See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Salicylic acid is 2-hydroxybenzoic acid. The hydroxyl (-OH) group is a powerful ortho/para director and strongly activates the ring toward electrophilic aromatic substitution. The -COOH group is a deactivator (meta director). When salicylic acid reacts with excess bromine water, two things happen: (1) the -OH group directs bromine to the ortho and para positions relative to itself, and (2) under the conditions of excess bromine water (aqueous, relatively mild), the -COOH group undergoes decarboxylation because the phenol formed after tribromination is highly activated and the carboxylate is lost (similar to how 2,4,6-tribromophenol precipitates in the bromine water test for phenol). Step 1 - Identify the directing effects: In salicylic acid (OH at C1, COOH at C2 by conventional phenol numbering, or OH at C2, COOH at C1 by carboxylic acid numbering), the -OH group strongly activates the ring and directs to ortho and para positions relative to itself. Step 2 - Reaction with excess bromine water: Bromine water (aqueous bromine) is used, which is the classic reagent for phenols. The -OH group directs Br to positions ortho and para to it. Since -COOH already occupies one ortho position, bromination occurs at the remaining ortho position and the para position relative to -OH. With excess bromine, all available activated positions get brominated. Step 3 - Decarboxylation: With excess bromine water, the intermediate tribrominated salicylic acid undergoes decarboxylation (loss of CO2) because the highly electron-rich tribromo phenol ring facilitates this process. This is well-known: bromine water with salicylic acid gives 2,4,6-tribromophenol as the final product. Step 4 - Product identification: 2,4,6-tribromophenol has OH at C1, Br at C2, Br at C4, Br at C6. This matches option (c), which shows a benzene ring with OH at one position and Br at both ortho positions and the para position (three Br atoms total), with no COOH group (decarboxylation has occurred). Why other options fail: - Option (a): Only one Br, incomplete reaction with excess bromine, and COOH retained - incorrect for excess bromine water conditions. - Option (b): Two Br atoms with COOH retained - still incomplete bromination and no decarboxylation. - Option (d): Three Br atoms but COOH retained and wrong positions - decarboxylation does not occur here, and the regiochemistry shown does not match expected ortho/para directing of -OH. Option (c) correctly represents 2,4,6-tribromophenol, the product of electrophilic aromatic substitution at all three positions activated by -OH, followed by decarboxylation. Therefore, the correct answer is C.