ChE 553 Lecture 17
Prediction of Mechanisms
1
Objectives
• Develop methods to predict
mechanisms
• Apply the ideas for a simple reaction
2
You Already Learned About the Mechanisms of
Reactions in Organic Chemistry
Organic view of mechanisms – things to
memorize
Masel view of mechanisms – things to
calculate
3
Key: Activation Barriers
Control Mechanisms
• Reaction goes by the pathway that has
the lowest activation barrier between
reactants and products
– Catalytic cycles used to lower barriers
4
The Idea of Computing a
Mechanism
1) write down all possible reactions
2) Use rules to make sure no important
reactions are missing.
3) Use rules to eliminate excess
reactions.
5
General Rules for
Mechanisms
All commercially important mechanisms are
basically the same !!
• Step 1 Create reactive species
• Step 2 Catalytic cycle to pump out product
• Step 3 Reactive species lost:
6
Example: H2 + Br2  2HBr
Br
H2
Br
2
HBr
1
X  Br2 
 2Br  X
2
Br  H 2  HBr  H
3
H  Br2  HBr  Br
4
X  2Br  Br2  X
5
H  HBr 
 H 2  Br
HBr
H
Br
2
Figure 5.1 A cycle for HBr
formation via reaction (5.3).
7
Initiation-Propagation
Mechanisms
• Initiation step: create reactive species
• Transfer step: convert initial radical into
a more reactive species
• Propagation step: go around cycle to
produce product
• Termination step: destroy radicals
8
Consider: H2 + Br2  2HBr
Br
1
X  Br2 
 2Br  X
2
Br  H 2  HBr  H
3
H  Br2  HBr  Br
H2
Br
2
HBr
4
X  2Br  Br2  X
5
H  HBr 
 H 2  Br
HBr
H
Br
2
Figure 5.1 A cycle for HBr
formation via reaction (5.3).
9
Discussion Problem: The reaction
CH3CH3CH2CH2 + H2
Goes By the Following Mechanism
CH3CH3 +X1 2CH3+X
CH3 + CH3CH3 2 CH4 + CH2CH3
CH2CH3 + X 3 CH2CH2 + H +X
H+ CH3CH3 4 H2 + CH2CH3
2 CH2CH3 +X 5 CH3 CH2CH2CH3 + X
Label each step as being a
a) initiation b) propagation c) termination
d) transfer.
10
Examples of Initiation
Propagation Mechanisms
Reaction
Example Mechanism
Combustion e.g.,
CH4 + O2  CO2 +2H2O + other products
O2 2O
O + CH4  CH3 + OH
OH + CH4  H2O + CH3
CH3 + O2  CH3 + O + O
CH3 + OH  CH2 + H2O + other products
OH  walls
CH3  walls
R22R
R+C2H4R(C2H4)
RC2H4+C2H4R(C2H4)2
R(C2H4)n+C2H4R(C2H4)n+1
R(C2H4)m+R(C2H4)nR(C2H4)m+nR
O2+h12O
O+O2+XO3
O3+h2O2+O
Cl+O3O2+ClO
ClO+OO2+Cl
X+CH3COHCH3+COH+X
CH3+CH3OHCH3CO+CH4
CH3CO+CH3OHCH4+CH3CO
COH+XCO+H+X
H+CH3COHCH4+COH
H+CH3COHCH3+CO+H2
2CH3+XC2H6+X
H+CH3+XCH4+X
H+CH3CO+XCH3COH+X
Free Radical Polymerization
e.g. ethylene  polyethylene with a free radical
catalyst, R2
Ozone Depletion
Hydrocarbon Pyrolysis
11
General Approach to Finding
a Mechanism
• Guess or predict all of the species that
are likely to form during the reaction.
• Write down all of the possible reaction
of those species (only include 7
generic types of reactions).
• Use various rules to pare down the list
to manageable of steps.
12
Rules for Initiation
Propagation Reactions
• There must be at least one initiation reaction
• The propagation reactions must occur in a
cycle where radicals react with the reactants
to form new radicals and then the new
radicals react to from the original radicals
again
• All of the steps in the catalytic cycle must have
low barriers
• There should be at least on termination
reaction where two radicals combine to yield
stable species
13
Example: The Reaction CH3CH3H2C=CH2+H2
Obeys the Following Mechanism:
CH 3CH 3 X2CH 3 X
2
CH 3 CH 3CH 3 CH 4 CH 2 CH 3
3
CH 2 CH 3 XCH 2 CH 2 HX
4
HCH 3CH 3 H 2 CH 2 CH 3
5
2CH 3 XCH 3CH 3
(+ other reactions)
1
Verify that it follows the rules
14
Step 1: Make a Diagram of the Reaction
Similar to That in Figure 5.3
CH 3CH 3 X2CH 3 X
CH 3 CH 3CH 3 2 CH 4 CH 2 CH 3
CH 2 CH 3 X3 CH 2 CH 2 HX
HCH 3CH 3 4 H 2 CH 2 CH 3
2CH 3 X5 CH 3CH 3
(+ other reactions)
CH3CH3
1
CH4
initiatio +X
n
CH3
Chai
Transf
n
er
CH3CH 3
X
H2
CH2 CH3
Catalyti
c Cycl
e
X+ H2C=CH 2
H
CH3CH 3
15
Step 2: Identify the Initiation Step, the Transfer Step, the
Propagation Steps, the Termination Steps
CH 3CH 3 X2CH 3 X
CH 3 CH 3CH 3 2 CH 4 CH 2 CH 3
CH 2 CH 3 X3 CH 2 CH 2 HX
HCH 3CH 3 4 H 2 CH 2 CH 3
2CH 3 X5 CH 3CH 3
(+ other reactions)
CH3CH3
1
CH4
initiatio +X
n
CH3
Chai
Transf
n
er
CH3CH 3
X
H2
CH2 CH3
Catalyti
c Cycl
e
X+ H2C=CH 2
H
CH3CH 3
16
Step 2: Continued
b)
Reaction 1 – initiation
Reaction 2 – chain transfer
Reaction 3 – propagation (hydrogen elimination)
Step 4 – propagation (hydrogen
transfer)
Reaction 5 - termination
17
The Mechanisms
Does the mechanisms follow the rules?
• There is an initiation step (step 1)
• There is a catalytic cycle (steps 3 and 4)
• There is a termination step (step 5)
• Still need to verify that the activation
barriers are low enough
18
Empirical Rules for Activation
Barriers
Practical reactions
Ea<0.15 Kcal T
initiation
Mole K
reactions
Kcal
Ea<0.05
T
Mole K
Kcal
Ea<0.07
T
Mole K
catalytic
cycle
Set minimum
T
For Reaction
Transfer reactions and
side reactions
19
Methods to Estimate Ea
Polanyi relationship
Ea  E0a   P Hr
Blowers Masel Equation
Hr
 1
when
0
0
4Ea
2
( w 0  0.5Hr )( VP  2w 0  Hr ) 
Hr
Ea  
when


1

1
2
2
2
0
( VP )  4( w 0 )  ( Hr )
4Ea



H
r
Hr
1
when
0
4Ea
Ea = Activation Energy
E0
a = Intrinsic Activation Barrier
 = Transfer coefficent
P
Hr = Heat of Reaction
Memorize this equation
(10.63)
 W0  E0a 

VP  2w 0 
0 
 w 0  Ea 
(10.65)
20
Intrinsic Barriers and Transfer. Coefficients for
Different Types of Neutral Species
Reaction
Example
Simple
bond AB+XA+B+X
scission
X=a collision partner
Recombination A+B+XAB+X
X=a collision partner
Exothermic
R x + R1  R + x-R1
atom
transfer
x = an atom
reaction
Endothermic
R- x + RR + x-R1
atom
x=an atom
transfer
reaction
Ligand transfer H+R-R1  HR + R1
reaction
to
hydrogen
Other
ligand x + R-R1  xR+ R1
transfer
x=an atom
reactions
P to assume
when
predicting
mechanisms
0-1
E OA to
Actual P
assume when
predicting
mechanisms
kcal/mole
1
1.0
0-1
1
1.0
1.0
8-16
12
0.2 to 0.6
0.3
8-16
12
0.4 to 0.8
0.7
40-50
45
0.4 to 0.6
0.5
50 or more
50
0.3 to 0.7
0.5
Actual E OA
kcal/mole
1.0
21
Next: Estimate the Activation
Barriers
Consider CH 3CH 3 X2CH 3X
First estimate H r
Next estimate EA using Table 5.4. This is a
simple bond scission reaction. From Table 5.4
E A 1H r 90.6 kcal /mole
From NIST Web book (http://webbook.nist.gov)
H f (CH 3CH 3 )20.0
Therefore
H f (CH 3 ) 34.8kcal /mole
H r 2(34.8)( 20.0)89.6kcal /mole
22
Next: CH3•+CH3CH3CH4+•CH2CH3
From the NIST web book
H f (CH 3CH 3 )20.0kcal /mole
H f (CH 3 )34.8kcal /mole
This is an atom transfer reaction. From Table 5.4
EA 10kcal/mole0.3(4.3)8.7kcal/mole
H f (CH 4 )17.9kcal /mole
H f (CH 2 CH 3 )28.4kcal /mole
Therefore
H r 17.928.434.8( 20.0)4.3 kcal /mole
23
CH2CH3+XCH3CH2+H+X
From Table 5.4
E A 150.7(36.2)40.3 kcal /mole
From the NIST web book
H f (CH 2 CH 3 )28.4kcal /mole
H f (CH 2 CH 2 )12.5kcal /mole
H f ( H )52.1kcal /mole
H52 .112 .528 .436 .2 kcal /mole
24
H•CH3CH3H2+•CH2CH3
This is a hydrogen transfer reaction. From Table
5.4
E A 100.3( 3.7)5.9 kcal /mole
From the NIST web book
H f ( H )52.1kcal /mole
H f (CH 3CH 3 )20.0 kcal /mole
H f (CH 2 CH 3 )28.4 kcal /mole
H f ( H 2 )0 kcal /mole
Therefore
H r 028.452.1( 20.0)3.7 kcal /mole
25
•CH3+ •CH3+X CH3 CH3+X
H r 89.6 kcal /mole (reverse reaction 1)
This is recombination reaction. From Table 5.4
E A 1kcal /mole
26
Next: Calculate Temperature
to Meet Constants
kcal
E A 0.15
for initiation

mol K
kcal
E A  0.07
mole   r K
T for all propagation
kcal
E A  0.05
T for initiation
moleK
27
Solution
For initiation
For propagation
E A 90.6 kcal /mole
T  40.3 / 0.05  806K
T 90 .6/0.15604 K
Therefore any temperature above
806K will satisfy all constraints.
28
Example 5.B Consider the Following Alternate
Mechanism for Ethylene Production from
Ethane
CH 3CH 3 X1 2CH 3 
CH 3 CH 3CH 3  CH 4 CH 2 CH 3
3
CH 2 CH 3 X H 2 CCH 2 HX
HCH 3CH 3 6 CH 4 CH 3 
2CH 3 X5 CH 3CH 3 X
2
a) Does this mechanism follow all of the rule at 810K?
b) Is this mechanism more or less likely than the
mechanism in example 5.A?
29
Solution
a) This does follow the rules!
1)There is an initiation step (step 1)
2) There is a catalytic cycle (steps 2,3,6)
3) There is a termination step (step 5)
Check all steps obey constraint in equation 5.36
Steps 1,2,3,5 do (see example 5.A)
Check step 6
30
H•CH3 CH3CH4+ •CH3
H r 34.8( 17.9)52.1( 20.0)15.2 kcal /mole
This is a ligand transfer reaction to hydrogen. From
Table 5.4
E A 45.00.5( 15.2)37.4 kcal /mole
From the Nist webbook
H f ( H)52.1kcal /mole
H f (CH 3CH 3 )20.0 kcal /mole
H f (CH 4 )17.9 kcal /mole
H f (CH 3 )34.8
31
This Reaction is a Reaction in
the Catalytic Cycle
kcal
kcal
E A  (0.05
)T  (0.05)(810k )  40.5
mole
mole
Therefore all constraints are satisfied
32
Which Mechanism is Better
HCH 3CH 3  H 2 CH 2 CH 3
6
HCH 3CH 3  CH 4  CH 3
4
Ea=8.9
Ea=37.4
33
Summary
Today derived a series of rules for reactions
•
•
•
•
Must be an initiation reaction
Must have a catalytic cycle
Should have termination
Barriers low enough
Next time: Use rules to predict mechanisms.
34