有机化学（英文原版）：Chapt07

259 CHAPTER 7 STEREOCHEMISTRY T he Greek word stereos means “solid,” and stereochemistry refers to chemistry in three dimensions. The foundations of organic stereochemistry were laid by Jacobus van’t Hoff* and Joseph Achille Le Bel in 1874. Independently of each other, van’t Hoff and Le Bel proposed that the four bonds to carbon were directed toward the cor- ners of a tetrahedron. One consequence of a tetrahedral arrangement of bonds to carbon is that two compounds may be different because the arrangement of their atoms in space is different. Isomers that have the same constitution but differ in the spatial arrangement of their atoms are called stereoisomers. We have already had considerable experience with certain types of stereoisomers—those involving cis and trans substitution patterns in alkenes and in cycloalkanes. Our major objectives in this chapter are to develop a feeling for molecules as three- dimensional objects and to become familiar with stereochemical principles, terms, and notation. A full understanding of organic and biological chemistry requires an awareness of the spatial requirements for interactions between molecules; this chapter provides the basis for that understanding. 7.1 MOLECULAR CHIRALITY: ENANTIOMERS Everything has a mirror image, but not all things are superposable on their mirror images. Mirror-image superposability characterizes many objects we use every day. Cups and saucers, forks and spoons, chairs and beds are all identical with their mirror images. Many other objects though—and this is the more interesting case—are not. Your left hand and your right hand, for example, are mirror images of each other but can’t be made to coin- cide point for point, palm to palm, knuckle to knuckle, in three dimensions. In 1894, William *Van’t Hoff was the recipient of the first Nobel Prize in chemistry in 1901 for his work in chemical dynam- ics and osmotic pressure—two topics far removed from stereochemistry. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Thomson (Lord Kelvin) coined a word for this property. He defined an object as chiral if it is not superposable on its mirror image. Applying Thomson’s term to chemistry, we say that a molecule is chiral if its two mirror-image forms are not superposable in three dimen- sions. The work “chiral” is derived from the Greek word cheir, meaning “hand,” and it is entirely appropriate to speak of the “handedness” of molecules. The opposite of chiral is achiral. A molecule that is superposable on its mirror image is achiral. In organic chemistry, chirality most often occurs in molecules that contain a car- bon that is attached to four different groups. An example is bromochlorofluoromethane (BrClFCH). As shown in Figure 7.1, the two mirror images of bromochlorofluoromethane cannot be superposed on each other. Since the two mirror images of bromochlorofluoromethane are not superposable, BrClFCH is chiral. The two mirror images of bromochlorofluoromethane have the same constitution. That is, the atoms are connected in the same order. But they differ in the arrangement of their atoms in space; they are stereoisomers. Stereoisomers that are related as an object and its nonsuperposable mirror image are classified as enantiomers. The word “enantiomer” describes a particular relationship between two objects. One cannot look at a single mole- cule in isolation and ask if it is an enantiomer any more than one can look at an individual human being and ask, “Is that person a cousin?” Furthermore, just as an object has one, and only one, mirror image, a chiral molecule can have one, and only one, enantiomer. Notice in Figure 7.1c, where the two enantiomers of bromochlorofluoromethane are similarly oriented, that the difference between them corresponds to an interchange of the positions of bromine and chlorine. It will generally be true for species of the type C(w, x, y, z), where w, x, y, and z are different atoms or groups, that an exchange of two of them converts a structure to its enantiomer, but an exchange of three returns the orig- inal structure, albeit in a different orientation. Consider next a molecule such as chlorodifluoromethane (ClF 2 CH), in which two of the atoms attached to carbon are the same. Figure 7.2 on page 262 shows two molecular models of ClF 2 CH drawn so as to be mirror images. As is evident from these drawings, it is a sim- ple matter to merge the two models so that all the atoms match. Since mirror-image repre- sentations of chlorodifluoromethane are superposable on each other, ClF 2 CH is achiral. The surest test for chirality is a careful examination of mirror-image forms for superposability. Working with models provides the best practice in dealing with mole- cules as three-dimensional objects and is strongly recommended. 7.2 THE STEREOGENIC CENTER As we’ve just seen, molecules of the general type x z w C y Cl±C±Br H W W F Bromochlorofluoromethane 260 CHAPTER SEVEN Stereochemistry Bromochlorofluoromethane is a known compound, and samples selectively enriched in each enantiomer have been described in the chemi- cal literature. In 1989 two chemists at Polytechnic Uni- versity (Brooklyn, New York) described a method for the preparation of BrClFCH that is predominantly one enan- tiomer. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website are chiral when w, x, y, and z are different substituents. A tetrahedral carbon atom that bears four different substituents is variously referred to as a chiral center, a chiral car- bon atom, an asymmetric center, or an asymmetric carbon atom. A more modern term is stereogenic center, and that is the term that we’ll use. (Stereocenter is synonymous with stereogenic center.) 7.2 The Stereogenic Center 261 (a) Structures A and B are mirror-image representations of bromochlorofluoromethane (BrClFCH). (b) To test for superposability, reorient B by turning it 180°. (c) Compare A and B. The two do not match. A and B cannot be superposed on each other. Bromochlorofluoromethane is therefore a chiral molecule. The two mirror-image forms are enantiomers of each other. B A AB Br Cl H F Br Cl H F Br Cl H F A Br Cl H F Br Cl H F B Br Cl H F turn 180° FIGURE 7.1 A molecule with four different groups attached to a single carbon is chiral. Its two mirror-image forms are not superposable. An article in the December 1987 issue of the Journal of Chemical Education gives a thorough discussion of molec- ular chirality and some of its past and present terminol- ogy. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Noting the presence of one (but not more than one) stereogenic center in a mole- cule is a simple, rapid way to determine that it is chiral. For example, C-2 is a stereo- genic center in 2-butanol; it bears a hydrogen atom and methyl, ethyl, and hydroxyl groups as its four different substituents. By way of contrast, none of the carbon atoms bear four different groups in the achiral alcohol 2-propanol. PROBLEM 7.1 Examine the following for stereogenic centers: (a) 2-Bromopentane (c) 1-Bromo-2-methylbutane (b) 3-Bromopentane (d) 2-Bromo-2-methylbutane SAMPLE SOLUTION A stereogenic carbon has four different substituents. (a) In 2-bromopentane, C-2 satisfies this requirement. (b) None of the carbons in 3- bromopentane have four different substituents, and so none of its atoms are stereogenic centers. Molecules with stereogenic centers are very common, both as naturally occurring substances and as the products of chemical synthesis. (Carbons that are part of a double bond or a triple bond can’t be stereogenic centers.) 4-Ethyl-4-methyloctane (a chiral alkane) CH 2 CH 3 CH 3 CH 2 CH 2 C CH 3 CH 2 CH 2 CH 2 CH 3 Linalool (a pleasant-smelling oil obtained from orange flowers) OH CHCH 2 CH 2 C(CH 3 ) 2 C CH 3 CH CH 2 H Br CH 2 CH 2 CH 3 CH 3 C 2-Bromopentane H Br CH 2 CH 3 CH 3 CH 2 C 3-Bromopentane 2-Butanol Chiral; four different substituents at C-2 OH CH 3 C H CH 2 CH 3 2-Propanol Achiral; two of the substituents at C-2 are the same OH CH 3 C H CH 3 262 CHAPTER SEVEN Stereochemistry Cl Cl H H F F F F FIGURE 7.2 Mirror- image forms of chlorodifluo- romethane are superposable on each other. Chlorodifluo- romethane is achiral. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A carbon atom in a ring can be a stereogenic center if it bears two different sub- stituents and the path traced around the ring from that carbon in one direction is differ- ent from that traced in the other. The carbon atom that bears the methyl group in 1,2- epoxypropane, for example, is a stereogenic center. The sequence of groups is O±CH 2 as one proceeds clockwise around the ring from that atom, but is CH 2 ±O in the anti- clockwise direction. Similarly, C-4 is a stereogenic center in limonene. PROBLEM 7.2 Identify the stereogenic centers, if any, in (a) 2-Cyclopenten-1-ol and 3-cyclopenten-1-ol (b) 1,1,2-Trimethylcyclobutane and 1,1,3-Trimethylcyclobutane SAMPLE SOLUTION (a) The hydroxyl-bearing carbon in 2-cyclopenten-1-ol is a stereogenic center. There is no stereogenic center in 3-cyclopenten-1-ol, since the sequence of atoms 1 → 2 → 3 → 4 → 5 is equivalent regardless of whether one proceeds clockwise or anticlockwise. Even isotopes qualify as different substituents at a stereogenic center. The stereo- chemistry of biological oxidation of a derivative of ethane that is chiral because of deu- terium (D H11005 2 H) and tritium (T H11005 3 H) atoms at carbon, has been studied and shown to proceed as follows: The stereochemical relationship between the reactant and the product, revealed by the isotopic labeling, shows that oxygen becomes bonded to carbon on the same side from which H is lost. One final, very important point about stereogenic centers. Everything we have said in this section concerns molecules that have one and only one stereogenic cen- ter; molecules with more than one stereogenic center may or may not be chiral. Mol- ecules that have more than one stereogenic center will be discussed in Sections 7.10 through 7.13. C T H D CH 3 C T HO D CH 3 biological oxidation H 4 OH 3 52 1 2-Cyclopenten-1-ol H 4 H11005 3 5 H11005 2 OH 3 H11005 4 2 H11005 5 1 3-Cyclopenten-1-ol (does not have a stereogenic carbon) H 2 C CHCH 3 O 1-2-Epoxypropane (product of epoxidation of propene) CH 3 H 3 C 26 5 4 1 CH 3 CH 2 Limonene (a constituent of lemon oil) 7.2 The Stereogenic Center 263 Examine the molecular models of the two enantiomers of 1,2-epoxypropane on Learn- ing By Modeling and test them for superposability. Back Forward Main Menu TOC Study Guide TOC Student OLC 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 BA (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 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 H9251, 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 H9251. (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 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 H11005 percent enantiomeric excess H11005 percent of one enantiomer H11002 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 (H11001), and rotation in the anticlockwise sense is taken as a negative (H11002) 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 (H11001)-2-butanol, (H11002)-2-butanol, and (H11006)-2-butanol, respectively. The observed rotation H9251 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 [H9251]. Specific rotation is calculated from the observed rotation according to the expression [H9251] H11005 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 [H9251] D 25 H11005H110013.8°. 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 H9251 of a 0.3-g sample of cholesterol in 15 mL of chloroform solution contained in a 10-cm polarimeter tube is H110020.78°. 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 [H9251] D 20 of H1100213°. What fraction of the mixture is nat- ural cholesterol? 100H9251 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 [H9251] H11005 H9251 cl Back Forward Main Menu TOC Study Guide TOC Student OLC 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 (H11001)-2-butanol and the other as (H11002)-2-butanol. Optically pure (H11001)-2-butanol has a specific rotation [H9251] D 27 of H1100113.5°; optically pure (H11002)-2-butanol has an exactly oppo- site specific rotation [H9251] D 27 of H1100213.5°. 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 (H11001)-2-butanol and the other is (H11002)-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 (H11001)-3-buten-2-ol. Hydrogenation of this compound yields (H11001)-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 (H11001)-3-buten-2-ol must be the same as that of the substituents in (H11001)-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 (H11002)-2-methyl-1-butanol with hydrogen bromide converts it to (H11001)-1-bromo-2-methylbutane. This reaction does not involve any of the bonds to the stereogenic center, and so both the starting alcohol (H11002) and the product bromide (H11001) have the same relative configura- tion. H11001 2-Methyl-1-butanol [H9251] D 25 H110025.8° CH 3 CH 2 CHCH 2 OH CH 3 1-Bromo-2-methylbutane [H9251] D 25 H110014.0° CH 3 CH 2 CHCH 2 Br CH 3 Hydrogen bromide HBr H11001 Water H 2 O H11001 3-Buten-2-ol [H9251] D 27 H1100133.2° OH CH 3 CHCH CH 2 2-Butanol [H9251] D 27 H1100113.5° OH CH 3 CHCH 2 CH 3 Hydrogen H 2 Pd C H H 3 C CH 3 CH 2 OH H CH 3 CH 2 CH 3 CHO 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 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 (H11001)-tartaric acid was determined, the absolute configurations of all the compounds whose configurations had been related to (H11001)-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 (H11001)-2-butanol or (H11002)-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 (H11001)-2-butanol. As outlined in Table 7.1, (H11001)-2-butanol has the S configuration. Its mirror image is (H11002)-2-butanol, which has the R configuration. C H H 3 C CH 3 CH 2 OH (S)-2-Butanol H CH 3 CH 2 CH 3 CHO (R)-2-Butanol and C H H 3 C CH 3 CH 2 OH H CH 3 CH 2 CH 3 CHO (H11001)-2-Butanol (H11002)-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 Often, the R or S configuration and the sign of rotation are incorporated into the name of the compound, as in (R)-(H11002)-2-butanol and (S)-(H11001)-2-butanol. PROBLEM 7.7 Assign absolute configurations as R or S to each of the following compounds: (a) (c) (b) (d) (H11001)-3-Buten-2-ol C H HO H 3 C CH CH 2 (H11001)-1-Fluoro-2-methylbutane C H CH 3 CH 2 H 3 C CH 2 F (H11001)-1-Bromo-2-methylbutane C CH 3 CH 3 CH 2 H CH 2 Br (H11001)-2-Methyl-1-butanol C H CH 3 CH 2 H 3 C CH 2 OH 7.6 The Cahn–Ingold–Prelog R–S Notational System 269 TABLE 7.1 Absolute Configuration According to the Cahn–Ingold–Prelog Notational System Step number 1. Identify the substituents at the stereogenic center, and rank them in order of decreasing precedence according to the system described in Section 5.4. Precedence is determined by atomic number, work- ing outward from the point of attachment at the stereogenic center. 2. Orient the molecule so that the lowest ranked sub- stituent points away from you. 4. If the order of decreasing precedence of the three highest ranked substituents appears in a clockwise sense, the absolute configuration is R (Latin rectus, “right,” “correct”). If the order of decreasing prece- dence is anticlockwise, the absolute configuration is S (Latin sinister, “left”). Example In order of decreasing precedence, the four substitu- ents attached to the stereogenic center of 2-butanol are As represented in the wedge-and-dash drawing at the top of this table, the molecule is already appro- priately oriented. Hydrogen is the lowest ranked sub- stituent attached to the stereogenic center and points away from us. The order of decreasing precedence is anticlockwise. The configuration at the stereogenic center is S. 3. Draw the three highest ranked substituents as they appear to you when the molecule is oriented so that the lowest ranked group points away from you. CH 3 CH 2 ± CH 3 ±HO± (highest) H± (lowest) H11022H11022H11022 CH 3 CH 2 OH CH 3 CH 3 CH 2 OH CH 3 (highest) (second highest) (third highest) (H11001)-2-Butanol C H H 3 C CH 3 CH 2 OHGiven that the absolute configuration of (H11001)-2-butanol is Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SAMPLE SOLUTION (a) The highest ranking substituent at the stereogenic cen- ter of 2-methyl-1-butanol is CH 2 OH; the lowest is H. Of the remaining two, ethyl outranks methyl. The lowest ranking substituent (hydrogen) points away from us in the drawing. The three highest ranking groups trace a clockwise path from CH 2 OH → CH 3 CH 2 → CH 3 . This compound therefore has the R configuration. It is (R)-(H11001)-2-methyl-1-butanol. Compounds in which a stereogenic center is part of a ring are handled in an anal- ogous fashion. To determine, for example, whether the configuration of (H11001)-4-methyl- cyclohexene is R or S, treat the right- and left-hand paths around the ring as if they were independent substituents. With the lowest ranked substituent (hydrogen) directed away from us, we see that the order of decreasing sequence rule precedence is clockwise. The absolute configuration is R. PROBLEM 7.8 Draw three-dimensional representations or make molecular mod- els of (a) The R enantiomer of (b) The S enantiomer of SAMPLE SOLUTION (a) The stereogenic center is the one that bears the bromine. In order of decreasing precedence, the substituents attached to the stereogenic center are When the lowest ranked substituent (the methyl group) is away from us, the order of decreasing precedence of the remaining groups must appear in a clockwise sense in the R enantiomer. Br H11022 O C H11022 CH 2 C H11022 CH 3 H F F H 3 CH 3 CBr O is treated as CH 3 H H H (H11001)-4-Methylcyclohexene Lower priority path Higher priority path CH 3 H CH 2 C C H 2 C H 2 C C C H H 3 CCH 2 OH CH 3 CH 2 Order of precedence: CH 2 OH CH 3 CH 2 CH 3 HH11022H11022H11022 270 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Since its introduction in 1956, the Cahn–Ingold–Prelog system has become the standard method of stereochemical notation. 7.7 FISCHER PROJECTIONS Stereochemistry deals with the three-dimensional arrangement of a molecule’s atoms, and we have attempted to show stereochemistry with wedge-and-dash drawings and computer-generated models. It is possible, however, to convey stereochemical informa- tion in an abbreviated form using a method devised by the German chemist Emil Fischer. Let’s return to bromochlorofluoromethane as a simple example of a chiral mole- cule. The two enantiomers of BrClFCH are shown as ball-and-stick models, as wedge- and-dash drawings, and as Fischer projections in Figure 7.6. Fischer projections are always generated the same way: the molecule is oriented so that the vertical bonds at the stereogenic center are directed away from you and the horizontal bonds point toward you. A projection of the bonds onto the page is a cross. The stereogenic carbon lies at the center of the cross but is not explicitly shown. It is customary to orient the molecule so that the carbon chain is vertical with the lowest numbered carbon at the top as shown for the Fischer projection of (R)-2-butanol. The Fischer projection HO H CH 2 CH 3 CH 3 (R)-2-Butanol corresponds to CH 3 CH 2 CH 3 HO C H Br O CH 2 C (R)-2-Bromo-2-methylcyclohexanone Br CH 3 O which leads to the structure 7.7 Fischer Projections 271 Br Cl H C F H H C H (R)-Bromochlorofluoromethane (S)-Bromochlorofluoromethane BrCl F Br Cl F BrCl F Fischer was the foremost or- ganic chemist of the late nineteenth century. He won the 1902 Nobel Prize in chemistry for his pioneering work in carbohydrate and protein chemistry. FIGURE 7.6 Ball-and- stick models (left), wedge- and-dash drawings (center), and Fischer projections (right) of the R and S enan- tiomers of bromochlorofluo- romethane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website When specifying a configuration as R or S, the safest procedure is to convert a Fischer projection to a three-dimensional representation, remembering that the horizontal bonds always point toward you. PROBLEM 7.9 Write Fischer projections for each of the compounds of Prob- lem 7.7. SAMPLE SOLUTION (a) The structure of (R)-(H11001)-2-methyl-1-butanol is shown in the structure that follows at the left. View the structural formula from a position chosen so that the HOCH 2 ±C±CH 2 CH 3 segment is aligned vertically, with the ver- tical bonds pointing away from you. Replace the wedge-and-dash bonds by lines to give the Fischer projection shown at the right. 7.8 PHYSICAL PROPERTIES OF ENANTIOMERS The usual physical properties such as density, melting point, and boiling point are iden- tical within experimental error for both enantiomers of a chiral compound. Enantiomers can have striking differences, however, in properties that depend on the arrangement of atoms in space. Take, for example, the enantiomeric forms of car- vone. (R)-(H11002)-Carvone is the principal component of spearmint oil. Its enantiomer, (S)-(H11001)-carvone, is the principal component of caraway seed oil. The two enantiomers do not smell the same; each has its own characteristic odor. The difference in odor between (R)- and (S)-carvone results from their different behavior toward receptor sites in the nose. It is believed that volatile molecules occupy only those odor receptors that have the proper shape to accommodate them. Because the receptor sites are themselves chiral, one enantiomer may fit one kind of receptor while the other enantiomer fits a different kind. An analogy that can be drawn is to hands and gloves. Your left hand and your right hand are enantiomers. You can place your left hand into a left glove but not into a right one. The receptor (the glove) can accommodate one enantiomer of a chiral object (your hand) but not the other. The term “chiral recognition” refers to the process whereby some chiral receptor or reagent interacts selectively with one of the enantiomers of a chiral molecule. Very high levels of chiral recognition are common in biological processes. (H11002)-Nicotine, for example, is much more toxic than (H11001)-nicotine, and (H11001)-adrenaline is more active in the (R)-(H11002)-Carvone (from spearmint oil) O CH 2 H 3 C CH 3 C (S)-(H11001)-Carvone (from caraway seed oil) O CH 2 H 3 C CH 3 C C H CH 3 CH 2 CH 3 CH 2 OH is the same as which becomes the Fischer projection CH 2 OH CH 2 CH 3 H CCH 3 HCH 3 CH 2 CH 3 CH 2 OH 272 CHAPTER SEVEN Stereochemistry An article entitled “When Drug Molecules Look in the Mirror” in the June 1996 is- sue of the Journal of Chemi- cal Education (pp. 481–484) describes numerous exam- ples of common drugs in which the two enantiomers have different biological properties. Edward Siloac, an under- graduate organic chemistry student at the University of Virginia, published a paper in the June 1999 issue of the Journal of Chemical Educa- tion (pp. 798–799) that de- scribed how to use your hands to translate Fischer projections to R and S configurations. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website constriction of blood vessels than (H11002)-adrenaline. (H11002)-Thyroxine is an amino acid of the thyroid gland, which speeds up metabolism and causes nervousness and loss of weight. Its enantiomer, (H11001)-thyroxine, exhibits none of these effects but is sometimes given to heart patients to lower their cholesterol levels. 7.8 Physical Properties of Enantiomers 273 CHIRAL DRUGS A recent estimate places the number of prescrip- tion and over-the-counter drugs marketed throughout the world at about 2000. Approx- imately one-third of these are either naturally occur- ring substances themselves or are prepared by chemi- cal modification of natural products. Most of the drugs derived from natural sources are chiral and are almost always obtained as a single enantiomer rather than as a racemic mixture. Not so with the over 500 chiral substances represented among the more than 1300 drugs that are the products of synthetic organic chemistry. Until recently, such substances were, with few exceptions, prepared, sold, and administered as racemic mixtures even though the desired therapeutic activity resided in only one of the enantiomers. Spurred by a number of factors ranging from safety and efficacy to synthetic methodology and econom- ics, this practice is undergoing rapid change as more and more chiral synthetic drugs become available in enantiomerically pure form. Because of the high degree of chiral recogni- tion inherent in most biological processes (Section 7.8), it is unlikely that both enantiomers of a chiral drug will exhibit the same level, or even the same kind, of effect. At one extreme, one enantiomer has the desired effect, and the other exhibits no biologi- cal activity at all. In this case, which is relatively rare, the racemic form is simply a drug that is 50% pure and contains 50% “inert ingredients.” Real cases are more complicated. For example, it is the S enan- tiomer that is responsible for the pain-relieving prop- erties of ibuprofen, normally sold as a racemic mix- ture. The 50% of racemic ibuprofen that is the R enantiomer is not completely wasted, however, be- cause enzyme-catalyzed reactions in our body con- vert much of it to active (S)-ibuprofen. O CHCOH(CH 3 ) 2 CHCH 2 CH 3 Ibuprofen A much more serious drawback to using chiral drugs as racemic mixtures is illustrated by thalidomide, briefly employed as a sedative and antinausea drug in Europe and Great Britain during the period 1959–1962. The desired properties are those of (R)- thalidomide. (S)-Thalidomide, however, has a very different spectrum of biological activity and was shown to be responsible for over 2000 cases of seri- ous birth defects in children born to women who took it while pregnant. Basic research directed toward understanding the factors that control the stereochemistry of chem- ical reactions has led to new synthetic methods that make it practical to prepare chiral molecules in enan- tiomerically pure form. Recognizing this, most major pharmaceutical companies are examining their exist- ing drugs to see which ones are the best candidates for synthesis as single enantiomers and, when prepar- ing a new drug, design its synthesis so as to provide only the desired enantiomer. In 1992, the United States Food and Drug Administration (FDA) issued guidelines that encouraged such an approach, but left open the door for approval of new drugs as racemic mixtures when special circumstances war- rant. One incentive to developing enantiomerically pure versions of existing drugs is that the novel pro- duction methods they require may make them eligi- ble for patent protection separate from that of the original drugs. Thus the temporary monopoly posi- tion that patent law views as essential to fostering in- novation can be extended by transforming a success- ful chiral, but racemic, drug into an enantiomerically pure version. N O O O N H O Thalidomide Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.9 REACTIONS THAT CREATE A STEREOGENIC CENTER Many of the reactions we’ve already encountered can yield a chiral product from an achi- ral starting material. Epoxidation of propene, for example, creates a stereogenic center by addition of oxygen to the double bond. In this, as in other reactions in which achiral reactants yield chiral products, the product is formed as a racemic mixture and is optically inactive. Remember, for a substance to be optically active, not only must it be chiral but one enantiomer must be present in excess of the other. Figure 7.7 shows why equal amounts of (R)- and (S)-1,2-epoxypropane are formed in this reaction. The peroxy acid is just as likely to transfer oxygen to one face of the double bond as the other, the rates of formation of the R and S enantiomers of the prod- uct are the same and a racemic mixture of the two results. CH 3 CH CH 2 Propene (achiral) CH 3 CH O CH 2 1,2-Epoxypropane (chiral) CH 3 CO 2 OH 274 CHAPTER SEVEN Stereochemistry Nicotine N N CH 3 Adrenaline (Can you find the stereogenic center in each of these?) OH HO HOCHCH 2 NHCH 3 Thyroxine I I HO O I I CH 2 CHCO 2 H11002 NH 3 H11001 50% 50% FIGURE 7.7 Epoxida- tion of propene produces equal amounts of (R)- and (S)-1,2-epoxypropane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website It is often helpful, especially in a multistep reaction, to focus on the step that cre- ates the stereogenic center. In the ionic addition of hydrogen bromide to 2-butene, for example, the stereogenic center is generated when bromide ion attacks sec-butyl cation. As seen in Figure 7.8, the bonds to the positively charged carbon are coplanar and define a plane of symmetry in the carbocation, which is achiral. The rates at which bromide ion attacks the carbocation at its two mirror-image faces are equal, and the product, 2-bromobutane, although chiral, is optically inactive because it is formed as a racemic mixture. It is a general principle that optically active products cannot be formed when opti- cally inactive substrates react with optically inactive reagents. This principle holds irre- spective of whether the addition is syn or anti, concerted or stepwise. No matter how many steps are involved in a reaction, if the reactants are achiral, formation of one enan- tiomer is just as likely as the other, and a racemic mixture results. When a reactant is chiral but optically inactive because it is racemic, any products derived from its reactions with optically inactive reagents will be optically inactive. For example, 2-butanol is chiral and may be converted with hydrogen bromide to 2- bromobutane, which is also chiral. If racemic 2-butanol is used, each enantiomer will react at the same rate with the achiral reagent. Whatever happens to (R)-(H11002)-2-butanol is mirrored in a corresponding reaction of (S)-(H11001)-2-butanol, and a racemic, optically inactive product results. CH 3 CH CHCH 3 (E)- or (Z)-2-butene (achiral) HBr 2-Bromobutane (chiral) CH 3 CHCH 2 CH 3 Br sec-Butyl cation (achiral) CH 3 CHCH 2 CH 3 H11001 via 7.9 Reactions That Create a Stereogenic Center 275 CH 3 CH?CHCH 3 H H11001 Br H11002 (S)-(H11001)-2-Bromobutane [ ] D H1100139H11034 (R)-(H11002)-2-Bromobutane [ ] D H1100239H11034 H11001 (50%) (50%) FIGURE 7.8 Elec- trophilic addition of hydro- gen bromide to (E) and (Z )-2-butene proceeds by way of an achiral carboca- tion, which leads to equal quantities of (R)- and (S)-2- bromobutane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Optically inactive starting materials can give optically active products if they are treated with an optically active reagent or if the reaction is catalyzed by an optically active sub- stance. The best examples are found in biochemical processes. Most biochemical reac- tions are catalyzed by enzymes. Enzymes are chiral and enantiomerically homogeneous; they provide an asymmetric environment in which chemical reaction can take place. Ordinarily, enzyme-catalyzed reactions occur with such a high level of stereoselectivity that one enantiomer of a substance is formed exclusively even when the substrate is achi- ral. The enzyme fumarase, for example, catalyzes the hydration of fumaric acid to malic acid in apples and other fruits. Only the S enantiomer of malic acid is formed in this reaction. The reaction is reversible, and its stereochemical requirements are so pronounced that neither the cis isomer of fumaric acid (maleic acid) nor the R enantiomer of malic acid can serve as a substrate for the fumarase-catalyzed hydration–dehydration equilibrium. PROBLEM 7.10 Biological reduction of pyruvic acid, catalyzed by the enzyme lactate dehydrogenase, gives (H11001)-lactic acid, represented by the Fischer projection shown. What is the configuration of (H11001)-lactic acid according to the Cahn–Ingold–Prelog R–S notational system? Making a molecular model of the Fis- cher projection will help. We’ll continue with the three-dimensional details of chemical reactions later in this chapter. First though, we need to develop some additional stereochemical principles con- cerning structures with more than one stereogenic center. 7.10 CHIRAL MOLECULES WITH TWO STEREOGENIC CENTERS When a molecule contains two stereogenic centers, as does 2,3-dihydroxybutanoic acid, how many stereoisomers are possible? CH 3 CHCHC HO OH OH O 3421 2,3-Dihydroxybutanoic acid O CH 3 CCO 2 H Pyruvic acid HO H CH 3 CO 2 H (H11001)-Lactic acid biological reduction H11001 HO 2 CH CO 2 HH CC Fumaric acid H 2 O C H HO 2 CCH 2 HO 2 C OH (S)-(H11002)-Malic acid fumarase HBr 2-Butanol (chiral but racemic) (H11006)-CH 3 CHCH 2 CH 3 OH 2-Bromobutane (chiral but racemic) (H11006)-CH 3 CHCH 2 CH 3 Br 276 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website We can use straightforward reasoning to come up with the answer. The absolute config- uration at C-2 may be R or S. Likewise, C-3 may have either the R or the S configura- tion. The four possible combinations of these two stereogenic centers are Figure 7.9 presents structural formulas for these four stereoisomers. Stereoisomers I and II are enantiomers of each other; the enantiomer of (R,R) is (S,S). Likewise stereoiso- mers III and IV are enantiomers of each other, the enantiomer of (R,S) being (S,R). Stereoisomer I is not a mirror image of III or IV, so is not an enantiomer of either one. Stereoisomers that are not related as an object and its mirror image are called diastereomers; diastereomers are stereoisomers that are not enantiomers. Thus, stereoisomer I is a diastereomer of III and a diastereomer of IV. Similarly, II is a diaste- reomer of III and IV. To convert a molecule with two stereogenic centers to its enantiomer, the config- uration at both centers must be changed. Reversing the configuration at only one stereo- genic center converts it to a diastereomeric structure. (2R,3R) (stereoisomer I) (2R,3S) (stereoisomer III) (2S,3S) (stereoisomer II) (2S,3R) (stereoisomer IV) 7.10 Chiral Molecules with Two Stereogenic Centers 277 (2R,3R) : [ ] D H110029.5H11034 CH 3 CO 2 H H I OH H HO CH 3 CO 2 H OH II H HO H 3 23 2 CH 3 CO 2 H H III H HO HO 3 2 CH 3 CO 2 H OH IV OH H H 3 2 Diastereomers (2S,3S) : [ ] D H110019.5H11034 (2R,3S) : [ ] D H1100117.8H11034 (2S,3R) : [ ] D H1100217.8H11034 Diastereomers Diastereomers Enantiomers Enantiomers FIGURE 7.9 Stereoisomeric 2,3-dihydroxybutanoic acids. Stereoisomers I and II are enan- tiomers. Stereoisomers III and IV are enantiomers. All other relationships are diastereomeric (see text). Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Enantiomers must have equal and opposite specific rotations. Diastereomeric sub- stances can have different rotations, with respect to both sign and magnitude. Thus, as Figure 7.9 shows, the (2R,3R) and (2S,3S) enantiomers (I and II) have specific rotations that are equal in magnitude but opposite in sign. The (2R,3S) and (2S,3R) enantiomers (III and IV) likewise have specific rotations that are equal to each other but opposite in sign. The magnitudes of rotation of I and II are different, however, from those of their diastereomers III and IV. In writing Fischer projections of molecules with two stereogenic centers, the mol- ecule is arranged in an eclipsed conformation for projection onto the page, as shown in Figure 7.10. Again, horizontal lines in the projection represent bonds coming toward you; vertical bonds point away. Organic chemists use an informal nomenclature system based on Fischer projec- tions to distinguish between diastereomers. When the carbon chain is vertical and like substituents are on the same side of the Fischer projection, the molecule is described as the erythro diastereomer. When like substituents are on opposite sides of the Fischer projection, the molecule is described as the threo diastereomer. Thus, as seen in the Fis- cher projections of the stereoisomeric 2,3-dihydroxybutanoic acids, compounds I and II are erythro stereoisomers and III and IV are threo. Because diastereomers are not mirror images of each other, they can have quite different physical and chemical properties. For example, the (2R,3R) stereoisomer of 3-amino-2-butanol is a liquid, but the (2R,3S) diastereomer is a crystalline solid. I erythro H OH H OH CO 2 H CH 3 HO H HO H CO 2 H CH 3 II erythro H OH HO H CO 2 H CH 3 III threo HO H H OH CO 2 H CH 3 IV threo 278 CHAPTER SEVEN Stereochemistry (a)(b)(c) HO H H OH CO 2 H CH 3 H H OH OH CO 2 H CH 3 H H OH OH CO 2 H CH 3 2 3 FIGURE 7.10 Representations of (2R,3R)-dihydroxybutanoic acid. (a) The staggered confor- mation is the most stable but is not properly arranged to show stereochemistry according to the Fischer projection method. (b) Rotation about the C-2±C-3 bond gives the eclipsed conformation, and projection of the eclipsed conformation onto the page gives (c) a correct Fischer projection. Erythro and threo describe the relative configuration (Section 7.5) of two stereo- genic centers within a single molecule. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 7.11 Draw Fischer projections or make molecular models of the four stereoisomeric 3-amino-2-butanols, and label each erythro or threo as appropriate. PROBLEM 7.12 One other stereoisomer of 3-amino-2-butanol is a crystalline solid. Which one? The situation is the same when the two stereogenic centers are present in a ring. There are four stereoisomeric 1-bromo-2-chlorocyclopropanes: a pair of enantiomers in which the halogens are trans and a pair in which they are cis. The cis compounds are diastereomers of the trans. 7.11 ACHIRAL MOLECULES WITH TWO STEREOGENIC CENTERS Now think about a molecule, such as 2,3-butanediol, which has two stereogenic centers that are equivalently substituted. Only three, not four, stereoisomeric 2,3-butanediols are possible. These three are shown in Figure 7.11. The (2R,3R) and (2S,3S) forms are enantiomers of each other and have equal and opposite optical rotations. A third combination of stereogenic centers, (2R,3S), however, gives an achiral structure that is superposable on its (2S,3R) mirror image. Because it is achiral, this third stereoisomer is optically inactive. We call achiral mole- cules that have stereogenic centers meso forms. The meso form in Figure 7.11 is known as meso-2,3-butanediol. CH 3 CHCHCH 3 HO OH 2,3-Butanediol Enantiomers Enantiomers H Br Cl H RR (1R,2R)-1-Bromo-2-chlorocyclopropane (1S,2S)-1-Bromo-2-chlorocyclopropane Cl H H Br SS H Br H Cl SR (1R,2S)-1-Bromo-2-chlorocyclopropane H Cl H Br SR (1S,2R)-1-Bromo-2-chlorocyclopropane H 2 N H 3 C H CH 3 H HO (2R,3R)-3-Amino-2-butanol (liquid) NH 2 H H 3 C CH 3 H HO (2R,3S)-3-Amino-2-butanol (solid, mp 49°C) 7.11 Achiral Molecules with Two Stereogenic Centers 279 A molecule framed in green is a diastereomer of one framed in red or blue. A molecule framed in black is an enantiomer of a green- framed one. Both are di- astereomers of their red or blue-framed stereoisomers. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website One way to demonstrate that meso-2,3-butanediol is achiral is to recognize that its eclipsed conformation has a plane of symmetry that passes through and is perpendicu- lar to the C-2±C-3 bond, as illustrated in Figure 7.12a. The anti conformation is achi- ral as well. As Figure 7.12b shows, this conformation is characterized by a center of symmetry at the midpoint of the C-2±C-3 bond. Fischer projection formulas can help us identify meso forms. Of the three stereoiso- meric 2,3-butanediols, notice that only in the meso stereoisomer does a dashed line through the center of the Fischer projection divide the molecule into two mirror-image halves. When using Fischer projections for this purpose, however, be sure to remember what three-dimensional objects they stand for. One should not, for example, test for superpo- sition of the two chiral stereoisomers by a procedure that involves moving any part of a Fischer projection out of the plane of the paper in any step. H OH HO H CH 3 CH 3 (2S,3S)-2,3-Butanediol HO H H OH CH 3 CH 3 (2R,3R)-2,3-Butanediol meso-2,3-Butanediol H OH H OH CH 3 CH 3 280 CHAPTER SEVEN Stereochemistry (a)(b)(c) (2R,3R)-2,3-Butanediol (2S,3S)-2,3-Butanediol meso-2,3-Butanediol Center of symmetry Plane of symmetry (a)(b) FIGURE 7.11 Stereo- isomeric 2,3-butanediols shown in their eclipsed con- formations for convenience. Stereoisomers (a) and (b) are enantiomers of each other. Structure (c) is a diastereo- mer of (a) and (b), and is achiral. It is called meso-2,3- butanediol. FIGURE 7.12 (a) The eclipsed conformation of meso-2,3-butanediol has a plane of symmetry. (b) The anti conformation of meso- 2,3-butanediol has a center of symmetry. In the same way that a Fischer formula is a projec- tion of the eclipsed confor- mation onto the page, the line drawn through its center is a projection of the plane of symmetry which is present in the eclipsed conformation of meso-2,3-butanediol. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 7.13 A meso stereoisomer is possible for one of the following com- pounds. Which one? 2,3-Dibromopentane; 2,4-dibromopentane; 3-bromo-2-pentanol; 4-bromo-2-pentanol Turning to cyclic compounds, we see that there are three, not four, stereoisomeric 1,2-dibromocyclopropanes. Of these, two are enantiomeric trans-1,2-dibromocyclo- propanes. The cis diastereomer is a meso form; it has a plane of symmetry. 7.11 Achiral Molecules with Two Stereogenic Centers 281 CHIRALITY OF DISUBSTITUTED CYCLOHEXANES D isubstituted cyclohexanes present us with a challenging exercise in stereochemistry. Con- sider the seven possible dichlorocyclohexanes: 1,1-; cis- and trans-1,2-; cis- and trans-1,3-; and cis- and trans-1,4-. Which are chiral? Which are achiral? Four isomers—the ones that are achiral because they have a plane of symmetry—are relatively easy to identify: ACHIRAL DICHLOROCYCLOHEXANES The remaining three isomers are chiral: CHIRAL DICHLOROCYCLOHEXANES Cl 1 Cl H H 2 cis-1,2 1 Cl Cl H H 2 trans-1,2 trans-1,3 H Cl 1 3 H Cl Cl Cl 4 1 1,1 (plane of symmetry through C-1 and C-4) H Cl Cl H 1 4 cis-1,4 (plane of symmetry through C-1 and C-4) Cl H H Cl 5 1 2 3 cis-1,3 (plane of symmetry through C-2 and C-5) Cl Cl H H 1 4 trans-1,4 (plane of symmetry through C-1 and C-4) Among all the isomers, cis-1,2-dichlorocyclo- hexane is unique in that the ring-flipping process typ- ical of cyclohexane derivatives (Section 3.8) converts it to its enantiomer. Structures A and AH11032 are nonsuperposable mirror im- ages of each other. Thus although cis-1,2-dichlorocy- clohexane is chiral, it is optically inactive when chair–chair interconversion occurs. Such interconver- sion is rapid at room temperature and converts opti- cally active A to a racemic mixture of A and AH11032. Since A and AH11032 are enantiomers interconvertible by a con- formational change, they are sometimes referred to as conformational enantiomers. The same kind of spontaneous racemization oc- curs for any cis-1,2 disubstituted cyclohexane in which both substituents are the same. Since such compounds are chiral, it is incorrect to speak of them as meso compounds, which are achiral by definition. Rapid chair–chair interconversion, however, converts them to a 1:1 mixture of enantiomers, and this mix- ture is optically inactive. Cl Cl H H A which is equivalent to Cl H Cl H AH11032 AH11032 H Cl H Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 7.14 One of the stereoisomers of 1,3-dimethylcyclohexane is a meso form. Which one? 7.12 MOLECULES WITH MULTIPLE STEREOGENIC CENTERS Many naturally occurring compounds contain several stereogenic centers. By an analy- sis similar to that described for the case of two stereogenic centers, it can be shown that the maximum number of stereoisomers for a particular constitution is 2 n , where n is equal to the number of stereogenic centers. PROBLEM 7.15 Using R and S descriptors, write all the possible combinations for a molecule with three stereogenic centers. When two or more of a molecule’s stereogenic centers are equivalently substituted, meso forms are possible, and the number of stereoisomers is then less than 2 n . Thus, 2 n represents the maximum number of stereoisomers for a molecule containing n stereogenic centers. The best examples of substances with multiple stereogenic centers are the carbo- hydrates (Chapter 25). One class of carbohydrates, called hexoses, has the constitution Since there are four stereogenic centers and no possibility of meso forms, there are 2 4 , or 16, stereoisomeric hexoses. All 16 are known, having been isolated either as natural products or as the products of chemical synthesis. PROBLEM 7.16 A second category of six-carbon carbohydrates, called 2-hexu- loses, has the constitution shown. How many stereoisomeric 2-hexuloses are pos- sible? Steroids are another class of natural products with multiple stereogenic centers. One such compound is cholic acid, which can be obtained from bile. Its structural for- mula is given in Figure 7.13. Cholic acid has 11 stereogenic centers, and so there are a total (including cholic acid) of 2 11 , or 2048, stereoisomers that have this constitution. Of A 2-hexulose O OH HOCH 2 CCH OH CH OH CHCH 2 OH A hexose OH H O HOCH 2 CH OH CH OH CH OH CH C 282 CHAPTER SEVEN Stereochemistry H Br Br H RR (1R,2R)-1,2-Dibromocyclopropane (1S,2S)-Dibromocyclopropane Br H H Br SS H Br H Br SR meso-1,2-Dibromocyclopropane Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website these 2048 stereoisomers, how many are diastereomers of cholic acid? Remember! Diastereomers are stereoisomers that are not enantiomers, and any object can have only one mirror image. Therefore, of the 2048 stereoisomers, one is cholic acid, one is its enantiomer, and the other 2046 are diastereomers of cholic acid. Only a small fraction of these compounds are known, and (H11001)-cholic acid is the only one ever isolated from natural sources. Eleven stereogenic centers may seem like a lot, but it is nowhere close to a world record. It is a modest number when compared with the more than 100 stereogenic cen- ters typical for most small proteins and the thousands of stereogenic centers that are pres- ent in nucleic acids. A molecule that contains both stereogenic centers and double bonds has additional opportunities for stereoisomerism. For example, the configuration of the stereogenic cen- ter in 3-penten-2-ol may be either R or S, and the double bond may be either E or Z. There are therefore four stereoisomers of 3-penten-2-ol even though it has only one ste- reogenic center. The relationship of the (2R,3E) stereoisomer to the others is that it is the enantiomer of (2S,3E)-3-penten-2-ol and is a diastereomer of the (2R,3Z) and (2S,3Z) isomers. H 3 C H HO H C H CH 3 CC (2R,3E)-3-Penten-2-ol (2R,3Z)-3-Penten-2-ol H H 3 C HO H C H CH 3 CC H 3 C H H H C OH CH 3 CC (2S,3E)-3-Penten-2-ol (2S,3Z)-3-Penten-2-ol H H 3 C H C CH 3 OH H CC 7.12 Molecules with Multiple Stereogenic Centers 283 HO OH H H H H HO CH 3 CH 2 CH 2 CO 2 H CH 3H CH 3 FIGURE 7.13 The structure of cholic acid. Its 11 stereogenic centers are those carbons at which stereochemistry is indicated in the diagram. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.13 REACTIONS THAT PRODUCE DIASTEREOMERS Once we grasp the idea of stereoisomerism in molecules with two or more stereogenic centers, we can explore further details of addition reactions of alkenes. When bromine adds to (Z)- or (E)-2-butene, the product 2,3-dibromobutane con- tains two equivalently substituted stereogenic centers: Three stereoisomers are possible: a pair of enantiomers and a meso form. Two factors combine to determine which stereoisomers are actually formed in the reaction. 1. The (E)- or (Z)-configuration of the starting alkene 2. The anti stereochemistry of addition Figures 7.14 and 7.15 depict the stereochemical relationships associated with anti addition of bromine to (E)- and (Z)-2-butene, respectively. The trans alkene (E)-2-butene yields only meso-2,3-dibromobutane, but the cis alkene (Z)-2-butene gives a racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane. Bromine addition to alkenes is an example of a stereospecific reaction. A stereo- specific reaction is one in which stereoisomeric starting materials yield products that are stereoisomers of each other. In this case the starting materials, in separate reactions, are the E and Z stereoisomers of 2-butene. The chiral dibromides from (Z)-2-butene are stereoisomers (diastereomers) of the meso dibromide formed from (E)-2-butene. Notice further that, consistent with the principle developed in Section 7.9, opti- cally inactive starting materials (achiral alkenes and bromine) yield optically inactive products (a racemic mixture or a meso structure) in these reactions. (Z)- or (E)-2-butene CH 3 CH CHCH 3 2,3-Dibromobutane Br Br CH 3 CHCHCH 3 Br 2 284 CHAPTER SEVEN Stereochemistry CH 3 meso CH 3 HBr H Br S R Br 2 50% CH 3 CH 3 H H E Br 2 50% CH 3 meso CH 3 BrH Br H R S FIGURE 7.14 Anti ad- dition of Br 2 to (E)-2-butene gives meso-2,3-dibromobu- tane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website PROBLEM 7.17 Epoxidation of alkenes is a stereospecific syn addition. Which stereoisomer of 2-butene reacts with peroxyacetic acid to give meso-2,3-epoxybu- tane? Which one gives a racemic mixture of (2R,3R)- and (2S,3S)-2,3-epoxybutane? A reaction that introduces a second stereogenic center into a starting material that already has one need not produce equal quantities of two possible diastereomers. Con- sider catalytic hydrogenation of 2-methyl(methylene)cyclohexane. As you might expect, both cis- and trans-1,2-dimethylcyclohexane are formed. The relative amounts of the two products, however, are not equal; more cis-1,2-dimethyl- cyclohexane is formed than trans. The reason for this is that it is the less hindered face of the double bond that approaches the catalyst surface and is the face to which hydro- gen is transferred. Hydrogenation of 2-methyl(methylene)cyclohexane occurs preferen- tially at the side of the double bond opposite that of the methyl group and leads to a faster rate of formation of the cis stereoisomer of the product. PROBLEM 7.18 Could the fact that hydrogenation of 2-methyl(methylene)cyclo- hexane gives more cis-1,2-dimethylcyclohexane than trans- be explained on the basis of the relative stabilities of the two stereoisomeric products? The hydrogenation of 2-methyl(methylene)cyclohexane is an example of a stereo- selective reaction, meaning one in which stereoisomeric products are formed in unequal amounts from a single starting material (Section 5.11). H11001 2-Methyl(methylene)cyclo- hexane CH 3 CH 2 H H 2 , Pt acetic acid cis-1,2-Dimethylcyclo- hexane (68%) CH 3 CH 3 H H trans-1,2-Dimethylcyclo- hexane (32%) CH 3 H H CH 3 7.13 Reactions That Produce Diastereomers 285 CH 3 2R, 3R H CH 3 Br H Br R R Br 2 50% CH 3 H CH 3 H Z Br 2 50% CH 3 H BrH 3 C Br H S S 2S, 3S Make molecular models of the reactant and both prod- ucts shown in the equation. FIGURE 7.15 Anti ad- dition of Br 2 to (Z)-2-butene gives a racemic mixture of (2R,3R)- and (2S,3S)-2,3-di- bromobutane. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A common misconception is that a stereospecific reaction is simply one that is 100% stereoselective. The two terms though have precise definitions that are indepen- dent of one another. A stereospecific reaction is one which, when carried out with stereoisomeric starting materials, gives a product from one reactant that is a stereoiso- mer of the product from the other. A stereoselective reaction is one in which a single starting material gives a predominance of a single stereoisomer when two or more are possible. Stereospecific is more closely connected with features of the reaction than with the reactant. Thus terms such as syn addition and anti elimination describe the stereo- specificity of reactions. Stereoselective is more closely connected with structural effects in the reactant as expressed in terms such as addition to the less hindered side. A stereo- specific reaction can also be stereoselective. For example, syn addition describes stereo- specificity in the catalytic hydrogenation of alkenes, whereas the preference for addition to the less hindered face of the double bond describes stereoselectivity. 7.14 RESOLUTION OF ENANTIOMERS The separation of a racemic mixture into its enantiomeric components is termed resolution. The first resolution, that of tartaric acid, was carried out by Louis Pasteur in 1848. Tartaric acid is a byproduct of wine making and is almost always found as its dex- trorotatory 2R,3R stereoisomer, shown here in a perspective drawing and in a Fischer projection. PROBLEM 7.19 There are two other stereoisomeric tartaric acids. Write their Fis- cher projections, and specify the configuration at their stereogenic centers. Occasionally, an optically inactive sample of tartaric acid was obtained. Pasteur noticed that the sodium ammonium salt of optically inactive tartaric acid was a mixture of two mirror-image crystal forms. With microscope and tweezers, Pasteur carefully sep- arated the two. He found that one kind of crystal (in aqueous solution) was dextrorota- tory, whereas the mirror-image crystals rotated the plane of polarized light an equal amount but were levorotatory. Although Pasteur was unable to provide a structural explanation—that had to wait for van’t Hoff and Le Bel a quarter of a century later—he correctly deduced that the enantiomeric quality of the crystals was the result of enantiomeric molecules. The rare form of tartaric acid was optically inactive because it contained equal amounts of (H11001)- tartaric acid and (H11002)-tartaric acid. It had earlier been called racemic acid (from Latin racemus, “a bunch of grapes”), a name that subsequently gave rise to our present term for an equal mixture of enantiomers. PROBLEM 7.20 Could the unusual, optically inactive form of tartaric acid stud- ied by Pasteur have been meso-tartaric acid? Pasteur’s technique of separating enantiomers not only is laborious but requires that the crystal habits of enantiomers be distinguishable. This happens very rarely. H OH HO H CO 2 H CO 2 H H HO 2 C OH CO 2 H OH H (2R,3R)-Tartaric acid (mp 170°C, [H9251] D H1100112°) 286 CHAPTER SEVEN Stereochemistry Note that the terms regiose- lective and regiospecific, however, are defined in terms of each other. A re- giospecific reaction is one that is 100% regioselective. A description of Pasteur’s work, as part of a broader discussion concerning crystal structure, can be found in the article “Molecules, Crys- tals, and Chirality” in the July 1997 issue of the Journal of Chemical Education, pp. 800–806. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Consequently, alternative and more general approaches for resolving enantiomers have been developed. Most are based on a strategy of temporarily converting the enantiomers of a racemic mixture to diastereomeric derivatives, separating these diastereomers, then regenerating the enantiomeric starting materials. Figure 7.16 illustrates this strategy. Say we have a mixture of enantiomers, which, for simplicity, we label as C(H11001) and C(H11002). Assume that C(H11001) and C(H11002) bear some func- tional group that can combine with a reagent P to yield adducts C(H11001)-P and C(H11002)-P. Now, if reagent P is chiral, and if only a single enantiomer of P, say, P(H11001), is added to a racemic mixture of C(H11001) and C(H11002), as shown in the first step of Figure 7.16, then the products of the reaction are C(H11001)-P(H11001) and C(H11002)-P(H11001). These products are not mirror images; they are diastereomers. Diastereomers can have different physical properties, which can serve as a means of separating them. The mixture of diastereomers is sepa- rated, usually by recrystallization from a suitable solvent. In the last step, an appropri- ate chemical transformation liberates the enantiomers and restores the resolving agent. Whenever possible, the chemical reactions involved in the formation of diastereo- mers and their conversion to separate enantiomers are simple acid–base reactions. For example, naturally occurring (S)-(H11002)-malic acid is often used to resolve amines. One such amine that has been resolved in this way is 1-phenylethylamine. Amines are bases, and malic acid is an acid. Proton transfer from (S)-(H11002)-malic acid to a racemic mixture of (R)- and (S)-1-phenylethylamine gives a mixture of diastereomeric salts. 7.14 Resolution of Enantiomers 287 C(+) C(–) C(+)-P(+) C(–)-P(+) 2P(+) Mixture of enantiomers Resolving agent (single enantiomer) Mixture of diastereomers C(+)-P(+) C(–)-P(+) Separate diastereomers Dissociate diastereomer to single enantiomer; recover resolving agent Dissociate diastereomer to single enantiomer; recover resolving agent C(+) C(–) + + P(+) P(+) FIGURE 7.16 The general procedure followed in resolving a chiral substance into its enan- tiomers. Reaction with a single enantiomer of a chiral resolving agent P(H11001) converts the racemic mixture of enantiomers C(H11001) and C(H11002) to a mixture of diastereomers C(H11001)-P(H11001) and C(H11002)-P(H11001). The mixture of diastereomers is separated—by fractional crystallization, for example. A chem- ical reaction is then carried out to convert diastereomer C(H11001)-P(H11001) to C(H11001) and the resolving agent P(H11001). Likewise, diastereomer C(H11002)-P(H11001) is converted to C(H11002) and P(H11001). C(H11001) has been sep- arated from C(H11002), and the resolving agent P(H11001) can be recovered for further use. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The diastereomeric salts are separated and the individual enantiomers of the amine lib- erated by treatment with a base: PROBLEM 7.21 In the resolution of 1-phenylethylamine using (H11002)-malic acid, the compound obtained by recrystallization of the mixture of diastereomeric salts is (R)-1-phenylethylammonium (S)-malate. The other component of the mixture is more soluble and remains in solution. What is the configuration of the more sol- uble salt? This method is widely used for the resolution of chiral amines and carboxylic acids. Analogous methods based on the formation and separation of diastereomers have been developed for other functional groups; the precise approach depends on the kind of chem- ical reactivity associated with the functional groups present in the molecule. The rapidly increasing demand for enantiomerically pure starting materials and intermediates in the pharmaceutical industry (see the boxed essay entitled Chiral Drugs in this chapter) has increased interest in developing methods for resolving racemic mixtures. 7.15 STEREOREGULAR POLYMERS Before the development of the Ziegler–Natta catalyst systems (Section 6.21), polymer- ization of propene was not a reaction of much value. The reason for this has a stereo- chemical basis. Consider a section of polypropylene: Representation of the polymer chain in an extended zigzag conformation, as shown in Figure 7.17, reveals several distinct structural possibilities differing with respect to the relative configurations of the carbons that bear the methyl groups. One structure, represented in Figure 7.17a, has all the methyl groups oriented in the same direction with respect to the polymer chain. This stereochemical arrangement is said to be isotactic. Another form, shown in Figure 7.17b, has its methyl groups alter- nating front and back along the chain. This arrangement is described as syndiotactic. CH 3 CH 2 CHCH 2 CHCH 2 CHCH 2 CHCH 2 CHCH 2 CH CH 3 CH 3 CH 3 CH 3 CH 3 H20898H20899 H11001 Hydroxide 2OH H11002 1-Phenylethylammonium (S)-malate (a single diastereomer) H11002 O 2 CCH 2 CHCO 2 H OH C 6 H 5 CHNH 3 CH 3 H11001 H11001H11001 1-Phenylethylamine (a single enantiomer) C 6 H 5 CHNH 2 CH 3 (S)-(H11002)-Malic acid (recovered resolving agent) H11002 O 2 CCH 2 CHCO 2 H11002 OH Water 2H 2 O H11001 1-Phenylethylamine (racemic mixture) C 6 H 5 CHNH 2 CH 3 (S)-(H11002)-Malic acid (resolving agent) HO 2 CCH 2 CHCO 2 H OH 1-Phenylethylammonium (S)-malate (mixture of diastereomeric salts) H11002 O 2 CCH 2 CHCO 2 H OH C 6 H 5 CHNH 3 CH 3 H11001 288 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Both the isotactic and the syndiotactic forms of polypropylene are known as stereoreg- ular polymers, because each is characterized by a precise stereochemistry at the carbon atom that bears the methyl group. There is a third possibility, shown in Figure 7.17c, which is described as atactic. Atactic polypropylene has a random orientation of its methyl groups; it is not a stereoregular polymer. Polypropylene chains associate with one another because of attractive van der Waals forces. The extent of this association is relatively large for isotactic and syndio- tactic polymers, because the stereoregularity of the polymer chains permits efficient pack- ing. Atactic polypropylene, on the other hand, does not associate as strongly. It has a lower density and lower melting point than the stereoregular forms. The physical prop- erties of stereoregular polypropylene are more useful for most purposes than those of atactic polypropylene. When propene is polymerized under free-radical conditions, the polypropylene that results is atactic. Catalysts of the Ziegler–Natta type, however, permit the preparation of either isotactic or syndiotactic polypropylene. We see here an example of how proper choice of experimental conditions can affect the stereochemical course of a chemical reaction to the extent that entirely new materials with unique properties result. 7.15 Stereoregular Polymers 289 (a) Isotactic polypropylene (b) Syndiotactic polypropylene (c) Atactic polypropylene FIGURE 7.17 Poly- mers of propene. The main chain is shown in a zigzag conformation. Every other carbon bears a methyl sub- stituent and is a stereogenic center. (a) All the methyl groups are on the same side of the carbon chain in isotac- tic polypropylene. (b) Methyl groups alternate from one side to the other in syndio- tactic polypropylene. (c) The spatial orientation of the methyl groups is random in atactic polypropylene. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.16 STEREOGENIC CENTERS OTHER THAN CARBON Our discussion to this point has been limited to molecules in which the stereogenic cen- ter is carbon. Atoms other than carbon may also be stereogenic centers. Silicon, like car- bon, has a tetrahedral arrangement of bonds when it bears four substituents. A large num- ber of organosilicon compounds in which silicon bears four different groups have been resolved into their enantiomers. Trigonal pyramidal molecules are chiral if the central atom bears three different groups. If one is to resolve substances of this type, however, the pyramidal inversion that interconverts enantiomers must be slow at room temperature. Pyramidal inversion at nitrogen is so fast that attempts to resolve chiral amines fail because of their rapid racemization. Phosphorus is in the same group of the periodic table as nitrogen, and tricoordi- nate phosphorus compounds (phosphines), like amines, are trigonal pyramidal. Phos- phines, however, undergo pyramidal inversion much more slowly than amines, and a number of optically active phosphines have been prepared. Tricoordinate sulfur compounds are chiral when sulfur bears three different sub- stituents. The rate of pyramidal inversion at sulfur is rather slow. The most common compounds in which sulfur is a stereogenic center are sulfoxides such as: The absolute configuration at sulfur is specified by the Cahn–Ingold–Prelog method with the provision that the unshared electron pair is considered to be the lowest ranking substituent. 7.17 SUMMARY Chemistry in three dimensions is known as stereochemistry. At its most fundamental level, stereochemistry deals with molecular structure; at another level, it is concerned with chemical reactivity. Table 7.2 summarizes some basic definitions relating to molec- ular structure and stereochemistry. Section 7.1 A molecule is chiral if it cannot be superposed on its mirror image. Non- superposable mirror images are enantiomers of one another. Molecules in which mirror images are superposable are achiral. 2-Chlorobutane (chiral) Cl CH 3 CHCH 2 CH 3 2-Chloropropane (achiral) Cl CH 3 CHCH 3 S H11001 CH 3 CH 3 CH 2 CH 2 CH 2 H11002 O (S)-(H11001)-Butyl methyl sulfoxide very fast N b a c a b c N 290 CHAPTER SEVEN Stereochemistry A detailed flowchart describ- ing a more finely divided set of subcategories of isomers appears in the February 1990 issue of the Journal of Chem- ical Education. Verify that CH 3 NHCH 2 CH 3 is chiral by trying to superpose models of both enantiomers. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Section 7.2 The most common kind of chiral molecule contains a carbon atom that bears four different atoms of groups. Such an atom is called a stereogenic center. Table 7.2 shows the enantiomers of 2-chlorobutane. C-2 is a ste- reogenic center in 2-chlorobutane. Section 7.3 A molecule that has a plane of symmetry or a center of symmetry is achi- ral. cis-4-Methylcyclohexanol (Table 7.2) has a plane of symmetry that bisects the molecule into two mirror-image halves and is achiral. The same can be said for trans-4-methylcyclohexanol. Section 7.4 Optical activity, or the degree to which a substance rotates the plane of polarized light, is a physical property used to characterize chiral sub- stances. Enantiomers have equal and opposite optical rotations. To be optically active a substance must be chiral, and one enantiomer must be present in excess of the other. A racemic mixture is optically inactive and contains equal quantities of enantiomers. Section 7.5 Relative configuration compares the arrangement of atoms in space to some reference. The prefix cis in cis-4-methylcyclohexanol, for example, 7.17 Summary 291 TABLE 7.2 Classification of Isomers* Definition 1. Constitutional isomers are isomers that differ in the order in which their atoms are connected. 2. Stereoisomers are isomers that have the same con- stitution but differ in the arrangement of their atoms in space. (b) Diastereomers are stereoisomers that are not enantiomers. Example There are three constitutionally isomeric compounds of molecular formula C 3 H 8 O: The two enantiomeric forms of 2-chlorobutane are The cis and trans isomers of 4-methylcyclohexanol are stereoisomers, but they are not related as an object and its mirror image; they are diastereomers. (a) Enantiomers are stereoisomers that are related as an object and its nonsuperposable mirror image. (R)-(H11002)-2-Chlorobutane C H CH 3 CH 2 H 3 C Cl (S)-(H11001)-2-Chlorobutane C H CH 2 CH 3 CH 3 Cland CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CHCH 3 OH 2-Propanol CH 3 CH 2 OCH 3 Ethyl methyl ether CH 3 HO cis-4-Methylcyclohexanol CH 3 HO trans-4-Methylcyclohexanol *Isomers are different compounds that have the same molecular formula. They may be either constitutional isomers or stereoisomers. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website describes relative configuration by referencing the orientation of the CH 3 group to the OH. Absolute configuration is an exact description of the arrangement of atoms in space. Section 7.6 Absolute configuration in chiral molecules is best specified using the pre- fixes R and S of the Cahn–Ingold–Prelog notational system. Substituents at a stereogenic center are ranked in order of decreasing precedence. If the three highest ranked substituents trace a clockwise path (high- est→second highest→third highest) when the lowest ranked substituent is held away from us, the configuration is R. If the path is anticlockwise, the configuration is S. Table 7.2 shows the R and S enantiomers of 2-chlorobutane. Section 7.7 A Fischer projection shows how a molecule would look if its bonds were projected onto a flat surface. Horizontal lines represent bonds coming toward you; vertical bonds point away from you. The projection is nor- mally drawn so that the carbon chain is vertical, with the lowest num- bered carbon at the top. Section 7.8 Both enantiomers of the same substance are identical in most of their physical properties. The most prominent differences are biological ones, such as taste and odor, in which the substance interacts with a chiral receptor site in a living system. Enantiomers also have important conse- quences in medicine, in which the two enantiomeric forms of a drug can have much different effects on a patient. Section 7.9 A chemical reaction can convert an achiral substance to a chiral one. If the product contains a single stereogenic center, it is formed as a racemic mixture. Optically active products can be formed from optically inactive starting materials only if some optically active agent is present. The best examples are biological processes in which enzymes catalyze the forma- tion of only a single enantiomer. Section 7.10 When a molecule has two stereogenic centers and these two stereogenic centers are not equivalent, four stereoisomers are possible. Enantiomers of erythro-3-bromo-2-butanol OHH BrH CH 3 CH 3 HHO HBr CH 3 CH 3 Enantiomers of threo-3-bromo-2-butanol HOH HBr CH 3 CH 3 HHO BrH CH 3 CH 3 Stearic acid HH CH 2 (CH 2 ) 13 CH 3 CH 2 CO 2 H (S)-3-Hydroxystearic acid HHO CH 2 (CH 2 ) 13 CH 3 CH 2 CO 2 H biological oxidation H CH 2 CH 3 CH 3 CCl H Cl CH 2 CH 3 CH 3 (S)-2-Chlorobutane(R)-2-Chlorobutane C H CH 3 CH 2 CH 3 Cl Cl H CH 2 CH 3 CH 3 292 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Stereoisomers that are not enantiomers are classified as diastereomers. Each enantiomer of erythro-3-bromo-2-butanol is a diastereomer of each enantiomer of threo-3-bromo-2-butanol. Section 7.11 Achiral molecules that contain stereogenic centers are called meso forms. Meso forms typically contain (but are not limited to) two equivalently substituted stereogenic centers. They are optically inactive. Section 7.12 For a particular constitution, the maximum number of stereoisomers is 2 n , where n is the number of structural units capable of stereochemical variation—usually this is the number of stereogenic centers, but can include E and Z double bonds as well. The number of stereoisomers is reduced to less than 2 n when there are meso forms. Section 7.13 Addition reactions of alkenes may generate one (Section 7.9) or two (Sec- tion 7.13) stereogenic centers. When two stereogenic centers are pro- duced, their relative stereochemistry depends on the configuration (E or Z) of the alkene and whether the addition is syn or anti. Section 7.14 Resolution is the separation of a racemic mixture into its enantiomers. It is normally carried out by converting the mixture of enantiomers to a mixture of diastereomers, separating the diastereomers, then regenerating the enantiomers. Section 7.15 Certain polymers such as polypropylene contain stereogenic centers, and the relative configurations of these centers affect the physical properties of the polymers. Like substituents appear on the same side of a zigzag carbon chain in an isotactic polymer, alternate along the chain in a syndiotactic polymer, and appear in a random manner in an atactic poly- mer. Isotactic and syndiotactic polymers are referred to as stereoregular polymers. Section 7.16 Atoms other than carbon can be stereogenic centers. Examples include those based on tetracoordinate silicon and tricoordinate sulfur as the stereogenic atom. In principle, tricoordinate nitrogen can be a stereogenic center in compounds of the type N(x, y, z), where x, y, and z are differ- ent, but inversion of the nitrogen pyramid is so fast that racemization occurs virtually instantly at room temperature. PROBLEMS 7.22 Which of the isomeric alcohols having the molecular formula C 5 H 12 O are chiral? Which are achiral? 7.23 Write structural formulas or make molecular models for all the compounds that are trichloro derivatives of cyclopropane. (Don’t forget to include stereoisomers.) Which are chiral? Which are achiral? meso-2,3-Dibromobutane BrH BrH CH 3 CH 3 H BrH Br CH 3 CH 3 (2R,3R)-2,3-Dibromobutane Br H H Br CH 3 CH 3 (2S,3S)-2,3-Dibromobutane Problems 293 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.24 In each of the following pairs of compounds one is chiral and the other is achiral. Identify each compound as chiral or achiral, as appropriate. (a) (b) (c) (d) 7.25 Compare 2,3-pentanediol and 2,4-pentanediol with respect to the number of stereoisomers possible for each constitution. Which stereoisomers are chiral? Which are achiral? 7.26 In 1996, it was determined that the absolute configuration of (H11002)-bromochlorofluoromethane is R. Which of the following is (are) (H11002)-BrClFCH? 7.27 Specify the configuration at R or S in each of the following. (a) (H11002)-2-Octanol (b) Monosodium L-glutamate (only this stereoisomer is of any value as a flavor- enhancing agent) H 3 N H11001 H CO 2 H11002 CH 2 CH 2 CO 2 H11002 Na H11001 Cl H F Br C Br H F Cl C Br H FCl H Cl F Br Cl and Cl H H 2 N H H NH 2 CH 3 CH 3 CH 3 H H NH 2 NH 2 CH 3 and CH 3 CH CHCH 2 Br and CH 3 CHCH CH 2 Br OH ClCH 2 CHCH 2 OH and Cl HOCH 2 CHCH 2 OH 294 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.28 A subrule of the Cahn–Ingold–Prelog system specifies that higher mass number takes prece- dence over lower when distinguishing between isotopes. (a) Determine the absolute configurations of the reactant and product in the biological oxidation of isotopically labeled ethane described in Section 7.2. (b) Because OH becomes bonded to carbon at the same side from which H is lost, the oxidation proceeds with retention of configuration (Section 6.13). Compare this fact with the R and S configurations you determined in part (a) and reconcile any appar- ent conflicts. 7.29 Identify the relationship in each of the following pairs. Do the drawings represent consti- tutional isomers or stereoisomers, or are they just different ways of drawing the same compound? If they are stereoisomers, are they enantiomers or diastereomers? (Molecular models may prove useful in this problem.) (a) (b) (c) (e) (f) H 3 CH HCl and andH OH CH 2 OH CH 2 OH HO H CH 2 OH CH 2 OH Br H CH 3 CH 2 CH 3 and C CH 3 CH 3 CH 2 H Br and C H Br H 3 C CH 2 CH 3 C CH 3 CH 3 CH 2 H Br C Br CH 3 CH 2 H CH 3 and C CH 3 HO H CH 2 Br C H Br H 3 C CH 2 OHand C T H D CH 3 C T HO D CH 3 biological oxidation Problems 295 (d) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) and H H Br CO 2 H CH 3 Br HBr CH 3 Br CO 2 H H and Br H CH 3 CO 2 H H Br H H Br CO 2 H CH 3 Br H H Br CO 2 H CH 3 H Br CH 3 Br and Br CO 2 H H and CH 3 CH 3 H H H H H 3 C CH 3 CH 3 H OH HO H CH 2 OH and OH CH 2 OH H H 3 C HO H and CH 2 OHHO CH 2 OH HO and CH 2 OHHO CH 2 OH HO and HHO OHH and HHO OHH and HHO OHH 296 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (q) (r) (s) (t) (u) (v) 7.30 Chemical degradation of chlorophyll gives a number of substances including phytol. The constitution of phytol is given by the name 3,7,11,15-tetramethyl-2-hexadecen-1-ol. How many stereoisomers have this constitution? 7.31 Muscarine is a poisonous substance present in the mushroom Amanita muscaria. Its struc- ture is represented by the constitution shown. (a) Including muscarine, how many stereoisomers have this constitution? (b) One of the substituents on the ring of muscarine is trans to the other two. How many of the stereoisomers satisfy this requirement? (c) Muscarine has the configuration 2S,3R,5S. Write a structural formula or build a molec- ular model of muscarine showing its correct stereochemistry. 7.32 Ectocarpene is a volatile, sperm cell-attracting material released by the eggs of the seaweed Ectocarpus siliculosus. Its constitution is HO CH 3 O 52 3 H11001 CH 2 N(CH 3 ) 3 HO H11002 H 3 CH and CH 3 H H 3 CH and CH 3 H CH 3 H 3 C and CH 3 and CH 3 I (CH 3 ) 3 C I (CH 3 ) 3 C and OHH 3 C and OH H 3 C Problems 297 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website All the double bonds are cis, and the absolute configuration of the stereogenic center is S. Write a stereochemically accurate representation of ectocarpene. 7.33 Multifidene is a sperm cell-attracting substance released by the female of a species of brown algae (Cutleria multifida). The constitution of multifidene is (a) How many stereoisomers are represented by this constitution? (b) Multifidene has a cis relationship between its alkenyl substituents. Given this informa- tion, how many stereoisomers are possible? (c) The butenyl side chain has the Z configuration of its double bond. On the basis of all the data, how many stereoisomers are possible? (d) Draw stereochemically accurate representations of all the stereoisomers that satisfy the structural requirements of multifidene. (e) How are these stereoisomeric multifidenes related (enantiomers or diastereomers)? 7.34 Streptimidone is an antibiotic and has the structure shown. How many diastereomers of streptimidone are possible? How many enantiomers? Using the E,Z and R,S descriptors, specify all essential elements of stereochemistry of streptimidone. 7.35 In Problem 4.26 you were asked to draw the preferred conformation of menthol on the basis of the information that menthol is the most stable stereoisomer of 2-isopropyl-5-methylcyclo- hexanol. We can now completely describe (H11002)-menthol structurally by noting that it has the R con- figuration at the hydroxyl-substituted carbon. (a) Draw or construct a molecular model of the preferred conformation of (H11002)-menthol. (b) (H11001)-Isomenthol has the same constitution as (H11002)-menthol. The configurations at C-1 and C-2 of (H11001)-isomenthol are the opposite of the corresponding stereogenic centers of (H11002)-menthol. Write the preferred conformation of (H11001)-isomenthol. 7.36 A certain natural product having [H9251] D H11001 40.3° was isolated. Two structures have been inde- pendently proposed for this compound. Which one do you think is more likely to be correct? Why? OH OHHO CO 2 H HO H H H OH H OH CH 2 OH CH 2 OH HO OH O NH O H OH H H H 3 C H H O H 3 C CH CH CHCH 2 CH 3 CH 2 CH 3 CH 2 CH CH 298 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.37 One of the principal substances obtained from archaea (one of the oldest forms of life on earth) is derived from a 40-carbon diol. Given the fact that this diol is optically active, is it com- pound A or is it compound B? 7.38 (a) An aqueous solution containing 10 g of optically pure fructose was diluted to 500 mL with water and placed in a polarimeter tube 20 cm long. The measured rotation was H110025.20°. Calculate the specific rotation of fructose. (b) If this solution were mixed with 500 mL of a solution containing 5 g of racemic fructose, what would be the specific rotation of the resulting fructose mixture? What would be its optical purity? 7.39 Write the organic products of each of the following reactions. If two stereoisomers are formed, show both. Label all stereogenic centers R or S as appropriate. (a) 1-Butene and hydrogen iodide (b) (E)-2-Pentene and bromine in carbon tetrachloride (c) (Z)-2-Pentene and bromine in carbon tetrachloride (d) 1-Butene and peroxyacetic acid in dichloromethane (e) (Z)-2-Pentene and peroxyacetic acid in dichloromethane (f) 1,5,5-Trimethylcyclopentene and hydrogen in the presence of platinum (g) 1,5,5-Trimethylcyclopentene and diborane in tetrahydrofuran followed by oxidation with hydrogen peroxide 7.40 The enzyme aconitase catalyzes the hydration of aconitic acid to two products: citric acid and isocitric acid. Isocitric acid is optically active; citric acid is not. What are the respective con- stitutions of citric acid and isocitric acid? 7.41 Consider the ozonolysis of trans-4,5-dimethylcyclohexene having the configuration shown. CH 3 CH 3 Aconitic acid C HHO 2 C CO 2 HHO 2 CCH 2 C Problems 299 HO CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 OH Compound A Compound B HO CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Structures A, B, and C are three stereoisomeric forms of the reaction product. (a) Which, if any, of the compounds A, B, and C are chiral? (b) What product is formed in the reaction? (c) What product would be formed if the methyl groups were cis to each other in the starting alkene? 7.42 (a) On being heated with potassium ethoxide in ethanol (70°C), the deuterium-labeled alkyl bromide shown gave a mixture of 1-butene, cis-2-butene, and trans-2-butene. On the basis of your knowledge of the E2 mechanism, predict which alkene(s), if any, contained deuterium. (b) The bromide shown in part (a) is the erythro diastereomer. How would the deuterium content of the alkenes formed by dehydrohalogenation of the threo diastereomer differ from those produced in part (a)? 7.43 A compound (C 6 H 10 ) contains a five-membered ring. When Br 2 adds to it, two diastereo- meric dibromides are formed. Suggest reasonable structures for the compound and the two dibromides. 7.44 When optically pure 2,3-dimethyl-2-pentanol was subjected to dehydration, a mixture of two alkenes was obtained. Hydrogenation of this alkene mixture gave 2,3-dimethylpentane, which was 50% optically pure. What were the two alkenes formed in the elimination reaction, and what were the relative amounts of each? D H H H H H H CH 3 CH CH 3 H O CH O A H CH 3 H H CH 3 H H H CH O CH O B H H H H CH 3 H H CH 3 CH O CH O C 300 CHAPTER SEVEN Stereochemistry Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 7.45 When (R)-3-buten-2-ol is treated with a peroxy acid, two stereoisomeric epoxides are formed in a 60:40 ratio. The minor stereoisomer has the structure shown. (a) Write the structure of the major stereoisomer. (b) What is the relationship between the two epoxides? Are they enantiomers or diaste- reomers? (c) What four stereoisomeric products are formed when racemic 3-buten-2-ol is epoxi- dized under the same conditions? How much of each stereoisomer is formed? 7.46 Verify that dibromochloromethane is achiral by superposing models of its two mirror image forms. In the same way, verify that bromochlorofluoromethane is chiral. 7.47 Construct a molecular model of (S)-3-chlorocyclopentene. 7.48 Construct a molecular model corresponding to the Fischer projection of meso-2,3-dibro- mobutane. Convert this molecular model to a staggered conformation in which the bromines are anti to one another. Are the methyl groups anti or gauche to one another in this staggered con- formation? 7.49 What alkene gives a racemic mixture of (2R,3S) and (2S,3R)-3-bromo-2-butanol on treat- ment with Br 2 in aqueous solution? (Hint: Make a molecular model of one of the enantiomeric 3-bromo-2-butanols, arrange it in a conformation in which the Br and OH groups are anti to one another, then disconnect them.) Problems 301 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website