Applications of Kinetics in Asymmetric Hydrogenation, Lecture notes of Chemistry

Download Applications of Kinetics in Asymmetric Hydrogenation and more Lecture notes Chemistry in PDF only on Docsity! 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.0 5.0x10 -7 1.0x10 -6 1.5x10 -6 2.0x10 -6 2.5x10 -6 ra te ( M s -1 ) [A] (M) experiment 1 experiment 2 E. Kwan Applications of Kinetics Chem 106 Asymmetric Hydrogenation This part of the course is essentially transcribed from a set of notes that Professor Clark Landis sent me. His approach is very | oe : (4) Implications: How does ee vary with pressure and clear, so | have made very few modifications. Things are somewhat abridged here, but Professor Landis presents a full account of this work in this article: "Asymmetric Hydrogenation of Methyl-(Z)-a-acetamidocinnate Catalyzed by...Rhodium(I)...". Landis, C.R.; Halpern, J. J. Am. Chem. Soc. 1987, 109, 1746-1754. Here is the catalytic reaction in a nutshell: N Om ot Ph DIPAMP is this chiral, bidentate phosphine: \ ° Me0,C. [Rh(DIPAMP)(COD)]” MeO, N Tt ~Ac Ph oat This reaction exemplifies a highly useful class of asymmetric hydrogenations with cationic rhodium(I). Instead of dealing with the complex atomistic basis of selectivity, | focus instead on what kinetics can tell us about ground vs. transition state stoichiometry and the mechanism of this reaction. Our strategy shall be as follows: (1) Ground State: identity? rates and equilibria of binding catalyst to olefins? mechanism of binding? (2) Transition State: rate law? mechanism? hydride or unsaturate pathway? : (3) Rate constants: study stoichiometric complexes to get information about individual steps of the catalytic cycle temperature? Surprisingly, such answers are not available for many catalytic : reactions. Over twenty years later, this remains one of the best : characterized systems! ‘ Let's look at the closely related DIPHOS catalyst in the ground : State. Treating the norbornadiene precursor with hydrogen results in a methanol disolvate: Pho Pho H POE 1 PL OMe vs +H, ~~ ~OMe Phy Ph, H How does this complex react with arenes? The disolvate forms 1:1 complexes, which can be followed by UV-vis spectrometry: 0.6 ABSORBANCE 2 a 2 w L [Rh(DIPHOS)CgH,)]*_ see Halpern, JACS 1977, 99, 8055 400 450 WAVELENGTH ,nm 500 E. Kwan Applications of Kinetics Chem 106 The first possibility is the unsaturate pathway, which involves ; Alternatively, one can consider a hydride pathway where olefin fast pre-equilibrium binding of olefin, followed by slow oxidative : binding is an unproductive, off-cycle pathway: addition of hydrogen (it is assumed the downstream steps are all fast): : C-olefin Pho H k Pho AS _ Ky eC) 1 PL® ‘ R Rh sOMe + aN ———— > SRh— lL K ‘ ~~ OMe ky “ R ‘ product He Ph H fast Ph ‘ Rh || + He 1 Seno * ‘ The 1+rate law becomes: P R slow Pp le R ‘ Pho Php H : k,[1-hexene]K,[H, ][Rh], As you can see, we already have the kinetic and thermodynamic | 1+X,[H,]+ X,[l-hexene] parameters for the first equation. Studying the behavior of a | | | step of a catalytic cycle separately is a good strategy for ‘ Because there is another possible state of the catalyst that is deconvoluting a complex manifold. : unproductive, we now have a new term in the denominator only. : Assuming that K[H9] << 1 + K,[1-hexene], we have the same To see what the rate law is, let me write this catalytic cycle ina __ functional form as before. Only the identity of the constants has more familiar form: | changed, and of course, when we do the experiment, we don't c AN ‘ know the microscopic origin of each of the constants, so the two product RS | pathways are kinetically indistinguishable. ky K ‘ Hp C-olefin : With MAC instead of 1-hexene, we find that: The corresponding 1+rate law is therefore: &,[H, ]K[I-hexene][Rh], ‘initial 1+ X[l-hexene] : rate Now, we can see why the rate is always first order in total catalyst and [H.], but displays "saturation kinetics" in [1-hexene]. When [1-hexene] is large (i.e., >>1), the rate will [ (i.€., ; IRhlror [He] [MAC] be [1-hexene]-independent (cancellation with the [1-hexene] in: . an the numerator). Conversely, when [1-hexene] is small (.e., : From the prior measurements of the binding constant, we know <<1), the rate will be first-order in [1-hexene]. Clearly, the data _; that catalyst is bound largely as a MAC/olefin complex. We have are consistent with this rate law. : "saturation" in catalyst: K;[MAC] >> 1.  [Rhlror [Ha] [MAC] ' : Parameters for the reductive elimination are: E. Kwan Applications of Kinetics Chem 106 ' Phy Phe H aa: PL @ 0-4 T<-40°C Sar 2 initial ' Ce A NH Hp rate e Br E ' 2 Ph -OMe ' H The alkyl hydride accumulates because the activation For the unsaturate pathway, we have a rate law of: 4(H,],[MAC][Rh],. GLH JKIMACIIRA], _ [HJIRhI, 1+K,[MAC] K, [MAC] For the hydride pathway, we have a rate law of (once again assuming the second denominator term is . 1+ K,[H,]+K [MAC] a This is rather unfortunate, because this means that we cannot distinguish between the two mechanisms. As Professor Landis notes, the lessons here are that: (1) the phenomenological catalytic rate law itself is often not very informative ' (2) knowledge of the rates and rate laws for the individual steps : is needed to provide further discriminatory evidence AH#* = 17 kcal/mol AS* = -6 eu | So at high temperatures, we have enough energy to overcome ‘ the enthalpic barriers of 7 and 17 kcal/mol for oxidative addition ' and reductive elimination, respectively. : lowered, we can get to a point where we can overcome the 7 ' kcal barrier, but not the 17 kcal barrier. So we undergo oxidative If the temperature is addition, but not reductive elimination, and we get "stuck" at the alkyl hydride stage of the cycle. The alkyl hydride is formed at a rate of rate = k [C-MAC][H,], k= 0.13 M's? @-80 °C At this temperature, and [H2]=0.001 M, we get a pseudo-first- order rate constant of k{Hz] = 0.0001 M ' or a half life of 100 minutes. If the hydride pathway is correct, : then the rate at which active catalyst is formed from C-MAC In this case, it is the temperature dependence of rate that rules out one mechanism. Measuring rate vs. temperature yields activation parameters of: AH* =7 kcal/mol AS* = -26 eu At low temperatures, we find a new intermediate in the reaction that is frozen before reductive elimination: ' (related to k.,) must be faster than the rate at which C-MAC. : gets turned into alkyl hydride. Mathematically, we have: k [Ha] < k That is because alkyl hydride formation is clearly rate-limiting at low temperatures. E. Kwan Applications of Kinetics k_, has already been estimated by arene trapping. Activation parameters give: AHt = 18 kcal/mol AS* = +2 eu At 25 °C, k; is 0.4 s'. Extrapolation to -80 °C gives a rate of k4=2x108s" or a half life of 1.1 years. This is inconsistent with the inequality : above, and therefore we can rule out the hydride pathway. | turn now to the energetics of enantioselectivity. Here are the key points that are relevant to chiral phosphine catalysts: (1) We propose that the reaction has two diastereomeric manifolds for the formation of the major enantiomer Pray and ' minor enantiomer Pynin: MAC Prin ro mal k,mai c- MAC major MAC Prnaj men ymin k, min c. Mac minor (2) The idea is that the catalyst and olefin can form two diastereomeric complexes. If the predominant ("major" diastereomer continues along the catalytic cycle, it will lead to the disfavored ("minor") enantiomer. Conversely, the minor catalyst/olefin complex will lead to the major enantiomer. This is a classic Curtin- Hammett scenario. More graphically: Chem 106 | Art Ey : PY oy Me0z,C._Nv : C ANS Me ——> : “Ac : aN ' Ar2* Ph C-olefin Prnin = R product ' major | He. pe Meo,c._N predominant : a 2! N wo N KC] — T Ac" product ‘Re * Ph 2 C-olefin Pimaj = S product : minor ‘ (3) Under typical conditions (1 equivalent of MAC + 0.01 M DIPAMP in MeOH at room temperature), we get an 11:1 ratio of diastereomeric complexes: K,mi_ C-MAC ZA major Km - Cc + MAC Kymin ~ “\. emac : Ki" minor Our job is now to extract the six rate constants that describe the : behavior of both diastereomeric manifolds. We can use UV-vis ‘ spectrometry to get the total binding constant: Ky + KT = Kip = 3700 M* : Stopped- -flow kinetics can tell us the forward rate constants, ' ; While trapping experiments can give the reverse rate constants. (4) How do we know that the major diastereomer is less ‘reactive? E. Kwan Applications of Kinetics Chem 106 The reaction progress approach is to monitor concentration or rate at conditions that are very similar or even exactly the same as those that are actually used in the reaction. It is worth asking what it means to measure concentration or rate vs. time. The idea of measuring concentration is easy to understand. Fora first-order reaction, we get a curve that looks like this: “integral measurement" [SM]=[SM], exp(—Kt) specify values [SMlo}----+ concentration t=0 The common ways to do this are IR, NMR, UV-vis, LC, or GC. All of them tell you how much stuff you have as a function of time. In contrast, if you use calorimetry, you get a direct measurement of rate. The faster the reaction goes, the more heat flow there is. Graphically, it amounts to specifying the instantaneous slope of the curve as it progresses from the initial boundary condition of having [SM]p starting material at time 0: time “differential measurement" [SM]o | -- ~~ d[SM [SMlo [SM] __,, dt concentration . specify instantaneous slopes t=0 time : This means that whether we specify things as concentrations or : rates doesn't make any difference mathematically. However, it : might affect things graphically. Let's use COPASI to see this. | have modeled a Heck reaction as follows: P c A [Alp = 0.16 ky = 0.01 ky = 100 [B]p = 0.24 B CA k,= [C]p = 0.001 ‘ You can imagine that A might represent aryl halide, which ' undergoes oxidative addition to form C-A, and then the usual ‘ insertion with olefin B and reductive elimination (subsumed into kz) to form product P. Here is concentration vs. time: 0.25 0.2 mmol /ml 2 0.05 T T T T 0 200,000 400,000 00,000 ' 00,000 ' te!06 s -f@l -i) -f€A -B) -Pl : Purple is product, red is the limiting reagent A, and cyan is the ' excess reagent B. It looks like the product curves up with some : dependence that's not clear. Is it a polynomial? Exponential? ' Rational function? Who knows! All| can see here is that the : graph is curved. The traditional approach would be to "linearize" ' the data to some model. However, looking at rate vs. : concentration is also useful: E. Kwan Notice that the graph gets "unsquashed"--a lot more of it has useful information in it: ‘2.5e-06 mmol /(ml*s) T 0.08 Ot mmol/ml —PRate|[A] The y-axis is d[P]/dt and the x-axis is [A]. Unlike concentration vs. time graphs, which are read from left (start of reaction) to right (end of reaction), these graphs are read from right (lots of starting material) to left (reaction is done). The reaction rate shoots up very quickly at the beginning of the reaction, decays through two linear regions. The first region has a non-zero intercept! What is going on? Here is the catalytic cycle again: P c A [Alo = 0.16 Ye 0.01 ys = 100 [Blo = 0.24 B CA k,=1 [C]p = 0.001 Since exit to product is slow, the pre-equilibrium condition is met here and we can write a 1+rate law: Applications of Kinetics Chem 106 y — MIBIK AICI 1+K [A] ‘ K, is ky/k., = 100. At the beginning of the reaction, [A] is 0.16, ‘ so K,[A]=16, which is a lot bigger than 1. This means the catalyst is "saturated" at the beginning of the reaction. The resting state of the catalyst early on in the reaction is C-A. We can see that in this graph (note the change in the x-axis): 0.0002 T 150,000 200,000 250,000 s -A -i -fcA -B) Pl With this approximation, the rate law reduces to: P| = k,{BIICk T T 0 50,000 100,000 What is [B]? Every time a molecule of A gets turned over, a molecule of B gets turned over, so we have: [B]-[A]=e eis the excess, a constant. This is strictly obeyed, as long as we have time-independent stoichiometry (i.e., the mechanism is : constant over the course of the reaction.) E. Kwan Applications of Kinetics Chem 106 So to understand the plot of rate vs. concentration (d[P]/dt vs. [A]), we can rewrite: AP) _ a ELBILCh =k, ([A]+ e)[C], =h[A]+e-k{Ch; slope _ intercept That is why we have a non-zero intercept: there is an excess of one reagent. Why are there two parts of the graph? As the reaction nears completion, [A] drops. At some point, the catalyst is no longer saturated, and the resting state returns to C. As [A] goes to zero, the rate law goes to: _ /IBIKIAICI, 1+K,[A] *k,[B]X[A][C], The curved part in the middle is a transition region where both catalyst states are important. That is why there is no one integer that can describe what is going on. This is called “saturation kinetics." Under more realistic conditions, we might imagine that catalyst decomposition can occur. Oxygen can get in, or palladium can black out. Here is my simulation: Cuead { Kdeact P c A [Alp = 0.16 kp = 0.01 k, = 100 [B]p = 0.24 B CA ky=1 [C]p = 0.001 ' ‘ Presumably decomposition is going to be more complicated than ‘a first-order irreversible process, but that doesn't matter for the ~~ mmol/(mls) ‘ purposes of this discussion. Here is a plot of rate vs. [A] for ' different values of Kgeact: 2.5€-06 increasing Kgeact —j- r T 0.02 0.04 0.06 0.08 0.1 0.2 out 0.16 mmol/ml —pratellAl ‘ Obviously, catalyst deactivation is bad. For high enough kgeact: ' this affects the yield of the reaction very significantly. The | x-intercept of the curve starts to move away from 0, meaning ' that not all of the [A] gets converted into product. But notice : that the shape of the curve doesn't change much. Sure, the ‘ curve we were just discussing gets truncated. ‘drops much, then we never leave the saturation regime. But If [A] never if we didn't already know that catalyst decomposition was ‘ occurring, all we that would be apparent is that we have a ' straight line and the yield isn't 100%--certainly not an unusual ' Situation! : The same excess experiment is designed to probe this very ' situation. The idea is to run the reaction with different initial : conditions related by having the same e. Why would we want ‘to do that? 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 -2.0x10 -7 0.0 2.0x10 -7 4.0x10 -7 6.0x10 -7 8.0x10 -7 1.0x10 -6 1.2x10 -6 1.4x10 -6 1.6x10 -6 ra te ( M s -1 ) [A] (M) experiment 1 experiment 2 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 -2.0x10 -7 0.0 2.0x10 -7 4.0x10 -7 6.0x10 -7 8.0x10 -7 1.0x10 -6 1.2x10 -6 1.4x10 -6 1.6x10 -6 1.8x10 -6 2.0x10 -6 2.2x10 -6 2.4x10 -6 ra te ( M s -1 ) [A] (M) experiment 1 experiment 2 E. Kwan Applications of Kinetics Chem 106 Measuring the concentrations vs. time amounts to measuring ‘ ‘ the solution is a line, not a point.) Therefore, we need two data rate (Vv) vs. concentration. In the pre-equilibrium case, we must! ‘ points from experiments with different excess values. In reality, fit this data to two independent parameters (K, and kz). Inthe —_: one takes thousands of data points from several experiments steady state case, we must fit this data to three independent ' to overspecify the system (like five equations and three parameters (k,, k.4, ko): unknowns). One then uses non-linear least squares analysis to ' : find the unknowns that will best fit all of the data. _ {BIA [AIC] __/*,[A]BIIC], = ~ ‘ Let's take a graphical approach to see how this works. Here are I+ KA] Ky +k, [A}+k [B] ‘ data from various Heck reactions (Blackmond JOC 2006, 77, We get a series of points: ' 4711). For each of the following, what is the resting state of the excess ' ‘ catalyst and what is the turnover-limiting step? time [A] [B] e=[B]-[A] [P] rate ty 1.0 1.2 0.2 0.0 wy : Example 1 ty 0.9 1.1 0.2 0.1 Vo ' 0.12 ts 0.5 0.7 0.2 0.5 v3 = 10°: a symbol _| [ArX}, (M) | [Nulo(M) | _[e] (M) ty 0.2 0.4 0.2 08 v4= 10%: 2 02 | 0.28 | 008 t 0.0 0.2 0.2 10 or o 03 | 038 | 008 5 aa . ao ; : _ ' A 0.16 0.39 0.23 Suppose we are in the pre-equilibrium regime. Assuming the oo reaction is exactly as modeled and the measurements are “ce 0.084 perfect, why can't | just take a mere two time points? That would : € be two equations and two unknowns. For example, consider ial time points 3 and 4 (suppose [C];=0.01): SS goed joe = £2 0-DK, (0.50.01) = : @ 1+ X,(0.5) Boos 107 = k,(0.4)K, (0.2)(0.01) 1+K,(0.2) 0.02 The problem is that taking points from the experiment mean that [A] and [B] are not independent. Specifically, they are related | by the excess: ' 0 T T T [B]-[A]=e ' 0.15 02 0.25 03 0.35 one equation and two unknowns. (This is like trying to solve As the excess is fixed throughout the experiment, this is really + [Nu] (M) x=2y and 2x=4y; because the equations are not independent, E. Kwan Applications of Kinetics Chem 106 It helps to consider the catalytic cycle and 1+rate law: ; That implies that the latter scenario: turnover-limiting insertion: P c ArX : 0.06 - Nu C-Arx | (All | did is change A to "ArX" and B to "Nu." B is the olefin.) What does the graph tell us? (1) There is overlay between curves of same excess, indicating that neither catalyst decomposition nor product inhibition are important here. 2 vats (2) What is meant by rate/[ArX]? The rate is a flat line, meaning | that it does not go up with increasing [Nu]. Therefore, the : 0.02 reaction is zero-order in [Nu]. The rate increases linearly . rate/[Nu] (min) with [ArX], so: ] symbol__| [ArXJp (M)|_[Nuls (M) | fe] (M) : oO 0.16 0.28 0.12 the reaction is first-order in [ArX]. ' J Oo 0.16 0.24 0.08 j A 12 | This is consistent with free resting catalyst and turnover- ' 1 0.12 02 | 008 limiting oxidative addition. As soon as oxidative addition ' 0 et occurs, the complex is whisked away to product. Thus, itis | not appropriate to apply the pre-equilibrium approximation ' 0 0.05 0.1 0.15 here. ' ' [ArX] (M) Example 2 . . . . What if the resting state changes from C-ArX at the beginning What if the reaction is first order in [Nu] but zero-order in [ArX]? : of the reaction to C later on in the reaction as all the ArX is It helps to consider the 1+rate law: : depleted (as in the earlier example)? y= INUIALArXIIC], —v _ ATAXIA IC], [Arx] "14K, [ArX] [Nu] 14+K,[ArX] 1 +e{ArX] ' The graph of rate vs. [Nu] will then be curved, and this is v~[Nu][ArX] _ Ifthe resting state is C, then 1>>K,[Arx]. ' "saturation kinetics" again. This is shown in this example: v~[Nu] If the resting state is C-ArX, then ' K,[ArX] >> 1. '