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
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.
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
Addition mechanism forms H CCCH or
2
O
H2CCCH
then ???
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
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)
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
O + propargyl  products
H2C=C=CH 
 H2C-C=CH
O
Addition mechanism forms H CCCH
2
or
O
H2CCCH
Testing our predictive ability
from first principle quantum mechanics
then ???
O + H2CCCH
H2CCC: + OH
HCCCH + OH
Energy (kcal/mol)
c-C3H2 + OH
INT1
INT2
INT2
O
(-60.3)
O
H2CCCH
O
H2CCCH
||
HC=CCH + H
H2C=C=C=O + H
INT2
INT1 Choi’s expts probed only the OH products.
Choi
(CBS-QB3)
His RRKM calcs indicated propynal + H dominates.
O + H2CCCH
H2CCC: + OH
HCCCH + OH
Energy (kcal/mol)
c-C3H2 + OH
INT1
INT2
INT2
O
(-60.3)
O
H2CCCH
O
H2CCCH
||
HC=CCH + H
H2C=C=C=O + H
INT2
vinyl + CO
INT1
Choi + Bowman
(CBS-QB3) (UB3LYP)
LM2
H2C-CHCO
Our expts produce each radical intermediate photolytically and disperse the radicals
by recoil ET and thus by internal energy
O
H
O
C
C
H
nozzle
Cl
C
193 nm
H
C
C
H
C
H
193 nm
Eint radical = hn-Do(C-Cl)-ET
.
H
Cl
ionization source
(electron impact at UofC, -30 kV Al
tunable VUV at ALS) 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
C-Cl fission gives H2CCHCO radicals
dispersed by internal energy
m/e = 35 (Cl+)
20o, 15.0 eV
Eint radical = hn + Eint,prec-Do(C-Cl)-ET
(81.9)*
O
H
O
C
C
H
Cl
C
193 nm
H
H
C
C
H
C
.
H
Cl
100
150
200
Time-of-Flight, t ( s)
250
High translational energy
C-Cl fission
P(E) T)
P(E
produces lowest
internal energy
radicals
T
Energy (kcal/mol)
50
(23.6)
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T)
0
10
20
30
ET(kcal/mol)
40
50
C-Cl fission gives H2CCHCO radicals
dispersed by internal energy
m/e = 35 (Cl+)
20o, 15.0 eV
Eint radical = hn + Eint,prec-Do(C-Cl)-ET
(81.9)
O
H
O
C
C
H
Cl
C
193 nm
H
H
C
C
H
C
.
H
Cl
100
150
200
Time-of-Flight, t ( s)
250
High translational energy
C-Cl fission
P(E) T)
P(E
produces lowest
internal energy
radicals
T
Energy (kcal/mol)
50
(23.6)
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T)
0
10
20
30
ET(kcal/mol)
40
50
All the H2CCHCO radicals
dissociate to vinyl + CO products
+
m/e = 27 (CH2CH )
20o, 12.0 eV
O
H
C
C
C
H
.
O
H
.
H
C
*
C
C
O
H
C
C
.
H
C
H
Energy (kcal/mol)
H
H
50
100
150
200
250
Time-of-Flight, t ( s)
300
m/e = 28 (CO+)
o
20 , 15.0 eV
(23.6)
H2C=CHCO
LM2 * H C-CHCO
2
vinyl + CO
50
CCSD(T)
100
150
200
Time-of-Flight, t (s)
250
300
Upper limit to barrier for
H2CCHCO
vinyl + CO
O
H
C
C
C
H
.
O
H
.
H
*
C
C
C
O
H
C
C
.
H
C
H
Energy (kcal/mol)
H
H
CCSD(T)
UB3LYP
Barrier too high?
(26.7)
(23.6)
(20.0)
H2C=CHCO
LM2 * H C-CHCO
2
vinyl + CO
(25.3) vinyl
H2C=CHCO
LM2 * H C-CHCO
2
+ CO
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)*
O
O
C
H
Cl
C
235 nm
Energy (kcal/mol)
H
H
C
C
H
C
.
H
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
(23.6)
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T)
R + Cl 2P
3/2
not normalized
C
R + Cl 2P
1/2
P(E )
T
H
0
10
20
30
E (kcal/mol)
T
40
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)*
O
H
O
C
C
H
Cl
C
235 nm
H
H
C
C
H
C
.
H
Cl
T
P(E )
Energy (kcal/mol)
all R + Cl
(23.6)
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T)
0
10
20
30
E (kcal/mol)
T
40
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)*
O
H
O
C
C
H
Cl
C
235 nm
C
C
H
C
.
H
Cl
Lowest internal energy at which the
H2CCHCO radicals dissociate is:
121.6+1.5-81.9-18=23 kcal/mol
all R + Cl
T
P(E )
Energy (kcal/mol)
H
H
stable
R + Cl
(23.6)
18 kcal/mol
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T)
0
5
10
15
20
25
E (kcal/mol)
T
30
35
40
CCSD(T)
Eint radical+Cl = hn + Eprec-Do(C-Cl)-ET
(81.9)*
CCSD(T) barrier = 23.6 kcal/mol
Energy (kcal/mol)
Expt’l dissociation onset at ET =18 kcal/mol
gives Expt’l barrier of 23.2 ±2 kcal/mol
UB3LYP
Barrier too high.
(26.7)
(25.3)
Is this because the UB3LYP
radical energy is too low
or the TS energy is too high?
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
CCSD(T) (G3//B3LYP good too)
Eint radical+Cl = hn + Eprec-Do(C-Cl)-ET
(81.9)*
UB3LYP
Eint radical+Cl = hn + Eprec-Do(C-Cl)-ET
(72.4)*
CCSD(T) barrier = 23.6 kcal/mol
Energy (kcal/mol)
Expt’l dissociation onset at ET =18 kcal/mol
gives Expt’l barrier of 23.2 ±2 kcal/mol
(26.7)
(23.6)
H2C=CHCO
LM2 * H C-CHCO
2
(25.3)
vinyl + CO
H2C=CHCO
LM2 * H C-CHCO
2
vinyl + CO
CH3O + CO  CH3OCO  CH3 + CO2
Bridging physical to organic chemistry
ORBITAL INTERACTIONS ALONG THE REACTION COORDINATE
OH + CO  HOCO  H + CO2
CH3O· + CO  CH3OCO  CH3 + CO2
30
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
ENERGY (kcal/mol)
20
10
0
22.1
23.1(JF)
17.3 (BW)
12.4 (ZZ)
k(T,P)
product branching
falloff behavior
0.0
CH 3 O + CO
-10
-20
-30
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
-40
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
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
Cl + CH3OCO*
CH3O(CO)Cl 193
+
m/e=35, (Cl )
Cl + CH3OCO
C-Cl fission P(ET )
o
19.5 , 14.8 eV
Einternal of CH3OCO
Cl + CH3OCO* (85%)
Cl + CH3OCO* (15%)
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
0
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
50
100
150
200
Time-of-Flight, t (sec)
P(ET)
0.0
CH 3 O + CO
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
0
CH 3 + CO
2
10
20
30
E (kcal/mol)
T
40
50
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
0.0
CH 3 O + CO
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
Expt. branching w. CO/CO2 signal
CH3OCO
+
m/e=28 (CO )
o
19.5 , 15.4 eV
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
CH 3 OCO* -> CH
3
CH 3 OCO -> CH
3
O + CO
O + CO
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
50
0.0
100
150
Time-of-Flight (m/e=44
s)
CH 3 O + CO
200
+
(CO2 )
19.5o, 14.8 eV
CH 3OCO -> CH 3 + CO2
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
OCOCl -> CO 2 + Cl
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
CH 3OCO* -> CH 3 + CO2
50
100
150
200
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
Expt. branching w. CO/CO2 signal
1
2.5
CH3OCO
+
m/e=28 (CO )
o
19.5 , 15.4 eV
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
CH 3 OCO* -> CH
3
CH 3 OCO -> CH
3
O + CO
O + CO
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
50
0.0
100
150
Time-of-Flight (m/e=44
s)
CH 3 O + CO
200
+
(CO2 )
19.5o, 14.8 eV
CH 3OCO -> CH 3 + CO2
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
OCOCl -> CO 2 + Cl
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
CH 3OCO* -> CH 3 + CO2
50
100
150
200
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
Expt. branching w. CO/CO2 signal
1
2.5
CH3OCO
+
m/e=28 (CO )
o
19.5 , 15.4 eV
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
CH 3 OCO* -> CH
3
CH 3 OCO -> CH
3
O + CO
O + CO
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
50
0.0
100
150
Time-of-Flight (m/e=44
s)
CH 3 O + CO
200
+
(CO2 )
19.5o, 14.8 eV
CH 3OCO -> CH 3 + CO2
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
OCOCl -> CO 2 + Cl
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
CH 3OCO* -> CH 3 + CO2
50
100
150
200
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
Expt. branching w. CO/CO2 signal
1
H3C…O
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
0.0
CH 3 O + CO
CH3OCO
C=O
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
I asked KC Lau to re-calculate
CH3 + CO2 barrier
G3//B3LYP and CCSD(T)
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
2.5
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
CH3O + CO
CH3OCO
CH3 + CO2
RRKM product branching BW TSs
280
1
CH3OCO
H3C…O
Expt. branching w. CO/CO2 signal
1
CH3OCO
2.5
C=O
6.0 (KC)
16.9 (KC)
5.8 (JF)
6.4 (BW)
4.9 (ZZ)
22.1 (JF)
23.1
17.3 (BW)
12.4 (ZZ)
0.0
CH 3 O + CO
-15.6 (KC)
-15.1 (JF) CH 3 OCO
-15.0 (BW)
-14.6 (ZZ)
Do=85.4 (G3//B3LYP)
-1.6 (KC)
-39.1 (KC)
-38.0 (JF)
-37.5 (BW)
-38.0 (ZZ)
CH 3 + CO
2
O
H3C…O
C
Glaude, Pitz, Thomson 2005
Good and Francisco 2000
60
Average RRKM product branching
50
over internal energies
in our expt.
CH3O + CO
EXPT.
E T (kcal/mole)
A
1
40
CH3OCO
CH3 + CO2
CH3OCO
2.5 ± 0.5
32.5
30
21.6
20
CH3O + CO
10
0
-10
-20
15.6
14.0
CH O + CO
8.1
3
0.0
transCH OCO
3
0.2
cisCH OCO
3
CH + CO
3
2
-23.5
CH3 + CO2
60
Average RRKM product branching
50
Over internal energies
in our expt.
CH3O + CO
CH3 + CO2
CH3OCO
30 CH3OCO
2.5 ± 0.5
1
40
1
280
E T (kcal/mole)
A
EXPT.
PRED.
CH3OCO
21.6
20
CH3O + CO
1
10
15.6
14.0
CH O + CO
8.1
3
CH3O + CO
0.8
0.0
0
transCH OCO
0.6
3
-10
0.4
-20
0.2
25
30
E
int
35
40
45
50
of CH OCO (kcal/mole)
3
cisCH OCO
3
CH + CO
CH3 + CO2
0
0.2
55
3
2
-23.5
CH3 + CO2
60
Average RRKM product branching
50
Over internal energies
in our expt.
CH3O + CO
EXPT.
40
E T (kcal/mole)
A
1
CH3OCO
CH3 + CO2
CH3OCO
2.5 ± 0.5
30
21.6
20
CH3O + CO
1
15.6
14.0
CH O + CO
10
8.1
3
CH3O + CO
0.8
0.0
0
transCH OCO
CH3 + CO2
0.6
3
-10
0.4
0.2
cisCH OCO
3
CH3O + CO
-20
0.2
CH + CO
CH3 + CO2
0
25
30
E
int
35
40
45
50
of CH OCO (kcal/mole)
3
55
3
2
-23.5
CH3 + CO2
60
Average RRKM product branching
50
Over internal energies
in our expt.
EXPT.
PRED.
E T (kcal/mole)
A
CH3O + CO
CH3OCO
CH3 + CO2
CH3OCO
30 CH3OCO
2.5 ± 0.5
2.1
40
1
1
21.6
20
CH3O + CO
1
15.6
14.0
CH O + CO
10
8.1
3
CH3O + CO
0.8
0.0
0
transCH OCO
CH3 + CO2
0.6
3
-10
0.4
0.2
cisCH OCO
3
CH3O + CO
-20
0.2
CH + CO
CH3 + CO2
0
25
30
E
int
35
40
45
50
of CH OCO (kcal/mole)
3
55
3
2
-23.5
CH3 + CO2
E T (kcal/mole)
A
40
Why is the cis barrier
so much
lower than the trans
one?
30
32.5
21.6
20
CH3O + CO
10
15.6
14.0
CH O + CO
0.0
transCH OCO
3
-10
O
H3C
C
0.2
cisCH OCO
3
cis-20
barrier is ~20 kcal/mol lower than trans
(CCSD(T))
O
H …O
(14.5)
8.1
3
0
…O
(34.2)
C cis barrier is ~7 kcal/mol lower than trans
Muckerman, FCC/CBS (2001)
CH + CO
3
2
-23.5
CH3 + CO2
E T (kcal/mole)
A
40
Why is the cis barrier
so much
lower than the trans
one?
30
32.5
21.6
20
CH3O + CO
10
15.6
14.0
CH O + CO
0.0
transCH OCO
-20
3
0.2
cisCH OCO
3
CH + CO
3
s*C-O
O
C
.
nC
(14.5)
8.1
3
Think about the0 interaction
between the radical orbital and
the H3C-OCO antibonding
orbital
-10
H3C … O
(34.2)
2
-23.5
CH3 + CO2
Radical energy lowers due to
interaction with s*C-O orbital
as H C-OCO bond stretches
Natural Bond Orbital analysis with Weinhold
Qu i c k T i m e ™ a n d a
T I F F (L Z W ) d e c o m p re s s o r
a re n e e d e d to s e e th i s p i c t u re .