1559T.ch05_70-9910/22/0520:19Page70 EQA 5 Stereoisomers 一g咖in This chpereoof te一 o 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. 52g 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. Diastereo ers n relationships between more complex molecules.Moves toward biological systems. 5-8 Resolution of Enantiomers Practical【echniques- Keys to the Chapter 5-1.Chiral Molecules In this 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 ror image.Methane is identical to its mirror imae:it is not chiral.The two possible shapes of a chiral mo ne two age shapesc ule are called enantiomers What mak of models and start manufacturing chiral molecules.Prove to yourself that the model of the mirror image o don the onginal.This is the frst step toward developing the ability to viualiz 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 ts th in its ability to perceive depth as well as left-right re For chiral n ecules.the co rpart to thi inter ith plane-po ized light:Th 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:.The two enantiomers of a chiral molecule each ate light by equal amou ts.butn opp of the large number of ne terms and ideas associated with this material,it merits careful 5-3.The R-S Sequence Rules Like all the rules of nomenclature,the R-S system for molecules containing stereocenters has one purpose cases tne s straightfor of the e ter so that you an con m their Ror S th you troub take it stepwise: 2.Starting at the lef.look at the first atom in each substituent chain and identify the atoms attached to it rity.orde nofd are the out from the first to the in the chain and repeat the the is direction of this move sh echosn 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
1559Tch0570-9810/22/0520:19.Pag972 ⊕ 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 CCls H- CHa CHaBr Procedure:H is obviously lowest priority:the other three need to be compared.Write them out side-by-side: @ -C1 一H一C一H -CH:Br to atomic number on each carbon.In -CCl it is Cl:in .All three are different therefore priorities can be assiged imme that ther 0 ⊕ the first Cl on -CCl and Br is "higger ☒ redraw the molecule with the highest priority group designated"a. becomes (d in back) CH.CH2B Counterclockwise S Erample:Determine R-S designation for H :The sr is a member of a ring.but the isnot really different.The H is lowest up is ethyl:that's iu eally just the quence of ring atoms attached to the stere start at the ringC(CHand move around the ring in the othercoo enter.For one group,start at the ring -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 (CH3)2C H2C H C CH2Br H CH3 CCl3 becomes C d c a b C c a b (d in back) Counterclockwise S C Cl Cl Cl –CCl3 C Br H H –CH2Br C H H H –CH3 H C CH2Br CH3 CCl3 1559T_ch05_70-98 10/22/05 20:19 Page 72
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--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
1559r_ch05_70-9810/22/0520:19Page74 74.Chapter 5 STEREOISOMERS O-CH2-CH; 0 CH2-CH -CH-CH; 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 onthesecon hihest is earbon.nthe astis hydrogen.No diference 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 tofdinca 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 压gher 0-H O-CH:-CH3 -CH-CH; VS. -C- CH2-CHs H H Group 1 Group 2
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