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276 Organic Chemistry Solutions Manual The two protons we can see in isomer A must be H¹ and H² as they have the largest shifts. The proton with only one coupling must be H¹ as it has only one neighbour (H²). Isomer A has a 9 coupling between H¹ and H² so both these protons must be axial. Isomer A is therefore the isomer. The signal for H² is a dt because it has two axial neighbours (H¹ and H³) and one neighbour (H⁴). Isomer B shows H¹ only and it is clearly equatorial (J₁₂ = 4). Though we can't H² it must be axial as B must be the cis-isomer since it is different from A. CN CN MeO CO₂Me MeO CO₂Me A = B = Problem 11 Muscarine, the poisonous principle of the death cap mushroom, has the following structure and proton NMR spectrum. Assign the spectrum. Can you see definite evidence for the 0 stereochemistry? All couplings in Hz; signals marked * exchange with D₂O. δH 1.16 (3H, d, J 6.5), 1.86 (1H, ddd, J 12.5, 9.5, 5.5), 2.02 (1H, ddd, J 12.5, 2.0, 6.0 3.36 (9H, s), 3.54 (1H, dd, J 13, 9.0), 3.74 (1H, dd, J 13, 1.0), 3.92 (1H, dq, J 2.5, 6.5 4.03 (1H, m), 4.30* (1H, d, J 3.5), and 4.68 (1H, m). Purpose of the problem Demonstrating that it can be very difficult to determine stereochemistry even with all information. Suggested solution The details are in J. Mulzer and Coupling constants around five-membered rings tend to be much the same whether they are group, Liebigs Ann. Chem., 1987, 7. (geminal), or (vicinal). Even so, the two diastereotopic CH₂ groups are easy to find their large couplings of 13 and 12.5. The one with one extra coupling must be in the side and the others in the ring. Here is the full analysis. You will see that it is, in fact, very difficult to conclusive evidence on stereochemistry though you should see that, in general, cis couplings tend be larger than trans. H 1.86 (1H, ddd, J 12.5, 9.5, and 5.5), 4.30* (1H, d, J3.5) 2.02 (1H, ddd, J 12.5, 2.0, and 6.0) H 4.03 (1H, m) or 4.68 (1H, m) H δ 3.36 (9H, s) δ 1.16 (3H, d, J 6.5 Hz) 0 H H H H 3.54 (1H, dd, and 9.0), δ 3.92 (1H, dq, J2.5 and 6.5) 3.74 (1H, dd, 13 and 1.0) 4.03 (1H, m) or 4.68 (1H, m) Problem 12 An antifeedant compound that deters insects from eating food crops has the gross structure shown below. Some of the NMR signals that can clearly be made out are also given. Since NMR coupling constants are clearly useless in assigning the stereochemistry, how would you 0 OAc OAc set about it? δH 2.22 (1H, d, J4), 2.99 (1H, dd, J4, 2.4), 4.36 (1H, d, J 12.3), 4.70 (1H, dd, J 4.7, 11.7). 4.88 (1H, d, J 12.3)