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CHM 8304 Kinetic analyses 1 CHM 8304 Physical Organic Chemistry Kinetic analyses Scientific method proposal of a hypothesis conduct experiments to test this hypothesis – confirmation refutation the trait of refutability is what distinguishes a good scientific hypothesis from a pseudo-hypothesis (see Karl Popper) 2

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Page 1: CHM 8304 Physical Organic Chemistry - University of Ottawamysite.science.uottawa.ca/jkeillor/English/Teaching_files/Kinetic... · CHM 8304! Kinetic analyses! 1! CHM 8304 Physical

CHM 8304

Kinetic analyses 1

CHM 8304 Physical Organic Chemistry

Kinetic analyses

Scientific method

•  proposal of a hypothesis •  conduct experiments to test this hypothesis

–  confirmation –  refutation

•  the trait of refutability is what distinguishes a good scientific hypothesis from a pseudo-hypothesis (see Karl Popper)

2

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Kinetic analyses 2

“Proof” of a mechanism

•  a mechanism can never really be ‘proven’ –  inter alia, it is not directly observable!

•  one can propose a hypothetical mechanism •  one can conduct experiments designed to refute certain hypotheses •  one can retain mechanisms that are not refuted •  a mechanism can thus become “generally accepted”

3

Mechanistic studies

•  allow a better comprehension of a reaction, its scope and its utility

•  require kinetic experiments –  rate of disappearance of reactants –  rate of appearance of products

•  kinetic studies are therefore one of the most important disciplines in experimental chemistry

4

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Kinetic analyses 3

Outline: Energy surfaces

•  based on section 7.1 of A&D –  energy surfaces –  energy profiles –  the nature of a transition state –  rates and rate constants –  reaction order and rate laws

5

Energy profiles and surfaces

•  tools for the visualisation of the change of energy as a function of chemical transformations –  profiles:

•  two dimensions •  often, energy as a function of ‘reaction coordinate’

–  surfaces: •  three dimensions •  energy as a function of two reaction coordinates

6

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Kinetic analyses 4

Example: SN2 energy profile

•  one step passing by a single transition state Fr

ee e

nerg

y

Reaction co-ordinate

X:- + Y-Z reactants

X---Y---Z transition state

δ- δ-

X-Y + :Z-

products

ΔG°

ΔG‡ activation energy

7

Reminder: activated complex •  ethereal complex formed at the transition state of the reaction

–  partially formed bonds –  partially broken bonds

+ ClC + ClClCHOδ−δ−

HOH

HH HH

H

HO CH

HH

transition state not isolable partial bonds

very unstable; <10-13 s

8

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Kinetic analyses 5

Free energy profiles vs surfaces

•  a reaction profile gives the impression that only one reaction pathway can lead to the formation of products

•  however, in reality, reactant molecules are free to ‘explore’ all available reaction pathways –  the pathway with the lowest energy is the one that determines the

predominant mechanism

•  free energy surfaces are in 3D, and take account of more than one event that must take place during a reaction (e.g. bond cleavage and formation) –  allow the consideration off ‘off-pathway’ interactions –  allow the visualisation of mechanistic promiscuity

9

Energy surfaces

•  allow the visualisation of several reaction pathways, or even several reactions that may take place

chem.wayne.edu reactant

product B

product A

activated complex for A activated complex for B

10

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Kinetic analyses 6

More-O’Ferrall and Jencks

•  Rory More-O’Ferrall –  University College, Dublin –  mechanisms et catalysis

•  William P. Jencks (1927-2007) –  Biochemistry, Brandeis University –  a father of modern physical organic chemistry

and biological chemistry –  author of classic textbook –  laureate of several awards

11

Profile to surface

X:- + Y-Z reactants

X-Y + :Z-

products

Cleavage of Y-Z bond

Form

atio

n of

X-Y

bon

d

X:- + Y+ + :Z- unstable

intermediates

X-Y--Z unstable

intermediate

Free

ene

rgy

Reaction co-ordinate

X:- + Y-Z reactants

X---Y---Z TS

δ- δ-

X-Y + :Z-

products

2D: rxn. coord. & energy

2D: 2 rxn. coord.

3D: 2 rxn. coord. & energy

12

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Kinetic analyses 7

Mountain terminology

13

summit (Helmet Pk.) summit (Mt. Robson)

Helmet/Robson col

‘Summit’ vs ‘col’

•  the predominant mechanism follows the reaction pathway of lowest activation energy

•  the ‘summit’ (TS) of this pathway is often a ‘col’ between peaks of even higher energy

Free

ene

rgy

Rxn. co-ord. 14

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Kinetic analyses 8

Exercise: Diagram to mechanism •  Draw the reaction profile that corresponds to the lowest energy

reaction pathway on the MOFJ diagram below. Draw the corresponding mechanism including its transition state(s).

Cleavage of C-LG bond

Form

atio

n of

C-Nu

c bon

d

+ Nuc+ GPEt LG

Nuc + GPLG

H3C CH2

Nuc

GPLG

GP + NucLG

Free

ene

rgy

Rxn. Co-ord.

15

R

R

R

R-

Multi-step reactions •  when a reaction takes place via several elementary chemical steps, one of

these steps will limit the global rate of transformation –  named the rate determining step (rds) or rate limiting step (rls)

•  the rate limiting step is always the step having the highest energy transition state –  even if this step does not have the largest microscopic activation barrier!

16

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Kinetic analyses 9

Rate-limiting step on a reaction profile

•  the rate-limiting step is the one with the highest energy transition state with respect to the ground state

Free

ene

rgy

Rxn. Co-ord.

reactants

products

TS1

int.

TS2

ΔG‡

Free

ene

rgy

Rxn. Co-ord.

reactants

products

TS1

int.

TS2 activation energy of global reaction; step 2 is rls

ΔG1‡

ΔG2‡

ΔG1‡ > ΔG2

‡ so step 1 is rls? NO!!

17

Rate-limiting step on a reaction profile

•  the rate-limiting step is the one with the highest energy transition state with respect to the ground state –  see A&D, Figure 7.3

Free

ene

rgy

Rxn. co-ord.

ΔG‡

Free

ene

rgy

Rxn. co-ord.

ΔG‡

Free

ene

rgy

Rxn. co-ord.

ΔG‡

Free

ene

rgy

Rxn. co-ord.

ΔG‡

Free

ene

rgy

Rxn. co-ord.

ΔG‡

Free

ene

rgy

Rxn. co-ord.

ΔG‡

18

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Kinetic analyses 10

Rate vs rate constant

•  the reaction rate depends on the activation barrier of the global reaction and the concentration of reactants, according to rate law for the reaction –  e.g. v = k[A]

•  the proportionality constant, k, is called the rate constant

19

Rate vs rate constant •  reaction rate at a given moment is the instantaneous slope of [P] vs time •  a rate constant derives from the integration of the rate law

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Temps

[Produit]

different initial rates

same half-lives; same rate constants

different concentrations of reactants; different end points

20

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Kinetic analyses 11

Rate law and molecularity •  each reactant may or may not affect the reaction rate, according to the

rate law for a given reaction •  a rate law is an empirical observation of the variation of reaction rate as

a function of the concentration of each reactant –  procedure for determining a rate law:

•  measure the initial rate (<10% conversion) •  vary the concentration of each reactant, one after the other •  determine the order of the variation of rate as a function of the concentration of

each reactant •  e.g. v ∝ [A][B]2

•  the order of each reactant in the rate law indicates the stoichiometry of its involvement in the transition state of the rate-determining step

21

Integers in rate laws

•  integers indicate the number of equivalents of each reactant that are found in the activated complex at the rae-limiting transition state –  e.g.:

•  reaction: A + B à P –  mechanism: A combines with B to form P

•  rate law: v ∝ [A][B] –  one equivalent of each of A and B are present at the TS of the rds

22

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Kinetic analyses 12

Fractions in rate laws

•  fractions signify the dissociation of a complex of reactants, leading up to the rds: –  e.g. :

•  elementary reactions: A à B + B; B + C à P –  mechanism: reactant A exists in the form of a dimer that must dissociate before

reacting with C to form P

•  rate law: v ∝ [A]½[C] –  true rate law is v ∝ [B][C], but B comes from the dissociation of dimer A –  observed rate law, written in terms of reactants A and C, reflects the dissociation of A –  it is therefore very important to know the nature of reactants in solution!

23

•  negative integers indicate the presence of an equilibrium that provides a reactive species: –  e.g. :

•  elementary reactions: A B + C; B + D à P –  mechanism: A dissociates to give B and C, before B reacts with D to give P

•  rate law: v ∝ [A][C]-1[D] –  true rate law is v ∝ [B][D], but B comes from the dissociation of A –  observed rate law, written in terms of reactants A and D, reflects the dissociation of A –  apparent inhibition by C reflects the displacement of the initial equilibrium

Negative integers in rate laws

24

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Kinetic analyses 13

Outline: Transition State Theory

•  based on section 7.2 of A&D –  Arrhenius approach –  Eyring approach –  temperature effects –  interpretation of thermodynamic parameters

25

Arrhenius •  Svante Arrhenius (1859-1927)

–  Swedish physicist and chemist (Stockholm) –  solution chemistry, activation energy –  also: panspermia, universal language,

greenhouse effect, racial biology –  one of the founders of the Nobel Institute –  Nobel Prize (1903) for the electrolytic theory of

of dissociation

26

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Kinetic analyses 14

Arrhenius equation •  in 1889, Arrhenius noted that the rate constant of a given reaction

increases exponentially with temperature :

•  the activation energy, Ea, was defined as the energy necessary to convert reactants into “high energy species”

•  however, this energy is not expressed in terms of thermodynamic parameters –  for this, it was necessary to develop the transition state theory

2a

RTE

Tln

=dkd

AlnRTEln a +−=k

27

Arrhenius plot

1 / T (K-1)

ln k

obs pente = - Ea / R

ln A

Graphique d'Arrhenius

AlnRTEln a +−=k

28

slope

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Kinetic analyses 15

Eyring •  Henry Eyring (Sr.) (1901-1981)

–  American professor of theoretical chemistry and reaction rates (Princeton, Utah)

–  laureate of several prizes: Wolf Prize, Priestley Medal et National Medal of Science

29

Transition State Theory (TST)

•  developed by Eyring to explain observed reaction rates in terms of standard thermodynamic parameters

•  in their transformation into products, reactants must attain a transition state, in the form of an activated complex –  the reaction rate is proportional to the concentration of this activated complex

K‡ k‡A•B PA B+

complexeactivé

[A][B]]B•A[ ‡

‡ =K [A][B]]B•A[ ‡‡ K= [A][B]]B•A[ ‡‡‡‡ Kkkv ==

30

quasi equilibrium

activated complex

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Kinetic analyses 16

Eyring equation

•  based on transition state theory, relating kinetic parameters to thermodynamic parameters –  k‡ equals the vibration frequency, ν, that leads to decomposition of the

activated complex :

•  kB is Boltzmann’s constant (kB = 1.3806 × 10-23 J K-1)

•  h is Planck’s constant (h = 6.626 × 10-34 J�s)

–  ΔG = -RT ln Keq so ΔG‡ = -RT ln K‡ and

K‡ k‡A•B PA B+

complexeactivé

[A][B]]B•A[ ‡‡‡‡ Kkkv ==

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

= RTΔG

eK

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

== RTGΔ

B‡‡obs

T ehkKkk

⎟⎟⎠

⎞⎜⎜⎝

⎛+

= RSΔ

RTHΔ

Bobs

‡‡

T ehkk

RSΔ

T1

RHΔ

Tln

‡‡

B

obs +⋅−

=⎟⎟⎠

⎞⎜⎜⎝

khk

hkk TB‡ ==

31

activated complex

Eyring plot

1 / T (K-1)

ln (k

obsh

/ k BT

)

pente = - ΔH‡ / R

Graphique d'Eyring

ΔS‡ / R

RSΔ

T1

RHΔ

Tln

‡‡

B

obs +⋅−

=⎟⎟⎠

⎞⎜⎜⎝

khk

32

slope =

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Kinetic analyses 17

Activation (free) energy, ΔG‡

•  directly related to reaction rate •  this parameter represents the sum of energetic factors that influence

reaction rate

33

Activation enthalpy, ΔH‡

•  a measure of the energy barrier that must be overcome by the reactants •  it is the amount of heat energy that the reactants must acquire during their

transformation into the activated complex –  principally related to the formation and cleavage of bonds

34

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Kinetic analyses 18

Activation entropy, ΔS‡

•  a measure of the fraction of all reactants having the necessary activation enthalpy to react that do, in reality, react

•  a measure of probability, this parameter takes account of concentration and solvation effects, steric hindrance and functional group orientation –  for example for a unimolecular reaction, ΔS‡ ≈ 0 –  for multimolecular reactions, ΔS‡ < 0

•  typically –5 to –40 cal mol-1 K-1 (“entropy units”, e.u.)

35

Exercise A: Eyring plot to mechanism •  use the Eyring plot to determine the enthalpy and entropy of activation

for the substitution reaction of iPr-Br –  are the data more consistent with SN1 or SN2 ?

Temperature dependence

-25.00

-20.00

-15.00

-10.00

-5.00

0.00

5.00

10.00

15.00

20.00

0.0E+00 5.0E-04 1.0E-03 1.5E-03 2.0E-03 2.5E-03 3.0E-03 3.5E-03 4.0E-03

1/T (K-1)

ln (k

obs h

/ k B

T)

R = 1.987 cal/mol/K

intercept = ΔS‡/R; ΔS‡ = -6 cal/mol/K

slope = -ΔH‡/R; ΔH‡ = 10 kcal/mol

more ordered TS

little change in bond order at TS

H

CH3H3C

BrNuδ− δ−

SN2 36

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Kinetic analyses 19

Temperature dependence

-50.00

-40.00

-30.00

-20.00

-10.00

0.00

10.00

20.00

0.0E+00 5.0E-04 1.0E-03 1.5E-03 2.0E-03 2.5E-03 3.0E-03 3.5E-03 4.0E-03

1/T (K-1)

ln (k

obs h

/ k B

T)

Exercise B: Eyring plot to mechanism

R = 1.987 cal/mol/K

intercept = ΔS‡/R; ΔS‡ = 10 cal/mol/K

slope = -ΔH‡/R; ΔH‡ = 25 kcal/mol less ordered TS

net bond cleavage

at TS

SN1

H

CH3H3C

Brδ+ δ−

37

•  use the Eyring plot to determine the enthalpy and entropy of activation for the substitution reaction of iPr-Br –  are the data more consistent with SN1 or SN2 ?

Effect of temperature •  an increase of 10 °C will double the rate of most reactions

–  due to the increase of the average energy of reactant molecules :

% o

f mol

cule

s hav

ing

a gi

ven

ener

gy

Energy

T1

T2

T2 > T1

ΔG‡

38

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Kinetic analyses 20

Rate constants and temperature

•  by analogy with ΔG° = -RT ln Keq, ΔG‡ = -RT ln(k/k0)

or

•  the more the temperature increases, the more the factor (ΔG‡/RT) decreases, and the more k increases

RTekk‡ΔG

0

−=

39

Outline: Postulates et principles

•  based on section 7.3 of A&D –  Hammond postulate –  reactivity vs selectivity –  Curtin-Hammett principle –  microscopic reversibility –  kinetic control vs thermodynamic control

40

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Kinetic analyses 21

Hammond •  George S. Hammond (1921-2005)

–  American chemistry professor –  studied the relation between kinetics and product

distribution –  laureate of Norris Award and Priestley Medal

41

Hammond Postulate •  the structure of the activated complex at the transition state of an

elementary reaction that is…

…endergonique (ΔG° > 0) resembles the products

…exergonique (ΔG° < 0) resembles the reactants

Free

ene

rgy

Reaction co-ordinate

42

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Kinetic analyses 22

Hammond Postulate •  to put it another way, the structure of the activated complex at the

transition step of an elementary reaction…

…varies according to ΔG° :

Free

ene

rgy

Reaction co-ordinate 43

Reactivity vs selectivity •  in a comparison of two different possible reactions, the selectivity

between these two reactions may be affected by the reactivity of the reactant: –  a very reactive molecule will pass by two TS having very similar structures,

without demonstrating great selectivity between them:

Free

ene

rgy

Reaction co-ordinate

reactant

product 2 product 1

small ΔΔG‡

44

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•  in a comparison of two different possible reactions, the selectivity between these two reactions may be affected by the reactivity of the reactant: –  a relatively unreactive molecule will pass by two TS having very different

structures, showing great selectivity between them:

Reactivity vs selectivity Fr

ee e

nerg

y

Reaction co-ordinate

reactant

product 2

product 1

large ΔΔG‡

45

Curtin •  David Y. Curtin (~1920 - )

–  American chemist (UIUC) –  studied the difference in reaction rate between

different isomers

46

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Kinetic analyses 24

Hammett •  Louis Hammett (1894-1987)

–  American physical chemist (Columbia) –  studied the correlation of structure and function –  credited with inventing the expression “physical

organic chemistry” –  laureate of awards from the National Academy of

Science, two Norris Awards, Priestley Medal

47

Curtin-Hammett Principle

•  for most organic compounds, conformational changes are more rapid than chemical transformations

•  consider the case where two conformers react to give two different products :

–  ... and where the conformational equilibrium is faster than the reactions that lead to the formation of these products :

A' A"kykx

X Y

k1

k1

k1 >> kx et k -1 >> ky

48

and

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Kinetic analyses 25

Curtin-Hammett Principle

•  a priori, one cannot predict the proportion of products X et Y, based exclusively on the relative stabilities of A' and A"

•  the following energy diagram illustrates that the the relative rates of formation of X and Y depend rather on the relative energies of the transition states leading to their formation

49

Curtin-Hammett Principle

X

Y

A' A"

Rxn. coord.

Free

ene

rgy

K = [A"]/[A'] = k1/k-1 = e-ΔG°/RT

ΔG°

ΔG‡y

ΔG‡x

ΔΔG‡

d[X]/dt = kx[A']

d[Y]/dt = ky[A"]

d[Y]/dt d[X]/dt

= (ky/kx) K

kx[A'] ky[A"]

=

‡ y e-ΔG /RT

= ‡ x e-ΔG /RT

× e-ΔG°/RT = e(ΔG -ΔG -ΔG°)/RT ‡ y

‡ x

= eΔΔG /RT ‡

50

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Kinetic analyses 26

Curtin-Hammett: Example #1 •  methylation of a tertiary bicyclic amine:

–  reactant conformational equilibrium does not determine product distribution

NMe

NMeH3C

13CH3IN

MeN

Me CH313CH3I

more stable

less stable

fast slow major minor

Rxn. co-ordinate

Free

ene

rgy

ΔΔG‡

51

Curtin-Hammett Conditions •  it is important to keep the following condition in mind: conformational

changes must be faster than chemical transformations –  in this case, [PA]/[PB] = kA/kB×Keq

•  however, at low temperatures, conformational changes are slow, and product ratios often reflect the composition of the initial equilibrium (“kinetic quench”) –  i.e. kx >> k1 and ky >> k-1, so [PA]/[PB] = Keq

–  without interconversion, A and B are essentially separate populations that react independently

•  all A is converted to PA and all B is converted to PB, regardless of the relative rates of these transformations

k1 >> kx et k -1 >> ky

52

and

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Kinetic analyses 27

N

O

Me

Me

CH3

CH3-BrN

O

Me

Me

CH3

CH3-Br

fast fast N

NMe

H

Me

OOMe

Me

H

N

O

Me

Me

s-BuLi

ΔG = -3.0 kcalby calculations

“Kinetic quench”: Example •  alkylation of pyrrolidinone at -78°C

JACS, 1997, 119, 4565

disfavoured

slow, at -78°C

Rxn. co-ordinate

Free

ene

rgy

ΔG‡A

99% 1%

ΔG‡B

[ ][ ]

[ ][ ]0

0

A

B

AB

PP

=

53

Curtin-Hammett: Conclusions

•  Curtin-Hammett principle is very important to keep in mind: –  a priori, one cannot correlate reactant equilibrium with product distribution

•  for fast conformational equilibria and relatively slow reactions, product ratio is determined by ΔΔG‡

•  however, for slow conformational equilibria and relatively fast reactions, product ratio is determined by Keq

•  in practice, it is difficult to apply this principle : •  it requires the measurement of the proportion of reactant conformers •  it requires independent measurement of equilibrium conformation and reaction

rates

54

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Kinetic analyses 28

Microscopic reversibility

•  the reaction pathway for a reverse reaction must be exactly the reverse of the reaction pathway for the forward reaction –  the transition states for the forward and reverse reactions are identical –  allows the prediction of the nature of a transition state for one reaction, based

on knowledge of the transition state for the reverse reaction :

general base catalysis

general acid catalysis

55

H3C OH

O

OEtH

H3C OH

O

OEt

BB

H

Microscopic reversibility

•  the reaction pathway for a reverse reaction must be exactly the reverse of the reaction pathway for the forward reaction –  the transition states for the forward and reverse reactions are identical –  allows the prediction of the nature of a transition state for one reaction, based

on knowledge of the transition state for the reverse reaction :

56

H3C OH

O

O

H3C OH

O

OO2N O2N

uncatalysed attack

uncatalysed expulsion

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Kinetic analyses 29

Kinetic vs thermodynamic control •  factors that can influence product ratios •  kinetic control:

–  ratio affected by ΔΔG‡, not by ΔΔGrxn –  irreversible reactions, Curtin-Hammett conditions

Reaction coordinate

Free

ene

rgy

ΔΔG‡

reactant product B product A

57

Kinetic vs thermodynamic control •  factors that can influence product ratios •  thermodynamic control:

–  ratio affected by ΔΔGrxn not by ΔΔG‡

–  reversible reactions (often at high temperatures)

58

Reaction coordinate

Free

ene

rgy

ΔΔGrxn

reactant product B product A

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Kinetic analyses 30

Outline: Experimental kinetics

•  based on A&D section 7.4 –  practical kinetics –  kinetic analyses

•  first order •  second order •  pseudo-first order

59

Practical kinetics

1.  development of a method of detection (analytical chemistry!) 2.  measurement of concentration of a product or of a reactant as a function

of time 3.  measurement of reaction rate (slope of conc/time; d[P]/dt or -d[A]/dt)

–  correlation with rate law and reaction order 4.  calculation of rate constant

–  correlation with structure-function studies

60

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Kinetic analyses 31

Kinetic assays

•  method used to measure the concentration of reactants or of products, as a function of time –  often involves the synthesis of chromogenic or fluorogenic reactants

•  can be continuous or non-continuous

61

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

Continuous assay •  instantaneous detection of reactants or products as the reaction is

underway –  requires sensitive and rapid detection method

•  e.g.: UV/vis, fluorescence, IR, (NMR), calorimetry

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time

[Pro

duct

]

62

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Kinetic analyses 32

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time[P

rodu

ct]

Discontinuous assay •  involves taking aliquots of the reaction mixture at various time points,

quenching the reaction in those aliquots and measuring the concentration of reactants/products –  wide variety of detection methods applicable

•  e.g.: as above, plus HPLC, MS, etc.

63

Continuous reaction assay

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

Time[P

rodu

ct]

Discontinuous

Initial rates •  the first ~10 % of a reaction is almost linear, regardless of the

order of a reaction •  ΔC vs Δt gives the rate, but the rate constant depends on the

order of the reaction

0 10 20 30 40 50 60 70 80 90 100

Temps (min)

0

10

20

30

40

50

60

70

80

90

100

[A] (

mM

)

k = 5 mM min- 1 k1 = 0.07 min- 1

k2 = 0.001 mM- 1 min- 1

tΩ = 10 min

Time

64

½

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Kinetic analyses 33

Calculation of a rate constant

•  now, how can a rate constant, k, be determined quantitatively? •  the mathematical equation to use to determine the value of k differs

according to the order of the reaction in question –  the equation must be derived from a kinetic scheme –  next, the data can be “fitted” to the resulting equation using a computer

(linear, or more likely, non-linear regression)

65

The language of Nature •  “Nul ne saurait comprendre la nature si

celui-ci ne connaît son langage qui est le langage mathématique”

- Blaise Pascal –  (‘None can understand Nature if one does not know its

language, which is the language of mathematics’)

•  Natural order is revealed through special mathematical relationships

•  mathematics are our attempt to understand Nature –  exponential increase: the value of e –  volume of spherical forms: the value of π

Blaise Pascal, French mathematician

and philosopher, 1623-1662

66

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Kinetic analyses 34

First order (simple)

A Pk1

= k1 [A]Vitesse = v = = -d[P]dt

d[A]dt

d[P] k ([P] [P])1dt= −∞

d[P] k ([A] [P])1 0dt= −

d dt[P][P] [P]

k1∞ −

=

∫∫ =−∞

t

01

P

0dtk

[P][P][P]d

ln ([P]∞ - [P]0) - ln ([P]∞ - [P]) = k1 t

ln t[A][A]

k01=

[A] = [A]0k1e− t

linear relation

mono-exponential decrease

ln t[A]½[A]

k0

01 ½=

t ln½

1

2k

=half-life

Rate

67

First order (simple)

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

kobs = k1

[A]∞

[A]0

0 10 20 30 40 50 60 70 80 90 100Temps

-7-6-5-4-3-2-101

ln ([

A]/[A

] 0)

pente = kobs = k1

mono-exponential decrease

Time

Time

slope

68

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Kinetic analyses 35

First order (reversible)

A Pk1

k-1

= k1 [A] - k-1 [P]Vitesse = v = = -d[P]dt

d[A]dt

( )( )[P]kk[P]k[A]k

[P]k[P][P][A]kdt[P]d

dt[A]d

1-1éq1éq1

1-éqéq1

+−+=

−−+==−

À l’équilibre, k1 [A]éq = k-1 [P]éq ( )

( )( )− = = + − +

= + −

ddt

ddt

[A] [P]k [P] k [P] k k [P]

k k [P] [P]

-1 éq 1 éq 1 -1

1 -1 éq

( )ddtt[P]

[P] [P]k k

éq0

P

1 -1 0−= +∫ ∫

( )( ) ( )ln t[P] [P]

[P] [P]k k

éq 0

éq

1 -1

−= +linear relation kobs = (k1 + k-1)

Rate

At equilibrium,

69

First order (reversible)

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]Èq

[P]Èq

kobs = k1

kobs = k1 + k-1

[A]∞

[A]0

[P]0

( )t

ln½

1

2k k

=+ −1

Kéq 1

-1

kk

=

Time

70

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Kinetic analyses 36

Second order (simple)

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

d[P] k ([A] [P])([B] [P])2 0 0dt= − −

d dt[P]([A] [P])([B] [P])

k0 0

2− −=

ddtt[P]

([A] [P])([B] [P])k

0 00

P

2 0− −=∫ ∫

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

a lot of error is introduced when [B]0 and [A]0 are similar

Rate

71

Second order (simplified)

If [A]0 = [B]0 :

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

ddtt[P]

([A] [P])k

020

P

2 0−=∫ ∫

1 1[A] [P] [A]

k0 0

2−− = t

1 1[A] [A]

k0

2− = tlinear relation

t½2 0

1k [A]

=half-life

1 1[A]2

[A]k

0 02 ½⎛

⎝⎜

⎞⎠⎟

− = t

Rate

72

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Kinetic analyses 37

Second vs first order

•  one must often follow the progress of a reaction for several (3-5) half-lives, in order to be able to distinguish between a first order reaction and a second order reaction :

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

premier ordre, k1

k2 = k1 ˜ 70

k2 = k1

first order,

73

Réactions cinétiques

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25

Temps

[Pro

duit]

Calcul de constante de vitesse

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

16.00

18.00

0 20 40 60 80 100 120

[A]

vite

sse

initi

ale

Example: first or second order? •  measure of [P] as a function of time •  measure of v0 as a function of time [A]

Reaction kinetics

first order [A] = [A]0 e –k1t

Time

[Pro

duct

]

Rate constant calculation

v0 = k1 [A]

In

itial

rate

74

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Kinetic analyses 38

Réactions cinétiques

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25

Temps

[Pro

duit]

Calcul de constante de vitesse

0.00

5.00

10.00

15.00

20.00

25.00

30.00

0 20 40 60 80 100 120

[A] = [B]

vite

sse

initi

ale

second order

•  measure of [P] as a function of time •  measure of v0 as a function of time [A]

Time

Reaction kinetics

[Pro

duct

]

Rate constant calculation

In

itial

rate

1/[A] – 1/[A]0 = k2t v0 = k2 [A][B]

Example: first or second order?

75

Pseudo-first order

•  in the case where the initial concentration of one of the reactants is much larger than that of the other, one can simplify the treatment of the experimental data

Vitesse = v = = k2[A][B]d[P]dt

A B P+k2

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t general solution:

second order:

76

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Kinetic analyses 39

Pseudo-first order •  consider the case where [B]0 >> [A]0 (>10 times larger)

–  the concentration of B will not change much over the course of the reaction; [B] = [B]0 (quasi-constant)

1[B]

[A] [B][B] ([A] [P])

k0

0 0

0 02ln

⎝⎜

⎠⎟ = t

1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

ln [A]([A] [P])

k [B]0

02 0−

= t

ln [A]([A] [P])

k où k k [B]0

01'

1'

2 0−= =t

ln t[A][A]

k01'=

[A] = [A]0k1'

e− tmono-exponential decrease

where

77

Pseudo-first order •  from a practical point of view, it is more reliable to determine a

second order rate constant by measuring a series of first order rate constants as a function of the concentration of B (provided that [B]0 >> [A]0)

0 10 20 30 40 50 60 70 80 90 100

Temps (min)

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]∞

[A]0

0.15 min- 10.10 min- 1

0.07 min- 1

0.035 min- 1

ko b s = k1'

[B]0 = 0.033 M[B]0 = 0.065 M[B]0 = 0.098 M[B]0 = 0.145 M

Time 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

[B]0 (M)

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

k obs

(min

-1)

pente = k2 = (1.02 ± 0.02) M- 1min- 1slope

78

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Kinetic analyses 40

Third order

•  reactions that take place in one termolecular step are rare in the gas phase and do not exist in solution –  the entropic barrier associated with the simultaneous collision of three

molecules is too high

A B P+k3

C+

Vitesse = v = = k3[A][B][C]d[P]dt

Rate

79

Third order, revisited

•  however, a reaction that takes place in two consecutive bimolecular steps (where the second step is rate-limiting) would have a third order rate law!

A + B AB

C P

k1

+ABk2

Vitesse = v =d[P]dt

= k3[A][B][C]Rate

80

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Kinetic analyses 41

Zeroth order

•  in the presence of a catalyst (organo-metallic or enzyme, for example) and a large excess of reactant, the rate of a reaction can appear to be constant

Vitesse = v = = kd[P]dt

= - d[A]dt

A Pk

catalyseur+ catalyseur+

− =∫ ∫d dtt

[A] kA

A

00[A]0 - [A] = k t [A] = -k t + [A]0 linear relation

[A]0 - ½ [A]0 = k t½

t½ 0[A]2 k

= half-life

catalyst catalyst

Rate

81

Zeroth order

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

pente = -k

not realistic that rate would be constant all the way to end of reaction...

Time

slope

82

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Kinetic analyses 42

Summary of observed parameters

Reaction order

Rate law Explicit equation Linear equation Half-life

zero d[P] kdt

= [A] = -k t + [A]0 [A]0 - [A] = k t t½ 0[A]2 k

=

first d[P] k [A]1dt= [A] = [A]0

k1e− t ln t[A][A]

k01= t ln

½1

2k

=

second d[P] k [A]22

dt= [A]

k[A]2

0

=+

11t

1 1[A] [A]

k0

2− = t t½2 0

1k [A]

=

d[P] k [A][B]2dt=

complex 1[B] [A]

[A] ([B] [P])[B] ([A] [P])

k0 0

0 0

0 02−

⎝⎜

⎠⎟ =ln t

complex

linear regression

non-linear regression

83

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

0.0

20.0

40.0

60.0

80.0

100.0

0 10 20 30 40 50 60 70

Time (min)

[A] (

mM

)

Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.000 10 55.6 0.59 0.008 20 38.5 0.96 0.016 30 29.4 1.22 0.024 40 23.8 1.44 0.032 50 20.0 1.61 0.040 60 17.2 1.76 0.048

84

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Kinetic analyses 43

0.00

0.50

1.00

1.50

2.00

2.50

0 10 20 30 40 50 60 70

Time (min)

ln[A

]0/[A

] y = 0.0008x

0.000

0.010

0.020

0.030

0.040

0.050

0.060

0 10 20 30 40 50 60 70

Time (min)

1/[A

] - 1

[A]0

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

linear non-linear

second order; kobs = 0.0008 mM-1min-1

85

Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

0.0

20.0

40.0

60.0

80.0

100.0

0 10 20 30 40 50 60 70

Time (min)

[A] (

mM

)

Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A]0 0 100.0 0.00 0.00 10 60.7 0.50 0.006 20 36.8 1.00 0.017 30 22.3 1.50 0.035 40 13.5 2.00 0.064 50 8.2 2.50 0.112 60 5.0 3.00 0.190

86

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Kinetic analyses 44

Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate

constant

y = 0.05x

0

0.5

1

1.5

2

2.5

3

3.5

0 10 20 30 40 50 60 70

Time (min)

ln ([

A]0/

[A]) y = 0.003x - 0.0283

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 10 20 30 40 50 60 70

Time (min)

1/[A

] - 1

/[A]0

linear non-linear

first order; kobs = 0.05 min-1

87

Outline: Complex reactions

•  based on A&D section 7.5 –  consecutive first order reactions –  steady state –  changes in kinetic order –  saturation kinetics –  rapid pre-equilibrium

88

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Kinetic analyses 45

First order (consecutive)

Ak1

B Pk2

= k2 [B]Vitesse = v = d[P]dt ≠ -

d[A]dt

− =d[A] k [A]1dt

[A] = [A]0k1e− t

d[B] k [A] k [B]1 2dt= −

d t[B] k [A] k [B]1 0k

21

dte= −− d t[B] k [B] k [A]2 1 0

k1

dte+ = −

solved by using the technique of partial derivatives

[B] k [A](k k )

( )1

2 1

k k=−

−− −0 1 2e et t

[P] [A] [A] k [A](k k )

( )k 1

2 1

k k= − −−

−− − −0 0

01 1 2e e et t t [P] [A]k

(k k )( )k 1

2 1

k k= − −−

−⎛

⎝⎜

⎠⎟− − −

0 1 1 1 2e e et t t

Rate

89

Induction period

•  in consecutive reactions, there is an induction period in the production of P, during which the concentration of B increases to its maximum before decreasing

A à B à P

•  the length of this period and the maximal concentration of B varies as a function of the relative values k1 and k2 –  consider three representative cases :

•  k1 = k2 •  k2 < k1

•  k2 > k1

90

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Kinetic analyses 46

Consecutive reactions, k1 = k2

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0

[B]0, [P]0

[P]∞

[A]∞ , [B]∞

[B]max

pÈriode d'induction

k2 ≈ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-1

0

1

2

3

4

5

ln([P

] ∞ -

[P] 0)

- ln

([P] ∞

- [P

])

pente = kobs = k2 ≈ k1

k2 ≈ k1

pÈriode d'induction

induction period

induction period slope

Time

Time

91

Consecutive reactions, k2 = 0.2 × k1

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0

[B]max

[A]∞

[P]

[B]

k2 = 0.2 ◊ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

ln([

P] ∞

- [

P] 0

) -

ln([

P] ∞

- [

P])

pÈriode d'induction

pente = kobs = k2

Time

induction period slope

Time

92

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Kinetic analyses 47

0 10 20 30 40 50 60 70 80 90 100

Temps

0

10

20

30

40

50

60

70

80

90

100

% [A

] 0

[A]0 [P]∞

[A]∞ , [B]∞

[B]max

k2 = 5 ◊ k1

0 10 20 30 40 50 60 70 80 90 100

Temps

-1

0

1

2

3

4

5

6

7

ln([P

] ∞ -

[P] 0)

- ln

([P] ∞

- [P

])

pÈriode d'induction

pente = kobs = k1

k2 = 5 ◊ k1

Consecutive reactions, k2 = 5 × k1

Time

induction period

slope

Time

93

Steady state •  often multi-step reactions involve the formation of a reactive intermediate

that does not accumulate but reacts as rapidly as it is formed •  the concentration of this intermediate can be treated as though it is

constant •  this is called the steady state approximation (SSA)

94

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Kinetic analyses 48

Steady State Approximation

•  consider a typical example (in bio-org and organometallic chem) of a two step reaction: –  kinetic scheme:

–  rate law:

–  SSA :

•  expression of [I] :

–  rate equation:

A I Pk1

k-1

k2[B]

[ ] [ ][ ]BItP

2kdd

=

[ ] [ ] [ ] [ ][ ] 0BIIAtI

211 =−−= − kkkdd

[ ] [ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BAI21

1

kkk

[ ] [ ][ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BBA

tP

21

21

kkkk

dd first order in A;

less than first order in B

95

Steady State Approximation

•  consider another example of a two-step reaction: –  kinetic scheme:

–  rate law:

–  SSA:

•  expression of [I] :

–  rate equation:

[ ] [ ]ItP

2kdd

=

[ ] [ ][ ] [ ] [ ] 0IIBAtI

211 =−−= − kkkdd

[ ] [ ][ ]⎟⎟⎠

⎞⎜⎜⎝

+=

− 21

1 BAIkk

k

[ ] [ ][ ] [ ][ ]BABAtP

obs21

21 kkk

kkdd

=⎟⎟⎠

⎞⎜⎜⎝

+=

A + B I Pk1

k-1

k2

kinetically indistinguishable from the mechanism with no intermediate!

96

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Kinetic analyses 49

Steady State Approximation

•  consider a third example of a two-step reaction: –  kinetic scheme:

–  rate law:

–  SSA:

•  expression of [I] :

–  rate equation:

[ ] [ ][ ]BItP

22 kdd

=

[ ] [ ] [ ][ ] [ ][ ] 0BIPIAtI

2111 =−−= − kkkdd

[ ] [ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPAI

211

1

kkk

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

A I + P1k1

k -1k2[B]

P2

first order in A, less than first order in B; slowed by P1

97

SSA Rate equations

•  useful generalisations: 1.  the numerator is the product of the rate constants and concentrations

necessary to form the product; the denominator is the sum of the rates of the different reaction pathways of the intermediate

2.  terms involving concentrations can be controlled by varying reaction conditions

3.  reaction conditions can be modified to make one term in the denominator much larger than another, thereby simplifying the equation as zero order in a given reactant

98

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Change of reaction order

•  by adding an exces of a reactant or a product, the order of a rate law can be modified, thereby verifying the rate law equation

•  for example, consider the preceding equation:

–  in the case where k-1 >> k2 , in the presence of excess B and (added) P1, the equation can be simplified as follows:

•  now it is first order in A and in B

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

[ ] [ ][ ][ ] ⎟⎟

⎞⎜⎜⎝

⎛=

− 11

212

PBA

tP

kkk

dd

99

Change of reaction order

•  however, if one considers the same equation:

–  but reaction conditions are modified such that k-1[P1] << k2[B] , the equation is simplified very differently:

•  now the equation is only first order in A

•  in this way a kinetic equation can be tested, by modifying reaction conditions

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BPBA

tP

211

212

kkkk

dd

[ ] [ ][ ][ ] ⎟⎟

⎞⎜⎜⎝

⎛=

BBA

tP

2

212

kkk

dd

[ ] [ ]AtP

12 kdd

=

100

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Saturation kinetics

•  on variation of the concentration of reactants, the order of a reactant may change from first to zeroth order –  the observed rate becomes “saturated” with respect to a reactant –  e.g., for the scheme

having the rate law

one observes:

A I Pk1

k-1

k2[B]

[ ] [ ][ ][ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− BBA

tP

21

21

kkkk

dd

v

[B]0

vmax = k1[A]

first order in B

~zero order in B

101

Example of saturation kinetics

•  for a SN1 reaction, we are taught that the rate does not depend on [Nuc], but this is obviously false at very low [Nuc], where vobsà0 all the same

•  in reality, for

one can show that •  but normally k2 >> k-1 and [-CN] >> [Br-]

so the rate law becomes:

•  i.e., it is typically already saturated with respect to [-CN]

Brk1

k -1+ Br-

k2 [-CN] CN

[ ] [ ][ ][ ] [ ]⎟

⎟⎠

⎞⎜⎜⎝

+=

− CNBrCNR

tP

21

21

kkkk

dd

[ ] [ ][ ][ ]

[ ]RCNCNR

tP

12

21 kkkk

dd

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

102

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Rapid pre-equilibrium

•  in the case where a reactant is in rapid equilibrium before its reaction, one can replace its concentration by that given by the equilibrium constant

•  for example, for the protonated alcohol is always in rapid eq’m with the alcohol and therefore

•  normally, k2[I-] >> k-1[H2O] and therefore

OHk1

k -1+ H2O

k2 [-I] IH

OHKeq[H+]

[ ] [ ][ ][ ][ ] [ ] ⎟⎟

⎞⎜⎜⎝

+

+=

− IOHIBuOHH

tP

221

21

kktKkk

dd eq

[ ] [ ][ ][ ][ ]

[ ][ ]BuOHHI

IBuOHHtP

12

21 tKkktKkk

dd

eqeq +=⎟⎟

⎞⎜⎜⎝

⎛ += first order in tBuOH,

zero order in I-, pH dependent

103

Outline: Multiple reaction co-ordinates

•  based on A&D section 7.8 –  variation of activated complexes –  More-O’Ferrall-Jencks diagrams –  vibrational state effects

104

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Nucleophilic substitution reactions •  SN1 and SN2 reactions both involve the cleavage of the C-LG bond and

the formation of the C-Nuc bond

•  if these events are assigned to the x and y axes of an energy surface, one can visualise how the structure of a substrate will determine which of the two mechanisms is favoured:

105

More-O’Ferrall-Jencks (MOFJ) diagram •  projection of an energy surface on a two-dimensional plot

–  (e.g. Figure 7.21 A&D) :

note the displacement of the TS from an SN2 reaction to an SN1

106

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Kinetic analyses 54

Displacement of a transition state •  on a MOFJ diagram, one raises or lowers the corners of the plot, as a

function of the change of structure of a species and then considers how the transition state would be displaced, according to two effects:

1. “Hammond effect”: if the corner of the reactants or products is lowered, the transition state is displaced along the diagonal of the pathway, towards the opposite corner, according to the Hammond postulate

2. “orthogonal effect” : if an “off-diagonal” corner is lowered, the transition state is displaced towards the lowered corner

‡ ‡

R

P

I1

I2 R

P

I1

I2 107

More-O’Ferrall-Jencks Diagram

‡ ‡

Et GP + Nuc Et Nuc + GPLG LG

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

GP

Nucδ−

δ−Me

H

H

LG

symmetric

108

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Kinetic analyses 55

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

More-O’Ferrall-Jencks Diagram

Et GP + Nuc Et Nuc + GPbetter

‡ ‡

LG LG

GP

Nucδ−

δ−Me

H

Hδ−

LG

109

“Hammond effect”

“orthogonal effect”

less C-LG cleavage

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

More-O’Ferrall-Jencks Diagram

worse

‡ ‡

Et GP + Nuc Et Nuc + GPLG LG

GP

Nucδ−

δ−

MeH

Hδ+

LG

110

more C-LG cleavage

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Kinetic analyses 56

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

More-O’Ferrall-Jencks Diagram

better

Et GP + Nuc Et Nuc + GPLG LG

GP

Nucδ−

δ−Me

H

Hδ+

LG

less C-Nuc formation

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

More-O’Ferrall-Jencks Diagram

better

worse Et GP + Nuc Et Nuc + GPLG LG

GP

Nucδ−

δ−

MeH

Hδ+

LG

112

dissociative

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Kinetic analyses 57

Cleavage of C-LG

Form

atio

n of

C-N

uc

GP + Nuc

Nuc + GP

+ Nuc+ GPEt

H3C CH2

Nuc

GP

LG

LG LG

LG

More-O’Ferrall-Jencks Diagram

worse

worse Et GP + Nuc Et Nuc + GP

LG LG

GP

Nucδ−

δ−

MeH

H

LG resembles products!

Vibrational effects

•  certain vibrational modes of certain bonds may assist in the formation of a given activated complex –  known as productive vibrations

•  on the other hand, other vibrational modes may impede attainment of the transition state –  known as non-productive vibrations

•  the impact of these vibrational effects depends on the reaction and the nature of its transition state(s)

•  the shape of the energy surface of the reaction reflects this dependence and helps to visualise the entropic effect of changes of the degrees of freedom on passing from a reactant to the transition state

114

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Shape of an energy surface •  the width of a “valley” on a surface is related to the

number of degrees of freedom, and therefore entropy, of a given molecule

1.  when the col is as narrow as the initial valley, the activated complex has the same shape (and degrees of freedom) as the reactant and ΔS‡ ≈ 0

2.  when the col is wider than the initial valley, there are more degrees of freedom at the transition state, and ΔS‡ > 0

3.  when the col is narrower than the initial valley, there are fewer degrees of freedom in the activated complex, and ΔS‡ < 0

1.

2.

3. Figure 7.23, A&D 115

Temperature and energy surfaces

•  at higher temperature, molecules are more excited and can navigate more easily through broad cols than through narrow cols

•  this analogy helps us understand how entropy contributes to the free energy of activation at higher temperatures, according to the following equation:

ΔG‡ = ΔH‡ - TΔS‡

116

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Summary: Kinetic approach to mechanisms

•  kinetic measurements provide rate laws –  ‘molecularity’ of a reaction

•  rate laws limit what mechanisms are consistent with reaction order –  several hypothetical mechanisms may be proposed

•  detailed studies (of substituent effects, etc) are then necessary in order to eliminate all mechanisms – except one! –  one mechanism is retained that is consistent with all data –  in this way, the scientific method is used to refute inconsistent

mechanisms (and support consistent mechanisms)

117