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

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, wag@wag.caltech.edu 316 Beckman Institute, x3093

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

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  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, wag@wag.caltech.edu 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 <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>

  2. Last time

  3. Bond H to O2 Bring H toward px on Left O Overlap doubly occupied (pxL)2 thus repulsive Overlap singly occupied (pxL)2 thus bonding Get HOO bond angle ~ 90º S=1/2 (doublet) Antisymmetric with respect to plane: A” irreducible representation (Cs group) Bond weakened by ~ 51 kcal/mol due to loss in O2 resonance 2A” state

  4. Bond 2nd H to HO2 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

  5. Compare bond energies (kcal/mol) O23Sg- 119.0 50.8 67.9 HO-O 68.2 H-O2 51.5 17.1 HO-OH 51.1 HOO-H 85.2 Interpretation: OO s bond = 51.1 kcal/mol OO p bond = 119.0-51.1=67.9 kcal/mol (resonance) Bonding H to O2 loses 50.8 kcal/mol of resonance Bonding H to HO2 loses the other 17.1 kcal/mol of resonance Intrinsic H-O bond is 85.2 + 17.1 =102.3 compare CH3O-H: HO bond is 105.1

  6. Bond O2 to O to form ozone Require two OO s bonds get States with 4, 5, and 6 pp electrons Ground state is 4p case Get S=0,1 but 0 better Goddard et al Acc. Chem. Res. 6, 368 (1973)

  7. Bond O2 to O to form ozone lose O-O p resonance, 51 kcal/mol New O-O s bond, 51 kcal/mol Gain O-Op resonance,<17 kcal/mol,assume 2/3 New singlet coupling of pL and pR orbitals Total splitting ~ 1 eV = 23 kcal/mol, assume ½ stabilizes singlet and ½ destabilizes triplet Expect bond for singlet of 11 + 12 = 23 kcal/mol, exper = 25 Expect triplet state to be bound by 11-12 = -1 kcal/mol, probably between +2 and -2

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

  9. Ring ozone Form 3 OO sigma bonds, but pp pairs overlap Analog: cis HOOH bond is 51.1-7.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2. 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 5599

  10. More on N2 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

  11. Ground state of C2 MO configuration Have two strong p bonds, but sigma system looks just like Be2 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

  12. Low-lying states of C2

  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 = NAkB NA = Avogadro constant; kB = Boltzmann constant Van der Waals equation of state (1873) [p + n2a/V2)[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)

  14. Noble gas dimers No bonding at the VB or MO level Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R6 • LJ 12-6 Force Field • E=A/R12 –B/R6 • = De[r-12 – 2r-6] • = 4 De[t-12 – t-6] • = R/Re • = R/s where s = Re(1/2)1/6 =0.89 Re Ar2 s Re De

  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/R3 , but since the average dipole is zero the first nonzero contribution is from 2nd order perturbation theory, which scales like -C/R6 (with higher order terms like 1/R8 and 1/R10)

  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/R3 , but since the average dipole is zero the first nonzero contribution is from 2nd order perturbation theory, which scales like -C/R6 (with higher order terms like 1/R8 and 1/R10) Consequently it is common to fit the interaction potentials to functional forms with a long range 1/R6 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)

  17. MO and VB view of He dimer, He2 MO view ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2 VB view ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2 Net BO=0 Pauli  orthog of R to L  repulsive Substitute sg = R +Land sg = R -L Get ΨMO(He2) ≡ ΨMO(He2)

  18. Remove an electron from He2 to getHe2+ MO view Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2 Two bonding and two antibonding  BO= 0 Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½ Get 2Su+ symmetry. Bond energy and bond distance similar to H2+, also BO = ½

  19. Remove an electron from He2 to getHe2+ - MO view Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2 Two bonding and two antibonding  BO= 0 Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½ Get 2Su+ symmetry. Bond energy and bond distance similar to H2+, also BO = ½ VB view Substitute sg = R +Land sg = L -R Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)] = (L)2(R) - (R)2(L)

  20. He2+ 2Sg+ (sg)1(su)2 2Su+ (sg)2(su) BO=0.5 - + He2 Re=3.03A De=0.02 kcal/mol No bond H2 Re=0.74xA De=110.x kcal/mol BO = 1.0 MO good for discuss spectroscopy, VB good for discuss chemistry H2+ Re=1.06x A De=60.x kcal/mol BO = 0.5 Check H2 and H2+ numbers

  21. Woodward-Hoffmann rulesorbital symmetry rulesFrontier Orbital rules Roald Hoffmann Certain cycloadditions occur but not others 2s+2s 2s+4s 4s+4s

  22. Woodward-Hoffmann rulesorbital symmetry rulesFrontier Orbital rules Certain cyclizations occur but not others conrotatory disrotatory disrotatory conrotatory

  23. 2+2 cycloaddition – Orbital correlation diagramground state 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 Forbidden

  24. 2+2 cycloaddition – Orbital correlation diagramexcited 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 Allowed ES

  25. Consider butadiene + ethene cycloaddition; Diehls-Aldor2+4 Ground State A A S A S S Ground state has S, S, and A occupied Product has S, A, and S occupied Thus transition state need not be high Allowed reaction Allowed

  26. WH rules – 2 + 4Excited State A A S Forbidden A S S

  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)

  28. S WH rules – cyclization-GS A A A S A A S Forbidden Allowed A S S A S A S S Rotation, C2 Reflection, s

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

  30. 2D Reaction Surface for H + CH4 H2 + CH3 Product: H2+CH3 H--C Reactant: H+CH4 H--H

  31. reaction surface of H + CH4 H2 + CH3 along reaction pat H + CH4 H2 + CH3 HF CCSD(T) XYG3 Energy (kcal/mol) SVWN HF_PT2 B3LYP BLYP SVWN Reaction Coordinate: R(CH)-R(HH) (in Å)

  32. GVB view reactions Reactant HD+T H D T Product H+DT Goddard and Ladner, JACS 93 6750 (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

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

  35. GVB analysis of cyclization (4 e case) Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 4 VB orbitals: A,B,C,D reactant φB φA φB φA φB φC 2 3 4 1 φD φC φA φD φC 2 3 φD 4 1 φB φA 2 3 Now ask how the CH2 groups 1 and 4 must rotate so that C and D retain positive overlap. Clearly 4n is conrotatory φC φD 1 4

  36. GVB analysis of cyclization (6 e case) φB φC φD φA 3 2 4 1

  37. Apply GVB model to 2 + 2 4 VB orbitals:A,B,C,D reactant Transition state: ignore C φB φA φA φB φC φD φD φB φD Nodal plane 4 VB orbitals product φA φC \ φC

  38. Transition state for 2 + 2 2 1 3 4 2 1 3 4 Transition state: ignore C 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 φB 2 1 φA 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. 3 4 φD Nodal plane φC Reaction Forbidden

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

  40. Next examine 2+4

  41. GVB 2+4 φC φB φD φA 2 3 1 4 6 5 φE φF φA φB φD φC 2 3 1 4 6 5 φE φF 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1

  42. GVB 2+4 2. Move EF bond; C changes phase again as it moves from 1 to 5 φA φB φD φC 2 3 1 4 φA φB φD 2 3 6 5 1 4 φE φF φE φC 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate 6 5 φF Reaction Allowed

  43. GVB 2+4 2. Move EF bond; C changes phase again as it moves from 1 to 5 φC φB φD φA 2 3 1 4 6 5 φE φF φA φB φD φC 2 3 1 4 φA φB φD 2 3 6 5 1 4 φE φF 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate φE 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 φC 6 5 φF Reaction Allowed

  44. 2D Reaction Surface for H + CH4 H2 + CH3 Product: H2+CH3 H--C Reactant: H+CH4 H--H

  45. reaction surface of H + CH4 H2 + CH3 along reaction pat H + CH4 H2 + CH3 HF CCSD(T) XYG3 Energy (kcal/mol) SVWN HF_PT2 B3LYP BLYP SVWN Reaction Coordinate: R(CH)-R(HH) (in Å)

  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

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

  48. GVB analysis of cyclization (4 e case) Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 4 VB orbitals: A,B,C,D reactant φB φA φB φA φB φC 2 3 4 1 φD φC φA φD φC 2 3 φD 4 1 φB φA 2 3 Now ask how the CH2 groups 1 and 4 must rotate so that C and D retain positive overlap. Clearly 4n is conrotatory φC φD 1 4

  49. Apply GVB model to 2 + 2 4 VB orbitals:A,B,C,D reactant Transition state: ignore C φB φA φA φB φC φD φD φB φD Nodal plane 4 VB orbitals product φA φC \ φC

  50. Transition state for 2 + 2 2 1 3 4 2 1 3 4 Transition state: ignore C 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 φB 2 1 φA 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. 3 4 φD Nodal plane φC Reaction Forbidden

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