See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Cold dilute KMnO4 (Baeyer's reagent) performs syn-dihydroxylation of alkenes, adding two OH groups to the same face of the double bond. Step 1: Identify the substrate. The structure shown is 1,2-dihydronaphthalene (a naphthalene-derived compound where one ring is aromatic/benzene and the other ring is a cyclohexene with one double bond at the 3,4-position relative to the saturated ring). More precisely, the compound is a bicyclic system with a benzene ring fused to a cyclohexene ring — this is 1,2-dihydronaphthalene, with the double bond between C3 and C4 of the non-aromatic ring. Step 2: Syn-dihydroxylation with cold dil. KMnO4 adds two -OH groups in a syn fashion across the double bond (C3 and C4), creating two new stereocenters at C3 and C4. Step 3: Analyze the stereochemistry of the product. The two carbons bearing -OH groups (C3 and C4) become stereocenters. The syn addition produces the cis-diol. Because the molecule is fused to a benzene ring, the fused bicyclic framework makes C3 and C4 non-equivalent in terms of their substituents relative to the ring junction — specifically, C3 and C4 each have different groups (one is adjacent to the ring fusion CH2 groups, and the connectivity differs). The key point is that the two stereocenters created are NOT related by an internal mirror plane in the product, so the compound cannot be meso. Step 4: The syn addition occurs from either the top face or the bottom face of the double bond with equal probability (the reagent attacks both faces equally since the molecule has no facial selectivity bias from the aromatic ring side vs. the open side — actually the benzene ring blocks one face partially, but in this context both faces are considered). The syn addition from the top face gives one stereoisomer, and syn addition from the bottom face gives its mirror image (enantiomer). Since C3 and C4 in 1,2-dihydronaphthalene are not identical substituent environments (they are part of a ring), we need to check if the product has an internal plane of symmetry. The product is cis-3,4-dihydroxy-1,2,3,4-tetrahydronaphthalene. At C3 and C4, both OH groups are on the same face (syn). C3 is bonded to: OH, H, C2(CH2), C4. C4 is bonded to: OH, H, C3, C4a(aromatic). Since C3 and C4 have different substituents (C3 connects to an aliphatic CH2, C4 connects to an aromatic carbon), there is no internal plane of symmetry — so the product is NOT meso. Step 5: Because the two faces of the double bond in 1,2-dihydronaphthalene are enantiotopic (not homotopic, not diastereotopic — the molecule itself is achiral but the two faces lead to enantiomeric products), syn attack from one face gives one enantiomer and syn attack from the other face gives the mirror image enantiomer. Both are produced in equal amounts (racemic mixture), constituting an enantiomeric pair. Why other options fail: - (a) Meso compound: Would require an internal plane of symmetry in the syn-addition product. Since C3 and C4 have different substituents (different ring environments), no internal mirror plane exists — not meso. - (c) Diastereomers: Diastereomers would result if two different syn-addition products with different relative configurations formed, which is not the case here since only syn addition occurs giving one relative configuration (cis-diol) that exists as two enantiomers. - (d) Optically pure enantiomer: Would require facial selectivity (asymmetric induction), which does not occur with achiral KMnO4 on this substrate. Therefore, the correct answer is B.