P4_Chemical Reactions - Chapter P4: Transition–State...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter P4: Transition–State Geometries and Activation Energies Up until this point, we have concerned ourselves with quantum chemical calculations on stable molecules, that is, molecules that at least in principle can be directly observed and characterized experimentally. There is, however, another class of “molecules” of great importance to chemists that cannot be observed, let alone directly characterized. These are transition states and correspond to locations on a potential energy surface that lie at the top of pathways connecting stable molecules. There is no reason why quantum chemical calculations should perform any poorer (or any better) for transition states than they do for stable molecules. The only difference is that we can not lean (directly) on experiment to gain confidence. However, we can use our prior experience with stable molecules to prejudge how well the calculations are likely to describe the geometries, energies and other properties of transition states. In one sense, calculations take on added value and become the primary (and sometimes only) means available. A Transition State in One Dimension In a one-dimension (a diatomic molecule), a transition state is a point on the potential energy curve where the energy is at a maximum, that is, a point for which the first derivative of the energy with respect to the bond distance, R, is zero and the second derivative (curvature) is negative. For comparison, a stable diatomic molecule is a minimum on the energy curve, that is, a point for which the first derivative is also zero but the second derivative is positive. transition state: dE/dR=0, d2E/dR2<0 minimum: dE/dR=0, d2E/dR2>0 The difference between the two will be reflected in the molecule’s infrared spectrum. Recall from the discussion in Chapter P2, that the vibrational frequency of a diatomic molecule is proportional to the square root of the ratio of curvature and reduced mass. This means that the frequency for a 1 transition state is an imaginary number (square root of a negative number). Of course, it is not possible to actually record the infrared spectrum of a transition state, but it is no more difficult and no less reliable to calculate such a spectrum than it is to calculate the spectrum of a normal molecule. A Transition State in Many Dimensions A non-linear molecule made up of N atoms is fully characterized by its location on a (3N-6) dimensional potential energy surface. (The origin of the coordinate system and orientation of the molecule in the coordinate system are arbitrary, removing six of the 3N Cartesian coordinates.) While it is not possible to actually visualize such a multi-dimensional surface (at least we don’t know how to visualize it), it is possible to represent it mathematically and to identify “interesting” points on the surface. These are points where the first derivative is zero for each of the coordinates, and are referred to as stationary points. The procedure required to decide whether a particular stationary point is a minimum or a maximum with regard to each of the coordinates has already been detailed in Chapter P2, and will only be briefly outlined here. First, the full set of 3N x 3N second energy derivatives (∂2E/∂xi∂xj) with respect to the Cartesian coordinates, x, are calculated. This is the costly (in terms of computer time) step in the overall procedure. Second, the second derivatives are mass weighted; diagonal terms (∂2E/∂xi2) are divided by the mass of the atom i, Mi, and off-diagonal terms (∂2E/∂xi∂xj) are divided by the product of the square root of the masses of the atoms i and j. Third, the Cartesian coordinates are replaced by a new set of normal coordinates, ζ, defined such that the matrix of mass-weighted second energy derivatives is diagonal. (∂2E/∂ ζ i∂ ζ j)/(√Mi√Mj) = δij (∂2E/∂ ζ i2)/Mi δij is 1 if i=j and 0 otherwise. Finally, the six normal coordinates corresponding to the three translations and three rotations of the molecule relative to its center of mass are removed, leaving 3N-6 coordinates corresponding to vibrational motions. This procedure can be fully automated and is no more demanding in terms of human time than energy calculation. However, it is an order of magnitude more demanding in terms of computer time. A transition state in one dimension corresponds to a maximum of the energy. Is it also an energy maximum in the general 3N-6 dimensional case? Yes and no. We hypothesize that a transition state is a stationary point on a 2 multidimensional potential energy surface for which only one of the second derivatives in normal coordinates is negative, and the second derivatives for all the remaining normal coordinates are positive. That is to say, a transition state is an energy maximum in only one dimension, the so-called reaction coordinate, and an energy minimum in the 3N-7 remaining dimensions. Liken a chemical reaction to driving across a mountain range (a twodimensional system). The “goal” is the same, to go from reactants to products in the case of the reaction and to go from one valley to another in the case of the drive. The difference is that the road traverses a twodimensional surface (of the mountain) whereas the chemical reaction may involve hundreds of dimensions. The important point is that there is no need to climb to the top of a mountain (a maximum in both dimensions) to move between two valleys, but it is sufficient to go through a valley between mountains (a maximum in one dimension but a minimum in the other). Similarly, it is only necessary to pass through an energy maximum in only one coordinate (the reaction coordinate) to move between reactants and products of a chemical reaction. figure Not all stationary points on a multidimensional potential energy surface that satisfy this requirement are likely to be transition states for the process that is actually of interest. In fact, it is likely that very few are. An additional “chemical” requirement needs to be imposed, mainly that the point lies on a path that actually connects the reactants and products. Finally, note that there may be more than one point on the energy surface that satisfies both the mathematical and chemical criteria. Without identifying and examining each and every one of them it will generally not be possible to say which is the lowest in energy and most likely to be pathway that is actually followed. While transition states cannot be directly observed, there are, however, indirect ways to suggest how their structures differ for those of the reactants. Most informative are kinetic isotope effects, that is, rate changes in response to changes in the masses of one or more atoms in the reactants. A large change in rate implies that the affected atoms are intimately involved in the reaction, whereas a negligible change implies that they are not significantly involved. Most common are isotope effects associated with substitution of deuterium for hydrogen, where the rate ratio is referred to as kH/kD. The effects are small but measureable, typically in the range of 1.03 to 1.07. kH/kD for a process that involves complete CH (CD) bond 3 dissociation is ~1.2. Effects resulting from mass changes for other elements, for example, 13C for 12C, are much smaller and are only rarely considered. What is the origin of kinetic isotope effects? While a full treatment is beyond the scope of this text, the simplified description provided here conveys the essential point. It is that the Born-Oppenheimer approximation removes nuclear mass from the Schrödinger equation (see Chapter xx), and the energy and wavefunction are not affected by changes in mass. However, vibrational frequencies change with mass; the heavier the atoms involved in a particular vibration the smaller the frequency (see Chapter P2). Properties that depend on frequencies also change with mass. This includes the zeropoint vibrational energy (ZPE), which needs to be added to the energy obtained from a quantum chemical calculation to account for the residual (0K) vibrational energy of a molecule (see Chapter P4). ZPE is positive and is proportional to the sum of the vibrational frequencies, meaning that it decreases with increasing mass, for example, as hydrogen atoms are replaced by heavier deuterium atoms. ZPEH ZPED To good approximation, a kinetic isotope effect may be seen to arise from differences in zero-point energies of different isotopic variants between the transition state and the reactants, ∆ZPE, for example, kH/kD. The expression below assumes energies in atomic units (hartrees) and that the imaginary frequency is not included in the calculation of ZPE for the transition states. kH/kD = exp[-1060(∆ZPEH – ∆ZPED)] ∆ZPE = ZPEtransition state – Σ ZPEreactants Do we want to give the “correct” expression from statistical mechanics? Ene Reaction: The ene reaction involves addition of a electron-poor double bond to an alkene with a allylic hydrogen. The hydrogen is transferred and a new carbon-carbon bond formed, for example, in the addition of maleic anhydride and propene. 4 A large kH/kD for the allylic hydrogen would suggest that its environment in the transition state is quite different from that in the reactant. On the other hand, a nearly unit kH/kD would suggest either that the hydrogen has yet to move from the reactant or has fully moved to the product. Structures and vibrational frequencies for the two reactants and for the transition state for the ene reaction of maleic anhydride and propene obtained from HF/6-31G* calculations are provided in ene reaction under the Chapter xx directory. Does the calculated transition-state structure show that bond making and bond breaking occurs simultaneously? Specifically, is the migrating hydrogen partially bonded to two carbons? Examine the vibrational motion corresponding to the imaginary frequency in the infrared spectrum of the transition state. This corresponds to the reaction coordinate. Would you describe this reaction as a concerted process? Elaborate. Calculate ∆ZPE for the reaction. (Do not count the imaginary frequency for the transition state in your calculation.) Next, change the “migrating hydrogen” to a “migrating deuterium” and repeat the calculation. This is very rapid as it requires only recalculation of the frequencies from the existing second energy derivatives and the new set of atomic masses. Calculate kH/kD. Is it consistent with a concerted process? 13 C/12C Isotope Effects on CH Bond Dissociation in Methane: How much does the CH bond dissociation energy in methane change as a result of changing the mass of carbon (from 12C to 13C)? Use the B3LYP/6-31G* model to calculate equilibrium geometries and vibrational frequencies for methane and methyl radical and evaluate the change in zeropoint energy for bond dissociation. Change the mass of carbon to (13C) for both molecules, rerun the calculation and recalculate the change in zero-point energy for bond dissociation. 5 Finding a transition state is likely to be more difficult than finding an equilibrium structure. One reason for this is that, in contrast to the enormous number of experimental equilibrium geometries (see Chapter P2), there are (and can be) no experimental transition-state geometries on which to base a guess. Another reason is that methods for locating transition states are less well developed than methods for finding minima. A third reason is that the energy surface in the vicinity of a transition state is likely to be more shallow than that in the vicinity of an energy minimum. After all, a transition state reflects a delicate balance between bond breaking and bond making, whereas an equilibrium structure reflects a situation in which bonds are formed to maximum extent. The reason that this may be a problem is that shallow energy surfaces are not likely to be properly described by a quadratic function, the form that is assumed in all common optimization algorithms. As a consequence, a search for a transition state will probably require more steps than a search for an equilibrium structure. The key to finding a transition state is to provide a good guess at its structure. Transitionstate geometries, like equilibrium geometries, are expected to exhibit a high degree of uniformity among closely-related systems, and the best way to do this is to base the guess on the transition state for a closely-related system. Spartan Student provides a database of transition-state geometries for a wide variety of simple reactions. A transition state is a stationary point and the procedure used to locate it is identical to that used to find an equilibrium structure. The only difference is that it is a stationary point which is an energy maximum in one dimension (and an energy minimum in each of the other dimensions). This needs to be enforced. The underlying algorithm, like that for finding an equilibrium geometry, involves an iterative process, which is terminated only when all energy first derivatives closely approach zero and all geometrical variables reach constant values. In practice, finding a transition state may be two or three times more costly in terms of computer time as finding an equilibrium geometry, although it is no more costly in terms of human time. After locating a transition state, it may be desirable to obtain its “infrared spectrum” in order to verify that it contains a single imaginary frequency, and to establish that the motion corresponding to this frequency is appropriate for the reaction of interest. 6 Limiting Behavior of Hartree-Fock, B3LYP and MP2 Models for Transition-State Geometries and Activation Energies We first set out to establish the limiting behavior of Hartree-Fock, B3LYP and MP2 models with regard to the geometries of transition states and energy differences between transition states and reactants (activation energies). This serves two purposes. First, it provides a set of “reference” transition-state geometries and associated activation energies with which to compare geometries and activation energies obtained from simpler and more practical models. Second, it separates any effects arising from the LCAO approximation (use of a finite basis set) from effects arising from replacement of the exact many-electron wavefunction by an approximate Hartree-Fock, B3LYP or MP2 wavefunction. In practice, it is not possible to actually reach the basis set limit of any of these models. However, it is possible to use a sufficiently large basis set such that the addition of further functions to the basis will have only small (and quantifiable) effects on calculated transition-state geometries and activation energies. We employ the cc-pVTZ basis set for this purpose. This is about as large a basis set that can presently be routinely applied for transition-state geometry and activation energy calculations on any but the simplest chemical reactions, that is, involving molecules with more than a few non-hydrogen elements. The cc-pVTZ is smaller and less flexible than the cc-pVQZ basis set previously used to establish limits for equilibrium geometries (Chapter P2) and reaction energies (Chapter P3). However, we have seen that HartreeFock, B3LYP and MP2 models using the cc-pVTZ basis set generally yield nearly identical equilibrium geometries and reaction energies with those obtained from the corresponding models with the cc-pVQZ basis set. There is no reason to expect that this behavior will not carry over to transition-state geometries and activation energies. Transition-State Geometries “Key” bond distances in transition states for a few simple chemical reactions obtained from the Hartree-Fock, B3LYP and MP2 models with the cc-pVTZ basis set are provided in Table P4-1. These include an intramolecular rearrangement, a pyrolysis reaction, a Diels-Alder reaction and examples of the Cope and Claisen rearrangements. Taken together, these represent a number of broad classes of important chemical reactions. A number of observations may be made. 7 i) Transition states, like stable molecules, show consistent structures, although variations in geometrical parameters appear to be larger. This is consistent with the expectation that the energy surface in the vicinity of a transition state is likely to be “flatter” than that in the vicinity of an energy minimum. ii) Where reaction leads to a change in bonds from single to double (or vice versa), the bonds seen in the transition state are intermediate in length. For example, transition states transforming single to double carboncarbon bonds typically show lengths around 1.4Å, intermediate between “normal” CC double bonds (1.30-1.32Å) and “normal” CC single bonds (1.50-1.54Å). Note that these lengths are quite similar to those found in typical aromatic compounds. 8 Table P4-1: “Limiting” (cc-pVTZ basis set) Hartree-Fock, B3LYP and MP2 Bond Lengths in Transition States (Å) transition state reaction a N Hartree-Fock B3LYP cc-pVTZ CH3 b C c MP2 a b c 1.85 1.80 1.19 1.87 1.75 1.20 a b c d e f 1.31 1.40 2.07 1.23 1.24 1.33 1.32 1.40 2.02 1.26 1.27 1.32 1.34 1.40 1.92 1.26 1.27 1.26 2.18 1.39 1.39 1.38 2.22 1.40 1.40 1.39 2.28 1.39 1.39 1.38 a b 1.38 2.08 1.40 2.05 1.42 1.84 a b c d e f H2 aC b CH2 c f O O eCd H 1.91 1.75 1.17 a b c d MeCN MeNC 1.37 1.25 1.94 1.38 1.37 2.29 1.38 1.28 1.95 1.39 1.38 2.36 1.38 1.30 1.79 1.40 1.38 2.20 H ! tOCHO H2C CH2 + HCO2H b + a d c a b b O O a d e O c f 9 iii) The lengths of forming (or dissociating) single bonds show wide variations. For example, forming (dissociating) CC bonds range from 1.80Å (in the methyl isocyanide rearrangement) to 2.36Å (in the Claisen rearrangement of allyl vinyl ether) according to the B3LYP/cc-pVTZ model. Most commonly, the lengths of forming (dissociating) single bonds obtained from the MP2/cc-pVTZ model are shorter than those obtained from the other two models. Forming Bonds in Transition States and van der Waals Radii: Van derWaals radii for hydrogen, carbon, nitrogen and oxygen are 1.2Ǻ, 1.92Ǻ, 1.55Ǻ, and 1.52Ǻ, respectively. How do the forming (dissociating) single bonds in the transition states given in Table P41 compare with the sum of the van der Waals radii for the two atoms involved in the bond? Does your conclusion alter with theoretical model? Activation Energies The rate of a chemical reaction is given by the product of a rate constant, k, and the reagent concentrations [A], [B] ... (a, b, … are integers or fractions). rate = k [A]a [B]b … The rate constant is given by the Arrhenius equation. k = A exp[∆E‡/RT] The pre-exponential factor, A, accounts for the efficiency of collisions among molecules and is generally assumed to be constant for reactions involving a single set of reactants going to different products or for reactions involving closely-related reactants. The activation energy, ∆E‡, is the energy of the transition state connecting the reactants and products of the reaction, referenced to the energy of the reactants. ∆E‡ = Etransition state – Ereactants Activation energies will always be positive numbers, meaning that the transition state will always be higher in energy than the reactants. R is the gas constant and T is the temperature (in K). A good rule of thumb is that reactions with activation energies >200 kJ/mol will not occur at normal temperatures, while those with activation energies <100 kJ/mol will be unstoppable under the same conditions. Since the rate also depends on temperature (through the rate constant), this means that a 10 reaction with a low activation energy can be slowed down or “stopped” by lowering the temperature, while a reaction with a high activation energy can be accelerated by raising the temperature. Some reactions proceed without energy barriers and discernible transition states. Radicals combine without a barrier, for example, two methyl radicals combine to form ethane, and radicals add to multiple bonds with little or no barrier, for example, methyl radical adds to ethylene to form 1-propyl radical. H3C• + •CH3 →H3C–CH3 H3C• + H2C=CH2 → CH3CH2CH2• In the gas phase, SN2 reactions involving anionic nucleophiles, for example, reaction of ethoxide anion and ethyl iodide leading to diethyl ether and iodide anion, proceed without an energy barrier. EtO- + EtI → EtOEt + IThe barrier observed in solution is a consequence of the solvent. Charge is already more spread out in the transition state than in the reactants, meaning that it the solvent has less of a role to play. The Arrhenius equation is assumes that all molecules pass through a single transition state on their way from reactants to products. While this is not unreasonable, it is also not necessarily true. While a reactant without sufficient energy will be unable to reach the transition state (and proceed to products) reactants with excess energy will be able to proceed to products over alternative (higher-energy) pathways. The previous analogy to a road through a mountain valley is also applicable here. Other roads that cross above the mountain valley could also be followed. Despite its limitations, the Arrhenius equation has proven of great value in interpreting measured reaction rates and in connecting variations in rates with changes in molecular structure. Activation energies from Hartree-Fock, B3LYP and MP2 models with the cc-pVTZ basis set for the same set of reactions used previously for bond distance comparisons are provided in Table P4-2. Activation energies estimated from experimental rate data by way of the Arrhenius equation are available for all but one of these reactions and have also been provided. For all the reasons, described above, comparisons between calculated and “experimental” activation energies must be viewed with healthy skepticism. 11 Table P4-2: “Limiting” (cc-pVTZ basis set) Hartree-Fock, B3LYP and MP2 and G3(MP2) Activation Energies (kJ/mol) Hartree-Fock B3LYP cc-pVTZ Reaction MP2 G3(MP2) expt. H2C CH2 + HCO2H 170 205 172 228 161 208 99 34 70 84 242 !tOCHO 186 283 184 MeCN MeNC 159 167-184 147 107 136 151 212 122 106 121 130 + O O 12 Also provided are activation energies obtained from the G3(MP2) recipe (based on equilibrium and transition-state geometries from the MP2/6-31G* model). A HF/6-31G* frequency calculation is part of the G3(MP2) recipe and, for the case of a transition state, leads to an imaginary frequency. We ignore this frequency in calculation of zero-point energy and temperature corrections terms. While the number of examples is too few to permit generalizations to be made from them, it is worthwhile to point out the obvious i)Activation energies obtained from the “limiting” (cc-pVTZ basis set) Hartree-Fock model are significantly larger than those from either the corresponding B3LYP or MP2 models or from the G3(MP2) recipe, as well as from activation energies obtained from experimental rate data using the Arrhenius equation. This suggests that electron correlation in (these) transition states is larger than that in corresponding reactants. The reason for this is not obvious, but suggests that correlation in delocalized systems is larger that in localized systems. ii)B3LYP and MP2 models and the G3(MP2) recipe yield smaller activation energies. The former are in good accord with activation energies obtained from experimental rates. On the other hand, activation energies from MP2/cc-pVTZ calculations for the Diels-Alder reaction, and the Cope and Claisen rearrangements appear to be much too small. That for the DielsAlder reaction is less than half of the estimated experimental value (and a third of the activation energy from B3LYP/cc-pVTZ calculations). iii)With one exception (isomerization of methyl isocyanide) results from limiting B3LYP calculations are closer to G3(MP2) results than results from limiting MP2 calculations. iv)In very general terms, the results for activation energies show much wider variation than those for reaction. Part of this is due simply to the fact that the activation energies dealt with here are typically larger numbers than the reaction energies dealt with in Chapter P3. However, part of the problem reflects the greater sensitivity to change in theoretical model for a transition state than for an equilibrium structure. 13 Practical Hartree-Fock, B3LYP and MP2 Models for Transition-State Geometries and Activation Energies Except for very small molecules, Hartree-Fock, B3LYP and MP2 models with large basis sets such as cc-pVTZ are not likely to be practical for routine calculations of transition-state geometries and activation energies. The G3(MP2) model is also not practical for any but very small molecules. While these and other models which involve very large basis sets and/or extended treatments of correlation are certainly of value to judge “limits”, simpler models smaller are needed for routine calculations. Hartree-Fock and B3LYP calculations with the 6-311+G** basis set may routinely be applied to molecules with molecular weights up to 400 amu, although MP2 calculations are more restricted. The 6-31G* basis set is significantly smaller and lacks diffuse functions. It may be routinely used for calculations on even larger molecules. The question to be addressed is whether they are able to match bond lengths from the corresponding cc-pVTZ basis set models to within ±0.02Å and activation energies to within ±4 kJ/mol. Transition-State Geometries “Key” bond distances in transition states for a few simple chemical reactions obtained from Hartree-Fock, B3LYP and MP2 models with 6-311+G** and 6-31G* basis sets are provided in Table P4-3. Mean absolute deviations for all six models relative to the corresponding models with the cc-pVTZ basis set are on the order of only 0.01A. Individual deviations for B3LYP and MP2 models in particular are significantly larger in some cases (see Table P4-1). The higher sensitivity of transition-state geometries obtained from correlated models relative to Hartree-Fock models to basis set is consistent to previous results for equilibrium geometries (see Chapter P2), and is not unexpected. Bond distances obtained with the 6-311+G** basis set are closer to those obtained from the cc-pVTZ basis set values than those obtained using the 6-31G* basis set, for both B3LYP and MP2 models. The differences are insignificant for Hartree-Fock models. The close agreement between “limiting” Hartree-Fock, B3LYP and MP2 models and the corresponding “practical” models for transition-state geometries is an important result as it means that quantum chemical methods can be applied to real problems and not just idealized systems. 14 Table P4-3: Bond Lengths in Transition States from Practical Hartree-Fock, B3LYP and MP2 Models (Å) Hartree-Fock B3LYP MP2 6-31G* 6-311+G** 6-31G* 6-311+G** 6-31G* 6-311+G** a N CH3 b C c a b c 1.91 1.75 1.17 1.83 1.79 1.20 1.84 1.84 1.19 1.86 1.75 1.21 1.87 1.76 1.20 a b c d e f 1.31 1.40 2.09 1.23 1.25 1.33 1.29 1.40 2.12 1.23 1.24 1.36 1.34 1.40 2.04 1.26 1.27 1.31 1.31 1.40 2.04 1.26 1.27 1.33 1.34 1.40 1.98 1.27 1.28 1.29 1.33 1.40 1.94 1.26 1.27 1.28 2.19 1.39 1.39 1.38 2.18 1.39 1.39 1.39 2.25 1.40 1.41 1.39 2.22 1.40 1.40 1.38 2.27 1.39 1.41 1.38 2.28 1.40 1.42 1.39 a b H2 aC b CH2 c f O O eCd + HCO2H H 1.90 1.75 1.17 a b c d MeCN MeNC 1.39 2.04 1.39 2.07 1.40 1.97 1.40 2.02 1.43 1.75 1.43 1.77 a b c d e f 1.37 1.26 1.92 1.39 1.38 2.27 1.38 1.26 1.96 1.39 1.38 2.31 1.38 1.29 1.91 1.40 1.38 2.31 1.38 1.28 1.96 1.40 1.38 2.39 1.38 1.31 1.80 1.41 1.39 2.20 1.39 1.30 1.80 1.41 1.39 2.22 H !tOCHO H2C CH2 b + a d c a b b O O a d e O c f 15 Transition States for Related Pyrolysis Reactions: Do transition states forcloselyrelated reactions have very similar geometries? Were this the case, it would mean that the transition state calculated for a simple reaction could be used as a guess at the transition state for a more complex but related reaction. A good example is provided by comparison of transition states for pyrolysis of tert-butyl formate (leading to formic acid and isobutene) and of formate (leading to formic acid and ethylene). Use the HF/631G* model to calculate the transition state for the former and compare it to the analogous transition structure for ethyl formate pyrolysis given in Table P4-3. Are the “key” bond lengths very similar? Rationalize any significant differences that you find. Transition States for Related Intramolecular Rearrangements: Another example is provided by rearrangements of methylisocyanide to acetonitrile and tert-butylisocyanide to tert-butyl cyanide. Use the HF/6-31G* model to calculate the transition state for the latter and compare lengths of both the CN bond that is being broken and the CC bond that is being formed with those in the analogous transition state for methylisocyanide (the geometry of which is given in Table P4-3. Rationalize any significant differences that you find. Transition States for “Chemically-Related” Reactions: The ene reaction of 1-pentene is closely related to the pyrolysis reaction of ethyl formate. Here the products are propene and ethylene rather than formic acid and propene. H H + Calculate the transition state for the ene reaction of 1-pentene and compare with that for ethyl formate pyrolysis given in Table P4-3. Use the HF/6-31G* model. In particular, focus on the CH bond length of the migrating hydrogen. Has it shortened, elongated or remained roughly the same from the ene reaction to the pyrolysis. If it has changed, is the direction of the change consistent with the Hammond Postulate? Elaborate. Do whatever additional calculations that may be required to answer the question. Localized vs. Delocalized Bonding: Diels-Alder cycloadditions as well as Cope and Claisen rearrangements all involve transition states with delocalized bonding The HF/631G* model typically overestimates activation energies for by 30-100 kJ/mol. The transition states for all of these reactions show “delocalized” bonding (single and double bonds are replaced by bonds of intermediate length), whereas the reactants exhibit normal single and double bonds. Does this suggest that the 6-31G* model always favors molecules with localized bonds? Use the HF/6-31G* model to obtain equilibrium geometries for toluene (a molecule with a delocalized π system) and its isomer, cycloheptatriene (a molecule with discernable single and double bonds). Is the energy difference smaller or larger than the known difference in heats of formation (toluene is 133 kJ/mol lower in energy)? 16 Activation Energies Activation energies for the same set of reactions obtained from HartreeFock, B3LYP and MP2 models with both 6-31G* and 6-311+G** basis sets are given in Table P4-4. Signed deviations from the corresponding models with the cc-pVTZ basis set are provided in parentheses alongside each entry. “Experimental” activation energies are also given. Significant differences appear between results with 6-31G* and cc-pVTZ basis sets for all three classes of models. The worst cases are the Diels-Alder reaction of cyclopentadiene and ethylene and (except for Hartree-Fock models) the ethyl formate pyrolysis reaction. These differences are significantly reduced upon moving from the 6-31G* to 6-311+G** basis sets. With one exception (the ethyl formate pyrolysis reaction from the MP2 model) deviations between activation energies calculated from the corresponding 6-311+G** and cc-pVTZ models are 4 kJ/mol or less. As with results for transition-state geometries, this is an important result and suggests that practical quantum chemical models are able to provide descriptions that are quite close to those from large-basis-set (“limiting”) models. 17 Table P4-4: Activation Energies from Practical Hartree-Fock, B3LYP and MP2 Models (kJ/mol) Hartree-Focka 6-31G* 6-311+G** reaction B3LYPa 6-31G* 6-311+G** MP2a 6-31G* 6-311+G** expt. 191 (5) 187 (1) 297 (14) 282 (-1) 171 (1) 221 (16) 167 (-3) 207 (2) 178 (6) 252 (24) 166 (-18) 180 (-4) 83 (-16) 97 (-2) 49 (-15) 37 (-3) 84 237 (-5) 241 (-1) 144 (-3) 148 (1) 119 (12) 110 (3) 151 205 (-7) 209 (-3) 121 (-1) 120 (-2) 109 (3) 107 (1) a) deviations from corresponding “limiting” (cc-pVTZ basis set) values are given in parenthesis 130 MeCN MeNC !tOCHO H2C CH2 + HCO2H 170 (-2) 159 236 (8) 167-184 + O O 18 Vinyl Alcohol: Even though vinyl alcohol, H2C=C(H)OH, is significantly less stable than its isomer acetaldehyde (see Chapter P3), it can be stored for long periods. This suggests that the activation energy for isomerization to acetaldehyde is likely to be substantial. Use the B3LYP/6-31G* model to calculate equilibrium geometries for acetaldehyde and vinyl alcohol. What is the room-temperature Boltzmann ratio of acetaldehyde and vinyl alcohol according to this model? Is vinyl alcohol likely to be detectable in an equilibrium mixture? Obtain the transition-state geometry for the isomerization and calculate the activation energy. Is it large enough that once formed, vinyl alcohol is likely to persist for long times? Elaborate. Dichlorocarbene Addition to Ethylene: Singlet dichlorocarbene adds to ethylene to yield 1,1-dichlorocyclopropane. Cl CCl 2 + H2C Cl CH2 Obtain an equilibrium geometry for singlet dichlorocarbene using the HF/6-31G* model and display an electrostatic potential map. This shows the distribution of charge on the accessible surface. Where is the electrophilic site, in the ClCCl plane or out of the plane? How would you expect dichlorocarbene to approach ethylene? Does this lead to a product with the proper three-dimensional geometry? Elaborate. Use the HF/6-31G* model to obtain the transition state for addition of dichlorocarbene to ethylene. Is the calculated structure consistent with the conclusions reached above regarding the electrophilic character of dichlorocarbene? Elaborate. Hydroxycarbene: Singlet hydroxycarbene (HCÖH) is a proton transfer isomer (or tautomer) of formaldehyde. The best experimental estimate is that it lies ~220 kJ/mol higher in energy than formaldehyde, which suggests that it would not be detectable were it in rapid equilibrium with formaldehyde. The only way that hydroxycarbene would be detectable were if the energy barrier leading to formaldehyde was sizable (>100 kJ/mol). Additionally, it is necessary that the barrier to dissociation to separated hydrogen and carbon monoxide molecules also be sizable. HCÖH H2CO HCÖH H2 + CO Use the B3LYP/6-311+G** model to calculate both the equilibrium geometries of all reactants and products of the two reactions above. Do the calculations reproduce the relative energies of formaldehyde and hydroxycarbene? Is the barrier to isomerization to formaldehyde sufficiently high to permit hydroxycarbene to be detected? Is dissociation to hydrogen and carbon monoxide competitive with isomerization? Dimerization of Borane: Borane (BH3) dimerizes to diborane. BH3 + BH3 B2H6 19 Is there an activation energy associated with dimerization? To see if there is, first obtain the equilibrium geometry of diborane and the elongate two “opposite” BH bonds (marked in the drawing above) from 1.3Ǻ (close to the equilibrium value) to 2.5Ǻ in steps of 0.1Ǻ. Use the B3LYP/6-31G* model. If you find an energy barrier, obtain the transition state. 20 Using Approximate Geometries for Activation Energy Calculations Finding a transition state geometry is likely to be more costly in terms of computation time than finding an equilibrium geometry. However, since the potential energy surface in the vicinity of a transition state would be expected to very flat, the small differences in transition-state geometries seen among different theoretical models are not likely to lead to significant changes in activation energies. Therefore, it may not always be necessary to utilize “exact” transition-state geometries in carrying out activation energy calculations. Hartree-Fock models may replace B3LYP and MP2 models and small basis sets may replace larger basis sets. Problems … 21 Relative Activation Energies It will not always be necessary to formulate reaction rate comparisons in terms of absolute activation energies. Differences in activation energies among closely-related reactions will often be sufficient to answer the question at hand. Important examples include rate changes due to substituents or to changes in regiochemistry or stereochemistry of the reaction. Table P4-5 compares activation energies for Diels-Alder reactions of 1,3butadiene with a series of cyanoalkenes, relative to the activation energy for reaction of 1,3-butadiene and acrylonitrile. Experimental relative rates for a series of closely-related Diels-Alder reactions involving cyclopentadiene instead of 1,3-butadiene are also provided and show both an increase in rate with increasing number of cyano groups and a sensitivity to the location of the groups. While both Hartree-Fock and MP2 models with both 6-31G* and 6-311+G** basis sets reproduce the ordering of reaction rates, neither of the B3LYP models is successful in this regard. In stead both show an increase in activation energy (suggesting a decrease in rate) from 1,1dicyanoethylene, to tricyanoethylene to tetracyanoethylene. The reason for the apparent failure is unclear. Discussion of Table P4-6 22 Table P4-5:Activation Energies for Diels-Alder Reactions of 1,3-Butadiene and Cyanoalkenes Relative to the Reaction of 1,3-Butadiene and Acrylonitrile from Practical Hartree-Fock, B3LYP and MP2 Models (kJ/mol)a Hartree-Fock 6-31G* 6-311+G** reaction acrylonitrile 0 0 B3LYP 6-31G* 6-311+G** MP2 6-31G* 6-311+G** expt. b 0 0 0 0 0 cis-1,2-dicyanoethylene -7 -9 0 -1 -14 -14 1.9 trans-1,2-dicyanoethylene -11 -13 -4 -6 -17 -18 1.9 1,1-dicyanoethylene -27 -29 -25 -26 -22 -21 4.6 tricyanoethylene tetracyanoethylene a) energies of reactions: -31 -33 -35 -37 -18 -12 -20 -14 -34 -46 -34 -49 5.7 7.6 b) log10 rate relative to reaction of cyclopentadiene acrylonitrile as a standard. Table P4-6: Room Temperature Boltzmann Ratios of Regio and Stereo Products from Practical Hartree-Fock, B3LYP and MP2 Models Hartree-Fock B3LYP MP2 6-31G* 6-311+G** 6-31G* 6-311+G** 6-31G* 6-311+G** regiochemistry dimethylborane + propene C2:C1 100 100 1,3-butadiene + acrylonitrile cyclopentadiene + acrylonitrile cyclopentadiene +acrolein cyclopentadiene + maleic anhydride stereochemistry endo:exo 30 24 endo:exo 49 49 endo:exo 82 80 endo:exo 97 97 23 99 99 98 98 18 35 56 84 19 42 60 88 43 46 71 98 53 55 75 99 Isomerization of Ethyl Isocyanide: The activation energy for isomerization of methyl isocyanide to acetonitile (methyl cyanide) is 191 kJ/mol according to the HF/6-31G* model, somewhat larger than the value of 159 kJ/mol obtained from experimental rate data. Use the same model to calculate the activation energy for the analogous reaction of ethyl isocyanide. Assuming that the error is the same as for the methyl isocyanide isomerization, what is your best guess for activation energy for reaction of ethyl isocyanide? Which reaction appears to be faster? Offer an explanation of your result. Hydroboration of Alkenes vs. Alkynes: Alkylboranes add not only to carbon-carbon double bonds but also to carbon-carbon triple bonds. Use the B3LYP/6-31G* model to obtain structures and energies for reactants and transition states for addition of dimethylborane (Me2BH) to both ethylene and acetylene. Which reaction is faster? Offer an explanation. Hydroboration of propene leads predominately to a product derived from attack of hydride onto the more highly substituted alkene carbon. Use the B3LYP/6-31G* model to obtain structures and energies for the two transition states for addition of dimethylborane to propyne. Which is lower in energy? Is the difference in transition-state energies small enough that you would expect to see both products? Elaborate. 24 Thermodynamic vs. Kinetic Control of Chemical Reactions We now have the computational tools in investigate both thermochemical and kinetic preferences of chemical reactions. For reactions where the two are different, this leads to the possibility changes in reaction conditions will lead to changes in product distributions. Long reaction times and high temperatures should favor thermodynamic products whereas short reaction times and low temperatures favor kinetic products. Consider the products resulting from heating 6-bromohexene in the presence of tri-n-butyl tin hydride (a radical initiator). The first step involves abstraction of bromine leading to hex-5-enyl radical. This can either pick up hydrogen giving 1-hexene, or rearrange either to cyclohexyl radical or to cyclopentylmethyl radical, which in turn may pick up hydrogen giving cyclohexane and methylcyclopentane, respectively. Bu3 SnH AlBN Br ! • hex-5-enyl radical 17% • rearrangement cyclopentylmethyl radical • cyclohexyl radical 81% 2% Certainly, cyclohexyl radical is more stable than cyclopentylmethyl radical, because six-membered rings are more stable than five-membered rings, and because 2° radicals are more stable than 1° radicals. However, the dominant (rearranged) product is methylcyclopentane. This suggests that the reaction is kinetically controlled, and that rearrangement to cyclopentylmethyl radical requires less energy than rearrangement to cyclohexyl radical. 25 Ring Opening of Hex-5-enyl Radical: Verify that cyclohexyl radical is lower in energy than cyclopentyl radical. Use the HF/6-31G* model. Were the ring opening reaction thermodynamically controlled, what would be the ratio of methylcyclopentane and cyclohexane at room temperature? Verify that the transition state leading from hex-5-enyl radical to cyclopentylmethyl radical is lower in energy than that leading to cyclohexyl radical. Were the ring opening reaction kinetically controlled, what would be the ratio of methylcyclopentane and cyclohexane at room temperature? Polymerization of Cyclopentadiene: Cyclopentadiene undergoes a Diels-Alder reaction with itself yielding either an endo or exo dimer. + + endo dimerization exo dimerization The process would be expected to continue, giving rise either to an endo or exo polymer (exo polymer depicted below), but it actually stops around the 20-mer. Space-filling models for short strands of exo (left) and endo (right) polymers of cyclopentadiene suggest why. The exo polymer is helical and there is nothing stopping it from continuing to grow, whereas the endo polymer closes back on itself (in about 20 units) and cannot continue. Short strands of both exo and endo polymers of cyclopentadiene are provided in cyclopentadiene polymers in the Chapter P4 directory. Use the HF/6-31G* model to establish whether the exo or endo cyclopentadiene dimer is lower in energy. Rationalize your result. Were the dimerization thermodynamically 26 controlled, what would be the ratio of exo and endo dimers at room temperature? Given the fact that polymerization stops at the 20-mer, would you conclude that the reaction is thermodynamically controlled? Elaborate. Next, establish which transition state is lower in energy. Rationalize your result. Were the dimerization kinetically controlled, what would be the ratio of exo and endo dimers at room temperature? Given the fact that polymerization stops at the 20-mer, would you conclude that the reaction is kinetically controlled? Elaborate. Suggest reaction conditions that would lead to polymerization beyond the 20-mer. 27 Charge Distributions in Transition States Electrostatic potential maps may be used to describe charge distributions in transition states just as they may be employed portray charge distributions in ordinary molecules (see discussion in Chapter P1). Pyrolysis of ethyl formate (leading to formic acid and ethylene) provides a good example of the kinds of information that may result. O O O O H H O + OH Here, the electrostatic potential map (center image) clearly shows that the hydrogen being transferred from carbon to oxygen is positively charged (it is an electrophile). An even clearer picture results by using a surface of much larger electron density in order to map the electrostatic potential (right hand image). 28 Charge Transfer in Diels-Alder Reactions: The most common Diels-Alder reactions are between electron-rich dienes and electron-deficient dienophiles, for example, between cyclopentadiene and tetracyanoethylene. NC CN CN + NC CNCN CN CN Do transition states for Diels-Alder reactions show evidence for charge transfer from the diene to the dienophile? If so, does the extent of transfer correlate with reaction rate? Use the HF/6-31G* model to obtain geometries for reactants and transition states for Diels-Alder reactions of cyclopentadiene as a diene, and ethylene and tetracyanoethylene as dienophiles. For each reaction, compare electrostatic potential maps of reactants and transition state. Is there a noticeable transfer of charge? Is it in the expected direction, that is, does the π system of the diene lose electrons and the double bond of the dienophile gain electrons in moving from reactants to the transition state? For which reaction is charge transfer greater? Is your result consistent with the observed change in rate with change in dienophile: tetracyanoethylene >> ethylene? SN2 Reaction of Chloride and Methyl Bromide: Electrostatic potential maps can be used to help explain why bromide is a better SN2 leaving group than chloride. Use the HF/6-31G* model to obtain an energy profile for reaction of chloride with methyl bromide, from separated reactants to separated products. Start with chloride 3.5Å from the carbon in methyl bromide and move it in steps of 0.1Å to 1.7Å. Plot both the energy and the energy corrected for the effect of aqueous environment vs. the CCl distance (the “reaction coordinate”). Is the reaction Cl– + CH3Br → CH3Cl + Br– endothermic or exothermic in the gas phase? How do the solvent corrections affect overall endo or exothermicity? Are your results consistent with the observation that iodide is a better leaving group than chloride? Elaborate. Examine electrostatic potential maps for the structures along the reaction profile, with particular focus on the reactants and products and on the transition state. For which (reactant, product or transition state) is the charge most localized? For which is it most delocalized? What does this say about the relative abilities of Cl and Br to act as leaving groups? Does the energy along the reaction pathway appear to correlate with the extent to which charge is delocalized? Elaborate. Does the change in energy due to the solvent correction correlate with the extent to which charge is delocalized? Elaborate. SN2 Reactions of Alkyl Bromides: The rates of SN2 reactions involving alkyl bromides decrease with each successive replacement of hydrogen by an alkyl group. H fast CH3 C H H Br > C H CH3 Br > C H H 29 CH3 CH3 Br > C CH3 CH3 Br slow The usual explanation centers around the increase in coordination of carbon from four in the reactants to five in an SN2 transition state, for example, for reaction of bromide and methyl bromide. – H Br– + CH3Br Br C H Br CH3Br + Br– H The resulting increase in (unfavorable) steric interactions should be least for reaction of the methyl halide and greatest for the tert-butyl halide, consistent with a decrease in rate with increased substitution. Obtain transition states for addition of bromide to both methyl bromide and to tert-butyl bromide using the HF/6-31G* model, and examine as space-filling models. Is the transition state for addition to tert-butyl bromide noticeably more crowded than that for addition to methyl bromide? If not, why not? Hint: compare the carbon-bromine bond distances in the two transition states. Calculate and compare electrostatic potential maps for the two SN2 transition states. These convey the distribution of charge on the accessible surface. Is the extent of charge separation comparable or is it significantly different between the two transition states? If the latter, is their a relationship between the carbon-bromine distance and degree of charge separation? Speculate on the origin of the change in rate of SN2 reactions with change in substitution at carbon. 30 Reaction Pathways As detailed above, a transition state connecting stable molecules (the reactants and products in a chemical reaction) is a well-defined location on the potential energy surface. Of course, there may be several transition states connecting the same reactants and products, just as there may be several passes leading across a mountain range, but each is a well-defined location. On the other hand, the reaction pathway leading upward from the reactants to the transition state and then downward to the products is not unique. Just as there are many possible roads leading up to and down from a particular mountain pass, there are many ways the reactants can reach a transition state and many ways to move away from the transition state to products. It is not obvious which pathway a reaction will follow or even that it will follow only one pathway. Several ways have been proposed to provide reaction pathways, but all are arbitrary in the sense that there are many (an infinite number) reaction pathways. A common procedure referred to as an intrinsic reaction coordinate smoothly connects reactants, transition state and products. While examination of the motion along an intrinsic reaction coordinate may be satisfying in that it “leaves an impression” of how a reaction proceeds, there is no guarantee that it at all reflects what is really happening. A simpler procedure is to follow the motion corresponding to the vibration with the imaginary frequency back from the transition state toward reactants and forward toward products. While this approach does not actually lead to either reactants or products, it is usually sufficient to tell whether the transition state is that for the reaction of interest. 31 Testing the Hammond Postulate The Hammond Postulate states that transition state in a one-step reaction more closely “resembles” the side of the reaction that is higher in energy. Thus, the transition state of an exothermic reaction more closely resembles the reactants while the transition state of an endothermic reaction more closely resembles the reactants. Thus, the closer the energy of a transition state to that of the reactant (or product), the greater its structure will resemble the structure of the reactant (or product). In the limit of an exothermic reaction with no energy barrier, the “transition state” is the reactant (and vice versa for an endothermic reaction). transition state transition state product Energy reactant endothermic reaction Energy exothermic reaction product reactant reaction coordinate reaction coordinate The Hammond Postulate should not be invoked for a reaction that is only slightly exo or endothermic. On the contrary, it should be valid for a reaction involving a high-energy reactive intermediate. Here, the reaction connecting the intermediate to product would be expected to be highly exothermic and have a low energy barrier. The transition state should closely resemble the intermediate. Because modeling allows chemists to explore the structures of both reactive intermediates and transition states, it is now possible to test the limits of the Hammond Postulate. 32 Kinetic Isotope Effects and Hydrogen Abstraction: Reaction of chlorine with d1dichloromethane can either involve hydrogen abstraction (leading to HCl) or deuterium abstraction (leading to DCl). Cl• Cl Cl H Cl• D + HCl hydrogen abstraction C• H + DCl deuterium abstraction C• Cl C D Cl Cl Cl The two reactions follow the same pathway, pass through the same transition state and lead to the same products. The only difference is whether a hydrogen or deuterium atom is abstracted. This affects the zero-point energies of the transition states and products, and leads to kinetic and equilibrium isotope effects, respectively. Obtain equilibrium geometries and infrared spectra for dichloromethyl radical and hydrogen chloride using the HF/6-31G* model. Use calculated zero-point energies to evaluate the equilibrium isotope effect, that is, the energy of the reaction. • Cl2CH + DC l • Cl2CD + HC l Use the Boltzmann equation to calculate the equilibrium distribution of products at 25°C. Obtain the structure of the transition state for hydrogen abstraction from dichloromethane by chlorine atom. Again use the HF/6-31G* model. Calculate zero-point energies for the two possible monodeuterated transition-states. Use the Boltzmann equation (with transition state zero-point energies instead of reaction product zero-point energies) to calculate the kinetic distribution of products at 25°C. 33 ...
View Full Document

This note was uploaded on 02/22/2010 for the course CHEM N/A taught by Professor Head-gordon during the Spring '09 term at University of California, Berkeley.

Ask a homework question - tutors are online