The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell
description
Transcript of The Butler Group Benj FitzPatrick Britni Ratliff Bridget Alligood Doran Bennett Justine Bell
The Butler Group
Benj FitzPatrickBritni RatliffBridget AlligoodDoran BennettJustine BellArjun Raman
Emily GlassmanDr. 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
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-Ea/kT
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.
O + propargyl products
H2C=C=CH H2C-C=CH
Addition mechanism forms or then ??? H2CCCHO
H2CCCHO
Many elementary bimolecular reactions proceed through addition/insertion,
so go through unstable radical intermediates along the bimolecular reaction coordinate
CH3O + CO CH3OCO CH3 + CO2
Traditional Crossed Molecular Beam Scattering or Imaging Exptsare a good way to probe “Direct” Chemical Reactions
Angular and Velocity Distribution of DF product shows Backward Scattered DF product
Eg. D2 + F D…D…F D + DF
D…D…F
D-D F
But how can one probe bimolecular reactions that proceed through long-lived radical intermediates?
Eg. C2D + HCCHDCCCCH + H
Forward/Backward symmetricproduct angular distributions indicate there is a long-livedintermediate in the reaction.
But what is happening along thereaction coordinate?
Kaiser et al., PCCP 4, 2950 (2002)
But how can one probe bimolecular reactions that proceed through long-lived radical intermediates?
Eg. C2D + HCCHDCCCCH + H
Kaiser et al., PCCP 4, 2950 (2002) UB3LYP/6-311+G** + ZPVE
O + propargyl products
H2C=C=CH H2C-C=CH
Addition mechanism forms or then ???
Testing our predictive ability from first principle quantum mechanics
H2CCCHO
H2CCCHO
H2C=C=C=O + H
c-C3H2 + OH
H2CCC: + OHHCCCH + OH
O ||HC=CCH + H
O + H2CCCHE
nerg
y (k
cal/m
ol)
Choi (CBS-QB3)
INT2
INT1
INT1
INT2
INT2
(-60.3)
H2CCCH
O
H2CCCH
O
His RRKM calcs indicated propynal + H dominates.Choi’s expts probed only the OH products.
H2C=C=C=O + H
c-C3H2 + OH
H2CCC: + OHHCCCH + OH
O ||HC=CCH + H
vinyl + CO
O + H2CCCH
Ene
rgy
(kca
l/mol
)
Choi (CBS-QB3)
+ Bowman (UB3LYP)
INT2
INT1
INT1
INT2
INT2
LM2
(-60.3)
H2CCCH
O
H2CCCH
O
H2C-CHCO
193 nmnozzle
skimmers
ionization source(electron impact at UofC, tunable VUV at ALS)
-30 kV Aldoorknob
quadrupole mass spec.
Scintillator
PMT
Eint radical = hDo(C-Cl)-ET
Our expts produce each radical intermediate photolytically and disperse the radicals by recoil ET and thus by internal energy
193 nm
Cl
CC
C
H
H
H
O
ClCC
C
H
H
H
.O
Measuring the velocities of the stable radicals and the velocities ofthe products from the unstable radicals can determine the barriers to each product channel and how product channel branching changes with internal energy
vinyl + COEne
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
50 100 150 200 250
m/e = 35 (Cl +)
20o, 15.0 eV
Time-of-Flight, t (μ )s
0 10 20 30 40 50
High translational energy C-Cl fission
ET(kcal/mol)
produces lowest internal energy
radicals
(23.6)
C-Cl fission gives H2CCHCO radicals dispersed by internal energy
Eint radical = h + Eint,prec-Do(C-Cl)-ET(81.9)*
193 nmCl
CC
C
H
H
H
O
ClCC
C
H
H
H
.O
P(E
T)
CCSD(T)*
vinyl + COEne
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
50 100 150 200 250
m/e = 35 (Cl +)
20o, 15.0 eV
Time-of-Flight, t (μ )s
0 10 20 30 40 50
High translational energy C-Cl fission
ET(kcal/mol)
produces lowest internal energy
radicals
(23.6)
C-Cl fission gives H2CCHCO radicals dispersed by internal energy
Eint radical = h + Eint,prec-Do(C-Cl)-ET(81.9)
193 nm
Cl
CC
C
H
H
H
O
ClCC
C
H
H
H
.O
P(E
T)
CCSD(T)*
vinyl + COEne
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
All the H2CCHCO radicals dissociate to vinyl + CO products
CCSD(T)
CC
C
H
H
H
.O
50 100 150 200 250 300
m/e = 27 (CH2CH+)
20o, 12.0 eV
Time-of-Flight, t (μ )s
CC
C
H
H
H
.O
CC
C
H
H
H
.O
50 100 150 200 250 300
m/e = 28 (CO +)
20o, 15.0 eV
Time-of-Flight, t (μ )s*
*
vinyl + COEne
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
CCSD(T)
vinyl + CO
LM2H2C-CHCO
H2C=CHCO
(26.7)
CC
C
H
H
H
.OCC
C
H
H
H
.O
CC
C
H
H
H
.O
UB3LYPBarrier too high?
**
*
(25.3)(20.0)
Upper limit to barrier for H2CCHCO vinyl + CO
vinyl + CO
Ene
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
C-Cl fission at 235 nm produces lower internal energy H2CCHCO radicals
Eint radical+Cl = h + Eint,prec-Do(C-Cl)-ET(81.9)*
235 nmCl
CC
C
H
H
H
O
ClCC
C
H
H
H
.O
CCSD(T)*
Cl 2P3/2 Cl 2P1/2 (Cl*)
0 10 20 30 40
R + Cl 2P3/2
ET (kcal/mol)
R + Cl 2P1/2
Add these two, correcting for 0.85Cl*/Cl line strength factor (Liyanage)to get total C-Cl fission P(ET)for producing all radicals
vinyl + CO
Ene
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
C-Cl fission at 235 nm produces lower internal energy H2CCHCO radicals
Eint radical+Cl = h + Eint,prec-Do(C-Cl)-ET(81.9)*
235 nmCl
CC
C
H
H
H
O
ClCC
C
H
H
H
.O
CCSD(T)*
Cl 2P3/2 Cl 2P1/2 (Cl*)
0 10 20 30 40
all R + Cl
ET (kcal/mol)
vinyl + COEne
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
Use 157 nm photoionization to detect all STABLE H2CCHCO radicals
Eint radical+Cl = h + Eprec-Do(C-Cl)-ET(81.9)*
235 nmCC
C
H
H
H
O
ClCC
C
H
H
H
.O
CCSD(T)*
(157 + 235) - (157 only)
0 5 10 15 20 25 30 35 40E
T (kcal/mol)
18 kcal/mol
Cl
Lowest internal energy at which the H2CCHCO radicals dissociate is: 121.6+1.5-81.9-18=23 kcal/mol
all R + Cl
stable R + Cl
vinyl + CO
Ene
rgy
(kca
l/mol
)
LM2H2C=CHCO
Eint radical+Cl = h + Eprec-Do(C-Cl)-ET(81.9)*
CCSD(T)
*
CCSD(T) barrier = 23.6 kcal/mol
(26.7)
UB3LYPBarrier too high.
(25.3)
Expt’l dissociation onset at ET =18 kcal/molgives Expt’l barrier of 23.2 ±2 kcal/mol
H2C-CHCO
Is this because the UB3LYP radical energy is too low or the TS energy is too high?
vinyl + CO
Ene
rgy
(kca
l/mol
)
LM2H2C-CHCO
H2C=CHCO
(23.6)
Eint radical+Cl = h + Eprec-Do(C-Cl)-ET(81.9)*
CCSD(T) (G3//B3LYP good too)
*
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
vinyl + CO
LM2H2C=CHCO
*
(26.7)
(25.3)
H2C-CHCO
Eint radical+Cl = h + Eprec-Do(C-Cl)-ET
UB3LYP
(72.4)*
CH3O + CO CH3OCO CH3 + CO2
Bridging physical to organic chemistry
ORBITAL INTERACTIONS ALONG THE REACTION COORDINATE
CH3O· + CO CH3OCO CH3 + CO2
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2JF: 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 + CO2
k(T,P)product branching
falloff behavior
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2
CH3O(CO)Cl 193
Cl + CH3OCO
Cl + CH3OCO*
C-Cl fission P(ET )
Do=85.4 (G3//B3LYP)
Einternal of CH3OCO
0 10 20 30 40 50E
T (kcal/mol)
0 50 100 150 200
m/e=35, (Cl+)
19.5o, 14.8 eV
Time-of-Flight, t (μ )sec
Cl+CH3
*OCO (15%)
Cl+CH3
*OCO (85%)
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
RRKM product branching BW TSs
CH3OCO 1280
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
RRKM product branching BW TSs
CH3OCO 1280 CH3OCO
50 100 150 200
m/e=28 (CO+)
19.5o, 15.4 eV
CH3OCO* -> CH
3O + CO
CH3OCO -> CH
3O + CO
Time-of-Flight (μ )s
50 100 150 200
m/e=44 (CO2
+)
19.5o, 14.8 eV
OCOCl -> CO2 + Cl
CH3OCO* -> CH
3 + CO
2
CH3OCO -> CH
3 + CO
2
Time-of-Flight (μ )s
Expt. branching w. CO/CO2 signal
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
RRKM product branching BW TSs
CH3OCO 1280 CH3OCO
50 100 150 200
m/e=28 (CO+)
19.5o, 15.4 eV
CH3OCO* -> CH
3O + CO
CH3OCO -> CH
3O + CO
Time-of-Flight (μ )s
50 100 150 200
m/e=44 (CO2
+)
19.5o, 14.8 eV
OCOCl -> CO2 + Cl
CH3OCO* -> CH
3 + CO
2
CH3OCO -> CH
3 + CO
2
Time-of-Flight (μ )s
1Expt. branching w. CO/CO2 signal
2.5
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
RRKM product branching BW TSs
CH3OCO 1280 CH3OCO
50 100 150 200
m/e=28 (CO+)
19.5o, 15.4 eV
CH3OCO* -> CH
3O + CO
CH3OCO -> CH
3O + CO
Time-of-Flight (μ )s
50 100 150 200
m/e=44 (CO2
+)
19.5o, 14.8 eV
OCOCl -> CO2 + Cl
CH3OCO* -> CH
3 + CO
2
CH3OCO -> CH
3 + CO
2
Time-of-Flight (μ )s
1Expt. branching w. CO/CO2 signal
2.5
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
RRKM product branching BW TSs
CH3OCO 1280 CH3OCO1
I asked KC Lau to re-calculateCH3 + CO2 barrier
G3//B3LYP and CCSD(T)
Expt. branching w. CO/CO2 signal
H3C…OC=O
2.5
23.1
-40
-30
-20
-10
0
10
20
30
0.0
5.8 (JF)6.4 (BW)4.9 (ZZ)
-15.1 (JF)-15.0 (BW)-14.6 (ZZ)
22.1 (JF)17.3 (BW)12.4 (ZZ)
-38.0 (JF)-37.5 (BW)-38.0 (ZZ)
CH3OCO
CH3O + CO
CH3 + CO2Do=85.4 (G3//B3LYP)
CH3OCOCH3O + CO CH3 + CO2
CH3OCO 1280 CH3OCO1
6.0 (KC)
-15.6 (KC)
16.9 (KC)
H3C…OC=O
-1.6 (KC)
-39.1 (KC)H3C…O
C
=O
RRKM product branching BW TSs Expt. branching w. CO/CO2 signal
2.5
Glaude, Pitz, Thomson 2005Good and Francisco 2000
23.1
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Average RRKM product branchingover internal energies in our expt.
CH3OCOCH3O + CO CH3 + CO2
CH3OCO 2.5 ± 0.51EXPT.32.5
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Average RRKM product branchingOver internal energies in our expt.
CH3O + CO
CH3 + CO20
0.2
0.4
0.6
0.8
1
25 30 35 40 45 50 55E
int of CH
3OCO (kcal/mole)
CH3OCOCH3O + CO CH3 + CO2
CH3OCO1EXPT.PRED. CH3OCO 1280
2.5 ± 0.5
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Average RRKM product branchingOver internal energies in our expt.
CH3O + CO
CH3 + CO2
CH3OCOCH3O + CO CH3 + CO2
CH3OCO1EXPT.
0
0.2
0.4
0.6
0.8
1
25 30 35 40 45 50 55E
int of CH
3OCO (kcal/mole)
CH3 + CO2
CH3O + CO
2.5 ± 0.5
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Average RRKM product branchingOver internal energies in our expt.
CH3O + CO
CH3 + CO2
CH3OCOCH3O + CO CH3 + CO2
CH3OCO1EXPT.PRED. CH3OCO 2.11
0
0.2
0.4
0.6
0.8
1
25 30 35 40 45 50 55E
int of CH
3OCO (kcal/mole)
CH3 + CO2
CH3O + CO
2.5 ± 0.5
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Why is the cis barrier so muchlower than the trans one?
H3C … OC
=O
32.5
cis barrier is ~20 kcal/mol lower than trans (CCSD(T))
H … OC
=O
(34.2)
(14.5)
cis barrier is ~7 kcal/mol lower than transMuckerman, FCC/CBS (2001)
-20
-10
0
10
20
30
40
50
60
70
ET
(kcal/mole)A
0.0trans-CH
3OCO
CH3O + CO
CH3 + CO
2
21.6
15.6
-23.5
14.0
0.2cis-CH
3OCO
8.1
CH3 + CO2
CH3O + CO
Why is the cis barrier so muchlower than the trans one?
Think about the interactionbetween the radical orbital and
the H3C-OCO antibonding orbital
H3C … OC
=O
nC
*C-O
Radical energy lowers due to interaction with *C-O orbitalas H3C-OCO bond stretches
32.5 (34.2)
(14.5)
.
QuickTime™ and aTIFF (LZW) decompressor
are needed to see this picture.
Natural Bond Orbital analysis with Weinhold