See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Find x - isomers of C8H10 that give ONLY aromatic dicarboxylic acid with hot alkaline KMnO4. C8H10 has degree of unsaturation = (2×8 + 2 - 10)/2 = 4, indicating an aromatic ring. These are alkylbenzenes. Possible structural isomers of C8H10: 1. Ethylbenzene (C6H5-CH2CH3) 2. 1,2-dimethylbenzene (o-xylene) 3. 1,3-dimethylbenzene (m-xylene) 4. 1,4-dimethylbenzene (p-xylene) When treated with hot alkaline KMnO4, alkyl groups on benzene are oxidized to -COOH groups. For ONLY a dicarboxylic acid product, the benzene ring must have exactly TWO alkyl substituents (giving two -COOH groups = phthalic acid derivatives), OR one substituent that gives two carboxylic acids is not possible from a single group. Actually, each alkyl group (regardless of length) is oxidized to one -COOH group. So: - Ethylbenzene: one alkyl group → monocarboxylic acid (benzoic acid) — NOT dicarboxylic - o-xylene: two methyl groups → phthalic acid (dicarboxylic) ✓ - m-xylene: two methyl groups → isophthalic acid (dicarboxylic) ✓ - p-xylene: two methyl groups → terephthalic acid (dicarboxylic) ✓ So x = 3 isomers of C8H10 give only aromatic dicarboxylic acid. Step 2: Find y - isomers of C4H8 that give CO2 with hot alkaline KMnO4. C4H8 has degree of unsaturation = (2×4 + 2 - 8)/2 = 1, so one degree: either one C=C (alkene) or one ring (cycloalkane). Hot alkaline KMnO4 cleaves C=C double bonds. CO2 is produced when a terminal =CH2 group is oxidized (i.e., RCH=CH2 gives RCOOH + CO2, and R2C=CH2 gives R2C=O (ketone) + CO2). Isomers of C4H8: Alkenes: 1. 1-butene: CH3CH2CH=CH2 → CH3CH2COOH + CO2 ✓ (terminal =CH2) 2. 2-butene (cis): CH3CH=CHCH3 → 2 CH3COOH (no CO2) ✗ 3. 2-butene (trans): same oxidation products as cis-2-butene → 2 CH3COOH ✗ 4. Isobutylene (2-methylpropene): (CH3)2C=CH2 → acetone + CO2 ✓ (terminal =CH2) Cycloalkanes: 5. Cyclobutane: ring, no C=C, hot KMnO4 would not easily cleave to give CO2 in standard conditions — cycloalkanes are generally resistant. ✗ 6. Methylcyclopropane: cyclopropane ring with methyl — cyclopropane can be oxidized by hot KMnO4 but does not give CO2 directly in a straightforward manner. ✗ Wait, let me reconsider. Cis and trans 2-butene are stereoisomers — the question asks for structural isomers typically, but they are counted as separate isomers in some contexts. However, for C4H8 structural isomers with C=C: - 1-butene ✓ - 2-butene (cis and trans are geometric isomers, counted as 2 isomers) — neither gives CO2 ✗ - 2-methylpropene ✓ So y = 2 (1-butene and 2-methylpropene give CO2). Step 3: x + y = 3 + 2 = 5. Therefore, the correct answer is 5.