See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic aromatic nitration is directed by substituents already present on the ring. The methoxy group (OMe) is a strong ortho/para director due to resonance donation. Step 1: Identify the substrate. The starting material is 6-methoxy-1,2,3,4-tetrahydronaphthalene (6-methoxytetralin). The aromatic ring bears a methoxy (OMe) group. The fused cyclohexane ring occupies positions 1-4 (aliphatic), and the aromatic ring spans positions 4a, 5, 6, 7, 8, 8a. Step 2: Determine directing effects. The OMe group at position 6 is an ortho/para director. The fused aliphatic ring (attached at C4a and C8a) acts as an alkyl substituent, which is also an ortho/para director relative to those ring junction positions. OMe is the dominant director. Step 3: Identify available positions on the aromatic ring. The aromatic ring positions are 5, 6 (OMe), 7, 8. Position 6 is occupied by OMe. Positions ortho to OMe (position 6) are positions 5 and 7. Position para to OMe would be outside the available ring positions given the ring fusion constraints. Step 4: Consider steric and electronic factors. Position 5 is ortho to OMe and also ortho to the ring junction (C4a/C8a). Position 7 is the other ortho position to OMe. Between positions 5 and 7, position 5 is flanked by two ring junctions (C4a and C8a on the fused ring side), making it more sterically hindered. However, in 6-methoxytetralin, the fused ring junction carbons at C4a and C8a are part of the aliphatic ring. Position 5 is adjacent to C4a (ring junction), and position 7 is adjacent to neither ring junction directly. Step 5: In 6-methoxytetralin, nitration with HNO3/H2SO4 (mild nitration, mono-nitration conditions) gives predominantly the product nitrated at position 5, which is ortho to OMe and also activated by the ring junction. The OMe directs strongly to the ortho positions (5 and 7). Position 5 is between the OMe and one ring junction, and experimental evidence shows mono-nitration gives 5-nitro-6-methoxy-tetralin. Step 6: Match with options. Option (c) shows 6-methoxy-5-nitro-1,2,3,4-tetrahydronaphthalene — the tetralin framework intact with NO2 adjacent to OMe on the aromatic ring. This is consistent with electrophilic aromatic substitution directed ortho to OMe. Why other options fail: - Option (a): Shows two nitro groups (dinitration), which requires harsher or excess conditions not implied here. - Option (b): Shows a naphthalene (fully aromatic) product, implying dehydrogenation/aromatization which does not occur under these conditions. - Option (d): Shows NO2 on the aliphatic portion of the ring, but EAS nitration occurs on the aromatic ring, not at sp3 carbons. Therefore, the correct answer is C.