See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the product after hydrogenation. The product shown is Carlina oxide, which consists of a benzene ring attached via a -CH2-CH2-CH2- chain to a furan-type ring (specifically a dihydrofuran or benzofuran-related moiety). The full structure of the product is: phenyl group + propyl chain + a five-membered oxygen-containing ring that appears to be 2,3-dihydrofuran fused or a butenolide-type structure. Looking more carefully, the product is (Z)-6-phenylhex-2-en-1-yl furan or more precisely the structure shown is phenyl-CH2-CH2-CH2 connected to a furan ring that has been partially reduced. The actual Carlina oxide is (E)-1-(furan-3-yl)-but-2-en-1-one or more precisely the natural product carlina oxide is phenylpropyl-dihydrobenzofuran. Re-examining: the product after adding 2H2 is benzene ring (6 carbons, already aromatic but the 2H2 reduces two double bonds elsewhere) connected to -CH2CH2CH2- connected to a 5-membered ring containing O. Step 2: Determine the structure of the product. Carlina oxide is known to be the compound with structure: a phenyl group connected through a -CH2CH2CH2- chain to a furanyl group (specifically (Z)-3-(but-2-en-1-yl)furan or similar). The product shown has a benzene ring + -CH2CH2CH2- + a dihydrofuran ring (5-membered ring with O, with one double bond remaining as shown by the wedge lines in the structure). Step 3: Count degrees of unsaturation (DoU) in the product. The product structure: benzene ring (4 DoU: 3 double bonds + 1 ring), the propyl chain (0 DoU), and the dihydrofuran ring (2 DoU: 1 ring + 1 double bond). Total DoU in product = 4 + 0 + 2 = 6. Wait - benzene has 4 DoU (1 ring + 3 double bonds). Dihydrofuran has 2 DoU (1 ring + 1 double bond). Total = 6 DoU in product. Step 4: Since compound (A) was hydrogenated with 2H2 (adding 2 moles of H2, reducing 2 double bonds), compound (A) has 2 more DoU than the product. DoU in (A) = 6 + 3 = 9. Wait - 2H2 means 2 moles of hydrogen added, which reduces 2 double bonds, so 2 additional DoU in (A). But the benzene ring in the product is already present and was not reduced (Pt with 2H2 at mild conditions selectively reduces non-aromatic double bonds). So DoU(A) = DoU(product) + 2 = 6 + 2 = 8? Let me recount product DoU more carefully. Step 5: The product has benzene (4 DoU) + dihydrofuran ring (1 ring DoU + 1 double bond DoU = 2 DoU) + propyl chain (0) = 6 DoU total. Compound A has 2 additional double bonds (from 2H2 reduction) = 6 + 3 = 9 DoU. Since 2 moles H2 reduces exactly 2 pi bonds, compound A has DoU = 6 + 2 + 1? Actually the furan (aromatic, 3 DoU) in compound A was partially reduced to dihydrofuran (2 DoU), accounting for 1 H2. And one other double bond in the chain used 1 H2. So A had furan ring (3 DoU) instead of dihydrofuran (2 DoU) = +1, and one extra double bond in the chain = +1. So DoU(A) = 6 + 1 + 1 + 1 (furan aromatic) = 9. This gives 9 units of unsaturation in compound (A). Why other options fail: 7 is too low (doesn't account for both additional unsaturations from 2H2), 8 is off by one, 10 overcounts. Therefore, the correct answer is C.