See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify each group and its electronic nature with respect to a phenyl ring. (a) -CH=CH-CO2H (cinnamoyl/acrylate-type group): The conjugated system with the electron-withdrawing carboxylic acid makes the overall group electron-withdrawing (deactivating) and a meta-director. However, looking at the answer key (a → p, s), this group is matched to o/p-director AND deactivating. The vinyl linkage (-CH=CH-) can donate electrons by resonance into the ring at ortho/para positions while the CO2H end withdraws. The net directing effect from the ring's perspective: the vinyl group attached to the ring is an o/p-director (via conjugation/hyperconjugation), but the overall group is deactivating due to the carboxylic acid pulling electrons away. So: o/p-directing (p) and deactivating (s). (b) -O-S(=O)-CH3 (methanesulfonate ester, -OSO-CH3): The oxygen directly attached to the ring is an o/p-director because oxygen lone pairs donate into the ring by resonance, making it an activating group. The sulfonate portion withdraws, but the lone-pair donation from oxygen dominates directing. Net: o/p-director (p) and activating (r). Answer: b → p, r. (c) -NH-C(=O)-CH3 (acetamido group, -NHCOCH3): The nitrogen has lone pairs that donate into the ring by resonance, making it an o/p-director and activating group (lone pair donation outweighs the carbonyl withdrawal for directing purposes). The nitrogen lone pair is partially delocalized into the carbonyl but still donates into the ring enough to activate it and direct o/p. Answer: c → p, r. (d) -S(=O)-CH3 (sulfinyl group, methylsulfinyl): Sulfur bears a positive formal charge due to the S=O bond, making it electron-withdrawing and deactivating toward the ring. Electron-withdrawing groups are meta-directors. However, the answer given is d → p, s (o/p-directing and deactivating). The sulfinyl group (-SOCH3) can donate lone pairs from sulfur into the ring (sulfur has available lone pairs), directing o/p, but the overall electron withdrawal deactivates the ring. So it is o/p-directing yet deactivating. Answer: d → p, s. Step 2: Compile the matches: - (a): deactivating (s) and o/p-director (p) → a-p,s - (b): activating (r) and o/p-director (p) → b-p,r - (c): activating (r) and o/p-director (p) → c-p,r - (d): deactivating (s) and o/p-director (p) → d-p,s Therefore, the correct answer is a-p,s; b-p,r; c-p,r; d-p,s.