教学课件高等有机化学lecture01

1、Lecture 1: Bonding and HybridizationE. KwanChem 106Course Details:Instructor: Dr. Eugene Kwan ()TF: Dr. Christian Gampe ()Bonding and HybridizationEugene E. KwanAugust 31, 2011.Textbook/Readings: Ref. 3 and online notess*Website: /icb/i

2、cb.do?keyword=k82275 (lecture videos available)Problem Sets: five questions every friday sGrading Scheme:Scope of Lectureshape of lone pairsproblem sets (10%)2x midterms (20%)writing assignment (20%)final exam (30%)multielectron atomsthe variational principlebonding and hybridizationPrerequisite: Ch

3、em 30 or equivalent (undergrads welcome)equivalent vs. inequivalent hybridizationLCAOmethodOverview of Topics:(1) Covalent vs. Non-covalent Bonding(2) Pericyclic Reactions(3) Conformational Analysis(4) Reaction Mechanisms (kinetics, computations, isotopes.)(5) Reactive Intermediates (cations, anions

4、, radicals, carbenes)(6) Catalysis (metal, secondary amine, hydrogen-bond, NHC)basis functions vs. orbitalsthe virial theoremHelpful References1. Coulsons Valence, 3rd ed. McWeeny, R. Oxford: Oxford University Press, 1979.Goals:2. Quantum Chemistry, 5th ed. Levine, I.N. Upper Saddle River, New Jerse

5、y: Prentice-Hall, Inc., 2000.By the end of the course, I would like you to be able to think and write intelligently about any organic reaction.3. Valency and Bonding: A Natural Bond Orbital Donor-Acceptor Perspective, Cambridge, Cambridge University Press, 2005.I thank Professor Frank Weinhold (Wisc

6、onsin) for many helpful discussions in the preparation of this lecture.Lecture 1: Bonding and HybridizationE. KwanChem 106Hydrogenlike AtomsLet us start at the beginning, with the hydrogen atom that you (hopefully) have come to know and love. The hydrogen atom has one proton and one electron. In hyd

7、rogenlike atoms, there are Z protons and one electron:1s radial distributionrone electron+Z protonsqThis includes H, He+, Li2+, etc.fHere, we imagine the nucleus is fixed at the origin. In spherical coordinates, the position of the electron is (r, q, f). Here are the key facts about the hydrogenlike

8、 atoms:In this state, we will find the electron at a radius of on average (but not every time).In quantum mechanics, we postulate that there is an abstract mathematical object called the wavefunction, which, when squared, gives the probability distribution (from which one can calculate the radial di

9、stribution function);(1) Radial Distribution FunctionsQ: Where is the electron?A: Dont ask.The electron isnt anywhere exactly, any more than the sound wave from a violin is anywhere specific. More precisely, ifwe prepare a bunch of hydrogen atoms in exactly the same way, and then we measure the posi

10、tion of the electron, then we will not get the same answer every time. Rather, we will get some distribution of answers. This is a probability distribution.(2) WavefunctionsDifferent systems have different wavefunctions. If the system is in a stationary state, then its probability density does not c

11、hange in time. The 1s state of the hydrogen atom is such a state. This means that if I prepare a lot of identical 1s H atoms and wait a while before measuring the position of their electrons, then the probability distribution I get doesnt depend on how long I wait. If the H atoms are in a superposit

12、ion of these stationary states, then the distribution will change with time in a complicated way.So what are these stationary states?To make this more concrete, here is an example. One of the states you can prepare the hydrogen atom in is the 1s state, which is its lowest energy, or ground, state. T

13、he radial probability distribution plots the probability of finding the electron in a spherical shells extending out from the nucleus.|Y|2 = probability densityLecture 1: Bonding and HybridizationE. KwanChem 106Hydrogenlike Atoms(3) Quantum NumbersEall positive energies allowedAs it turns out, these

14、 stationary states cant be just anything.Inr0fact, they are labelled by three quantum numbers, which have to be integers. Because these numbers can only take on certain discrete values, the stationary states are said to be quantized.n=4 and up.n=33s3p3dn=22,0,02s2,1,-12,1,0 2,1,1principal quantum nu

15、mber: n = 1, 2, 3.energies: E a 1 / n2angular momentum quantum number: l = 0, 1, 2, ., n-1l=0 (s); l=1 (p); l=2 (d); l=3 (f); .magnetic quantum number: m = -l, -l+1, ., 0, ., l-1, l- essentially give different orientations of similar orbitals2py(n, l, m)- for example, for n=2, there are three possib

16、le solutions:y(n, l, m)y(2, 0, 0) = 2s-13.6 eVn=11,0,01sn,l,mActually, not all the energies are quantized. Zero energy is defined as the electron and the proton at an infinite distance where they dont interact. If the electron has more than zero energy, then you can interpret that as an ionized atom

17、: the electron has escaped the potential of the hydrogen atom, and is free to be anywhere it wants, with any energy (more or less).y(2, 1, -1), y(2, 1, 0), y(2, 1, 1):linear combinations are the 2px, 2py, and 2pz(4) EnergiesNotice that the energy depends only on n. That means the 2s and 2p wavefunct

18、ions (any one electron wavefunction is called an orbital) are of the same energy. The 3s, 3p, and 3d states are also of the same energy (but dont have the same energy as the 2s/2p).Notice that the levels get closer and closer together as n increases, but the s/p/d/f energies remain degenerate on eac

19、h level. This is only true for hydrogenlike atoms.(for Z=1)En = -13.6 eV / n2Lecture 1: Bonding and HybridizationE. KwanChem 106Hydrogenlike Atoms(5) OrthogonalityThus, even though the 1s and 2s orbitals occupy the same space, they still have zero overlap:A key idea is that of orthogonality. The Pau

20、li Exclusion Principle says that no two electrons can occupy the same state. More precisely, the electrons have to be in orthogonal orbitals, which means their overlap integral= f1f2 dtS12has to be zero. For example, 2s orbitals are orthogonal to 2p ones:Even though the 2s and 2p orbitals occupy rou

21、ghly the same space, in that theres a reasonable probability of finding the electron within a common set of coordinates, they are still orthogonal because the 2p orbital has an angular node where the wavefunction changes sign.In general, there are:n-1 total nodes,l angular nodes (m in the xy-plane;

22、l-m on the z-axis), and n-l-1 radial nodes.Multielectron AtomsOnce there is more than one electron, interelectronic repulsions makes energy depend on both n and l:Although the hydrogenlike atom only has one electron, the idea is that atoms with more electrons will look similar, with electrons piling

23、 on top of each other in successively higher energy levels.So how do the 1s and 2s orbitals remain orthogonal? This time, there is a radial node in the 2s orbital:2p2s2p 2shydrogen (Z=1)helium (Z=2)The ordering of the levels is very complicated, as shown on the following page. Nonetheless, theory an

24、d experiment agree very well on the energies of atoms.credit: The Orbitron Univ. of SheffieldLecture 1: Bonding and HybridizationE. KwanChem 106Multielectron AtomsThis chart (Levine, pg 312-315) shows the intricate ordering of energies. Eventually, if the atomic number Z gets high enough, the energi

25、es once again only depend on n (and not l) as nucleus-electron attractionsoverwhelm electron-electron repulsions. These energies are for neutral, isolated atoms only; real energies and orbital shapes depend on the specific chemical environment of the atom.at Z=7 (N), the4s becomes lowerin energy tha

26、n the 3dhydrogen atomatomic number1 HydrogenH2 HeliumHe3 LithiumLi4 BerylliumBe5 BoronB6 CarbonC7 NitrogenN8 OxygenO9 FluorineF10 NeonNe11 SodiumNa12 Magnesium Mg13 AluminiumAl14 SiliconSi15 Phosphorus P16 SulfurS17 ChlorineCl18 ArgonAr19 PotassiumK20 CalciumCa21 ScandiumSc22 TitaniumTi23 VanadiumV2

27、4 ChromiumCr25 Manganese Mn26 IronFe27 CobaltCo28 NickelNi29 CopperCu30 ZincZn31 GalliumGa32 Germanium Ge33 ArsenicAs34 SeleniumSe35 BromineBr36 KryptonKr37 RubidiumRb38 StrontiumSr39 YttriumY40 ZirconiumZrLecture 1: Bonding and HybridizationE. KwanChem 106The Hydrogen MoleculeThe hydrogen molecule

28、has two protons and two electrons:(Note that f has nothing to do with the spherical coordinate.)So, for any guess f, the ground state energy of the system E1 is no higher than W. This suggests we should try to vary f, and seek a form of it that minimizes W:Translation: 2 e HHWhat we want is to obtai

29、n the energy levels and wavefunctions for the hydrogen molecule.Unfortunately, the interelectronic repulsions make it impossible to do this analytically, so we need to resort to approximate methods. Despite the fact that they are not analytic in nature, they are still good enough to describe chemist

30、ry!The Variational TheoremIf a system has a ground state energy of E1, and f is a normalized, well-behaved function of the systems, thenE1 is no higher than the variational integral W: The Linear Combination of Atomic Orbitals (LCAO)Lets try this for the hydrogen molecule. A reasonable guess for f m

31、ight be 1s orbitals of the hydrogen atom. One might guess that the wavefunctions are linear combinations of the form:f * Hfdt E W =1Proof: Levine, pg 208-209.a guess for the wavefunction; has to be reasonable (satisfy the boundary conditions of the problem being considered)the complex conjugate of f

32、 (f need not be real)ff = c f + c f1 12 2f1 = 1s orbital on hydrogen Af2 = 1s orbital on hydrogen BHBHAf*Thus, we seek values of c1 and c2 that minimize W. More precisely:the Hamiltonian operator, which you can think of as a widget which gives the energy of the guess fHWc2Wc1= 0= 0dta notation that

33、means integrate over all spaceThis is advantageous because its hard to vary a function, but its easy to vary the coefficients c1 and c2 in a linear combination.Wthe variational energyRefine f and try again.Compute the integral W.The real ground state energyis below the value of the integral.Make a r

34、easonable guess, f.Exact solution unknown.Lecture 1: Bonding and HybridizationE. KwanChem 106The Secular EquationsAs it turns out, if you vary c1 and c2 to minimize W, then c1 and c2 have to satisfy the secular equations:c1 (H11 - S11W )+ c2 (H12 - S12W ) = 0energyY2 = N2 (f1 - f2)first excited stat

35、e, anti-bondingW2c ( H- S W ) + c ( H- S W ) = 01212122222H= ff=f Hf d tHamiltonian matrix element/ resonance integraloverlap integralH*where121212ffH1112S= ff=f f dt*W112121 2ground state, bondingY1 = N1 (f1 + f2)S= S(overlap of i with j = overlap of j with i)Note:ijji1s orbitalH= H*(the Hamiltonia

36、n is Hermitian)ijji(The factors N are to normalize the wavefunctions. This amounts to saying that the electron has a 100% chance of being somewhere.)For details, please see Levine, 220-223.For the moment, dont worry too much about these equations.(1) We are not saying that each hydrogen atom has a 1

37、s orbital on it. Clearly, it doesnt, since a 1s orbital is what an isolated hydrogen atom looks like, not a hydrogen in the hydrogen molecule. Rather, we are just using the 1s orbitals in the hydrogen atom as a convenient guess.Instead, let us return to the hydrogen atom. A feature of the solution t

38、o these equations is that if n atomic orbitals fi go into the calculation, then n molecular orbitals Yj come out. By convention, the molecular orbital Yj has energy Ej. For example, the ground state wavefunction is Y1 and has energy E1.(2) A technical, perhaps controversial detail: In this diagram,

39、it looks like the anti-bonding level goes up by more than the bonding orbital goes down. It turns out this is true only if S12 0. But for orthogonal orbitals, S12=0!This is the idea behind the Hartree-Fock method.If this seems too abstract, lets try the hydrogen molecule as an example. Here is the e

40、nergy diagram:Lecture 1: Bonding and HybridizationE. KwanChem 106Unequal Energy LevelsWhat about something like hydrogen fluoride (HF), where the atoms are of different electronegativities? As you can see from slide 8, the energy of the 1s orbital declines as Z increases. This isnt too surprising, s

41、ince, there are more protons in the nucleus to attract the 1s electrons, which lowers them in energy. Thus, we get a diagram that looks like:Basis SetsA natural question is: what do people use for these atomic orbitals fi in actual computations? The set fi is called the basis set: the collection of

42、all the atomic orbitals. A really common basis set type was developed by Boys and Pople.Instead of using hydrogenlike orbitals as above, they use Gaussian functions of the Cartesian form:= Nxi y j zk exp(-ar2 )energygijkY2 = c12f1 + c22f2W2normalization constanta Gaussianindicates angular momentumH2

43、2f2i + j + k = 0: s-type Gaussian i + j + k = 1: p-type Gaussian i + j + k = 2: d-type Gaussianless electronegative atomfH111Each gijk is a primitive Gaussian. Special linear combinations of primitives are selected to look like hydrogenlike orbitals.These are called contracted Gaussian-type orbitals

44、 (CGTO). These orbitals are used because integrating Gaussians is very easy computationally.Y1 = c11f1 + c21f2W1more electronegative atom, lower in energyThis time, we need four coefficients to describe the molecular orbitals. For example, c12 means the coefficient of atomic orbital 1 in molecular o

45、rbital 2.Lecture 1: Bonding and HybridizationE. KwanChem 106The 6-31g(d) Basis SetThis is used in many, many computational papers (its also called 6-31G*). Lets look at whats inside it. Each atom has different basis orbitals associated with it. For example, each hydrogen in a molecule gets two 1s-ty

46、pe CGTOS:Basis Set Description STO-3G3-21G6-31G6-31G*6-311G*6-311+G*6-311+G*6-311+G(2df,2pd)6-311+G(3df,3pd)cc-pVTZ cc-pVQZaug-cc-pCVQZNumber of Functions2648487290106130226264204400712Rel. Energy (kcal/mol)1577.0728.7108.954.728.826.220.411.40.02x1shydrogens:Carbons, however, have a lot m

47、ore electrons, and so need many more basis orbitals:COREVALENCEcarbons:2But wait! How come carbon has d orbitals on it? All sorts of chemical evidence suggests that carbon does not use 3d orbitals for bonding. So if thats the case, then should we expect to find that ad

48、ding 3d polarization functions does little to improve the variational energy?Clearly, the polarization functions are necessary!Does this mean that the carbon in acetone is using d orbitals? No, because the GTOs are not real orbitals.Lets try it for acetone. Here are the results of a Hartree-Fock cal

49、culation on its equilibrium geometry with different basis sets. The * notation denotes the presence of d orbitals;as you go down the list, bigger basis sets (with correspondingly larger and larger shells of atomic orbitals) are shown.In fact, if you try to ask how many electrons are in GTO X, you ca

50、n actually get numbers that are less than 0 or greater than 2!So whats the problem?Lecture 1: Bonding and HybridizationE. KwanChem 106Constant Orbital SizeCGTOs: for the same element, the same size, regardless of nuclear environmentPhysical Issues with CGTOsNo NodesAccording to the Pauli Principle,

51、two electrons of the same spin cannot occupy the same space.reality: more positively charged atoms attract their electrons more, and have larger orbitals1s and 2s CGTOs for the neutral lithium atom:2s (Li anion)The CGTOs have no nodes, which is unphysical.The real 1s and 2s orbitals (again for a neu

52、tral Li atom)do have nodes (for the precise meaning of real, see below):2s (Li cation)2s (neutral Li)Lecture 1: Bonding and HybridizationE. KwanChem 106Symmetry and Molecular OrbitalsIn general, there will be an interaction between any two orbitals with a non-zero Hamiltonian off-diagonal element:fi

53、 interacts with fj if Hij is nonzeroRegardless of the distance between the hydrogen atoms, this symmetry-adapted representation applies.r = 1 r = 1 kmHHHHan integralbonding and antibonding MOs are reasonable descriptorsthese are isolated H atoms, not bonded to each otherhigh symmetry environmentmany

54、 integrals are zeroIn the low-symmetry environment of organic chemistry, this approach is less useful. With fast modern computers, we are free to consider all possible combinations of Hij and let the form of H take care of the symmetry automatically.Another criticism:Methane and ethane are generally

55、 thought of as quite similar chemically, but their HOMOs look quite different:reduce amount workgroup theory A criticism: Just because Hij can be nonzero, does not mean that it actually is nonzero. The requirement for symmetry adaptation can lead to spurious results. Consider H2:energyY2 = N2 (f1 -

56、f2)LUMOW2f1f2H11W1HOMOSo MOs are not always transferable from one molecule to another. How can we recover the idea of a chemical bond from this approach?Y1 = N1 (f1 + f2)Lecture 1: Bonding and HybridizationE. KwanChem 106Delocalized vs. Localized OrbitalsNatural Atomic OrbitalsOne question you might ask