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Adv Physical Chemistry

by: Mr. Clementine Gottlieb

Adv Physical Chemistry CHEM 6320

Marketplace > Tennessee Tech University > Chemistry > CHEM 6320 > Adv Physical Chemistry
Mr. Clementine Gottlieb
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This 73 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 6320 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 35 views. For similar materials see /class/225694/chem-6320-tennessee-tech-university in Chemistry at Tennessee Tech University.


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Date Created: 10/21/15
CHEM 6320 Fall 2005 Concerted Pericyclic Reactions A Electrocyclic Reactions 1 Introduction a Concerted reactions are singlestep processes and they can be both unimolecular and bimolecular reactions b Pericyclic reactions are cyclic unimolecular electronic rearrangements occurring through cyclic transition states c Another name is valence tautomerizations 2 electronic isomerizations d Woodward and Hoffmann proposed that the pathways of the concerted pericyclic reactions were determined by the symmetry properties of the orbitals directly participating in the reaction e Electrocyclic reactions are a class of pericyclic reactions in which formation of a single bond occurs between the ends of a linear conjugated system of electrons as well as the reverse process f Electrocyclic reactions involve redistributing 7rand abonds but the total number of bonds remains constant 2 Common Electrocyclic Reactions a The case of two bonds lt gt bond abond This is a fourelectron system A 3 B c E 27v The question is what diastereoisomer will be formed According to microscopic reversibility principle both forward and reverse reaction will follow the same mechanism If the mechanism were concerted one would expect a single stereochemical result ll ll n6 147 CHEM 6320 Fall 2005 Example 24hexadiene lt gt 34dimethylcyclobutene reaction Me Me H A E Z Me H H H cis E E A Me H AV Me Z Z trans Each reaction is highly stereospeci c no other stereoisomer formed Additional example The EZ isomer cannot be obtained sterically impossible The ZZ isomer cannot be obtained sterically possible but electronically forbidden X CHEM 6320 b Fall 2005 The case of three 7zbonds lt gt two 7zbonds abond This is a sixelectron system This is a different system from the ones above because there are two extra nelectrons The rules for this system will be different Example 246octatriene gt 56dimethyl13cyclohexadiene E Me Z 395 r E Me E Me Z I3 h Me n H Z trans Rules for Predicting the Products a The question is how the molecule ends should twist to make the product The two options are disrotatory and conrotatory motion Electronic rules do not address the equilibrium constant ie the preference of the forward or the reverse reaction Analysis of the symmetry of reactant HOMO HOMO for butadiene has one node and one predicts conrotatory motion both clockwise and counterclockwise For hexatiiene lt gt cyclohexadiene reaction looking at 113 two nodes one predicts disrotatory motion CHEM 6320 Fall 2005 d Analysis of the topology of the transition state Like drawing resonance for cyclic conjugated systems Disrotatory motion for 411 2 systems 0 Transition state has Huckel topology A Huckel system has zero or any even number of phase changes around an orbital array A Huckel system is aromatic for 411 2 electrons antiaromatic for 411 electrons Conrotatory motion for 411 systems Transition state has Mobius topology A Mobius system has one or any odd number of phase changes around an orbital array A Mobius system is aromatic for 411 electrons antiaromatic for 4n 2 electrons e Using correlation diagrams It is based on analyzing orbital symmetry of reactants and products The procedure Find a symmetry element preserved R gt I gt P Analyze all reacting orbitals orbitals changing during the reaction as symmetric S or antisymmetric A with respect to the symmetry element Fill the orbitals with electrons Identify correlations between the reactant and product orbitals 150 CHEM 6320 Fall 2005 f Correlation diagram for cyclobutene lt gt butadiene disrotatory motion plane of symmetry H H H H 7 VP conrotatory motion aXis of symmetry Symmetry properties of cyclobutene and butadiene orbitals classification 039 7r 7 0 with respect to V V V 7 9 V 7 V 96 A 94 59 96 59 96 A A A A A A A A A W1 W2 W3 the plane the axis W4 classification with respect to the plane the axis 151 CHEM 6320 Fall 2005 Correlation diagram for the conrotatory motion The element of symmetry is the axis of symmetry cyclobutene butadiene 0 V4 7r 13 W2 0 W1 This is the symmetryallowed conrotatory reaction The symmetry is conserved Correlation diagram for the disrotatory motion The element of symmetry is the plane of symmetry cyclobutene butadiene 0 V4 7r 113 W2 0 Vi This is the symmetryforbidden disrotatory reaction The HOMO in butadiene antisymmetric corresponds to LUMO in cyclobutene and reverse g Correlation diagram for hexatriene lt gt cyclohexadiene ll W6 914426 This is an example of siXelectron system that is allowed for disrotatory motion maintaining a plane of symmetry 152 CHEM 6320 Fall 2005 Correlation diagram for the disrotatory motion hexatriene cyclohexadiene 116 0 14 V4 V4 W3 W3 W2 112 V4 V4 a 4 More examples a Dewar benzene Dewar benzene is unexpectedly stable b Cycloheptatriene Disrotatory motion allowed for H H SIX electrons Sterically allowed and electronically allowed H Could not make the trans isomer 153 CHEM 6320 Fall 2005 F Conformational Effects on Reactivity l Cyclohexane Conformer Reactivity a The question is which conformation reacts faster and an example is which conformation reacts faster with hydride ion H or R 7 MBT H CHO VS H eq ax CHZO This is not the same question as which conformer is more stable for which the answer is equatorial b The issue here is to assess AG between reactant state and transition state I ie AG c The question is Is the substituent getting more crowded or less crowded in going from the reactant state to the transition state If getting more crowded eg Sp2 gt sp3 then equatorial conformer reacts faster If getting less crowded eg Sp3 gt spz then axial conformer reacts faster 93 CHEM 6320 Fall 2005 2 Stereoelectronic Effects a They are related to anomeric effects b In reactions that involve Sp2 lt gt sp3 formation and collapse of Sp3 state is favored when the leavingentering group is antiperiplanar to lone pairs on both of the other heteroatoms QO CH3 0 CH3 H Bow 2 fastex lower O O For axial conformer stabilization is coming from only one pair of electrons c More examples OCH3 OCH3 CW i 0 CH30H 0 QOCH3 O O O O Dlabeling shows exclusive loss of axial methoxy The reason is maximization of 7zoverlap at the reaction transition states 94 CHEM 6320 Fall 2005 G Reactivity of Cyclic Compounds 1 Effect of Angle Strain on Reactivity a Manifested especially for cyclopropane that reacts with ringopening with electrophiles X2 Hl H2cat that normally react only with double bond CC and not other cycloalkanes 2 Stability of IransCycloalkanes a Iranscycloheptene has only been trapped as a transient intermediate b Iranscyclooctene is known is known but is unstable and reacts rapidly to relieve the strain It reacts with X2 faster than cyclohexene Note that transcyclooctene can be resolved into and enantiomers atropisomerism H I H H 3 Brendt s Rule 3 Brendt s rule says that no CC is permitted at the bridgehead position of bicyclic compounds zbzb possible not possible H b SiX atoms cannot all be coplanar unless bridgehead loops are long enough 4 StereoelectronicSteric Inhibition of Resonance a First example 0 0 c lt gt Co A gt9 R39 NRZ R39 ifoR2 3 If O O CHEM 6320 Fall 2005 Nitrogen s lone pair is perpendicular to carbonyl 7rsystem resonance form is a Brendt s rule violation b Second example H lt gt o o o o o o M M M L 7 39H L 794 7 0 o o o o o 5 Ring Closure Rates Rate of Cyclization Versus Ring Size a The general trend is 5 gt 6 gt 3 gt 7 gt 4 gt 810 Ring size of 3 has unfavorable AH1t but favorable AS Ring sizes of 810 have normal AH1t but large negative AS This rule applies to for example A BrCH2nNH2 gt CH2Il NH V but varies with type of reaction and hybridization of reacting ends Exceptions from general trend DAR often occur faster than CHEM 6320 Fall 2005 b Baldwin s rules Baldwin s rules provide a systematic analysis on ring formation by taking into account ring size hybridization at reaction center Whether reacting bond is endocyclic or exocyclic to forming ring Y O C C X C E CCR J C gt Nu Nu Nu nexotet nexotrig nexodi g S193 S192 Sp OK for any ring size OK for any ring size OK for n 2 5 Z l CNu J gN111 nendotri g nendodi g S192 Sp OK for n 2 6 OK for any ring size Example kl 5end0trig CH CHPh not allowed HO O 0 Ca CPh 5end0di g Ph allowed HOJ O 97 CHEM 6320 Fall 2005 H Torsional and Stereoelectronic Effects on Reactivity l Torsional Effects on Reactivity a For reactions of cyclic compounds Sp2 gt Sp3 transformation is more favorable for 6 than 5 membered ring Sp3 gt Sp2 transformation is more favorable for 5 than 6 membered ring b Example 0 OH O OH 6 i C Q i 0 0 VS 0 For reduction reaction involving Sp2 gt sp3 the 6membered ring is faster than the 5membered ring For oxidation reaction involving Sp3 gt spz the 5membered ring is faster than the 6membered ring 2 Stereoelectronic Effects of Carbonyl Reactivity a Example of nucleophilic attack on cyclohexanone Nu ax H H O H V Nu eq H H b Equatorial attack requires eclipsing of C 0 bond with two C Heq bonds at transition state Nu O H axial H equatorial H g atack 0 atack a ENH H H H Axial attack is favored for small nucleophiles Equatorial attack is preferred for large bulky nucleophiles because of 13diaxial problem 98 CHEM 6320 Fall 2005 3 Rules for Predicting the Major Product 3 Cram s Rule Predict the correct major and minor diastereoisomers H L I IL R CC R C M M I S S Nu Nu When the three substituents at the Vicinal C differ in size called them Llarge Mmedium Ssmall and the molecule is oriented such that the largest group is anti to the carbonyl oxygen the major product arises from addition of the nucleophile syn to the smaller substituent M o M OH M OH H HI39Nu 7 401 L R L R L Nu major minor Transitionstate theorybased explanation the molecule is oriented such that the largest group is oriented perpendicularly to the carbonyl group the nucleophilic attack occurs from the opposite side c Example CH3 CH3 CH3 H3 n exo attack f exo attack 0 C endo attack Cenalo attack 99 I CHEM 6320 Fall 2005 Problems and Exercises 1 Estimate AHO for each of the following conformational equilibria CH CH H icng chj gcng H H H H 3 CH3 H H CH3 chj icn3 CH3 CH3 ch CH3 ch n b H H CH3 O CH3 CH3 BIACH3 C 2 Draw a clear threedimensional representation showing the preferred conformation of ciscislrans perhydro9bphenalenol A 3 The trans cis ratio at equilibrium for 4Ibutylcyclohexanol has been established for several solvents near 80 C Solvent trans cis Cyclohexane 700 300 Tetrahydrofuran THF 725 275 i Propyl Alcohol 790 210 From these data calculate the conformational energy of the hydroxyl group in each solvent Explain any correlation between the observed conformational preference and the properties of the solvent 100 CHEM 6320 Fall 2005 Explain the basis for the selective formation of the product shown over the alternative product H3O CH3 H3O CH3 H3O CH3 RCO3H 0 gt versus a 88 O 12 H3B39 H3C H3C N BH3 N preferred TN H30quot I to H3B quotquot b CH3 CH3 CH3 0 LiAlH4 gE rather EE 0 CH3 CH3 than CH3 0 0 11 13 I I so N 3 5 S CH3 RCO3H N 9 5 3 CH3 gt COZCH3 COZCH3 rather than the d stereoisomeric sulfoxide O CH2CH2 A Sncl rather 4gt than OH 6 CHZOH 101 CHEM 6320 Fall 2005 For the following pairs of reactions indicate which you would expect to be more favorable and explain the basis of your prediction a Which isomer will solvolyze more rapidly in acetic acid OSOgPh 01 A B b Which isomer will be converted to a quaternary salt more rapidly NCH32 NCH32 930 or CH33CN C D c Which lactone will be formed more rapidly 0 O 3902CCH22CH2Br M f or o o 3902CCH22CH2Br M F d Which compound will undergo hydrolysis more rapidly o o mom or OQNOZ 43 G H e Which compound will aromatize more rapidly by loss of ethoxide ion 0 OC2H5 O OC2H5 N N N02 N02 N02 N02 I J 102 CHEM 6320 Fall 2005 6 Predict the most stable conformation for each of the following molecules and explain the basis of your prediction COZCH3 I a Cl 0 CH3 0 CH3 FIG 6 F O O 0 g I 103 CCH33 i OH i b CCH33 O COCH3 61 O CO CH3 0 x I o co CH3 d 1393r OCH3 O O f 0C E h FCHZCHZ39 CHEM 6320 Fall 2005 Conformational Steric and Stereoelectronic Effects A Conformational Analysis in Saturated Compounds 1 Strain Energy excess energy caused by nonideal geometry Estrain EU E9 E05 Ed a Bond stretching Er 051r Ar2 b Bond angle distortion E6 05keA62 c Torsional strain E 05V0 l cos 3 for tetrahedral carbon 1 Nonbonded repulsions or van der Waals repulsions E d when two nonbonded groups are closer than the sum of the van der Waals radii 2 Linear Alkanes For ethane and propane the strain energy inherent in the eclipsed as opposed to staggered conformation is a quantum mechanical repulsion between pairs of bonding electrons H CH3 H CH3 H HH HH b For nbutane an additional factor comes into play when one considers rotation about the central C C a 29 kcalmol E kcalmol 0 60 120 180 2 300 360 Tors1on angle degree 60 50 40 30 20 E kcalmol 10 00 l0 79 CHEM 6320 Fall 2005 3 Conformational Equilibrium a The relative distribution of a molecule among various conformations can be calculated if the energy differences are known AG 2 RT an where K represents the conformational equilibrium constant b For example to determine the distribution of nbutane between the gauche and anti conformations anti gauche AG 2 08 kcalmol RTln 2 where R71n 2 term is a statistical correction due two gauche to one anti conformers At room temperature AG 2 08 041 039 kcalmol gt K 19 gt 66 anti34 gauche c Compositionequilibriumfreeenergy relationships More stable isomer Equilibrium Free energy AGozgg constant K kcalmol 50 100 00 55 122 0 1 19 60 150 0240 65 186 0367 70 233 0 502 75 300 0 651 80 400 0 821 85 567 1 028 90 900 1 302 95 1900 1 744 98 4900 2 306 99 9900 2 722 99 9 99900 4 092 CHEM 6320 Fall 2005 4 Rotational Energy Barrier a Rotational energy barriers of compounds of the type CH3 X Compound Barrier height kcalmol CH3 CH3 288 CH3 CH2CH3 34 H CH3 CHCH32 39 1 R2 CH3 CCH33 47 CH3 SiH3 17 R3 H CH3 NH2 20 CH3 OH 11 b Observations and explanations 5 Another Example 3 The case of lchloropropane CH3 C One predicts the same conformer as in nbutane anti favored over gauche CHEM 6320 Fall 2005 B Conformational Analysis in Compounds with Double Bond 1 Propene and Acetaldehyde 3 There are two possible conformers H H H eclipsed bisected b The eclipsed conformer is more stable c The bisected conformer is less stable due to unfavorable interactions repulsion between the system and the two C H abonds 1 Note that the rotational barriers in these cases are less than the case of ethane 2 More Examples 3 lbutene have a methyl group connected to C3 so there are two different eclipsed conformations CH3 H H H H3C CH2 is more stable than H CH2 H H due to more favorable van der Waals interactions b If another methyl group is present at C2 the two conformations are similar in energy H CH3 3 CH2 VS39 CH2 CH3 CH3 Note the gauche interaction energetically less favored between methyl groups in the first conformer 82 CHEM 6320 Fall 2005 c A similar case but for a ketone H R H Hn R H O O R39 R39 minor major In the major conformer the repulsion between R and O is not as great and the R and R39 are in anti position preferred over gauche 3 Compounds with Conjugated Double Bonds 3 13 butadiene has two conformers in equilibrium with the sIrans conformer being preferred b Acrolein has also two conformers in equilibrium sIrans conformer is still being preferred but less than the case of 13 butadiene The 16 interaction is less important in acrolein than in 13 butadiene c 3methyl acrolein has also two conformers in equilibrium sIrans conformer is still being preferred but less than the case of acrolein CHEM 6320 Fall 2005 C Conformational Analysis in Small Rings 1 Cyclopropane a The molecule is planar ie the carbon atoms form a plane 2 Cyclobutane a If the molecule would be planar there would be a smaller angle strain But all the C H bonds are eclipsed so there is bigger torsion strain b The molecule is not planar ie one carbon atom is above or below the plane made by the other three carbon atoms In this way the angle strain in increased the C C C bond angle is decreased but the torsion strain is decreased c cisl3dimethylcyclobutane is more stable than trans13 dimethylcyclobutane because cis has two pseudoequatorial bonds 3 Cyclopentane a Again a planar molecule would have little angle strain 108 vs 1095 degrees but it would have large torsion strain due to eclipsed bonds The most stable conformations have the carbon atoms not in the same plane reducing therefore the torsion strain CHEM 6320 Fall 2005 D Conformational Analysis in 6Membered Rings 1 Cyclohexane 55 kcal Potential Energy CAWMV chair halfchair twistboatl boat twistboat H conformer conformer conformer a Ring ip of chair conformation of cyclohexane interchanges all equatorial groups and aXial groups A B B C C 2 Monosubstituted Cyclohexane There is equilibrium between the two conformers aXial and equatorial CH3 H a CHEM 6320 Fall 2005 b The equatorial conformer is energetically preferred In the axial conformed of methylcyclohexane there are two unfavorable gauche butane interactions and unfavorable 13 diaXial interactions CH3 CH3 H3 C The Newman projection about C1C2 bond C H Conformational restriction makes this gauche interaction a little worse than normal 09 rather than 08 kcalmol Equatorial methyl is favored over aXial methyl by 18 kcalmol c The free energy difference between conformers is referred to as the conformational free energy AGO Values of AGquot in kcalmol for axialequatorial equilibrium CH3 18 Ph 29 CH2CH3 18 OCH3 06 z39Pr 21 COOC2H5 12 IBu 45 CEN 02 Relative AGO values for Me Et iPr IBu H CH3 H3C CH3 H CH3 H c CH3 H CltCH3 M For isopropylcyclohexane the van der Waals repulsion is no worse than in the case of methyl but there is restricted rotation therefore an entropy loss For Ibutylcyclohexane the van der Waals repulsion cannot be relived by rotation and AGquot increases to 45 kcalmol Equatorial preference for I Bu has comparable AGquot to that between chair and twistboat conformations of cyclohexane 86 CHEM 6320 Fall 2005 1 Measurements of axialequatorial equilibrium Take NMR at low enough temperature Where conformational interconversion is slow on the NMR time scale then integrate the separated signals to get the equilibrium constant K allax Measure AGO for two diastereomers Where the only difference is Whether the substituent is equatorial or axial X w x locked locked 3 Disubstituted Cyclohexane a trans12dimethylcyclohexane The axialaxial conformer has four gauche interactions The equatorialequatorial conformer has one gauche interaction CHEM 6320 Fall 2005 b cisl4dimethylcyclohexane CH3 H X A The two conformers are the same c trans14dimethylcyclohexane CH3 CH3 H CH3 H CH3 CH3 CH3 H H CH3 C m U m A is less stable than B by 19 kcalmol same as AGO for one axial CH3 C is less stable than B by 37 kcalmol close to twice AGO for axial CH3 2 about 4 gauche interactions 1 cisl3dimethylcyclohexane CH3 CH3 H CH3 H The axialaxial conformer is less stable than B by about 55 kcalmol the additional 18 kcalmol is due to particularly unfavorable 13 MeMe diaXial interactions CH3 33 CHEM 6320 Fall 2005 e cisl4diI butylcyclohexane The presence of I butyl group in an axial position is so energetically unfavorable that the conformation of the ring changes to a twistboat to accommodate both Ibutyl groups in pseudoequatorial positions H H H W M 4 Decalin H H H H cis trans The trans isomer has the ring locked b The cis isomer has three additional gauche interactions One would predict 3 X 09 kcalmol 27 kcalmol less stable c The cis isomer can do ring ip as shown above EH H a The two Sp2 carbon atoms and the four atoms directly bonded to these are in one plane while the other two carbon atoms C4 and C5 are one above the plane and one below the plane b There are four pseudoaXial and four pseudoequatorial valences for the Sp3 carbon atoms in the ring 9 5 Cyclohexene 89 CHEM 6320 Fall 2005 6 Cyclohexanone and Derivatives a In cyclohexanone and methylenecyclohexane there is less torsional strain associated Sp2 compared with sp3 therefore a more facile ring 1nvers1on X Xo CH2 b Methylcyclohexanone For 2methylcyclohexanone the conformer with methyl in equatorial position is more stable that the conformer with methyl in aXial position by about 18 kcalmol same as methylcyclohexane o H CH H CH3 3 O For 3methylcyclohexanone the conformer with methyl in equatorial position is more stable that the conformer with methyl in aXial position by only about 1314 kcalmol because in this case there is only one unfavorable 13 diaXial interaction H 0 CH3 VS k CHgm H H O c Chloromethylcyclohexanone This is an example of the conformational preference is in uenced than other factors than steric O H 4 Egg 39 WiH Cl 90 CHEM 6320 Fall 2005 The conformer with methyl in equatorial position has a very large dipole moment The conformer with methyl in axial position has the unfavorable axial interaction between Cl and two H but has a smaller dipole moment and is the favored form in nonpolar solvents 7 13 Dioxan and Derivatives a The presence of oxygen makes the ring more exible and the 13 diaxial interactions less significant b The C O C angles are also different than the C C C angles in cyclohexane c The equilibrium between conformers is shifted Example 8 More Comments a When considering conformational preferences of multisubstituted cyclohexanes one has to consider dipoledipole andor stereoelectronic factors in addition to simple steric effects b Axialequatorial preferences can change in complicated molecules CH3 CH3 H The equatorial conformer has unfavorable bad van der Waals repulsion c One cannot necessarily predict exactly which conformation is the most stable in a molecule but one can make correlations and comparisons with similar compounds 91 CHEM 6320 Fall 2005 G Problems and Exercises 1 For the molecule CXZYZ if carbon uses Sp2 orbitals for bonding to X what hybrid orbitals does it use for bonding to Y 2 The H N H angle in ammonia is 10730 Calculate the hybrid orbital N uses to bond to H Then calculate the hybridization of the lonepair and the interorbital angle between the H and the lone pair The strain energy of spiropentane 625 kcalmol is considerably greater than twice that of cyclopropene 275 kcalmol Suggest an explanation 2 E 3 x 1 The fractional s character in bonds to carbon in organic molecules may be estimated by its relation to 13C BC coupling constants as determined by NMR Estimate the fractional s character of C1 in its bond to C3 of spiropentane given the following information S16 J13C13C K330 where K is a constant equal to 550 Hz the 13C BC coupling constant J between C1 and C3 is observed to be 202 Hz and s30 is the s character at C3 in its bond to C1 4 Provide electronic configurations Lewis dot structures for the following species also indicating the appropriate hybridization of the heavy atoms everything but H Give resonance forms and the resonance hybrid when resonance is important you may use line bonds to indicate electron pairs a 99 C0327 HC03 N05 N027 N37 azide 47 CHEM 6320 Fall 2005 f 8042 g 5032 h szof thiosulfate i NCO isocyanate j CNO cyanate k CH3NC methyl isocyanate 1 C1047 Write resonance forms for the following when appropriate a 2cyanoimidazole b 2 and 3 nitrofuran c 2 3 and 4aminobenzaldehyde d 2 3 and 4methoxypyridine Suggest a possible explanation for the following observations a The dipole moment of the hydrocarbon calicene has been estimated to be as large as 56 D l calicene b The dipole moment of furan is smaller than and in opposite direction from that of pyrrole U U Predict the energetically preferred site of protonation for each of the following molecules and eXplain the basis of your prediction C6H5CHN C6H5 o CH3 C b NHCH3 a 48 CHEM 6320 Fall 2005 Use thermochemical relationships to obtain the required information a The heats of formation of cyclohexane cyclohexene and benzene are respectively 295 11 and 198 kcalmol Estimate the resonance energy of benzene b The heats of formation of 2methyllpentene and 2methylpenthane are respectively 136 and 4l7 kcalmol Calculate the heat of hydrogenation of 2methyllpentene Identify the point group and the elements of symmetry for the following molecules a orthodichlorobenzene b metadichlorobenzene c paradichlorobenzene d naphthalene e 15dichloronaphthalene f anthracene Consider construction of full MO depiction of acetylene HCECH the same way ethylene was handled in the lecture notes There are 10 total atomic orbitals The 1s orbitals of carbon are not included Draw all the molecular orbitals considering nodal symmetry requirements Describe how many nodes each molecular orbital has and show where these nodes are located Calculate the energy levels and coefficients for 13 butadiene using Huckel MO theory 49 CHEM 6320 Fall 2005 B Sigmatropic Rearrangements 1 Introduction a Sigmatropic rearrangements are reactions that involve a concerted reorganization of electrons during which a group attached to a abond migrates to the terminus of an adjacent 7relectron system with a simultaneous shift of the 7relectrons b Sigmatropic rearrangements are labeled ij where i represents the number of atoms in the migrating fragment and j represents the number of atoms in the 7rsystem directly involved in the bonding changes c Topology of sigmatropic migration Suprafacial or supra migration involves a process in which the migrating group remains associated with the same face of the conjugated 7rsystem throughout the process Antarafacial or antara migration involves a process in which the migrating group moves to the opposite face of the 7rsystem during the migration d Examples an 0 a quotquot0 an 0 a H HvY a WH 11W a b d b d b d b d 13suprafacial shift of hydrogen 13antarafacial shift of hydrogen A R R I l R R 13shift of alkyl group 15shift of alkyl group O O 33sigmatropic rearrangement of Cope rearrangement A a 5hexadiene 33sigmatropic rearrangement of allyl vinyl ether oxiCope rearrangement V CH3 CH2 CH2 H3C CH2R CH2 CH2 RCH2 17sigmatropic shift of hydrogen 17sigmatropic shift of alkyl group A A A A R s o R S O RzNio gt R2N O 23sigmatropic rearrangement of allyl sulfoxide 23sigmatropic rearrangement of amine oxide 155 CHEM 6320 Fall 2005 2 Classification of Sigmatropic HydrogenAlkyl Shifts suprafacial 13 13 suprafacial retention 13 suprafacial inversion Hi39ickel system Hi39ickel system Mobius system 4 electrons 4 electrons 4 electrons suprafacial 15 15 suprafacial retention 15 suprafacial inversion Hi39ickel system Hi39ickel system Mobius system 6 electrons 6 electrons 6 electrons antarafacial 17 suprafacial 17 Mobius system Hi39ickel system 8 electrons 8electrons 156 CHEM 6320 Fall 2005 Selection Rules for Sigmatropic Shifts of Order ij Order 1j 1 j supraret suprainv antararet antarainv 4n forbidden allowed allowed forbidden 4n 2 allowed forbidden forbidden allowed Order ij i 39 suprasupra supraantara antaraantara 4n forbidden allowed forbidden 4n 2 allowed forbidden allowed ret retention inv inversion Examples a 13 alkyl shift occurs thermally with inversion of configuration of migrating carbon H H D H A H OAc H I l H D H H H H D OAc product if retention product if inversion not observed observed This reaction is not possible for hydrogen b 15alkyl shift occurs thermally with retention of configuration of migrating carbon Me H Merl 4 CHEM 6320 Fall 2005 c Cope rearrangement 33sigmatropy on 15diene gt COOEt COOEt The product is stabilized because it allows conjugation Reaction proceeds usually through chairlike transition state Boatlike transition state is less stable by about 6 kcalmol Me Me Ph gt Ph Me Me maj or Ph Ph Me 46 f Me Me Z minor Both transformations are electronically allowed Chirality in reactant is translated into the product This is a general feature of sigmatropic shifts d Degenerate rearrangement H H fast at high T H H The molecule homotropilidene transforms into itself At low temperature the transformation is slow so the 13C NMR gives five signals five different carbons At high temperature the transformation is fast so the 13C NMR gives three signals three different carbons Another example is bullvalene CHEM 6320 Fall 2005 C Cycloaddition Reactions 1 DielsAlder Reaction 3 b Cycloaddition reactions involve two molecules as reactants DielsAlder reaction also called the 47r 27 cycloaddition or 4 2 cycloaddition is the addition of an alkene or dienophile to a diene DielsAlder reaction involves 6 electrons so the transition state has a Huckel topology The reaction is stereospeci c syn suprafacial with respect to both diene and alkene Correlation diagram for 47z 27 cycloaddition Define a symmetry element preserved R gt 3 gt P In this case this is a plane of symmetry Assign all reacting orbitals as symmetric S or antisymmetric A with respect to the symmetry element E Draw the correlation diagram and identify correlations between the reactant and product orbitals Fill the orbitals with electrons and decide if the reaction is allowed HHo H 1 E m QQGE 05 Same arguments show that 2 2 cycloaddition is forbidden 159 CHEM 6320 Fall 2005 2 Selection Rules for m n Cycloadditions m n suprasupra supraantara antaraantara 4n forbidden allowed forbidden 4n 2 allowed forbidden allowed 3 Stereochemistry of DielsAlder Reaction a Alder Rule If two isomeric adducts are possible the one that has an unsaturated substituents on the alkene oriented toward the newly formed cyclohexene double bond is the preferred product In other words endo stereoisomer is preferred over exo stereoisomer endo addition exo addition Example 0 gt O lt O R R H Ile 2o R R H X gt XE U R endo product exo product CHEM 6320 Fall 2005 b Regioselectivity of the DielsAlder reaction It gives the preferred major product of a Diels Alder reaction when both diene and dienophile are not symmetric ie are substituted The reactivity is enlarged when the substituent on one reactant is electronwithdrawing group EWG while the substituent on the other reactant is electronreleasing group ERG The four possibilities and the major products are given below ERG ERG E ty A 6 W WG EWG pe gt B ERG ERG type l gt EWG EWG EWG EWG ERG ERG type C l gt EWG EWG type D l gt ERG ERG The preferred orientation is governed by matching largest coefficients on interacting frontier orbitals HOMO of one molecule and LUMO of the other molecule 161 CHEM 6320 Fall 2005 D Problems and Exercises 1 Show by constructing a correlation diagram whether each of the following disrotatory cyclizations is symmetry allowed a pentadienyl cation to cyclopentenyl cation b pentadienyl anion to cyclopentenyl anion 2 Which of the following reactions are allowed according to the orbital symmetry conservation rules Explain H A H H COOCH3 gt COOCH3 a COOCH3 COOCH3 A O b O i H O C CH3 CH3 CHZOSiCH33 Agt gti H C CH3CH2 d 3 OSiCH33 162 3 CHEM 6320 Fall 2005 Suggest a mechanism by which each transformation could occur More than one step can be involved Ho CH3 0 CH3CHC CHO 100 C I CH3 a 370 C Predict the regiochemistry and stereochemistry of the following cycloadditions reactions and indicate the basis of your prediction 0 CH3 H O H CHCH2 a O CH302CNH H H2CCHCOOCH3 b H CHCH2 CH3 H gt ltH ampN H c H CN 163 CHEM 6320 Fall 2005 E Hiickel Molecular Orbital Theory 1 a e Method Description Only 7relectrons are treated In the LCAO equation above as is a 2p orbital on atom s so that our basis set consists on n such 2p orbitals in the 7 system The Hamiltonian integral Hrs r i s also called resonance integral which represents the interaction energy of the electron in or with the electron in as is assumed to be 0 when r and s are not directly bonded 8 some average nonzero value when r and s are directly bonded Both of these assumptions are severe approximations The Hamiltonian integral Hm representing the energy of the electron in the basis set atomic orbital or is the same value 05 if all the atoms in the 7rsystem are carbons Altogether remembering what restrictions our use of an orthogonal set of orbitals places on the overlap integral S S 1 S H r a the Coulomb integral Hm 8 if atoms r and s are adjacent 201fris H m 0 if atoms r and s are neither identical nor adjacent Note 1 Both 05 and 8 are negative numbers Note 2 The Coulomb and resonance integrals Hrr a and Hrs 8 assume different values for 2p orbitals on oxygen nitrogen etc With these approximations the secular determinant can be reduced according to diagonal elements are 05 E offdiagonal elements are either 8 or 0 depending upon the connectivity of the molecule 27 CHEM 6320 Fall 2005 2 Examples 3 The case of ethylene CH2 2 CH2 The two molecular orbitals are given by 11 01A 2pA 01302pB W2 2 02A 2pA 02302paa The secular determinant will be Introduce a new variable x There are two solutions x 1 and x 1 that are associated with two orbitals E1 058 andE2 05 8 After getting the energy levels plug each solution for x into the original set of secular equations and solve using normalization requirements that icy 1 i1 For x 1 solution gt E 05 8 The two secular equations for this solution HAA EISAA clA HA3 EISABCIB 0 05 E101A 5413 2 0 33901A IB39013 0 CIA 013 c where the two carbon atoms are labeled A and B 28 CHEM 6320 Fall 2005 To obtain the values for CM and 01 B one need to put the normality condition for the molecular orbital W1 2 c1A 2pA C1B 2pB C02pA 02103 2 WW107 01Al 2pA02pAdT 2 2011011 l 2pA 2pBdT 013 l 2pB02pBdT 2c2 1 c 2145 1 1 W1 321 3021003 For x 1 solution 3 E2 a The two secular equations for this solution HAA EZSAA C2A HAB EZSAB 023 0 06 E2C2A 339023 0 339021 IB39C2B 0 02A 2 023 c To obtain the values for CM and 023 one need to put the normality condition for the molecular orbital W2 2 02A 2pA C2B 2pB C02pA 02103 WW1077 c22Al 2pA02pAdT 2021021l1 2p1C02p13aquot 023 l 2pB02pBdT 2c2 1 c 1wE 1 1 W2 2pA 33 29 CHEM 6320 Fall 2005 Molecular orbital energy diagram The energy of an electron in a pure 2p orbital is 05 The total 7renergy of ethylene is 205 8 205 28 More general for ground state E Z niEl The values of 05 and 8 are not calculated but rather obtained empirically by comparison with experiment By doing so 8 75 kJmol b The case of butadiene CH2 2 CH CH 2 CH2 The secular determinant is a E 8 0 0 8 05 E 8 0 0 0 8 oc E Introducing x lt3 E a Bx x 1 0 0 1 x 1 0 0 0 1 x 1 0 0 1 x 30 CHEM 6320 Fall 2005 7Expand x 1 0 1 1 0 x1 x 1 10 x 10xx3 2x x2 1 0 1 x 0 1 x x4 3x21 7 Solutions are xz li 2 2 7 Molecular orbital energy diagram 71618 061806181618 a71618 a 061 S a 1 618 4r J J a 7 I 1321 M a 7 The wavefunctions for butadiene 02p quot 7W I a 061 1quot a r anu 7 The energy of the 7z39e1ectrons total 7z39e1ectronic energy 7 Compared with two ethylenes 22a Z there is an additional 0472 of energy This is the delocalization energy Edeloc CHEM 6320 Fall 2005 c The case of benzene 7 The secular determinant is x 1 0 0 0 1 1 1 x 0 0 0 6 2 0 1 x 1 0 0 0 0 0 1 x 1 0 5 3 0 0 0 1 x 1 4 1 0 0 0 1 x 7 There are six solutions x i1 i1 i2 7 Molecular orbital energy diagram 0H 0H 2 7 The energy of the 7z39e1ectrons 7 The delocalization energy 7 The wavefunctions for benzene CHEM 6320 Fall 2005 d The case of naphthalene 8 l 7 9 2 6 10 3 5 4 The secular determinant is x 00000010 1x10000000 01x1000000 001x000001 0000x10001 00001x10000 000001x100 0000001x10 10000001xl 000110001x e The case of allyl The secular determinant is x10 0 CHZ CH CHZ 1x120 O O 01x x3 2x 0 The three solutions are x 5 0 5 Obtaining the molecular orbitals for solution x wE E1 05 xE m 02 0 c1 kz c3 0 02 45 0 From rst equation 0 2012 From rst and third equations 01 2 c3 33 CHEM 6320 Fall 2005 Normalization condition 012c c32 1 gt 0122012c12 1 gt 4012 1 L l l 01 5303 3 Ecz W1 2 2pl 2pagt 2p3 8 8 no node Obtaining the molecular orbitals for solution x 0 E2 05 02 0 cl c3 0 02 0 s 020gtcl 03 1 1 W2 31 2p3 g g one node Obtaining the molecular orbitals for solution x xE E3 0 J23 Find as above 01 c3 02 f 1 1 1 W3 21270 E 021220 51021176 8 g 8 two nodes 34 CHEM 6320 Fall 2005 Molecular orbital energy diagram or 7 xE a or wE Energy of the 7relectrons allyl species 7r electrons E Edamc cation 2 205 Z 0828 radical 3 3a z 082 8 anion 4 4a2J 0828 The extra energy compared with pure p electrons and the delocalization energy are the same for the allyl cation radical and anion 3 Limitations and Improvements a Filling orbitals with electrons in this way ignores the fact that there will be increased repulsion as we add successive electrons to the system Such repulsion is considered only in some average way incorporated into the parameters 05 and 8 Again the solutions are numerically inaccurate but relative comparisons consistently predict physical properties We assumed that the overlap integral Slj 0 for i 7 For non neighboring 2p orbitals this assumption is valid but overlap for neighboring 2p orbitals is about 025 How would the calculation change if we took this into account The case of ethylene H11ES11 H12ES12 0 H21ES21 H22ES22 oc E ES B ES a E where S is the adjacent 2p 2p overlap 35 CHEM 6320 Fall 2005 Introduce anew variable x a ES E a Bx l Sx 0 1 x There are two solutions x i1 and the associated energies are E1 a and E2 1 S 1 S and a new Define a new nonbonding level 0539 Olh SZ B 8 S 05 1 5239 With these definitions the energy levels are given by E 0539 i B39 the same form as before bond resonance integral 839 c Since the exact choices of 05 and 8 are numerically inaccurate arbitrary anyway one can make believe that the used 05 and 8 incorporate an average correction for S17 when i and j are adjacent The spacing between energy levels will be slightly different but are reasonably good for relative comparisons 4 Simplified Solutions of Energy Levels and Coefficients a For linear polyene of 11 atoms Secular determinant solutions x 2cosi where j 123n n1 The coefficient of atom i in jth molecular orbital wavefunction l 2 ij cl s1n n1 n1 36 CHEM 6320 b For rings ofn atoms Secular determinant solutions 2 397 x j 2 cos L n n1 2 n where j 0ili2 Fall 2005 for 11 odd for 11 even Trigonometric solutions can be obtained through use of Frost s circle trick For a ring of 11 atoms inscribe a regular polygon of 11 side in a circle of diameter 4 8 with one comer at the lowest point The lowest energy is always E 05 28 The points of the polygon will then correspond to the appropriate energy levels Note Frost s circle works for complete 7rsystems It should not be used for cyclopentadiene that should be treated as butadiene 37 CHEM 6320 Fall 2005 5 Qualitative Pictures for Molecular Orbitals in Linear Polyene a Procedure Draw a straight line connecting 2 points at one C C bond length on each terminus Draw vibrating strings with increasing number of nodes but keep the same amplitude Draw the atomic orbital at each C to touch the string b Example butadiene The coefficients of atomic orbitals in molecular orbitals are given by geometric condition above Observations 38 CHEM 6320 Fall 2005 6 Molecular Orbital Analysis a Electron densities and charge densities Total electron density at atom i 2 q Z njcij Where is the number of MO nj is the number of electrons in that orbital and cf is the coefficient of an atomic orbital on atom i in jth MO 11 2 if the molecular orbital is doubly occupied 11 1 if the molecular orbital is singly occupied 11 0 if the molecular orbital is empty Charge density at atom 139 Charge density 1 q The case of ethylene 11 12 Rf 1 Charge density 0 The case of allyl cation There are only two electrons in VA 139 01 012 q charge density 1 172 14 172 12 2 1M 12 1 0 3 12 14 12 12 E z1 The case of allyl radical and anion For allyl radical one additional electron in 12 For allyl anion two additional electrons in 12 39 CHEM 6320 Fa112005 Z cl cl qiradjcal qianion anion charge density 1 12 14 12112 1221215 12 2 0 0 10 10 0 3 12 14 12112 1221215 12 23 g g Note that charge densities match those predicted by valence bond theory resonance forms b Bond orders Partial 7zbond order in jth MO between atoms r and s pigs ancjrcjs J The case of ethylene p12 2 2lx lx 1 as expected The case of allyl cation radical anion 712 723 2121J 1JE 0707 Note that the value is in contrast to our usual sense of a half 7rbond order from valence bond resonance forms The bond is stronger than given by resonance theory 40 CHEM 6320 Fall 2005 7 Summary a Delocalization energy is calculated relative to 205 8 which is the energy of a double bond in ethylene and 05 which is the energy of a pure p orbital not participating in a double bond if the molecule is a radical Edeloc E system n2a 28 ma where n is the number of double bonds and m is the number of unpaired electrons eg m 1 in allyl radical b Resonance energy is calculated relative to 05 which is the energy of a pure p orbital in an atom c Procedure for Huckel method Write the molecule and label the atoms Write the secular determinant Solve the secular determinant to calculate the energy levels E a x where highest value of x corresponds to the lowest energy Plug in the values of x into secular equations Solve secular equations including the normalization condition to determine the coef cient of atomic orbitals in each molecular orbital Represent draw each molecular orbital Fill in the molecular orbital energy diagram with the appropriate number of electrons Calculate 7relectron energy E Z niEl Calculate delocalization energy and resonance energy Calculate electron densities charge densities and bond orders 41 CHEM 6320 Fall 2005 F Application of Molecular Orbital Theory to Reactivity and Stability l Hyperconjugation Hyperconjugative Stabilization Effect 3 b It explains Why ethyl cation is more stable than methyl cation It is sometimes referred to as nobond resonance H H9 he H c CH2 gt H CCH2 The CH3CH2 ion has a plane of symmetry 39 b The three bonding C H orbitals of the methyl group in CH3CH22 The symmetry adapted C H orbitals Y s and 0 0 w 12 3 3 Interaction of the methyl group orbital with the p orbital of the cationic carbon stabilizes a filled orbital therefore the molecule 42 CHEM 6320 Fall 2005 2 Bimolecular Substitutions 3 Reactivity can be explained in most cases by looking at the HOMOLUMO interaction in the transition state HOMO stands for the highest occupied molecular orbital LUMO stands for the lowest unoccupied molecular orbital b Molecular orbital diagram for bimolecular electrophilic substitution The substitution can occur from the front because symmetry permits HOMOLUMO interaction and stabilization A LUMOLUMO interaction is not possible due to symmetry and would not produce any stabilization anyway c Molecular orbital diagrams for bimolecular nucleophilic substitution l Eu Frontside attack C X Nu LUMO HOMO quotquotquot HOMO Symmetry prevents HOMOLUMO interaction so the only interaction is between filled orbitals The reaction will not take this path 43 CHEM 6320 Fall 2005 Backside attack LUNlO C X HOMO ll lt u H HOMO Nu Now the HOMOLUMO interaction is possible and the transition state is stabilized 3 Substituent Effects or Heteroatom Replacements in Conjugated System 3 Example ethylene vs formaldehyde or 7f a C70 Ol 7139 n 7 As electronegative increases the energy of a 2p atomic orbital goes down in energy There is a greater contribution of the atomic orbital of 0 than that of C in the bonding molecular orbital and reverse for the antibonding one 44 CHEM 6320 Fall 2005 b Example ethylene vs vinylamine A 9 CH2 CH QHQ lt gt CH2 CH gHQ Vinylamine is isoelectronic with allyl anion ai a7 H a J55 l39l a The shape of vinylamine HOMO 112 4 Symmetry Approach to Interaction between Reactants Molecular Orbitals a Use perturbation MO analysis of reaction probability between two molecules each defined in MO terms Transition state I for reaction involves partial bonding therefore partial interaction between orbitals The most important interaction will be between HOMO of nucleophile and LUMO of electrophile frontier orbital control The strength of interaction increases as interacting orbitals are closer in the energy Only molecular orbitals of matching symmetry can interact To do that de ne a symmetry element which is preserved from the beginning to the end of the reaction R gt I gt P One needs to assign reactant and product orbitals as symmetric or antisymmetric with respect to chosen symmetry element 45 CHEM 6320 Fall 2005 Principles of Stereochemistry A Definitions 1 Isomers different compounds with the same molecular formula 2 Stereoisomers isomers with the same connectivity but different configuration noninconvertible by single bond rotation Note distinction between con guration and conformation different spatial orientation interconvertible by single bond rotation 3 Atropisomers stereoisomers that result from restricted single bond rotation 4 Enantiomers stereoisomers that are mirror images of one another a Chirality the property of nonsuperimposability on mirror image b Homochiral the property of being enantiomerically pure 5 Diastereomers nonmirror image stereoisomers stereoisomers in which the distance between atoms ie nuclei is different 6 Optical Activity property of being able to rotate the plane of plane polarized light 3 Optical rotation a physical property but varies with wavelength of light xl conditions solvent temperature and concentration b Specific rotation 05 the associated physical constant quoted for a given solvent temperature and concentration to exclude complications of nonBeer s law behavior D refers to the sodium D line at 589 nm c Optical purity de ned for a mixture of enantiomers the enantiomers have equal but opposite in sign 05 values as a of mixture 0 tical uri V p p ty 0 aof pure enantiomers CHEM 6320 Fall 2005 d Enantiomeric excess ee an equivalent property defined as 100 mole fraction of mole fraction of X major enantiomer minor enantiomer Optical Rotatory Dispersion 0RD relates to how 05 changes with the wavelength 1 and is a re ection of configuration Circular Dichroism CD relates to the ability to show differential absorption of circularly polarized light The constant analogous to a is 6 called the molecular ellipticity Chiral Center typically a tetrahedral sp3 carbon with 4 nonidentical ligands Other examples are unsymmetrical sulfoxides and unsymmetrical tetraalkylammonium salts but not unsymmetrical amines O S CH3 Phs but N lllll mMe fast Ph Ph 0 0 quotCH3 Ph Et 0 lMe a Unsymmetrical amines b Unsymmetrical tetraalkylammonium salts c Unsymmetrical sulfoxides CHEM 6320 Fall 2005 B Conventions l CahnIngoldPrelog Convention a Use R rectus and S sinister descriptors that are assigned using the sequence rule to assign a priority order to the substituents on the atom to which a con guration is being assigned The R and S descriptors are not related to and clockwise and counterclockwise rotation of polarized light respectively b Procedure Assign decreasing priority in order of decreasing atomic number When two or more of the substituent atoms are the same element the priority is assign based on the next attached atom in those substituents An atom that is multiply bonded is counted once for each formal bond When two isotopes of the same elements are substituents the isotope with larger atomic mass takes priority over the isotope with lower atomic mass D has priority over H 13C has priority over 12C Lone pair is equivalent with atomic number 0 Look at the molecule with the substituent with the lowest priority 4 going away going back and the substituents with priorities 1 2 and 3 toward you Assign the con guration R if the substituents decrease in priority 1 gt2 gt3 in a clockwise fashion or assign the configuration S if the substituents decrease in priority in a counterclockwise sense CHEM 6320 c Examples Fall 2005 Assign priorities for the following substituents El aim El What is the smallest hydrocarbon that can have a chiral carbon Assign R or S configuration for all chiral centers below H quotMon CH3CH2 2 Fisher Projections a CH3 S quotquot39 CH3 0 Ph A Fisher projection is a twodimensional representation of a three dimensional structure in which the horizontal bonds are directed toward the Viewer toward front while the vertical bonds are directed away from the Viewer toward back CHEM 6320 Fall 2005 b It is used mainly for carbohydrates c Rotating a Fisher projection by 90 gives the enantiomer of the initial molecule and rotating a Fisher projection by 180 gives back the initial molecule CH3 Hc1 Cl Br H Br CH3 D C1 3 Fisher DL Convention 3 It was initially developed and is used mainly for carbohydrates b Procedure Align the molecule with the highest oxidation state C at top Write the molecule in a Fisher projection Look at the chiral center closest to the bottom If the substituent eg OH projects to right gt D isomer If the substituent eg OH projects to left gt L isomer Note The D and L descriptors are not related to and clockwise and counterclockwise rotation of polarized light respectively OH b Examples Glyceraldehydes HO CH2 CH CH0 CHEM 6320 Fall 2005 Erythrose and threose OH OH l HO CHg CH CH CH 0 The presence of another chiral center increase the numbers of stereoisomers There are 2quot possible stereoisomers for n chiral centers Derythrose and Dthreose are diastereoisomers so they have different names Derythrose and Lerythrose are enantiomers mirror image of each other IUPAC names Derythrose E Lerythrose E Dthreose E Lthreose E 4 ErythroThreo Terminology a It can be used only for the case of two chiral centers b The erythro and threo are defined based on erythrose and threose as A A H X X H H X H X B B erythro threo CHEM 6320 Fall 2005 c The case ofA i B There are two chiral centers therefore 4 stereoisomers an erythro enantiomeric pair optically active a threo enantiomeric pair optically active 1 The case ofA B There are still two chiral centers but because the substituents for the these centers are the same there are only 3 different stereoisomers a threo enantiomeric pair optically active an achiral molecule due to symmetry the molecule has a plane of symmetry called meso CHEM 6320 Fall 2005 C Stereochemistry of Selected Classes of Compounds 1 Amino Acids a Natural amino acids found in proteins have the Lconfiguration COOH COOH Lalanine H2N H Lserine H2N H CH3 CHZOH 2 Cycloalkanes Introduction 3 Many cycloalkanes present different possible conformations that are easily interchangeable into one another Example cyclohexane b If the conformers are easily interchanged in investigating the stereochemistry of the molecule it is more convenient to look at the carbon ring as being planar c In general if a molecule has a plane of symmetry or a center of inversion that molecule is optically inactive even though it has chiral centers two or more 3 Monosubstituted Cycloalkanes a b CHEM 6320 Fall 2005 4 Disubstituted Cycloalkanes a Disubstituted cyclopropane 12dimethyl cyclopropane b Disubstituted cyclobutane 12dimethyl cyclobutane 13 dimethyl cyclobutane Two achiral stereoisomers not meso cis and trans CHEM 6320 Fall 2005 c Disubstituted cyclopentane 12dimethyl cyclopentane 13 dimethy1 cyclopentane 60 CHEM 6320 Fall 2005 d Disubstituted cyclohexane 12dimethyl cyclohexane 13dimethyl cyclohexane 14dimethyl cyclohexane 61


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