D. A. EvansAn Introduction to Frontier Molecular Orbital Theory-1 Problems of the DayChem 206http://www.courses.fas.harvard.edu/~chem206/ http://evans.harvard.edu/problems/Chemistry 206 Advanced Organic ChemistryLecture Number 1The molecule illustrated below can react through either Path A or Path B to form salt 1 or salt 2. In both instances the carbonyl oxygen functions as the nucleophile in an intramolecular alkylation. What is the preferred reaction path for the transformation in question? O O Br N H O BrPath B Path A 1Br+N O H Br O2+N O H Br BrIntroduction to FMO Theory General Bonding Considerations The H2 Molecule Revisited (Again!) Donor & Acceptor Properties of Bonding & Antibonding States Hyperconjugation and "Negative" Hyperconjugation Anomeric and Related EffectsThis is a "thought" question posed to me by Prof. Duilo Arigoni at the ETH in Zuerich some years ago(First hr exam, 1999)The three phosphites illustrated below exhibit a 750fold span in reactivity with a test electrophile (eq 1) (Gorenstein, JACS 1984, 106, 7831).(RO)3P OMe O P O A O Reading Assignment for week:Kirby, Stereoelectronic Effects Carey & Sundberg: Part A; Chapter 1 Fleming, Chapter 1 & 2 Fukui,Acc. Chem. Res. 1971, 4, 57. (pdf) Curnow, J. Chem. Ed. 1998, 75, 910 (pdf) Alabugin & Zeidan, JACS 2002, 124, 3175 (pdf) D. A. Evans1-01-Cover Page 9/15/03 8:56 AM+El(+)O P O B+ (RO)3PEl(1)O P OMe O CRank the phosphites from the least to the most nucleophilic and provide a concise explanation for your predicted reactivity order.Monday, September 15, 2003D. A. EvansAn Introduction to Frontier Molecular Orbital Theory-1 Stereoelectronic EffectsChem 206Universal Effects Governing Chemical Reactions There are three: Steric EffectsNonbonding interactions (Van der Waals repulsion) between substituents within a molecule or between reacting moleculesMe S N2 Me Nu C R RO H H Me Me2CuLi RO Me H H RGeometrical constraints placed upon ground and transition states by orbital overlap considerations. Fukui Postulate for reactions:"During the course of chemical reactions, the interaction of the highest filled (HOMO) and lowest unfilled (antibonding) molecular orbital (LUMO) in reacting species is very important to the stabilization of the transition structure."Nu:RC RBrBr:RO HOOmajor General Reaction TypesOp. p.minorRadical Reactions (~10%): Polar Reactions (~90%):A + B A(:) + B(+)Lewis AcidA AB B Electronic Effects (Inductive Effects):The effect of bond and through-space polarization by heteroatom substituents on reaction rates and selectivities Inductive Effects: Through-bond polarization Field Effects: Through-space polarizationLewis BaseFMO concepts extend the donor-acceptor paradigm to non-obvious families of reactions Examples to considerMe R C R BrS N1+R R C Me + Br:H2+2 Li(0) Mg(0)2 LiH CH3MgBrCH3I +rate decreases as R becomes more electronegative "Organic chemists are generally unaware of the impact of electronic effects on the stereochemical outcome of reactions." "The distinction between electronic and stereoelectronic effects is not clear-cut."1-02-Introduction-19/12/03 4:44 PMD. A. EvansSteric Versus Electronic Effects; A time to be careful!!O R3SiO EtO O R3Si OSiR3 O TiCl4 Nu OSiR3Chem 206 Steric Versus electronic Effects: Some Case StudiesWhen steric and electronic (stereoelectronic) effects lead to differing stereochemical consequencesWoerpel etal. JACS 1999, 121, 12208.diastereoselection >94:6 OSiR3 H O diastereoselection 93:7 H OSiR3OOAcSnBr4 Me SiMe3O MeMeAlCl3stereoselection 99:1 stereoselection >95:5 OSiR3OOAcSnBr4 BnOOO Danishefsky et al JOC 1991, 56, 387 BnOp. p.BnOO EtO2C O EtO2C (R)2CuLi Bu OTBS O OTBS diastereoselection 8:1 O Ph N O only diastereomer Bu OTBS OAc OAc O H Ph N O H N N N N O OAc OAc H H N R R O TBS Al R O Yakura et al Tetrahedron 2000, 56, 7715 O Ph 60-94% AcO AcO H H N O only diastereomer PhBu3AlR3 O Al OEtO2CEtOYakura's rationalization:Mehta et al, Acc Chem. Res. 2000, 33, 278-2861-03-Introduction-1a9/15/03 8:14 AMD. A. EvansThe H2 Molecular Orbitals & Antibonds The H2 Molecule (again!!)Chem 206Linear Combination of Atomic Orbitals (LCAO): Orbital Coefficients Rule Two: Each MO is constructed by taking a linear combination of the individual atomic orbitals (AO): Bonding MO Antibonding MO = C11 + C22 = C*11 C*22Let's combine two hydrogen atoms to form the hydrogen molecule. Mathematically, linear combinations of the 2 atomic 1s states create two new orbitals, one is bonding, and one antibonding: Rule one: A linear combination of n atomic states will create n MOs. (antibonding) E Energy H 1 p. p. E (bonding) 1s 1s H 2The coefficients, C1 and C2, represent the contribution of each AO. Rule Three: (C1)2 + (C2)2 = 1The squares of the C-values are a measure of the electron population in neighborhood of atoms in question Rule Four: bonding(C1)2 + antibonding(C*1)2= 1 In LCAO method, both wave functions must each contribute one net orbital Consider the pibond of a C=O function: In the ground state pi-CO is polarized toward Oxygen. Note (Rule 4) that the antibonding MO is polarized in the opposite direction.Let's now add the two electrons to the new MO, one from each H atom: (antibonding) E1EnergyCO (antibonding)H 11s E21sH 2 Energy (bonding)CONote that E1 is greater than E2. Why?C O (bonding)1-04-Introduction-2 9/15/03 8:38 AMD. A. EvansBonding GeneralizationsChem 206 Bond strengths (Bond dissociation energies) are composed of a covalent contribution ( Ecov) and an ionic contribution ( Eionic). Bond Energy (BDE) = Ecovalent + Eionic (Fleming, page 27) Orbital orientation strongly affects the strength of the resulting bond.For Bonds: A B Better than A BWhen one compares bond strengths between CC and CX, where X is some other element such as O, N, F, Si, or S, keep in mind that covalent and ionic contributions vary independently. Hence, the mapping of trends is not a trivial exercise.For Bonds:ABBetter thanABUseful generalizations on covalent bonding Overlap between orbitals of comparable energy is more effective than overlap between orbitals of differing energy.p. p.This is a simple notion with very important consequences. It surfaces in the delocalized bonding which occurs in the competing anti (favored) syn (disfavored) E2 elimination reactions. Review this situation. An anti orientation of filled and unfilled orbitals leads to better overlap. This is a corrollary to the preceding generalization. There are two common situations.For example, consider elements in Group IV, Carbon and Silicon. We know that C-C bonds are considerably stronger by Ca. 20 kcal mol-1 than C-Si bonds.C C C CC CSi Cbetter thanCSiCSiCase-1: Anti Nonbonding electron pair & CX bondXSi-SP3X Alone pair HOMOC-SP3 CCC-SP3A C* CX LUMOX Better thanlone pair HOMO* CX LUMOC-SP3 CSiCACH3CCH3 BDE = 88 kcal/mol Bond length = 1.534 H3CSiH3 BDE ~ 70 kcal/mol Bond length = 1.87 Case-2: Two anti sigma bondsX C CY HOMOThis trend is even more dramatic with pi-bonds: CC = 65 kcal/mol CSi = 36 kcal/mol SiSi = 23 kcal/mol Weak bonds will have corresponding low-lying antibonds.Formation of a weak bond will lead to a corresponding low-lying antibonding orbital. Such structures are reactive as both nucleophiles & electrophiles A YY* CX LUMOX CCBetter thanX C CY HOMO* CX LUMOCY1-05-Introduction-3 9/12/03 4:36 PMD. A. EvansDonor-Acceptor Properties of Bonding and Antibonding StatesChem 206Donor Acceptor Properties of C-C & C-O Bonds Consider the energy level diagrams for both bonding & antibonding orbitals for CC and CO bonds.* C-C * C-OHierarchy of Donor & Acceptor States Following trends are made on the basis of comparing the bonding and antibonding states for the molecule CH3X where X = C, N, O, F, & H.-bonding States: (CX)CH3CH3 CH3H CH3NH2 CH3OHC-SP3C-SP3 O-SP3very close!!decreasing -donor capacity C-C C-OCH3Fpoorest donorp. p. The greater electronegativity of oxygen lowers both the bonding & antibonding C-O states. Hence: CC is a better donor orbital than CO CO is a better acceptor orbital than CC-anti-bonding States: (CX)CH3HFor the latest views, please read Alabugin & Zeidan, JACS 2002, 124, 3175 (pdf)CH3CH3 CH3NH2 CH3OH CH3FDonor Acceptor Properties of CSP3-CSP3 & CSP3-CSP2 Bonds* CC * CC better acceptor Increasing -acceptor capacitybest acceptorC-SP3C-SP3 C-SP2The following are trends for the energy levels of nonbonding states of several common molecules. Trend was established by photoelectron spectroscopy.Nonbonding States CCbetter donor CCH3P: H2S: H3N: H2O: HCl: decreasing donor capacitypoorest donor The greater electronegativity of CSP2 lowers both the bonding & antibonding CC states. Hence: CSP3-CSP3 is a better donor orbital than CSP3-CSP2 CSP3-CSP2 is a better acceptor orbital than CSP3-CSP31-06-donor/acceptor states 9/12/03 5:16 PMD. A. EvansHybridization vs ElectronegativityChem 206Electrons in 2S states "see" a greater effective nuclear charge than electrons in 2P states. This becomes apparent when the radial probability functions for S and P-states are examined: The radial probability functions for the hydrogen atom S & P states are shown below.100 % 100 %There is a linear relationship between %S character & Pauling electronegativity5N4.5SPRadial ProbabilityPauling ElectronegativityRadial Probability1 S Orbital4N N3.5SP3SP2C 3SP2 S Orbital2 S Orbital 2 P OrbitalCp. p.3 S Orbital 3 P Orbital2.5SP2CSP322025303540455055% S-CharacterThere is a direct relationship between %S character & hydrocarbon acidity60CH (56)S-states have greater radial penetration due to the nodal properties of the wave function. Electrons in S-states "see" a higher nuclear charge. Above observation correctly implies that the stability of nonbonding electron pairs is directly proportional to the % of S-character in the doubly occupied orbitalLeast stable Most stablePka of Carbon Acid5545045C H (44)6 640CSP3CSP2CSP35PhCC-H (29)30The above trend indicates that the greater the % of S-character at a given atom, the greater the electronegativity of that atom.1-07-electroneg/hybrization 9/12/03 4:49 PM25 20 25 30 35 40 45 50 55% S-CharacterD. A. EvansHyperconjugation: Carbocation StabilizationChem 206 The interaction of a vicinal bonding orbital with a p-orbital is referred to as hyperconjugation. This is a traditional vehicle for using valence bond to denote charge delocalization.R H H CPhysical Evidence for Hyperconjugation Bonds participating in the hyperconjugative interaction, e.g. CR, will be lengthened while the C(+)C bond will be shortened. First X-ray Structure of an Aliphatic Carbocation+CR+ H H H H C C H HThe graphic illustrates the fact that the C-R bonding electrons can "delocalize" to stabilize the electron deficient carbocationic center. Note that the general rules of drawing resonance structures still hold: the positions of all atoms must not be changed.p. p.+1.431 [F5SbFSbF5]+CStereoelectronic Requirement for Hyperconjugation: Syn-planar orientation between interacting orbitals100.6 1.608 MeMe MeThe Molecular Orbital Description CR CRT. Laube, Angew. Chem. Int. Ed. 1986, 25, 349+CH H+CH HThe Adamantane Reference (MM-2)H1.528 CR CR110 Me Me Me1.530 Take a linear combination of CR and CSP2 p-orbital: "The new occupied bonding orbital is lower in energy. When you stabilize the electrons is a system you stabilize the system itself."1-08-Hyperconj (+)-1 9/12/03 4:53 PMD. A. Evans"Negative" HyperconjugationSyn OrientationRChem 206antibonding CR R: H H C X+ H H antibonding CR R: R X+ H H C filled hybrid orbital X Delocalization of nonbonding electron pairs into vicinal antibonding orbitals is also possible R H H CR C XR H H H HH X H H HCXfilled hybrid orbitalXCThis decloalization is referred to as "Negative" hyperconjugationAnti OrientationRSince nonbonding electrons prefer hybrid orbitals rather that P orbitals, this orbital can adopt either a syn or anti relationship to the vicinal CR bond.H HCXH HCThe Molecular Orbital Description CR Overlap between two orbitals is better in the anti orientation as stated in "Bonding Generalizations" handout.XNonbonding e pairThe Expected Structural PerturbationsChange in Structure Shorter CX bond Longer CR bondSpectroscopic ProbeX-ray crystallography X-ray crystallography Infrared Spectroscopy Infrared Spectroscopy NMR Spectroscopy NMR Spectroscopy CRAs the antibonding CR orbital decreases in energy, the magnitude of this interaction will increase Note that CR is slightly destabilized Stronger CX bond Weaker CR bond Greater e-density at R Less e-density at X1-09-Neg-Hyperconj 9/12/03 4:53 PMD. A. EvansLone Pair Delocalization: N2F2The trans IsomerChem 206Now carry out the same analysis with the same 2 orbitals present in the trans isomer.The interaction of filled orbitals with adjacent antibonding orbitals can have an ordering effect on the structure which will stabilize a particular geometry. Here are several examples: Case 1: N2F2 F N N This molecule can exist as either cis or trans isomers F F N N F There are two logical reasons why the trans isomer should be more stable than the cis isomer. The nonbonding lone pair orbitals in the cis isomer will be destabilizing due to electron-electron repulsion. The individual CF dipoles are mutually repulsive (pointing in same direction) in the cis isomer. In fact the cis isomer is favored by 3 kcal/ mol at 25 C. Let's look at the interaction with the lone pairs with the adjacent CF antibonding orbitals. The cis Isomer Ffilled N-SP2 F antibonding NF NF (LUMO) filled N-SP2 (HOMO)filled N-SP2N FNF antibonding NF filled N-SP2 (HOMO) NF (LUMO) In this geometry the "small lobe" of the filled N-SP2 is required to overlap with the large lobe of the antibonding CF orbital. Hence, when the new MO's are generated the new bonding orbital is not as stabilizing as for the cis isomer.Conclusions Lone pair delocalization appears to override electron-electron and dipole-dipole repulsion in the stabilization of the cis isomer. This HOMO-LUMO delocalization is stronger in the cis isomer due to better orbital overlap. Important Take-home Lesson Orbital orientation is important for optimal orbital overlap.NNABforms stronger pi-bond thanAB Note that by taking a linear combination of the nonbonding and antibonding orbitals you generate a more stable bonding situation. Note that two such interactions occur in the molecule even though only one has been illustrated.1-10- N2F2 9/12/03 4:59 PMABforms stronger sigma-bond thanABThis is a simple notion with very important consequences. It surfaces in the delocalized bonding which occurs in the competing anti (favored) syn (disfavored) E2 elimination reactions. Review this situation.D. A. EvansLone Pair Delocalization: The Gauche EffectChem 206The interaction of filled orbitals with adjacent antibonding orbitals can have an ordering effect on the structure which will stabilize a particular conformation. Here are several examples of such a phenomon called the gauche effect: HydrazineH N H N H HantiThe closer in energy the HOMO and LUMO the better the resulting stabilization through delocalization. Hence, N-lone pair NH delocalization better than NH NH delocalization. Hence, hydrazine will adopt the gauche conformation where both N-lone pairs will be anti to an antibonding acceptor orbital. The trend observed for hydrazine holds for oxygen derivatives as well Hydrogen peroxideH O O H anti Hydrazine can exist in either gauche or anti conformations (relative to lone pairs). H N HH HH N H Hgaucheobserved HNNH dihedral angle Ca 90HThere is a logical reason why the anti isomer should be more stable than the gauche isomer. The nonbonding lone pair orbitals in the gauche isomer should be destabilizing due to electron-electron repulsion. In fact, the gauche conformation is favored. Hence we have neglected an important stabilization feature in the structure. HOMO-LUMO Interactions Orbital overlap between filled (bonding) and antibonding states is best in the anti orientation. HOMO-LUMO delocalization is possible between: (a) N-lone pair NH; (b) NH NHH filled N-SP3 (HOMO) N N NH (LUMO) NH (HOMO) H NH (LUMO) N N H NH (LUMO)H2O2 can exist in either gauche or anti conformations (relative to hydrogens). The gauche conformer is prefered. H OO Hgauche Hobserved HOOH dihedral angle Ca 90H Major stabilizing interaction is the delocalization of O-lone pairs into the CH antibonding orbitals (Figure A). Note that there are no such stabilizing interactions in the anti conformation while there are 2 in the gauche conformation. Figure A Figure B OH (LUMO) (HOMO) filled O-SP3filled N-SP3 (HOMO)(HOMO) filled O-SP3 Note that you achieve no net stabilization of the system by generating molecular orbitals from two filled states (Figure B). better stabilization NH (HOMO)Problem: Consider the structures XCH2OH where X = OCH3 and F. What is the most favorable conformation of each molecule? Illustrate the dihedral angle relationship along the CO bond.1-11 Gauche Effect 9/11/01 11:27 PMD. A. EvansThe Anomeric Effect: Negative HyperconjugationUseful LIterature ReviewsChem 206http://www.courses.fas.harvard.edu/~chem206/Kirby, A. J. (1982). The Anomeric Effect and Related Stereoelectronic Effects at Oxygen. New York, Springer Verlag. Box, V. G. S. (1990). The role of lone pair interactions in the chemistry of the monosaccharides. The anomeric effect. Heterocycles 31: 1157. Box, V. G. S. (1998). The anomeric effect of monosaccharides and their derivatives. Insights from the new QVBMM molecular mechanics force field. Heterocycles 48(11): 2389-2417. Graczyk, P. P. and M. Mikolajczyk (1994). Anomeric effect: origin and consequences. Top. Stereochem. 21: 159-349. Juaristi, E. and G. Cuevas (1992). Recent studies on the anomeric effect. Tetrahedron 48: 5019. Plavec, J., C. Thibaudeau, et al. (1996). How do the Energetics of the Stereoelectronic Gauche and Anomeric Effects Modulate the Conformation of Nucleos(t)ides? Pure Appl. Chem. 68: 2137-44. Thatcher, G. R. J., Ed. (1993). The Anomeric Effect and Associated Stereoelectronic Effects. Washington DC, American Chemical Society.Chemistry 206 Advanced Organic ChemistryLecture Number 2Stereoelectronic Effects-2 Anomeric and Related Effects Electrophilic & Nucleophilic Substitution Reactions The SN2 Reaction: Stereoelectronic Effects Olefin Epoxidation: Stereoelectronic Effects Baeyer-Villiger Reaction: Stereoelectronic Effects Hard & Soft Acid and Bases (Not to be covered in class)Problem 121http://evans.harvard.edu/problems/Sulfonium ions A and B exhibit remarkable differences in both reactivity and product distribution when treated with nucleophiles such as cyanide ion (eq 1, 2). Please answer the questions posed in the spaces provided below. KCNReading Assignment: Kirby, Chapters 1-3S BF4 EtArel. rate = 8000SEt+ PhCH2CN(1)D. A. EvansWednesday, September 17, 2003KCN S Et B rel. rate = 1 S+MeCH2CN(2)2-00-Cover Page 9/17/03 8:35 AMD. A. EvansThe Anomeric EffectThe Anomeric Effect: Negative HyperconjugationChem 206It is not unexpected that the methoxyl substituent on a cyclohexane ring prefers to adopt the equatorial conformation. H OMe HOMe Gc = +0.6 kcal/mol What is unexpected is that the closely related 2-methoxytetrahydropyran prefers the axial conformation: H O OMe Gp = 0.6 kcal/mol O H OMe Since the antibonding CO orbital is a better acceptor orbital than the antibonding CH bond, the axial OMe conformer is better stabilized by this interaction which is worth ca. 1.2 kcal/mol. Other electronegative substituents such as Cl, SR etc also participate in anomeric stabilization. H 1.781 H O Cl H O O Cl 1.819 ClThis conformer preferred by 1.8 kcal/mol Why is axial CCl bond longer ?axial O lone pair CCl CClThat effect which provides the stabilization of the axial OR conformer which overrides the inherent steric bias of the substituent is referred to as the anomeric effect. Let anomeric effect = A Gp = A = Gc + A Gp GcO ClH O HOMOThe Exo-Anomeric Effect CClA = 0.6 kcal/mol 0.6 kcal/mol = 1.2 kcal/mol Principal HOMO-LUMO interaction from each conformation is illustrated below: H There is also a rotational bias that is imposed on the exocyclic COR bond where one of the oxygen lone pairs prevers to be anti to the ring sigma CO bond H O O R O R favored R O OOOOMeOH OMeaxial O lone pair CHaxial O lone pair COA. J. Kirby, The Anomeric and Related Stereoelectronic Effects at Oxygen, Springer-Verlag, 1983 E. Jurasti, G. Cuevas, The Anomeric Effect, CRC Press, 19952-01-Anomeric Effect-1 9/16/03 2:40 PMD. A. EvansThe Anomeric Effect: Carbonyl GroupsAldehyde CH Infrared Stretching FrequenciesChem 206Do the following valence bond resonance structures have meaning?R C X O X R C OPrediction: The IR CH stretching frequency for aldehydes is lower than the closely related olefin CH stretching frequency. For years this observation has gone unexplained. R R C H CH = 2730 cm CF3-1R C C RPrediction: As X becomes more electronegative, the IR frequency should increase O Me CH3 Me O CBr3 Me OO H CH = 3050 cm -1C=O (cm-1)Sigma conjugation of the lone pair anti to the H will weaken the bond. This will result in a low frequency shift.172017501780Infrared evidence for lone pair delocalization into vicinal antibonding orbitals.The NH stretching frequency of cis-methyl diazene is 200 cm-1 lower than the trans isomer. Me H N N Mefilled N-SP2 H antibonding NHPrediction: As the indicated pi-bonding increases, the XCO bond angle should decrease. This distortion improves overlap.R C X * CX O lone pair OR C X ONN NH = 2188 cm -1 Me N N H NH = 2317 cm -1Me Nfilled N-SP2Nantibonding NH HEvidence for this distortion has been obtained by X-ray crystallography Corey, Tetrahedron Lett. 1992, 33, 7103-7106.. The low-frequency shift of the cis isomer is a result of NH bond weakening due to the anti lone pair on the adjacent (vicinal) nitrogen which is interacting with the NH antibonding orbital. Note that the orbital overlap is not nearly as good from the trans isomer. N. C. Craig & co-workers JACS 1979, 101, 2480.2-02-Anomeric Effect-2 9/16/03 2:41 PMD. A. EvansThe Anomeric Effect: Nitrogen-Based SystemsCMe3 Me3C CH N N N CMe3 Me3C N N N Me3CChem 206Observation: CH bonds anti-periplanar to nitrogen lone pairs are spectroscopically distinct from their equatorial CH bond counterpartsH H H HN HMe3C G = 0.35kcal/molN HOMO CHA. R. Katritzky et. al., J. Chemm. Soc. B 1970 135Favored Solution Structure (NMR)Me MeN NMe NMe Me N Me J. E. Anderson, J. D. Roberts, JACS 1967 96 4186 N N N Me MeNSpectroscopic Evidence for Conjugation Infrared Bohlmann Bands Characteristic bands in the IR between 2700 and 2800 cm-1 for C-H4, C-H6 , & C-H10 stretch Bohlmann, Ber. 1958 91 2157 Reviews: McKean, Chem Soc. Rev. 1978 7 399 L. J. Bellamy, D. W. Mayo, J. Phys. Chem. 1976 80 1271 NMR : Shielding of H antiperiplanar to N lone pair H10 (axial): shifted furthest upfield H6, H4: = Haxial - H equatorial = -0.93 ppm Protonation on nitrogen reduces to -0.5ppm H. P. Hamlow et. al., Tet. Lett. 1964 2553 J. B. Lambert et. al., JACS 1967 89 37612-03-Anomeric Effect-3 9/16/03 2:43 PMFavored Solid State Structure (X-ray crystallography)1.484 1.453Bn N Me1.457Me1.453N N Bn1.459NA. R. Katrizky et. al., J. C. S. Perkin II 1980 1733D. A. EvansAnomeric Effects in DNA PhosphodiestersChem 206Calculated Structure of ACGTGC DuplexThe Phospho-Diesters Excised from Crystal StructureGuanine Cytosine Cytosine1B2B 1A Phosphate-1A p. p.Thymine AdeninePhosphate-1BThe Anomeric EffectAcceptor orbital hierarchy: * POR * > * POR OR OP OO O RPO O RO R P O O RPhosphate-2A Phosphate-2BGauche-Gauche conformation OR P OO O R O OAnti-Anti conformation Gauche-Gauche conformation affords a better donor-acceptor relationshipOxygen lone pairs may establish a simultaneous hyperconjugative relationship with both acceptor orbitals only in the illustrated conformation.Plavec, et al. (1996). How do the Energetics of the Stereoelectronic Gauche & Anomeric Effects Modulate the Conformation of Nucleos(t)ides? Pure Appl. Chem. 68: 2137-44.2-04-DNA Duplex/Anomeric 9/17/03 9:25 AMD. A. EvansCarboxylic Acids (& Esters): Anomeric Effects Again? Hyperconjugation: (Z) ConformerO O R' O Me H O MeChem 206 Conformations: There are 2 planar conformations.O O O O H R' RLet us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane. * COO R R O(Z) Conformer Specific Case: Methyl FormateR(E) ConformerRG = +4.8 kcal/molOC ROIn the (Z) conformation this lone pair is aligned to overlap with * CO.The (E) conformation of both acids and esters is less stable by 3-5 kcal/mol. If this equilibrium were governed only by steric effects one would predict that the (E) conformation of formic acid would be more stable (H smaller than =O). Since this is not the case, there are electronic effects which must also be considered. These effects will be introduced shortly. Rotational Barriers: There is hindered rotation about the =COR bond. These resonance structures suggest hindered rotation about =COR bond. This is indeed observed:O R O R' R O O R'Energy(E) ConformerR R O C OIn the (E) conformation this lone pair is aligned to overlap with * CR.* CROR O Cbarrier ~ 10-12 kcal/molO R O R O RG ~ 2-3 kcal/molSince * CO is a better acceptor than * CR (where R is a carbon substituent) it follows that the (Z) conformation is stabilized by this interaction.RO RO R R OEsters versus Lactones: Questions to Ponder.Esters strongly prefer to adopt the (Z) conformation while small-ring lactones such as 2 are constrained to exist in the (Z) conformation. From the preceding discussion explain the following: 1) Lactone 2 is significantly more susceptible to nucleophilic attack at the carbonyl carbon than 1? Explain.Rotational barriers are ~ 10-12 kcal/mol. This is a measure of the strength of the pi bond.O 1 Et CH3CH2 O O O 2 Lone Pair Conjugation: The oxygen lone pairs conjugate with the C=O.versus R OC ROThe filled oxygen p-orbital interacts with pi (and pi*) C=O to form a 3-centered 4-electron bonding system.2) Lactone 2 is significantly more prone to enolization than 1? In fact the pKa of 2 is ~25 while ester 1 is ~30 (DMSO). Explain. 3) In 1985 Burgi, on carefully studying O O O the X-ray structures of a number of lactones, noted that the O-C-C () & O O O O-C-O () bond angles were not equal. Explain the indicated trend in bond = 12.3 = 6.9 = 4.5 angle changes.SP2 Hybridization Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridization is driven by system to optimize pi-bonding.2-05 RCO2R Bonding 9/16/03 2:50 PMD. A. EvansThree-Center BondsCase 3: 2 p-Orbitals; 1 s-orbitalChem 206Consider the linear combination of three atomic orbitals. The resulting molecular orbitals (MOs) usually consist of one bonding, one nonbonding and one antibonding MO. Case 1: 3 p-Orbitals pi-orientationantibondingantibonding2+nonbondingEnergy3nonbondingbondingCase 4: 2 s-Orbitals; 1 p-orbitalbondingNote that the more nodes there are in the wave function, the higher its energy. H 2C H 2C CH CH + CH2 Allyl carbonium ion: both pi-electrons in bonding stateDo this as an exerciseExamples of three-center bonds in organic chemistry A. H-bonds: (3center, 4electron)O CH3 O H O H O CH3CH2Allyl Radical: 2 electrons in bonding obital plus one in nonbonding MO.The acetic acid dimer is stabilized by ca 15 kcal/molH 2CCH Allyl Carbanion: 2 electrons in bonding obital plus 2 in CH2 nonbonding MO.B. H-B-H bonds: (3-center, 2 electron)H B H B H H H H H B H H B H HCase 2: 3 p-Orbitals sigma-orientation 3nonbonding antibondingHdiborane stabilized by 35 kcal/molEnergyC. The SN2 Transition state: (3center, 4electron)H Nu C H H BrThe SN2 transition state approximates a case 2 situation with a central carbon p-orbital The three orbitals in reactant molecules used are: 1 nonbonding MO from Nucleophile (2 electrons) 1 bonding MO CBr (2 electrons) 1 antibonding MO * CBrbonding2-06 3-center bonds/review 10/28/03 12:00 PMD. A. EvansSubstitution Reactions: General ConsiderationsChem 206Why do SN2 Reactions proceed with backside displacement?R Nu: H H C X R Nu XElectrophilic substitution at saturated carbon may occur with either inversion of retention Inversion + MR Nu C H H X:C H HRa El(+) H Rb C M+ NuRa C HRa Nu C Rb H M+Given the fact that the LUMO on the electrophile is the CX antibonding orblital, Nucleophilic attack could occur with either inversion or retention.RbInversionR NuRetentionRa El(+) Ra C M H Rb C M+ RetentionR X H H CRa H Rb C El M+C H HLUMOXH RbEl +HOMOConstructive overlap between Nu & *CXNu Overlap from this geometry results in no net bonding interaction El(+)LUMORa Ra C H RbC M H RbMExpanded view of *CXHOMOInversionLUMOEl(+)Cantibonding HOMOXbonding Br2 H BrRetention ExamplesLi HCO2 CO2Li HNuFleming, page 75-76predominant inversionpredominant retentionStereochemistry frequently determined by electrophile structureSee A. Basu, Angew. Chem. Int. Ed. 2002, 41, 717-7382-07-SN2-1 9/18/03 12:38 PMD. A. EvansThe reaction under discussion:R R C H XSN2 Reaction: Stereoelectronic EffectsThe use of isotope labels to probe mechanism. C H H X R Nu C H HChem 206Nu: HNuX: 1 and 2 containing deuterium labels either on the aromatic ring or on the methyl group were prepared. A 1:1-mixture of 1 and 2 were allowed to react. If the rxn was exclusively intramolecular, the products would only contain only three deuterium atoms:O S O O CH3 SO3 D3C Nu The NuCX bonding interaction is that of a 3-center, 4-electron bond. The frontier orbitals which are involved are the nonbonding orbital from Nu as well as CX and CX: CXD 3CNu:exclusively intramolecular(CD3ArNuCH3)CH31O SO O CD3SO3 H 3C Nu (CH3ArNuCD3energyNu: C XMeNu:exclusively intramolecularCD3CX RCH2XNu Experiments have been designed to probe inherent requirement for achieving a 180 NuCX bond angle: Here both Nu and leaving group are constrained to be part of the same ring.R R2 If the reaction was exclusively intermolecular, products would only contain differing amounts of D-label depending on which two partners underwent reaction. The deuterium content might be analyzed by mass spectrometry. Here are the possibilities: 1 + 1 D3-product 2 CD3ArNuCH3 D'3-product 2 + 2 2 CH3ArNuCD3 1 CD3ArNuCD3 1 CH3ArNuCH3 Hence, for the strictly intermolecular situation one should see the following ratios D0 : D3 : D'3 : D6 = 1 : 2 : 2 : 1. 1 + 2 The product isotope distribution in the Eschenmoser expt was found to be exclusively that derived from the intermolecular pathway! Other Cases: exclusively intermolecular(CH3)2N SO3CH3 (CH3)3NNu:C XC H H XH HNuD6-product D0-product"tethered reactants""constrained transition state"The Eschenmoser Experiment (1970): Helv. Chim Acta 1970, 53, 2059 The reaction illustrated below proceeds exclusively through bimolecular pathway in contrast to the apparent availability of the intramolecular path.O S O O CH3 SO3+SO3CH316% intramolecular 84% intermolecularSO3CH3 N(CH3)2SO3Nu:N(CH3)3+Nu2-08-The SN2 RXN-FMO 9/16/03 2:56 PMHence, the NuCX 180 transition state bond angle must be rigidly maintained for the reaction to take place.D. A. EvansSO3CH3 N(CH3)2Intramolecular methyl transfer: Speculation on the transition structures(CH3)2NChem 206+SO3SO3SO3CH3N(CH3)3+(CH3)3N16% intramolecular; 84% intermolecular9- membered cyclic transition stateexclusively intermolecular8- membered cyclic transition state174 1742-09-Intra alk TS's 9/16/03 2:56 PM00000 00000 00000 00000 00000est CO bond length 2.1 000000 000000 000000 000000 000000est CO bond length 2.1 000000 00000 000000 00000 00 00000 00 00000 00 00 00 00 0000 00 00 0000 00000 0000 00000est CN bond length 2.1 est CN bond length 2.1 Approximate representation of the transition states of the intramolecular alkylation reactions. Transition state CO and CN bond lengths were estimated to be 1.5x(CX) bond length of 1.4 D. A. Evans The General Reaction:R R OOlefin Epoxidation via Peracids: An Introduction Per-arachidonic acid EpoxidationOChem 206R O OH RR+R R RR+R OHMe OO HHOMO CCLUMO *OOnote labeled oxygen is transferfedO-O bond energy: ~35 kcal/mol Reaction rates are governed by olefin nucleophilicity. The rates of epoxidation of the indicated olefin relative to cyclohexene are provided below: OHOH OAc1.00.60.050.4 The indicated olefin in each of the diolefinic substrates may be oxidized selectively.Me Me Me Me Me Me MeH Me Me The transition state:MeO HOE. J. Corey, JACS 101, 1586 (1979) For a more detailed study see P. Beak, JACS 113, 6281 (1991) View from below olefin2-10 Epoxidation-1 9/16/03 2:58 PMFor theoretical studies of TS see R. D. Bach, JACS 1991, 113, 2338 R. D. Bach, J. Org. Chem 2000, 65, 6715D. A. Evans The General Reaction:R ROlefin Epoxidation with DioxiranesAsymmetric Epoxidation with Chiral KetonesR R O R RChem 206+R RR ROReview: Frohn & Shi, Syn Lett 2000, 1979-2000Me O Me O O O Me O +R R note labeled oxygen is transferfedHOMO CCLUMO *OOchiral catalystO MeOO-O bond energy: ~35 kcal/molR2 H R1 R R O R2 Synthesis of the Dioxirane OxidantK+ O R O O S O O H R R O O SO3O R2 R1 R2oxone, CH3CN-H2O pH 7-8Me Me PhO R O(Oxone)PhPhPhPh>95% ee84% ee92% eeSynthetically Useful Dioxirane SynthesisO Me O F 3C CF3 MeQuestion: First hour Exam 2000 (Database Problem 34)Question 4. (15 points). The useful epoxidation reagent dimethyldioxirane (1) may be prepared from "oxone" (KO3SOOH) and acetone (eq 1). In an extension of this epoxidation concept, Shi has described a family of chiral fructose-derived ketones such as 2 that, in the presence of "oxone", mediate the asymmetric epoxidation of di- and tri-substituted olefins with excellent enantioselectivities (>90% ee) (JACS 1997, 119, 11224). Me O Me R2 R1 R2KO3SOOH CH3CN-H2O pH 10.5 1 equiv 2 oxone, CH3CN-H2O pH 10.5oxoneO MeO Me O CF3co-distill to give ~0.1 M soln of dioxirane in acetone co-distill to give ~0.6 M soln of dioxirane in hexafluoroacetoneoxoneO F 3CMeMe MeO O1 (1) 2Me OCurci, JOC, 1980, 4758 & 1988, 3890; JACS 1991, 7654. Transition State for the Dioxirane Mediated Olefin EpoxidationO R R O O R ROOO R2(2)O Me O MeOR1R2planar>90% eeOspirorotate 90Part A (8 points). Provide a mechanism for the epoxidation of ethylene with dimethyldioxirane (1). Use three-dimensional representations, where relevant, to illustrate the relative stereochemical aspects of the oxygen transfer step. Clearly identify the frontier orbitals involved in the epoxidation. Part B (7 points). Now superimpose chiral ketone 2 on to your mechanism proposed above and rationalize the sense of asymmetric induction of the epoxidation of trisubstituted olefins (eq 2). Use three-dimensional representations, where relevant, to illustrate the absolute stereochemical aspects of the oxygen transfer step.stabilizing Olp * C=C cis olefins react ~10 times faster than trans Houk, JACS, 1997, 12982.2-11 Epoxidation-2 9/16/03 3:01 PMD. A. EvansO C RL RS + RCO3H RCO2HThe Baeyer-Villiger Reaction: Stereoelectronic EffectsO RL C O major RS O Me O minor RS Me O + RCO3H CMe3 Me3C O H O O O- MeCO2HChem 206CMe3 Me O H O R O OLet RL and RS be Sterically large and small substituents.+C RLRThe major product is that wherein oxygen has been inserted into theRLCarbonyl bond. O kR kR / KMe RR O O C R Me + CF3CO3H O C MeHFavoredCMe3 O Me O OH O RMigrating groupmajorCH3CH2 CH3(CH2)2 (CH3)3C72 150 830 >2000 Conformer AHO Me O CMe3kMeC R OMeminorPhCH2RS OH C RL O ODisfavoredMe OMigrating groupThe IntermediateRO OH Me O CMe3OMe3C OThe important stereoelectronic components to this rearrangement:RO1. The RLCOO dihedral angle must be180 due to the HOMO LUMO interaction -RLCOO. 2. The COOC' dihedral angle will be ca. 60 due to the gauche effect (O-lone pairsCO). This gauche geometry is probably reinforced by intramolecular hydrogen bonding as illustrated on the opposite page:2-12- Baeyer Villiger Rxn 9/16/03 5:33 PMConformer BThe destabilizing gauche interactionSteric effects destabilize Conformer B relative to Conformer A; hence, the reaction is thought to proceed via a transition state similar to A. For relevant papers see: Crudden, Angew. Chem. Int. Ed 2000, 39, 2852-2855 (pdf) Kishi, JACS 1998, 120, 9392 (pdf)D. A. EvansThe Baeyer-Villiger Reaction: Stereoelectronic EffectsCMe3 MeChem 206Conformer A in three dimensionsO Me+ RCO3H CMe3Me3C O HOMe O- MeCO2HO O H O R OORFavoredHMigrating group CMe3 O O OH O R O Me O CMe3MeConformer AHDisfavoredMe O Me3C O R OH OMigrating groupO Me1OCMe32 3Conformer BThe destabilizing gauche interaction4Steric effects destabilize Conformer B relative to Conformer A; hence, the reaction is thought to proceed via a transition state similar to A. For relevant papers see: Crudden, Angew. Chem. Int. Ed 2000, 39, 2852-2855 (pdf) Kishi, JACS 1998, 120, 9392 (pdf)2-13- Baeyer Villiger Rxn-2 9/16/03 5:41 PM23 dihedral angle ~ 178 from Chem 3DB. BreitFMO-Theory/HSAB Principle 1Hard and Soft Acids and Bases (HSAB-Principle)Reading Assignment: Fleming, Chapter 3, p33-46 Pearson, JACS 1963, 85, 3533.Chem 206FMO-Theory and Klopman-Salem equation provide an understanding of this empirical principle: Hard Acids have usually a positive charge, small ion radii (high charge density), energy rich (high lying) LUMO. Soft Acids are usually uncharged and large (low charge density), they have an energy poor (low lying ) LUMO (usually with large MO coefficient). Hard Bases usually have a negative charge, small ion radii (high charge density), energy poor (low lying) HOMO. Soft Bases are usually uncharged and large (low charge density), energy rich (high lying) HOMO (usually with large MO coefficient).Hard Acids prefer to interact with hard bases Soft acids prefer to interact with soft bases.Softness: Polarizability; soft nucleophiles have electron clouds, which can be polarized (deformed) easily. Charged species with small ion radii, high charge density.Hardness:Qualitative scaling possible: Molecular Orbital Energies of anidealized Soft SpeciesE E small HOMO/LUMO gapidealized Hard Specieslarge HOMO/LUMO gapFMO-Theory for interaction:Soft-SoftE EHard-HardAcidBase Acid BaseSignificant Energy gain through HOMO/LUMO interactionOnly neglectable energy gain through orbital interaction.2-14-FMO HSAB 1 9/20/00 8:30 AMB. BreitFMO-Theory/HSAB Principle 2Chem 206Klopman-Salem Equation for the interaction of a Nucleophile N (Lewis-Base) and an Electrophile E (Lewis-Acid).E =QNQE RNE Coulomb Term2(cNcE)2 EHOMO(N) - ELUMO(E) Frontier Orbital TermQ: Charge density : Dielectricity constant R: distance (N-E) c: coefficient of MO : Resonance Integral E: Energy of MOSoft-Soft Interactions: Coulomb term small (low charge density). Dominant interaction is the frontier orbital interaction because of a small E(HOMON/LUMOE). formation of covalent bonds Hard-Hard Interactions: Frontier orbital term small because of large E(HOMON/LUMOE). Dominant interaction is described by the Coulomb term (Q is large for hard species), i.e. electrostatic interaction. formation of ionic bonds Hard-Soft Interactions: Neither energy term provides significant energy gain through interaction. Hence, Hard-Soft interactions are unfavorable.2-15-FMO HSAB 2 9/20/00 8:27 AMB. BreitFMO-Theory/HSAB Principle 3HSAB principle - Application to Chemoselectivity Issues(c) SN2 vs E2Chem 206(a) Enolate AlkylationO softCO2Rhard O soft C Csoft MeIC-AlkylationMe OTMSHC(COOR)2 H Br hard OC2H5CO2RS N2hard TMSClE2O-Alkylation(d) Ambident Nucleophiles (b) 1,2- vs. 1,4-addition to ,-unsaturated carbonyl compounds + 0.29O H H- 0.48Osoft Ag S C N hard Nasoft MeIH3 CSC ONS-Alkylation+ 0.01 Charge density+ 0.62 LUMO-coefficientshard RCOXSCNRN-Acylationhard softOhard MeLiOH Me O Ag soft MeI H3C NO2N-Alkylation1,2-Addition soft Me2CuLi Conjugate AdditionMe OO hardN Na soft hard t-BuCl ONOO-Alkylation2-16-FMO HSAB 3 9/20/00 8:27 AMD. A. EvansRules for Ring Closure: IntroductionChem 206http://www.courses.fas.harvard.edu/~chem206/The Primary Literature Baldwin, J. Chem. Soc., Chem. Comm. 1976, 734, 736. Baldwin, J. Chem. Soc., Chem. Comm. 1977 233. Baldwin, J. Org. Chem. 1977, 42, 3846. Baldwin, Tetrahedron 1982, 38, 2939.Chemistry 206 Advanced Organic ChemistryLecture Number 3Stereoelectronic Effects-3 "Rules for Ring Closure: Baldwin's Rules"Kirby, "Stereoelectronic Effects" Chapters 4, 5Useful LIterature Reviews Johnson, C. D. (1993). Stereoelectronic effects in the formation of 5and 6-membered rings: the role of Baldwin's rules. Acc. Chem. Res. 26: 476-82. (Handout) Beak, P. (1992). Determinations of transition-state geometries by the endocyclic restriction test: mechanisms of substitution at nonstereogenic atoms. Acc. Chem. Res. 25: 215. (Handout) Problems of the DayPropose mechanisms for the following reactionsR O R HO H+ R R O O+HO OD. A. Evans3-00-Cover Page 9/19/03 8:36 AMFriday, September 19, 2003MeNH2NH2 OMe MeHN NH OD. A. Evans, J. JohnsonRules for Ring Closure: IntroductionChem 206Ring Closure and Stereoelectronic Connsiderations An Examination of Baldwin's Rules"Baldwin's Rules" provides a qualitative set of generalizations on the probability of a given ring closure. There are circumstances where the "rules" don't apply. They do not apply to non-first-row elements participating in the cyclization event. The longer bond lengths and larger atomic radii of 2nd row elements result in relaxed geometrical constraints. For example, a change in a heteroatom from O to S could result in relaxation of a given geometric constraint. X = O vs S X YendoC. Nucleophilic ring closures sub-classified according to hybridization state of electrophilic component: (tetrahedral = tet; trigonal = trig; digonal = dig) D. Nucleophilic ring closures further subclassified according to size of the fomed ring. For example:5-exo-tetX YXY5-exo-trigX Y5-exo-digXYXYX YXY The "rules" do not apply to electrocyclic processes.NomenclatureClasses of Ring Closing Processes A. Exo-cyclization modes identified by the breaking bond being positioned exocyclic to the forming cycle.exoRequired trajectories (Baldwin):X Y = 180 X YX XXYY Y B. Endo-cyclization modes identified by the breaking bond being positioned endocyclic to the forming cycle.endo = 109 XYWill come back to this case laterX Y X = first-row element N, OX * 120 *XYXY YBaldwin, J. Chem. Soc., Chem. Commun., 1976, 734.3-01-Baldwin Rules-1 9/18/03 3:38 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP3 Carbon & Related SystemsFRST-PLATTNER RULEChem 206Tetrahedral CarbonAll exo cyclization modes are allowed: (n-exo-tet, n = 3)exoIn this simple model, the transition-state leading to 1 involves the diaxial orientation of nucleophile and leaving group. This orientation affords the best overlap of the anti-bonding CY orbital and the nonbonding electron pairs on the nucleophile O. In the formation of the diastereomeric epoxide 2, the proper alignment of orbitals may only be achieved by cyclization through the less-favored boat conformer. Accordingly, while both cyclizations are "allowed", there are large rate differences the the rates of ring closure. While the FRST-PLATTNER RULE deals wilth the microscopic reverse, in the opening of epoxides by nucleophiles, the stereoelectronic arguments are the same.XC YXC YThere are stereoelectronic issues to consider for n-exo-tet cyclizations Formation of 3-Membered Rings (3-exo-tet)H H HYX C H2CYXCXCH2 C H2C H2H+ YStereoelectronic Effects in Epoxide Ring CleavageNu Me3C NuO H O Me3C H HO HO NuH Me Me3C H H Nu Me H HConformational Effects in Epoxide Ring Formation/cleavageThose stereoelectronic effects that operate in ring cleavage also influence ring formation. Consider a rigid cyclohexene oxide system:Y H H O H Y H O H O HY fasterH H H Me3C OY 1OslowerH ONu-Nu HOO H Hchairboat2"The diaxial nucleophilic ring cleavage of epoxides" For more information on epoxide cleavage see Handout 03A.3-02-Baldwin Rules-2 9/18/03 3:39 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP3 Carbon & Related SystemsCase 2: King, J.C.S. Chem. Comm., 1979, 1140.O S O Me NMe2 O OChem 206Tetrahedral CarbonEndo cyclization modes that are disallowed (n-endo-tet, n = 39)8-endo-tet disfavored Rxn exclusively intermolecular 8-endo-tet disfavored Rxn exclusively intermolecular 9-endo-tet borderline 84% intermolecular, 16% intramolecularO OS_NMe3+endoXYCY C(SP3)XSO2OMe NMe2SO3 NMe3+The stereoelectronic requirement for a 180 XCY bond angle is only met when the endo cyclization ring size reaches 9 or 10 members. Case 1: Eschenmoser, Helvetica Chim. Acta 1970, 53, 2059.O S O O CX3 O S O O S O OCX3 O S O SO2OMe NMe2SO3 NMe3+NaH 6-endo-tet disfavored Rxn exclusively intermolecular (lecture 2)ConclusionsAllowed endo cyclization modes will require transition state ring sizes of at least nine members.Intramolecular epoxidation has also been evaluatedCY3CY3O Cl OOH n-Beak, JACS 1991, 113, 6281. 8-endo-tet disfavoredCl CO2HCyclization exclusively intermolecular. However the exocyclic analog is exclusively intramolecularO S O O CX2I O S O O S O OnONaH 6-exo-tet favored Rxn exclusively intramolecularO SCX2 On = 1: rxn exclusively intermolecular n = 9: rxn is intramolecularBeak states that the conclusions made with carbon substitution also hold for oxygen atom transfer.Beak, P. (1992). Determinations of transition-state geometries by the endocyclic restriction test: mechanisms of substitution at nonstereogenic atoms. Acc. Chem. Res. 25: 215.CY3CY33-03-Baldwin Rules-3 9/18/03 4:07 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP2 Carbon & Related SystemsChem 206Trigonal CarbonEndo cyclization modes that are disallowed (3 to 5-endo-trig)n-endo-trig X Y X C Y MeO2C NH2 CO2MeCX = first-row element The 5-endo-trig cyclization is a watershed case distance from reacting centers: 2.77 Case 1: Baldwin, J. Chem. Soc., Chem. Commun., 1976, 734.CO2Me OH baseXCO2Me OhoweverCO2Me SH5-endo-trig Disfavored baseIt is possible that a "nonvertical" trajectory is operational like that suspected in C=O additionCO2MeSSecond row atom relaxes the cyclization geometrical requirement Case 2: Baldwin, J. Chem. Soc., Chem. Commun., 1976, 736.MeO2C NH2 CO2MeXMeO2C HNCO2Me5-endo-trig 0%MeO2C HN O5-exo-trig 100%3-04-Baldwin Rules-4 9/18/03 4:07 PMD. A. Evans, J. JohnsonCase 2: continued...MeO2C NH2 CO2MeXRules for Ring Closure: SP2 Carbon & Related SystemsChem 206Apparent exceptions to disallowed 5-endo-trig cyclization processMeO2C HN CO2MeO5-endo-trig 0%MeO2C HN ONCH3CO2H+ NOHNOFiler, J. Am. Chem. Soc. 1979, 44, 285.5-exo-trig 100%R HC N1CO2Me KO Bu R2 CO2MetCO2Me R1 HN R CO2Me2CO2Me R 3:11Control experiment: Intermolecular reaction favors conjugate addtion.Me CO2Me PhCH2NH2 H N Me CO2Me 100% Me H N O 0% PhHNCO2Me R2PhR1 = aryl, R2 = aryl, alkylGrigg, J. Chem. Soc., Chem. Commun. 1980, 648.Case 3:O Ph OMe NH2NH2 65 oC Ph HN NH ODoes the illustrated ketalization process necessarily violate "the rules"?R O R(CH2OH)2 H+R RO OO Ph MeI OK1) EtO2CCl, pyridine 2) NH2NH2Ph NHOR O R(CH2OH)2R HOR OOHH+ H2OR R OOH( )2H 2N 200 oC O Ph OMe NH2NH2 65 oC HN X+( )25-endo-trig5-endo-trigH+R Odisfavored ?R R O OR Ph CO2Me NH2 Ph O HN NH HO+OH 5-exo-tet( )2favored ?5-exo-trigJohnson, C. D. (1993). Stereoelectronic effects in the formation of 5- and 6-membered rings: the role of Baldwin's rules. Acc. Chem. Res. 26: 476-82.3-05-Baldwin Rules-5 9/18/03 4:08 PMD. A. Evans, J. JohnsonMore ExceptionsRules for Ring Closure: SP2 Carbon & Related SystemsBuChem 206BuZard, Org. Lett. 2002, 4, 1135MeO O MeO N S S OEt N OX YNaHHODMF, 60 C Y F H Cl Br CondY OROOR heat 80% OOX F FYield 80 17 15DMF, 60 C, 2 h DMF, 80 C, 43 h DMF, 60 C, 8 h DMF, 60 C, 5 hO MeOOO MeO NMeO O MeO NCl Br5-endo-trigIchikawa, et al Synthesis 2002, 1917-1936, PDF on Course Website Numerous other cases are provided in this review.OOOOBrH ORevisiting Case 2 with FluorinesO O MeO2CBu3SnH AIBN 82%O HOMeO2C N OOMeMeO2C N TsCO2MeTs5-exo-trigTsHN5-endo-trigFavoredH ONot Observed5-endo-trigO OO MeO2C N Ts O CF2 MeO2C TsHNO OMe CF2 MeO2C N Ts F CO2Me5-exo-trig5-endo-trigChem. Comm 2088, 28 Review: "5-Endo-Trig Radical Cyclizatons" Ishibashi, et al Synthesis 2002, 695-713, PDF on Course WebsiteNot ObservedFavored3-06-Baldwin Rules-6 9/18/03 5:10 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP2 Carbon & Related SystemsChem 206Trigonal Carbon: Exocyclic Enolate Alkylationexodistance between reacting centers: 3.37C O C C Y-OC CC YBrX By definition, an exo-tet cyclization, but stereoelectronically behaves as an endo trig.Me Me O Me Me MO BrXMOOMe Me O(1)The overlap for C-alkylation is poor due to geometrical constraints of 5-membered ringonly observed productHowever:Me Me O Me BrKOt-Bu or LDAdistance between reacting centers: 3.04Me Me O > 95% by NMRThe relaxed geometrical constraint provided by the added CH2 group now renders the 6-membered cyclization possibleBaldwin, J. Chem. Soc., Chem. Commun. 1977, 233. Given the failure of the enolate alkylation shown above (eq 1), explain why these two cyclizations are successful.Br NHAr O R O Ar NH R OMs base O N Ar base N O R Ar RMOBrOFavorskii Rearrangement (Carey, Pt B, pp 609-610) Your thoughts on the mechanismO Cl MeO HCl O MeOCO2Me3-07-Baldwin Rules-7 9/18/03 4:09 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP2 & SP Carbon & Related SystemsChem 206Trigonal Carbon: Intramolecular Aldol CondensationsBaldwin, Tetrahedron 1982, 38, 2939Digonal Carbon: Cyclizations on to AcetylenesDIGONAL: Angle of approach for attack on triple bondsMO RX Y(Enolendo)-Exo-trigO RX YMNu120 120 E+Baldwin: - 3 and 4-Exo-dig are disfavored - 5 to 7-Exo-dig are favored - 3 to 7-Endo-dig are favoredFavored: 6-7-(enolendo)-exo-trig Disfavored: 3-5-(enolendo)-exo-trigX MO R Y(Enolexo)-exo-trigO RX YMH _Ab initio SCF 4-31G calculations for the interaction of hydride with acetylene:H2.13 127 o 148o HFavored: 3-7-(enolexo)-exo-trigH O Me OI4-31G basis set Houk, J.ACS.1979, 101, 1340.H5-(Enolendo)-Exo-trigMe Me O6-(Enolendo)-Exo-trigMe O HH 156o 1.22CC1.5-2.0110o -120oSTO-3G minimal basis set H Dunitz, Helv Chim. Acta 1978, 61, 2538.favoredMe O Me O MeIIIH OCCMe Me OIICrystal Structures do not support BaldwinOStatistical Distribution, (I + II)/III = 2:1 Experimental Distribution, = 0:100NN2.92 104o(KOH, MeOH, r.t., 5 min, 77% y.)ON+N+NO932.44oNoCaution: Baldwin's conclusions assume that the RDS is ring closure; however, it is well known (by some!) that the rate determining step is dehydration in a base-catalyzed aldol condensation.86J. Dunitz and J. Wallis J. C. S. Chem. Comm. 1984, 671.3-08-Baldwin Rules-8 9/18/03 5:49 PMD. A. Evans, J. JohnsonRules for Ring Closure: SP Carbon & Related Systems Indole synthesis:CH32 equiv. LDA 2 equiv. RX -78 oC R = Me, Bu, CO2Me LiTMP+Chem 206Endo Digonal versus Endo Trigonal Cyclizations5-endo-trigY X:CH2R N+NC-C-In-plane approach; nucleophile lone pair is orthogonal to *Out-of-plane approach; nucleophile lone pair can't achieve Brgi-Dunitz angleRSaegusa, J. Am. Chem. Soc. 1977, 99, 3532. _N5-endo-dig:X Spiro dihydrofuranones:O Li HO MeOn n = 1,2Li+YAllowed due to in-plane pi orbitalsOMeKOtBuOOMeXFor an opposing viewpoint to Baldwin's view of nucleophile trajectories, see Menger's article on directionality in solution organic chemistry: Tetrahedron 1983, 39, 1013.O Me Me HO Ph NaOMe MeOH Me Me O Ph OnnDeveloping negative charge on the central allenic carbon is in the same plane as the OMe groupMagnus, J. Am. Chem. Soc. 1978, 100, 7746.5-endo-digO5-exo-digO R OHR = H, OMeNaOMeX5-endo-trigR O PhLiPhLiPhhowever, the acid catalyzed version does cyclizeBaldwin, J. Chem. Soc., Chem. Commun., 1976, 736. Johnson, Can. J. Chem. 1990, 68, 1780 J. Am. Chem. Soc. 1983, 105, 5090 J. Chem. Soc., Chem. Commun. 1982, 36.4-endo-digXLiLiPh3-09-Baldwin Rules-9 9/19/03 8:38 AMD. A. Evans, J. JohnsonRules for Ring Closure: SP Carbon & Related SystemsChem 206Digonal Cyclizations: Interesting ExamplesMeO2C CN Et3N, Toluene, reflux 12 h, 65-70% y. CN Trost, J. Am. Chem. Soc., 1979, 101, 1284. Proposes E-olefin geometry, E/Z > 95:55-exo-digO O CO2Me:R R'30-40 kcal/mol ?R R':HOHHO2C HHirsutic Acid COConclusions and CaveatsOTBSO MeLiCH2NC; TBS-Cl 71%+ Baldwin's Rules are an effective first line of analysis in evaluating the stereoelectronics of a given ring closure Baldwin's Rules have provided an important foundation for the study of reaction mechanism Competition studies between different modes of cyclization only give information about relative rates, and are not an absolute indicator of whether a process is "favored" or "disfavored" Structural modifications can dramatically affect the cyclization mode; beware of imines and epoxidesEXO Tet Trig Dig X X Tet ENDO Trig X X X X X X DigOTBS 1) RCOCl2) AgBF4 86%Me C-N+ C OMe RNWorks for varying ring sizes and R groups; acylnitrilium ion can also work as an electophile in a Friedel-Crafts type of reaction 5-endo-dig Livinghouse, Tetrahedron 1992, 48, 2209.O H N O Me R3 4 5 6 7 3-10-Baldwin Rules-10 9/18/03 5:21 PMD. A. EvansAcyclic Conformational Analysis-1Professor Frank Weinhold Univ. of Wisconsin, Dept of Chemistry B.A. 1962, University of Colorado, BoulderChem 206http://www.courses.fas.harvard.edu/~chem206/Chemistry 206 Advanced Organic ChemistryLecture Number 4A.M. 1964, Harvard University Ph.D. 1967, Harvard University Physical and Theoretical Chemistry.Useful LIterature ReviewsEliel, E. L., S. H. Wilen, et al. (1994). Stereochemistry of Organic Compounds. New York, Wiley. Juaristi, E. (1991). Introduction to Stereochemistry and Conformational Analysis. New York, Wiley. Juaristi, E., Ed. (1995). Conformational Behavior of Six-Membered Rings: Analysis, Dynamics and Stereochemical Effects. (Series: Methods in Stereochemical Analysis). Weinheim, Germany, VCH. Schweizer, W. B. (1994). Conformational Analysis. Structure Correlation, Vol 1 and 2. H. B. Burgi and J. D. Dunitz. Weinheim, Germany, V C H Verlagsgesellschaft: 369-404. Kleinpeter, E. (1997). Conformational Analysis of Saturated Six-Membered Oxygen-Containing Heterocyclic Rings. Adv. Heterocycl. Chem. 69: 217-69. Glass, R. R., Ed. (1988). Conformational Analysis of Medium-Sized Ring Heterocycles. Weinheim, VCH. Bucourt, R. (1973). The Torsion Angle Concept in Conformational Analysis. Top. Stereochem. 8: 159.Acyclic Conformational Analysis-1 Ethane, Propane, Butane & Pentane Conformations Simple Alkene Conformations Reading Assignment for week A. Carey & Sundberg: Part A; Chapters 2 & 3R. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout)The Ethane Barrtier ProblemF. Weinhold, Nature 2001, 411, 539-541 "A New Twist on Molecular Shape" (handout) F. M. Bickemhaupt & E. J. Baerends, Angew. Chem. Int. Ed. 2003, 42, 4183-4188,"The Case for Steric Repulsion Causing the Staggered Conformation in Ethane" (handout) F. Weinhold,, Angew. Chem. Int. Ed. 2003, 42, 4188-4194,"Rebuttal of the BikelhauptBaerends Case for Steric Repulsion Causing the staggered Connformation of Ethane" (handout) Problems of the DayO Predict the most stable conformation of the indicated dioxospiran? OD. A. Evans4-00-Cover Page 9/22/03 9:01 AMMonday, September 22, 2003D. A. EvansAcyclic Conformational Analysis-1Chem 206The following discussion is intended to provide a general overview of acyclic conformational analysisEthane Rotational Barrier: The FMO ViewF. Weinhold, Angew. nature 2001, 411, 539-541"A New Twist on Molecular Shape"Ethane & PropaneThe conformational isomerism in these 2 structures reveals a gratifying level of internal consistency.H H RH C H HOne can see from the space-filling models that the Van der Waals radii of the hydrogens do not overlap in the eclipsed ethane conformation. This makes the steric argument for the barrier untenable. One explanation for the rotational barrier in ethane is that better overlap is possible in the staggered conformation than in the eclipsed conformation as shown below. In the staggered conformation there are 3 anti-periplanar CH Bondseclipsed conformationH E = +3.0 kcal mol-1 (R = H) Van derWaals radii of vicinal hydrogens do not overlap in ethane E = +3.4 kcal mol-1 (R = Me) C H C CH HOMOHC C* CH LUMO CHp. p.H CHR H C H HIn propane there is a discernable interactionH HIn the eclipsed conformation there are 3 syn-periplanar CH Bonds staggered conformationH C H C CH HOMO CHHCHC* CH LUMO CHFor purposes of analysis, each eclipsed conformer may be broken up into its component destabilizing interactions. Incremental Contributions to the Barrier. Structure ethane propane Eclipsed atoms E (kcal mol -1) 3 (HH) 2 (HH) 1 (HMe) +1.0 kcal mol -1 +1.0 kcal mol -1 +1.4 kcal mol -1Following this argument one might conclude that: The staggered conformer has a better orbital match between bonding and antibonding states. The staggered conformer can form more delocalized molecular orbitals. J. P. Lowe was the first to propose this explanation"A Simple Molecuar Orbital Explanation for the Barrier to Internal Rotation in Ethane and Other Molecules" J. P. Lowe, JACS 1970, 92, 3799 Me Me MeCalculate the the rotational barrier about the C1-C2 bond in isobutane4-01-introduction 9/22/03 8:28 AMD. A. EvansAcyclic Conformational Analysis: Butane The 1,2-Dihaloethanes ButaneChem 206XH C H H H H HH H CX XXX = Cl; H = + 0.91.3 kcal/mol X = Br; H = + 1.41.8 kcal/mol X = F; H = 0.6-0.9 kcal/molUsing the eclipsing interactions extracted from propane & ethane we should be able to estimate all but one of the eclipsed butane conformationsMe C H Me HObservation: While the anti conformers are favored for X = Cl, Br, the gauche conformation is prefered for 1,2-difluroethane. Explain. Discuss with class the origin of the gauche stabiliation of the difluoro anaolg. Recent Article: Chem. Commun 2002, 1226-1227 (handout)staggered conformationHHH HMe H C MeHeclipsed conformationE=?Eclipsed atoms 1 (HH) 2 (HMe) E (kcal mol -1) +1.0 kcal mol -1 +2.8 kcal mol -1Relationship between G and Keq and pKap. p. Recall that: or G = RT Ln K G = 2.3RT Log10K2.3RT = 1.4 (G in kcal Mol1 ) E est = 3.8 kcal mol -1 The estimated value of +3.8 agrees quite well with the value of +3.6 reported by Allinger (J. Comp. Chem. 1980, 1, 181-184)At 298 K: G298 = 1.4 Log10KeqSincen-Butane Torsional Energy ProfileH H HH C Me Me H C HpKeq = Log10KeqE1E2 G298 = 1.4 pKeqenergyH HMe H C MeH H HHence, pK is proportional to the free energy changeH HMe CHKeq 1.0 10 100pKeq 0 1 2G 0 1.4 2.8 kcal /molH Me ARef = 0+3.6Me Me G+5.1+0.884-02-introduction-2 9/22/03 8:33 AMD. A. Evans Butane continuedAcyclic Conformational Analysis: ButaneNomenclature for staggered conformers:H H Me Me C H H H H H Me C H H Me MeChem 206Me C H H HFrom the torsional energy profile established by Allinger, we should be able to extract the contribution of the MeMe eclipsing interaction to the barrier:H H C H H Me H H Me Me C HHstaggered conformation Meeclipsed conformationtrans or t or (anti) Conformer population at 298 K: 70%gauche(+) or g+ 15%gauche(-) or g15% E = +5.1 kcal mol-1Let's extract out the magnitide of the MeMe interaction 2 (HH) + 1 (MeMe) = +5.1 1 (MeMe) = +5.1 2 (HH) p. p. 1 (MeMe) = +3.1General nomenclature for diastereomers resulting from rotation about a single bond (Klyne, Prelog, Experientia 1960, 16, 521.)RR C 0 R R C -60R CRsp sc sc+60Incremental Contributions to the Barrier. Eclipsed atoms 2 (HH) 1 (MeMe) E (kcal mol -1) +2.0 +3.1R Eclipsed Butane conformationacR C -120ac+120 R C R 180 R C RapFrom the energy profiles of ethane, propane, and n-butane, one may extract the useful eclipsing interactions summarized below: Hierarchy of Eclipsing InteractionsXX YY H E kcal mol -1 +1.0 +1.4 +3.1Energy Maxima Energy Minima Torsion angle 0 30 +60 30 +120 30 180 30 -120 30 -60 30HHDesignation syn periplanar + syn-clinal + anti-clinal antiperiplanar - anti-clinal - syn-clinalSymbol sp + sc (g+) + acH HCC HH Me Me Men-Butane Conformer E2 G E1 A E1 Gap (anti or t) - ac - sc (g-)4-03-butane 9/26/03 1:51 PMD. A. Evans n-PentaneAcyclic Conformational Analysis: Pentane The double-gauche pentane conformationThe new high-energy conformation: (g+g)Me Hg-g-Chem 206Rotation about both the C2-C3 and C3-C4 bonds in either direction (+ or -):Me Me H Me Hg+g+ g+tMe H Me Metg-H Meg+g-Me HH MeMe Me Ht,tMe Me Me H HH MeEstimate of 1,3-Dimethyl Eclipsing Interaction YMetg+Me HXg-t 1 g-g+p. p.135351 (t,t) Anti(2,3)-Anti(3,4)2 (g+t) Gauche(2,3)-Anti(3,4) G = +5.5 kcal mol -1 G = X + 2Y where: X = 1,3(MeMe) & Y = 1,3(MeH) 1,3(MeH) = Skew-butane = 0.88 kcal mol -1 1,3(MeMe) = G 2Y = 5.5 1.76 = + 3.7 kcal mol -11 1 5 31,3(MeMe) = + 3.7 kcal mol -1Estimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing Interactions34 (g+g) Gauche(2,3)-Gauche'(3,4) double gauche pentane3 (g+g+)5MeMeMeMeMeMeMeMeGauche(2,3)-Gauche(3,4) 3.1 ~ 3.7 ~3.9 ~ 7.6From prior discussion, you should be able to estimate energies of 2 & 3 (relative to 1). On the other hand, the least stable conformer 4 requires additional data before is relative energy can be evaluated.It may be concluded that in-plane 1,3(MeMe) interactions are Ca +4 kcal/mol while 1,2(MeMe) interactions are destabliizing by Ca 2.2 kcal/mol.4-04-pentane 9/26/03 11:23 AMD. A. EvansAcyclic Conformational Analysis: Natural Products Lactol & Ketol Polyether AntibioiticsChem 206The syn-Pentane Interaction - ConsequencesR. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout)The conformation of these structures are strongly influenced by the acyclic stereocentersMe O HO R O OH H Me Me OH O H O O Me OH Et Et OH MeR Me R Me Me MeR'RR'orMe H RR'g -g -Me H R'Me H H MettMeMeEtR'R MeorMeMe H H R'tgH H R HgtFerensimycin B, R = Me Lysocellin, R = HConsequences for the preferred conformation of polyketide natural products p. p.Analyze the conformation found in the crystal state of a bourgeanic acid derivative!The conformation of these structures are strongly influenced by the acyclic stereocenters and internal H-bondingAlborixin R = Me; X-206 R = HInternal H-BondingOH Me Me Me Me O O OR Bourgeanic acid Me Me C O OH R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OHMetal ion ligation sites (M = Ag, K)Me Me O C O O R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OHM4-05-Natural Products 9/22/03 8:42 AMD. A. EvansConformational Analysis: Ionophore X-206/X-rays X-ray of Ionophore X-206 H2OChem 206X-ray of Ionophore X-206 - Ag+ - ComplexMetal ion ligation sites (M = Ag, K)H Me Me OH Me O C O O R Me OH H Me O OH O OH OH O Me Me Me H O Me Me OH O Et Me OHInternal H-BondingMe Me O C O OH Me OH H Me O OH O OH OH O Me Me Me O Me Me OH O EtMp. p."The Total Synthesis of the Polyether Antibiotic X-206". Evans, D. A.; Bender, S. L.; Morris, J. J. Am. Chem. Soc. 1988, 110, 2506-2526.4-06-X-206 conformation 9/22/03 8:42 AMD. A. EvansConformational Analysis: Ionophore X-206/X-ray overlayChem 206p. p.4-07-X-206 overlay 9/19/01 11:57 AMD. A. EvansStabilized Eclipsed Conformations in Simple Olefins Butane versus 1-ButeneMe H C H H H Me H H Me MeChem 206Simple olefins exhibit unusal conformational properties relative to their saturated counterpartsPropane versus Propene109 H H H H Me H H H H H 120 CH2staggered conformationC H Heclipsed conformation G = +4 kcal mol-1MeHybridilzation change opens up the CCC bond angle The Propylene Barrierp. p.H H C H CH2 H H CH2 C Hstaggered conformationCH2 C H H HH HMe CH2 C Heclipsed conformation = 50 staggered conformation G = 0.83 kcal mol-1=0The Torsional Energy Profile = 50Me H C H H C H H+2.0 kcal/mol = 180H H C H H C Me Heclipsed conformationH=0 New destabilizing effectH C H Me CH H H +1.32 kcal H C H H C H Me H +1.33 kcalH X H C H Hrepulsive interaction between CX & CH+0.49 kcal = 120 = 180H X C H H=0Conforms to ab initio (3-21G) values: Wiberg, K. B.; Martin, E. J. Am. Chem. Soc. 1985, 107, 5035. Acetaldehyde exhibits a similar conformational biasO H H H H Me H H O H H H H O Me Me H H O MeK. Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109, 6591-6600The low-energy conformation in each of above cases is eclisped4-08-simple alkenes 9/22/03 8:54 AMUseful Destabilizing Interactions to RememberHierarchy of Vicinal Eclipsing InteractionsXX YY H Me Me E kcal mol -1 +1.0 +1.4 +3.1HHH HCC HH MeEstimates of In-Plane 1,2 &1,3-Dimethyl Eclipsing InteractionsMe Me Me Me Me Me Me Me~ 3.1~ 3.7~3.9~ 7.6It may be concluded that in-plane 1,3(MeMe) interactions are Ca +4 kcal/mol while 1,2(MeMe) interactions are destabliizing by Ca +3 kcal/mol.0000 0000 000 0000 0000 000 0000 0000 0000 0000 000 0000 00 0 00 0 0 00 0 0 00 0 0000 0000 0000 0000 0000 0000 0000 0000 00000000 0000 0000minimized structure4-09-Destabilizing Effects 9/20/01 5:33 PMD. A. EvansAcyclic Conformational Analysis-2 Problems of the Day:Chem 206http://www.courses.fas.harvard.edu/~chem206/Can you predict the stereochemical outcome of this reaction? O OTs MeLiNR2OLi EtO n-C4H9 H MeOTs 1Chemistry 206 Advanced Organic ChemistryLecture Number 5EtO n-C4H9 H+ 98:22D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943.Acyclic Conformational Analysis-2 Conformations of Simple Olefinic Substrates Introduction to Allylic Strain Introduction to Allylic Strain-2: Amides and EnolatesO Me CH2OBn MeB2H6 H2O2OH O Me CH2OBnMediastereoselection 8:1 Reading Assignment for week A. Carey & Sundberg: Part A; Chapters 2 & 3R. W. Hoffmann, Angew. Chem. Int. Ed. Engl. 2000, 39, 2054-2070 Conformation Design of Open-Chain Compounds (handout) R. W. Hoffmann, Chem. Rev. 1989, 89, 1841-1860 Allylic 1-3-Strain as a Controlling Element in Stereoselective Transformations (handout) F. Weinhold, Nature 2001, 411, 539-541 "A New Twist on Molecular Shape" (handout)Y. Kishi & Co-workers, J. Am. Chem. Soc. 1979, 101, 259.MeMePhNCO Et3NMeMeO N only one isomerNO2A. Kozikowski & Co-workers, Tetrahedron Lett. 1982, 23, 2081.D. A. EvansWednesday, September 24, 20015-00-Cover Page 9/24/03 8:46 AMD. A. EvansStabilized Eclipsed Conformations in Simple Olefins Butane versus 1-ButeneMe H C H H H Me H H Me MeChem 206Simple olefins exhibit unusal conformational properties relative to their saturated counterpartsPropane versus Propene109 H H H H Me H H H H H 120 CH2staggered conformationC H Heclipsed conformation G = +4 kcal mol-1MeHybridilzation change opens up the CCC bond angle The Propylene Barrierp. p.H H C H CH2 H H CH2 C Hstaggered conformationCH2 C H H HH HMe CH2 C Heclipsed conformation = 50 staggered conformation G = 0.83 kcal mol-1=0The Torsional Energy Profile = 50Me H C H H C H H+2.0 kcal/mol = 180H H C H H C Me Heclipsed conformationH=0 New destabilizing effectH C H Me CH H H +1.32 kcal H C H H C H Me H +1.33 kcalH X H C H Hrepulsive interaction between CX & CH+0.49 kcal = 120 = 180H X C H H=0Conforms to ab initio (3-21G) values: Wiberg, K. B.; Martin, E. J. Am. Chem. Soc. 1985, 107, 5035. Acetaldehyde exhibits a similar conformational biasO H H H H Me H H O H H H H O Me Me H H O MeK. Wiberg, JACS 1985, 107, 5035-5041 K. Houk, JACS 1987, 109, 6591-6600The low-energy conformation in each of above cases is eclisped5-01-simple alkenes 9/22/03 8:54 AMEvans, Duffy, & Ripin5Conformational Barriers to Rotation: Olefin A-1,2 Interactions5Chem 2061-butene42-propen-1-olH H H C C OH H HH H H C C MeH H4E (kcal/mol)3E (kcal/mol)322110 -180-900901800 -180-90090180 (Deg)The Torsional Energy Profile = 50Me H H H C H C H H H C H (Deg)The Torsional Energy ProfileH H C H C OH H = 180H C Me H H HH = 60HO C H C H H = 180=0C H Me H CH H H = 120H C H COH H H+1.32 kcalH H C H CMe H H+1.33 kcalH+2.00 kcal=0C H HO CH H HH+1.18 kcal+0.49 kcal=0 = 120 = 180+0.37 kcal = 180=0Conforms to ab initio (3-21G) values: Wiberg, K. B.; Martin, E. J. Am. Chem. Soc. 1985, 107, 5035.5-02-1-butene & 2-propen-1-ol 9/23/03 2:59 PMEvans, Duffy, & Ripin5Conformational Barriers to Rotation: Olefin A-1,2 Interactions-25Chem 2062-methyl-1-butene42-methyl-2-propen-1-ol4H H H C C MeH MeH H H C C OHH MeE (kcal/mol)E (kcal/mol)3322110 -180-900901800 -180-90090180 (Deg)The Torsional Energy Profile = 180 = 50Me H H H C H C Me H H C H H H C H C H Me Me (Deg)The Torsional Energy Profile = 180 = 60HO C Me H H H H C H C OH Me = 120OH=0C H Me =0 H C = 110H Me H+1.39 kcalH H C H CMe Me H+2.68 kcal=0C H HO = 180 =0 H CH Me H+1.16 kcalH H+2.01 kcalMeC HC H+0.06 kcal+0.21 kcal = 1805-03-methylbutene etal 9/23/03 3:00 PMEvans, Duffy, & Ripin5Conformational Barriers to Rotation: Olefin A-1,3 Interactions5Chem 206(Z)-2-pentene4 4(Z)-2-buten-1-olH Me H C C OH H HH MeHE (kcal/mol)3E (kcal/mol)CC MeHH322110 -180-900901800 -180-90090180 (Deg)H Me C H Me C H H (Deg)The Torsional Energy Profile =0Me C H HO C H H H MeThe Torsional Energy Profile =0 = 180H C H C H OH H+3.88 kcal = 180 = 90Me Me H C H C H H Me H H C H C Me H+1.44 kcalMe = 120OH C H C H H+0.86 kcal+0.52 kcal = 180 =0H=0 = 180Values calculated using MM2 (molecular mechanics) force fields via the Macromodel multiconformation search. 5-04-2-pentene/ z-2-buten-1-ol 9/23/03 3:00 PMReview: Hoffman, R. W. Chem. Rev. 1989, 89, 1841.Evans, Duffy, & Ripin5Conformational Barriers to Rotation: Olefin A-1,3 Interactions-2OH Me Rotate clockwise C Me C H HChem 206(Z)-2-hydroxy-3-pentene4Me OH MeH4.6 kcal/molHOH C Me HE (kcal/mol)Me3CHMe100 H OH2Me HC HC0.3-0.4 kcal/mol1H Me H C C OH-90 0 90 180Me H30 0 -180Lowest energy conformerMe Me Me OH H H C H (Deg)The Torsional Energy ProfileOH60 Me=0H Me C H HO C H MeMe C H Me2.7 kcal/molC OHCH H = -140Needs to be redone = 110Me Me C H H H OH C H OH+2.72????? = -80HO Me C H C Me H H+4.68A(1,3) interaction 4.0 kcal/molR23HO H H C C MeH Me30 = 80 MeMe C C H HY X2R1 = 150H C C Me HR3*1R large+0.66HMeR smallLowest energy conformer+0.34=0+0.40 kcalHO = 180A(1,2) interaction 2.7 kcal/mol (MM2)5-05-z-2-hydroxy-3-pentene 9/23/03 6:22 PMD. A. EvansAcyclic Conformational Analysis: Allylic Strain The Definition of Allylic StrainChem 206Can you predict the stereochemical outcome of this reaction?D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943. O EtO R large 1 R small n-C4H9 H Me OTsLiNR2F. Johnson, Chem. Rev. 1968, 68, 375; Allylic Strain in Six-Membered Rings R. W. Hoffmann, Chem. Rev. 1989, 89, 1841-1860 (handout) Allylic 1-3-Strain as a Controlling Element in Stereoselective Transformations Houk, Hoffmann JACS 1991, 113, 5006Consider the illustrated general structure where X & Y are permutations of C, N, and O:R23 Y X 2R1OLi EtO n-C4H9 O H Me MeOTs 1+ 98:22R3Typical examples:R2 R3 R1R2 R large NR1R2 R large R + NR1R2 R largeR1 N Relevant enolateconformationsR large 1EtO n-C4H9 HmajorO +p. p.R small OlefinR small ImineR small Imonium ionR small Nitrone TsO(H2C)4 Me C Bu H C OR OLi Me Bu C C H (CH2)4OTs OR OLi Bu Me (CH2)4OTs OR C C OLiIn the above examples, the resident allylic stereocenter () and its associated substituents frequently impart a pronounced bias towards reactions occuring at the pi-bond. A(1,3) interaction R2 3 R1 Nonbonding interactions between the allylic Y substituents (Rlarge, Rsmall) & substituents at X R large the 2- & 3-positions play a critical role in R3 2 1 defining the stereochemical course of such R small reactions A(1,2) interaction Representative Reactions controlled by Allylic Strain InteractionsHO H Me HO O R OBnHg(OAc)2 NaBH4A1B1critical conformationsC1 HH OR Me TsO(H2C)4 C C Bu OLi MeH CBu CBu OR OLi Me H C C OR OLi (CH2)4OTs(CH2)4OTsA2B2C2HOR EtO2C OBn 2 n-C4H9MeHO MeminorHdiastereoselection 10:1 M. Isobe & Co-workers, Tetrahedron Lett. 1985, 26, 5199.5-06-Allylic Strain-1 9/23/03 6:24 PMD. A. EvansO EtO n-C4H9 H O EtO n-C4H9 HLiNR2 MeIAllylic Strain & Enolate Diastereoface SelectionOTs MeLiNR2Chem 206PhNH4ClO EtO n-C4H9 O EtO n-C4H9Me diastereoselection 98:2 H Me3SiPhOM OMe RR-substituent R = Me diastereoselection 87:13 80:20 40:60O OMe RMe3SiMe diastereoselection 89:11 H Ph Me3Si OR = Et R = CHMe2major diastereomer opposite to that shownI. Fleming & Co-workers, Chem. Commun. 1985, 318. Y. Yamamoto & Co-workers, Chem. Commun. 1984, 904.D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943.PhLiNR2HO OBn diastereoselection 90:10 at C3one isomer at C2OBnMeCHO Me3Si 71% yieldMeOp. p.RO2C H CO2MeOMeOLiNR2RO2C H H CO2MeO "one isomer" Me Bn N Boc O N S SSn(OTf)2 RCHO 91-95%MeOHI. Fleming & Co-workers, Chem. Commun. 1986, 1198.Br95% yieldMe Bn N Boc RHO N OHS S diastereoselection >95%G. Stork & Co-workers, Tetrahedron Lett. 1987, 28, 2088.TBSOCH2Me CH2LiNR2 MeIMe CH2 TBSOCH2 Me H H CO2Me "one isomer" Me OPh(MeS)2CLiT. Mukaiyama & Co-workers, Chem. Letters 1986, 637H CO2MeOMeMeI 86%MeS MeS MeSMeO OMe diastereoselection 99:1T. Money & Co-workers, Chem. Commun. 1986, 288.MeK. Koga & Co-workers, Tetrahedron Letters 1985, 26, 3031.R PhMe2SiOM OEtRMeIO OEt MeR = Me: diastereoselection 99:1 R = Ph: diastereoselection 97:3PhMe2SiI CO2Et ROLi O-t-BuHCO2EtR = H: one isomerI. Fleming & Co-workers, Chem. Commun. 1984, 28.KOt-Bu THF -78 CCO2-t-Bu R = Me: > 15 :1 H RY. Yamaguchi & Co-workers, Tetrahedron Letters 1985, 26,1723.5-07-A-strain enolates 9/20/01 4:45 PMD. A. Evanss The basic processS H HRAllylic Strain & Olefin HydroborationHydroborations dominated by A(1,3) StrainS HRChem 206HROHBH HRBH2 BRH C RRO Me MeCH2 OBnB2 H6 H2 O2O Me MeCH2 OBnCC RC R RCC RRdiastereoselection 8:1 OMe Me O B2 H6 H2 O2 OH O Me Me Me diastereoselection 12:1 OH OMe OHs Response to steric effects: Here is a good calibration system:ACH2 Me3 C HMeMeOxidantMCPBA BH3, H2 O2 ERatio, A:E69:31 34:66ReferenceJOC, 1967, 32, 1363 JOC, 1970, 35, 2654 BnO MeY. Kishi & Co-workers, J. Am. Chem. Soc. 1979, 101, 259.OH OH Me Me B2 H6 H2 O2 BnO Me Me Me OHs Acyclic hydroboration can be controlled by A(1,3) interactions:OH RL RM Me OH R2 BH H2 O2 RL RM Me R R B C RL OH RL RM Me R2 BH H2 O2 OH RL RM Me R R RM B C RL H H C Me TrO CH2 OR OH H H C CH2 OR TrODiastereoselection = 3:1 C. H. Heathcock et. al. Tetrahedron Lett 1984 25 243. OHmajor diastereomerMe Mecontrol elementsA(1,3) allylic strain Steric effects; RL vs RM Staggered transition statesRM HMe ThexylBH2 , OH then BH3 OTr TrO OHMeMe OTr OH OH 5:1Diastereoselection;major diastereomerMeMeMeMeThexylBH2 , then BH3 TrOMeMeMeMe OTrOHOHOHOH OTrOHOHOH 4: 1OHDiastereoselection;See Houk, Tetrahedron 1984, 40, 2257HStill, W.C.; Barrish, J. C. J. Am. Chem. Soc. 1983, 105, 2487.5-07a-A-strain hydroboration 9/24/03 9:45 AMD. A. EvansConsider the resonance structures of an amide:O R3 C N1Allylic Strain & Amide ConformationR2 R33Chem 206Y X2R1 R large1The selection of amide protecting group may be done with the knowledge that altered conformational preferences may result:O O H HR1 R RO R3C N +R1 R1R smallFavored for R = H, alkylH O HRN RHN R OFavored for R = CORA(1,3) interactions between the "allylic substituent" and the R1 moiety will strongly influence the torsion angle between N & C1. FavoredO Me C N MeN H R H O N O H C R H H RNHDisfavoredMe MeODisfavoredNH C O RFavoredp. p. conformations of cyclic amidesO R C N + R R C O H Me N R C Me O H R C O H N N R A(1,3)Chow Can. J. Chem. 1968, 46, 2821H R C O NRH strongly favored Me MeR O 1 A(1,3) interaction between the C2 & amide C R substituents will strongly influence the torsion N angle between C1 & C2. R R2R O1 2CRN R + RH strongly favoredAs a result, amides afford (Z) enolates under all conditionsH HO Me Ph N OO published X-ray structure of this amide shows chair Me diaxial conformationQuick, J. Org. Chem. 1978, 43, 2705O MeCN HL LOM base favored O Me C N H L L HMeN LL Problem: Predict the stereochemical outcome of this cyclization.OH D. Hart, JACS 1980, 102, 397 Ph N O HOCOHCO2H(Z)-EnolateHH O H C L N Me L base disfavored O HH LOM C N Me L H Me N L LN Ph O diastereoselection >95%identify HOMO-LUMO pair5-08-A-strain Amides-1 9/23/03 6:09 PM(E)-EnolateD. A. EvansAllylic Strain & Amide ConformationChem 206A(1,3) Strain and Chiral Enolate DesignO Me N Bn H L L El O Me N Bn O LDA O or NaNTMS2 Me O M O OPolypropionate Biosynthesis: The Acylation EventO O SR O SR Me O SR Me OH O SR MeN Bnenolization selectivity >100:1HOR OAcylation CO2RReductionREl(+) JACS. 1982,104, 1737.O OO MeCN HFirst laboratory analogue of the acylation eventO Et Cl O Me Li O N R O Me Me R O N O O Ofavored enolization geometry p. p. In the enolate alkylation process product epimerization is a serious problem. Allylic strain suppresses product enolization through the intervention of allylic strainH O El C N Me L L O Me C N H El L L Me O C H El N L Lwith M. Ennis JACS 1984, 106, 1154.Diastereoselection ~ 97 : 3Why does'nt the acylation product rapidy epimerize at the exocyclic stereocenter?? R O Me CfavoredBACH N H R R O R C N Me R RWhile conformers B and C meet the stereoelectronic requirement for enolization, they are much higher in energy than conformer A. Further, as deprotonation is initiated, A(1,3) destabilization contributes significantly to reducing the kinetic acidity of the system These allylic strain attributes are an integral part of the design criteria of chiral amide and imide-based enolate systemsO O Me N Bn CH2OH Me N O O O Me N Me OH MeEvans Tetr Lett. 1977, 29, 2495Evans JACS 1982,104, 1737.Myers JACS 1997, 119, 6496X-ray structure5-09-A-strain Amides-2 9/20/01 5:22 PMD. A. EvansDiscodermolideChem 206hinge Me HO O O H16Me17MeMe Me Me OH MeOHO ONH2Me OH- immunosuppressive activity - potent microtubule-stabilizing agent (antitumor activity similar to that of taxol)The epimers at C16 and C17 have no or almost no biological activity.The conformation about C16 and C17 is critical to discodermolide's biological activity.S. L. Schreiber et al. JACS 1996, 118, 11061.5-10-Discodermolide 9/19/01 12:14 AMD. A. EvansConformational Analysis - Discodermolide X-ray 1Me HO Me Me Me OH Me OH O O NH2 Me MeChem 206OO HMe OH5-11-Discodermolide X-ray1 9/21/01 8:34 AMD. A. EvansConformational Analysis - Discodermolide X-ray 2Me Me MeChem 206HO O O H16 Me Me Me OH Me OH O O NH2Me OH165-12-Discodermolide X-ray2 9/21/01 8:40 AMD. A. EvansConformational Analysis: Part3Chem 206http://www.courses.fas.harvard.edu/~chem206/Conformational Analysis of Cyclic SystemsThree Types of Strain:Chemistry 206 Advanced Organic ChemistryPrelog Strain: van der Waals interactions Baeyer Strain: bond angle distortion away from the ideal Pitzer Strain: torsional rotation about a sigma bondLecture Number 6Baeyer Strain for selected ring sizes"angle strain" size of ring Ht of Combustion Total StrainStrain per CH2 (kcal/mol) (kcal.mol) deviation from 10928' (kcal/mol)Conformational Analysis-3 Conformational Analysis of C4 C6 Rings Reading Assignment for week A. Carey & Sundberg: Part A; Chapter 3Eliel & Wilen, "Stereochemistry of Organic Compounds, "Chapter 11, Configuration and Conformation of Cyclic Molecules, Wiley, 19943 4 5 6 7 8 9 10 11 12 13 14 15499.8 656.1 793.5 944.8 1108.3 1269.2 1429.6 1586.8 1743.1 1893.4 2051.9 2206.1 2363.527.5 26.3 6.2 0.1 6.2 9.7 12.6 12.4 11.3 4.1 5.2 1.9 1.99.17 6.58 1.24 0.02 0.89 1.21 1.40 1.24 1.02 0.34 0.40 0.14 0.132444' 944' 044' -516'Eliel, E. L., Wilen, S. H. Stereochemistry of Organic Compounds Chapter 11, John Wiley & Sons, 1994.Ribeiro & Rittner, "The Role of Hyperconjugation in the Conformational Analysis of Methylcyclohexane and Methylheterocyclohexanes" J. Org. Chem., 2003 , 68 , 6780-6787 (handout)de Meijere, "Bonding Properties of Cyclopropane & their Chemical Characteristics" Angew Chem. Int. Ed. 1979, 18, 809-826 Baeyer "angle strain" is calculated from the deviation of the planar bond angles from the ideal tetrahedral bond angle. Discrepancies between calculated strain/CH2 and the "angle strain" results from puckering to minimize van der Waals or eclipsing torsional strain between vicinal hydrogens. Why is there an increase in strain for medium sized rings even though they also can access puckered conformations free of angle strain? The answer is transannular strain- van der Waals interactions between hydrogens across the ring.D. A. Evans6-00-Cover Page 9/26/03 8:44 AMFriday, September 26, 2003Evans, Kim, BreitCyclopropane: Bonding, Conformation, Carbonium Ion Stabilization CyclopropaneHChem 206Carbocation Stabilization via CyclopropylgroupsH H H Necessarily planar. Subtituents are therefore eclipsed. Disubstitution prefers to be trans.C = 120 Almost sp2, not sp3 = 3080 cm-1MeWalsh Model for Strained Rings: Rather than and * c-c bonds, cyclopropane has sp2 and p-type orbitals instead.A rotational barrier of about 13.7 kcal/mol is observed in following example:HMeNMR in super acids (CH3) = 2.6 and 3.2 ppm OX-ray Structures support this orientation 1.302 1.517 1.478 1.222 R1.474 (antibonding) H (antibonding) Nonbondingside view (antibonding)1.464 1.541 1.409 H3H1.444 1.534 OPh (bonding) (bonding)1 (bonding) de Meijere, "Bonding Properties of Cyclopropane & their Chemical Characteristics" Angew Chem. Int. Ed. 1979, 18, 809-826 (handout)de Meijere, A.; Wessjohann, L. "Tailoring the Reactivity of Small Ring Building Blocks for Organic Synthesis." Synlett 1990 , 20.R. F. Childs, JACS 1986, 108, 16926-01-Cyclopropane 9/25/03 7:55 PMEvans, Kim, Breit145-155ax eq eq eq eq ax ax axConformational Analysis: Cyclic Systems-2 CyclobutaneH H HChem 206H H H H H H H H H H H H H H H H HCyclopentaneH H H H H H Eclipsing torsional strain overrides increased bond angle strain by puckering. Ring barrier to inversion is 1.45 kcal/mol.H H H H = 28 CsEnvelopeC2 Half-ChairCsEnvelope Two lowest energy conformations (10 envelope and 10 half chair conformations Cs favored by only 0.5 kcal/mol) in rapid conformational flux (pseudorotation) which causes the molecule to appear to have a single out-of-plane atom "bulge" which rotates about the ring.(MM2) Since there is no "natural" conformation of cyclopentane, the ring conforms to minimize interactions of any substituents present.HCsEnvelope (MM2)H HHH HHH H G = 1 kcal/mol favoring R = Me equatorial 1,3 Disubstitution prefers cis diequatorial to trans by 0.58 kcal/mol for di-bromo cmpd. A single substituent prefers the equatorial position of the flap of the envelope (barrier ca. 3.4 kcal/mol, R = CH3). 1,2 Disubstitution prefers trans for steric/torsional reasons (alkyl groups) and dipole reasons (polar groups). Me Me XX 1,3 Disubstitution: Cis-1,3-dimethyl cyclopentane 0.5 kcal/mol more stable than trans. 1,2 Disubstitution prefers trans diequatorial to cis by 1.3 kcal/mol for diacid (roughly equivalent to the cyclohexyl analogue.)H H A carbonyl or methylene prefers the planar position of the half-chair (barrier 1.15 kcal/mol for cyclopentanone). X6-02-Conform/cyclic-2 9/25/03 7:55 PMEvans, Kim, BreitConformational Analysis: Cyclic Systems-3Chem 206Methylenecyclopentane and CyclopenteneStrain trends: > > Decrease in eclipsing strain more than compensates for the increase in angle strain.Cyclohexane Energy Profile (kcal/mol)Half-ChairBoat +1.01.5 10.711.5Relative to cyclohexane derivatives, those of cyclopentane prefer an sp2 center in the ring to minimize eclipsing interactions. "Reactions will proceed in such a manner as to favor the formation or retention of an exo double bond in the 5-ring and to avoid the formation or retention of the exo double bond in the 6-ring systems." Brown, H. C., Brewster, J. H.; Shechter, H. J. Am. Chem. Soc. 1954, 76, 467.+5.5 +5.5 Chair Twist Boat Inverted ChairExamples:H H H H H H H H O O NaBH4 HH OHk6H H HHNaBH4H H OH Hk6 = 23 k5k5Brown, H. C.; Ichikawa, K. Tetrahedron 1957, 1, 221.O OE = +6.5-7.0 E = +5.5OOhydrolyzes 13 times faster thanConan, J-Y.; Natat, A.; Priolet, D. Bull. Soc. Chim., Fr. 1976, 1935.O O OEt OH O OEtThe barrier: +10.7-11.5 E=095.5:4.5 keto:enol76:24 enol:ketoBrown, H. C., Brewster, J. H.; Shechter, H. JACS 1954, 76, 467.6-03-Conform/cyclic-3 9/25/03 7:57 PMEvans, BreitConformational Analysis: Cyclic Systems-4Chem 206Monosubstituted Cyclohexanes: A ValuesR HA Values depend on the relative size of the particular substituent.H H H H Me H H H Me Me H Me Me Me HKeqHRG = RTlnKeq Meaxial has 2 gauche butane interactions more than Meequatorial. Expected destabilization: 2(0.88) kcal/mol = ~1.8 kcal/mol; Observed: 1.74 kcal/molMe H C H H H H Me H H C Me H C HHAValue1.741.802.155.0The "relative size" of a substituent and the associated A-value may not correlate. For example, consider the CMe3 and SiMe3 substituents. While the SiMe3 substituent has a larger covalent radius, it has a smaller A-value:Me C Me H Me Si Me H Me Sn Me H The A Value, or -G, is the preference of the substituent for the equatorial position.MeMeMeAValue4.5-5.02.51.1Can you explain these observations? The impact of double bonds on A-values:Lambert, Accts. Chem. Res. 1987, 20, 454R H H Rsubstituent R = Me R = OMe R = OAcG 0.8 0.8 0.6A-value (cyclohexane) 1.74 0.6 0.71The Me substituent appears to respond strictly to the decrease in nonbonding interactions in axial conformer. With the more polar substituents, electrostatic effects due to the trigonal ring carbon offset the decreased steric environment.6-04-Conform/cyclic-4 9/25/03 7:59 PMRigberio & Rittner, The Role of Hyperconjugation in the Conformational Analysis of Methylcyclohexane and Methylheterocyclohexanes JOC 2003, 68, 6780 Commentary by Ken Houk University of California, Los Angeles Department of Chemistry Dear David, The calculations in the Ribeiro article look fine, but I am not convinced by the interpretation. It does seem to work pretty well for many systems, but not obviously for the isomeric 1,3-dioxane cases they note early on. There seems no explanation of why CH hyperconjugates better than CC. Further, the results with alkyls larger than methyl still require traditional steric arguments. I would say that the equatorial methyl preference has been attributed in part to hyperconjugative effects that occur when the CH bonds are anti-periplanar. But I would not yet go much beyond that! Best regards, (b. 1943) A.B. 1964, Ph.D. 1968, Harvard University; Assistant-Full Professor, Louisiana State University, 1968-1980; Alfred P. Sloan Fellow, 1975-1977; Camille and Henry Dreyfus Teacher-Scholar, 1972-1977; LSU Distinguished Research Master, 1978; Professor, University of Pittsburgh, 1980-1985; Alexander von Humboldt Senior U.S. Scientist Award, 1982; Akron Section, American Chemical Society Award, 1983; Arthur C. Cope Scholar Award, 1988; Director, Chemistry Division, National Science Foundation, 1988-1990; James Flack Norris Award in Physical Organic Chemistry, 1991; Schrdinger Medal, World Association of Theoretically Oriented Chemists, 1998; Tolman Medal, Southern California Section, American Chemical Society, 1999; Fellow of the American Academy of Arts and Sciences, 2002; American Chemical Society Award for Computers in Chemical and Pharmaceutical Research, 2003; International Academy of Quantum Molecular Science, 2003.D. A. EvansBond Lengths and A-Values of Methyl HalidesChem 206CF: 1.39 CCl: 1.79 CBr: 1.95 CI: 2.16 F A-value: 0.250.42Cl A-value: 0.5364Br Avalue: 0.48-0.67I A-value: 0.470.61p. p.Chem 3D Pro (Verson 5.0)6-04a methyl halides 9/26/01 8:26 AMEvans, BreitConformational Analysis: Cyclic Systems-5Chem 206The impact of trigonal Carbon Let's now compare look at the carbonyl analog in the 3-positionMe H O O H MePolysubstituted Cyclohexane A Values As long as the substituents on the ring do not interact in either conformation, their A-values are roughly additive 1,4 Disubstitution: A Values are roughly additive.Me Me Me Me Me Me Me MeG = 1.36 kcal/mol versus 1.74 for cyclohexane Let's now compare look at the carbonyl analog in the 2-positionMe Me3C O H base epimerization Me3C O H MeG = 0 kcal/molG = 2(1.74) = 3.48 kcal/molG = 1.56 kcal/mol versus 1.74 for cyclohexane However, the larger alkyl groups do not follow the expected trend. Can you explain? (see Eliel, page 732)CHMe2 Me3C O H base epimerization Me3C O H CHMe21,3 Disubstitution: A Values are only additive in the trans diastereomerH X H Me H Me H XG = A(Me) A(X)HH X H Me X HG = 0.59 kcal/mol versus 2.15 for cyclohexaneCMe3 Me3C O H base epimerization Me3C O HMeThe new interaction! For X = MeH CMe3 Me H Me H H H H Me MeG = 1.62 kcal/mol versus 5.0 for cyclohexane6-05-Conform/cyclic-5 9/25/03 8:01 PM+ 0.88+ 0.88G = 2(.9) + 1(+3.7)= 5.5 kcal/mol + 3.7Evans, BreitMe PhConformational Analysis: Cyclic Systems-6 Let's now consider vicinal substitutionMe H PhChem 206Let's now consider geminal substitutionMeCase 1:MeHH Me HMeThe prediction:G = A(Ph) A(Me) G = +2.8 1.7 = +1.1 kcal/molObserved:G = 0.32 kcal/molThe prediction:G = 1 gauche butane 2A(Me) G = +0.88 2(1.74) = +2.6 kcal/molHence, when the two substituents are mutually interacting you can predict neither the magnitude or the direction of the equilibrium. Let's analyze this case. Allinger, Tet. Lett. 1971, 3259Observed:G = +2.74 kcal/molIf the added gauche butane destabilization in the di-equatorial conformer had not been added, the estimate would have been off.6-06-Conform/cyclic-6 9/25/03 8:13 PM00 000000 000000 0 0 0 000000 00 000A G = +2.80000 00 00000 00 0 0 00 000 00 000B G = 0.32Case 2:H Me Me OH H OH H H Me MeHHThe conformer which places the isopropyl group equatorial is much more strongly preferred than would be suggested by A- Values. This is due to a syn pentane OH/Me interaction. Problem: Can you rationalize the stereochemical outcome of this reaction?00 0000000 0 00 0 000000 00000000 0 000000 0 000 0O EtO LiNR2 MeI H EtOOMeCDn-C4H9n-C4H9Hdiastereoselection 89:11Note the difference in the Ph substituent in A & B.D. Kim & Co-workers, Tetrahedron Lett. 1986, 27, 943.Evans, BreitConformational Analysis: Cyclic Systems-7Chem 206Heteroatom-Substituted 6-Membered Rings A-values at the 2-position in both the O & N heterocycles are larger than expected. This is due to the shorter CO (1.43 ), and CN (1.47 ) bond lengths relative to carbon (CC; 1.53 ). The combination of bond length and bond angle change increases the indicated 1,3-diaxial interaction (see eq 1, 4).A-Values