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98 Chapter 5 STEREOISOMERS Next determine how many stereoisomers can exist. A molecule with two stereocenters can have up to four stereoisomers. This is the case here. If a plane of symmetry were present in any possible stereoisomeric structure, the total number of stereoisomers would drop, because two of them would become identical as the meso form. That's not possible with this molecule, so there will be four structures for us to draw and name in full. We make the remainder of the problem as easy as possible by first drawing a structure for which determination of R and S is simplified by putting the H atoms on the stereocenters below their carbons (on hashed bonds): Br H C2 is R 3 2 C3 is also R The name is H We derive the remaining three stereoisomers and their names from this one by switching the stereochemistries of each of the stereocenters one at a time and switching the R/S designations to match: Br H H Br H Br 3 2 3 2 3 2 H H H (2S,3R) (2R,3S) 47. Use (specific rotation) = α (observed rotation) conc. (g mL⁻¹) X path length (dm) (a) C = 0.4 g/10 mL = 0.04 g mL⁻¹; α = -0.56°, and = 10 cm = 1 dm; so = 14.0° (b) = +66.4°, C = 0.3 g mL⁻¹, / = 1 dm; α = +19.9° (c) Rearranging gives C = = 57.3°/23.1° = 2.48 g mL⁻¹ 48. C = 1 g/20 mL = 0.05 g mL⁻¹, so X l) = = 50°, which is identical to the actual Therefore the epinephrine is optically pure and, presumably, safe to use. c 49. (a) H₂N CO₂H Priorities: a b H d (b) 8°/24° = 0.33 or 33% optically pure. corresponding to a mixture of 33% pure S + 67% racemate, or 67% S and 33% R. (c) 16°/24° = 0.67 or 67% optical which equals 67% pure S + 33% racemate. That's the same as 83% S and 17% R.