Experiment 7: Stereochemistryand Molecular ModelsDepartment of Chemistry University of Ottawa 10 Marie Curie Ottawa Ontario K1N 6N5.June 2008ABSTRACTIn this experiment you will construct models of compounds and investigate their stereochemical preferences. Youmust provide your own models for this experiment (see the introduction to this manual for suggested types).A significant amount of background information isprovided in the following sections. More detail is found inthe text in chapters 1 4 and 5. You are expected to readthe experiment and relevant sections of the text beforeentering the lab.A key aspect of modern organic chemistry is theconsideration and analysis of the three dimensionalstructure of molecules. Shape and stereochemistry areextremely important in medicinal chemistry andbiochemistry because enzymes proteins and othermolecular machines inside our bodies are threedimensional entities. Any interactions between biologicalmolecules or between biological molecules and drugs willbe based on overall geometry. Stereochemistry is alsovery important in chemical reactivity. Chemical reactionsrequire that molecules and atoms interact to achievefavorable orbital overlap. This can only be done if theorbitals have the proper geometry to overlap.Understanding the three dimensional shape andreactivity of organic compounds is very difficult to dousing two dimensional drawings. To properly understandorganic structure and reactivity it is necessary to considerthe molecules in three dimensions. Today this can bedone on a computer; however the best way to work inthree dimensions is to use plastic molecular models.Models are cheaper more rugged and are easier tovisualize with.An example of the types of reactivity differences youcan expect when stereochemistry is considered is shownExperiment 7above. On top is a cyclization reaction in which anepoxide is formed by displacement of a halogen in an SN2process. When seen in two dimensions the reaction issimple. An example of a flat representation of a reactionis the conversion of 1 to 2. The introduction ofstereochemistry complicates matters. Notice that the transproduct 3 is readily closed to produce epoxide 4 whereasthe cis product 5 does not easily react. The reason for thisis that the nucleophile and leaving group cannot becomeantiperiplanar to one another in structure 5 because of therelative configurations of the groups. Antiperiplanaralignment is necessary in this reaction so that theunpaired electrons of the nucleophile can interact with the* orbital of the leaving group to produce a new bond.On paper this relationship is difficult to see but becomesobvious when molecular models are used.OH1 BrOHbaseO2base3 BrO4OHO5 Brbase61 Darling Molecular ModelsIn this experiment you will use your molecular modelsto construct a set of compounds from a list yourdemonstrator will provide. For best results Darlingmolecular models should be used although other typesmay be permitted. Consult your instructor before enteringthe lab if you wish to use an alternate model kit.The models you will use for the experiment are madeof variously colored plastic parts. The most common“organic” elements and the colors associated with themare shown below. Other elements can be various colorsdepending on need. Hydrogen is assumed on the “empty”bonds of carbon. On heteroatoms (O N etc.) it is usuallyconvenient to use one of the small ball markers torepresent a hydrogen. Empty bonds on heteroatoms arenormally interpreted as lone pairs of electrons. Whenmaking models the bonds should slide together easily.Do not force the models or try to make small rings (3 or4 atoms). See the instruction booklet that came with yourmodels.ElementCarbonOxygenNitrogenSulfurHydrogenColorBlackRedBlueYellowSmall ball markeror “empty” bondYou will be using framework models. In this type ofmolecular model the “sticks” represent bonds betweenatoms or lone pairs of electrons. Atoms are assumed to lieat the point at which the “sticks” come together (vertices)or at the ends of “empty sticks”. Most of the atom parts inyour kits are designed to form tetrahedral atoms thatconsist of the central atom surrounded by 4 equallydistributed bonds. It is important to always buildtetrahedral atoms so that overall molecular geometriescan be visualized. This includes heteroatoms. The bondsconverge at the vertices of a tetrahedron centered on eachatom.You create a tetrahedral atom by joining together twoV-shaped parts labeled sp3. Slide the pieces together atright angles so that the V-shaped openings point towardseach other. Pinch the pieces together until they clicktwice. Double bonds are made using the sp2 pieces whichlook like elongated oval rings. Triple bonds arerepresented by long arrow shaped pieces.It is usually easiest to assemble models by preparingthe required number of tetrahedral atoms and then joiningthem together. As much as possible try to make yourExperiment 7models look like your drawings this will make it easier tobuild models. Similarly making drawings look likemodels makes it easier to transfer information to paper.Structural RepresentationsSome molecular projections are shown below. Theserepresent some of the various ways to depict thestructures of compounds in two dimensions. Each of thesenotations has been designed to convey structuralinformation for different purposes. More detail on theserepresentations can be found in your textbook.CH3CH2CH2CH2CH3condensedzig-zagHC2H5HHHNewmanCH3Condensed structures show no three-dimensionalstructure. They are used when composition is important or when you are trying to convey a structure using aword-processor. These structures provide informationabout how each atom in a compound is connected.Condensed formulas make assumptions about how manyatoms can be connected to a given atom. To understandthese formulas you need to know how many connectionseach type of atom can make. Carbons normally form fourbonds nitrogen forms three oxygen two and halogensone. All of the hydrogens that are attached to an atom arewritten immediately after that atom. Atoms following thehydrogens are assumed to be attached to the precedingnon-hydrogen atom up to the maximum number of atomsthat can be bonded to the proceeding element. Groups thatare attached to a given atom are frequently surrounded byparentheses. Some examples are shown below.CH3CH2CH2CH3CH3CHClCH2CH3CH3CH2OHCH3OCH3OHCH(CH3)2In the first example above all 4 carbons are connectedin a linear chain. The first carbon is surrounded by threehydrogens the second and third carbons by two2 hydrogens each and the fourth carbon by threehydrogens. Note that each carbon is bonded to four otheratoms. In the second example the first carbon isconnected to three hydrogens and is also connected to thesecond carbon. This second carbon is connected to onehydrogen a chlorine atom and to the third carbon. Thethird carbon is connected to two more hydrogens and tothe last carbon which is bonded to three hydrogens.The third and fourth compounds illustrate the bondingof heteroatoms. In the third molecule the first carbon isconnected to three hydrogens and to the second carbon.This carbon is connected to two hydrogens and to theoxygen. The final hydrogen is connected to the oxygen.In the fourth structure the first carbon is connected tothree hydrogens and to the oxygen. The oxygen isconnected to the second carbon which is connected tothree hydrogens.The fifth compound shows an example of groups. Inthis case the oxygen is connected to a hydrogen and tothe first carbon. This carbon is connected to a hydrogenand to two CH3 groups. The connection to these groups isa carbon-carbon bond in each case.The structure shown below provides an example of acondensed formula containing a double bond. The firstcarbon is connected to three hydrogens and to the secondcarbon. The second carbon is connected to one hydrogenand to the third carbon. This third carbon is connected toa hydrogen and to the final CH3 group. Drawing thisstructure with bonds clearly shows that the second andthird carbons each now have three bonds. Since it isunderstood that carbon normally makes four bonds it isimplied that these two carbons must be connected by adouble bondCH3CHCHCH3H 3CHCCHCH 3Zig-zag structures convey stereochemical informationin a compact manner. In these structures carbons andhydrogens are not explicitly written. Instead only thebonds connecting the carbons are drawn. Carbons areassumed at each vertex and terminus. The atomic symbolsfor heteroatoms are shown. Hydrogens are not explicitlyshown unless bonded to a drawn atom. In these structures it is important to remember that carbon always forms amaximum of four bonds. Any time a vertex or terminushas less than four bonds the remaining bonds (to amaximum of four) are assumed to be to hydrogens.Shown below are some condensed structures with thecorresponding zig-zag structure on the right.In the first example below three carbons are connectedin a linear chain. This chain is drawn as a zig-zag. Thetermini and vertexes represent the carbons. It is assumedExperiment 7that the first and third carbons each carry three hydrogens while the middle carbon is bonded to two hydrogens(total of four bonds).CH 3CH2 CH3H2CH2 CCH 2H2 C CH2OHCH3 CH 2COOHOIn the second example five carbons are connected in aring. Each vertex represents a carbon each line representsa bond. The number of hydrogens bonded to each carboncan be deduced by remembering that each carbon canbond to a maximum of four other atoms.In the final example three carbons are connected in achain. The third is connected to two oxygens. Note thatbecause the oxygens are heteroatoms that the symbol foroxygen is explicitly drawn. The final hydrogen is writtenafter an oxygen and so must be connected to the oxygen.The first oxygen must be doubly bonded since it is notfollowed by a hydrogen. This double bond can only be tothe carbon preceding it.The models in your kit naturally form zig-zags becauseof the tetrahedral geometry of the atoms. This can behelpful when drawing structures or building modelsbecause you can make your models look like yourdrawings and vice-versa. In the sections below you willbe asked to build models from condensed structures andthen draw the zig-zags. Build each model and place it onthe desk-top in a zig-zag fashion then simply draw whatyou see. This method helps to reduce errors when youconvert models to drawings and vice-versa.Zig-zag structures often feature bonds that look likewedges or hashes. These bonds are used to show thethree-dimensional tetrahedral shape of an atom. Bondsdrawn as simple lines are assumed to be in the plane ofthe paper. Wedge bonds are assumed to project above theplane of the paper. Hash bonds are assumed to projectbelow the plane of the paper.OHOHHashOHWedgeTetrahedronIncorrect(see model)When drawing wedges and hashes it is important torespect the tetrahedron shape. A tetrahedron is formedfrom two V shapes joined at their vertexes and twisted3 90° to each other. When drawing stereochemistry(wedges or hashes) you should always draw the atoms toshow the two V shapes joining at the vertex (see above).Two of the bonds should lie in the plane of the paper(single lines) one should be up and the other down.A Newman projection is a cartoon showing thedihedral angle of a bond. The dihedral angle is defined by4 atoms and the bond in question is assumed to be theone between the second and third atom of the four atomsspecified. The Newman is very useful in determiningreactivity and overall shape.In the line structure below ethane is depicted with allthe hydrogens drawn in stereochemical notation. If youlook along the axis of the carbon-carbon bond you willsee the projection of this bond and the dispersal of thehydrogen atoms. To illustrate this dispersion of atoms theNewman projection is used.In the Newman projection the carbon atom that is inback is represented as a circle. Each of the hydrogens thatare bonded to this carbon are represented by lines spaced120° apart. Notice that these lines stop at the outside ofthe circle that represents the rear carbon.FrontCarbonHBackCarbonHHHHHrotate molecule 90°HHHHBackCarbonHput bothHFrontCarbonHtogetherHHHHHNewmanProjectionThe three carbon-hydrogen bonds connected to thefront carbon are depicted as lines spaced 120° apart. Thecarbon is assumed to be the vertex formed by these lines.The Newman projection is formed by combining therepresentation of the front and back carbons. This showsthe conformation that the molecule is in. The anglebetween a hydrogen on the front carbon and a hydrogenon the back carbon is called the dihedral angle (). It isdefined by the four atoms that make up the angle (H-C-CH).Your model kit contains small plastic spheres that aredesigned to fit over tetrahedral atoms. The sphere can beused to make Newman projections easier to see. Place asphere on the back atom of the bond you are viewing. Asyou look down the bond you will see the Newmanprojection. The sphere represents the circle shown in theNewman projection. Using a sphere that contrasts thecolor of the atom in front can often make the Newmanprojection easier to see.Experiment 7HHHHHHStaggeredHHHHHHEclipsedThe dihedral angle defined by atoms gives rise tospecific conformations in a molecule. A conformation issimply a shape that a compound can adopt by the rotationof bonds. The example above shows two types ofconformation that are of particular importance. In the firstexample the dihedral angle shown is 60°. Thisconformation in which the hydrogens are maximallyseparated is called a staggered conformation. The secondstructure shows a conformation in which the dihedralangle is 0°. This conformation in which the carbonhydrogen bonds of the front and back carbons overlap iscalled an eclipsed conformation. In addition to theseconformations there are two types of staggeredconformer the anti-staggered and the gauche-staggered.You should consult your text for descriptions of theseconformations.Absolute configuration. The shape of chiralcompounds is conveyed by a naming system that providesthe absolute configuration of each of the stereogeniccenters in the compound. This system is analogous tonaming our hands right or left. The system that is used iscalled the Cahn-Ingold-Prelog system or simply the R Ssystem. Details are given in the course text. To name astereogenic center using this system the following stepsare carried out.1. Each of the four groups attached to a stereogenicatom is assigned a priority (1 2 3 4) based on theatomic number of the directly attached atoms.Priority #1 goes to the atom with the highest atomicnumber.2. In the case of a tie (same atomic number) the nextdirectly attached atom in each of the groups inquestion is examined. This process continues examining each of the atoms at the branch point untilthe tie is broken. Atoms are examined according tothe order highest-to-lowest atomic number.3. For double or triple bonds the group is re-drawnduplicating the atoms at each end of the -bond. Notethat only the directly attached atoms are duplicated not entire groups.4. For isotopes break ties using atomic masses (highestmass = highest priority).5. Rotate the structure so that the lowest priority grouppoints away from you. The molecule is namedaccording to the direction of rotation of the threehighest priority groups in the order of 1-2-3.Clockwise rotation means that the stereogenic centeris called R counter-clockwise rotation gives the Sdesignation.4 Chair Structures. Cyclohexane is a special structure inorganic chemistry because this is the only cyclichydrocarbon that has a strain-free conformation. Thisconformation is called a chair conformation because aline-representation resembles a reclining chair. Asshown below when cyclohexane is viewed from the top the molecules looks like a hexagon. However if thisstructure is rotated slightly and viewed from the side athree-dimensional shape appears that is called a chair.As part of this lab you will be expected to construct amolecule of cyclohexane in the chair conformation andto identify the component hydrogens. Details on chairconformations is given in the text on pages 156 -168.2.3.4.The ExperimentEnantiomers and Diastereomers. You will be given alist of compounds that must be made from the molecularmodels. For each compound on your list:1. Construct a model of the compound.2. Make the model of the mirror image of thecompound.3. Carefully draw the models that you have madeusing proper stereochemical notation. The mirrorimages should be drawn side-by-side indicatingthe mirror plane. Models should be drawn usingthe zig-zag style.4. Determine if the mirror images can besuperposed and therefore if they are enantiomersor not.5. Identify the stereogenic centre(s) of the molecule6. Determine the configuration at each stereogeniccentre7. For compounds that have more than onestereogenic centre prepare all the possiblestereoisomers of the compound including allmirror images. You should try to prepare all ofthe isomers of a given compound at the sametime.8. Carefully draw the models that you have madeusing proper stereochemical notation. The mirrorimages should be drawn side-by-side indicatingthe mirror plane. Models should be drawn usingthe zig-zag style.9. Determine if the mirror images can besuperposed and therefore if they are enantiomersor not.10. Identify the stereogenic centre(s) of the moleculeand determine the configuration at eachstereogenic centre.Newman projections.1. Construct a model of n-butane. Notice that thebonds are flexible and permit rotation. Youwill be looking at the bond between C1 andC2. To help you see the Newman projection Experiment 7add a sphere to C2 of the model. This will bethe rear atom in your Newman projection.Using your model construct an energydiagram to show the variation in the freeenergy of n-butane as the dihedral angle madeby C1-C2-C3-C4 is changed from 0º to 360º in60º increments. Begin with a staggeredconformation. Rotating by 60° produces aneclipsed conformation that is 3.4 kcal·mol-1higher in energy. Continue rotating in 60°increments drawing the energy change.Draw Newman projections for eachconformation on the bottom of your diagram.What is the energy difference between the twomost stable conformations?Cyclohexanol.1. Build a model of cyclohexanol in the moststable chair form.2. Draw your model properly using the chairconvention.3. Indicate all of the axial and equatorialsubstituents (including hydrogens).4. Without breaking any bonds convert yourmodel to the other chair form. Be careful notto rotate the overall molecule.5. Draw your model properly and indicate all ofthe axial and equatorial substituents (includinghydrogens).6. Without breaking any bonds convert yourmodel to a boat form.7. Draw your model properly and indicate all ofthe pseudo-axial and pseudo-equatorialsubstituents (including hydrogens).8. Can the model be converted to other boatforms?12. Place one of the Newman spheres on C1 ofyour cyclohexanol model.9. With the model in a chair conformation drawthe Newman projection defined by the sphereand one of the adjacent carbons.10. Flip the chair draw the new Newmanprojection and note the changes between thetwo projections.Questions1) Draw the zig-zag structure of the R isomer ofCH(OH)ClBr.2) Draw the zig-zag structures of all the isomers of2 3-diaminopentane. Draw the Newman projectionalong the C2-C3 bond of the major conformers ofeach isomer.5 3) How many stereoisomers are possible for thefollowing compound? Draw them.NH2OHReport NotesReports for this lab consist only of your drawings ofthe required structures together with the answers to thequestions.CH3Experiment 76
