See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Concept: Toluene undergoes electrophilic aromatic substitution (EAS). The methyl group is an ortho/para director. We count distinct constitutional isomers (structural products) for each level of nitration. Step 2 – Mono-nitration (A): Toluene has positions 2 (ortho), 3 (meta), and 4 (para) as unique positions due to the plane of symmetry. Nitration can give 2-nitrotoluene (ortho), 3-nitrotoluene (meta), and 4-nitrotoluene (para). Total = 3 distinct products. A = 3. ✓ Step 3 – Di-nitration (B): Starting from toluene, two NO2 groups are added to the ring. The ring has 5 available H positions (positions 2, 3, 4, 5, 6 relative to CH3). We choose 2 of these 5 positions for NO2, but symmetry reduces the count. Systematically listing all unique pairs of positions on a monosubstituted benzene (positions labeled 2,3,4,5,6 with mirror symmetry making 2≡6, 3≡5): The unique dinitro isomers are: 2,3- ; 2,4- ; 2,5- ; 2,6- ; 3,4- ; 3,5-. That gives 6 distinct dinitrotoluene isomers. B = 6. ✓ Step 4 – Tri-nitration (C): Three NO2 groups are placed on the ring (5 remaining H positions, choose 3). Using symmetry on a monosubstituted benzene, systematically list unique trinitro combinations from positions {2,3,4,5,6}: {2,3,4}; {2,3,5}; {2,3,6}; {2,4,5}; {2,4,6}; {3,4,5}. Applying mirror symmetry (2↔6, 3↔5): {2,3,4}≡{2,3,6} wait—let's recheck. Position 4 is unique (para), positions 2&6 are equivalent, 3&5 are equivalent. Unique trinitro sets: (2,3,4), (2,3,5), (2,3,6)→same as (2,4,6) after symmetry? Careful enumeration: mirror maps 2↔6 and 3↔5 while 4 stays. So {2,3,4}↔{4,5,6}={4,5,6}; {2,3,5}↔{3,5,6}; {2,3,6}↔{2,4,6}... wait these are different sets. Full list without symmetry reduction from {2,3,4,5,6} choose 3: C(5,3)=10 combinations. Applying the one mirror symmetry (reflection): pairs that are mirror images: {2,3,4}↔{4,5,6}, {2,3,5}↔{3,5,6}, {2,3,6}↔{2,4,6}, {2,4,5}↔{3,4,6}, and self-symmetric sets: {2,4,6} is mirror of {2,4,6}? 2↔6,4↔4,6↔2 → {6,4,2}={2,4,6} yes, self-symmetric. {3,4,5} is mirror of itself: 3↔5,4↔4,5↔3 → {5,4,3}={3,4,5} yes. So unique isomers: from 10 combinations, 2 are self-symmetric (count once each), and 8 form 4 mirror pairs (count 4). Total unique = 2 + 4 = 6. C = 6. ✓ Step 5 – Matching answer: A=3, B=6, C=6 corresponds to option (b). Step 6 – Why other options fail: (a) says C=8, (c) says C=10, (d) says B=4—all incorrect based on the symmetry analysis above. Therefore, the correct answer is B.