See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify functional group from solubility data: Compound A (C7H8O) is insoluble in water, dilute HCl, and aqueous NaHCO3, but dissolves in dilute NaOH. This behavior is characteristic of a phenol: phenols are weak acids (pKa ~10) that do not react with NaHCO3 (too weak) but do dissolve in NaOH by forming phenoxide salts. Option (a), anisole (methoxybenzene), has no acidic OH and would NOT dissolve in NaOH, so it is eliminated. Options (b), (c), and (d) are all cresols (methylphenols), all with formula C7H8O, all phenols. Step 2 - Use bromination data to distinguish among cresols: When A reacts rapidly with Br2 water to give C7H5OBr3, three bromine atoms are introduced. In phenols, electrophilic aromatic bromination with Br2/water occurs at positions ortho and para to the OH group (strongly activating, ortho/para director). The number of positions available for bromination depends on the substitution pattern. Step 3 - Analyze each cresol: - o-Cresol (2-methylphenol, option b): OH directs to ortho (C3, C5) and para (C5 or C1 blocked, actually para to OH is C4, and C6 is ortho to OH but adjacent to CH3). Available positions ortho/para to OH: C3, C4, C6. However C2 has CH3, so three positions are available. But C3, C4, and C6 are the reactive sites — tribromo product is possible but the methyl group at C2 blocks one ortho position reducing reactivity slightly. - m-Cresol (3-methylphenol, option c): OH is at C1, CH3 is at C3. Positions ortho to OH: C2 and C6; para to OH: C4. The CH3 at C3 also activates C2 and C4 (ortho/para to CH3). Thus C2 is activated by both OH (ortho) and CH3 (ortho); C4 is activated by both OH (para) and CH3 (ortho); C6 is activated by OH (ortho). All three positions (C2, C4, C6) are activated and unhindered, making tribromination rapid and straightforward to give 2,4,6-tribromo-3-methylphenol = C7H5OBr3. - p-Cresol (4-methylphenol, option d): OH at C1, CH3 at C4 (para to each other). Ortho to OH: C2 and C6; para to OH: C4 (blocked by CH3). So only two unblocked reactive positions (C2 and C6) are available for bromination directed by OH, giving a dibromo product, not tribromo. Therefore p-cresol cannot give C7H5OBr3. Step 4 - Confirm option (c): m-Cresol has three open, activated positions (C2, C4, C6) for bromination, consistent with rapid formation of a tribrominated product C7H5OBr3 (tribromo-m-cresol). o-Cresol could also give three bromines but the CH3 at C2 sterically hinders one ortho position of OH, making complete tribromination slower and less clean. m-Cresol is the classic textbook example where mutual activation by both OH and CH3 at three positions leads to rapid tribromination. Step 5 - Eliminate remaining options: (a) eliminated by NaOH solubility test. (b) o-cresol has one position blocked by CH3, making full tribromination less favorable. (d) p-cresol has para position blocked by CH3, only two positions available, giving dibromo product. (c) m-cresol gives tribromo product rapidly. Therefore, the correct answer is C.