© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 1 Nature of the Chemical Bond with applications to catalysis, materials.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Lecture 14 February 7, 2011 Reactions O2, Woodward-Hoffmann William A. Goddard, III, wag@wag.caltech.eduwag@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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 3 Bond H to O 2 Bring H toward px on Left O Overlap doubly occupied (  xL ) 2 thus repulsive Overlap singly occupied (  xL ) 2 thus bonding 2 A” state 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 O 2 resonance

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 5 Compare bond energies (kcal/mol) O2 3g-O2 3g- 119.0 HO-O HO-OH51.1 68.2H-O 2 51.5 HOO-H85.2 17.1 67.950.8 Interpretation: OO  bond = 51.1 kcal/mol OO  bond = 119.0-51.1=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 85.2 + 17.1 =102.3 compare CH 3 O-H: HO bond is 105.1

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 6 Bond O 2 to O to form ozone Goddard et al Acc. Chem. Res. 6, 368 (1973) Require two OO  bonds get States with 4, 5, and 6 p  electrons Ground state is 4  case Get S=0,1 but 0 better

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 7 Bond O 2 to O to form ozone New O-O  bond, 51 kcal/mol lose O-O  resonance, 51 kcal/mol Gain O-O  resonance,<17 kcal/mol,assume 2/3 New singlet coupling of  L and  R 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 8 Alternative view of bonding in ozone Start here with 1-3 diradical Transfer electron from central doubly occupied p  pair to the R singly occupied p . Now can form a  bond the L singly occupied p . Hard to estimate strength of bond

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 9 Ring ozone Form 3 OO sigma bonds, but p  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 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 5599

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 11 Ground state of C 2 MO configuration Have two strong  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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 14 Noble gas dimers Ar 2 ReRe DeDe  LJ 12-6 Force Field E=A/R 12 –B/R 6 = D e [   12 – 2   6 ]  D e [   12 –   6 ]  = R/Re  = R/  where  = R e (1/2) 1/6 =0.89 R e No bonding at the VB or MO level Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R 6

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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 )

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 17 MO and VB view of He dimer, He 2 VB view MO view Ψ MO (He 2 ) = A [(  g  )(  g  )(  u  )(  u  )]= (  g ) 2 (  u ) 2 Ψ VB (He 2 ) = A [(  L  )(  L  )(  R  )(  R  )]= (  L ) 2 (  R ) 2 Substitute  g =  R +  L  and  g =  R -  L Get Ψ MO (He 2 ) ≡ Ψ MO (He 2 ) Net BO=0 Pauli  orthog of  R to  L  repulsive

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 18 Remove an electron from He 2 to get He 2 + Ψ(He 2 ) = A [(  g  )(  g  )(  u  )(  u  )]= (  g ) 2 (  u ) 2 Two bonding and two antibonding  BO= 0 Ψ(He 2 + ) = A [(  g  )(  g  )(  u  )]= (  g ) 2 (  u )  BO = ½ Get 2  u + symmetry. Bond energy and bond distance similar to H 2 +, also BO = ½ MO view

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 19 Remove an electron from He 2 to get He 2 + Ψ(He 2 ) = A [(  g  )(  g  )(  u  )(  u  )]= (  g ) 2 (  u ) 2 Two bonding and two antibonding  BO= 0 Ψ(He 2 + ) = A [(  g  )(  g  )(  u  )]= (  g ) 2 (  u )  BO = ½ Get 2  u + symmetry. Bond energy and bond distance similar to H 2 +, also BO = ½ VB view MO view Substitute  g =  R +  L  and  g =  L -  R Get Ψ VB (He 2 ) ≡ A [(  L  )(  L  )(  R  )] - A [(  L  )(  R  )(  R  )] = (  L ) 2 (  R ) - (  R ) 2 (  L ) -

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 20 He 2 + He 2 R e =3.03A D e =0.02 kcal/mol No bond 2u+2u+ 2g+2g+ (g)2(u)(g)2(u) (g)1(u)2(g)1(u)2 - + BO=0.5 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 = 0.5 Check H2 and H2+ numbers MO good for discuss spectroscopy, VB good for discuss chemistry

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 21 Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Certain cycloadditions occur but not others 2 s +2 s 2 s +4 s 4 s +4 s Roald Hoffmann

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 22 Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Certain cyclizations occur but not others conrotatorydisrotatory conrotatory

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 23 2+2 cycloaddition – Orbital correlation diagram ground state GS Forbidden 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 24 2+2 cycloaddition – Orbital correlation diagram excited state ES Allowed 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 25 Consider butadiene + ethene cycloaddition; Diehls-Aldor 2+4 Ground State S S A A 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 27 Summary WH rules cycloaddition 2n + 2mn+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)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 29 Summary WH rules cyclization 2nn odd: thermal disrotatory Photochemical conrotatory n even: Thermal conrotatory Photochemical disrotatory n=2  butadiene n=3  hexatriene

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 31 HF HF_PT2 SVWN B3LYP BLYP XYG3 CCSD(T) SVWN H + CH 4  H 2 + CH 3 Reaction Coordinate: R(CH)-R(HH) (in Å) Energy (kcal/mol) reaction surface of H + CH 4  H 2 + CH 3 along reaction pat

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 33 GVB view reactions Reactant HD+T HDT Product H+DT 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 34 GVB view reactions Reactant HD+T HDT Product H+DT Transition state HDT Bond pair keeps high overlap while flipping from reactant to product nonbond orbital keeps orthogonal, hence changes sign

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 35 GVB analysis of cyclization (4 e case) φAφA φBφB φCφC φDφD 4 VB orbitals: A,B,C,D reactant 1 2 4 3 φAφA φBφB φCφC φDφD 1 2 4 3 Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 Now ask how the CH 2 groups 1 and 4 must rotate so that C and D retain positive overlap. Clearly 4n is conrotatory φAφA φBφB φCφC φDφD φAφA φBφB φCφC φDφD 1 2 4 3

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 37 Apply GVB model to 2 + 2 \ φAφA φBφB φCφC φDφD 4 VB orbitals:A,B,C,D reactant Transition state: ignore C φBφB φAφA φDφD φCφC φBφB φAφA φDφD φCφC 4 VB orbitals product Nodal plane

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 38 Transition state for 2 + 2 Transition state: ignore C φBφB φAφA φDφD φCφC 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 1 2 4 3 43 1 2 1 2 4 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 Nodal plane

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 39 GVB model fast analysis 2 + 2 φAφA φBφB φCφC φDφD 4 VB orbitals:A,B,C,D reactant \ φBφB φAφA φDφD φCφC 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 1 2 4 3 C and D start with positive overlap and end with negative overlap. Thus break bond  forbidden

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 41 GVB 2+4 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φCφC 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 42 GVB 2+4 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φCφC 2. Move EF bond; C changes phase again as it moves from 1 to 5 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 43 GVB 2+4 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φCφC 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 2. Move EF bond; C changes phase again as it moves from 1 to 5 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 45 HF HF_PT2 SVWN B3LYP BLYP XYG3 CCSD(T) SVWN H + CH 4  H 2 + CH 3 Reaction Coordinate: R(CH)-R(HH) (in Å) Energy (kcal/mol) reaction surface of H + CH 4  H 2 + CH 3 along reaction pat

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 46 GVB view reactions Reactant HD+T HDT Product H+DT 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 47 GVB view reactions Reactant HD+T HDT Product H+DT Transition state HDT Bond pair keeps high overlap while flipping from reactant to product nonbond orbital keeps orthogonal, hence changes sign

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 48 GVB analysis of cyclization (4 e case) φAφA φBφB φCφC φDφD 4 VB orbitals: A,B,C,D reactant 1 2 4 3 φAφA φBφB φCφC φDφD 1 2 4 3 Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 Now ask how the CH 2 groups 1 and 4 must rotate so that C and D retain positive overlap. Clearly 4n is conrotatory φAφA φBφB φCφC φDφD φAφA φBφB φCφC φDφD 1 2 4 3

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 49 Apply GVB model to 2 + 2 \ φAφA φBφB φCφC φDφD 4 VB orbitals:A,B,C,D reactant Transition state: ignore C φBφB φAφA φDφD φCφC φBφB φAφA φDφD φCφC 4 VB orbitals product Nodal plane

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 50 Transition state for 2 + 2 Transition state: ignore C φBφB φAφA φDφD φCφC 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 1 2 4 3 43 1 2 1 2 4 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 Nodal plane

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 51 GVB model fast analysis 2 + 2 φAφA φBφB φCφC φDφD 4 VB orbitals:A,B,C,D reactant \ φBφB φAφA φDφD φCφC 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 1 2 4 3 C and D start with positive overlap and end with negative overlap. Thus break bond  forbidden

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 53 GVB 2+4 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φCφC 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 54 GVB 2+4 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φCφC 2. Move EF bond; C changes phase again as it moves from 1 to 5 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 55 GVB 2+4 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φBφB φAφA φDφD 1 2 4 3 φFφF φEφE 56 φAφA φBφB φCφC φDφD 1 2 4 3 φFφF φEφE 56 φCφC 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 2. Move EF bond; C changes phase again as it moves from 1 to 5 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate Reaction Allowed

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 59 More on resonance like structure 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.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 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. 1.5x A 1.34 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 61 graphene This is referred to as 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 62 Graphene band structure Unit cell has 2 carbon atoms Bands: 2p  orbitals per cell  2 bands of states each with N states where N is the number of unit cells 2  electrons per cell  2N electrons for N unit cells The lowest N MOs are doubly occupied, leaving N empty orbitals. 1x1 Unit cell 1 st band 2 nd band The filled 1 st band touches the empty 2 nd band at the Fermi energy Get semi metal

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 63 Graphite Stack graphene layers as ABABAB Can also get ABCABC Rhombohedral AAAA stacking much higher in energy Distance between layers = 3.3545A CC bond = 1.421 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 = 168.6 Thus average in-plane bond = (2/3)168.6 = 112.4 kcal/mol 112.4 = sp 2  + 1/3  Diamond: average CCs = 85 kcal/mol   = 3*27=81 kcal/mol

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 1 Nature of the Chemical Bond with applications to catalysis, materials.

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© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L14 1 Nature of the Chemical Bond with applications to catalysis, materials.

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