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Chapter 32 Suggested solutions for Chapter 32 271 Suggested solution We can make some preliminary assignments from a combination of shift and coupling. Signal Integral and splitting Comments Assignment 3H, d, 6 3H di must be CHMe Me⁷ Me 161* 1H, broad S exchanges must be OH OH 1.87 1H, ddd, J 14, 3, 3.5 14 Hz looks like geminal coupling H² or 1H, ddd, J 14, 3, 1.5 2.21 and 1.87 are group H² or H³ MeO OMe 1H, dd, J 10, 3 must be axial proton (10 Hz) or 3.40 3H, S one OMe group OMe HO 3.47 3H, S the other OMe group OMe 1H, dq, 10, 6 q means H7, must be axial (10 Hz) 4.24 1H, ddd, J 3, 3, 3.5 all small J, must be equatorial or 4.79 1H, dd, J 3.5, 1.5 all small J, must be equatorial H1 We don't mind which is H² or as they don't affect the stereochemistry, but we do mind which A. K. Mallams and group, J. Am. or H⁵. Since is a 10 Hz doublet with H⁵, we know that H⁵ is at 2.87 and is axial. This gives Chem. Soc., 1981, 103, 3938. the entire assignment and the stereochemistry: and are axial; H¹ and are equatorial. That is why there are no large vicinal couplings to the diastereotopic CH₂ group (H²and H³). δH 1.33 Hz δH 2.87 H H Me Me H δH 1.87/2.21 H 0 14 Hz δH 3.40/3.47 MeO J 10 Hz MeO 2.87 H H δH 4.79 H H IV δH 3.99 10 H H H OH OMe δH 3.40/3.47 OH OMe δH 1.61* all couplings not shown are