Lecture 14 February 7, 2011 Reactions O2, Woodward-Hoffmann

Transcription

1 Lecture 14 February 7, 2011 Reactions O2, Woodward-Hoffmann Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Wei-Guang Liu Caitlin Scott 1

2 Last time 2

3 Bring H toward px on Left O Bond H to O 2 2 A state Overlap doubly occupied (π xl ) 2 thus repulsive Get HOO bond angle ~ 90º S=1/2 (doublet) Overlap singly occupied (π xl ) 2 thus bonding Antisymmetric with respect to plane: A irreducible representation (Cs group) Bond weakened by ~ 51 kcal/mol due to loss in O 2 resonance 3

4 Bond 2 nd H to HO 2 to form hydrogen peroxide Bring H toward py on right O Expect new HOO bond angle ~ 90º Expect HOOH dihedral ~90º Indeed H-S-S-H: HSS = 91.3º and HSSH= 90.6º But H-H overlap leads to steric effects for HOOH, net result: HOO opens up to ~94.8º HOOH angle 111.5º trans structure, 180º only 1.2 kcal/mol higher 4

5 Compare bond energies (kcal/mol) O 3 2 Σ - g HO-O H-O HO-OH 51.1 HOO-H 85.2 Interpretation: OO σ bond = 51.1 kcal/mol OO π bond = =67.9 kcal/mol (resonance) Bonding H to O 2 loses 50.8 kcal/mol of resonance Bonding H to HO 2 loses the other 17.1 kcal/mol of resonance Intrinsic H-O bond is =102.3 compare CH 3 O-H: HO bond is

6 Bond O 2 to O to form ozone Require two OO σ bonds get States with 4, 5, and 6 pπ electrons Ground state is 4π case Get S=0,1 but 0 better Goddard et al Acc. Chem. Res. 6, 368 (1973) 6

7 Bond O 2 to O to form ozone lose O-O π resonance, 51 kcal/mol New O-O σ bond, 51 kcal/mol Gain O-Oπ resonance,<17 kcal/mol,assume 2/3 New singlet coupling of π L and π R orbitals Total splitting ~ 1 ev = 23 kcal/mol, assume ½ stabilizes singlet and ½ destabilizes triplet Expect bond for singlet of = 23 kcal/mol, exper = 25 Expect triplet state to be bound by = -1 kcal/mol, probably between +2 and -2 7

8 Alternative view of bonding in ozone Start here with 1-3 diradical Transfer electron from central doubly occupied pπ pair to the R singly occupied pπ. Now can form a π bond the L singly occupied pπ. Hard to estimate strength of bond 8

9 Ring ozone Form 3 OO sigma bonds, but pπ pairs overlap Analog: cis HOOH bond is =43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O 2. Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane. Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol, But if formed it might be rather stable with respect various chemical reactions. Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN

10 More on N 2 The elements N, P, As, Sb, and Bi all have an (ns)2(np)3 configuration, leading to a triple bond Adding in the (ns) pairs, we show the wavefunction as This is the VB description of N2, P2, etc. The optimum orbitals of N2 are shown on the next slide. The MO description of N2 is Which we can draw as 10

11 Ground state of C 2 MO configuration Have two strong π bonds, but sigma system looks just like Be 2 which leads to a bond of ~ 1 kcal/mol The lobe pair on each Be is activated to form the sigma bond. The net result is no net contribution to bond from sigma electrons. It is as if we started with HCCH and cut off the Hs 11

12 Low-lying states of C2 12

13 Van der Waals interactions For an ideal gas the equation of state is given by pv =nrt where p = pressure; V = volume of the container n = number of moles; R = gas constant = N A k B N A = Avogadro constant; k B = Boltzmann constant Van der Waals equation of state (1873) [p + n 2 a/v 2 )[V - nb] = nrt Where a is related to attractions between the particles, (reducing the pressure) And b is related to a reduced available volume (due to finite size of particles) 13

14 Noble gas dimers No bonding at the VB or MO level Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R 6 Ar 2 σ R e D e LJ 12-6 Force Field E=A/R 12 B/R 6 = D e [ρ 12 2ρ 6 ] = 4 D e [τ 12 τ 6 ] ρ= R/Re τ= R/σ where σ = R e (1/2) 1/6 =0.89 R e 14

15 London Dispersion The weak binding in He2 and other noble gas dimers was explained in terms of QM by Fritz London in 1930 The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R 3, but since the average dipole is zero the first nonzero contribution is from 2 nd order perturbation theory, which scales like -C/R 6 (with higher order terms like 1/R 8 and 1/R 10 ) 15

16 London Dispersion The weak binding in He2 and other nobel gas dimers was explained in terms of QM by Fritz London in 1930 The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R 3, but since the average dipole is zero the first nonzero contribution is from 2 nd order perturbation theory, which scales like -C/R 6 (with higher order terms like 1/R 8 and 1/R 10 ) Consequently it is common to fit the interaction potentials to functional forms with a long range 1/R 6 attraction to account for London dispersion (usually referred to as van der Waals attraction) plus a short range repulsive term to account for short Range Pauli Repulsion) 16

17 MO and VB view of He dimer, He 2 MO view Ψ MO (He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 VB view Net BO=0 Ψ VB (He 2 ) = A[(χ L α)(χ L β)(χ R α)(χ R β)]= (χ L ) 2 (χ R ) 2 Substitute σ g = χ R + χ L and σ g = χ R - χ L Get Ψ MO (He 2 ) Ψ MO (He 2 ) Pauli orthog of χ R to χ L repulsive 17

18 Remove an electron from He 2 to get He 2 + MO view Ψ(He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 Two bonding and two antibonding BO= 0 Ψ(He 2+ ) = A[(σ g α)(σ g β)(σ u α)]= (σ g ) 2 (σ u ) BO = ½ Get 2 Σ u+ symmetry. Bond energy and bond distance similar to H 2+, also BO = ½ 18

19 MO view Remove an electron from He 2 to get He 2 + Ψ(He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 Two bonding and two antibonding BO= 0 Ψ(He 2+ ) = A[(σ g α)(σ g β)(σ u α)]= (σ g ) 2 (σ u ) BO = ½ Get 2 Σ u+ symmetry. Bond energy and bond distance similar to H 2+, also BO = ½ VB view Substitute σ g = χ R + χ L and σ g = χ L - χ R Get Ψ VB (He 2 ) A[(χ L α)(χ L β)(χ R α)] - A[(χ L α)(χ R β)(χ R α)] = (χ L ) 2 (χ R ) - (χ R ) 2 (χ L ) - 19

20 He Σ g + (σ g ) 1 (σ u ) 2 2 Σ u + (σ g ) 2 (σ u ) - BO=0.5 MO good for discuss spectroscopy, VB good for discuss chemistry Check H2 and H2+ numbers He 2 R e =3.03A D e =0.02 kcal/mol No bond H 2 R e =0.74xA D e =110.x kcal/mol BO = 1.0 H 2+ R e =1.06x A D e =60.x kcal/mol BO =

21 Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Certain cycloadditions occur but not others Roald Hoffmann 2 s +2 s 2 s +4 s 4 s +4 s 21

22 Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Certain cyclizations occur but not others conrotatory disrotatory disrotatory conrotatory 22

23 2+2 cycloaddition Orbital correlation diagram ground state Forbidden GS Start with 2 ethene in GS Occupied orbitals have SS and SA symmetries Now examine product cyclobutane Occupied orbitals have SS and AS symmetry Thus must have high energy transition state: forbidden reactions 23

24 2+2 cycloaddition Orbital correlation diagram excited state Start with 1 ethene in GS and one in ES Open shell orbitals have SA and AS symmetries Now examine product cyclobutane Open shell orbitals have AS and SA symmetry Thus orbitals of reactant correlate with those of product Thus photochemical reaction allowed ES Allowed 24

25 Consider butadiene + ethene cycloaddition; Diehls-Aldor 2+4 Ground State A A S A S Allowed Ground state has S, S, and A occupied Product has S, A, and S occupied Thus transition state need not be high Allowed reaction S 25

26 WH rules Excited State A A S Forbidden A S S 26

27 Summary WH rules cycloaddition 2n + 2m n+m odd: Thermal allowed Photochemical forbidden n+m even: Thermal forbidden Photochemical allowed n=1, m=1: ethene + ethene n=1, m=2: ethene + butadience (Diels-Alder) 27

28 A S Allowed A S S A S A WH rules cyclization-gs A A Forbidden S S A S A S Rotation, C 2 Reflection, σ 28

29 Summary WH rules cyclization 2n n odd: thermal disrotatory Photochemical conrotatory n even: Thermal conrotatory Photochemical disrotatory n=2 butadiene n=3 hexatriene 29

30 2D Reaction Surface for H + CH 4 H 2 + CH 3 Product: H 2 +CH 3 H--C Reactant: H+CH 4 H--H 30

31 reaction surface of H + CH 4 H 2 + CH 3 along reaction pat H + CH 4 H 2 + CH 3 HF HF_PT2 HF Energy (kcal/mol) XYG3 CCSD(T) B3LYP BLYP SVWN CCSD(T) XYG3 B3LYP HF_PT2 SVWN 5.00 BLYP SVWN Reaction Coordinate: Reaction R(CH)-R(HH) coordinate (in Å) 31

32 GVB view reactions Reactant HD+T H D T Product H+DT 32 Goddard and Ladner, JACS (1971)

33 GVB view reactions Reactant HD+T H D T During reaction, bonding orbital on D stays on D, Bonding orbital on H keeps its overlap with the orbital on D but delocalizes over H and T in the TS and localizes on T in the product. Thus highly overlapping bond for whole reaction Nonbonding Orbital on free T of reactant becomes partially antibonding in TS and localizes on free H of product, but it changes sign Product H+DT 33

34 GVB view reactions Reactant HD+T H D T Transition state Bond pair keeps high overlap while flipping from reactant to product nonbond orbital keeps orthogonal, hence changes sign Product H+DT H D T 34

35 GVB analysis of cyclization (4 e case) 4 VB orbitals: A,B,C,D reactant φ B φ C φ D Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 1 φ φ B B 2 3 φ φ A A 4 φ A φ φ C C φ φ D D φ B φ A Now ask how the CH 2 groups 1 and 4 must rotate so that C and D retain positive overlap. φ C 2 3 φ D Clearly 4n is conrotatory

36 GVB analysis of cyclization (6 e case) φ B φ C φ A φ D 4 36

37 Apply GVB model to VB orbitals:a,b,c,d reactant Transition state: ignore C φ B φ A φ B φ A φ C φ D φ D φ B φ D φ C φ A \ 4 VB orbitals product φ C Nodal plane 37

38 Transition state for Orbitals A on 1 and B on 2 keep high overlap as the bond moves from 12 to 23 with B staying on 2 and A moving from 1 to 3 Orbital D must move from 3 to 1 but must remain orthogonal to the AB bond. Thus it gets a nodal plane The overlap of D and C goes from positive in reactant to negative in product, hence going through 0. thus break CD bond. Reaction Forbidden φ A φ C Transition state: ignore C φ D Nodal plane φ B 38

39 GVB model fast analysis VB orbitals:a,b,c,d reactant 1 2 Move A from 1 to 3 keeping overlap with B φ A φ C 4 3 φ B φ D Simultaneously D moves from 3 to 1 but must change sign since must remain orthogonal to A and B C and D start with positive overlap and end with negative overlap. Thus break bond forbidden φ D φ C φ B φ A \ 39

40 Next examine

41 GVB 2+4 φ A 1 φ B φ C φd φ C 1 φ B φ A φd φ F 6 5 φ E φ F 6 5 φ E 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 41

42 GVB 2+4 φb φ C φ A 4 φ D 2. Move EF bond; C changes phase again as it moves from 1 to 5 φ B φ A φ F 6 5 φ E φ D 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed φ E φ F 6 5 φ C 42

43 φ A 1 φ B φ C φd GVB 2+4 φ C φ B φ A φd Move EF bond; C changes phase again as it moves from 1 to 5 φ B φ A φ F 6 5 φ E 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 φ F 6 φ E 5 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate 1 φ E φ F φ C Reaction Allowed φ D 43

44 2D Reaction Surface for H + CH 4 H 2 + CH 3 Product: H 2 +CH 3 H--C Reactant: H+CH 4 H--H 44

45 reaction surface of H + CH 4 H 2 + CH 3 along reaction pat H + CH 4 H 2 + CH 3 HF HF_PT2 HF Energy (kcal/mol) XYG3 CCSD(T) B3LYP BLYP SVWN CCSD(T) XYG3 B3LYP HF_PT2 SVWN 5.00 BLYP SVWN Reaction Coordinate: Reaction R(CH)-R(HH) coordinate (in Å) 45

46 GVB view reactions Reactant HD+T H D T During reaction, bonding orbital on D stays on D, Bonding orbital on H keeps its overlap with the orbital on D but delocalizes over H and T in the TS and localizes on T in the product. Thus highly overlapping bond for whole reaction Nonbonding Orbital on free T of reactant becomes partially antibonding in TS and localizes on free H of product, but it changes sign Product H+DT 46

47 GVB view reactions Reactant HD+T H D T Transition state Bond pair keeps high overlap while flipping from reactant to product nonbond orbital keeps orthogonal, hence changes sign Product H+DT H D T 47

48 GVB analysis of cyclization (4 e case) 4 VB orbitals: A,B,C,D reactant φ B φ C φ D Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 1 φ φ B B 2 3 φ φ A A 4 φ A φ φ C C φ φ D D φ B φ A Now ask how the CH 2 groups 1 and 4 must rotate so that C and D retain positive overlap. φ C 2 3 φ D Clearly 4n is conrotatory

49 Apply GVB model to VB orbitals:a,b,c,d reactant Transition state: ignore C φ B φ A φ B φ A φ C φ D φ D φ B φ D φ C φ A \ 4 VB orbitals product φ C Nodal plane 49

50 Transition state for Orbitals A on 1 and B on 2 keep high overlap as the bond moves from 12 to 23 with B staying on 2 and A moving from 1 to 3 Orbital D must move from 3 to 1 but must remain orthogonal to the AB bond. Thus it gets a nodal plane The overlap of D and C goes from positive in reactant to negative in product, hence going through 0. thus break CD bond. Reaction Forbidden φ A φ C Transition state: ignore C φ D Nodal plane φ B 50

51 GVB model fast analysis VB orbitals:a,b,c,d reactant 1 2 Move A from 1 to 3 keeping overlap with B φ A φ C 4 3 φ B φ D Simultaneously D moves from 3 to 1 but must change sign since must remain orthogonal to A and B C and D start with positive overlap and end with negative overlap. Thus break bond forbidden φ D φ C φ B φ A \ 51

52 Next examine

53 GVB 2+4 φ A 1 φ B φ C φd φ C 1 φ B φ A φd φ F 6 5 φ E φ F 6 5 φ E 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 53

54 GVB 2+4 φb φ C φ A 4 φ D 2. Move EF bond; C changes phase again as it moves from 1 to 5 φ B φ A φ F 6 5 φ E φ D 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed φ E φ F 6 5 φ C 54

55 φ A 1 φ B φ C φd GVB 2+4 φ C φ B φ A φd Move EF bond; C changes phase again as it moves from 1 to 5 φ B φ A φ F 6 5 φ E 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 φ F 6 φ E 5 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate 1 φ E φ F φ C Reaction Allowed φ D 55

56 Benzene and Resonance referred to as Kekule or VB structures 56

57 Resonance 57

58 Benzene wavefunction is a superposition of the VB structures in (2) benzene as + 58

59 More on resonance That benzene would have a regular 6-fold symmetry is not obvious. Each VB spin coupling would prefer to have the double bonds at ~1.34A and the single bond at ~1.47 A (as the central bond in butadiene) Thus there is a cost to distorting the structure to have equal bond distances of 1.40A. However for the equal bond distances, there is a resonance stabilization that exceeds the cost of distorting the structure, leading to D 6h symmetry. 59

60 Cyclobutadiene For cyclobutadiene, we have the same situation, but here the rectangular structure is more stable than the square. That is, the resonance energy does not balance the cost of making the bond distances equal A 1.5x A The reason is that the pi bonds must be orthogonalized, forcing a nodal plane through the adjacent C atoms, causing the energy to increase dramatically as the 1.54 distance is reduced to 1.40A. For benzene only one nodal plane makes the pi bond orthogonal to both other bonds, leading to lower cost 60

61 graphene Graphene: CC=1.4210A Bond order = 4/3 Benzene: CC=1.40 BO=3/2 Ethylene: CC=1.34 BO = 2 CCC=120 Unit cell has 2 carbon atoms 1x1 Unit cell This is referred to as graphene 61

62 1x1 Unit cell Graphene band structure Unit cell has 2 carbon atoms Bands: 2pπ orbitals per cell 2 bands of states each with N states where N is the number of unit cells 2 π electrons per cell 2N electrons for N unit cells The lowest N MOs are doubly occupied, leaving N empty orbitals. The filled 1 st band touches the empty 2 nd band at the Fermi energy Get semi metal 2 nd band 1 st band 62

63 Graphite Stack graphene layers as ABABAB Can also get ABCABC Rhombohedral AAAA stacking much higher in energy Distance between layers = A CC bond = Only weak London dispersion attraction between layers D e = 1.0 kcal/mol C Easy to slide layers, good lubricant Graphite: D 0K =169.6 kcal/mol, in plane bond = Thus average in-plane bond = (2/3)168.6 = kcal/mol = sp 2 σ + 1/3 π Diamond: average CCs = 85 kcal/mol π = 3*27=81 kcal/mol 63

64 energetics 64

65 Allyl Radical 65

66 Allyl wavefunctions It is about 12 kcal/mol 66

67 stop 67