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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)

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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”

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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

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Outline: Energy surfaces

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

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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

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Example: SN2 energy profile

• one step passing by a single transition state

δ- δ- X---Y---Z transition state

ΔG‡ energy X:- + Y-Z reactants

energy Free ΔG° X-Y + :Z- products

Reaction co-ordinate

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Reminder:

• ethereal complex formed at the transition state of the reaction – partially formed bonds -13 – partially broken bonds very unstable; <10 s

H H δ− δ− H HO + C Cl HO C Cl HO C + Cl HH HH H H

not isolable partial transition state bonds

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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 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

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Energy surfaces

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

activated complex for A activated complex for B

A

product B

reactant chem.wayne.edu

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More-O’Ferrall and Jencks

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

• 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

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Profile to surface

δ- δ- X---Y---Z 2D: TS rxn. coord. & energy

X:- + Y-Z X-Y + :Z-

energy Free reactants products Reaction co-ordinate

- X:- + Y+ + :Z- X: + Y-Z Cleavage of Y-Z bond reactants unstable intermediates

‡ 2D: 2 rxn. coord.

- X-Y -Z 3D: unstable bond FormationX-Y of X-Y + :Z- 2 rxn. coord. & energy intermediate products 12

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Mountain terminology

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

Helmet/Robson col

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‘Summit’ vs ‘col’

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

energy Free

Rxn. co-ord.

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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). R GLGP + Nuc Cleavage of C-LG bond

bond

Nuc Formation of C- + GP REt LG + Nuc

Nuc - H 3 C RCH 2 GLGP energy Free

Rxn. Co-ord. R Nuc + GLGP 15

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!

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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

activation energy TS2 TS2 of global reaction; step 2 is rls TS1 ‡ TS1 ΔG2 int. int.

products products ‡ ‡ ‡ ‡ ΔG energy Free energy Free ΔG1 ΔG1 > ΔG2 so step 1 is rls? reactants NO!! reactants

Rxn. Co-ord. Rxn. Co-ord.

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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

G‡ Δ ΔG‡ ΔG‡ energy Free energy Free energy Free Rxn. co-ord. Rxn. co-ord. Rxn. co-ord.

‡ ΔG ‡ ΔG‡ ΔG energy Free energy Free energy Free

Rxn. co-ord. Rxn. co-ord. Rxn. co-ord. 18

Kinetic analyses 9 CHM 8304

Rate vs rate constant

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

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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

90 different 80 70 of reactants; 60 different 50

[Produit] 40 end points different 30 initial 20 rates 10 0 0 5 10 15 20 Temps same half-lives; same rate constants 20

Kinetic analyses 10 CHM 8304

Rate law and • 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 of its involvement in the transition state of the rate-determining step

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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

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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 !

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Negative integers in rate laws

• 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

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Outline:

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

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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

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Arrhenius equation

• in 1889, Arrhenius noted that the rate constant of a given reaction increases exponentially with temperature : E d lnk a Ea = 2 lnk = − + lnA dT RT RT

• 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

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Arrhenius plot Graphique d'Arrhenius

ln A E lnk = − a + lnA RT

slopepente = - Ea / R obs k ln

1 / T (K-1)

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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

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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 A•B P cactivatedomplexe complex quasi activé equilibrium [A • B‡ ] K ‡ = [A • B‡ ] = K ‡[A][B] v = k ‡[A • B‡ ] = k ‡K ‡[A][B] [A][B]

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Eyring equation

K‡ k‡ A + B A•B P v = k ‡[A • B‡ ] = k ‡K ‡[A][B] coactivatedmplexe complexactivé • 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 : k T k ‡ = = B h

-23 -1 • kB is Boltzmann’s constant (kB = 1.3806 × 10 J K ) • h is Planck’s constant (h = 6.626 × 10-34 JŸs) ⎛ ΔG‡ ⎞ ⎜ − ⎟ ⎜ RT ⎟ ‡ ‡ ‡ ⎝ ⎠ – ΔG = -RT ln Keq so ΔG = -RT ln K and K = e

‡ ‡ ‡ ⎛ −ΔH ΔS ⎞ ⎛ −ΔG ⎞ ⎜ ⎟ ‡ ‡ ⎜ ⎟ ⎜ + ⎟ ⎜ ⎟ kBT RT R ⎛ k h ⎞ − ΔH 1 ΔS ‡ ‡ kBT RT ⎝ ⎠ obs k k K e⎝ ⎠ kobs = e ln⎜ ⎟ = ⋅ + obs = = h ⎜ ⎟ h ⎝ kBT ⎠ R T R

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Eyring plot Graphique d'Eyring

‡ ΔS / R ‡ ‡ ⎛ kobsh ⎞ − ΔH 1 ΔS ln⎜ ⎟ = ⋅ + ⎝ kBT ⎠ R T R T) B h k / ‡ obs slopepente = = - ΔH / R k ln ln (

1 / T (K-1)

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Activation (free) energy, ΔG‡

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

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Activation , 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

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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.)

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Exercise A: Eyring plot to mechanism • use the Eyring plot to determine the enthalpy and for the substitution reaction of iPr-Br

– are the data more consistent with SN1 or SN2 ?

Temperature dependence R = 1.987 cal/mol/K 20.00

15.00 ‡ ‡ 10.00 intercept = ΔS /R; ΔS = -6 cal/mol/K more ordered TS

T) 5.00 B ‡ ‡ 0.00 slope = -ΔH /R; ΔH = 10 kcal/mol little change in h / k -5.00 bond order at TS obs -10.00 ‡ ln (k -15.00 H -20.00

-25.00 δ− δ− 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 Nu Br 1/T (K-1) H3C CH3

SN2 36

Kinetic analyses 18 CHM 8304

Exercise B: 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 R = 1.987 cal/mol/K 20.00 ‡ ‡ less ordered TS 10.00 intercept = ΔS /R; ΔS = 10 cal/mol/K ‡ ‡ 0.00 slope = -ΔH /R; ΔH = 25 kcal/mol net bond cleavage T) B at TS -10.00

h / k ‡ -20.00 obs H

ln (k -30.00

-40.00 δ+ δ− Br -50.00 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) H3C CH3

SN1 37

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 :

‡ having having ΔG T1

T2 molcules energy given a

T2 > T1 % of

Energy

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Rate constants and temperature

• by analogy with ΔG° = -RT ln Keq, ‡ ΔG = -RT ln(k/k0) or ΔG‡ − RT k = k0e

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

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Outline: Postulates et principles

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

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Hammond • George S. Hammond (1921-2005) – American chemistry professor – studied the relation between kinetics and product distribution – laureate of Norris Award and Priestley Medal

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

…exergonique …endergonique (ΔG° < 0) (ΔG° > 0) resembles the resembles the reactants products energy Free

Reaction co-ordinate

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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° : energy Free

Reaction co-ordinate

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

small ΔΔG‡

reactant

energy Free product 1 product 2

Reaction co-ordinate

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

large ΔΔG‡ product 2

product 1 energy Free reactant

Reaction co-ordinate

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

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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

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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 :

k1 k x ky X A' A" Y k1

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

k1 >> kx andet k-1 >> ky

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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

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Curtin-Hammett Principle

k [A"] -ΔGy /RT ‡ ‡ d[Y]/dt y e -ΔG°/RT (ΔG -ΔG -ΔG°)/RT = = (k /k ) K = × e = e x y y x ‡ d[X]/dt k [A'] -ΔGx /RT x e ‡ = eΔΔG /RT

ΔΔG‡

‡ d[X]/dt = kx[A'] ΔG x

‡ ΔG y d[Y]/dt = ky[A"]

A" ΔG° Freeenergy A'

K = [A"]/[A'] Y = k1/k-1 X = e-ΔG°/RT

Rxn. coord.

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

H C Me Me Me Me CH 3 N N N N 3 13 13 CH3I CH3I

fast less more slow major stable stable minor ΔΔG‡

Freeenergy

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 >> kx andet k-1 >> ky

• however, at low , 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

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

Me N

O s-BuLi Me disfavoured Me Me Me Me N CH3-Br CH -Br Me N H 3 Me N H fast Me ΔG = -3.0 kcal CH3 N fast CH3 O by calculations O O O slow, at -78°C 1% 99%

[PB ] [B]0 = P A ‡ [ A ] [ ]0 ΔG A ‡ ΔG B Freeenergy

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

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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

H3C OH H3C OH

Et O Et O general acid H H catalysis general base catalysis B B

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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

H3C OH H3C OH

O N O O2N O 2 uncatalysed expulsion uncatalysed attack

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

ΔΔG‡

product A reactant product B Freeenergy

Reaction coordinate

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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)

Freeenergy product A reactant product B

ΔΔGrxn

Reaction coordinate

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Outline: Experimental kinetics

• based on A&D section 7.4 – practical kinetics – kinetic analyses • first order • second order • pseudo-first order

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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

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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

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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 70 60 50 40 [Product] [Product] 30 20 10 0 0 5 10 15 20 Time

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Discontinuous assay • involves taking aliquots of the reaction 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.

DiscontinuousContinuous reaction assay

100 90 80 70 60 50 40 [Product] [Product] 30 20 10 0 0 5 10 15 20 Time

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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

70

60

50 t = 10 min Ω½ 40 [A] (mM) [A] 30

20 - 1 - 1 k2 = 0.001 mM min

10 - 1 k = 5 mM min- 1 k1 = 0.07 min 0 0 10 20 30 40 50 60 70 80 90 100 TempsTime (min)

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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)

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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’)

Blaise Pascal, • Natural order is revealed through special French mathematician mathematical relationships and philosopher, • mathematics are our attempt to understand Nature 1623-1662 – exponential increase: the value of e – volume of spherical forms: the value of π

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First order (simple)

k1 A P

d[P] d[A] Vitesse Rate = v = = - = k [A] dt dt 1 d[P] d[P] d[P] = k1([A]0 −[P]) = k1([P]∞ −[P]) = k1 dt dt dt [P]∞ −[P] P d[P] t = k1 dt ∫0 ∫0 ln ([P]∞ - [P]0) - ln ([P]∞ - [P]) = k1 t [P]∞ −[P]

[A]0 [A]0 linear relation ln = k1 t ln = k1 t½ [A] ½[A]0

−k1 t mono-exponential decrease [A] = [A]0 e ln2 half-life t½ = k1

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First order (simple)

1 0 100 [A]0 -1 ) 0

90 ] -2

A slopepente = k = k [ obs 1 / -3 80 ]

mono-exponential A [ -4 ( decrease 70 n

l -5 60 -6 0 ]

A 50 -7 [

% 40 0 10 20 30 40 50 60 70 80 90 100 TempsTime 30

20 kobs = k1 10 t tΩ½ 0 [A]∞ 0 10 20 30 40 50 60 70 80 90 100 TempsTime

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Kinetic analyses 34 CHM 8304

First order (reversible)

k1 A P k-1

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

d[A] d[P] − = = k1 ([A]éq +[P]éq −[P])− k -1[P] dt dt

= k1[A]éq + k1[P]éq − (k1 + k -1 )[P]

d[A] d[P] At Àequilibrium, l’équilibre, k [A] = k [P] − = = k [P] + k [P] − k + k [P] 1 éq -1 éq dt dt -1 éq 1 éq ( 1 -1 ) k k [P] [P] = ( 1 + -1 )( éq − ) P d[P] t = k + k dt ∫0 ( 1 -1 )∫0 [P]éq −[P]

[P] − [P] linear relation ( éq 0 ) ln = (k1 + k -1 )t k = (k + k ) [P] [P] obs 1 -1 ( éq − )

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First order (reversible)

100 [A]0 90 ln 2 t = 80 ½ (k 1 + k −1 ) 70

60 [P]Èq 0 ] k1 A 50 kobs = k1 + k-1 [ Kéq = k

% -1 40 [A]Èq

30

20 kobs = k1 10 [A] 0 [P]0 ∞ 0 10 20 30 40 50 60 70 80 90 100 TempsTime

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Second order (simple)

k2 A + B P

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

d[P] d[P] P d[P] t = k2 dt = k dt = k2 ([A]0 −[P])([B]0 −[P]) ∫0 2 ∫0 dt ([A]0 −[P])([B]0 −[P]) ([A]0 −[P])([B]0 −[P])

1 ⎛ [A]0 ([B]0 −[P])⎞ ⎜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)

k2 A + B P

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

P d[P] t = k dt If [A] = [B] : ∫0 2 2 ∫0 0 0 ([A]0 −[P]) 1 1 − = k2t [A]0 −[P] [A]0

1 1 1 1 linear relation − = k2t − = k 2 t ½ [A] [A] ⎛[A]0 ⎞ [A]0 0 ⎜ ⎟ ⎝ 2 ⎠ 1 half-life t½ = k 2[A]0

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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

60 0 ]

A 50 [

pfirstremi eorder,r ordre , k1 % 40

30

20 k2 = k1 ˜ 70

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]

v0 = k1 [A] –k1t [A] = [A]0 e first order RéactionsReaction cinétiques kinetics Calcul Rate de constant constante calculation de vitesse

100 18.00 90 16.00 80 14.00

70 12.00 60 10.00 50 8.00 40 [Produit] [Product] 6.00 30 Initial rate Initial vitesse initiale vitesse 4.00 20 10 2.00 0 0.00 0 5 10 15 20 25 0 20 40 60 80 100 120 TempsTime [A]

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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 second order v0 = k2 [A][B] RéactionsReaction cinétiques kinetics Calcul Rate de constant constante calculation de vitesse

100 30.00 90 80 25.00 70 20.00 60 50 15.00 40 [Produit] [Product] 10.00 30 Initial rate Initial vitesse initiale vitesse 20 5.00 10 0 0.00 0 5 10 15 20 25 0 20 40 60 80 100 120 TempsTime [A] = [B]

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Pseudo-first order

k2 A + B P second order: d[P] Vitesse = v = = k [A][B] dt 2

1 ⎛ [A]0 ([B]0 −[P])⎞ general solution: ⎜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)

1 ⎛ [A]0 ([B]0 −[P])⎞ 1 ⎛ [A]0 [B]0 ⎞ ⎜ln ⎟ = k 2 t ⎜ln ⎟ = k 2 t [B]0 −[A]0 ⎝ [B]0 ([A]0 −[P])⎠ [B]0 ⎝ [B]0 ([A]0 −[P])⎠

[A]0 ln = k2[B]0 t ([A]0 −[P])

[A] ' ' 0 where ln = k1t où k1 = k2[B]0 ([A]0 −[P]) [A] ln 0 = k' t [A] 1

k' t − 1 mono-exponential decrease [A] = [A]0 e

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)

100 [A]0 0.16 90

80 0.14 ko b s = k1' - 1 - 1 slopepent e = k2 = (1.02 ± 0.02) M min 70 0.12 ) 1 60 - 1 - 0.035 min n 0.10 i 0 ] m A

50 ( [

s 0.08

[B] = 0.033 M b % 40 0 [B]0 = 0.065 M o [B] = 0.098 M k 0 0.06 30 [B]0 = 0.145 M - 1 20 0.07 min 0.04 0.10 min- 1 10 0.02 0.15 min- 1 0 [A]∞ 0.00 0 10 20 30 40 50 60 70 80 90 100 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

TimeTemps (min) [B]0 (M)

78

Kinetic analyses 39 CHM 8304

Third order

k3 A + B + C P

d[P] VitesseRate = v = = k [A][B][C] dt 3

• 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!

k1 A + B AB

k2 AB + C P

d[P] VRateitess e = v = = k [A][B][C] dt 3

80

Kinetic analyses 40 CHM 8304

Zeroth order

k A + c acatalysttalyseur P + c catalystatalyseur

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

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

A t − d[A] = k dt [A]0 - [A] = k t [A] = -k t + [A]0 linear relation ∫A0 ∫0

[A]0 - ½ [A]0 = k t½

[A] t = 0 half-life ½ 2 k

81

Zeroth order

100

90

80

70 slopepente = -k

60 0 ]

A 50 not realistic [ that rate would % 40 be constant all 30 the way to end 20 of reaction... 10

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

82

Kinetic analyses 41 CHM 8304

Summary of observed parameters

Reaction Rate law Explicit equation Linear equation Half-life order

zero d[P] [A] = -k t + [A]0 [A]0 - [A] = k t [A] = k t = 0 dt ½ 2 k

first d[P] −k1 t [A] ln2 [A] = [A]0 e 0 = k1[A] ln = k1 t t½ = dt [A] k1

second d[P] 2 1 1 1 1 = k2[A] [A] = − = k2t t½ = dt 1 [A] [A]0 k 2[A]0 k2t + [A]0

d[P] complex 1 ⎛ [A]0 ([B]0 − [P])⎞ complex = k2[A][B] ⎜ln ⎟ = k2t dt [B]0 − [A]0 ⎝ [B]0 ([A]0 − [P])⎠

non-linear linear regression regression 83

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

100.0

80.0

Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A] 0 60.0 0 100.0 0.00 0.000 10 55.6 0.59 0.008 [A] (mM) [A] 40.0 20 38.5 0.96 0.016 30 29.4 1.22 0.024

20.0 40 23.8 1.44 0.032 50 20.0 1.61 0.040

0.0 60 17.2 1.76 0.048 0 10 20 30 40 50 60 70 Time (min)

84

Kinetic analyses 42 CHM 8304

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

2.50 0.060

0.050 2.00 non-linear linear 0.040 1.50 y = 0.0008x

0.030 ln[A]0/[A] 1.00 1/[A] - 1[A]0 0.020

0.50 0.010

0.00 0.000 0 10 20 30 40 50 60 70 0 10 20 30 40 50 60 70 Time (min) Time (min)

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

85

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

100.0

80.0 Time (min) [A] (mM) ln ([A]0/[A]) 1/[A] – 1/[A] 0 0 100.0 0.00 0.00

60.0 10 60.7 0.50 0.006 20 36.8 1.00 0.017 [A] (mM) [A] 40.0 30 22.3 1.50 0.035 40 13.5 2.00 0.064 50 8.2 2.50 0.112 20.0 60 5.0 3.00 0.190

0.0 0 10 20 30 40 50 60 70 Time (min)

86

Kinetic analyses 43 CHM 8304

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

3.5 0.25 y = 0.05x 3 0.2 2.5 linear non-linear 0.15 2 y = 0.003x - 0.0283

0.1 1.5 ln ([A]0/[A]) 1/[A] - 1/[A]0 1 0.05

0.5 0 0 10 20 30 40 50 60 70 0 0 10 20 30 40 50 60 70 -0.05 Time (min) Time (min)

first order; -1 kobs = 0.05 min

87

Outline: Complex reactions

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

88

Kinetic analyses 44 CHM 8304

First order (consecutive)

k1 k2 A B P

d[P] d[A] Vitesse Rate = v = = k2 [B] - dt ≠ dt

d[A] −k1 t − = k [A] [A] = [A]0 e dt 1

d[B] d[B] d[B] −k1 t −k1 t = k [A]− k [B] = k1[A]0 e − k2[B] + k2[B] = k1[A]0 e dt 1 2 dt dt solved by using the technique of partial derivatives k [A] [B] = 1 0 (e−k1t − e−k2t ) (k2 − k1)

k [A] ⎛ k ⎞ −k1t 1 0 −k1t −k2t −k1t 1 −k1t −k2t [P] = [A]0 −[A]0 e − (e − e ) [P] = [A]0 ⎜1− e − (e − e )⎟ (k 2 − k1) ⎝ (k 2 − k1 ) ⎠

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

Kinetic analyses 45 CHM 8304

Consecutive reactions, k1 = k2

100 [A] 0 [P]∞ 90 pÈriodeinduction d'induction period 80

5 k2 ≈ k1 ) 70 ] P [ 4 -

∞ ] P [

60 ( 3 n l k ≈ k induction period

- pÈriode d'induction

0 2 1 ) 0 ] 2 ] slopepent e = kobs = k2 ≈ k1 P

50 [ A

-

[ 1 ∞

] P [ % 40 ( n [B]max l 0

-1 30 0 10 20 30 40 50 60 70 80 90 100 Temps 20 Time

10 [A]∞ , [B]∞ 0 [B]0, [P]0 0 10 20 30 40 50 60 70 80 90 100 TempsTime

91

Consecutive reactions, k2 = 0.2 × k1

1.2 ) ]

100 [A] P 1.0 [

0 -

] 0.8 P

90 [ k2 = 0.2 ◊ k1 ( n 0.6 l

pente = k = k

- slope obs 2

)

0 pÈriode d'induction 80 ] 0.4 induction period P [

- 0.2

max ∞ 70 [B] ]

[P] P [

( 0.0 n l 60 -0.2 0

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

A 50 Temps [ Time

% 40

30 [B]

20

10 [A] 0 ∞ 0 10 20 30 40 50 60 70 80 90 100 TempsTime

92

Kinetic analyses 46 CHM 8304

Consecutive reactions, k2 = 5 × k1

100 [A] 0 [P]∞ 90

k = 5 ◊ k 80 2 1 7 ) ]

P 6 [

70 -

∞ 5 ] pÈriodeinduction d'induction period P [

60 (

n 4 l

0 pente = k = k - slope obs 1

] )

0 3 ]

A 50 P [ k = 5 ◊ k [

2 1 - 2

∞ ] %

40 P 1 [ ( n l 30 0 -1 0 10 20 30 40 50 60 70 80 90 100 20 max Temps [B] Time 10 [A]∞ , [B]∞ 0 0 10 20 30 40 50 60 70 80 90 100 TempsTime

93

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

Kinetic analyses 47 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: A I P k-1 – rate law: d[P] = k2 [I][B] dt – SSA : d[I] = k1[A]− k 1[I]− k2 [I][B]= 0 dt −

⎛ k A ⎞ • expression of [I] : 1[ ] [I]= ⎜ ⎟ ⎝ k−1 + k2 [B]⎠

: d[P] ⎛ k1k2 [A][B]⎞ first order in A; = ⎜ ⎟ dt ⎝ k−1 + k2 [B]⎠ less than first order in B

95

Steady State Approximation

• consider another example of a two-step reaction:

– kinetic scheme: k1 k2 A + B I P k-1 d[P] – rate law: = k2 [I] dt d[I] – SSA: = k1[A][B]− k 1[I]− k2 [I]= 0 dt − ⎛ k A B ⎞ • expression of [I] : 1[ ][ ] [I]= ⎜ ⎟ ⎝ k−1 + k2 ⎠ kinetically – rate equation: d P ⎛ k k A B ⎞ [ ] 1 2 [ ][ ] indistinguishable = ⎜ ⎟ = kobs[A][B] dt ⎜ k k ⎟ ⎝ −1 + 2 ⎠ from the mechanism with no intermediate!

96

Kinetic analyses 48 CHM 8304

Steady State Approximation

• consider a third example of a two-step reaction:

– kinetic scheme: k 1 A I + P1 k -1 – rate law: d[P2 ] k 2[B] = k2 [I][B] dt

P2

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

⎛ k1[A] ⎞ • expression of [I] : [I]= ⎜ ⎟ ⎝ k−1[P1]+ k2 [B]⎠ – rate equation: first order in A, d[P2 ] ⎛ k1k2 [A][B] ⎞ = ⎜ ⎟ less than first order in B; dt ⎝ k−1[P1]+ k2 [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

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Kinetic analyses 49 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] << k2[B] , the equation is simplified very differently:

d[P2 ] ⎛ k1k2 [A][B]⎞ = ⎜ ⎟ dt ⎝ k2 [B] ⎠

d[P2 ] = k1[A] dt • now the equation is only first order in A

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

100

Kinetic analyses 50 CHM 8304

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 k1 k2[B] A I P k-1

d[P] ⎛ k1k2 [A][B]⎞ having the rate law = ⎜ ⎟ dt ⎜ k + k B ⎟ ⎝ −1 2 [ ]⎠ one observes: vmax = k1[A]

v ~zero order in B

first order in B

[B]0 101

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 k - • in reality, for 1 k 2 [ CN] Br CN

k -1 + Br-

d[P] ⎛ k1k2 [R][CN] ⎞ one can show that = ⎜ ⎟ dt ⎝ k−1[Br]+ k2[CN]⎠

- - • but normally k2 >> k-1 and [ CN] >> [Br ]

d P ⎛ k k R CN ⎞ so the rate law becomes: [ ] 1 2 [ ][ ] k R = ⎜ ⎟ = 1[ ] dt k2 [CN] ⎝ ⎠ • i.e., it is typically already saturated with respect to [-CN]

102

Kinetic analyses 51 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

K [H+] H k - eq 1 k 2 [ I] • for example, for OH OH I k -1 + H2O the protonated alcohol is always in rapid eq’m with the alcohol and therefore d[P] ⎛ k1k2Keq [H +][tBuOH][I]⎞ = ⎜ ⎟ dt ⎝ k−1[H2O]+ k2 [I] ⎠

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

d P ⎛ k1k2Keq [H +][tBuOH][I]⎞ [ ] k K H tBuOH first order in tBuOH, = ⎜ ⎟ = 1 eq [ +][ ] dt ⎝ k2 [I] ⎠ 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 52 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 53 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:

R 1. “Hammond effect”: if the corner of the I1 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 I2 P R I1 2. “orthogonal effect” : if an “off- diagonal” corner is lowered, the transition state is displaced towards ‡ the lowered corner ‡

I2 P 107

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

δ− Nuc

H Me δ− H GLGP symmetric ‡ ‡ Formation of C-Nuc of Formation

Nuc

H3C CH2 Nuc + GLGP GLGP

108

Kinetic analyses 54 CHM 8304

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP better

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

“Hammond effect”

‡ ‡ ‡ δ− “orthogonal effect” Nuc δ− H Me

Formation of C-Nuc of Formation δ− GLGP H

Nuc less C-LG

H3C CH2 cleavage Nuc + GLGP GLGP

109

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP worse

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

δ− Nuc H Me δ+ ‡ ‡ H δ− ‡ GLGP more C-LG

Formation of C-Nuc of Formation cleavage

Nuc

H3C CH2 Nuc + GLGP GLGP

110

Kinetic analyses 55 CHM 8304

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP better

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

δ− Nuc ‡ δ+ H ‡ Me δ− H GLGP less C-Nuc Formation of C-Nuc of Formation formation Nuc

H3C CH2 Nuc + GLGP GLGP

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP better worse

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

‡ δ− Nuc

δ+ H Me ‡ H ‡ δ− GLGP dissociative Formation of C-Nuc of Formation

Nuc

H3C CH2 Nuc + GLGP GLGP

112

Kinetic analyses 56 CHM 8304

More-O’Ferrall-Jencks Diagram

Et GLGP + Nuc Et Nuc + GLGP worse worse

GLGP + Nuc Cleavage of C-LG Et + GLGP + Nuc

δ− Nuc Me H H ‡ δ− GLGP resembles ‡

Formation of C-Nuc of Formation ‡ products!

Nuc

H3C CH2 Nuc + GLGP GLGP

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

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Kinetic analyses 57 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. 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.

2. 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 58 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