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The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell Arjun Raman Emily Glassman Dr. Xiaonan Tang Molecular Beam.

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Presentation on theme: "The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell Arjun Raman Emily Glassman Dr. Xiaonan Tang Molecular Beam."— Presentation transcript:

1 The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell Arjun Raman Emily Glassman Dr. Xiaonan Tang Molecular Beam Studies of the the Electronic and Nuclear Dynamics of Chemical Reactions: Accessing Radical Intermediates 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 branching k(T)=Ae -E a /kT

3 We use a combination of state-of-the-art experimental techniques and theoretical analysis Molecular Beam analysis of product velocities and angular distributions State-selective velocity map imaging Electronic 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 O + propargyl  products H 2 C=C=CH   H 2 C-C=CH Addition mechanism forms or then ??? H 2 CCCH O O Many elementary bimolecular reactions proceed through addition/insertion, so go through unstable radical intermediates along the bimolecular reaction coordinate CH 3 O + CO  CH 3 OCO  CH 3 + CO 2

5 Traditional Crossed Molecular Beam Scattering or Imaging Expts are a good way to probe “Direct” Chemical Reactions Angular and Velocity Distribution of DF product shows Backward Scattered DF product Eg. D 2 + F  D … D … F  D + DF D…D…FD…D…F D-D  F

6 But how can one probe bimolecular reactions that proceed through long-lived radical intermediates? Eg. C 2 D + 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. C 2 D + HCCH  DCCCCH + H Kaiser et al., PCCP 4, 2950 (2002) UB3LYP/6-311+G** + ZPVE

8 O + propargyl  products H 2 C=C=CH   H 2 C-C=CH Addition mechanism forms or then ??? Testing our predictive ability from first principle quantum mechanics H 2 CCCH O O

9 H 2 C=C=C=O + H c-C 3 H 2 + OH H 2 CCC: + OH HCCCH + OH O || HC=CCH + H O + H 2 CCCH Energy (kcal/mol) Choi (CBS-QB3) INT2 INT1 INT2 (-60.3) H 2 CCCH O O His RRKM calcs indicated propynal + H dominates. Choi’s expts probed only the OH products.

10 H 2 C=C=C=O + H c-C 3 H 2 + OH H 2 CCC: + OH HCCCH + OH O || HC=CCH + H vinyl + CO O + H 2 CCCH Energy (kcal/mol) Choi (CBS-QB3) + Bowman (UB3LYP) INT2 INT1 INT2 LM2 (-60.3) H 2 CCCH O O H 2 C-CHCO

11

12 193 nm nozzle skimmers ionization source (electron impact at UofC, tunable VUV at ALS) -30 kV Al doorknob quadrupole mass spec. Scintillator PMT E int radical = h  D o (C-Cl)-E T Our expts produce each radical intermediate photolytically and disperse the radicals by recoil E T and thus by internal energy 193 nm Cl 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

13 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) C-Cl fission gives H 2 CCHCO radicals dispersed by internal energy E int radical = h + E int,prec -D o (C-Cl)-E T (81.9)* 193 nm Cl P(E T ) CCSD(T) *

14 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) C-Cl fission gives H 2 CCHCO radicals dispersed by internal energy E int radical = h + E int,prec -D o (C-Cl)-E T (81.9) 193 nm Cl P(E T ) CCSD(T) *

15 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) All the H 2 CCHCO radicals dissociate to vinyl + CO products CCSD(T) * *

16 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) CCSD(T) vinyl + CO LM2 H 2 C-CHCO H 2 C=CHCO (26.7) UB3LYP Barrier too high? * * * (25.3) (20.0) Upper limit to barrier for H 2 CCHCO vinyl + CO

17

18 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) C-Cl fission at 235 nm produces lower internal energy H 2 CCHCO radicals E int radical+Cl = h + E int,prec -D o (C-Cl)-E T (81.9)* 235 nm Cl CCSD(T) * Cl 2 P 3/2 Cl 2 P 1/2 (Cl*) Add these two, correcting for 0.85 Cl*/Cl line strength factor (Liyanage) to get total C-Cl fission P(E T ) for producing all radicals

19 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) C-Cl fission at 235 nm produces lower internal energy H 2 CCHCO radicals E int radical+Cl = h + E int,prec -D o (C-Cl)-E T (81.9)* 235 nm Cl CCSD(T) * Cl 2 P 3/2 Cl 2 P 1/2 (Cl*)

20 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) Use 157 nm photoionization to detect all STABLE H 2 CCHCO radicals E int radical+Cl = h + E prec -D o (C-Cl)-E T (81.9)* 235 nm CCSD(T) * (157 + 235) - (157 only) Cl Lowest internal energy at which the H 2 CCHCO radicals dissociate is: 121.6+1.5-81.9-18=23 kcal/mol all R + Cl stable R + Cl

21 vinyl + CO Energy (kcal/mol) LM2 H 2 C=CHCO E int radical+Cl = h + E prec -D o (C-Cl)-E T (81.9)* CCSD(T) * CCSD(T) barrier = 23.6 kcal/mol (26.7) UB3LYP Barrier too high. (25.3) Expt’l dissociation onset at E T =18 kcal/mol gives Expt’l barrier of 23.2 ±2 kcal/mol H 2 C-CHCO Is this because the UB3LYP radical energy is too low or the TS energy is too high?

22 vinyl + CO Energy (kcal/mol) LM2 H 2 C-CHCO H 2 C=CHCO (23.6) E int radical+Cl = h + E prec -D o (C-Cl)-E T (81.9)* CCSD(T) (G3//B3LYP good too) * CCSD(T) barrier = 23.6 kcal/mol Expt’l dissociation onset at E T =18 kcal/mol gives Expt’l barrier of 23.2 ±2 kcal/mol vinyl + CO LM2 H 2 C=CHCO * (26.7) (25.3) H 2 C-CHCO E int radical+Cl = h + E prec -D o (C-Cl)-E T UB3LYP (72.4)*

23 CH 3 O + CO  CH 3 OCO  CH 3 + CO 2 Bridging physical to organic chemistry ORBITAL INTERACTIONS ALONG THE REACTION COORDINATE

24 CH 3 O· + CO  CH 3 OCO  CH 3 + CO 2 JF: Francisco, J. Chem. Phys. 237, (1998) 1-9. QCISD(T) BW: Wang, B. et al. JPCA 103, (1999) 8021-9. G2(B3LYP/MP2/CC) ZZ: Zhou, Z. et al. Chem. Phys. Lett. 353, (2002) 281-9. B3LYP OH + CO  HOCO  H + CO 2 k(T,P) product branching falloff behavior 23.1

25 CH 3 O(CO)Cl 193 Cl + CH 3 OCO Cl + CH 3 OCO* C-Cl fission P( E T ) D o =85.4 (G3//B3LYP) E internal of CH 3 OCO 23.1

26 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 RRKM product branching BW TSs CH 3 OCO 1280 23.1

27 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 RRKM product branching BW TSs CH 3 OCO 1280 CH 3 OCO Expt. branching w. CO/CO 2 signal 23.1

28 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 RRKM product branching BW TSs CH 3 OCO 1280 CH 3 OCO 1 Expt. branching w. CO/CO 2 signal 2.5 23.1

29 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 RRKM product branching BW TSs CH 3 OCO 1280 CH 3 OCO 1 Expt. branching w. CO/CO 2 signal 2.5 23.1

30 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 RRKM product branching BW TSs CH 3 OCO 1280 CH 3 OCO 1 I asked KC Lau to re-calculate CH 3 + CO 2 barrier G3//B3LYP and CCSD(T) Expt. branching w. CO/CO 2 signal H3C…OH3C…O C=O 2.5 23.1

31 D o =85.4 (G3//B3LYP) CH 3 OCO CH 3 O + CO CH 3 + CO 2 CH 3 OCO 1280 CH 3 OCO 1 6.0 (KC) -15.6 (KC) 16.9 (KC) H3C…OH3C…O C=O -1.6 (KC) -39.1 (KC) H3C…OH3C…O C = O RRKM product branching BW TSs Expt. branching w. CO/CO 2 signal 2.5 Glaude, Pitz, Thomson 2005 Good and Francisco 2000 23.1

32 CH 3 + CO 2 CH 3 O + CO Average RRKM product branching over internal energies in our expt. CH 3 OCO CH 3 O + CO CH 3 + CO 2 CH 3 OCO 2.5 ± 0.5 1 EXPT. 32.5

33 CH 3 + CO 2 CH 3 O + CO Average RRKM product branching Over internal energies in our expt. CH 3 O + CO CH 3 + CO 2 CH 3 OCO CH 3 O + CO CH 3 + CO 2 CH 3 OCO 1 EXPT. PRED. CH 3 OCO 1280 2.5 ± 0.5

34 CH 3 + CO 2 CH 3 O + CO Average RRKM product branching Over internal energies in our expt. CH 3 O + CO CH 3 + CO 2 CH 3 OCO CH 3 O + CO CH 3 + CO 2 CH 3 OCO 1 EXPT. CH 3 + CO 2 CH 3 O + CO 2.5 ± 0.5

35 CH 3 + CO 2 CH 3 O + CO Average RRKM product branching Over internal energies in our expt. CH 3 O + CO CH 3 + CO 2 CH 3 OCO CH 3 O + CO CH 3 + CO 2 CH 3 OCO 1 EXPT. PRED. CH 3 OCO 2.11 CH 3 + CO 2 CH 3 O + CO 2.5 ± 0.5

36 CH 3 + CO 2 CH 3 O + CO Why is the cis barrier so much lower than the trans one? H 3 C … O C = O 32.5 cis barrier is ~20 kcal/mol lower than trans (CCSD(T)) H … O C = O (34.2) (14.5) cis barrier is ~7 kcal/mol lower than trans Muckerman, FCC/CBS (2001)

37 CH 3 + CO 2 CH 3 O + CO Why is the cis barrier so much lower than the trans one? Think about the interaction between the radical orbital and the H 3 C-OCO antibonding orbital H 3 C … O C = O nCnC  * C-O Radical energy lowers due to interaction with  * C-O orbital as H 3 C-OCO bond stretches 32.5(34.2) (14.5).

38 Natural Bond Orbital analysis with Weinhold

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