CHAPTER 26 STEREOISOMERISM SOLUTIONS TO REVIEW QUESTIONS. ƒ C Cl ƒ

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1 EINS v1.qxd 11/9/07 1:13 PM Page 400 APTER 26 STEREOISOMERISM SOLUTIONS TO REVIEW QUESTIONS 1. A chiral carbon atom is one to which four different atoms or groups are attached and is a center of asymmetry in a molecule. In the following three compounds, the chiral carbon atoms are marked with an asterisk. (These are merely three examples; there are an infinite number of compounds which contain one chiral carbon atom.) l * Br Br l * l 2. When the axes of two pieces of polaroid film are parallel, you have maximum brightness of the light passing through both. When one piece has been rotated by 90 the polaroid appears black, indicating very little light passing through. 3. A necessary and sufficient condition for a compound to show enantiomerism is that the compound not be superimposable on its mirror image. 4. Enantiomers are nonsuperimposable mirror image isomers. Diastereomers are stereoisomers that are not (not mirror image isomers). 5. A diastereomer is a non-mirror image stereoisomer. A meso compound contains chiral carbons but is not optically active; it may also be a diastereomer. 6. Physical properties of a pair of (+) 2-methyl-1-butanol (-) 2-methyl-1-butanol l l l * 3 (O) 2 O specific rotation boiling point density g>ml g>ml 7. Most chiral molecules are stereospecific in their biological activity. Therefore a racemic mixture of a drug provides only half the bioactive material prescribed. By using a single isomer of a compound, the dosage can be cut in half and possible side effects can be avoided from its enantiomer

2 EINS v1.qxd 11/9/07 1:13 PM Page hapter A meso compound contains chiral carbons but is not optically active. A racemic mixture contains two optically active compounds that exactly cancel out each other s optical activity (the mixture is not optically active). 9. The enantiomer will have a specific rotation of The specific rotation for the racemic mixture will be zero

3 EINS v1.qxd 11/9/07 1:13 PM Page 402 APTER 26 SOLUTIONS TO EXERISES 1. Enantiomers have identical physical properties except their effect on polarized light, so they cannot be separated by ordinary chemical and physical means. 2. Diastereomers do not have identical physical properties, so the differences form a basis for chemical or physical separation. Differences of boiling point, freezing point, and solubilities would be most commonly used. 3. The objects that are chiral: (a) your ear; (b) a pair of pliers; (c) a coiled spring; (d) the letter b. 4. The objects that are chiral: (a) a wood screw; (c) the letter g; (d) this textbook. 5. Number of chiral carbon atoms. (a) 1 (b) 1 (c) 2 (d) 1 6. Number of chiral carbon atoms. (a) 0 (b) 0 (c) 3 (d) 2 7. Which compounds will show optical activity? (a) l no optical activity (b) 2 l 2 no optical activity (c) 2 2 l no optical activity (d) * l 2 will show optical activity (* chiral carbon) 8. Which compounds will show optical activity? (a) 2 2 * l will show optical activity (* chiral carbon) (b) 2 2 * 2 l will show optical activity (* chiral carbon)

4 EINS v1.qxd 11/9/07 1:13 PM Page hapter 26 - (c) 2 * l will show optical activity (* chiral carbon) (d) 2 l 2 no optical activity 9. Glucose, which has four chiral carbon atoms will have 16 possible stereoisomers. This can be determined from 2 n = 2 4 = Fructose, which has three chiral carbon atoms, will have 8 possible stereoisomers. This can be determined from 2 n = 2 3 = The two projection formulas (A) and (B) are the same compound, for it takes two changes to make (B) identical to (A). Br l Br l F F F F (A) (B) 1st change in (B) 2nd change in (B) ( and Br) ( and l) 12. The two projection formulas (A) and (B) are the same compound, for it takes two changes to make (B) identical to (A). Br l Br Br F Br Br Br F F F (A) (B) 1st change in (B) 2nd change in (B) ( and ) (F and ) l OO OO 13. (-)-lactic acid is O (+)-lactic acid is O (a), (e), and (f) are (-)-lactic acid (b), (c), and (d) are (+)-lactic acid

5 EINS v1.qxd 11/9/07 1:13 PM Page hapter 26 - OO OO 14. (+)-alanine is 2 N (-)-alanine is N 2 (b), (c), and (d) are (+)-alanine (a), (e), and (f) are (-)-alanine 15. l N 2 OO l N 2 OO N O OO 2 O N 2 O OO 17. O O Br

6 EINS v1.qxd 11/9/07 1:13 PM Page hapter 26-2 O 18. l Br l OO 19. All possible stereoisomers of the following compounds, with and meso compounds labeled. 2 Br 2 Br (a) 1,2-dibromopropane Br Br (b) 2-butanol O 2 O (c) 3-chlorohexane There are no meso compounds. l l

7 EINS v1.qxd 11/9/07 1:13 PM Page hapter All possible stereoisomers of the following compounds with and meso compounds labeled. (a) 2,3-dichlorobutane l l l l l l meso (b) 2,4-dibromopentane Br Br Br Br Br Br meso 2 2 (c) 3-hexanol O O There are no meso compounds

8 EINS v1.qxd 11/9/07 1:13 PM Page hapter All the stereoisomers of 1,2,3-trihydroxybutane: 2 O 2 O 2 O 2 O O O O O O O A B D ompounds A and B, and and D are pairs of. There are no meso compounds. Pairs of diastereomers are A and, A and D, B and, and B and D. O O 22. All the stereoisomers of 3,4-dichloro-2-methylpentane: ( ) 2 ( ) 2 ( ) 2 ( ) 2 l l l l A B D ompounds A and B, and and D are pairs of. There are no meso compounds. Pairs of diastereomers are A and, A and D, B and, and B and D. 23. The four stereoisomers of 2-hydroxy-3-pentene: l l l l O O O O cis cis trans trans The two cis compounds are and the two trans compounds are

9 EINS v1.qxd 11/9/07 1:13 PM Page hapter The four stereoisomers of 2-chloro-3-hexene. l l l cis cis trans trans The two cis compounds are and the two trans compounds are. l 25. (a) 2 2 l 2 2 l 2 l l 2 2 l A B 2 l l 2 l 2 ll D E F l 2 l 2 l 2 l 2 l G I (b) B is chiral 2 2 l l is chiral l 2 2 l 2 2 l l 2 2 l l

10 EINS v1.qxd 11/9/07 1:13 PM Page hapter 26 - F can be both chiral and meso l l l l l l meso A, D, E, G,, and I are achiral. 26. (a) (b) 2 Br 2 A Br 2 B is chiral 2 Br Br 2 Br B 2 Br 2 2 Br D 2 Br A,, and D are achiral; there are no meso compounds. 27. Assume (+)-2-bromopentane is Br Br Br 2 2 All possible isomers formed when (+)-2-bromopentane is further brominated to dibromopentanes:

11 EINS v1.qxd 11/9/07 1:13 PM Page hapter 26-2 Br Br 2 Br Br 2 Br Br Br Br 2 2 A B D 2 2 Br 2 Br 2 Br 2 Br Br 2 2 Br E F G ompounds A,, D, F, G would be optically active; B has no chiral carbon atom; E is a meso compound. 28. Assume (+)-2-chlorobutane is 2 l All possible isomers formed when (+)-2-chlorobutane is further chlorinated to dichlorobutane: 2 l 2 l 2 l l 2 l l l l l 2 l A B D E ompounds A, B, and E would be optically active; does not have a chiral carbon atom; D is a meso compound

12 EINS v1.qxd 11/9/07 1:13 PM Page hapter Neither of the products, 1-chloropropane or 2-chloropropane, have a chiral carbon atom so neither product would rotate polarized light. 30. If 1-chlorobutane and 2-chlorobutane were obtained by chlorinating butane, and then distilled, they would be separated into the two fractions, because their boiling points are different. 1-chlorobutane has no chiral carbon, so would not be optically active. 2-chlorobutane would exist as a racemic mixture (equal quantities of ) because substitution of l for on carbon-2 gives equal amounts of the two. Distillation would not separate the because their boiling points are identical. The optical rotation of the two of the 2-chlorobutane fraction would exactly cancel, and thus would not show optical activity. 31. ompounds (a) and (d) are meso. Make two changes on -3 in compound (a) to prove that it is meso. (a) (d) OO O O OO l Br l 32. ompound (d) is meso. Br l Br

13 EINS v1.qxd 11/9/07 1:13 PM Page hapter (b) is chiral Br Br Br Br 34. (c) is chiral l l l l 35. (a) diasteromers (b) nonisomers (c) diasteromers 36. (a) diasteromers (b) (c) diasteromers 37. If four different groups were attached to a central carbon atom in a planar arrangement, it would not rotate polarized light because there would be a plane of symmetry in the molecule. No such plane of symmetry is possible when the four different groups are arranged in a tetrahedral structure. 38. (a) and (b) ( ) 2 2 OO OO chiral carbon 2 ( )

14 EINS v1.qxd 11/9/07 1:13 PM Page hapter (a) A chiral primary alcohol of formula 5 12 O. (b) 40. (a) (b) 2 2 O A compound with three primary alcohol groups is chiral, and has the formula 6 14 O O O 2 O 2 O chiral carbon caraway The spearmint molecule is the optical isomer of the caraway molecule and differs from it in structure at the chiral carbon atom. 41. Ephedrine has two chiral carbons and can have four stereoisomers. This number is calculated using 2 n = 2 2 = O O 2 N chiral carbon O O N 2 mirror image 43. (-) placed in front of a name is used to indicate the rotation of plane-polarized light to the left, and (+) for rotation to the right. Therefore, we can write (-)-methorphan for levomethorphan and (+)-methorphan for dextromethorphan. There is no obvious correlation between the structures of and the direction in which they rotate plane polarized light

15 EINS v1.qxd 11/9/07 1:13 PM Page hapter Stereoisomer structures (a) 2-bromo-3-chlorobutane Br l Br l l Br Br l (b) 2,3,4-trichloro-1-pentanol 2 O l l 2 O l 2 O l l l 2 O l l l l l l l l 2 O l 2 O l l 2 O l 2 O l l l l l l There are no meso compounds

16 EINS v1.qxd 11/9/07 1:13 PM Page hapter Meso structures for alkanes (a) 8 18 (b) 9 20 (c)

17 EINS v1.qxd 11/9/07 1:13 PM Page hapter Optically active alcohols of 6 14 O (one enantiomer of each structure) O O 2 O 2 O 2 2 O O 2 3 O O

18 EINS v1.qxd 11/9/07 1:13 PM Page hapter O OO N 2 OO 2 N 2 N N 2 OO O The similarity is that the chiral carbon in each compound is bonded to an N 2,a OO, and an group. 48. A compound of formula 3 8 O 2 : (a) is chiral; contains two O groups 2 O O (b) is chiral; contains one O group O O (c) is achiral; contains two O groups O O