See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is p-cymene (4-isopropyltoluene), which has a methyl group (-CH3) at C1 and an isopropyl group (-CH(CH3)2) at C4 on a benzene ring. Step 2 - Reaction P: Cl2/hv (free radical halogenation). Under free radical conditions, chlorination occurs preferentially at the most stable radical intermediate. The isopropyl group has a tertiary C-H bond (the CH of -CH(CH3)2), which forms a more stable tertiary radical than the primary C-H bonds of the methyl group. Therefore, the major product P is chlorination at the tertiary carbon of the isopropyl group, giving a benzene ring with methyl at C1 and -C(CH3)2Cl at C4 (tertiary chloride). Step 3 - Reaction Q: Cl2/AlCl3 (electrophilic aromatic substitution, Friedel-Crafts type chlorination). Both methyl and isopropyl groups are ortho/para directors. The ring positions activated by both groups are considered. The methyl group directs ortho and para; the isopropyl group also directs ortho and para. The position ortho to the methyl group (C2) is activated by the methyl group and also relatively accessible. The major product of EAS on p-cymene with Cl2/AlCl3 is chlorination ortho to the methyl group (at C2), giving a ring with methyl at C1, Cl at C2, and isopropyl at C4. This is because the C2 position is ortho to methyl and also activated by the para-isopropyl group's ortho/para directing effect relative to C4 placing activation at C1, C3 — but the dominant directing effect from methyl gives ortho (C2) as the major ring-substituted product. Step 4 - Match to options: Option (c) shows P = tertiary chloride on isopropyl side chain (C(CH3)2Cl) and Q = ring chlorination ortho to methyl with intact isopropyl group. This matches the analysis. Step 5 - Why other options fail: - Option (a): P shows CH2Cl (primary benzylic chloride from methyl group), which is less favored than tertiary radical chlorination. - Option (b): Q shows a secondary chloride on the isopropyl in addition to ring Cl, which is not the expected EAS product. - Option (d): Q shows tertiary chloride on isopropyl AND ring Cl, meaning both radical and EAS occurred in one step, which is not the expected single major product of EAS. Therefore, the correct answer is C.