教学课件高等有机化学lecture2

1、Lecture 2: Donor-Acceptor InteractionsE. KwanChem 106Can you explain the following?NMRDonor-Acceptor InteractionsEugene E. KwanSeptember 2, 2011.1JCH,ax = 157.4 HzHOOMeMeH1JCH,eq = 167.5 Hzs*IROOOO sMeRNRSMeRMeRClMe 1660169017101735Scope of Lecturerotational barriers in substituted ethanesCoulsons T

2、heoremweakerp-bondstrongerp-bondcm-1Bents Ruledonor-acceptor interactionsThermodynamicsamide resonancegaucheeffectmore stable thanOOROnon-Lewis corrections to Lewis structuresanomeric effectORhyper- conjugationKineticsOMeRHelpful References1. Coulsons Valence, 3rd ed. McWeeny, R. Oxford: Oxford Univ

3、ersity Press, 1979.RMgXOOMeMeOOMeMe2. Quantum Chemistry, 5th ed. Levine, I.N. Upper Saddle River, New Jersey: Prentice-Hall, Inc., 2000.RMgXOOMeMeOMeno reaction3. Valency and Bonding: A Natural Bond Orbital Donor-Acceptor Perspective, Cambridge, Cambridge University Press, 2005.I thank Professor Fra

4、nk Weinhold (Wisconsin) for many helpful discussions in the preparation of this lecture.Lecture 2: Donor-Acceptor InteractionsE. KwanChem 106The NBO Hierarchy/nbo5/The NBO algorithim converts the wavefunctions/CMOs into other basis sets that are mathematically equivalent (span

5、 the same space), but easier to understand:Review: Conceptual Frameworkstandard global approach:The standard approach is to use computations to predict bond lengths, transition state energies, vibrational frequencies, etc. These calculations correlate well with experimental observations. Therefore,

6、it is reasonable to think that other useful information might be contained in the wavefunctions produced by these calculations. However, the canonical molecular orbitals (CMOs) are relatively delocalized entities that are hard to interpret chemically.new localized approach:If we want to talk about n

7、ormal chemical ideas, like bonds or lone pairs, then we need to ask questions about localized one- electron subsystems. The natural bond orbital (NBO) approach is a very reasonable way to do this.Example: Hydrogen Moleculeenergys* antibondHHBA1s1so bondNBO picture:o and s* bonds are linear combinati

8、onsof natural atomic orbitalsMO picture: bonds are delocalized blobs: accurate, butconfusingThe various natural orbitals are constructed such that every occupied orbital has as close to two electorns as possible (maximum occupancy).contracted Gaussian-type orbitals (CGTOs)linear combinations of Gaus

9、sian chosen to look like orbitals in the hydrogen atom; no physical significancenatural atomic orbitals (NAOs)1-center orbitals (1s, 2s, 2p.)natural hybrid orbitals (NHOs) linear combinations of NAOs (eg., sp3 hybrids on carbon)natural bond orbitals (NBOs)1 or 2-center orbitals (bonds, antibonds, lo

10、ne pairs)canonical molecular orbitals (CMOs)fully delocalized orbitals (HOMO, LUMO)Lecture 2: Donor-Acceptor InteractionsE. KwanChem 106NAO/NBO TerminologyBy convention, orbitals are given abbreviations to indicate their type. Asterisks (*) mean the orbital is mostly empty. You can think of these un

11、occupied orbitals, such as antibonds, as a molecules capacity to respond to change, like irradiation by light or the approach of a reagent.e.g. neonRydberg orbitals (RY*):empty, and of littleOrthogonalityThe Pauli Principle requires that no two electrons can occupy the same state, and therefore, any

12、 reasonable orbitals must be orthogonal. The NAOs, NHOs, and NBOs are all orthogonal. But orthogonal means overlap is zero!However, we can recover the idea of overlap but considering the pre-orthogonal versions of each:PNBOs: overlap corresponds to donor-acceptor interactions (e.g., hyperconjugation

13、 from a lone pair into an antibond)3sphysical significancePNHOs/PNAOs: overlap corresponds to strength of chemical bondingvalence orbitals (VAL) these are the ones involved in bonding2pQ: How do you think about the difference between the pre-orthogonal orbitals and the fully orthogonal ones?2sA: Her

14、e is an approximate picture. Imagine having an electron in a molecule where the nuclei are frozen in place. Now take away all the other electrons. The one remaining electron sits in orbitals whose shapes reflect the chemical environment ofthe nuclei around it. However, these orbitals dont have to ha

15、ve as many nodes as they would if other electrons were around (because of the Pauli principle).core electrons (CR)1se.g. methanelone pairs (LP):one-center valence electrons (none here)RY*3s Thus, it not surprising that NAOs dont look like hydrogenlike orbitals! (Electrons in molecules are largely un

16、der the influence of the local nuclear potential, so their orbitals look kind of hydrogenlike, but are also influenced by the other electrons around them.) antibonds (BD*)s*C-Hbonds (BD)sC-HCR1sLecture 2: Donor-Acceptor InteractionsChem 106E. KwanEquivalent vs. Inequivalent HybridsConsider the Lewis

17、 structure of methyl fluoride:assumingAlthough these hybrids are superficially similar in appearance their NHO composition is very different:sCH has 2 e in itHH equivalent sp3 hybrids on C and FFF1.(1.99926)( 71.99%)( 28.01%)(1.99832)( 59.90%)( 40.10%)BD ( 1) F0.8485*0.5293*BD ( 1) C0.7739*0.6333*1-

18、 C2F1 s( 26.20%)p 2.81( 73.64%)C2 s( 19.91%)p 4.01( 79.87%)2- H3C2 s( 26.74%)p 2.74( 73.15%)H3 s(100.00%)carbon is using an sp2.74 hybridHHHHThe naive VSEPR-type interpretation assumes that both carbon and fluorine form their bonds with equivalent sp3 hybrids, which suggests the lone pairs are equiv

19、alent.Is this correct?2.the H hybrid isweighted 40% in the NBOHybridization at CarbonPNBO: sCHThe difference in s-character is not trivial!Rationalization: s orbitals are lower in energy than p orbitals, so hybrids of higher s-character will be directed towards more electropositive ligands. This is

20、Bents Rule.ref: Bent, H.A. Chem. Rev. 1961, 61, 275.Consider a hypothetical A-L bond. The real bond can be thought of as a linear combination of fully covalent and fully ionic structures:PNBO: sCFA Lfully covalent+A :Lfully ionicAL=As the ligand L gets more and more electronegative, the emphasis shi

21、fts to the fully ionic resonance structure, leaving less and less electron density on A.Lecture 2: Donor-Acceptor InteractionsChem 106E. KwanBents RuleIf A has two ligands B and X, where X is much more electronegative than B, then there will be a largely covalent A-B bond and a largely ionic A-X bon

22、d:With inequivalent hybrids, A directs a hybrid of higher s- character towards B. In the limit,lower energy sand more electronshigher energy p, but fewer electronsXABionic-like covalent-likeSo there is hardly any electon density on A pointed at X, and quite a lot on A pointed at B. Looking just at t

23、he hybrids on A:ABp-rich s-rich hybrid hybridNow, a small amount of density will resides in the higher energy p-rich hybrid, and a larger density resides in the lower energy s-rich hybrid.XABsmall contributionlarge contributionEquivalent vs. Inequivalent HybridsIn the equivalent hybrid picture, A wi

24、ll direct NHOs of equal s-character towards both X and B:One can consider a lone pair to be a bond to an infinitely electropositive substituent.Although there are some subtleties to this rule in the periodic table, it is generally correct.ABCoulsons TheoremWhat is the relationship between hybridizat

25、ion and observed bond angles?equal s-characterBecause there is hardly any electron density on A in the A-X bond, the energy of the bonding is largely determined by the composition of the A-B bond. Here, half of the low energy s- character is being wasted on the A-X bond.In a typical situation, we ha

26、ve an atom with orthonormal set of NAOs, such as:valence NAOs = s, px , py , pz j = dijiThe atom forms four orthonormal hybrids hi which can be written as:hi = ci1s + ci2 px + ci3 py + ci4 pzi=1.4Lecture 2: Donor-Acceptor InteractionsChem 106E. KwanMethyl FluorideCoulsons TheoremEach hybrid can be c

27、haracterized by a hybridization parameter:assuming equivalent sp3 hybrids on C and FHHc+ c+ c222FFl =i 2i3i 4HHHHic2i1In other words, the hybrid hi is spl hybridized. that:It can be shownIf the equivalent hybrid picture is incorrect, and atoms use inequivalent hybrids in accordance with Bents Rule,

28、thenQ1: What is the qualitative form of the NHOs on fluorine?cos(w )=- 1Coulsons Theorem(ll)1/ 2ijQ2: Predict the shape of the lone pairs on fluorine.ijKey Point: Hybridization only occurs in response to bonding.angle between hybridsFor equivalent sp3 hybrids, li=lj=3, so wij=109.47.First, imagine t

29、hat fluorine is sp-hybridized, with two lone pairs residing in empty p-orbitals orthogonal to the s-bond:Coulsons Theorem reflects the fact that the relationship between the NAO and NHO bases is a unitary transformation; the total fraction of s-character in all the hybrids must be exactly 25%. For e

30、xample, it is impossible for a carbon atom to form four sp hybrids to its ligands.Example:Hp lone pairFHsp lone pairHWill fluorine:(a) form equivalent hybrids for the s-bond and lone pair;(b) emphasize s-character in the s-bond hybrid and p-character in the lone pair hybrid; or(c) emphasize p-charac

31、ter in the s-bond hybrid and s-character in the lone pair hybrid?Suppose a carbon uses an sp4 NHO to form a bond to ligand A and and an sp3 NHO to form a bond to ligand B. Please estimate the A-C-B bond angle.-1cos(w)=w=106.8oABAB4(3)Lecture 2: Donor-Acceptor InteractionsChem 106E. KwanQ: Will fluor

32、ine:(a) form equivalent hybrids for the s-bond and lone pair;(b) emphasize s-character in the s-bond hybrid and p-character in the lone pair hybrid; or(c) emphasize p-character in the s-bond hybrid and s-character in the lone pair hybrid?PNBO 7:This is borne out by the NBO analysis:PNBO 8:1. (1.9992

33、6) BD ( 1) F1- C2( 71.99%)0.8485*F1 s( 26.20%)p C2 s( 19.91%)p2.81( 73.64%)4.01( 79.87%)( 28.01%)0.5293*7. (1.99479) LP ( 1) F1F2 s( 73.82%)p1.0000*0.35( 26.17%)(100.00%)The orthogonal lone pairs have complete p-character:8. (1.97018) LP ( 2) F(100.00%)1.0000*9. (1.97018) LP ( 3) F(100.00%)1.0000*1F

34、2 s(1F2 s(0.00%)p 1.00( 99.94%)0.00%)p 1.00( 99.94%)PNBO 9:The same thing is presented pictorially below:PNBO 1:Lecture 2: Donor-Acceptor InteractionsChem 106E. KwanThe Shape of Lone Pairs(acknowledgement: this discussion borrowed heavily from Prof. Weinhold and coworkers)Indeed, NBO analysis shows:

35、 (PNBOs, B3LYP/6-31g(d):ref: Laing, M. J. Chem. Educ. 1987, 64, 124.Consider the Lewis structure of water:OHHIf one naively imagines oxygen to have equivalent hybrids, then this textbook picture results:lone pairs are in equivalent sp3 rabbit earsOs-rich (-0.62 au)44% pp-rich (-0.29 au)99% pHHHoweve

36、r, photoelectron spectroscopy clearly shows water to have three, not two peaks:Similar results are obtained if one computes the full MO diagram: the lone pairs are not equivalent. What is going on?Put the water molecule in the x-y plane. There are no ligands that can interact with the pz orbital, so

37、 it remains unmixed:zout of plane: pzyOHHin plane: combination of s, px, pyxHybridization only occurs in response to bonding.eVLecture 2: Donor-Acceptor InteractionsChem 106E. KwanVB vs. MO vs. NBO TheoriesThis NBO treatment is very similar to classical valence bond (VB) theory developed in 1927 by

38、Heitler and London. Forthe hydrogen molecule, the first step is to describe the wavefunction as that for two non-interacting hydrogen atoms:Instead, Pauling and Wheland consider polar bonds to be described by a resonance hybrid of the form:Y= cY+ cYcovcov covion ionMore sophisticated MO treatments i

39、nclude more terms and parameters (see Levine, pg 414-416).HHpurely covalent Heitler-London functionIn contrast, the NBO analysis follows Mulliken in treating the polar bond A-B as:YNBO = cAhA(1) + cBhB(1)cAhA(2) + cBhB(2)Here, hA and hB are NHOs. This represents a doubly occupied NBO, where covalenc

40、y/ionicity are represented by the ratio of |cA|2 to |cB|2, not |ccov|2 to |cion|2 as in VB theory. In either case, the transition from polar to ionic bonding is a continuous one.DifluorineLet us contrast the VB and NBO for homopolar difluorine:means:Y= 1s (1)1s (2) +1s (1)1s (2)covABABputs electron

41、1 into the 1s orbital on hydrogen atom AActually, these wavefunctions need to be antisymmetrized to satisfy the Pauli principle. But let us ignore that for this discussion.However, this Heitler-London function only really works for atoms of equal electronegativity. To describe a polar bond A-B, wher

42、e A is more electronegative, one can imagine:purely ionic Heitler-London functionABYVB = hA(1)hB(2) +hA(1)hB(2)one-center limit:no interaction between atoms; inappropriate when R is near equilibriumFFYion = hA(1)hA(2)FFIn the simplest LCAO-MO treatment, the bond is described by equal weightings of e

43、ach permutation:singlet diradicalYcov =hA(1)hA(2) + hB(1)hB(2)ionic termsYNBO = cAhA(1) + cBhB(1) cAhA(2) + cBhB(2)covalent terms+ hA(1)hB(2) + hB(1)hA(2)FFThis treats the ionic and covalent terms as equally likely, which means that the hydrogen molecule would be equally likely to dissociate to two

44、neutral H atoms as a hydride and a proton.two-center interactions: incorporates donor-acceptor interactions; realistic FFsinglet pairLecture 2: Donor-Acceptor InteractionsChem 106E. KwanIn reality, nobody uses the literal Heiter-London function. The more sophisticated treatments incorporate more ter

45、ms which circumvent these formal problems. The conclusions of semi- empirical VB theory do agree qualitatively with those of NBO:(1) Relaxation EffectsE2E2E2 hnrelaxKTbonds are localized, transferable two-electron entities.IPE1E1neutral speciesE1Koopmans TheoremIf the NBO/VB picture is so reasonable

46、, then why do photoelectron measurements seem to show that electrons are ionized from the canonical molecular orbitals (CMOs) of MO theory?frozen cationreal cationUpon ionization, a hole is created that other orbitals can interact with in a donor-acceptor sense. So, Koopmans Theorem uses fully relax

47、ed orbitals for the neutral species but frozen orbitals for the cationic species.(2) Correlation EffectsIn the single-determinantal HF picture, we draw one set of energy levels with a single configuration. But this ignores electron correlation. In configuration interaction (CI) we consider small con

48、tributions from excited states:Koopmans Theorem (KT)IPn = -EnKoopmans Theorem (Koopmans Physica 1933 1 104) says that the n-th ionization energy of a molecule is equal to the energy of its n-th CMO. Unfortunately, this is not a very good approximation:E2E2E2CIalso considerE1E1E1The relative errors f

49、rom (1) and (2) tend to cancel, so in some cases, Koopmans Theorem works well for IPs.However, for electron affinities, the errors are additive and KT is entirely inaccurate.So regardless of whether NBO or MO energies are used, the approximation is quite poor. Why does/nt KT work?Thus, orbital energ

50、ies should not be taken to mean literal energies; proper ionization energies and electron affinities should require analysis of both the ground and excited states.first ionization energy (au)SpeciesKTMOKTNBOExp.CO0.38750.48680.5150N20.44090.47680.5726H2O0.32300.31990.4638CH40.39540.50440.4636Lecture

51、 2: Donor-Acceptor InteractionsE. KwanChem 106FormamideEach component in the Lewis structure has an NBO analog:Non-Lewis Corrections to Lewis StructuresWe can imagine that the Lewis structure corresponds to a variational guess for the wavefunction in the NBO basis which has all the Lewis-type NBOs f

52、ully occupied (2 electrons) and all the non-Lewis-type NBOs completely empty (0 electrons):nNsNH, sCH, sCOOHHNHNBOFull BasisIdealizedpCONBOs..1.BD BD BD BD BD BDCR CR CRLP LP LP(1)1)2)1)1)1)1)1)1)1)1)2)1)1)2)1)1)1)C C C C N NC N ON O O1-1-1-1-2-2-1252551-1-1-

53、1-2-2-N O O H H H2556341.997731.998691.996961.983921.991551.992081.999581.999561.999821.746711.982881.851710.066400.248810.006680.067540.014220.010292.000002.000002.000002.000002.000002.000002.000002.000002.000002.000002.000002.000000.000000.000000.000000.000000.000000.00000Lewis StructureHowever, t

54、he NBO antibonds have no Lewis analogs:energyY *= cf + cf212 122 2no Lewis analogBD*(C C C C N NN O O H H H255634f2BD*(BD*(B3LYP/6-31g(d); not all NBOs shownOne finds that the energy of the idealized Lewis structure guess is 171 kcal/mol too high, which means that describing the structure of formamide with one Lewis structure is not