See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Hot aqueous KMnO4 oxidizes alkyl side chains on a benzene ring to carboxylic acids (-COOH), provided there is at least one benzylic hydrogen. A tert-butyl group (no benzylic H) is resistant to oxidation. The product ortho-benzenedicarboxylic acid (phthalic acid) requires two carboxylic acid groups at the 1,2 (ortho) positions of the benzene ring. Step 1 – Identify what structure gives ortho-dicarboxylic acid: We need a compound that, after oxidation of its side chains or aliphatic portion, yields a benzene ring bearing -COOH groups at the 1 and 2 positions. Step 2 – Analyze option (b): Option (b) is 1,2-dihydronaphthalene (more precisely, tetralin or a partially saturated naphthalene analog). The structure shows a benzene ring fused to a six-membered ring that contains a double bond and CH2 groups. When treated with hot aqueous KMnO4, the non-aromatic ring is cleaved/oxidized. The two benzylic carbons (C1 and C2 of the saturated ring, which are attached to the aromatic ring) get oxidized to -COOH groups. Because these two carbons are at the 1,2 positions relative to each other on the benzene ring, the product is ortho-benzenedicarboxylic acid (phthalic acid). This is the correct answer. Step 3 – Why other options fail: (a) ortho-tert-butyl toluene: The tert-butyl group has no benzylic hydrogen and cannot be oxidized by KMnO4 to -COOH. Only the methyl group would be oxidized to give a mono-acid (ortho-toluic acid derivative with one -COOH), not a dicarboxylic acid. (c) 1,4-diethylbenzene: Both ethyl groups are at para positions (1,4). Oxidation would give 1,4-benzenedicarboxylic acid (terephthalic acid), which is the para isomer, not the ortho isomer. (d) 1-methyl-3-propylbenzene: The substituents are at meta positions (1,3). Oxidation would give 1,3-benzenedicarboxylic acid (isophthalic acid), the meta isomer, not the ortho isomer. Therefore, the correct answer is B.