1559T_ch05._70-9810/22/0520:19Page70 EQA 5 Stereoisomers e teedime水is daper epom一 yet,do n't wait any longe Outline of the Chapter 5-1 Chiral Molecules The key definitions:what makes an object different from its miror image. of chiral molecules 5-3 The R-S Sequence Rules How to name chiral molecules 5-4 Fischer Projections How to represent chiral molecules in two dimensions. relationships between more complex molecules.Moves toward biological systems. 5-8 Resolution of Enantiomers Practical【echniques- Keys to the Chapter wil be introduced to many new terms.Follow their definitions alon with the structures given as examples.The first new term is the chapter title:stereoisomer.In brief.stereoisomers are molecules 70
5 Stereoisomers By now you are well aware that molecules are three-dimensional objects. This chapter explores some of the more subtle, but extremely critical, consequences of this fact. If you have not done so yet, don’t wait any longer to obtain a set of models to aid you in visualizing the structures described in this chapter. For many students, the isomeric relationships discussed here are the most difficult ones encountered in organic chemistry, and they are important later on in descriptions of several types of compounds and reactions. The implications for biological chemistry are especially significant. Outline of the Chapter 5-1 Chiral Molecules The key definitions: what makes an object different from its mirror image. 5-2 Optical Activity Physical properties of chiral molecules. 5-3 The R–S Sequence Rules How to name chiral molecules. 5-4 Fischer Projections How to represent chiral molecules in two dimensions. 5-5 and 5-6 Molecules Incorporating Several Stereocenters: Diastereomers Further elaboration, including the introduction of several new terms describing relationships between more complex molecules. Moves toward biological systems. 5-7 Stereochemistry in Chemical Reactions Ground rules. 5-8 Resolution of Enantiomers Practical techniques. Keys to the Chapter 5-1. Chiral Molecules In this chapter you will be introduced to many new terms. Follow their definitions along with the structures given as examples. The first new term is the chapter title: stereoisomer. In brief, stereoisomers are molecules 70 1559T_ch05_70-98 10/22/05 20:19 Page 70
1559T_ch05_70-9810/22/0520:19Pa9e71 ⊕ EQA Keys to the Chapter·71 ty)but that do not cules that are chiral.Chiral molecules can exist as either of two stereoisomeric shapes.which are related t er as an ooject is t mage Before going any further.le al is that y to it rorimage Methane sidentical imae:it is not chiral.The wo possible shapes of a ne two ule are called enantiomers What mal of models and start manufacturing chiral molecules.Prove to yourself that the model of the mirror image o cdon the onginal.This is the frst step towd developing the ability to visai Optical Activity enantiomers is so subtle that they end up for the something else that t is,itself,alre cady"handed."By analogy.a right and a left glove will have the same weight and text ce,a ng igals from our binocular visual system is"handed in its ability to perceive depth as well as left-right re For chiral m ecules.the co rpart to thi inter ith plane-po zed light:Ih of a chiral molecule.This ptical motation is the most common way of dete ecting chira onsiderable detail in this sec Molecul s possessing the ility to rotate th the two enantiomers of a chiral molecule:racemic mirure.The two enantiomers of a chiral molecule each ate light by equal amou displays no optical activity be Again,of the large number of ne terms and ideas associated with this material,it merits careful study,with a good set of models close at hand 5-3.The R-S Sequence Rules Like all the rules of nomenclature,the R-S system for molecules containing stereocenters has one purpose o cases tne s of the e ter so that you an con m their R or S th difference”is well un tood.If this gives you trouble.take it stepwise: 2.Starting at the lef.look at the first atom in each substituent chain and identify the atoms attached to it are the out from the first to the cond atom in the chain and repeat the process.If ther is direction of this move sh be chosn to examine the highest priority atom for
that have the same atoms linked in the same order (i.e., identical connectivity), but that do not have identical three-dimensional shapes. The first example in Section 5-1, 2-bromobutane, is one of a vast number of molecules that are chiral. Chiral molecules can exist as either of two stereoisomeric shapes, which are related to each other as an object is to its mirror image. Before going any further, let’s make one point clear: Every molecule has a mirror image, obviously. What makes a chiral molecule special is that it is not identical to its mirror image. Methane is identical to its mirror image; it is not chiral. The two possible shapes of a chiral molecule differ in the same way that a right-handed object differs from a left-handed object, as gloves, shoes, and hands do. Chirality, therefore, is “handedness” on a molecular level. The two mirror-image shapes of a chiral molecule are called enantiomers. What makes a molecule chiral? The most common of several types of structural features that can make a molecule chiral is the presence of a carbon atom attached to four different atoms or groups (an asymmetric carbon atom, an example of what is called a stereocenter). At this point it is worthwhile to dust off your set of models and start manufacturing chiral molecules. Prove to yourself that the model of the mirror image of one cannot be superimposed on the original. This is the first step toward developing the ability to visualize this relationship clearly. 5-2. Optical Activity The physical difference between enantiomers is so subtle that they end up for the most part displaying identical physical and chemical properties. They can be distinguished from each other only after interacting with something else that is, itself, already “handed.” By analogy, a right and a left glove will have the same weight, color, and texture. However, interaction with, for instance, a right hand will immediately distinguish them. The fact that we can tell them apart just by looking at them reflects the fact that our brain’s interpretation of the signals from our binocular visual system is “handed” in its ability to perceive depth as well as left–right relationships. For chiral molecules, the counterpart to this is their interaction with plane-polarized light: The plane of polarization of plane-polarized light is rotated when it passes through a solution of one enantiomer of a chiral molecule. This phenomenon, labeled optical rotation, is the most common way of detecting chiral molecules and is described in considerable detail in this section. Molecules possessing the ability to rotate the plane of polarized light are said to be optically active, or to display optical activity; and another term for enantiomers is optical isomers. One further term of importance is the one given to a mixture of equal amounts of the two enantiomers of a chiral molecule: racemic mixture. The two enantiomers of a chiral molecule each rotate light by equal amounts, but in opposite directions, so the racemic mixture displays no optical activity because the two components exactly cancel out each other’s rotations. Again, because of the large number of new terms and ideas associated with this material, it merits careful study, with a good set of models close at hand. 5-3. The R–S Sequence Rules Like all the rules of nomenclature, the R–S system for molecules containing stereocenters has one purpose: the concise, unambiguous description of a single chemical structure. In most cases the system is not particularly difficult to apply, because the assignment of priorities to the groups on an asymmetric carbon is usually straightforward and use of models to view the stereocenter properly takes care of the rest. Models should be made of the examples in the chapter so that you can confirm their R or S designations. Priority rule 2 is an occasional source of trouble until the concept of “first point of difference” is well understood. If this gives you trouble, take it stepwise: 1. Write the substituent groups to be compared side-by-side, with the bond of attachment to the asymmetric atom on the left. 2. Starting at the left, look at the first atom in each substituent chain and identify the atoms attached to it in descending order of priority. If the highest priority atoms in each are the same, work your way down in priority, looking for the first nonidentical atoms (first point of difference). If no differences are found at this stage, move out from the first to the second atom in the chain and repeat the process. If there is branching here, the direction of this move should be chosen to examine the highest priority atom for Keys to the Chapter • 71 1559T_ch05_70-98 10/22/05 20:19 Page 71
72.Chapter 5 STEREOISOMERS a point of difference.If no difference is found,then examine the second highest priority group.and It is actually easier to do than it is to describe.so let's analyze three examples Example:Determine R-S designation for Procedure:H is obviously lowest priority:the other three need to be compared.Write them out side-by-side: Identify the highest to atomic number on each carbon.In -CCl it is Cl:in -CH2Br,it is Br; is H.All three are different:therefore priorities can be assigned imme The fa that the 80 the first Cl on CCl and Br is "higge thanCl.Once the first point of diference. redraw the with the highest priority group designated"" CHCH.CH :The sr is a member of a ring.but the isnot really different.The H is lowest up:that's ovious. eally just the quence of ring atoms attached to the stere enter.For one group,start at the ring CH and move around the rstart at move around thering nthe other direction.So,we com -CHz-CH3. -CH-C(CHa)-etc..and -C(CHa)-CH,-ctc
72 • Chapter 5 STEREOISOMERS a point of difference. If no difference is found, then examine the second highest priority group, and so on. It is actually easier to do than it is to describe, so let’s analyze three examples. Example: Determine R–S designation for Procedure: H is obviously lowest priority; the other three need to be compared. Write them out side-by-side: Identify the highest priority atom according to atomic number on each carbon. In OCCl3, it is Cl; in OCH2Br, it is Br; and in OCH3, it is H. All three are different; therefore priorities can be assigned immediately. Br Cl H, therefore OCH2Br OCCl3 OCH3. The fact that there are three Cl’s on OCCl3 and only one Br on OCH2Br is irrelevant. The first point of difference is the Br on OCH2Br vs. the first Cl on OCCl3, and Br is “bigger” than Cl. Once the first point of difference is identified, nothing else matters. With priorities now assigned, we can redraw the molecule with the highest priority group designated “a,” the second “b,” the third “c,” and the lowest “d”: Example: Determine R–S designation for Procedure: The stereocenter is a member of a ring, but the procedure is not really different. The H is lowest priority, and the other three groups have to be compared. The problem is interpreting what it means to have the stereocenter in a ring. One group is ethyl; that’s obvious. The other two “groups” are really just the sequence of ring atoms attached to the stereocenter. For one group, start at the ring OCH2 and move around the ring; for the other, start at the ring OC(CH3)2, and move around the ring in the other direction. So, we compare the “groups” OCH2OCH3, OCH2OC(CH3)2Oetc., and OC(CH3)2OCH2Oetc. C CH2CH3 (CH
1559T_ch05_70-9810/22/0520:19Pa9e73 EQA Keyso the Chopter·73 nical (carbon).We move out to the atom attached to cach of these carbons for First point of differenc ② ©H ©H ⊙CHh-ctc. ©H,-ctc CH: Ring CHa Ring C(CH)2 n each case the la orities are determined by moving one atom further out,as shown (circled): ©H -cH2-C-④-cH,-C-etc ⊙H Ethvl Ring CH Example:Determine R-Sdesignation for the marked carbon in OCH,CH, CHCH(CH3)2 OCH>CH
Keys to the Chapter • 73 where the left-hand bond goes to the stereocenter and “etc.” means “continuing around the ring.” Priorities are assigned in accordance with rule 2 again, because in all three groups the first atom in the “chain” attached to the asymmetric carbon is identical (carbon). We move out to the atoms attached to each of these carbons for comparison (circled): In each case the largest of these atoms is carbon. No difference. However, in the case of the group at the right, the second largest is also carbon, whereas the second largest for the other two groups is hydrogen. The highest priority of these three groups is therefore the ring-contained OC(CH3)2OCH2Oetc. Second and third priorities are determined by moving one atom further out, as shown (circled): Comparison here is straightforward. The CH2 in the ethyl is connected only to a simple CH3 group, whereas the CH2 of the ring is attached to a OC(CH3)2Oetc. group. So the latter one is higher in priority (C larger than H). Therefore we have: Example: Determine R–S designation for the marked carbon in C CHCH2CH3 CHCH(CH3)2 OCH2CH3 OCH2CH3 Cl H * C CH2CH3 (CH3)2C H2C H becoming C b a b a c c d C (d in back) Counterclockwise S C H3 H3 CH2 C C H H H CH2 C etc. First point of difference Ethyl Ring CH2 C H C H3 C H3 C H3 H C H C (CH3)2 etc. H C C H2 etc. First point of difference Ethyl Ring CH2 Ring C(CH3)2 1559T_ch05_70-98 10/22/05 20:19 Page 73
1559rch0570-9810/22/0520:19Page74 74.Chapter 5 STEREOISOMERS O-CH2-CH; 0 CH2-CH -CH-CH3 vs. -CH2-CH H First point of differenc Group1 Group 2 Start at the carbon atom labeled"a"in each gro os.the highes priority atom attached to car oethesecond hihest is earonn the atishydroen.No difference is found. to the highe priority atom on carl for the next companison:Fo oup attached to "a. we look to carbon"b.and see that here the tie can be broken:In group 1.this atom is att With priorities done.the groups can all be labeled and RorSassigned: (din back Clockwise =R If,on the other hand,the oxygen in group I were attached to an H instead of a C,that would become the rstonfrnceand group2would be higher in priority The differeces theooul Lower Higher 0-H O-CH:-CH3 -CH-CH; VS. -C -CH2-CHs H Group 1 Group2
74 • Chapter 5 STEREOISOMERS Procedure: Highest priority is Cl, and lowest is H. A priority choice between the two outlined groups needs to be made, however. We write the groups out side-by-side: Start at the carbon atom labeled “a” in each group. In both groups, the highest priority atom attached to carbon “a” is oxygen, the second highest is carbon, and the last is hydrogen. No difference is found, so move out to the highest priority atom on carbon “a” for the next comparison: Follow the arrow and move to the oxygen (not carbon atom “b”—oxygen is higher priority, so it is evaluated first). Both groups 1 and 2 have identical CH2’s attached directly to O, so they are still tied. Turning to the second largest group attached to “a,” we look to carbon “b,” and see that here the tie can be broken: In group 1, this atom is attached to two carbons and one hydrogen, whereas in group 2, it is attached to just one carbon and two hydrogens. Group 1 therefore is higher in priority than group 2 at atom “b,” the first point of difference. With priorities done, the groups can all be labeled and R or S assigned: If, on the other hand, the oxygen in group 1 were attached to an H instead of a C, that would become the first point of difference, and group 2 would be higher in priority. The differences at the “b” carbons would become irrelevant. The above three examples have each been designed to contain a “trick”—an unusual feature that is not often encountered but illustrates the application of the rules in detail. Most chapter and exam problems you see will not be “tricky.” However, by understanding the toughies, proper application of the procedure becomes quicker and simpler in all cases, because you’ve now seen what to do when things get complicated. H C CH3 CH3 CH O H Group 1 H C O CH2 CH3 CH2 CH3 Group 2 vs. Lower Higher C d a c b C a c b (d in back) Clockwise R H C CH3 CH3 CH O CH2 CH3 a b Group 1 H C O CH2 CH3 CH2 CH3 a b Group 2 vs. First point of difference 1559T_ch05_70-98 10/22/05 20:19 Page 74
1559T_ch05_70-9810/22/0520:19Pa9e75 ⊕ EQA Keys to the Chapter·75 . For eample.dob ode other ab chged i bnded t s the carbon actually present,and the other is invented.So 只H ©H becomes CC-H Atoms actually presen another carbe purpos bonds to two carbons (one real,one invented).So CH. CH -Real atoms ⊕ -C=0 becomes Invented atom projections is to simplify the on-paper drav ing of asymmetric carbons by using a sim The rules are straightf wing the text with a set of mo stereoisomers.In particular,s ereocenters give rise o as many as 2"stereoisomers.If we consider the cas where n relatedBeca it about the other six isomers?They are also stere ed by a w term:dlastereome each other.they have different physical p perties and can therefore be separated by standard tory techni This is very important.This feature distinguishes diastereomers from enantiomers.which one that you shoud make a poin of understanding well before you leave this section.Notice that both "cnantiomer"and"diastereomer"really
Keys to the Chapter • 75 Before leaving this section, note also that there is a bit of a trick to rule 3 as well. For priority determination purposes, double and triple bonds are changed so that both atoms involved are doubled or tripled. For example, a carbon doubly bonded to another carbon is changed to a carbon singly bonded to two carbons. One is the carbon actually present, and the other is invented. So The procedure is similar for a carbon–oxygen double bond: The carbon bonds are changed and converted to single bonds to two oxygens (one real, one invented), and likewise the oxygen bonds are changed to single bonds to two carbons (one real, one invented). So 5-4. Fischer Projections The purpose of Fischer projections is to simplify the on-paper drawing of asymmetric carbons by using a simple convention to represent the three-dimensional structural details. The rules are straightforward, but again, following the text with a set of models handy will help you master this material more readily. 5-5 and 5-6. Molecules Incorporating Several Stereocenters: Diastereomers When a molecule has more than one stereocenter, as 2-bromo-3-chlorobutane does, it will have more than two stereoisomers. In particular, n stereocenters give rise to as many as 2n stereoisomers. If we consider the case where n 3, how are all the 2n 8 stereoisomers related? Because an object has only one mirror image, if we pick any one of these (stereoisomer A), it may have no more than one enantiomer (stereoisomer B). What about the other six isomers? They are also stereoisomers of A, but they can’t be its mirror images. The relationship of any of these other six molecules with A is described by a new term: diastereomer. Diastereomers are stereoisomers that are not mirror images of each other. Because diastereomers are not mirror images of each other, they have different physical properties and can therefore be separated by standard laboratory techniques. This is very important. This feature distinguishes diastereomers from enantiomers, which cannot readily be separated from one another. Diastereomer is a very important term, as important as enantiomer, and one that you should make a point of understanding well before you leave this section. Notice that both “enantiomer” and “diastereomer” really O C Real atoms O CH3 C Invented atoms O CH3 C becomes Carbons doubly bonded to another carbon C H H C H These atoms are invented for priority evaluation purposes Atoms actually present C H H C C C becomes H 1559T_ch05_70-98 10/22/05 20:19 Page 75
15597_ch0570-9010/22/0s20:19Page76 76.Chapter 5 STEREOISOMERS describe relation reAs described above.is the antio mer of B:A is also a di can be called both n enantiomer and dastereomer at the same time.However.if you remember that these terms really describe relationships between pairs of structures,it makes more sense.The following illustra- p you get the i nations of hands and feet raised.The relation ship betwe en right hand-right foot raised and left hand-left foot raised is a mirror-image.enantiomeric on ither is The mage of say.right hand- -lelt foot raised,so the latter is ral carbons drawing and orienting the pictures of the molecules so that the stereocenters can be compared in the first place Let's look at an ample two compounds. CH; CH. H -C CI- -H Br -H CH
describe relationships between structures. As described above, A is the enantiomer of B; A is also a diastereomer of each of the other six isomers we talked about. It may seem odd at first that a single molecule can be called both an enantiomer and a diastereomer at the same time. However, if you remember that these terms really describe relationships between pairs of structures, it makes more sense. The following illustration is not a perfect analogy, but it might help you get the idea. Imagine a dancer who at different times has various combinations of hands and feet raised. The relationship between right hand–right foot raised and left hand–left foot raised is a mirror-image, enantiomeric one. Neither is the mirror image of, say, right hand–left foot raised, so the latter is diastereomerically related to the first two. The illustration below shows how the four possible combinations of “one hand up, one hand down, one foot up, one foot down” are related in a way similar to the four stereoisomers of a molecule with two chiral carbons. As the figures in this section show, molecules work similarly. Actually the real trouble in determining, for instance, whether two structures are enantiomers or diastereomers of each other stems from the difficulty in drawing and orienting the pictures of the molecules so that the stereocenters can be compared in the first place. Let’s look at an example. Anyone can look at the two compounds, CH3 CH3 H Cl Br and H CH3 CH3 H H Cl Br 76 • Chapter 5 STEREOISOMERS 1559T_ch05_70-98 10/22/05 20:19 Page 76
1559T_ch05_70-9810/22/0520:19Pa9e77 EQA Keyso the Chopter·7刀 al ze the mirror plane between them,and recognize them as mers.Similarly. CH H Br- and H Br- clearly lack Br without very thoroughly.The one important precaution here is never operate on more than one stereocenterat a time when interconverting Fischer projections. Examole:What is the relationship between the following structures:identical.enantiomeric.or diastereomeric? CI Br CH:- -Br and C1- CH: c- Procedure 1.Operate on the top stereocenter. Br CHCICH CHCICH CHCICH Same as Structure 2 Twoswitches made the top stereocenr of1 identical to that of 2:therefore these carbons are identical in configuration
Keys to the Chapter • 77 immediately visualize the mirror plane between them, and recognize them as enantiomers. Similarly, clearly lack a mirror-image relationship and must be diastereomers. The hard part is determining, for example, the relationship between without getting all messed up (or taking forever to do it, and then getting all messed up). In short, you need to be able to move quickly and accurately among Newman, dashed-wedged line, and Fischer structures. This capability takes practice at visualizing three-dimensional structures from flat drawings and requires application of a couple of specific techniques. The text covers comparison and interconversion of Fischer projections very thoroughly. The one important precaution here is never operate on more than one stereocenter at a time when interconverting Fischer projections. Example: What is the relationship between the following structures: identical, enantiomeric, or diastereomeric? Procedure: 1. Operate on the top stereocenter. Two switches made the top stereocenter of 1 identical to that of 2; therefore these carbons are identical in configuration. Cl CHClCH3 CH3 Br Structure 1 CH3 CHClCH3 Cl Br Br CHClCH3 Cl CH3 Same as Structure 2 Switch Cl, CH3 Switch CH3, Br Cl CH3 CH3 Br H Structure 1 Structure 2 and Cl Br H CH3 CH3 Cl Cl Cl Br CH3 H H CH3 Br C C and H H CH3 CH3 Cl CH3 CH3 H Cl H and CH3 CH3 H H Cl Br Br 1559T_ch05_70-98 10/22/05 20:19 Page 77
1559r_ch05.70-9810/22/0520:19Page78 78.Chapter 5 STEREOISOMERS 2.Then do the bottom stereocenter. CCIBrCH CI--H -CH; CH3- Structure Answer:The two structures are diastereomers(not completely identical and not mirror images). Let's now examine our problem of comparing a dashed-wedged line formula with a Fischer projection.The obemshowterone-wedged The res of a mo rotation will do,such as rotation of the left-hand methyl group up out of the page.toward you: cH Look at these two conformations of the same,identical molecule carefully-with the help of a model,if necessary-to convince yourself that the pictures on the page are what I've said they are. CH CH C CI-C-H CI- H H Br-C-CHs Br- H H H fyou look at it like this This is what you sce Which is this Now you can compare the Fischer projection above with the one we wrote earlier -H H with Br- CHH Br Structure3 Structure4
2. Then do the bottom stereocenter. One switch made the bottom stereocenter of 1 identical to that of 2; therefore these carbons are opposite in configuration. Answer: The two structures are diastereomers (not completely identical and not mirror images). Let’s now examine our problem of comparing a dashed-wedged line formula with a Fischer projection. The actual problem is how to interconvert dashed-wedged line and Fischer formulas. The key to this is recognizing that Fischer projections are pictures of a molecule in an eclipsed conformation. Very important. So the first step in our comparison is to get the dashed-wedged line formula into an eclipsed conformation. Any 60° rotation will do, such as rotation of the left-hand methyl group up out of the page, toward you: Look at these two conformations of the same, identical molecule carefully—with the help of a model, if necessary—to convince yourself that the pictures on the page are what I’ve said they are. We can now convert the eclipsed formula, above, into a Fischer projection. Imagine looking at it from a direction such that the carbon–carbon bond is vertical, and the groups horizontal to it point toward us; for example: Now you can compare the Fischer projection above with the one we wrote earlier: CH3 H Cl H CH3 Structure 3 Structure 4 with Br Cl Br CH3 H H CH3 If you look at it like this This is what you see Which is this Br C C H H CH3 Rotate CH3 Cl towards you 60 Becomes CH3 C C H H CH3 Br Cl CClBrCH3 CH3 Cl H Structure 1 Switch H, CH3 CClBrCH3 H Cl CH3 Same as Structure 2 78 • Chapter 5 STEREOISOMERS 1559T_ch05_70-98 10/22/05 20:19 Page 78
1559T_ch05_70-9810/22/0520:19Pa9e79 EQA Keys to the Chapter·79 1.Top stereocenter H 一CH Same as Structure CHCICH, B-CH →CH3 H →H Br Br Structure 3 Same as Structure4 ⊕ Two switches;therefore also identical. So the two structures shown arlier were identical to each other H -CH 一H Br are exactly the same molecule. It is useful to at this point that comparing pictorial formulas is justn of sever ways to deter and to determine the k or S co guration of ea Once you hay come com reochemical relationship is obvious.For two stereocenters.R.R and S.S are enantiomers of each other.R.S S.R are at Ir eacn other,and any other combination is a pair ays of drav mhdmcinacre.Wth prctice,hee comeprience ad onf need come exam time
Keys to the Chapter • 79 1. Top stereocenter Two switches; therefore identical. 2. Bottom stereocenter Two switches; therefore also identical. So the two structures shown earlier were identical to each other. are exactly the same molecule. It is useful to recognize at this point that comparing pictorial formulas is just one of several ways to determine the relationship between possible stereoisomeric structures. If you have the time, making models is always useful, especially for visualization purposes. An even better way is to apply the rules of nomenclature to each structure and to determine the R or S configuration of each stereocenter. Once you have become comfortable with this technique as it applies to the different types of structural pictures, you may find that this is the quickest method of all: Once R–S assignments have been made to structures under consideration, their stereochemical relationship is obvious. For two stereocenters, R,R and S,S are enantiomers of each other, R,S and S,R are enantiomers of each other, and any other combination is a pair of diastereomers. Notice that I never said this was easy. It takes practice to do and to extend to other common ways of drawing molecules that imply three-dimensional structure. With practice, however, comes experience and confidence, just what you will need come exam time. Cl Br CH3 H H CH3 Br C C and H H CH3 CH3 Cl CHClCH3 H Br CH3 Structure 3 CHClCH3 Br H CH3 CHClCH3 Br CH3 H Same as Structure 4 Switch Br, H Switch H, CH3 CH3 CHBrCH3 Cl H Structure 3 Cl CHBrCH3 CH3 H Cl CHBrCH3 H CH3 Same as Structure 4 Switch Cl, CH3 Switch CH3, H 1559T_ch05_70-98 10/22/05 20:19 Page 79