《有机化学》课程教学资源（教材文献，英文版）CHAPTER 7 STEREOCHEMISTRY

CHAPTER SEVEN Stereochemistry 7. 3 SYMMETRY IN ACHIRAL STRUCTURES Certain structural features can sometimes help us determine by inspection whether a mol ecule is chiral or achiral. For example, a molecule that has a plane of symmetry or a cen ter of symmetry is superposable on its mirror image and is achiral a plane of symmetry bisects a molecule so that one half of the molecule is the mirror image of the other half. The achiral molecule chlorodifluoromethane, for exam- ple, has the plane of symmetry shown in Figure 7.3 A point in a molecule is a center of symmetry if any line drawn from it to som element of the structure will, when extended an equal distance in the opposite direction encounter an identical element. The cyclobutane derivative in Figure 7. 4 lacks a plane of symmetry, yet is achiral because it possesses a center of symmetry. PROBLEM 7. 3 Locate any planes of symmetry or centers of symmetry in each of the following compounds. Which of the compounds are chiral? Which are achiral? (a)(Er-1, 2-Dichloroethen (c)cis-1, 2-Dichlorocyclopropane (b)(Z)-1, 2, Dichloroethene (d) trans-1, 2-Dichlorocyclopropane SAMPLE SOLUTION (a)(E)-1, 2-Dichloroethene is planar. The molecular plane is plane of syi Furthermore, (E)-1, 2-dichloroethene has a center of symmetry located at the mid point of the carbon-carbon double bond. It is achira FIGURE 7. 3 A plane f symmetry defined atoms H-C-Cl chlorodifluoromethane into Br Br FIGURE 7. 4(a)Struc tural formulas a and b are drawn as mirror images. (b) The two mirror images are superposable by rotating form b 180 about an axis passing through the center B≡A f the molecule. the center of the molecule is a center of (a) (b) Back Forward Main MenuToc Study Guide ToC Student o MHHE Website

7.3 SYMMETRY IN ACHIRAL STRUCTURES Certain structural features can sometimes help us determine by inspection whether a mol￾ecule is chiral or achiral. For example, a molecule that has a plane of symmetry or a cen￾ter of symmetry is superposable on its mirror image and is achiral. A plane of symmetry bisects a molecule so that one half of the molecule is the mirror image of the other half. The achiral molecule chlorodifluoromethane, for exam￾ple, has the plane of symmetry shown in Figure 7.3. A point in a molecule is a center of symmetry if any line drawn from it to some element of the structure will, when extended an equal distance in the opposite direction, encounter an identical element. The cyclobutane derivative in Figure 7.4 lacks a plane of symmetry, yet is achiral because it possesses a center of symmetry. PROBLEM 7.3 Locate any planes of symmetry or centers of symmetry in each of the following compounds. Which of the compounds are chiral? Which are achiral? (a) (E)-1,2-Dichloroethene (c) cis-1,2-Dichlorocyclopropane (b) (Z)-1,2,Dichloroethene (d) trans-1,2-Dichlorocyclopropane SAMPLE SOLUTION (a) (E)-1,2-Dichloroethene is planar. The molecular plane is a plane of symmetry. Furthermore, (E)-1,2-dichloroethene has a center of symmetry located at the mid￾point of the carbon–carbon double bond. It is achiral. 264 CHAPTER SEVEN Stereochemistry F F Cl H Br Br Cl Cl Br Br Cl Cl A B (a) Br Br Cl Cl B (b) Br Br Cl Cl BPA FIGURE 7.4 (a) Struc￾tural formulas A and B are drawn as mirror images. (b) The two mirror images are superposable by rotating form B 180° about an axis passing through the center of the molecule. The center of the molecule is a center of symmetry. FIGURE 7.3 A plane of symmetry defined by the atoms H±C±Cl divides chlorodifluoromethane into two mirror-image halves. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website

4 Properties of Chiral Molecules: Optical Activity Any molecule with a plane of symmetry or a center of symmetry is achiral, but their absence is not sufficient for a molecule to be chiral. a molecule lacking a center of symmetry or a plane of symmetry is likely to be chiral, but the superposability test should be applied to be certain 7. 4 PROPERTIES OF CHIRAL MOLECULES: OPTICAL ACTIVITY The experimental facts that led vant Hoff and Le bel to propose that molecules having The phenomenon of optical the same constitution could differ in the arrangement of their atoms in space concerned ctivity was discovered by the physical property of optical activity. Optical activity is the ability of a chiral sub- the French physicist Jean- stance to rotate the plane of plane-polarized light and is measured using an instrument Baptiste Biot in 1815 alled a polarimeter.( Figure 7.5) The light used to measure optical activity has two properties: it consists of a sin- le wavelength and it is plane-polarized. The wavelength used most often is 589 nm (called the D line ), which corresponds to the yellow light produced by a sodium lamp Except for giving off light of a single wavelength, a sodium lamp is like any other lamp in that its light is unpolarized, meaning that the plane of its electric field vector can have any orientation along the line of travel. A beam of unpolarized light is transformed to plane-polarized light by passing it through a polarizing filter, which removes all the waves except those that have their electric field vector in the same plane. This plane polarized light now passes through the sample tube containing the substance to be exam ined, either in the liquid phase or as a solution in a suitable solvent(usually water, ethanol, or chloroform). The sample is"optically active "if it rotates the plane of polar ized light. The direction and magnitude of rotation are measured using a second polar izing filter (the"analyzer")and cited as a, the observed rotation. To be optically active, the sample must contain a chiral substance and one enantiomer must be present in excess of the other: A substance that does not rotate the plane of polar ized light is said to be optically inactive. All achiral substances are optically inactive. What causes optical rotation? The plane of polarization of a light wave undergoes a minute rotation when it encounters a chiral molecule. enantiomeric forms of a chiral molecule cause a rotation of the plane of polarization in exactly equal amounts but in Sample tube with solution of optically Angle rotatIon ctive substance Analyzer light oscillates Plane-polarized 180° FIGURE 7.5 The mp emits light moving in all When the light passes through stance. The plan contains a solu econd polarizing filter(called the analyzer)is ed in degrees that is used to measure the angle of rotation (Adapted from M. Silberberg, Chemistry, 2d edition, McGraw-Hill Higher Education, New York, 1992,p.616) Back Forward Main MenuToc Study Guide ToC Student o MHHE Website

Any molecule with a plane of symmetry or a center of symmetry is achiral, but their absence is not sufficient for a molecule to be chiral. A molecule lacking a center of symmetry or a plane of symmetry is likely to be chiral, but the superposability test should be applied to be certain. 7.4 PROPERTIES OF CHIRAL MOLECULES: OPTICAL ACTIVITY The experimental facts that led van’t Hoff and Le Bel to propose that molecules having the same constitution could differ in the arrangement of their atoms in space concerned the physical property of optical activity. Optical activity is the ability of a chiral sub￾stance to rotate the plane of plane-polarized light and is measured using an instrument called a polarimeter. (Figure 7.5). The light used to measure optical activity has two properties: it consists of a sin￾gle wavelength and it is plane-polarized. The wavelength used most often is 589 nm (called the D line), which corresponds to the yellow light produced by a sodium lamp. Except for giving off light of a single wavelength, a sodium lamp is like any other lamp in that its light is unpolarized, meaning that the plane of its electric field vector can have any orientation along the line of travel. A beam of unpolarized light is transformed to plane-polarized light by passing it through a polarizing filter, which removes all the waves except those that have their electric field vector in the same plane. This plane￾polarized light now passes through the sample tube containing the substance to be exam￾ined, either in the liquid phase or as a solution in a suitable solvent (usually water, ethanol, or chloroform). The sample is “optically active” if it rotates the plane of polar￾ized light. The direction and magnitude of rotation are measured using a second polar￾izing filter (the “analyzer”) and cited as , the observed rotation. To be optically active, the sample must contain a chiral substance and one enantiomer must be present in excess of the other. A substance that does not rotate the plane of polar￾ized light is said to be optically inactive. All achiral substances are optically inactive. What causes optical rotation? The plane of polarization of a light wave undergoes a minute rotation when it encounters a chiral molecule. Enantiomeric forms of a chiral molecule cause a rotation of the plane of polarization in exactly equal amounts but in 7.4 Properties of Chiral Molecules: Optical Activity 265 The phenomenon of optical activity was discovered by the French physicist Jean￾Baptiste Biot in 1815. 0° 180° 270° 90° Analyzer Rotated polarized light Plane-polarized light oscillates in only one plane Sample tube with solution of optically active substance α Polarizing filter Unpolarized light oscillates in all planes Light source Angle of rotation FIGURE 7.5 The sodium lamp emits light moving in all planes. When the light passes through the first polarizing filter, only one plane emerges. The plane-polarized beam enters the sam￾ple compartment, which contains a solution enriched in one of the enantiomers of a chiral sub￾stance. The plane rotates as it passes through the solution. A second polarizing filter (called the analyzer) is attached to a movable ring calibrated in degrees that is used to measure the angle of rotation . (Adapted from M. Silberberg, Chemistry, 2d edition, McGraw-Hill Higher Education, New York, 1992, p. 616.) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website

CHAPTER SEVEN Stereochemistry pposite directions. A solution containing equal quantities of enantiomers therefore exhibits rotation because all the tiny increments of clockwise produced by molecules of one"handedness"are canceled by an equal number of increments of anticlockwise rotation produced by molecules of the opposite handedness Mixtures containing equal quantities of enantiomers are called racemic mixtures. Racemic mixtures are optically inactive. Conversely, when one enantiomer is present in excess, a net rotation of the plane of polarization is observed. At the limit, where all the molecules are of the same handedness, we say the substance is optically pure Optical purity, or percent enantiomeric excess, is defined Optical purity percent enantiomeric excess Thus, a material that is 50%o optically pure contains 75%o of one enantiomer and 25%o of Rotation of the plane of polarized light in the clockwise sense is taken as positive (+) and rotation in the anticlockwise sense is taken as a negative(-)rotation. The clas ical terms for positive and negative rotations are dextrorotatory and levorotatory, from the Latin prefixes dextro-("to the right")and levo-(to the left), respectively. At one time, the symbols d and l were used to distinguish between enantiomeric forms of a sub- stance. Thus the dextrorotatory enantiomer of 2-butanol was called d-2-butanol, and the levorotatory form 1-2-butanol; a racemic mixture of the two was referred to as dl-2- butanol. Current custom favors using algebraic signs instead, as in(+)-2-butanol, (-)-2-butanol, and(+)-2-butanol, respectively The observed rotation a of an optically pure substance depends on how many mol- ecules the light beam encounters. A filled polarimeter tube twice the length of another produces twice the observed rotation, as does a solution twice as concentrated. To account for the effects of path length and concentration, chemists have defined the term specific rotation, given the symbol [a]. Specific rotation is calculated from the observed rotation according to the expression ams per milliliter of so la lution instead of grams per where c is the concentration of the sample in grams per 100 mL of solution, and l is the length of the polarimeter tube in decimeters. (One decimeter is 10 cm Specific rotation is a physical property of a substance, just as melting point, boil ing point, density, and solubility are. For example, the lactic acid obtained from milk is exclusively a single enantiomer. We cite its specific rotation in the form [a]=+3.8% The temperature in degrees Celsius and the wavelength of light at which the measure ment was made are indicated as superscripts and subscripts, respectively PRoBLEM 7. 4 Cholesterol, when isolated from natural sources, is obtained as a single enantiomer. The observed rotation a of a 0.3-g sample of cholesterol in 15 L of chloroform solution contained in a 10-cm polarimeter tube is -0.78. Cal culate the specific rotation of cholesterol PROBLEM 7. 5 A sample of synthetic cholesterol was prepared consisting entirely of the enantiomer of natural cholesterol. a mixture of natural and synthetic cho- lesterol has a specific rotation [a] of -13. What fraction of the mixture is nat- ural cholesterol? Back Forward Main MenuToc Study Guide ToC Student o MHHE Website

opposite directions. A solution containing equal quantities of enantiomers therefore exhibits no net rotation because all the tiny increments of clockwise rotation produced by molecules of one “handedness” are canceled by an equal number of increments of anticlockwise rotation produced by molecules of the opposite handedness. Mixtures containing equal quantities of enantiomers are called racemic mixtures. Racemic mixtures are optically inactive. Conversely, when one enantiomer is present in excess, a net rotation of the plane of polarization is observed. At the limit, where all the molecules are of the same handedness, we say the substance is optically pure. Optical purity, or percent enantiomeric excess, is defined as: Optical purity percent enantiomeric excess percent of one enantiomer percent of other enantiomer Thus, a material that is 50% optically pure contains 75% of one enantiomer and 25% of the other. Rotation of the plane of polarized light in the clockwise sense is taken as positive (), and rotation in the anticlockwise sense is taken as a negative () rotation. The clas￾sical terms for positive and negative rotations are dextrorotatory and levorotatory, from the Latin prefixes dextro- (“to the right”) and levo- (“to the left”), respectively. At one time, the symbols d and l were used to distinguish between enantiomeric forms of a sub￾stance. Thus the dextrorotatory enantiomer of 2-butanol was called d-2-butanol, and the levorotatory form l-2-butanol; a racemic mixture of the two was referred to as dl-2- butanol. Current custom favors using algebraic signs instead, as in ()-2-butanol, ()-2-butanol, and ()-2-butanol, respectively. The observed rotation of an optically pure substance depends on how many mol￾ecules the light beam encounters. A filled polarimeter tube twice the length of another produces twice the observed rotation, as does a solution twice as concentrated. To account for the effects of path length and concentration, chemists have defined the term specific rotation, given the symbol []. Specific rotation is calculated from the observed rotation according to the expression [] where c is the concentration of the sample in grams per 100 mL of solution, and l is the length of the polarimeter tube in decimeters. (One decimeter is 10 cm.) Specific rotation is a physical property of a substance, just as melting point, boil￾ing point, density, and solubility are. For example, the lactic acid obtained from milk is exclusively a single enantiomer. We cite its specific rotation in the form [] D 25 3.8°. The temperature in degrees Celsius and the wavelength of light at which the measure￾ment was made are indicated as superscripts and subscripts, respectively. PROBLEM 7.4 Cholesterol, when isolated from natural sources, is obtained as a single enantiomer. The observed rotation of a 0.3-g sample of cholesterol in 15 mL of chloroform solution contained in a 10-cm polarimeter tube is 0.78°. Cal￾culate the specific rotation of cholesterol. PROBLEM 7.5 A sample of synthetic cholesterol was prepared consisting entirely of the enantiomer of natural cholesterol. A mixture of natural and synthetic cho￾lesterol has a specific rotation [] D 20 of 13°. What fraction of the mixture is nat￾ural cholesterol? 100 cl 266 CHAPTER SEVEN Stereochemistry If concentration is expressed as grams per milliliter of so￾lution instead of grams per 100 mL, an equivalent ex￾pression is []  cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website

7.5 Absolute and Relative Configuration It is convenient to distinguish between enantiomers by prefixing the sign of rota- tion to the name of the substance. For example, we refer to one of the enantiomers of 2-butanol as (+)-2-butanol and the other as(-)-2-butanol. Optically pure(+)-2-butanol has a specific rotation [a]D of +13.5 optically pure (-)-2-butanol has an exactly oppo- site specific rotation [a]D of -13.59 7.5 ABSOLUTE AND RELATIVE CONFIGURATION The spatial arrangement of substituents at a stereogenic center is its absolute configu ration. Neither the sign nor the magnitude of rotation by itself can tell us the absolute configuration of a substance. Thus, one of the following structures is(+)-2-butanol and he other is(-)-2-butanol, but without additional information we can't tell which is which In several places throughout CH, CH H the chapter we will use red CH Although no absolute configuration was known for any substance before 1951 organic chemists had experimentally determined the configurations of thousands of com pounds relative to one another(their relative configurations)through chemical inter- conversion. To illustrate, consider (+)-3-buten-2-o1. Hydrogenation of this compound :lds(+)-2-butanol CH3 CHCH=CH2+ H2 CH3CHCH2CH3 Make a molecular model buten-2-ol and the 2-butanol 3-Buten-2-ol Hvdrogen 2-Butanol alb+135° Since hydrogenation of the double bond does not involve any of the bonds to the stereo- genic center, the spatial arrangement of substituents in(+)-3-buten-2-ol must be the same as that of the substituents in (+)-2-butanol. The fact that these two compounds have the same sign of rotation when they have the same relative configuration is established by the hydrogenation experiment; it could not have been predicted in advance of the experiment Sometimes compounds that have the same relative configuration have optical tions of opposite sign. For example, treatment of (-)-2-methyl-1-butanol with hydr bromide converts it to(+)-l-bromo-2-methylbutane of one of the enantiomers of 2. bromo-2-methylbutane formed CH3 CH,CHCH,OH HBr →CHCH2CHCH2Br+H2O from it lethyl-1-butane I-Bromo-2-methylbutane Water aB3-5.8° ab3+4.0° This reaction does not involve any of the bonds to the stereogenic center, and so both the starting alcohol (-)and the product bromide(+) have the same relative configura- Back Forward Main MenuToc Study Guide ToC Student o MHHE Website

It is convenient to distinguish between enantiomers by prefixing the sign of rota￾tion to the name of the substance. For example, we refer to one of the enantiomers of 2-butanol as ()-2-butanol and the other as ()-2-butanol. Optically pure ()-2-butanol has a specific rotation [] D 27 of 13.5°; optically pure ()-2-butanol has an exactly oppo￾site specific rotation [] D 27 of 13.5°. 7.5 ABSOLUTE AND RELATIVE CONFIGURATION The spatial arrangement of substituents at a stereogenic center is its absolute configu￾ration. Neither the sign nor the magnitude of rotation by itself can tell us the absolute configuration of a substance. Thus, one of the following structures is ()-2-butanol and the other is ()-2-butanol, but without additional information we can’t tell which is which. Although no absolute configuration was known for any substance before 1951, organic chemists had experimentally determined the configurations of thousands of com￾pounds relative to one another (their relative configurations) through chemical inter￾conversion. To illustrate, consider ()-3-buten-2-ol. Hydrogenation of this compound yields ()-2-butanol. Since hydrogenation of the double bond does not involve any of the bonds to the stereo￾genic center, the spatial arrangement of substituents in ()-3-buten-2-ol must be the same as that of the substituents in ()-2-butanol. The fact that these two compounds have the same sign of rotation when they have the same relative configuration is established by the hydrogenation experiment; it could not have been predicted in advance of the experiment. Sometimes compounds that have the same relative configuration have optical rota￾tions of opposite sign. For example, treatment of ()-2-methyl-1-butanol with hydrogen bromide converts it to ()-1-bromo-2-methylbutane. This reaction does not involve any of the bonds to the stereogenic center, and so both the starting alcohol () and the product bromide () have the same relative configura￾tion.  2-Methyl-1-butanol []D 25 5.8° CH3CH2CHCH2OH CH3 1-Bromo-2-methylbutane []D 25 4.0° CH3CH2CHCH2Br CH3 Hydrogen bromide HBr  Water H2O  3-Buten-2-ol []D 27 33.2° OH CH3CHCH CH2 2-Butanol []D 27 13.5° OH CH3CHCH2CH3 Hydrogen H2 Pd C H H3C CH3CH2 OH H CH3 CH2CH3 HO C 7.5 Absolute and Relative Configuration 267 In several places throughout the chapter we will use red and blue frames to call at￾tention to structures that are enantiomeric. Make a molecular model of one of the enantiomers of 3- buten-2-ol and the 2-butanol formed from it. Make a molecular model of one of the enantiomers of 2- methyl-1-1-butanol and the 1- bromo-2-methylbutane formed from it. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website

CHAPTER SEVEN Stereochemistry An elaborate network connecting signs of rotation and relative configurations was developed that included the most important compounds of organic and biological chemist When, in 1951, the absolute configuration of a salt of (+)-tartaric acid was determined, the absolute configurations of all the compounds whose configurations had been related to (+)-tartaric acid stood revealed as well. Thus, returning to the pair of 2-butanol enantiomers that introduced this section, their absolute configurations are now known to be as shown CHaO CHCH OH CH (+)-2-Butano (一)2- Butanol PROBLEM 7.6 Does the molecular model shown represent (+)-2-butanol or ()-2-butanol? 7.6 THE CAHN-INGOLD-PRELOG R-S NOTATIONAL SYSTEM Just as it makes sense to have a nomenclature system by which we can specify the con stitution of a molecule in words rather than pictures, so too is it helpful to have one that lets us describe stereochemistry. We have already had some experience with this idea when we distinguished between E and Z stereoisomers of alkenes. In the E-Z system, substituents are ranked by atomic number according to a set of rules devised by R. S. Cahn, Sir Christopher Ingold, and Vladimir Prelog(Section 5.4) Actually, Cahn, Ingold, and Prelog first developed their ranking system to deal with the problem of the absolute configuration at a stereogenic center, and this is the systems major application. Table 7.1 shows how the Cahn-Ingold-Prelog system, called the sequence rules, is used to specify the absolute configuration at the stereogenic center in(+)-2-butanol The January 1994 issue of As outlined in Table 7.1,(+)-2-butanol has the S configuration. Its mirror image is(-)-2-butanol, which has the R configuration cation contains an article that describes how to your hands to assign R and S configurations. CH: CH H H CH,CH3 -OH Ho-C CH (S)-2-Butane (R)-2-Butano Back Forward Main MenuToc Study Guide ToC Student o MHHE Website

An elaborate network connecting signs of rotation and relative configurations was developed that included the most important compounds of organic and biological chemistry. When, in 1951, the absolute configuration of a salt of ()-tartaric acid was determined, the absolute configurations of all the compounds whose configurations had been related to ()-tartaric acid stood revealed as well. Thus, returning to the pair of 2-butanol enantiomers that introduced this section, their absolute configurations are now known to be as shown. PROBLEM 7.6 Does the molecular model shown represent ()-2-butanol or ()-2-butanol? 7.6 THE CAHN–INGOLD–PRELOG R–S NOTATIONAL SYSTEM Just as it makes sense to have a nomenclature system by which we can specify the con￾stitution of a molecule in words rather than pictures, so too is it helpful to have one that lets us describe stereochemistry. We have already had some experience with this idea when we distinguished between E and Z stereoisomers of alkenes. In the E–Z system, substituents are ranked by atomic number according to a set of rules devised by R. S. Cahn, Sir Christopher Ingold, and Vladimir Prelog (Section 5.4). Actually, Cahn, Ingold, and Prelog first developed their ranking system to deal with the problem of the absolute configuration at a stereogenic center, and this is the system’s major application. Table 7.1 shows how the Cahn–Ingold–Prelog system, called the sequence rules, is used to specify the absolute configuration at the stereogenic center in ()-2-butanol. As outlined in Table 7.1, ()-2-butanol has the S configuration. Its mirror image is ()-2-butanol, which has the R configuration. C H H3C CH3CH2 OH (S)-2-Butanol H CH3 CH2CH3 HO C (R)-2-Butanol and C H H3C CH3CH2 OH H CH3 CH2CH3 HO C ()-2-Butanol ()-2-Butanol 268 CHAPTER SEVEN Stereochemistry The January 1994 issue of the Journal of Chemical Edu￾cation contains an article that describes how to use your hands to assign R and S configurations. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website