Experimental kinetics.pptx
Transcription
Experimental kinetics.pptx
CHM 8304 Enzyme Kinetics Experimental Kinetics Outline: Experimental kinetics • • • • rate vs rate constant rate laws and molecularity practical kinetics kinetic analyses (simple equations) – first order – second order – pseudo-first order • kinetic analyses (complex reactions) – consecutive first order reactions – steady state – changes in kinetic order – saturation kinetics – rapid pre-equilibrium see A&D sections 7.4-7.5 for review 2 Experimental Kinetics 1 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 3 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 4 Experimental Kinetics 2 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 (<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 5 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 6 Experimental Kinetics 3 CHM 8304 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! 7 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 8 Experimental Kinetics 4 CHM 8304 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 9 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 10 Experimental Kinetics 5 CHM 8304 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 11 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 12 Experimental Kinetics 6 CHM 8304 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) 13 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) 14 Experimental Kinetics 7 CHM 8304 Outline: Kinetic analyses (simple eqns) – – – – – first order reactions second order reactions pseudo-first order third order reactions zeroth order 15 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 16 Experimental Kinetics 8 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 17 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 18 Experimental Kinetics 9 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) 19 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 20 Experimental Kinetics 10 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 21 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 ⎠ 22 Experimental Kinetics 11 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 23 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] 24 Experimental Kinetics 12 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] 25 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 26 Experimental Kinetics 13 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 27 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) 28 Experimental Kinetics 14 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 29 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] 30 Experimental Kinetics 15 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 31 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 32 Experimental Kinetics 16 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 33 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) 34 Experimental Kinetics 17 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 35 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) 36 Experimental Kinetics 18 CHM 8304 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 Time (min) 10 20 30 40 50 60 70 -0.05 Time (min) first order; kobs = 0.05 min-1 37 Outline: Kinetic analyses (complex rxns) – – – – – consecutive first order reactions steady state changes in kinetic order saturation kinetics rapid pre-equilibrium 38 Experimental Kinetics 19 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 ) ⎝ ⎠ 39 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 40 Experimental Kinetics 20 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 41 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 42 Experimental Kinetics 21 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 43 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) 44 Experimental Kinetics 22 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] ⎟⎠ 45 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! 46 Experimental Kinetics 23 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 47 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 48 Experimental Kinetics 24 CHM 8304 Change of reaction order • by adding an excess 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 B and excess (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 49 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 50 Experimental Kinetics 25 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 k [B] A having the rate law k-1 I 2 P d [P] ⎛ k1k2 [A][B] ⎞ ⎟ = ⎜ dt ⎜⎝ k−1 + k2 [B] ⎟⎠ vmax = k1[A] one observes: ~zero order in B v first order in B [B]0 51 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 k1 k 2 [-CN] • in reality, for Br CN k -1 one can show that + Br- d [P] ⎛ k1k2 [R ][CN] ⎞ ⎟ = ⎜ dt ⎜⎝ k−1 [Br ]+ k2 [CN] ⎟⎠ • but normally k2 >> 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] 52 Experimental Kinetics 26 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 53 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) 54 Experimental Kinetics 27