Stereochemistry. Conformers: Compounds that differ by orientation of atoms in space. They are interconvertible via rotation about single bonds.

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1 Stereochemistry Terms onformers: ompounds that differ by orientation of atoms in space. They are interconvertible via rotation about single bonds. onstitutional isomers (also called structural isomers): ompounds with the same chemical formula, but differ in their connectivity. They cannot be interconverted without breaking covalent bonds. Stereoisomers: ompounds with the same constitution (same numbering in name), but differ by the spatial arrangement of atoms. They cannot be interconverted without breaking covalent bonds. Enantiomers (optical isomers): A pair of molecules that are mirror images of each other, but are not identical (no matter how they are rotated). Enantiomers are one type of stereoisomer. Stereogenic enter: A carbon atom (or other atom) in a molecule that is substituted in such a way that a new stereoisomer is produced when two substituents (on the same carbon/atom) are switched. A hirality center is a carbon atom (or other atom) that is attached to four different substituents. All chirality centers are stereogenic centers. onfiguration: A chirality center can have one of two configurations, or S (to be defined later). Switching any two substituents on a chirality center changes the configuration (from to S or from S to ). Likewise, the mirror image of an center is S, and the mirror image of an S center is. ommon abbreviations: Me = methyl; Et = ethyl; Pr = propyl; Ph = phenyl (a benzene ring). hirality enters? Stereochemistry and rings: Treat rings a flat; ignore conformations because they interconvert rapidly.

2 hirality enter?: ings Treat the rings themsleves as flat; ignore conformers as they interconvert rapidly ortisone: circle all the chirality centers.

3 Be sure you draw molecules well and clearly: Good Drawings: ' ' ' Bad Drawings: ' ' ' hiral (handedness) compounds: ompounds that exhibit enantiomerism. hiral compounds are optically active (optically activity will be discussed later). mirror A B B A D D Achiral compounds: ompounds whose mirror images are superimposable. Achiral compounds are optically inactive. hiral Molecules: Is Molecule X hiral? A chiral molecule: (1) as handedness, (2) can be separated into enantiomers, and (3) is not superimposable on its mirror image. A molecule that contains only one stereogenic center is chiral. A molecule that contains a plane of symmetry (or mirror plane; not to be confused with mirror image) is achiral. A molecule with two or more stereogenic centers may or may not be chiral. [A compound that contains no stereogenic centers is likely to be achiral, but could be chiral: helical molecules and atropisomers are examples of chiral molecules with no stereogenic centers.]

4 3? ? Allenes l l Atropisomers Me F F Me l l Problem: Draw all the constitutional and stereoisomers named bromochlorocyclopentane. n an exam, you will lose points if you draw the same structure twice. Structural Determination: ow do you know what you have? If you make a compound or isolate a compound, how do you know what its structure is? ne large set of methods of structural analysis or characterization is by spectroscopic methods. Photons over the range of the electromagnetic spectrum interact with molecules in various ways. X-rays can be diffracted, which gives direct structural information, based on electron densities. UV absorption relates to electronic excitation (e.g., raising an electron from a pi to a pi* orbital). Microwave absorption relates to rotational excitation. Infrared

5 relates to vibrational excitation. adiowave absorption relates to nuclear excitation (NM) and distinguishes different types of hydrogens (e.g., N 3 from 3 ). The photons can be absorbed, scattered, or emitted, among other forms of interactions. Enantiomers of a molecule can only be distinuished by their structure (visually), by recognition by other molecules, or by optical rotation. Nearly all other properties are the same for two enantiomers: NM, I, UV, rotation, mp, bp, etc. Light interacts with chiral molecules in such as way that the light wave rotates. Picture a sine wave along an x-axis rotating about this axis. Enantiomers of a molecule rotate light in equal but opposite directions. To observe the rotation, the light must be plane polarized. therwise, the randomness of the light waves would appear to average out to zero rotation. The more concentrated the solution, the more chiral molecules each photon will be rotated by, the more overall rotation will be observed. Likewise for the pathlength: the longer the pathlength, the more molecules each photon will encounter. Both pathlength and sample concentration are normalized by specific rotation. ptical Activity: otation of plane polarized light that occurs on passing through a sample containing one (or a majority of one) enantiomer. otation is measured using a polarimeter. otation of light in the right-hand direction (clockwise) is called dextrorotatory, d, or (+). otation of light in the left-hand direction (counter-clockwise) is called levorotatory, l, or (-). Specific otation: Defined for an optically pure sample as: α [ ] D 20 = α lx[x] 20: 20 (temperature of the experiment) D: D-line of sodium (wavelength of light used in the experiment) α: observed rotation from polarimeter l: pathlength of cell containing sample (often 10 cm) [X]: concentration of sample (in g/ml) A acemic mixture is a 1:1 mixture of corresponding enantiomers. otation = 0 because enantiomers have equal but opposite optical activity. ptically Pure: ne and only one enantiomer is present. ptically active: a majority of one enantiomer is present. ptically inactive: either (1) no chiral compounds are present, or (2) there is a racemic mixture. Problem: [ ]-α-carvone has a specific rotation of 72. A solution of α-carvone has a rotation of +50. What is the % of [ } and [+] enantiomers?

6 3 3 2 [ ]-α-carvone Solution: Let x = the fraction of [ ]; 1 x = fraction of [+]. 72x + 72(1 x) = x x = x = 22 x = 0.153; the solution contains 15.3% [ ] and 84.7% [+]. heck: 15.3%( 72) $(+72) =? (+50 ) Another method: = 22; 22/(72+72) = 15.3% ahn-ingold Prelog ules: Enantiomers can be distinguished on paper (and thus, by name) using a system of nomenclature called the ahn-ingold-prelog rules. These rules assign onfiguration (i.e., or S; see below) to each chiral carbon. First, assign priority to each substituent on the stereogenic carbon using rules 1 and 2: 1. ompare the atomic number of the atoms that are attached directly to the stereogenic carbon. The atoms with the highest atomic number have the highest priority. The second-highest atomic number has the next highest priority, etc. 2. If there is a tie, go to the atoms that are two bonds away from the stereogenic carbon. Again, assign priorities by atomic numbers. For example, ethyl has higher priority than methyl: There is a tie at the first carbon, but then there are 3 next (i.e., two bonds away) in the case of the methyl group, but 2 and a next in the case of the ethyl group. The in the ethyl case "beats" all three 's in the methyl case: >, >, > ; the 's are not "added together". 3. Draw the molecule such that the lowest-priority substituent (usually ) is sticking into the page and the other three substituents form a steering wheel. ount going either clockwise or counter-clockwise, depending on the priorities. lockwise is assigned "" (for rectus, Latin for right) and counter clockwise is assigned "S" (for sinister, Latin for left).

7 F l 1 1 F l F 3 4 l 2 F 3 4 l 2 For the compound below, ties in priorities have to be broken. We can do this with a table (eventually, you can do this in your head): l l l Substituent l Methyl Ethyl At one bond l away: Two bonds away: ,,,, We can see from one bond away that l is the number 1 priority and is number 4. For methyl and ethyl, we need to go two bonds away. The one carbon (the green one) of ethyl beats each of the three s of the methyl. The center is. The compound is 2-()-chlorobutane or ()-2-chlorobutane. Problem: draw (S)-2-chlorobutane. 1 l arbonyls: For assigning the priorities of carbonyls, treat the carbonyl carbon as if it is bound to two oxygens: Treat: as: Assign /S configuration to alanine, below: N

8 N Substituent N 2 2 Methyl At one bond N away: Two bonds away: ,,,, The nitrogen wins in round one as does the hydrogen lose. ound two goes to the carboxylic acid. The compound is (S)-alanine. 1 N ings: Assign priorities the same way. 3 Substituent Methyl At one bond away: Two bonds away:,, ---,,,, loses. Methyl is third. For the ring, going to the right as drawn we get to 2, then an. Going to the left, we get to 2 then another 2. From the table, we can see that the winner is the side with the.

9 What if there are two or more chirality centers in a molecule? Simply assign each center, independently For the compound above, first assign prioritied to the carbon on the left (in red). Note that the other chirality center is included as the second priority group (dotted circle). The configuration will be S because the hydrogen (not drawn) is pointing toward you. Next, assign priorities for the second carbon (middle structure, carbon in red). Likewise, this carbon is S. ptions on viewing the 4 th priority group: If it is pointing away from you, the molecule is setup for assigning configuration. If not, there a few things you can do; choose the option you find easiest. (1) Use a model: be sure the model is consistent with the drawing, then assign configuration. You can simplify the model to a tetrahedron with four different colored balls, where you assign appropriate priorities to the colors. This way all such assignments can be done using one or two models. You can bring these pre-built to exams; just don t write on them. (2) edraw the molecule so that the 4 th substituent is pointing away from you. Be careful to redraw the molecule properly. (3) Visualize the molecule so that you are looking down the 4 th bond properly. This may entail visualizing from behind. (4) Simply switch the rule: if looking down the 4 th bond the wrong way (i.e., it is pointing toward you), the rule becomes: clockwise = S; counterclockwise =. Enantiomerism in nature. Most physical and chemical properties of enantiomers are identical (optical activity being a key exception). owever, nearly all reactions of chiral compounds that occur in nature involve only one enantiomer. Enzymes are chiral and recognize one enantiomer over the other. The recognition is like two right hands shaking together. A right and a left just won t do. The enantiomers of α-carvone, which we encountered earlier, have different properties that we can detect: one smells like caraway seeds (rye bread) and the other smells like spearmint. Some people S S

10 have a mutation and cannot make the distinction. Such folk usually do not like rye bread as it tastes minty to them [ ]-α-carvone A much more somber story is that of thalidomide. This compound was used as an anti-nausea drug in the 1950 s. Sadly, it led to birth defects. Turns out that one enantiomer is responsible for the antinausea activity while the other is responsible for the birth defects. The drug was sold as a racemic mixture. Later there was the idea that had only the active enantiomer be used, no side-effects would have occurred. owever, the two enantiomers interconvert in the body, so there was no way around the side effects other than avoiding the drug altogether. N Thalidomide N "" and "S" configuration cannot be used to predict optical rotation. d(+) and l(-) are observables. and S are conventions. There is no relationship between /S and d/l, except if a molecule has only one chiral center and if = d, then S = l (and vice versa). [D and L refer to amino acid and carbohydrate nomenclature and are not a large part of hem 260.] In a pair of enantiomers, the configurations about all stereogenic centers are opposite. In creating the mirror image of a molecule, all configurations become S, and all S become. If you switch two substituents on a chiral center, becomes S, and S becomes. Diastereomers: Stereoisomers that are not enantiomers. Diastereomers have different properties. Meso compounds: ompounds that have more than one stereogenic center, but are achiral (and have stereoisomers that are chiral).

11 2 N 2 2 N 2 2 N 2 2 N A B D Enantiomers Enantiomers Diastereomers: A and ; A and D; B and ; B and D. Tartaric Acid (+) ( ) Enantiomers (meso)-tartaric Acid Identical, Achiral, Meso (+) and ( ) tartaric acid are enantiomers. (+) and meso tartaric acid are diastereomers. ( ) and meso tartaric acid are diastereomers. amifications of hanging onfiguration of Stereogenic enters: (1) If a compound has only one stereogenic center, and you change the configuration (from S, or from S ), you create the enanatiomer of that compound. (2) If a chiral compound has more than one stereogenic center, and you change the configuration of all of its stereogenic centers, you create the enantiomer of that compound. (3) If a compound has more than one stereogenic center, and you change the configuration of any but not all of its stereogenic centers, you create a diastereomer of that compound. is-trans isomers: Substituted alkenes can exist as either cis or trans stereoisomers (sometimes called geometrical isomer), as the substituents can be either on the same side (cis) of the double bond or on opposite sides (trans). is and trans stereoisomers are a type of diastereomers. As the substituents get more complex and/or more than one double bond exists in a molecule, complications arise. A system of nomenclature based on the ahn-ingold-prelog rules has been developed. First, assign priorities for each substituent on each alkene carbon. If the two highest-priority substituents are on the same side of the double bond (as in cis), it is called a Z double bond for zusammen, together ("Zee same side"); if the substituents are on opposite sides (as in trans), the double bond is E for entgegen, opposite. For alkenes, always use E/Z nomenclature, not cis/trans. Note that E/Z isomers are indeed isomers, as the energy barrier for interconversion (breaking of a π-bond) is 65 kcal/mol.

12 For the example below, the ethyl has the higher priority on the left hand carbon; the methyl has the higher priority on the right hand carbon (middle figure). Next (structure on right), the two winners are on the same side of the double bond (on the top, as drawn), so the compound is (Z)-3-methyl-2- pentene Disubstituted cycloalkanes can have the two substituents either on the same side of the ring, cis, or on opposite sides of the ring, trans. The isomers cannot interconvert without breaking a σ-bond, which is around 80 kcal/mol. Always use cis/trans for cycloalkanes. For stereochemical determinations (e.g., is compound X chiral?), treat cycloalkanes as flat since the conformers interconvert rapidly. If compounds differ in their numbering (and have the same name otherwise), they are onstitutional isomers. If they have the same numbering (and name, outside of stereochemical designations), they are stereoisomers. ow many stereoisomers are there for dimethylcyclohexane? All Isomers Named Dimethylcyclohexane onstitutional Isomer Stereoisomers with denoted onstitution Number of Stereoisomers 1,1 1 1,2 3 1,3 3 1,4 2 All 9 There is only one 1,1-isomer. There are three 1,2-isomers, one cis and two trans. The two trans are enantiomers; the cis is a diastereomer of the trans. The case for 1,3 is similar to 1,2. For 1,4, there are only one cis and one trans isomer. ow many stereoisomers are there for 1-ethyl-3-methylcyclohexane? (ans: 4) ow many for 1-ethyl-4- methylcyclohexane? (ans: 2)

13 Naming: If there are two or more stereocenters, name each or S with the appropriate number indicating which carbon refers to the designation. For example, (1,2)-trans-1,2- dimethylcyclohexane, (1S,2S)-trans-1,2-dimethylcyclohexane, and cis-1,2-dimethylcyclohexane. Although it doesn t matter with the first two examples where you start the numbering, you should include the numbers. For the cis example, the stereocenters are,s, which is the same as S, (you can start at either stereocenter), but you don t need to indicate this because the molecule is achiral (meso). More detail: The compounds below are both trans-1,2-dimethylcyclohexane. They have opposite configuration at both stereogenic centers. They have the same constitution, are mirror images of one another, and are not identical: they are enantiomers. They are named trans-(,)-1,2- dimethylcyclohexane and trans-(s,s)-1,2-dimethylcyclohexane. You can leave out the trans if you like, as that is defined in the -(,)- or -(S,S)-. You can write -(1,2)- or -(1S,2S)-, but that is unnecessary. The numbering can go before or after the word trans. You do need to indicate the configuration, as that is how to distinguish one enantiomer from the other. In principle, the word trans can be left out in this case, but it is left in for convenience. S S The compounds below are cis-1,2-dimethylcyclohexane. They have the same constitution and although they are mirror images of one another, they are identical, so they cannot be enantiomers. You can include (,S), (1,2S), or (1S,2) in the name, but it is unnecessary. Note that cis-1,2- dimethylcyclohexane is a meso compound. S S For the compound below, the has priority, so it is carbon-1. You can name it 2S-methyl-bromocyclopentane or (1, 2S)-2-methylbromocyclopentane. 3 For each constitutional and stereoisomer of dimethylcyclohexane that contains the name dimethylcyclohexane, draw the chair conformations. Draw the chair conformations from different perspectives (e.g., attach the first substituent to each of the 12 positions).

14 For the 1,1-isomer, there is only one stereoisomer. Upon ring-flipping, the new chair is indistinguishable from the first. The conformers of cis-1,2-dimethylcyclohexane are in rapid equilibrium, just as any two chair conformations are. The two conformers of cis-1,2-dimethylcyclohexane can be considered to be conformational enantiomers, and as such, a kind of racemic mixture. But due to the rapid equilibration, we simply consider the molecule to be achiral. The two conformers are of equal energy and thus exist as a 50:50 mixture. For the 1,2-trans isomer, the analysis for each enantiomer is the same. Draw either conformer first. Generate the other conformer by switching axial to equatorial, and equatorial to axial. Keep up as up and down as down. The two conformers in this case are different in energy by 2.7 kcal/mol. This is the difference between two axial versus two equatorial substituents (3.6 kcal) less 0.9 kcal. What accounts for the 0.9 kcal? For the 1,3-isomer, let s start with the cis-stereoisomer. Two axial versus two equatorial substituents yields a ΔG = 3.6 kcal/mol. For the 1,3-trans, the two conformers are indistinguishable. Keep looking until you can see this. The case is the same for the enantiomer.

15 For 1,4-trans, ΔG = 3.6 kcal/mol. For cis-1,4, the two conformers are indistinguishable.