1559T_ch05_70-9810/22/0520:19Pa9e75 ⊕ EQA Keys to the Chapter·75 Formco bondedtootherchmged tobonded to 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 ing of asymmetric carbons by using a sim The rules are straight 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
1559r_ch05.70-9810/22/0520:19Page76 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 and 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 immediately visualize the mirror plane between them,and recognize them as mers.Similarly. CH H- Br- and H Br- H 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? 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 eacn other,and any other combination is a pair ays of drav ehredimre Wth prctice.hever comes eperienceand co 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