CHEMISTRY ORGANIC 307 LECTURE V Stereoisomerism: Already encountered. Two compounds are stereoisomers when they differ only in the spatial relationship of their parts. The composition is the same, the connectivity is the same. It's only when we consider three dimensions that the difference manifests itself. Define stereoisomerism in general and contrast with constitutional isomerism. Define conformation and configuration. Emphasize difference, conformations are interconvertible by free rotation about single bonds, configurations are not. A. Restriction of rotation can lead to isomerism. 1.This kind of isomerism is generally called cis-trans or geometric isomerism. We have already seen examples in the cycloakanes. We will see more examples when we get to alkenes. 2. How do we recognize it? Identify a plane, which leads to a top and a bottom. Then there must be two (or more) substituents on at least two different atoms. Then same side = cis and opposite side = trans. 3. What can restrict rotation at RT? a. Rings, tying up the two ends b. bonds, which have a stereoelectronic requirement. If you rotate then the pi overlap weakens because the p-orbitals are no longer parallel.
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In general it is hard to break bonds at RT, so isomers that could interconvert by bond breaking and then free rotation don't. B. We have seen that there can be two ways to arrange atoms in 3D (keeping connections the same) when there is a plane. As we know from experience, there are other situations that lead to more than one way to arrange objects in 3D. 1. The classic is your hands. Discuss. a. How can we tell they are really different? Notice they are both 5 fingers attached to a palm etc. b. The superimposability test. Discuss and mention models. c. But hands are also related. How? Mirror images, but nonsuperimposable.
2. Whenever an object is different from its mirror image, the object is said to be chiral. The difference can be described as "handedness" by analogy to left and right hands. 3. How can we tell when an object is chiral? Symmetry. If it is not symmetrical, it is chiral. If it is symmetrical then it is not chiral (achiral). How can we tell if an object is symmetrical? Look for a symmetry element. If you can find one then the object is symmetrical and superimposable on (identical with) its mirror image. If there is none then it is chiral and nonsuperimposable on its mirror image.
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a. Mirror plane. Define: slices object imagine it is a mirror Examples: cube, many, beaker one, any flat molecule has at least one, example trans-1,2-dichloroethene; cis-1,2dichlorocyclobutane, has one, (extend to methyls) which spin but trans-1,2-dichlorocyclobutane has none. [Figure 5.4 for more examples] b. Center of symmetry (not in book) define point in center, extend any line, it meets the same thing at the same distance in either direction. Example cube, trans-1,3-dibromotrans-2,4-dichlorocyclobutane
4. How can we tell when a molecule is achiral without analyzing the symmetry? Look for certain structural features. a. "Asymmetric Carbon", also called a stereocenter. Most common, most important. It is sp3 and has 4 different substituents. The differences can be very minor (even H and D). A compound with one asymmetric carbon is chiral. The two mirror images are different compounds. They are stereoisomers and they are called enantiomers. Of course there can be only one pair of enantiomers. 5, Drawing achiral compounds. Because this is a 3-D phenomenon we must use a projection of some kind. We have already learned a number of different ones. Generally the dottedline wedge will be best for showing the 3-D relationships in chiral compounds. Let's draw dotted line wedge projection of the two mirror images of 2-chlorobutane. To simplify, we (unlike the book) will follow the convention of putting the main carbon chain, drawn as a
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zig zag, in the plane of the paper and the H (if there is one) on the dotted line going back from the paper, whenever possible.
Cl Cl
H
H
6. Absolute Configuration. Naming Enantiomers. System based on prioritizing (by a method used for other purposes also) the four different substituents and then indicating their arrangement with respect to each other in 3-D. The system is called the R,S system. One enantiomer has the R configuration and the other the S configuration. Remember configuration means 3-D arrangement and that configurations do not interconvert at room temperature. a. Prioritizing can be relatively simple, the higher the atomic number of the atom, the higher its priority. Thus C is higher than H and Br is higher than Cl. b. What if there is a tie?(usually between carbon atoms) Now it's not so simple. Look at the substituents on the tied atoms. The one with the first point of difference in priority of substituents wins. Compare methyl, the C has 3 H's, to ethyl, the C has 2 H's and a C (the point of difference) and propyl, the attached C has one C and two H's, and isopropyl, the attached C has one H and two C's (the point of difference) Note we don't add, just look for the first difference (somewhat analogous to numbering alkane chains) CH2Br over CCl3 and -CH2NH2 over -C(t-Bu)3
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c. How are multiple bonds handled? We use "phantom atoms." One atom attached by a double bond counts as two and by a triple bond counts as 3. So (ethynyl) (not yet covered, draw), where C is attached to 3 C's over vinyl (not yet covered, draw), where C is attached to two C's over ethyl, where C is attached to one C. d. In the unusual event that there is isotopic substitution (D, T 13C etc) ties are broken based on atomic weight. T>D>H and 13C>12C e. Now let's prioritize the 4 substituents. For 2-chlorobutane, it is Cl, CH2CH3, CH3, and H. f. Orient the projection so that the group of lowest priority (H in this case) is in back of the paper. g. Locate the group with the highest priority and go from there to the second highest priority. (In this case Cl to CH2CH3) If the path is clockwise we call the stereoisomer the R-enantiomer. If the path is counterclockwise we call it the S-enantiomer (Alphabetical mnemonic Rcl; Sco.)
Cl Cl
H
H
R
S
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You should build [models]to be sure you understand the relationships. 7. Properties of Enantiomers: Because enantiomers are identical in every way except for their "handedness" which results from their asymmetry, differences in properties are only revealed by asymmetric probes. There are many such probes. The most well known is plane polarized light. a. When ordinary light (which consists of waves oscillating in all directions and is therefore symmetrical) is passed through a prism it becomes plane polarized. It oscillates in only one direction. [Figure 5.5] b. When plane polarized light is passed through a solution of a chiral compound the plane of the light is rotated by a certain angle in a certain direction, clockwise or counterclockwise. Chiral compounds are thus said to be optically active. c. A compound that rotates light clockwise is called dextrorotatory (to the right) and is designated by a (+) or d in front of its name. A compound that rotates light to the left is called levorotatory (to the left) and is designated by a (-) or l in front of its name. There is no direct connection between d and l and R and S. d. We can measure optical rotation, including the actual direction and angle of rotation, using a device called a polarimeter. i. The actual value of the angle of rotation depends on a number of factors including the
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concentration of the solution and the distance light travels through the solution. ii. We can correct for these factors by defining the specific rotation, which refers to a solution of concentration 1.0 g/mL and a path length of 1.0 dm. The specific rotation is represented by []. [] = /l x c where l is the path length in decimeters (1 dm = 10 cm) and c is the concentration in g/mL iii. The specific rotation also depends on the temperature and the wavelength of the light. So we must also specify those. e. Whenever we detect optical activity it signals the existence of a chiral compound. But, failure to observe optical activity does not mean that we do not have a chiral compound. Why? i. Consider rotation of light by a compound with one asymmetric C atom. This compound has two enantiomers. Each one rotates light and has the same magnitude of specific rotation. But they rotate light in opposite directions, one to the right (+) and one to the left (-). Which is which? We can't predict. ii. Suppose we have a 50-50 mixture of the pair of enantiomers. What will we observe? No rotation because the two rotations, which are
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equal in magnitude but opposite in direction, cancel each other. iii. A 50-50 mixture of two enantiomers is called a racemic mixture or a racemate. It is represented by ( ). Racemic mixtures are very commonly encountered because whenever a reaction that provides a symmetrical environment takes place at a chiral carbon a racemic mixture is obtained. Why? (entropy not enthalpy). The conversion of one enantiomer into a racemic mixture is called racemization. f. Optical Purity: Sometimes we may have a mixture of enantiomers that is not racemic a mixture because an asymmetric environment didn't operate 100% or because racemization was stopped before completion. In such a case we can determine the composition of the mixture if we know the specific rotation of the enantiomers. The observed rotation is due to the excess of one of the enantiomers, which can be determined by comparing the observed rotation to the rotation of the pure enantiomers. If it is 30% of the observed for pure it must mean that 30% of the mixture is the pure enantiomer and the other 70% is racemic mixture. So there is 30% + of 70% or 65% total of that enantiomer in the nonracemic mixture. 8. Compounds with more than one asymmetric carbon; If there are no other stereocenters (eg rings or double bonds) then the maximum number of stereoisomers is 2n where n is the number of asymmetric carbons.
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Let's first focus on the case where n =2 and the two atoms are adjacent in the chain. Example 2-bromo-3chlorobutane. Let's draw a dotted line wedge projection of one isomer and its mirror image:
Br Cl Br H H R H S R H S Cl
The two mirror images are nonsuperimposable. So we have a pair of enantiomers, which have different absolute configurations. Let's change the configuration at one of the carbons in one of the dotted line projections by switching the H and Cl (now the H is in front). Note that when the H is in the front and the priority 1 group is in the back we can find the absolute configuration in the same way as when the H is in the back and the priority 1 group is in the front except that everything is opposite. We pretend the H is priority 1 instead of priority 4 and the Cl is priority 4 instead of priority 1. Then R is counterclockwise from 1 to 2 to 3 and S is clockwise from 1 to 2 to 3.
Br H Br H R H
S
Cl H
S
Cl R
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Now this isomer is not the mirror image of either of the first two. It is a diastereomer of them. Two diastereomers are stereoisomers that are not mirror images of each other. But this isomer also has a mirror image, which is its enantiomer. Note the absolute configuration relationship. In enantiomers the pair is RR-SS or RS-SR, while in diastereomers it is RR-RS or RR-SR or SS-RS or SS-SR
So two asymmetric carbons give rise to four stereoisomers. There are two pairs of enantiomers that are diastereomers of each other. b. This kind of stereoisomerism was first studied in detail by Emil Fischer in connection with his work on sugars. Such stereoisomers are related to sugars and are sometimes named after sugars. For many compounds of this type, we can identify the two pairs by noting that in the dotted line-wedge projections of one pair, similar or identical substituents are on the same side of the carbon chain, and in the other pair they are on opposite sides. When the substituents are both on a wedge (same side) we call this pair of enantiomers syn. When one is on a wedge and the other is on a dotted line we call the pair of enantiomers anti. (not in book)
c. There can be fewer than 4 stereoisomers when there are two asymmetric carbons if there is an overall
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symmetry to the molecule due to identical substituents on the two carbons. In the specific situation where the two asymmetric carbons have the identical substituents, there are only three stereoisomers. There is one pair of enantiomers and a third isomer, which is superimposable on its mirror image because of the overall symmetry that results. This isomer is called the meso and it is a diastereomer of either isomer of the enantiomeric pair. You can see the overall symmetry in the dotted line wedge projections by rotating around the 2-3 bond into the other eclipsed conformation.
Cl H Cl H S Cl H S R Cl R H
Cl Cl H H
Me Me
d. Fischer Projections. Just a brief mention. Do not use these projections for anything having to do with absolute configurations at this point. They are projections that have been traditionally used in carbohydrate chemistry, but are not much used anymore except in intro biochem courses and perhaps a bit on mcats.
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A Fischer projection is a dotted-line wedge projection without the dotted lines and wedges. We know which lines are dotted lines and which are wedges by the way the projection is oriented on the paper. The projection consists of intersecting perpendicular lines representing the four bonds of an asymmetric carbon. The carbon is at the intersection. The two verticals are dotted lines, the two horizontals are wedges. In general Fischer projections are used for compounds of more than one carbon. For such compounds the main carbon chain is always shown oriented vertically with the lowest numbered carbon at the top. Your textbook does not always follow this convention. Here are Fischer projections of the two enantiomers of 2-chlorobutane
Cl
Cl
Note we draw it with the main chain vertical, although the book does not. We test for superimposability only with 180 rotations in the plane of the paper. Fischer projections are used primarily for compounds with more than one asymmetric carbon. In such cases it is especially important that the main carbon chain be oriented vertically with the lowest numbered carbon at the top. From the way it is defined a Fischer projection must correspond to an eclipsed
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conformation. This fact is important to remember when converting between Fischer projections and dotted-line wedge projections, Fischer projections can help us see symmetry relationships in compounds with more than one asymmetric carbon. For example
Show mirror images and superimposability of the two meso mirror images by180 rotation e. Now let's look at rings. A Monosubstituted cycloalkane does not have an asymmetric carbon unless it is in the substituent. Let's ignore that. Disubstituted cycloalkanes may have 0, 1, or 2 asymmetric carbon atoms depending on the pattern of substitution, the ring size, and which geometric isomer we look at. Assume ring is flat for this purpose and look for symmetry elements. Draw projections and mirror images using the flat sawhorse. i. Two asymmetric carbons, four stereoisomers. Show the four isomers of 1-bromo-2chlorocyclohexane.
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Br Cl H H
Br Cl H
H Br Br H H Cl Cl H H
ii. Two asymmetric carbons, three stereoisomers. Show the three stereoisomers of 1,2dichlorocyclohexane and mention meso.
Cl Cl H H H Cl Cl H
Cl
Cl H H Cl H
H
Cl
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iii. Two substituents 1,4, no optical isomerism, only geometric. Will always be the case in opposite positions in even numbered rings. There is a mirror plane perpendicular to the plane of the ring that bisects C-1 and C-4 and the substituents, which are all spherical on those carbons.
Cl Br Cl
Br
Cl Br
Cl
Br
f. Naming isomers with more than one asymmetric carbon. Name each carbon separately. May require good visualization. C. Reactions of Compounds with one or more Asymmetric Carbons. 1. Case 1. No bonds of the C are broken in the reaction. The relative configuration of the four substituents is maintained although the R,S designation may change because of a change in the priority of the substituents.
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2. Case 2. One or more bonds at the asymmetric carbon are broken and new bonds are formed. What happens depends on the mechanism of the reaction. a. mechanism is stereospecific, meaning a given stereoisomer forms another given stereoisomer or meaning the spatial relations of all the participants in the reaction are specified and there are no options. We will see our first example of such a reaction in Chapter 6. Then there are two choices. i. the new bond is formed from the same direction as the old bond is broken. In this case the product has the same absolute configuration as the starting material unless priorities change. Show simple general example.
ii the new bond is formed from the opposite side. In this case the product has the opposite absolute configuration as the starting material unless priorities change. Show simple general example. b. mechanism is not stereospecific. For example it goes through a planar intermediate (e.g. free radical). Then even if the product has an asymmetric carbon, the product will be a racemic mixture. [Figure 5.13] Even if we start with a compound that has a chiral center, generation of the free radical at that carbon results in the formation of a racemic mixture.
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B. Reactions that form one or more asymmetric carbons. What happens depends on the mechanism, the reactant, and sometimes even the conditions under which the reaction is carried out. A reaction that forms one of a pair of stereoisomers preferentially is said to be stereoselective. A. Addition Reactions that form one asymmetric center 1. If we start with an optically inactive compound and carry out a reaction that forms an asymmetric carbon we get both enantiomers. So if that is the only asymmetric carbon in the molecule, the product is a racemic mixture. 2. But if start with a molecule that already has an asymmetric carbon and we start with one of the enantiomers of that molecule then we get two products that are diastereomers of each other. Suppose the enantiomer we start with is R then the product will be R, R and R, S, which are diastereomers. But we do not necessarily form these in equimolar amounts. [Figure 5.14]
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