 4 = 1$ , which corresponds to 1 aliphatic carbon-carbon double bond (confirmed by decolorisation of alkaline $H I$ ). Total number of $π$ bonds = 3 (from benzene ring) + 1 (from alkene) = 4.
Verify: Total unsaturation is 5 (1 ring + 4 $π$ bonds). A benzene ring has 3 $π$ bonds and 1 ring. The remaining unsaturation is 1 alkene $π$ bond. Thus, total $π$ bonds = 4.

Determining the Degree of Unsaturation and $π$ Bonds in an Unknown Aromatic Compound

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Compound 'A' has a degree of unsaturation of 5, comprising 1 benzene ring (3 $π$ bonds, 1 ring) and 1 alkene double bond (1 $π$ bond), giving a total of 4 $π$ bonds.

Why each option works or fails

Solution

01Given

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