ChE 553 Lecture 17 Prediction of Mechanisms

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Objectives

• Develop methods to predict mechanisms

• Apply the ideas for a simple reaction

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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

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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

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The Idea of Computing a Mechanism

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1) write down all possible reactions2) Use rules to make sure no important

reactions are missing.3) Use rules to eliminate excess reactions.

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:

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Example: H2 + Br2 2HBr

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X Br 2Br X

Br H HBr H

H Br HBr Br

X 2Br Br X

H HBr H Br

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22

23

42

52

Br

H

HBr

Br 2HBr

H 2

Br2

Figure 5.1 A cycle for HBr formation via reaction (5.3). 

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

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X Br 2Br X

Br H HBr H

H Br HBr Br

X 2Br Br X

H HBr H Br

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22

23

42

52

Br

H

HBr

Br 2HBr

H 2

Br2

Figure 5.1 A cycle for HBr formation via reaction (5.3).

Discussion Problem: The reaction CH3CH3CH2CH2 + H2

Goes By the Following Mechanism

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Label each step as being a a) initiation b) propagation c) termination d) transfer.

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

Examples of Initiation Propagation Mechanisms

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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 Free Radical Polymerization

e.g. ethylene polyethylene with a free radical catalyst, R2

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

Ozone Depletion O2+h1 2O

O+O2+X O3 O3+h2

O2+O Cl+O3

O2+ClO ClO+O O2+Cl

Hydrocarbon Pyrolysis 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

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.

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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

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Example: The Reaction CH3CH3H2C=CH2+H2 Obeys the Following Mechanism:

XCH2XCHCH 31

33 3242

333 CHCHCHCHCHCH XHCHCHXCHCH 223

32 3224

33 CHCHHCHCHH 335

3 CHCHXCH2 (+ other reactions)

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Verify that it follows the rules

Step 1: Make a Diagram of the Reaction Similar to That in Figure 5.3

XCH2XCHCH 31

33 3242

333 CHCHCHCHCHCH XHCHCHXCHCH 223

32 3224

33 CHCHHCHCHH 335

3 CHCHXCH2 (+ other reactions)

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H

H 2 X

2 H C=CH 2

3 CH CH 2

X+ 3 CH CH 3

CH 4

3 CH CH 3

CH 3

3 CH CH 3

+X

Catalytic Cycl

e

Chain Transf

er

initiation

Step 2: Identify the Initiation Step, the Transfer Step, the Propagation Steps, the Termination Steps

XCH2XCHCH 31

33 3242

333 CHCHCHCHCHCH XHCHCHXCHCH 223

32 3224

33 CHCHHCHCHH 335

3 CHCHXCH2 (+ other reactions)

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H

H 2 X

2 H C=CH 2

3 CH CH 2

X+ 3 CH CH 3

CH 4

3 CH CH 3

CH 3

3 CH CH 3

+X

Catalytic Cycl

e

Chain Transf

er

initiation

Step 2: Continued

b) Reaction 1 – initiation

Reaction 2 – chain transfer

Reaction 3 – propagation (-hydrogen elimination)

Step 4 – propagation (hydrogen transfer)

Reaction 5 - termination

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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

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Empirical Rules for Activation Barriers

Practical reactions Ea<0.15 T initiation

reactions Set minimum T

Ea<0.05 T catalytic For Reaction

cycle

Ea<0.07 T Transfer reactions and side reactions 19

KMole

Kcal

KMole

Kcal

KMole

Kcal

Methods to Estimate Ea

Polanyi relationship

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rP0aa HEE

Blowers Masel Equation

when

when

when

1E4

H0a

r

1E4

H1

0a

r

1E4

H0a

r

(10.63)

0a0

0a0

0PEw

EWw2V

(10.65)

0

H

2r

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2P

2r0Pr0

a

r

)H()w(4)V(

)Hw2V)(H5.0w(E

Ea= Activation Energy

Ea0= Intrinsic Activation Barrier

P= Transfer coefficent

Hr= Heat of Reaction Memorize this equation

Intrinsic Barriers and Transfer. Coefficients for Different Types of Neutral Species

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Reaction Example Actual EAO

kcal/mole EA

O to assume when predicting mechanisms kcal/mole

Actual P P to assume

when predicting mechanisms

Simple bond scission

AB+X A+B+X X=a collision partner

0-1 1 1.0 1.0

Recombination A+B+X AB+X X=a collision partner

0-1 1 1.0 1.0

Exothermic atom transfer reaction

R x + R1 R + x-R1

x = an atom

8-16 12 0.2 to 0.6 0.3

Endothermic atom transfer reaction

R- x + R R + x-R1 x=an atom

8-16 12 0.4 to 0.8 0.7

Ligand transfer reaction to hydrogen

H+R-R1 HR + R1 40-50 45 0.4 to 0.6 0.5

Other ligand transfer reactions

x + R-R1 xR+ R1

x=an atom 50 or more 50 0.3 to 0.7 0.5

Next: Estimate the Activation Barriers

Consider

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XCH2XCHCH 333 First estimate rH

Next estimate EA using Table 5.4. This is a simple bond scission reaction. From Table 5.4 mole/kcal6.90H1E rA

From NIST Web book (http://webbook.nist.gov) 0.20)CHCH(H 33f

Therefore mole/kcal8.34)CH(H 3f

mole/kcal6.89)0.20()8.34(2Hr

Next: CH3•+CH3CH3CH4+•CH2CH3

From the NIST web book mole/kcal0.20)CHCH(H 33f

mole/kcal8.34)CH(H 3f

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This is an atom transfer reaction. From Table 5.4

mole/kcal7.8)3.4(3.0mole/kcal10EA

mole/kcal9.17)CH(H 4f mole/kcal4.28)CHCH(H 32f

Therefore mole/kcal3.4)0.20(8.344.289.17Hr

CH2CH3+XCH3CH2+H+X

From the NIST web book mole/kcal4.28)CHCH(H 32f mole/kcal5.12)CHCH(H 22f

mole/kcal1.52)H(Hf mole/kcal2.364.285.121.52H

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From Table 5.4 mole/kcal3.40)2.36(7.015EA

H•CH3CH3H2+•CH2CH3

From the NIST web book mole/kcal1.52)H(Hf

mole/kcal0.20)CHCH(H 33f mole/kcal4.28)CHCH(H 32f

mole/kcal0)H(H 2f Therefore

mole/kcal7.3)0.20(1.524.280Hr 25

This is a hydrogen transfer reaction. From Table 5.4

mole/kcal9.5)7.3(3.010EA

•CH3+ •CH3+X CH3 CH3+X

mole/kcal6.89Hr (reverse reaction 1) This is recombination reaction. From Table 5.4

mole/kcal1EA

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Next: Calculate Temperature to Meet Constants

Kmolkcal

15.0EA for initiation

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Kmole

kcal07.0E

rA T for all propagation

moleK

kcal05.0EA T for initiation

Solution

For initiation

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For propagation

K80605.0/3.40T

Therefore any temperature above 806K will satisfy all constraints.

mole/kcal6.90EA

K60415.0/6.90T

Example 5.B Consider the Following Alternate Mechanism for Ethylene Production from

Ethane

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31

33 CH2XCHCH

3242

333 CHCHCHCHCHCH

XHCHCHXCHCH 223

32

346

33 CHCHCHCHH

XCHCHXCH2 335

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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?

Solution

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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

H•CH3 CH3CH4+ •CH3

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From the Nist webbook mole/kcal1.52)H(Hf

mole/kcal0.20)CHCH(H 33f

mole/kcal9.17)CH(H 4f 8.34)CH(H 3f

mole/kcal2.15)0.20(1.52)9.17(8.34Hr This is a ligand transfer reaction to hydrogen. From Table 5.4

mole/kcal4.37)2.15(5.00.45EA

This Reaction is a Reaction in the Catalytic Cycle

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mole

kcal5.40)k810)(05.0(T)

mole

kcal05.0(EA

Therefore all constraints are satisfied

Which Mechanism is Better

33

3224

33 CHCHHCHCHH

346

33 CHCHCHCHH

Ea=8.9Ea=37.4

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