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The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell

Molecular Beam Studies of the the Electronic and Nuclear Dynamics of Chemical Reactions: Accessing Radical Intermediates. The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell Arjun Raman Emily Glassman Dr. Xiaonan Tang.

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The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell

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  1. Molecular Beam Studies of the the Electronic and Nuclear Dynamics of Chemical Reactions: Accessing Radical Intermediates The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell Arjun Raman Emily Glassman Dr. Xiaonan Tang National Science Foundation, Chemistry Division Department of Energy, Basic Energy Sciences

  2. Understanding Chemical Reactions:What is the nuclear dynamics during the reaction? (vibration and rotation in the colliding molecules)What is happening to the electrons in the system? (do they adjust instantaneously, or lag behind and cause nonadiabatic suppression of the reaction rate?)How can we get predictive ability from first principle quantum mechanics?How does this change our qualitative understanding of chemical reaction rates and product branchingk(T)=Ae-Ea/kT

  3. We use a combination of state-of-the-art experimental techniques and theoretical analysisMolecular Beam analysis of product velocities and angular distributionsState-selective velocity map imagingElectronic structure calculations of minima and transition states along each reaction coordinate(e.g. G3//B3LYP or CCSD(T) )Analyzing the change in electronic wavefunction along the reaction coordinates.

  4. Many elementary bimolecular reactions proceed through addition/insertion, so go through unstable radical intermediates along the bimolecular reaction coordinate CH3O + CO CH3OCO CH3 + CO2 O + propargyl  products  H2C-C=CH H2C=C=CH  O O Addition mechanism forms or then ??? H2CCCH H2CCCH

  5. Traditional Crossed Molecular Beam Scattering or Imaging Expts are a good way to probe “Direct” Chemical Reactions Eg. D2 + F  D…D…F  D + DF D…D…F D-D  F Angular and Velocity Distribution of DF product shows Backward Scattered DF product

  6. But how can one probe bimolecular reactions that proceed through long-lived radical intermediates? Eg. C2D + HCCHDCCCCH + H Forward/Backward symmetric product angular distributions indicate there is a long-lived intermediate in the reaction. But what is happening along the reaction coordinate? Kaiser et al., PCCP 4, 2950 (2002)

  7. But how can one probe bimolecular reactions that proceed through long-lived radical intermediates? Eg. C2D + HCCHDCCCCH + H Kaiser et al., PCCP 4, 2950 (2002) UB3LYP/6-311+G** + ZPVE

  8. O + propargyl  products  H2C-C=CH H2C=C=CH  O O Addition mechanism forms or then ??? Testing our predictive ability from first principle quantum mechanics H2CCCH H2CCCH

  9. O + H2CCCH H2CCC: + OH HCCCH + OH c-C3H2 + OH INT1 INT2 INT2 O || HC=CCH + H Energy (kcal/mol) (-60.3) O H2C=C=C=O + H H2CCCH O INT2 H2CCCH INT1 Choi’s expts probed only the OH products. His RRKM calcs indicated propynal + H dominates. Choi (CBS-QB3)

  10. O + H2CCCH H2CCC: + OH HCCCH + OH c-C3H2 + OH INT1 INT2 INT2 O || HC=CCH + H Energy (kcal/mol) (-60.3) O H2C=C=C=O + H H2CCCH O INT2 H2CCCH vinyl + CO INT1 Choi (CBS-QB3) + Bowman (UB3LYP) LM2 H2C-CHCO

  11. Our expts produce each radical intermediate photolytically and disperse the radicals by recoil ET and thus by internal energy Eint radical= hn-Do(C-Cl)-ET 193 nm Cl 193 nm nozzle ionization source (electron impact at UofC, tunable VUV at ALS) -30 kV Al doorknob skimmers quadrupole mass spec. Scintillator PMT Measuring the velocities of the stable radicals and the velocities of the products from the unstable radicals can determine the barriers to each product channel and how product channel branching changes with internal energy

  12. C-Cl fission gives H2CCHCO radicals dispersed by internal energy Eint radical= hn + Eint,prec-Do(C-Cl)-ET (81.9)* 193 nm Cl P(ET) Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO LM2 * H2C-CHCO CCSD(T)

  13. C-Cl fission gives H2CCHCO radicals dispersed by internal energy Eint radical= hn + Eint,prec-Do(C-Cl)-ET (81.9) 193 nm Cl P(ET) Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO LM2 * H2C-CHCO CCSD(T)

  14. All the H2CCHCO radicals dissociate to vinyl + CO products * Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO LM2 * H2C-CHCO CCSD(T)

  15. Upper limit to barrier for H2CCHCO vinyl + CO * CCSD(T) UB3LYP Barrier too high? Energy (kcal/mol) (26.7) (23.6) vinyl + CO (25.3) vinyl + CO (20.0) H2C=CHCO H2C=CHCO LM2 * LM2 * H2C-CHCO H2C-CHCO

  16. C-Cl fission at 235 nm produces lower internal energy H2CCHCO radicals Cl 2P3/2 Cl 2P1/2 (Cl*) Eint radical+Cl= hn + Eint,prec-Do(C-Cl)-ET (81.9)* 235 nm Cl Add these two, correcting for 0.85 Cl*/Cl line strength factor (Liyanage) to get total C-Cl fission P(ET) for producing all radicals Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO * LM2 H2C-CHCO CCSD(T)

  17. C-Cl fission at 235 nm produces lower internal energy H2CCHCO radicals Cl 2P3/2 Cl 2P1/2 (Cl*) Eint radical+Cl= hn + Eint,prec-Do(C-Cl)-ET (81.9)* 235 nm Cl Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO * LM2 H2C-CHCO CCSD(T)

  18. Use 157 nm photoionization to detect all STABLE H2CCHCO radicals (157 + 235) - (157 only) Eint radical+Cl= hn + Eprec-Do(C-Cl)-ET (81.9)* 235 nm Cl Lowest internal energy at which theH2CCHCO radicals dissociate is: 121.6+1.5-81.9-18=23 kcal/mol all R + Cl stable R + Cl Energy (kcal/mol) (23.6) vinyl + CO H2C=CHCO * LM2 H2C-CHCO CCSD(T)

  19. CCSD(T) Eint radical+Cl= hn + Eprec-Do(C-Cl)-ET (81.9)* CCSD(T) barrier = 23.6 kcal/mol Expt’l dissociation onset at ET =18 kcal/molgives Expt’l barrier of 23.2 ±2 kcal/mol Is this because the UB3LYP radical energy is too low or the TS energy is too high? UB3LYP Barrier too high. (26.7) (25.3) Energy (kcal/mol) vinyl + CO H2C=CHCO * LM2 H2C-CHCO

  20. CCSD(T) (G3//B3LYP good too) UB3LYP Eint radical+Cl= hn + Eprec-Do(C-Cl)-ET Eint radical+Cl= hn + Eprec-Do(C-Cl)-ET (81.9)* (72.4)* CCSD(T) barrier = 23.6 kcal/mol Expt’l dissociation onset at ET =18 kcal/molgives Expt’l barrier of 23.2 ±2 kcal/mol (26.7) (23.6) Energy (kcal/mol) vinyl + CO (25.3) vinyl + CO H2C=CHCO H2C=CHCO * * LM2 LM2 H2C-CHCO H2C-CHCO

  21. CH3O + CO CH3OCO CH3 + CO2 Bridging physical to organic chemistry ORBITAL INTERACTIONS ALONG THE REACTION COORDINATE

  22. OH + CO  HOCO H + CO2 CH3O· + CO  CH3OCO CH3 + CO2 23.1 k(T,P) product branching falloff behavior JF: Francisco, J. Chem. Phys.237, (1998) 1-9. QCISD(T) BW: Wang, B. et al. JPCA103, (1999) 8021-9. G2(B3LYP/MP2/CC) ZZ: Zhou, Z. et al. Chem. Phys. Lett.353, (2002) 281-9. B3LYP

  23. Cl + CH3OCO* CH3O(CO)Cl 193 Cl + CH3OCO C-Cl fission P(ET ) Einternal of CH3OCO 23.1 Do=85.4 (G3//B3LYP)

  24. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs 280 1 CH3OCO 23.1 Do=85.4 (G3//B3LYP)

  25. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs Expt. branching w. CO/CO2 signal 280 1 CH3OCO CH3OCO 23.1 Do=85.4 (G3//B3LYP)

  26. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs Expt. branching w. CO/CO2 signal 280 1 1 2.5 CH3OCO CH3OCO 23.1 Do=85.4 (G3//B3LYP)

  27. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs Expt. branching w. CO/CO2 signal 280 1 1 2.5 CH3OCO CH3OCO 23.1 Do=85.4 (G3//B3LYP)

  28. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs Expt. branching w. CO/CO2 signal 280 1 2.5 CH3OCO CH3OCO 1 C=O H3C…O 23.1 I asked KC Lau to re-calculate CH3 + CO2 barrier G3//B3LYP and CCSD(T) Do=85.4 (G3//B3LYP)

  29. CH3OCO CH3 + CO2 CH3O + CO RRKM product branching BW TSs Expt. branching w. CO/CO2 signal 280 1 2.5 CH3OCO CH3OCO 1 C=O H3C…O 6.0 (KC) 16.9 (KC) 23.1 -1.6 (KC) O = C -15.6 (KC) -39.1 (KC) H3C…O Glaude, Pitz, Thomson 2005 Good and Francisco 2000 Do=85.4 (G3//B3LYP)

  30. Average RRKM product branching over internal energies in our expt. CH3OCO CH3 + CO2 CH3O + CO EXPT. 2.5 ± 0.5 CH3OCO 1 32.5 CH3O + CO CH3 + CO2

  31. Average RRKM product branching Over internal energies in our expt. CH3OCO CH3 + CO2 CH3O + CO EXPT. 2.5 ± 0.5 CH3OCO 1 PRED. 280 1 CH3OCO CH3O + CO CH3O + CO CH3 + CO2 CH3 + CO2

  32. Average RRKM product branching Over internal energies in our expt. CH3OCO CH3 + CO2 CH3O + CO EXPT. 2.5 ± 0.5 CH3OCO 1 CH3O + CO CH3O + CO CH3 + CO2 CH3O + CO CH3 + CO2 CH3 + CO2

  33. Average RRKM product branching Over internal energies in our expt. CH3OCO CH3 + CO2 CH3O + CO EXPT. 2.5 ± 0.5 CH3OCO 1 PRED. 1 2.1 CH3OCO CH3O + CO CH3O + CO CH3 + CO2 CH3O + CO CH3 + CO2 CH3 + CO2

  34. Why is the cis barrier so much lower than the trans one? 32.5 (34.2) CH3O + CO (14.5) O = cis barrier is ~20 kcal/mol lower than trans (CCSD(T)) C CH3 + CO2 H3C … O O = C cis barrier is ~7 kcal/mol lower than trans Muckerman, FCC/CBS (2001) H … O

  35. Why is the cis barrier so much lower than the trans one? 32.5 (34.2) CH3O + CO (14.5) Think about the interaction between the radical orbital and the H3C-OCO antibonding orbital s*C-O CH3 + CO2 O = . C nC H3C … O Radical energy lowers due to interaction with s*C-O orbital as H3C-OCO bond stretches

  36. Natural Bond Orbital analysis with Weinhold

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