Chapter 5 tereoisomerism eview of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 5. Each of the sentences below appears verbatim in the section entitled eview of Concepts and Vocabulary. isomers have the same connectivity of atoms but differ in their spatial arrangement. Chiral objects are not superimposable on their. The most common source of molecular chirality is the presence of a, a carbon atom bearing different groups. A compound with one chirality center will have one non-superimposable mirror image, called its. The Cahn-Ingold-Prelog system is used to assign the of a chirality center. A polarimeter is a device used to measure the ability of chiral organic compounds to rotate the plane of light. uch compounds are said to be active. A solution containing equal amounts of both enantiomers is called a mixture. A solution containing a pair of enantiomers in unequal amounts is described in terms of enantiomeric (ee). For a compound with multiple chirality centers, a family of stereoisomers exists. Each stereoisomer will have at most one enantiomer, with the remaining members of the family being. A compound contains multiple chirality centers but is nevertheless achiral because it possesses reflectional symmetry. projections are drawings that convey the configuration of chirality centers, without the use of wedges and dashes. eview of kills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 5. The answers appear in the section entitled killbuilder eview. killbuilder 5.1 Identifying cis-trans tereoisomerism AIG TE CFIGUATI F TE FLLWIG DUBLE BD A CI TA killbuilder 5.2 Locating Chirality Centers CICLE TE CIALITY CETE I TE FLLWIG CMPUD
CAPTE 5 79 killbuilder 5.3 Drawing an Enantiomer W TEE WAY T DAW TE EATIME F TE FLLWIG CMPUD. PLACE YU AWE I TE BXE W. 2 killbuilder 5.4 Assigning Configuration AIG TE CFIGUATI F TE CIALITY CETE I TE FLLWIG CMPUD 2 killbuilder 5.5 Calculating specific rotation CALCULATE TE PECIFIC TATI GIVE TE FLLWIG IFMATI: 0.300 grams sucrose dissolved in 10.0 ml of water sample cell = 10.0 cm observed rotation = +1.99 specific rotation = [ ] = c l = = killbuilder 5.6 Calculating % ee CALCULATE TE EATIMEIC EXCE GIVE TE FLLWIG IFMATI: The specific rotation of optically pure adrenaline is -53. A mixture of ()- and ()- adrenaline was found to have a specific rotation of - 45. Calculate the % ee of the mixture % ee = = observed [ ] [ ] of pure enantiomer = killbuilder 5.7 Determining tereoisomeric elationship IDETIFY TE TEEIMEIC ELATIIP BETWEE TE FLLWIG TW CMPUD killbuilder 5.8 Identifying so Compounds DAW ALL PIBLE TEEIME F 1,2-CYCLEXAEDIL (W LEFT), AD TE LK F A PLAE F YMMETY I AY F TE DAWIG. TE PEECE F A PLAE F YMMETY IDICATE A ME CMPUD + = EATIME ME
80 CAPTE 5 killbuilder 5.9 Assigning configuration from a Fischer projection AIG TE CFIGUATI F TE CIALITY CETE I TE FLLWIG CMPUD C 2 olutions 5.1. trans not stereoisomeric trans trans e) trans f) not stereoisomeric g) cis 5.2. 2 CCC 2 C 2 C 2 CC 2 = either double bond exhibits stereoisomerism, so this compound does not have any stereoisomers. 5.3. 5.4. All chirality centers are highlighted below: e)
CAPTE 5 81 f) 5.5. chirality center 5.6. The phosphorus atom has four different groups attached to it (a methyl group, an ethyl group, a phenyl group, and a lone pair). This phosphorous atom therefore represents a chirality center. This compound is not superimposable on its mirror image, as can be seen clearly by building and comparing molecular models. 5.7. C3 e) f) g) 5.8.
82 CAPTE 5 5.9. e) f) 2 5.10. 3 2 1 5.11. 4 1 P 2 3 5.12. specific rotation = [] = 5.13. specific rotation = [] = c l ( + 1.47º ) (0.0575 g / ml) (1.00 dm) = +25.6 c l ( 2.99º ) (0.095 g / ml) (1.00 dm) = -31.5 5.14. specific rotation = [] = c l = ( + 0.57º ) (0.260 g / ml) (1.00 dm) = +2.2 5.15. This compound does not have a chirality center, because two of the groups are identical: Accordingly, the compound is achiral and is not optically active.
CAPTE 5 83 5.16. [] = c l = [] c l = (+13.5)(0.100 g / ml)(1.00 dm) = +1.35 º 5.17. % ee = observed [ ] [ ] of pure enantiomer = ( - 37 ) ( - 39.5 ) = 94 % 5.18. % ee = observed [ ] [ ] of pure enantiomer = ( - 6.0 ) ( - 6.3 ) = 95 % 5.19. % ee = observed [ ] [ ] of pure enantiomer = ( 85 ) ( 92 ) = 92 %
84 CAPTE 5 5.20. bserved [] = c l = ( + 0.78º ) (0.350 g / ml) (1.00 dm) = +2.2 % ee = observed [ ] [ ] of pure enantiomer = ( 2.2 ) ( 2.8 ) = 79 % 5.21. enantiomers diastereomers diastereomers diastereomers e) diastereomers f) enantiomers 5.22. There are three chirality centers, and only one of these chirality centers has a different configuration in these two compounds. The other two chirality centers have the same configuration in both compounds. Therefore, these compounds are diastereomers. 5.23. yes yes no yes e) yes f) no 5.24. 5.23f has three planes of symmetry. 5.25. e) f)
CAPTE 5 85 5.26. e) 5.27. Each of these compounds is a meso compound and does not have an enantiomer. 5.28 There are only four stereoisomers: not a chirality center (see problem 5.5) meso meso not a chirality center (see problem 5.5) 5.29.
86 CAPTE 5 5.30. C 2 C 2 C 2 5.31. C 2 C 2 C 2 5.32. F 5.33. C 3 5.34. Paclitaxel has eleven chirality centers. The enantiomer of paclitaxel is shown below:
CAPTE 5 87 5.35. trans not stereoisomeric not stereoisomeric 5.36. enantiomers same compound constitutional isomers constitutional isomers e) diastereomers f) same compound g) enantiomers h) diastereomers i) same compound j) same compound k) same compound l) same compound 5.37. 8 3 16 3 e) 3 f) 32 5.38. Et F F e) f) g) C 3 h) C 2 i) j) k) l)
88 CAPTE 5 5.39. Et 2 Et e) F f) g) h) i) 5.40. 96% ee 5.41. diastereomers diastereomers enantiomers same compound e) enantiomers f) diastereomers g) enantiomers h) diastereomers i) enantiomers j) same compound k) enantiomer l) diastereomers 5.42. % ee = observed [ ] [ ] of pure enantiomer = ( -55 ) ( -61 ) = 90 %
CAPTE 5 89 5.43. True. False. True. 5.44. specific rotation = [] = c l = ( 0.47º ) (0.0075 g / ml) (1.00 dm) = -63 5.45. ()-limonene ()-limonene ()-limonene ()-limonene 5.46. 3 C C 3 5.47. C 3 C 3 C 3 C 3 5.48. The first compound has three chirality centers: chirality center three chirality centers two chirality centers This is apparent if we assign the configuration at C1 and C3 of the cyclohexane ring. In the first compound, the configuration at C1 is different than the configuration at C3. As a result, there are four different groups attached to the C2 position. That is, C1 and C3 represent two different groups: one with the configuration and the other with the configuration. In contrast, consider the configuration at C1 and C3 in the second compound. Both of these positions have the same configuration, and therefore, the C2 position in that compound does not have four different groups. Two of the groups are identical, so C2 is not a chirality center.
90 CAPTE 5 5.49. enantiomers diastereomers enantiomers same compound e) enantiomers f) diastereomers g) same compound h) constitutional isomers i) diastereomers j) diastereomers k) same compound l) enantiomers 5.50. -61 90 % ee 95 % of the mixture is ()-carvone 5.51. chiral chiral achiral achiral e) chiral f) achiral g) achiral h) chiral i) chiral j) achiral k) chiral l) chiral l) achiral m) chiral n) achiral o) achiral 5.52. [] = c l = [] c l = (+24)(0.0100 g / ml)(1.00 dm) = +0.24 º 5.53. optically inactive (meso) optically active optically active optically inactive e) optically active f) optically inactive (3-methylpentane has no chirality centers) g) optically inactive (meso) h) optically inactive 5.54. e)
CAPTE 5 91 5.55. o. A racemic mixture is not optically active. Yes, because d and e are not enantiomers. They are diastereomers, which are not expected to exhibit equal and opposite rotations. 5.56. C 2 C 2 C 2 5.57. 3-methylpentane and 2-methylpentane are constitutional isomers. trans-1,2-dimethylcyclohexane and cis-1,2-dimethylcyclohexane are diastereomers. 5.58. The following two compounds are enantiomers because they are nonsuperimposable mirror images. You may find it helpful to construct molecular models to help visualize the mirror image relationship between these two compounds. C C C C C C 5.59. This compound will be achiral. 5.60. This compound cannot be completely planar because steric hindrance prevents the two ring systems from rotating with respect to each other. The compound is locked in a particular conformation that is chiral. This ring system cannot be planar because of steric hindrance, and must therefore adopt a spiral shape (like a spiral staircase). The spiral can be right handed or left handed, and the relationship between these two forms is enantiomeric. 5.61. The compound is chiral because it is not superimposable on its mirror image. C 3 3 C 5.62. This compound has a center of inversion, which is a form of reflection symmetry. As a result, this compound is superimposable on its mirror image and is therefore optically inactive.