Modelling of Reaction Mechanisms KJE-3102

Transcription

1 Modelling of Reaction Mechanisms KJE-3102 Kathrin H. Hopmann

2 Outline Potential energy surfaces Transition state optimization Reaction coordinates Imaginary frequencies Verification of TS (e.g. IRC) Level of theory Barriers and reaction energies Gibbs free energy, enthalpy, and entropy and how they are computed Transition State Theory Modelling of enzymatic reactions The consequence of freezing How to compute the surroundings Effect of model size

3 Potential Energy Surface (PES) First order saddle points are transition states between two local minima (for example reactant and product of a chemical reaction). Transition State (TS) First-order saddlepoint Relative Energy Reactant Local Minimum Barrier Reaction energy Product Local Minimum Reaction Coordinate

4 Potential Energy Surface (PES) Can also look at this in several dimensions! Minima and first-order saddle points are stationary points (gradient = 0)! BUT: First-order saddle points have ONE negative force constant.

5 Transition State Optimization Two general approaches: 1) Interpolation: Feed system with information on reactant and product. Find intermediate structure by some form of interpolation. 2) Local method: Newton-Raphson (NR) optimization scheme will converge to closest stationary point (gradient = 0). Start optimization sufficiently close to TS structure. Typically done after interpolation or directly (e.g. by providing starting guess of TS structure based on chemical intuition/knowledge/experience).

6 TS Optimization Schemes Typical interpolation scheme: Linear Synchronous Transit (LST) Quadratic Synchronous Transit (QST) TS NR IRC (F. Jensen, Introduction to Computational Chemistry, p. 328). TS optimization using interpolation + local method: Do LST QST Start NR optimization TS optimization using interpolation : Give geometry of R and P LST max Minimize in perpendicular direction QST min QST max (close to TS) Alternative: TS optimization using local method only Guess geometry of TS Start NR optimization How can we guess the TS? Need to identify a reaction coordinate..

7 Reaction Coordinate, I Coordinate that represents transformation of reactant into product. Can be multiple bond lengths/angles! Example: Rotation of methyl group in ethanol 60º Relative Energy (kcal/mol) Start TS opt from here 60º 0º -60º Reaction Coordinate: Dihedral angle O-C-C-H

8 Reaction Coordinate, II Reaction catalyzed by the enzyme Limonene Epoxide Hydrolase (LEH): O OH LEH active site 1 H 2 O H 1 2 OH Arg99 Asp101 Asp132 Wat Asn55 Tyr53 Substrate Start TS opt from here Relative Energy (kcal/mol) 3,10 2,90 2,70 2,50 2,30 2,10 Fixed Distance, O to C (Å) 1,90 1,70 K. H. Hopmann et al., JACS 127, , 2005

9 Transition Structure Arg99 Asp101 Asp Asn Frequencies Tyr53 Optimized TS for LEH reaction. How do I know this is a TS? One Imaginary Frequency cm -1 (355.6 i )

10 Imaginary Frequency, I Force constant k: 2 nd derivative of energy with respect to coordinates k > 0 k < 0 All other coordinates are in minimum (k > 0) Frequency (harmonic oscillator): All coordinates in minimum (k > 0) Reaction coordinate is in maximum (k < 0) ω = k m r All coordinates in minimum (k > 0) If k < 0, ω k m i = = r k m r i (F. Jensen, Introduction to Computational Chemistry, p.297).

11 Imaginary Frequency, II Magnitude of the imaginary frequency depends on the reduced mass (function of the atomic masses involved in the motion). 1218i cm i cm i cm i cm -1 74i cm -1 53i cm -1

12 How do I verify that I have the right TS?, I H H F H C H Cl F H C H Reactant TS Cl H F C H Cl H Product + I: Visual inspection (!!!!!!!!!!!!!!!!) II: IRC calculations Requirements S N 2 reaction: F C-Cl angle = ~ 180 º CH 3 = ~ planar Intrinsic Reaction Coordinate (IRC) = the minimum energy reaction pathway between the transition state of a reaction and its reactants and products. 15 Relative Energy (kcal/mol) Reac IRC = reverse TS IRC = forward Input: Geometry of TS and force constants. The IRC algorithm walks along the reaction coordinate until a minimum is reached. Two directions are possible, forward and reverse. -25 Prod

13 How do I verify that I have the right TS?, II Reactant Possible TS Product Bruno Cardey, unpublished results Barrier -3 kcal/mol? Barrier ~25 kcal/mol

14 How do I verify that I have the right TS?, III Reactant Possible TS Product IRC = reverse IRC = forward Wrong reactant! 0 Right product Wrong reactant Wrong TS Right product

15 How do I verify that I have the right TS?, IV Reactant Possible TS Product (spin density: ρα(r) ρβ(r)) 10 5 IRC = reverse IRC = forward 0 Right reactant! -5 Intermediate Reactant TS1 Intermediate TS2 Product

16 TS Optimizations: Some typical errors, I 1. Basis sets H H F H C H Cl F H C H Reactant TS Cl H F C H Cl H Product + Reaction coordinate: F-C distance Reaction coordinate: F-C distance Start TS opt. from here 85i cm Relative Energy (kcal/mol) G(d,p) 401i cm G(d,p) Relative Energy (kcal/mol) B3LYP/6-31G(d,p), IEFPCM(water) The PES has no well-defined maximum. The optimized TS looks incorrect. What did I do wrong? The basis set! The reaction involves anionic species (F - and Cl - ). These typically NEED diffuse functions for a correct description.

17 TS Optimizations: Some typical errors, II 2. RHF versus UHF Bond dissociation RHF = Restricted HF UHF = unrestricted HF ( (F. Jensen, Introduction to Computational Chemistry) RHF UHF

18 TS Optimizations: Some typical errors, II Open-shell singlets Reactant Intermediate Product (R)-B3LYP intermediate Closed-shell singlet (β) (β) U-B3LYP intermediate Open-shell singlet (α) C: O: O: Remaining atoms: = 0 Closed-shell singlet: Number of α and β electrons the same, all paired! Open-shell singlet: Number of α and β electrons the same, not all paired. U-B3LYP optimized TS -585i cm -1 (reaction studied in T. Saito et al. J. Phys. Chem. A 2010, 114, )

19 Broken Symmetry DFT Open-shell singlet calculations are also referred to as Broken symmetry (BS) DFT Many systems require BS descriptions, in particluar transition metal complexes often do. Note: BS-DFT introduces some errors in the geometry (too long bond lengths) and in the energy (too high) due to spin contamination. O O Fe 3+ 2 unpaired electrons Fe 3+ S S Fe unpaired electrons NO- Fe 3+ O O O NO- Fe 3+ - ON - ON S 2- Fe 3+ NO - 3+ Fe NO - S 2- Fe 3+ Fe 3+ S 2- NO - NO - - NO 14 unpaired electrons 34 unpaired electrons

20 Summary 1. Part Transition states are saddle points on the PES. Minima have only real frequencies, TS have 1 (!) imaginary frequency TS structures often optimized through linear transit schemes (e.g. based on approximate reaction coordinate) The nature of the TS should be verified through inspection and IRC calculations! The level of theory is important! Consider basis set, method and effect of surroundings carefully!

21 Outline Potential energy surfaces Transition state optimization Reaction coordinates Imaginary frequencies Verification of TS (e.g. IRC) Level of theory Barriers and reaction energies Gibbs free energy, enthalpy, and entropy and how they are computed Transition State Theory Modelling of enzymatic reactions The consequence of freezing How to compute the surroundings Effect of model size

22 Barriers and Reaction Energies Transition State (TS) Relative Energy Reactant ΔG Gibbs free energy of activation/barrier: ΔG = G TS G Reac Gibbs free reaction energy: ΔG r = G Prod G Reac ΔG r Product Reaction Coordinate (RC) The barrier gives information about how fast a reaction proceeds. The reaction energy determines if a reaction can occur spontaneously (ΔG r < 0)

23 Gibbs Free Energy, Enthalpy and Entropy G = H T S H = enthalpy, S = entropy, T = temperature What is entropy? Disorder of the system. More disorder leads to lower energy. Example: there is an entropy gain in the reaction: AB A + B What is enthalpy? H = U + PV (P = pressure, V = volume, U = internal energy, sum of all kinetic + potential energy in system) If P is constant, H = heat exchange. Exothermic: heat is lost. Endothermic: heat is gained from surroundings. The reaction AB A + B involves bond breaking and is typically endothermic. Enthalpy and entropy and Gibbs free energy depend on conditions (T, P, V).

24 Computing Gibbs Free Energy If we do a geometry optimization, all we have is the electronic energy at 0 Kelvin! Electronic energy is NOT the same as Gibbs free energy! Relative Energy (kj/mol) Reac Gaussian 03/B3LYP/6-311G(d,p), K. H. Hopmann et al. Organometallics 2009, 28, 3710 TS1 ΔE ΔG Inter TS2 Prod Electronic Energy, 0 K Gibbs Free Energy, K

25 Statistical Mechanics To obtain G: Compute H and S. Include contributions from rotation, vibration, translation, and electronic energy. H tot = H vib + H rot + H trans + H elec S tot = S vib + S rot + S trans + S elec G = H tot TS tot With H, S and G for Reac, Prod and TS, we can compute the differences: ΔG = ΔH TΔS ΔG r = ΔH r TΔS r Partition function is the fundamental equation from which macroscopic observables are computed. q = allstates i ei k e / B T Typically computed for the canonical ensemble (N, V, T) constant number of molecules, volume, temperature. Standard: 1 mole, K, 24.5 L (F. Jensen, Introduction to Computational Chemistry, p. 298ff).

26 Computing Enthalpy, I H tot = H elec + H trans + H rot + H vib We are typically interested in the enthalpy of activation, i.e. H(TS)- H(Reac) Enthalpy difference: H tot (TS) - H tot (Reac) = ΔH tot = ΔH elec + ΔH trans + ΔH rot + ΔH vib Transition State (TS) Electronic: Energy of stationary point on PES => ΔH elec = ΔE PES PES Reactant Translational and Rotational (1 mole non-linear molecules): Reaction Coordinate (RC) H rot = 3/2 RT H trans = 5/2 RT => ΔH rot = 0, ΔH trans = 0 Constants!

27 Computing Enthalpy, II Vibrational: H vib = R 3N 6(7) i= 1 hωi + 2kB hωi k B e 1 hω / k T i B 1 = ZPVE + H vib (T) ZPVE = zero-point vibrational energy, 0 K => ΔH vib = ΔZPVE + ΔH vib (T) ZPVE e 0 = ½ hω i Terms contributing to relative enthalpy: Enthalpy of Activation: ΔH tot = ΔE PES + ΔZPVE + ΔH vib (T) Approximation (0 K): ΔH tot = ΔE PES + ΔZPVE

28 Computing Entropy Entropy difference: S tot = S elec + S vib + S rot + S trans S tot (TS) - S tot (Reac)= ΔS tot = ΔS elec + ΔS vib + ΔS rot + ΔS trans Electronic: S elec = R ln(2s+1) (S on r.h.s = spin!, 2S + 1 = degeneracy factor) Translational: S trans = 5/ 2R + R ln V N A 2πMk 2 h B T 3/ 2 (M = molecular mass, N A = Avogardros number) => => ΔS elec = 0 ΔS trans = 0 Rotational: R 3 + ln 2 2 π 8π k 2 σ h T 3/ 2 B S rot = I1I2I3 => ΔS rot = S rot (Reac) S rot (TS) (I = principal moments of inertia, σ = order of the rotational subgroup in the molecular point group)

29 Computing Entropy, continued Vibrational: = = 6(7) 3 1 / / ) ln(1 1) ( N i T k h T k h B i B i B i e e T k h R ω ω ω S vib lim 0 ω 1 if S then G = H - T S - ΔS vib = S vib (Reac) S vib (TS) => Important: Entropy is VERY sensitive to very small frequencies! Entropy of Activation: ΔS tot = ΔS rot + ΔS vib = = 6(7) 3 1 / / ) ln(1 1) ( N i T k h T k h B i B i B i e e T k h R ω ω ω S vib Terms contributing to relative entropy:

30 Example Thermodynamics Product O O O Ti 4+ O Cl Cl HO HO O O O Ti O O O 74i cm -1 Reactant Product Output (Gaussian), Freq calculation Sum of electronic and thermal Enthalpies= = H tot Sum of electronic and thermal Free Energies= = G tot E (Thermal) CV S KCal/Mol Cal/Mol-Kelvin Cal/Mol-Kelvin Total = S tot S(Reac) = cal mol -1 K -1, S(TS) = cal mol -1 K -1 S = cal mol -1 K -1 = J mol -1 K -1 H kj mol - 1 S J mol - 1 K - 1 G kj mol - 1 Computed Exp G03/B3LYP/6-311G, CPCM methanol J. Conradie, K. H. Hopmann, unpublished results

31 Transition State Theory (TST), I Transition State (TS) A k r k r = B k B T h The rate is related to the speed of the reaction ΔG exp RT Relative Energy A (reactant) ΔG ΔG r B (product) k r = rate constant (s -1 ) h = Plancks constant, ( J s) R = gas constant, ( JK -1 mol -1 ) k B = Boltzmans constant, ( JK 1 ) Reaction Coordinate (RC) Strictly only valid of all molecules that pass over the TS become products. Sometimes a κ ( transmission coefficient ) is used (to allow for molecules to reflect back to the reactant side, and to allow for tunneling effects). κ is usually close to 1 and therefore, usually, omitted. Using TST, the equilibrium constant can be calculated from the reaction energy: A B K eq = [B] [A] Keq ΔG = exp RT r

32 Transition State Theory (TST), II TST used to compare experimental rates to computed ΔG values. k r = s kcal/mol = 77.7 kj /mol Note: Barrier increase of ~1.4 kcal/mol rate is reduced ~10 times! ΔG = 14 kcal/mol k r = 340 s -1 ΔG = 15.4 kcal/mol k r = 32 s -1 Typical error in DFT calculations of barriers: 1 to 3 kcal/mol (with metals more, maybe up to 5 kcal/mol) => Errors in predicted rates of 100 times or more. Therefore it is not possible to predict experimental rates from DFT! Always translate exp. rates to barriers and compare to experiment, not opposite. If there is no experimental rate or barrier to compare to, how do I know what is a reasonable value for a barrier? Almost anything can be OK, depending on the type of reaction. For enzymes a barrier above 20 kcal/mol is typically considered too high! For organic reactions it can often be more, 30 or 40 kcal/mol.

33 Summary 2. Part Remember the difference bewteen reaction energy and barrier! ΔE PES ΔG Thermochemical quantities only valid for stationary points! (g = 0, min: k > 0 TS: 1 k < 0). Have to use same PES as optimizations (no change of basis set or method or solvent model allowed) Entropy (S) is very sensitive to small frequencies makes ΔG very sensitive. Computation of ΔH is most reliable. TST is used to compare experiment and theory Cannot predict rates from theory (due to inherent error in calculation), only barriers!

34 Modelling of Enzymatic Reactions QM methods can be used to model enzymatic reactions these typically take place in the active site involving a limited number of residues (= amino acids). Crystal structure QM model Energies Asp101 TS Arg99 Valpromide Substrate Asp132 Wat Asn55 Reactant ε = 4 Tyr53 Active site Truncation Substrate Hydrogen Fixation addition Surroundings Extraction Product

35 The Consequence of Freezing Limonene Epoxide Hydrolase: A H A O OH B H O H BH OH 292.1i cm -1 Arg99 Asp101 Substrate Asp132 Tyr53 Wat Asn55 ΔG value highly affected by small frequencies (entropy contribution). Often only ΔH ( Δ E + ZPVE) is reported. 12.4i cm -1 K. H. Hopmann et al., JACS 127, , 2005

36 How to treat the Surroundings? Rest of enzyme typically modelled with PCM. Value of ε = 4 is often used for the dielectric constant (mixture of water and protein) somewhat abitrary! Haloalcohol dehalogenase HheC: ΔH kcal mol - 1 (ΔE + ΔZPVE ) Vacuum (ε = 1) 23.0 ε = ε = ε = ε = ε = Asp80 Phe186 Arg149 Tyr145 Ser132 ΔH (kcal/mol) Wat Leu178 Tyr177 ΔH is reduced in more polar environment! (better solvation of liberated Cl -. NB: effect depends on type of reaction) Most of the solvation effect already at e = 4. This is the case for any QM model! K. H. Hopmann, F. Himo, JCTC 4, , 2008

37 Effect of Model Size Haloalcohol dehalogenase HheC: Asp80 Phe186 Wat Leu178 Tyr177 Arg atoms. Charge 0. Barrier (ΔH ): Vacuum: 23.0, PCM = 15.0 kcal/mol, Sol-effect = 8.0 kcal/mol Reaction energy (ΔH r ) : Vacuum: 17.5, PCM = 3.4 kcal/mol, Sol-effect= 14.1 kcal/mol Tyr145 Ser132 RCPE Phe12 Asp80 Pro175 Leu178 Arg149 Tyr145 Ser1 32 Phe186 Wa t Asn176 Tyr177 RCPE 161 atoms. Charge 0. Tyr187 Barrier (ΔH ): Vacuum: 18.2, PCM = 17.4 kcal/mol, Sol-effect = 0.8 kcal/mol Reaction energy (ΔH r ) : Vacuum: 5.5, PCM = 4.5 kcal/mol, Sol-effect= 1.0 kcal/mol K. H. Hopmann, F. Himo, JCTC 4, , 2008

38 Summary 3. Part Conclusions on enzymatic reactions: Cut-off atoms are kept frozen (except if there are stong interactions, such as metal complexes). ΔG values cannot be computed (freezing introduces small imaginary frequencies) Surrounding treated with PCM, eps = 4 abitrary value! Often large QM models (up to 200 atoms). Bigger models reduce solvation effect. Effects of model size, freezing and solvation are also valid for smaller (nonenzymatic) models

39 Overall Conclusions Transition states are saddle points on the PES TS structures have one negative eigenvalue 1 imaginary freq. (frozen structures can exhibit small unrelated imaginary frequencies). Gibbs Free barriers and reaction energies are computed from enthalpy and entropy contributions using statistical mechanics. Computed Entropy is very sensitive to low frequency modes, therefore enthalpy is more reliable! Transition State Theory can be used to compare computed barriers to experimental rates. Model size, freezing, dielectric constants, etc. can have big effects on results!