Organic Chemistry

Richard F. Daley and Sally J. Daleywww.ochem4free.comOrganicChemistryChapter 11Chirality11.1 Symmetry and Asymmetry 53211.2 Nomenclature of Stereocenters 53811.3 Properties of Asymmetric Molecules 544Sidebar - Chiral Recognition 54411.4 Optical Isomerism 54611.5 Fisher Projections 54911.6 Molecules with Two Stereocenters 55311.7 Resolution of Enantiomers 55811.8 Stereocenters Other than Carbon 561Key Ideas from Chapter 11 564

Organic Chemistry - Ch 11 531 Daley & DaleyCopyright 1996-2005 by Richard F. Daley & Sally J. DaleyAll Rights Reserved.No part of this publication may be reproduced, stored in a retrieval system, ortransmitted in any form or by any means, electronic, mechanical, photocopying,recording, or otherwise, without the prior written permission of the copyrightholder.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 532 Daley & DaleyChapter OutlineChapter 11Stereochemistry11.1 Symmetry and AsymmetryStructural features that differentiate betweensymmetric and asymmetric molecules11.2 Nomenclature of StereocentersNaming molecules with stereogenic centers11.3 Properties of Asymmetric MoleculesThe various properties of asymmetric molecules11.4 Optical ActivityOptical activity as an indicator of an asymmetricmolecule11.5 Fischer ProjectionsWorking with Fischer projections of asymmetricmolecules11.6 Molecules with Two AsymmetricCentersA description of asymmetric molecules with twostereogenic centers11.7 Resolution of EnantiomersChemical separation of enantiomers from aracemic mixture11.8 Asymmetric Centers Other thanCarbonStereogenic centers involving atoms other thancarbonwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 533 Daley & DaleyObjectives✔ Be able to find molecules that are asymmetric and recognize if anystereogenic centers are present✔ Understand that nonsuperimposible mirror image compounds haveidentical physical and chemical properties in a symmetricenvironment, but are different in an asymmetric environment✔ Know the nomenclature of stereogenic centers✔ Be able to work with Fischer projections of stereogenic centers✔ Know the difference between enantiomers and diastereomers✔ Recognize the techniques used to separate enantiomersTh' wurruld has held a lookin'-glass in front iv ye fr'mth' day ye were born an' compelled ye to make facesin it.—Finley Peter DunneMr. Dooley Says (1910)Chapter 11 continues the study begun in Chapter 3 of thethree-dimensional structural features of molecules. Thiss tudy, known as stereochemistry, is important because the waymolecules react with each other depends, in part, on their threedimensionalstructure. The chemical reactions discussed in Chapter12, and onward, require an understanding of the stereochemistry ofthe molecules. Stereochemistry is especially important withbiologically significant molecules such as amino acids, proteins,carbohydrates, and nucleic acids because they react and behave inways highly dependent on their stereochemistry.So far, you have studied two categories of isomers: structuralisomers and geometric isomers. Structural isomers differ from eachother in the bonding sequences of the atoms they contain. Butane and2-methylpropane are examples of structural isomers. Geometricisomers have identical bond sequences, but the spatial relationshipbetween those bonds is different. For example, cis-2-butene and trans-2-butene are geometric isomers. This chapter introduces a third typeof isomer called stereoisomers. Stereoisomers are molecules that havethe same bond sequence, but different three-dimensional structures.If you have not yet developed the ability to visualize complexobjects in three-dimensions from a two-dimensional representation,www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 534 Daley & Daleymake molecular models of the examples. Using molecular modelshelps to clarify the molecular relationships discussed in this chapter.11.1 Symmetry and AsymmetryA symmetric moleculecan be superimposedon its mirror image. Anasymmetric moleculecannot besuperimposed on itsmirror image.All three-dimensional objects are either symmetric orasymmetric. When the structure of an object allows you tosuperimpose it upon its mirror image, that object is symmetric, oridentical, to its mirror image. The rectangular box on the left in Figure11.1 is superimposible upon its mirror image, so the box is symmetric.However, if you take the same box and paint an X in one corner, youcan no longer superimpose it on its mirror image. The box is thenasymmetric. Note that the difficulty in superimposing the box with theX onto its mirror image is the impossibility of placing one X on top ofthe other X.MirrorsFigure 11.1. The box on the left is superimposible on its mirror image. Painting an Xin one corner, as shown on the right, makes the box nonsuperimposible on its mirrorimage.A molecule possesseschirality if it cannot besuperimposed on itsmirror image.The nonsuperimposibility characteristic of an object upon itsmirror image, a trait common to all asymmetric objects, is calledchirality. The term chiral comes from the Latin word chiro or chirthat means “hand” or “handedness.” The most obvious example ofchirality is your own hands. Examine them. A typical human handhas four fingers and a thumb. Although all hands have these samebasic structural components, hands are asymmetric. That is, a lefthand is a nonsuperimposible mirror image of a right hand. Placingyour palms together makes this mirror image relationship clearlyvisible. Many other common objects also have chirality: golf clubs, thethreaded base of a light bulb, shoes, and even this book.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 535 Daley & DaleyA plane of symmetry isan internal mirrorplane.As with all other three-dimensional objects, organic moleculesare either symmetric or asymmetric. A way to determine whether aparticular molecule is symmetric or asymmetric is to determinewhether the molecule contains a plane of symmetry. If there is aplane of symmetry, then the molecule is symmetric. If there is noplane of symmetry, then the molecule is asymmetric. A plane ofsymmetry bisects the molecule into two halves that are exact mirrorimages of each other. Ethyne (or acetylene) contains an infinitenumber of planes of symmetry. Figure 11.2 shows three of thoseplanes of symmetry, but make a molecular model to help yourselfvisualize them.Figure 11.2. Acetylene has an infinite number of planes of symmetry.Other symmetric molecules have only one or two planes ofsymmetry. Molecules with a C=C or a C=O double bond have at leastone and often two planes of symmetry. Because these functionalgroups are planar, they impart only symmetry, never asymmetry, to amolecule. However, other portions of the molecule may make amolecule with a double bond or carbonyl group asymmetric. Make amolecular model of propene (Figure 11.3). Rotate the methyl groupuntil you place one of the hydrogens in the same plane as the threehydrogens bonded to the double bonded carbons. Hold the molecule soyou are looking at this plane. Notice that the plane bisects all fourhydrogens within the plane and that the other two hydrogens areidentical mirror images of each other giving the whole molecule aplane of symmetry. This plane is the only plane of symmetry inpropene.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 537 Daley & Daleyc) Spiro[4.3]octaned) trans-DecalinSample Solutionb) 1-Methylcyclohexanol has one plane of symmetry that bisects the —OH and —CH 3 groups on C-1 as well as C-1 and C-4.OHPlane of symmetryCH 3An enantiomer is one ofa pair ofnonsuperimposiblemirror imagemolecules.Molecular asymmetryis the result of amolecule’s shape.An organic molecule without a plane of symmetry or center ofsymmetry is asymmetric. A characteristic of all asymmetric moleculesis that they possess chirality. That is, for that molecule there exists anonsuperimposible mirror image molecule. All chiral, or asymmetric,molecules occur in pairs with a “right hand” and a “left hand”molecule. Each member of the pair is called an enantiomer. Forexample, there are two enantiomers of 2,3-pentadiene. To confirmthis, make a model of Figure 11.5 and then another of its mirrorimage. As you examine these two stereoisomers, note that they do nothave a plane of symmetry and that you cannot superimpose onemolecule upon the other. Their asymmetry is due to their shape, ormolecular asymmetry.HCH 3CCCHCH 3Figure 11.5. 2,3-Pentadiene is an example of a chiral molecule that does not have aplane or center of symmetry.A stereogenic center isa specific atom in amolecule that isresponsible for some orall of the molecularasymmetry.Although all chiral molecules have molecular asymmetry,many asymmetric molecules have a stereogenic center or, morespecifically, an asymmetric carbon atom. An asymmetric carbon atomis often called a stereocenter. An asymmetric carbon atom is bonded tofour nonequivalent substituents. This asymmetry is a propertyallowed by the tetrahedral structure of a carbon atom.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 538 Daley & DaleyA symmetricalmolecule has nostereogenic centers andhas an internal planeor center of symmetry.Examining the symmetric properties of the tetrahedral carbonatom will help you understand a stereogenic center. For example,make a molecular model of dibromomethane (Figure 11.6) which is acarbon atom with two pairs of equivalent substituents. Notice that themolecule contains two planes of symmetry. One bisects the carbonatom and lies between the two bromine atoms; the other also bisectsthe carbon atom, but it lies between the hydrogen atoms. The portionsof the molecule on either side of either plane of symmetry are mirrorreflections of each other. Construct the mirror image ofdibromomethane. Notice that the two are superimposible. That is,dibromomethane is symmetrical.HHCBrBrPlanes ofsymmetryBrHCHBrFigure 11.6. Planes of symmetry in dibromomethane.Now substitute chlorine for one of the bromines and fluorine forone of the hydrogens. The name of the new molecule isbromochlorofluoromethane (Figure 11.7).HFCClBrBrHCFClFigure 11.7. Two different three-dimensional representations of the same enantiomerof bromochlorofluoromethane. This molecule has no plane of symmetry.With four different substituents attached to one carbon atom,bromochlorofluoromethane contains no element of symmetrywhatsoever. The molecule is asymmetric. Because the molecule isasymmetric, a nonsuperimposible mirror image molecule ofbromochlorofluoromethane exists as its enantiomer.Convince yourself that the mirror image molecules ofbromochlorofluoromethane are not identical structures by trying tosuperimpose one model on the other. Orient the two enantiomers intoa mirror image position and slide them together placing the hydrogenwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 539 Daley & Daleyand the fluorine together. As you move the two structures together tosuperimpose them, note that the bromine and the chlorine point inopposite directions. Next flip the molecules so that the bromine andchlorine are lined up, but now the hydrogen and fluorine point inopposite directions. No matter how you twist, slide or turn thestructures, you cannot obtain an overlap of the two isomers. You canmake any two of the atoms coincide, but then the other two do not.The structures are asymmetric because they are nonsuperimposible.Mirror planeHFCClBrClBrCHFMirror images of bromochlorofluoromethaneHFHCFCl BrCBr ClAttempting to superimpose the enantiomers of bromochlorofluoromethaneExercise 11.2Construct each of the following molecules and its mirror image. Thendetermine whether one is superimposible upon the other.a)H 3 CHCCH 3CH 2 Clb)HOCH 3c)OHd)CH 3CH 3e)HBrCCH 3f)CH 2 ClCH 3CH 3www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 540 Daley & DaleySample solutiona) The molecule is symmetrical; therefore the mirror images aresuperimposible. A plane of symmetry exists between the two methylgroups.CH 3CCH 3H CH2Cl11.2 Nomenclature of StereocentersThe Cahn-Ingold-Prelog convention is amethod of designatingthe placement of groupsabout a stereogeniccenter.After examining several pairs of enantiomers in Section 11.1,you should now know that the structure of each enantiomer within apair is different. However, based on the rules of nomenclature inChapter 2, they have the same name. Although the basic name of thetwo enantiomers is indeed the same, you need a way to differentiatebetween the two. The most widely accepted procedure, also adopted bythe IUPAC committee, is the Cahn-Ingold-Prelog convention. Thismethod designates each stereocenter with a letter, either (R) or (S). Todetermine whether to use an (R) or an (S) follows this three-stepprocedure.Step 1 Assign a priority number to each atom, or group of atoms,bonded to the stereocenter.Step 2 Redraw the structure using the Wedge and Dash method.Orient the lowest priority group away from you and the otherthree groups in the plane of the paper.Step 3 Then draw an arrow from the first priority group, past thesecond towards the third. If the arrow circles clockwise, call thestereocenter atom (R) (rectus, Latin for right-handed). If thearrow circles counterclockwise, call the stereocenter atom (S)(sinister, Latin for left-handed).To determine the priority numbers, look at the atomic numberof the four atoms—the higher the atomic number, the higher thepriority number. For example, bromochlorofluoromethane has a Br,Cl, F, and H attached to the carbon. Assign the highest prioritynumber, 1, to Br, as its atomic number is 35. The next prioritywww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 541 Daley & Daleynumber, 2, goes to Cl, which has an atomic number of 17. Then comesF, priority number 3, with an atomic number of 9, and finally H,priority number 4, with an atomic number of 1. If the atoms aredifferent isotopes of the same element, prioritize them according toatomic weight, giving the heaviest isotope the highest prioritynumber. For example, tritium ( 3 H) comes before deuterium ( 2 H),which comes before hydrogen ( 1 H).If there is a tie, use the atoms bonded to the tied atoms todetermine which gets the higher priority number. For example,consider a stereogenic carbon attached to an isopropyl (—CH(CH 3 ) 2 )group and an ethyl (—CH 2 CH 3 ) group. Before going any further, makea molecular model. For the other two groups attached to thestereocenter, use any two different atoms (H and Br, for example). Thename of such a molecule is 3-bromo-2-methylpentane. Br and H getpriority numbers 1 and 4 respectively, but because the other twogroups begin with a carbon, there is a tie for numbers 2 and 3. Tobreak the tie, you must compare the atomic numbers of the otheratoms attached to each of the two carbons. The first carbon of theethyl group is bonded to two hydrogens and one other carbon. In theisopropyl group, the first carbon is bonded to one hydrogen and twoother carbons. Thus, the isopropyl group has the higher prioritybecause it has two other carbons in contrast to only one other carbonin the ethyl group.When considering groups bonded to a stereogenic carbon in acyclic compound, start one group with the atom immediately to theright of the stereocenter and start another group with the atomimmediately to the left of the stereocenter. In the case of a tie in acyclic compound, continue around the ring until you find a difference.In the example below, the first point of difference is the directionaround the ring that comes to the double bond first.First point ofdifference21OH3CH 34To determine the priority number of atoms involved in a doubleor triple bond, treat the multiple bond as if it were broken into thecorresponding number of single bonds with the atoms at each end ofthe multiple bond doubled or tripled.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 542 Daley & DaleyH HThus, C C becomesCCbecomes C CHHH, and CHCCO becomesHCCH, C C HHC O. Thus, forC CO Cexample, a carbon double bonded to another carbon is considered asthough it was bonded to two carbonsAfter you have determined the priority of the groups label thegroup with the highest priority atom 1. The group with the nexthighest priority label 2, the group with the third highest priority label3, and the group with the lowest priority label 4.Finally, draw the molecule with the lowest priority group backand the other three groups in front of the plane of the paper.21C34When the molecule is drawn with this orientation, add a curved arrowpointing from number 1, past number 2, and on to number 3.21C34If the curved arrow is clockwise, the isomer drawn is the (R) isomer. Ifthe arrow curves counterclockwise the isomer is the (S) isomer.The amino acid serine has one stereocenter with —NH 2 , —CH 2 OH, —COOH, and —H bonded to it. The highest priority group isthe —NH 2 group because nitrogen has the highest atomic number.The carboxylic acid comes next because it has two oxygens, with one ofthem doubly bonded to carbon. The —CH 2 OH is third with only oneoxygen. Last is the hydrogen because it has the lowest atomic number.The figure below shows the two enantiomers of serine, the naturallyoccurring serine on the left and the synthesized serine on the right.The structure of the naturally occurring serine molecule is (S); thestructure of the synthesized serine is (R).www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 543 Daley & DaleyHOOCHOH 2 CCHNH2HH 2 NCCOOHCH 2 OHNatural serineSynthetic serineSome groups bonded to a stereocenter carbon are long chains.To determine whether the stereocenter is (R) or (S), continuesearching along these chains away from the stereocenter to the firstdifference between similar chains. Give the chains their prioritynumbers according to this difference.Solved Exercise 11.1Determine whether the following molecule has any stereogenic centers. Markall asymmetric carbon atoms and determine whether they have an (R) or (S)configuration.ClHCH 3OHSolutionThis molecule has three stereogenic carbon atoms. The carbon bearing the —OH group, the one bearing the —Cl and the one bearing the —CHClCH 3group. These three are starred in the following structure.Cl**HCH 3*OHThe carbon atom bearing the —CHClCH 3 has an (R) configuration.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 544 Daley & DaleyHClHRCH 3OHYou need to rotate the bond between the six-membered ring and the carbonbearing the —Cl group. Doing so helps to see that the atom has an (R)configuration.HClHCH 3HCH 3RClHOHOHThe carbon bearing the —OH group also has an (R) configuration. Flippingthe molecule 180 o helps to more clearly see the configuration of this carbonatom.ClH 3 CHRHO H(R,R,R)-3-(1-Chloroethyl)cyclohexanolExercise 11.3For each of the following examples, identify any stereogenic carbonatoms, and then determine whether they have (R) or (S)configurations.a)b)H 3 CHCOHCH 2 CH 3CH 3 CH 2BrCCH 3Hwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 545 Daley & Daleyc)HCH 3d)HCH 3CH 3ClCle)Clf)Og)h)OHOHCH 3Sample solutionb) Assign the priority to the four groups this way.2 3CH 3 CH 2 CH 3CBrH41Then redraw with the lowest priority group back. This structure hasan (S) configuration because the arrow curves counterclockwise. Themolecule is (S)-2-bromobutane.34HCH 3CH 2 CH 3Br1211.3 Properties of Asymmetric Moleculeswww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 546 Daley & DaleyA chiral probe is atechnique that uses aninstrument or chemicalcompound todistinguish betweenindividualenantiomers. Thechiral probe must bechiral in some fashion.A polarimeter is aninstrument thatmeasures the numberof degrees a moleculeturns a plane ofpolarized light.An enzyme is a largechiral moleculeconsisting of chiralamino acids, such asserine (page 000), withchiral reactive sites.Most of the physical properties of both pure enantiomers areidentical. For example, both molecules have identical melting points,boiling points, densities, NMR, IR, MS, and indexes of refraction. Theydo differ, however, in two significant ways. 1) Both enantiomers of apair rotate a plane of polarized light and rotate it the same number ofdegrees, but they rotate the light in opposite directions: one clockwiseand the other counterclockwise. Thus, they are called optical isomers.2) Each enantiomer of a pair reacts in an individual way in anasymmetric chemical environment to form another asymmetriccompound. Because of their identical physical properties, the only wayto distinguish between enantiomers is by using a chiral probe.Polarimeters are chiral probes, as are chiral, or asymmetric,molecules.The two molecules in a pair of enantiomers react differentlyfrom each other with another asymmetric compound because of themolecular asymmetry of the pair of enantiomers. Reactions take placewhen two different molecules physically come in contact with eachother. Thus, their shapes determine how they fit together and how or,in some cases, whether or not they will react with each other.Your hands represent a very simplistic example of thisprinciple. As mentioned previously, your hands are chiral objects, solet them represent a pair of enantiomers. React that pair ofenantiomers with a symmetric compound, represented by a pair ofgreen flat mittens. Because both mittens will easily fit on either hand,they illustrate how two enantiomers react in the same way with asymmetric compound. Now suppose you react a pair of enantiomerswith a pair of asymmetric compounds represented by one red mittenand one green mitten. In this case, either mitten can fit on eitherhand, but the right hand wearing the green mitten is different thanwhen it is wearing the red mitten. There are four possiblecombinations with a red or green mitten on the left or right hands.To extend this example further, replace the mittens withgloves. In this case, the right glove fits only on your right hand, andthe left glove fits only on your left hand. An example of a glove-likecompound is an enzyme. Enzymes catalyze the reactions that takeplace within the cells of your body and in all other living organisms.Each enzyme is specifically shaped and fits only with thecorrespondingly shaped enantiomer making up the nutrients from thefood you eat, the medicines you take, or the hormones your bodyproduces. This situation is analogous to fitting your hand into a glove.Because the reaction takes place during the time the enzyme and theenantiomer are fitted together, the two won't react if they don't fit.SIDEBAR—Chiral Recognitionwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 547 Daley & DaleyOpen a jar of spearmint oil. Smell it. Then taste it. Next, do thesame with a container of caraway seed. As you can observe, their odorand taste are very different. However, the principal component for theflavor and odor of both are enantiomers of each other. (R)-carvone isthe principal component of spearmint oil. Its enantiomer, (S)-carvone,is the principal component of caraway oil. The only difference betweenthe two enantiomers is the arrangement of their atoms in space.CH 3CH 2CH 3CH 2OOCH 3(R)-CarvoneCH 3(S)-CarvoneThe distinctive odor of each is the result of the odor receptorsin the nose that appear to be based on chiral receptor sites. Thus, thenose perceives odors based on the fit of the odor-causing molecule intothe receptor sites located in the nose. This behavior is similar tofitting a key into a lock—only the key designed for a specific lock fits.Likewise, the odor-causing enantiomers fit only into the appropriatelyshaped odor receptors in the nose. Therefore, (R)-carvone fits into adifferent receptor than does its enantiomer, (S)-carvone.Biological systems commonly exhibit high levels of chiralrecognition. Chiral recognition takes place when a chiral receptor orreagent interacts specifically with one enantiomer of a chiral molecule.A striking example of chiral recognition occurred with the drugthalidomide, a sedative and mild hypnotic introduced to the Europeanmarketplace in 1956.O ONHOThalidomideNHOImmediately, thalidomide became popular with pregnant women tocontrol morning sickness because, even in large doses, it was notwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 548 Daley & Daleylethal and testing had indicated no undesirable side effects. At aboutthe same time, however, the number of congenitally deformed babiesincreased. The most seriously deformed babies were born withphocomelia, or seal limbs. That is, they either had no limbs or hadshortened ones causing the hands and feet to grow close to the body.Other serious defects included absent or malformed external ears,fusion defects of the eye, and the absence of the normalgastrointestinal tract openings. After five years, medical researchersfinally traced the problem to the use of thalidomide during the firstthree months of pregnancy, and thalidomide was removed from themarket.Thalidomide had been tested, so what happened to cause thisproblem? True, the (R) isomer of thalidomide, the medically effectiveisomer, was made in a laboratory and was tested for negative sideeffects, and there were none. However, when manufactured underproduction conditions, the reaction temperature was increased tospeed up the reaction time, thus producing a mixture of the (R) and (S)isomers. This was discovered later through more testing along withthe fact that the (S) isomer of thalidomide was a teratogen. Ateratogen is a chemical substance that causes birth defects. Not allbabies whose mothers took thalidomide were born with defects,however. The fetus was susceptible only at specific points in the firstthree months of its development. Therefore, if the mother did not usethe drug at those specific points, there were no problems. If just thepure (R) isomer of thalidomide had been manufactured and ingested,thousands of babies would not have been born deformed.11.4 Optical IsomerismOne way to differentiate between the two enantiomers in a pairof enantiomers is to observe the differences in how they react withanother asymmetric compound. Another way is to contrast theirinteractions with a plane of polarized light. With ordinary light, thelight waves vibrate at all angles in an infinite number of planes ofvibration. Passing ordinary light through some polarizing materialallows only one of these orientations to pass through. This light isplane-polarized light. See Figure 11.8.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 549 Daley & DaleySlitPolarizerLampCross section of light beamFigure 11.8. Polarization of a light beam. Before passing through the polarizingfilter, the light beam has random orientations of the light waves. After passingthrough the polarizing filter, the light beam contains only one orientation of the lightwaves.A polarimeter is aninstrument thatmeasures opticalactivity.An optically activesubstance rotates aplane of polarizedlight.A dextrorotatory isomerrotates a plane ofpolarized lightclockwise or to theright.A levorotatorysubstance rotates thepolarized lightcounterclockwise or tothe left.If you pass this plane-polarized light through a polarimetercontaining a solution of a pure enantiomer, the plane of polarizationvisibly rotates a certain number of degrees. If you place a solution ofthe other enantiomer with the same concentration in the sameinstrument, the plane of polarized light rotates the same number ofdegrees in the opposite direction. The amount of rotation staysconstant for a given pair of enantiomers with the same concentration,using polarimeters of the same wavelength of light and the same tubelength. Substances that exhibit this type of behavior are opticallyactive, and the number of degrees that the plane of polarized lightrotates is a measure of optical activity.An enantiomer that rotates light clockwise, as you look towardthe light source is dextrorotatory. Designate a dextrorotatoryenantiomer as a d or (+) isomer. An enantiomer that rotates lightcounterclockwise is levorotatory. Designate a levorotatoryenantiomer as an l or (–) isomer. For example, the d form of 2-bromobutane has a rotation of +39.4 o when you measure it with the589 nm wavelength band of a sodium vapor lamp. This wavelength iscalled the sodium D line. Its enantiomer, l-2-bromobutane, has arotation of –39.4 o . Remember, there is no connection between the d or lisomers and the (R) and (S) designations. The d or l designation comesfrom the experimental observation of the sign of the optical rotationand the (R) or (S) designation comes from the Cahn-Ingold-Prelogrules.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 550 Daley & DaleySpecific rotation is howmuch a plane ofpolarized light rotatesunder certainconditions.The normalized measurement of the optical activity of anoptical isomer is called its specific rotation. Normalization isnecessary to provide a constant measurement of a molecule's opticalactivity because not all polarimeters use the same tube length anddifferent concentrations of the isomers are used. The specific rotationof a molecule is a physical property of that molecule; thus, itsmeasurement is always the same. The symbol used to indicate specificrotation is [α] T λ . The [α] is the specific rotation, the T is thetemperature of the solution, and the λ is the wavelength of light beingused. Use the following expression to calculate the specific rotation ofa molecule:[α] T λ = [α]l [X]Where [α] is the observed rotation, l is the length of the tube indecimeters (1 dm = 10 cm), and [X] is the concentration of the solute ing/mL. Although [α] T λ is the term used for specific rotation, anothersymbol commonly used is [α] D. This symbol indicates the sodium D lineas the light source at a temperature of 25 o C.Exercise 11.4Researchers isolated a pure enantiomer of an unknown chemicalstructure from a rare flower. They then dissolved it in chloroform at aconcentration of 0.22 g/mL. When they placed the mixture in apolarimeter, it gave a rotation of +2.697 o in a 1 dm cell using thesodium D line. Calculate the value of [α] D.A racemic mixture, orracemate, is a mixturecontaining equalamounts of bothenantiomers.An equimolar mixture of two enantiomers has no effect on aplane of polarized light because the two isomers have an identical butopposite effect on the plane of polarized light. An equal mixture of apair of enantiomers is called a racemic mixture, or a racemate.When doing a chemical reaction that produces a molecule with astereocenter using symmetrical materials, you always obtain aracemic mixture of the product. In other words, you cannot produceasymmetric compounds directly from symmetric starting materials.For example, reacting phenyl magnesium bromide with 2-butanoneproduces racemic 2-phenyl-2-butanol.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 551 Daley & DaleyO1)MgBrHO2) H 3 O+OHRacemic2-phenyl-2-butanolExercise 11.5A series of solutions contains mixtures of the two enantiomers of 2-bromobutane. The specific rotations for these compounds are +39.4 ofor the d isomer and –39.4 o for the l isomer. Calculate the observedrotations of [α] using a 1 dm cell at 25 o C for each of the followingsolutions.a) 0.1 g/mL of d and 0.1 g/mL of lb) 0.01 g/mL of d and 0.1 g/mL of lc) 0.5 g/mL of d and 0.3 g/mL of ld) 1.0 g/mL of d and 1.5 g/mL of lSample solutionb) The rotation of the d isomer isFor the l isomer, the rotation is= 39.4 o x 1 x 0.01= 0.394 o= –39.4 o x 1 x 0.1= –3.94 oBecause the observed rotation would be the sum of the two, then= 0.394 o + (–3.94 o )= –3.55 o11.5 Fischer Projectionswww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 552 Daley & DaleyTo this point, you have used only the wedge-and-dashstructures to indicate a three-dimensional perspective when drawingmolecules. As the complexity of the molecules you study increases,these drawings will become more difficult to understand and moretime consuming to draw. Similarities and differences between complexstereoisomers are difficult to see with the wedge-and-dashstereochemical drawings.Around 1900, as Emil Fischer studied the stereochemistry ofsugars that contain up to seven stereocenters each, he begandeveloping a simple way to represent these complex molecules so hecould draw them more rapidly. Later, Victor Meyer worked to simplifyFischer's method of drawing molecules. This method of representingmolecules is called Fischer projections. Fischer projections alsofacilitate the comparison of stereoisomers by emphasizing thedifferences in the complex stereochemistry of enantiomers with morethan one stereocenter.Fischer projections use perpendicular lines to represent astereocenter. The point where the lines intersect represents theasymmetric carbon. The horizontal lines represent the bonds thatproject forward, and the vertical lines represent the bonds that projectbackward. For example, to convert a wedge-and-dash representationof (R)-bromochlorofluoromethane to a Fischer projection begin bymaking a model to help yourself visualize the conversion. Rotate themodel until the atoms projecting back in the wedge-and-dashrepresentation, the hydrogen and the fluorine, are in the verticalpositions, as in Figure 11.8, and the atoms projecting forward, thebromine and the chlorine, are in the horizontal positions. (The centerdrawing shows why this representation is sometimes called the “bowtie” representation.) Replace the wedges and dashes with a pair ofperpendicular lines.HFCClBrBrHCFWedge-and-dash projectionsClHBr ClFFischer projectionFigure 11.8. Converting a wedge-and-dash structure to a Fischer projection. TheFischer projection uses perpendicular lines to represent a stereogenic carbon atom.The horizontal lines project toward the viewer, and the vertical lines project awayfrom the viewer.When working with a Fischer projection, you may rotate theprojection 180 o in the plane of the paper because the vertical (dashed)lines are still vertical and the horizontal (wedge) lines are stillhorizontal.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 553 Daley & DaleyHCHOC OHCH 2 OHHCHOOHCH 2 OHRotateo180HOCH 2 OHCHOHHOCH 2 OHC HCHOHowever, rotating the projection by 90 o is NOT allowed, as it changesthe configuration of the projection. A 90 o rotation no longer fits theconvention of horizontal lines forward and vertical lines backward. A90 o rotation reverses this convention.HCHOC OHCH 2 OHHCHOOHCH 2 OHRotate90oHOH 2 CHOHCHO/HOH 2 CHCOHCHOYou may not turn the figure over, because this also gives an incorrectrepresentation of the structure.CHOCHOCHOCHOHCOHHOHFlip HO H /HOCHCH 2 OHCH 2 OHCH 2 OHCH 2 OHBoth the 90o rotation and flip operations drawn above actually aredrawing the enantiomers of the initial structures.When drawing a longer chain with the stereocenter in themiddle of a chain, the convention is to arrange the chain verticallywith the longer part of the chain at the bottom of the structure.CH 3CH 3HHC OH H OHCH 2 CH 3 CH 2 CH 3NotCH 3OHCH 2 CH 3Solved Exercise 11.2Consider the following pairs of Fischer projections. What operation was doneto go from one projection to another: flip or rotate 90 o , 180 o , or 270 o . Are thetwo structures the same molecule or are they enantiomers?a)www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 554 Daley & DaleyHCH 3?BrCH 2 CH 3CH 3 CH 2HBrCH 3(S)-2-BromobutaneSolutionThe structure on the right is rotated clockwise by 90 o compared to thestructure on the left. Converting the structure on the right to a threedimensionalstructure shows it is (R)-2-bromobutane. Thus, they areenantiomers.HHCH 3 CH 2 CH 3CH 3 CH 2 CH 3BrBr(R)-2-Bromobutaneb)CH 2 CH 3Br CH 3?H 3 CCH CH 2BrCH CH 2CH 2 CH 3(R)-3-Bromo-3-methyl-1-penteneSolutionThis molecule is rotated 180 o and thus the two Fischer projections areidentical.Exercise 11.6For each of the following sets of structures, state how the firststructure relates to each of the following structures in the set. Makemodels to help yourself. The relationships may include being the samecompound, a structural isomer, or an enantiomer.a)CH 3CH 3COOHCOOHH NH 2COOHHNH 2COOHNH 2CH 3HHCH 3NH 2b)www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 555 Daley & Daleyc)d)COOHHCH 3COOHCH 3 OHCOOHHH OH H 3 C COOH H OH H 3 C OHCH 3 CH 2 CH 3 CH 2 CH 3CH 3CH 3CH 2 CH 3 CH 3 2 CH CH 3Cl H Cl H H Cl H ClCH 3CH 3HOCH 3CH 2 CH 3CH 2 CH 3CH 3CH 2 CH 3HOCH 3HCH 3OCH 2 CH 3HOCH 3CH 3Sample Solutiona) The first and second structures are enantiomers, the first and thirdare identical, and the first and fourth are enantiomers.11.6 Molecules with Two StereocentersDiastereomers arestereoisomers that arenot mirror images ofeach other.The atoms in stereoisomers bond in the same sequence buthave different spatial arrangements. Enantiomers are stereoisomersthat are mirror images of each other. All stereoisomers that are notenantiomers are diastereomers. When a molecule contains morethan one stereocenter, its stereoisomers exist in pairs that exhibit anenantiomeric relationship; that is, they are mirror images of eachother. They also exist in pairs that exhibit a diastereomericrelationship. That is, they are not mirror images of each other.Consider the following two structures of 3-chloro-2-butanol.HCH 3OHHCH 3OHHClClHCH 3CH 3To follow this discussion, make molecular models of the two structuresshown above. In these structures, both C2 and C3 are stereocenters.The atoms making up these two molecules have the same bondingorder yet are different in their spatial arrangement, therefore they arewww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 556 Daley & Daleystereoisomers. They are not enantiomers, however, because they arenot mirror images of each other. C2 has an (S) configuration in bothstructures, and C3 has an (R) configuration in the structure on the leftand an (S) configuration on the right. Carbon C3 has a mirror imagerelationship in the molecules, but C2 does not. Because these twomolecules are stereoisomers, but they do not have a mirror-imagerelationship, they are diastereomers.Both diastereomers are stereogenic, and each has anenantiomer giving a total of four stereoisomeric 3-chloro-2-butanols.To help visualize this, make molecular models of the two mirror-imagemolecules.CH 3CH 3HHCH 3OHCH 3ClHOClCH 3HHHClCH 3OHCH 3HHOHHCH 3ClWith the 2n rule youcan calculate themaximum possiblenumber ofstereoisomers with nstereogenic centers.You now have two pairs of enantiomers, and either member of onepair of enantiomers is a diastereomer of both members of the otherpair. Because each stereocenter can have the atoms in an (R) or an (S)configuration arrangement, a molecule with n stereocenters has 2 npossible stereoisomers. This rule, called the 2 n rule, allows you tocalculate the maximum number of stereoisomers possible for amolecule.However, not all molecules have this maximum number ofstereoisomers. 2,3-Butanediol, for example, does not. It has twostereocenters, C2 and C3 and, according to the 2 n rule, it should havefour stereoisomers.CH 3CH 3HHCH 3OHCH 3OHHOHOCH 3HHHHOCH 3OHCH 3HHOHCH 3HOHHowever, if you draw all four of the isomers, or make molecularmodels of them, you will find that the apparent enantiomeric pair onthe left above is actually different representations of the samemolecule. (Remember that rotating a Fischer projection 180 o in theplane of the page is acceptable.)www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 557 Daley & DaleyHHCH 3OHCH 3OHHOHOCH 3CH 3HHApparent pair of enantiomersRotate 180oin the planeof the pageHHCH 3OHCH 3OHIdentical to thestructure on the left.These two structures are the same because there is an internal planeof symmetry in the Fischer projection. This internal plane ofsymmetry reflects one half onto the other half. The molecule has twostereogenic centers: one is (R) and the other is (S). It is almost as if themolecule were a racemic mixture within itself.HHCH 3OHCH 3OHInternal planeof symmetryA meso compoundcontains stereogeniccenters and an internalplane of symmetry.Compounds, such as 2,3-butanediol, which exhibit an internalplane of symmetry and are symmetrical, even though they havestereocenters, are called meso compounds. Meso compounds aresymmetric structures with the two halves of the molecule havingidentical substituents, but opposite configurations. The symmetricdiastereomer is called the meso diastereomer. A meso compound is anexample of a compound with stereogenic centers, but does not exhibitoptical activity. The compounds that are diastereomers of the mesocompound and are enantiomers of each other are called the (±) or dlpair.Exercise 11.7For the following compounds mark any stereocenters and determinewhether they are (R) or (S). Mark any meso compounds. Determinewhich are optically active.a)HCH 3Clb)CH 3BrCH 3HCH 3www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 558 Daley & Daleyc)d)H 3 CCH 3OHOHe)CH 3f)BrHHHOHOOHOHHHClCH 3Sample Solutionb) Redraw the structure as a Fischer projection using your models tohelp you see the transformation.CH 3CH 3HHPlane of symmetryThis compound does have an internal plane of symmetry making it ameso compound with no optical activity. The top stereocenter is (S)and bottom one is (R).An erythro moleculehas two stereocenterswith similar groupsnext to each other inthe eclipsedconformation.A threo molecule hastwo stereocenters withtwo pairs of the similargroups on oppositesides of the molecule.When naming a molecule with two stereocenters, consider eachcenter separately and assign each stereocenter the appropriate (R) or(S) designation along with the appropriate carbon number. Forexample, the name of the compound in Exercise 11.7d is (1R,2R)-1,2-cyclopentanediol. Another common way of naming molecules with twostereocenters is to use the terms erythro and threo. The termserythro and threo are derived from the simple sugar molecules 2,3,4-trihydroxybutanal that are commonly known as erythrose andthreose.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 559 Daley & DaleyCHOCHOHOHHOHHOHHOHCH 2 OHCH 2 OHErythroseThreoseUse the term erythro when the similar groups are adjacent to eachother on the molecule. When these groups are identical, the moleculeis meso. Then name the molecule with the meso name. Use the termthreo when the similar groups are on opposite sides of the molecule.When these groups are identical, they are (±) or dl molecules. Thenuse (±), or dl. For example, the terms meso and dl are appropriate forsymmetric molecules such as 2,3-dichlorobutane or tartaric acid.COOHCOOHCOOHHOHHOHHOHHOHHOHHOHCOOHCOOHCOOHmeso-Tartaric acid(dl)-Tartaric acidFor molecules such as 2-bromo-3-chlorobutane or 3-chloro-2-butanol,erythro and threo are more appropriate. While the threo, erythro, meso,and dl terms are widely used, none of these terms are a part of IUPACnomenclature rules.CH 3CH 3HClClCH 3HHClClCH 3HHOHHOHHOHHOHCH 3CH 3Erythro 3-chloro-2-butanolCH 3CH 3Threo 3-chloro-2-butanolSolved Exercise 11.3There are two stereocenters in 1,3-dimethylcyclohexane and thus, accordingto the 2 n rule, there are four stereoisomers. Two of these stereoisomers havethe methyl groups cis and in the other two the methyl groups are trans. Whatis the stereochemical relationship between the two chair conformations of cis-1,3-dimethylcyclohexane? What about trans-1,3-dimethylcyclohexane?www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 560 Daley & DaleySolutionThe two chair forms of cis-1,3-dimethylcyclohexane are a pair ofnonsuperimposible nonmirror images. Thus, they are diastereomers.The two chair forms of trans-1,3-dimethylcyclohexane are a pair ofnonsuperimposible mirror images. Thus, the two are enantiomers.Exercise 11.8Draw the threo and erythro isomers of 2-bromo-3-chlorobutane.11.7 Resolution of EnantiomersResolving enantiomersis a method ofseparating a pair ofenantiomers.A resolving agent is achiral compound orapparatus used as anaid in separatingenantiomers.Because enantiomers have identical physical and chemicalproperties in a symmetric environment, separating, or resolving,them using normal separation methods such as distillation,chromatography, or crystallization do not work. Resolution ofenantiomers requires the use of a resolving agent. The firstresolution of a pair of enantiomers occurred in 1848, and the resolvingagents were a microscope and a pair of tweezers. After crystallizingsalts of (±) tartaric acid, Louis Pasteur observed through hismicroscope that there were two forms of crystals.COOHCOOHHOHHOHHOHHOHCOOHCOOHwww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 561 Daley & DaleySee Section 1.7, page000, for more on van’tHoff and Le Bel.Section 7.7, page 000discusses the reactionof a ketone with ahydride nucleophile.Then, while looking through the microscope, he used a pair oftweezers to separate the crystals into two piles. When he comparedthe properties of the two piles, the only difference he found was theirinteraction with a plane of polarized light. The crystals in one pileturned the light clockwise; the crystals in the other pile turned thelight counterclockwise. He had managed to affect the first separationof enantiomers. It was this work by Pasteur that led to van't Hoff'sand Le Bel's suggestion in 1874 that four atoms bonded to carbonresulted in a tetrahedral structure.Enantiomers rarely crystallize in different forms, as doestartaric acid, so chemists must use other methods to separate them.The most common method is to react the racemic mixture with asingle pure enantiomer. Single pure enantiomers of optically activecompounds are generally isolated from biological sources becausebiological systems usually contain only one enantiomer of a pair. Thisreaction produces a pair of diastereomers that is separated by physicalmethods.The synthesis of (R)-2-butanol shows the resolution of a pair ofenantiomers. A convenient synthetic method for the synthesis of (R)-2-butanol is the addition of hydrogen to the carbonyl group of 2-butanone. The product is a racemic mixture of 2-butanol becauseneither starting material nor reagents are asymmetric. Unless you usean optically active starting material or reagent when synthesizing anasymmetric compound, the result is a racemic mixture of enantiomers.OCH 3 CH 2 CCH 3CH 31) LiAlH 4H OH2) H 3 OCH 2 CH 3+HOCH 3HCH 2 CH 3(S)-2-Butanol(R)-2-ButanolTo resolve this racemic mixture, you need an asymmetric reagent thatreacts with the alcohol to produce a pair of diastereomers. Anasymmetric carboxylic acid, such as Louis Pasteur’s (+)-tartaric acid,is a logical choice because it will react with the alcohol to form a pairof diastereomeric esters which can then be separated and hydrolyzed.Additionally, (+)-tartaric acid is readily available because wineriesproduce large quantities of it as a side product.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 562 Daley & DaleyCOOHHHOOHHCOOH(+)-Tartaric acidFigure 11.9 shows the esterification reaction of the resultingdiastereomeric esters. Because diastereomers generally have differentphysical properties, they can be separated easily via distillation,chromatography, crystallization, or other physical methods. Forexample, the two diastereomeric esters of tartaric acid have meltingpoints that differ by about 20 o C. They also have different watersolubilities and different acid strengths. These differences allowseparation using ordinary physical methods. The separation is socomplete that you end up with each of the diastereomeric esters in thepure form.OCH 3HCOHOCH 3HCOOHHHOOHHCH 2 CH 3HCH 2 CH 3CH 3OHCH 2 CH 3+H OH HHO HCOOHHCOOH(R)-2-Butyl (+)-tartrateCH 3 OO CRacemic 2-butanol(+)-Tartaric acidCH 2 CH 3HOHHOHCOOH(S)-2-Butyl (+)-tartrateFigure 11.9. Reaction of (R)- and (S)-2-butanol with (+)-tartaric acid to produce (R)-and (S)-2-butyl (+)-tartrate. The two diastereomeric esters of tartaric acid can beseparated because they have different physical properties.A hydrolysis reaction is a good way to isolate the enantiomericalcohols. Hydrolysis of (R)-2-butyl tartrate gives (R)-2-butanol.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 563 Daley & DaleyHHOOC OOHHCOOHCH 3COOHCH 3OHH OHCH 2 CH 3 HO H +HO HCH 2 CH 3COOHydrolysis of the (S)-2-butyl tartrate gives (S)-2-butanol, completingthe resolution of the enantiomers.HCH 3OCOOO CCH 3OHH OHCH 2 CH 3 H OHH OH +HO HHO HCH 2 CH 3COOCOOHExercise 11.9The same concept of resolution of enantiomers also applies to amines.An amine reacts with a carboxylic acid to produce a salt.R'COOH +CarboxylicacidRNH 2AmineR'COO NH 3 RSaltUsing (+)-tartaric acid, accomplish the resolution of 2-aminobutane.Draw the diastereomeric salts that result from this reaction.11.8 Stereocenters Other than CarbonAlthough carbon is the most commonly found stereocenter in amolecule, a number of other atoms can be stereocenters, too. One ofthese atoms is silicon. Silicon forms the same sort of tetrahedralstructure as carbon; thus, a silicon atom with four different groupsbonded to it can be asymmetric, and indeed there are many examplesof compounds with silicon stereocenters.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 564 Daley & DaleyH 3 CHCH 3 CH 2SiHCH 3 CH 2ClSiCH 3(R)-Ethyl methyl phenyl silane(S)-Ethyl methyl silyl chlorideAnother possible stereocenter atom is nitrogen. Nitrogen formsamines that are derivatives of ammonia (NH 3 ) in which carboncontaininggroups replace one or more of the hydrogens. For example,ethylmethylamine has four different groups around the nitrogen: anethyl group, a methyl group, a hydrogen, and an electron pair.Because the geometry of this molecule is nearly tetrahedral,ethylmethylamine should be an asymmetric molecule with nitrogen asits stereocenter.NCHN3CHHH3CH 3 CH 2CH 2 CH 3Amine inversion flipsthe groups about theamine nitrogen fromone enantiomer to theother.A pair of enantiomers should exist, but they don't because a processcalled amine inversion prevent them from being separated. If youcompare the amine structure to an umbrella, then amine inversion issimilar to an umbrella turning inside out in the wind. At roomtemperature, this inversion process proceeds so rapidly that resolutionof the enantiomers is impossible.CH 3CH 3 CH 2NHCH 3 CH 2CH 3NHNitrogen is a stereocenter when ammonia forms derivativeswith cyclic structures and bulky groups. In these cases, inversion doesnot occur making it possible to separate the enantiomers. Some stericarrangements of the atoms within a compound also make aminewww.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 565 Daley & Daleyinversion impossible. For example, 1-azabicyclo[3.2.1]octane cannotundergo amine inversion, thus allowing the enantiomers to beseparated.N1-Azabicyclo[3.2.1]octanePhosphorus is in the same group on the periodic table asnitrogen and, like nitrogen, it also forms trigonal pyramidalstructures. However, unlike the amines formed by nitrogen,phosphines do not undergo rapid inversion. Thus, phosphorus is astereocenter that can form a number of optically active phosphines.For example, ethyl methyl phenyl phosphine is asymmetric and hasbeen resolved.••PCH 3CH 2 CH 3Ethyl methyl phenyl phosphineWhen sulfur has three different substituents, it also formscompounds that are asymmetric. The rate of inversion for asymmetricsulfur compounds is quite slow, allowing for ready resolution of theenantiomers. An example of a compound with sulfur as thestereocenter is a dialkyl sulfoxide such as the ethyl methyl sulfoxideshown below.••CH 3CH 3 CH 2SOEthyl methyl sulfoxideThe Cahn-Ingold-Prelog priority rules specify the absoluteconfiguration of any of these centers. These rules assign the unsharedpair of electrons a lower priority than hydrogen. Application of thesepriority rules to ethyl methyl sulfoxide is shown below.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 566 Daley & Daley••43 SCH 3 OCH 3 CH 212(S)-Ethyl methyl sulfoxideExercise 11.10What is the isomeric relationship between the two forms of thefollowing compound that are interconverted by amine inversion?Could you isolate this compound in an optically active form?CH 3H••C N CH3C N CH 3CH ••CH 3 CH 32CH 2 CH 3HCH 3 CH 2CH 2 CH 3Key Ideas from Chapter 11❏ All three-dimensional objects are either symmetric orasymmetric. You can superimpose a symmetric object onto itsmirror image, but you cannot superimpose an asymmetricobject onto its mirror image.❏❏Stereoisomers are molecules that have the same sequence ofbonds, but have a different arrangement of the atoms in space.There are two types of stereoisomers.1. Enantiomers: Compounds that have a mirror imagerelationship.2. Diastereomers: Stereoisomers that are not enantiomers.❏A carbon atom bonded to four different groups is stereogeniccenter. A molecule with a single stereogenic center isasymmetric. Molecules with more than one stereocenter may bechiral molecules or they may be meso molecules. Mesomolecules are symmetrical molecules with even numbers ofstereogenic center. Meso isomers occur with molecules thathave an internal plane of symmetry.www.ochem4free.com 5 July 2005

Organic Chemistry - Ch 11 567 Daley & Daley❏❏❏❏The pure components of an enantiomeric pair have the samemelting points, boiling points, and other physical properties,but they rotate the plane of polarized light in equal butopposite directions.Diastereomers, in general, differ in their physical properties.That is, they have different melting points, boiling points, etc.A chemical reaction capable of producing an asymmetricproduct from a symmetric substrate produces both enantiomersequally with symmetric reagents. Asymmetric reagents usuallyprefer one enantiomer rather than the other.Atoms other than carbon (e.g. nitrogen, phosphorus, andsulfur) can also be stereocenters. The four different groups caninclude a non-bonding pair of electrons as one of the groups.www.ochem4free.com 5 July 2005