CHM 8304

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

Kinetic analyses

1

CHM 8304

“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

Kinetic analyses

2

CHM 8304

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

Kinetic analyses

3

CHM 8304

Example: SN2 energy profile •  one step passing by a single transition state

δ-

δ-

Free energy

X---Y---Z transition state

activation energy

ΔG‡ X:- + Y-Z reactants

ΔG°

X-Y + :Zproducts

Reaction co-ordinate

7

Reminder: activated complex •  ethereal complex formed at the transition state of the reaction –  partially formed bonds very unstable; so step 1 is rls? NO!! ΔG1‡

ΔG2‡

Free energy

Free energy

TS1

activation energy of global reaction; step 2 is rls

int.

ΔG‡

products

reactants

Rxn. Co-ord.

Rxn. Co-ord.

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

ΔG‡

Rxn. co-ord.

Kinetic analyses

Free energy

Rxn. co-ord.

Free energy

Free energy

Rxn. co-ord.

ΔG‡

ΔG‡ Rxn. co-ord.

ΔG‡ Rxn. co-ord.

Free energy

ΔG‡

Free energy

Free energy

–  see A&D, Figure 7.3

ΔG‡ Rxn. co-ord.

18

9

CHM 8304

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 100

different concentrations of reactants; different end points

90 80

different initial rates

[Produit]

70 60 50 40 30 20 10 0 0

5

10

15

20

Temps

same half-lives; same rate constants 20

Kinetic analyses

10

CHM 8304

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 ( T1

Energy

38

Kinetic analyses

19

CHM 8304

Rate constants and temperature •  by analogy with ΔG° = -RT ln Keq, ΔG‡ = -RT ln(k/k0) or

k = k0e

ΔG ‡ − RT

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

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

Kinetic analyses

20

CHM 8304

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

Free energy

…exergonique (ΔG° < 0) resembles the reactants

Reaction co-ordinate

42

Kinetic analyses

21

CHM 8304

Hammond Postulate •  to put it another way, the structure of the activated complex at the transition step of an elementary reaction…

Free energy

…varies according to ΔG° :

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:

Free energy

–  a very reactive molecule will pass by two TS having very similar structures, without demonstrating great selectivity between them:

small ΔΔG‡ reactant

product 1

product 2

Reaction co-ordinate

44

Kinetic analyses

22

CHM 8304

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:

Free energy

–  a relatively unreactive molecule will pass by two TS having very different structures, showing great selectivity between them:

large ΔΔG‡

product 2

product 1 reactant

Reaction co-ordinate

45

Curtin •  David Y. Curtin (~1920 - ) –  American chemist (UIUC) –  studied the difference in reaction rate between different isomers

46

Kinetic analyses

23

CHM 8304

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 : X

kx

k1 A'

k1

A"

ky

Y

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

k 1 >> kx and et k -1 >> k y

48

Kinetic analyses

24

CHM 8304

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 d[Y]/dt = ky[A"] = (k /k ) K y x d[X]/dt kx[A']

‡

=

e-ΔGy /RT e

‡ -ΔGx /RT

‡

‡

× e-ΔG°/RT

= e(ΔGx -ΔGy -ΔG°)/RT = eΔΔG

‡

/RT

ΔΔG‡

Free energy

d[X]/dt = kx[A'] ΔG‡x ΔG‡y A"

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

ΔG°

A'

X

K = [A"]/[A'] = k1/k-1 = e-ΔG°/RT

Y

Rxn. coord. 50

Kinetic analyses

25

CHM 8304

Curtin-Hammett: Example #1 •  methylation of a tertiary bicyclic amine: –  reactant conformational equilibrium does not determine product distribution

H 3C

N

Me 13CH I 3

fast major

N

Me

Me

N

more stable

less stable

Me 13CH I 3

N

CH 3

slow minor

Free energy

ΔΔG‡

Rxn. co-ordinate 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

k1 >> k x and et k -1 >> k y

•  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

52

Kinetic analyses

26

CHM 8304

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

disfavoured

Me Me

O

s-BuLi

Me

CH3 -Br

N

CH 3 O

1% Free energy

N

fast

Me N

H Me

N

O

ΔG = -3.0 kcal by calculations

Me

Me

CH 3-Br

H

N CH 3 O

slow, at -78°C

ΔG‡A

Me

fast

O

99%

[PB ] = [B]0 [PA ] [A]0

ΔG‡B Rxn. co-ordinate

JACS, 1997, 119, 4565

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

Kinetic analyses

27

CHM 8304

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 :

O

O

H 3C Et

general base catalysis

OH

H 3C Et

O H

OH

general acid catalysis

O H

B

B

55

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 :

O H 3C O2N

uncatalysed attack

O OH

O

H 3C O2N

OH O

uncatalysed expulsion

56

Kinetic analyses

28

CHM 8304

Kinetic vs thermodynamic control •  factors that can influence product ratios •  kinetic control: –  ratio affected by ΔΔG‡, not by ΔΔGrxn –  irreversible reactions, Curtin-Hammett conditions

Free energy

ΔΔG‡

product A

reactant

product B

Reaction coordinate 57

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

Free energy

–  ratio affected by ΔΔGrxn not by ΔΔG‡ –  reversible reactions (often at high temperatures)

product A

reactant

product B

ΔΔGrxn Reaction coordinate 58

Kinetic analyses

29

CHM 8304

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

Kinetic analyses

30

CHM 8304

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 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 100 90 80

[Product]

70 60 50 40 30 20 10 0 0

5

10

15

20 20

Time

62

Kinetic analyses

31

CHM 8304

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. Continuous reaction assay Discontinuous 100 90 80

[Product]

70 60 50 40 30 20 10 0 0

5

10

15

20

Time

63

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 100 90 80

[A] (mM)

70 60 50

tΩ = 10 min ½

40 30 20 10 0

k 2 = 0.001 mM- 1 min- 1 k 1 = 0.07 min- 1

k = 5 mM min- 1 0

10

20

30

40

50

60

70

80

90

100

Temps Time (min) 64

Kinetic analyses

32

CHM 8304

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

Blaise Pascal, French mathematician and philosopher, 1623-1662

–  exponential increase: the value of e –  volume of spherical forms: the value of π

66

Kinetic analyses

33

CHM 8304

First order (simple) k1

A

Vitesse Rate

= v =

P

0

d[P] d[A] = = k1 [A] dt dt d [P] = k 1 dt [P]∞ − [P]

d[P] = k1 ([P]∞ − [P]) dt

d[P] = k1 ([A]0 − [P]) dt

∫

P

t d[P] = k 1 ∫ dt 0 [P]∞ − [P]

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

ln

linear relation

[A]0 = k1 t [A]

ln

[A]0 = k 1 t½ ½[A]0

[A] = [A]0 e − k1 t

mono-exponential decrease

t½ =

half-life

ln 2 k1 67

First order (simple) 1 0

[A]0

-1

ln ([A]/[A]0)

100 90

mono-exponential decrease

80

% [A]0

70

-2

pente = kobs = k1 slope

-3 -4 -5

60

-6

50

-7 0

40

10 20 30 40 50 60 70 80 90 100

Temps Time

30 20

kobs = k1

10

ttΩ½

0 0

10

20

[A]∞ 30

40

50

60

70

80

90

100

Temps Time 68

Kinetic analyses

34

CHM 8304

First order (reversible) k1

A

Vitesse Rate

= v =

d[P] d[A] = = k1 [A] - k-1 [P] dt dt −

l’équilibre, k1 [A]éq = k-1 [P]éq At À equilibrium,

P

k-1



d[A] d[P] = = k 1 ([A]éq + [P]éq − [P]) − k -1[P] dt dt = k 1[A]éq + k 1[P]éq − (k 1 + k -1 )[P]

−

d[A] d[P] = = k -1 [P] éq + k 1 [P] éq − (k 1 + k -1 )[P] dt dt

(

)

= (k 1 + k -1 ) [P] éq − [P]

∫

P

0

linear relation

ln

t d[P] = (k 1 + k -1 )∫ dt 0 [P] éq − [P]

([P] ([P]

éq éq

) = (k − [P])

− [P] 0

1

+ k -1 ) t

kobs = (k1 + k-1)

69

First order (reversible) [A] 0

100 90

t½ =

80

ln 2

(k

1

+ k −1 )

70

[P] Èq

% [A]0

60

Kéq =

k obs = k 1 + k -1

50

[A] Èq

40

k1 k -1

30 20

k obs = k 1

10

[A] ∞

[P] 0

0 0

10

20

30

40

50

60

70

80

90

100

Temps Time

70

Kinetic analyses

35

CHM 8304

Second order (simple) +

A

B

Vitesse Rate

= v =

k2

P

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

d [P] = k 2 dt ([A]0 − [P])([B]0 − [P])

d[P] = k 2 ([A]0 − [P])([B]0 − [P]) dt

∫

P

0

t d[P] = k 2 ∫ dt 0 ([A] 0 − [P])([B] 0 − [P])

⎛ [A] 0 ([B] 0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠

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

71

Second order (simplified) A

+

B

Vitesse Rate

= v =

If [A]0 = [B]0 :

∫

P

0

k2

P

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

t d[P] = k 2 ∫ dt 0 ([A] 0 − [P]) 2

1 1 − = k 2t [A]0 − [P] [A]0 linear relation

half-life

1 1 − = k 2t [A] [A]0

t½ =

1 k 2 [A]0

1 1 − = k2 t½ ⎛ [A] 0 ⎞ [A] 0 ⎜ ⎟ ⎝ 2 ⎠

72

Kinetic analyses

36

CHM 8304

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 : 100 90 80 70

% [A]0

60 50

premier ordre, k 1 first order,

40 30

k 2 = k 1 ˜ 70

20 10

k2 = k1

0 0

10

20

30

40

50

60

70

80

90

100

Temps 73

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

[A] = [A]0 e

–k1t

v0 = k1 [A]

first order

Réactions Reactioncinétiques kinetics

Calcul constante de vitesse Ratede constant calculation 18.00

90

16.00

80

14.00

vitesse initiale Initial rate

100

[Produit] [Product]

70 60 50 40 30 20

12.00 10.00 8.00 6.00 4.00 2.00

10

0.00

0 0

5

10

15

Time Temps

20

25

0

20

40

60

80

100

120

[A]

74

Kinetic analyses

37

CHM 8304

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

1/[A] – 1/[A]0 = k2t

v0 = k2 [A][B]

second order

Réactions Reactioncinétiques kinetics

Calcul constante de vitesse Ratede constant calculation

100

30.00

90

25.00

Initial rate vitesse initiale

80

[Produit] [Product]

70 60 50 40 30 20

20.00 15.00 10.00 5.00

10

0.00

0 0

5

10

15

20

0

25

20

Temps Time

40

60

80

100

120

[A] = [B]

75

Pseudo-first order A

+

B

second order:

Vitesse = v =

general solution:

k2

P

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

⎛ [A] 0 ([B] 0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠

•  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

76

Kinetic analyses

38

CHM 8304

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) ⎛ [A] 0 ([B] 0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B] 0 − [A] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠

⎞ [A] 0 [B] 0 1 ⎛ ⎜ ln ⎟ = k 2 t [B] 0 ⎝ [B] 0 ([A] 0 − [P]) ⎠ ln

ln

[A]0 = k 2 [B]0 t ([A]0 − [P])

[A]0 = k 1't where

où k 1' = k 2 [B]0 ([A]0 − [P])

ln

mono-exponential decrease

[A]0 = k 1' t [A]

[A] = [A]0 e

− k '1 t

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)

[A]0

100

0.16

90 80

0.14

k o b s = k 1'

70

kobs (min-1)

% [A]0

60

0.035 min- 1

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

40 30

0.07 min- 1

20

10

[A]∞ 20

30

40

0.08 0.06

0.02

0.15 min- 1 0

0.10

0.04

0.10 min- 1

10 0

slope pente = k 2 = (1.02 ± 0.02) M- 1min- 1

0.12

50

60

Temps (min) Time

70

80

90

100

0.00 0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

[B]0 (M)

78

Kinetic analyses

39

CHM 8304

Third order A

+

B

+

Rate

= v = Vitesse

k3

C

P

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

•  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

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

Rate

= v = Vitesse

d[P] dt

k1 k2

AB P

= k3[A][B][C]

80

Kinetic analyses

40

CHM 8304

Zeroth order k

catalyseur catalyst

+

A

P

catalyseur catalyst

+

d[P] - d[A] = = k dt dt

Vitesse Rate

= v =

•  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 A

t

A0

0

− ∫ d[A] = k ∫ dt

[A]0 - [A] = k t



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

[A]0 - ½ [A]0 = k t½

t½ =



linear relation



[A]0 2k

half-life

81

Zeroth order

100 90 80

pente = -k slope

70

% [A]0

60

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

50 40 30 20 10 0 0

10

20

30

40

50

60

70

80

90

100

Temps Time

82

Kinetic analyses

41

CHM 8304

Summary of observed parameters Reaction order

Rate law

Explicit equation

Linear equation

zero

d[P] =k dt

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

[A]0 - [A] = k t

first

d[P] = k1[A] dt

[A] = [A]0 e − k1 t

second

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

ln

1 1 − = k 2t [A] [A]0

1

[A] =

k 2t +

[A]0 = k1 t [A]

1 [A]0

complex

⎛ [A]0 ([B]0 − [P]) ⎞ 1 ⎜ ln ⎟ = k 2 t [B]0 − [A]0 ⎝ [B]0 ([A]0 − [P]) ⎠

non-linear regression

linear regression

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

Half-life t½ =

[A]0 2k

t½ =

ln 2 k1

t½ =

1 k 2 [A]0

complex

83

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 100.0

80.0

[A] (mM)

60.0

40.0

20.0

0.0 0

10

20

30

40

50

60

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

70

Tim e (m in)

84

Kinetic analyses

42

CHM 8304

Exercise A: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 0.060

2.50

non-linear

2.00

linear

0.050

0.040 1/[A] - 1[A]0

ln[A]0/[A]

1.50

1.00

y = 0.0008x 0.030

0.020

0.50

0.010

0.000

0.00 0

10

20

30

40

50

60

0

70

10

20

30

40

50

60

70

Tim e (m in)

Tim e (m in)

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

100.0

80.0

[A] (mM)

60.0

40.0

20.0

Time (min)

[A] (mM)

ln ([A]0/[A])

0

100.0

0.00

1/[A] – 1/[A]0 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

0.0 0

10

20

30

40

50

60

70

Tim e (m in)

86

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Exercise B: Data fitting •  use conc vs time data to determine order of reaction and its rate constant 0.25

3.5 y = 0.05x 3

linear

0.2

non-linear

0.15

2

1/[A] - 1/[A]0

ln ([A]0/[A])

2.5

1.5

y = 0.003x - 0.0283 0.1

0.05

1 0.5

0 0

0 0

10

20

30

40

50

60

70

10

20

30

40

50

60

70

-0.05

Time (min)

Time (min)

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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CHM 8304

First order (consecutive) k1

A

Rate

= v = Vitesse −

d[A] = k 1[A] dt

B

k2

P

d[P] d[A] = k2 [B] ≠ dt dt

[A] = [A]0 e − k1 t

d[B] = k1[A] − k 2 [B] dt

d [B] = k 1[A]0 e − k1 t − k 2 [B] dt

d [B] + k 2 [B] = k 1[A]0 e − k1 t dt

solved by using the technique of partial derivatives

[B] =

[P] = [A]0 − [A]0 e − k1t −

k 1[A]0 (e − k1t − e − k 2t ) (k 2 − k 1 )

k 1[A]0 (e − k1t − e − k 2 t ) (k 2 − k 1 )

⎛ ⎞ k1 [P] = [A] 0 ⎜1 − e − k1t − (e − k1t − e − k 2 t )⎟ (k 2 − k 1 ) ⎝ ⎠ 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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CHM 8304

Consecutive reactions, k1 = k2 100

[A]0

90

[P]∞

induction period pÈriode d'induction

80

5

% [A]0

60

ln([P]∞ - [P]0) - ln([P]∞ - [P])

70

k2 ≈ k1

50 40

[B]max

30

k2 ≈ k1

4 3

induction period pÈriode d'induction

2

pente = k obs = k 2 ≈ k 1 slope

1 0

-1

0

10

20

30

40

50

60

70

80

90

100

Temps

Time

20 10

[A]∞ , [B]∞

[B]0, [P]0

0 0

10

20

30

40

50

60

70

80

90

100

Temps Time 91

Consecutive reactions, k2 = 0.2 × k1

[A]0

1.0

90

0.8

k 2 = 0.2 ◊ k 1

0.6

80

0.4

[B]max

70

[P]

60

% [A]0

ln([P]∞ - [P]0) - ln([P]∞ - [P])

1.2

100

pente = kobs = k2 slope pÈriode d'induction induction period

0.2 0.0

-0.2 0

50

10

20

30

40

50

60

70

80

90

100

Temps Time

40 30

[B]

20 10

[A]∞

0 0

10

20

30

40

50

60

70

80

90

100

Temps Time

92

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CHM 8304

Consecutive reactions, k2 = 5 × k1

100

[A]0

[P]∞

90 80 ln([P]∞ - [P]0) - ln([P]∞ - [P])

70 60

% [A]0

k2 = 5 ◊ k1

7

50

k2 = 5 ◊ k1

40 30

6

4

pente = k obs = k 1 slope

3 2 1 0

-1

20

induction period pÈriode d'induction

5

0

10

[B]max

20

30

40

50

60

70

80

90

100

Temps

Time

10

[A]∞ , [B]∞

0 0

10

20

30

40

50

60

70

80

90

100

Temps 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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CHM 8304

Steady State Approximation •  consider a typical example (in bio-org and organometallic chem) of a two step reaction: k1 k2[B] –  kinetic scheme: I A P k-1

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

–  rate law: –  SSA : d [I] dt

= k1 [A] − k−1 [I] − k 2 [I][B] = 0

•  expression of [I] :

–  rate equation:

k1 [A] ⎞ ⎟⎟ k ⎝ −1 + k 2 [B] ⎠ ⎛

[I] = ⎜⎜

first order in A; less than first order in B

d [P] ⎛ k1k2 [A][B] ⎞ ⎟ = ⎜ dt ⎜⎝ k−1 + k2 [B] ⎟⎠

95

Steady State Approximation •  consider another example of a two-step reaction: k1 –  kinetic scheme: k2 A+B

–  rate law:

k-1

I

P

d [P] = k 2 [I] dt

d [I] = k1 [A][B] − k −1 [I] − k2 [I] = 0 dt ⎛ k [A][B] ⎞ ⎟⎟ •  expression of [I] : [I] = ⎜⎜ 1 ⎝ k −1 + k2 ⎠

–  SSA:

–  rate equation:

d [P] ⎛ k1k2 [A][B] ⎞ ⎟ = kobs [A][B] = ⎜ dt ⎜⎝ k−1 + k2 ⎟⎠

kinetically indistinguishable from the mechanism with no intermediate!

96

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CHM 8304

Steady State Approximation •  consider a third example of a two-step reaction: k1 –  kinetic scheme: A

–  rate law: d [P2 ] dt

k -1

I + P1 k 2[B]

= k 2 [I][B]

P2

–  SSA: d [I] = k1 [A] − k−1 [I][P1 ] − k2 [I][B] = 0 dt

•  expression of [I] :

⎞ k1 [A] ⎟⎟ [ ] [ ] k P + k B 2 ⎝ −1 1 ⎠ ⎛

[I] = ⎜⎜

–  rate equation: d [P ] ⎛ k k [A][B] ⎞ 2 1 2 ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

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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CHM 8304

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: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

–  in the case where k-1 >> k2 , in the presence of excess B and (added) P1, the equation can be simplified as follows: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] ⎠ •  now it is first order in A and in B

99

Change of reaction order •  however, if one considers the same equation: d [P2 ] ⎛ k1k2 [A][B] ⎞ ⎟⎟ = ⎜⎜ dt ⎝ k−1 [P1 ] + k2 [B] ⎠

–  but reaction conditions are modified such that k-1[P1] > k-1 and [-CN] >> [Br-] so the rate law becomes:

d [P] ⎛ k1k2 [R ][CN ] ⎞ ⎟ = k1 [R ] = ⎜ dt ⎜⎝ k2 [CN ] ⎟⎠

•  i.e., it is typically already saturated with respect to [-CN] 102

Kinetic analyses

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CHM 8304

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 OH

•  for example, for

K eq[H +]

H OH

k1

k 2 [-I]

I

k -1 + H2 O

the protonated alcohol is always in rapid eq’m with the alcohol and therefore d [P] ⎛ k1k2 K eq [H +][tBuOH][I] ⎞ dt

= ⎜⎜ ⎝

k−1 [H 2O] + k2 [I]

⎟⎟ ⎠

•  normally, k2[I-] >> k-1[H2O] and therefore d [P] ⎛ k1k2 K eq [H +][tBuOH][I] ⎞ ⎟⎟ = k1K eq [H +][tBuOH] = ⎜ dt ⎜⎝ k2 [I] ⎠

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

Kinetic analyses

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CHM 8304

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

Kinetic analyses

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CHM 8304

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 “offdiagonal” corner is lowered, the transition state is displaced towards the lowered corner

R

I1

‡

‡

I2

P

R

I1

‡

‡

I2

P

107

More-O’Ferrall-Jencks Diagram LG Et GP LG + Nuc GP

+

Nuc

Cleavage of C-LG

Et Nuc Et +

+

LG GP

GP LG + Nuc

Formation of C-Nuc

δ− Nuc

Nuc H3 C CH2 GP LG

Me

H δ− H GP LG

symmetric

‡

‡

LG Nuc + GP

108

Kinetic analyses

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CHM 8304

More-O’Ferrall-Jencks Diagram LG Et GP

+

Nuc better

Cleavage of C-LG

LG + Nuc GP

Et Nuc Et +

+

GP LG

GP LG + Nuc

Formation of C-Nuc

“Hammond effect”

‡

‡

δ− Nuc Me

δ−

‡

“orthogonal effect”

H

δ−

LG H GP

less C-LG cleavage

Nuc H3 C CH2 GP LG

LG Nuc + GP

109

More-O’Ferrall-Jencks Diagram LG Et GP

Formation of C-Nuc

LG + Nuc GP

Nuc H3 C CH2 GP LG

+

Nuc worse

Cleavage of C-LG

Et Et +

Nuc

+

LG GP

GP LG + Nuc

δ− Nuc

‡

‡

Me δ+

H H

δ−

GP LG

more C-LG cleavage

‡

LG Nuc + GP

110

Kinetic analyses

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CHM 8304

More-O’Ferrall-Jencks Diagram Et GP + LG better LG + Nuc GP

Nuc

Et Nuc

Cleavage of C-LG

Et +

+

GP LG

GP LG + Nuc

Formation of C-Nuc

‡

δ−

Nuc

‡

Me

δ+

‡

H

δ− H LG GP

less C-Nuc formation

Nuc H3 C CH2 GP LG

LG Nuc + GP

More-O’Ferrall-Jencks Diagram LG Et GP better LG + Nuc GP

+

Nuc worse

Cleavage of C-LG

Et Nuc Et +

Formation of C-Nuc

LG GP

GP LG + Nuc

‡

Nuc H3 C CH2 GP LG

+

δ−

Nuc

‡

Me

δ+

H H

δ−

‡

GP LG

dissociative

LG Nuc + GP

112

Kinetic analyses

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CHM 8304

More-O’Ferrall-Jencks Diagram Et GP + LG worse

Formation of C-Nuc

LG + Nuc GP

Nuc worse

Cleavage of C-LG

Et Nuc Et +

+

LG GP

GP LG + Nuc

δ−

Nuc H

Me

H

‡

δ− LG GP

‡

Nuc H3 C CH2 GP LG

resembles products!

‡

LG Nuc + GP

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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CHM 8304

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. 

2. 

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

1.

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

3. 

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

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

Kinetic analyses

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CHM 8304

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

Kinetic analyses

59