See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic aromatic substitution (nitration) rates depend on the electron density of the aromatic ring. Electron-donating groups increase the rate; electron-withdrawing groups decrease it. Step 1: Identify the electronic effect of each substituent on the central benzene ring. - In structure (i): Both flanking rings contain lactone groups (C=O-O) where the carbonyl is directly conjugated with the benzene ring (alpha,beta-unsaturated lactone arrangement). This means the C=O groups are in conjugation with the aromatic ring via the ring fusion, withdrawing electron density by resonance from the benzene ring. Two such strongly electron-withdrawing conjugated ester/lactone groups are present. This makes the benzene ring most electron-poor. - In structure (ii): One ring has a conjugated lactone (C=O directly conjugated with aromatic ring) and the other has a saturated lactone (C=O not directly conjugated because a CH2 breaks conjugation). So only one strongly withdrawing group is in conjugation; the other withdraws only by induction. The benzene ring is moderately electron-poor. - In structure (iii): Both flanking rings are saturated lactones where CH2-CH2 groups break direct conjugation of C=O with the aromatic ring. The carbonyl groups withdraw only inductively (weakly), not by resonance conjugation. The benzene ring is least electron-poor (most electron-rich among the three). Step 2: Rank the electron density of the benzene ring. - Structure (i): Lowest electron density (two conjugated EWG carbonyls) → slowest nitration. - Structure (ii): Intermediate electron density (one conjugated, one non-conjugated EWG) → intermediate rate. - Structure (iii): Highest electron density (both EWGs non-conjugated, only inductive withdrawal) → fastest nitration. Step 3: Increasing order of rate of nitration: i < ii < iii Step 4: Why other options fail: - (a) iii < ii < i: Incorrect, this is decreasing order. - (b) ii < iii < i: Incorrect, places i as fastest which contradicts its most electron-poor ring. - (c) i < iii < ii: Incorrect, iii should be faster than ii, not slower. Therefore, the correct answer is D.