P3_Reaction Energies

P3_Reaction Energies - Chapter P3: Energies of Chemical...

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Unformatted text preview: Chapter P3: Energies of Chemical Reactions In addition to its geometry, the internal energy of a molecule is its most fundamental property. Differences in internal energies between the products and reactants of a reaction (reaction enthalpies) indicate whether the reaction will be favorable (exothermic) or unfavorable (endothermic). Combined with the entropy, which may be calculated from knowledge of equilibrium geometry and vibrational frequencies, reaction enthalpy may be used to provide the Gibbs energy. Our first objective in this chapter will be to establish the ability of each of the three classes of models that have now been introduced: Hartree-Fock, B3LYP density functional and MP2 models, to reproduce the energies for different types of chemical reactions. We will employ a very-large basis set to explore the limits of each model and to attempt to separate the LCAO approximation from the underlying treatment of electron correlation. While we cannot hope to actually achieve these limits, by repeating the calculations with a somewhat smaller basis set we can hope to put bounds on the magnitude of the error caused by the LCAO approximation. Our second objective will be to establish the performance of the same three classes of methods with smaller, more practical basis sets. This will allow us to provide much greater numbers and diversity of examples and to construct meaningful problems. The third objective will be to define combinations of theoretical models (“recipes”) that are able to provide accurate heats of formation to be used in thermochemical comparisons. We will consider both recipes with and without empirical parameters. The final objective of the chapter will be to address the calculation of entropy and subsequent determination of the Gibbs energy. Problems throughout this chapter are intended to exemplify the types of questions that may be posed to modern quantum chemical methods and the types and quality of responses that can be expected. Some will need only a few seconds or minutes of computer time, while others may require several hours. 1 Total Energy vs. Heat of Formation The heat of formation typically reported in an experimental study differs in three ways from the total energy obtained from a quantum chemical calculation. The obvious difference is that enthalpy and energy are not the same, but are related through a pressure-volume term. ∆H = ∆E + ∆(PV) or at constant pressure ∆H = ∆E+ P∆V However, except where comparisons involve experiments that have been carried out under very high pressures, the energy and enthalpy of a reaction can be safely assumed to be identical. Reactions can be (and have been) carried out as a function of pressure and provide information about differences in the volumes of reactants and products. A more interesting involves extracting the volumes of transition states (relative to those of reactants) from the pressure dependence of activation energies. This will be discussed in Chapter P4. Pressure Affects Reaction Enthalpy: Ring opening of cyclobutene to 1,3-butadiene is exothermic by ~56 kJ/mol. Use the HF/6-31G* model to calculate equilibrium geometries for 1,3-butadiene and cyclobutene. Based on space-filling models as a measure, which molecule takes up less volume? If it is cyclobutene, what pressure would be required to reverse the direction of the isomerism. Need information on units. Instead of space-filling models, use surfaces that enclose 99% of the total electron density as a measure of volume. Do your results change significantly? Another difference is that experimental measurements are always carried out at finite temperature (typically at or near room temperature) whereas quantum chemical calculations refer to systems at 0K. The change in enthalpy from 0K to a finite temperature (T) is given by: ΔH(T) = ΔHtr(T) + ΔHrot(T) +ΔHvib(T) + RT ΔHtr(T) = 3/2 RT ΔHrot(T) = 3/2 RT (RT for a linear molecule) 2 This requires the vibrational frequencies, νi. R is the gas constant, k is Boltzmann’s constant, h is Planck’s constant and N is Avogadro’s number. Were it actually possible to measure the enthalpy at 0K, it would still not correspond to the enthalpy (energy) obtained from a quantum chemical calculation. This is because the calculated energy calculated from a quantum chemical model refers to a molecule resting at the bottom of a potential energy well, whereas a measured enthalpy refers to a molecule in its lowestenergy (“zeroeth”) vibrational state. The difference, referred to as the zero-point-vibrational energy, is given by. The summation is carried out over the vibrational frequencies, υ. h is Planck’s constant. Zero-point vibrational energies are quite large, but largely cancel in a chemical reaction. Zero-Point Energies Affect Reaction Energy: Use the HF/6-31G* model to calculate equilibrium geometry and vibrational frequencies for acetonitrile, CH3CN, and its isomer, methyl isocyanide, CH3NC. Evaluate the zero-point energy for each. Does its explicit inclusion change in the energy of isomerism by more than 10%? Repeat your calculations and analysis for propene and its isomer cyclopropane. Temperature Affects Reaction Energy: Using the results from the calculations you performed in the previous problem, evaluate the change in the energy of acetonitrile and methyl isocyanide with change in temperature from 0 K to 298 K. Combine this with the difference in zero-point energies for the two isomers (see previous problem). Does the full correction (zero-point energy + temperature) change in the energy of isomerism by more than 10%? Repeat your calculations and analysis for propene and its isomer cyclopropane. Mass Affects Reaction Energy: As detailed in Chapter xx, the Born-Oppenheimer approximation eliminates nuclear mass from the Schrödinger equation. However, because of differences in zero-point energies, reaction energies actually change with change in nuclear mass (isotope). The most common mass substitution is deuterium for hydrogen. Use the B3LYP/6-31G* 3 model to determine the difference in zero-point energy (and the difference in bond energies) between H2 and D2. Is the change in bond energy likely to be noticed? Assume that a change in bond energy of 5% is detectable. Repeat the calculations and the analysis for the change from molecule. 14 N to 15 N in nitrogen A trivial difference is one of units. Heats of formation are most commonly reported in kJ/mol, whereas total energies are most commonly reported in atomic units (hartrees). 1 hartree = 2625 kJ/mol kJ/mol will be used throughout this text, replacing the more familiar but now defunct kcal/mol (1 kcal/mol = 4.184 kJ/mol). The reason for the difference in units is simply a matter of convention and to some extent convenience. Heats of formation are typically only a few tenths of a hartree while total energies are typically tens of thousands to several million kJ/mol. In their respective units, both lie in a more convenient range from a few tens to a few thousands. Finally, and perhaps most important, while both total energy and heat of formation refer to the energies (heats) of specific chemical reactions, the reactions are different. Heat of formation refers to a balanced chemical reaction in which a molecule is converted to a set of standard products, each corresponding to the most stable form of the element at room temperature. The heat of formation of each standard is defined as zero. For example, the heat of formation of ethylene is defined by the reaction. C2H4 → 2C (graphite) + 2H2 (gas) Graphite and hydrogen molecule are the carbon and hydrogen standards, respectively. Of course, the experimental measurement is not actually carried out for this reaction, but more typically (but not necessarily) for a combustion reaction (reaction with O2), for example, for ethylene: C2H4 + 3O2 → 2CO2 + 2H2O In some cases, combustion leads to products that cannot be fully characterized. The most conspicuous case involves combustion of molecules incorporating silicon where polymeric silicon oxide polymers (sand) are formed. The clever solution is to “burn” the molecule in fluorine rather than oxygen, leading to gaseous SiF4 as a product. 4 Heats of formation may be either positive or negative quantities and their (absolute) values will generally span a range of only a few hundred kJ/mol. Molecules with positive heats of formation much greater than this are likely to the thermodynamically unstable and not likely to “stick around” long enough to be detected let alone characterized. Heats of formation may not be obtained directly from quantum chemical calculations simply because some of the standards are not isolated species on which calculations may be performed. A suitable alternative is to use a hypothetical reaction that splits a molecule into isolated nuclei (not atoms) and electrons, for example, for ethylene: C2H4 → 2C+6 + 4H+ + 16e – Each of the products (H+, C+6 and e-) contains but a single particle, meaning that its energy is zero. Total energies, as the energies of such reactions are termed, are very large negative numbers, (several tens of thousands to several million kJ/mol), but only a few tens to a few thousands of hartrees. There are two important points to be made. First, is it is straightforward to obtain heats of formation indirectly from total energies (or vice versa), either by using experimental atomic data or a mixture of experimental and theoretical atomic data. In fact, combinations of theoretical models (“recipes”) have been developed in order to provide accurate heats of formation. These will be discussed later in this chapter. The second point is more important. Either heats of formation or total energies are suitable for calculations of the energies (heats) of mass-balanced chemical reactions. Here, the “standards” cancel. 5 Sources and Quality of Experimental Thermochemical Data Experimental heats of formation have been reported for approximately three thousand compounds, a large fraction of which are hydrocarbons and oxycarbons. Among the most extensive compilations is the NIST database, freely available on line (http://webbook.nist.gov). A snapshot of part of the NIST database for closed-shell neutral organic molecules has been provided with Spartan Student, and heats of formation where available are displayed in the Molecule Properties dialog. No attempt has been made to identify suspect experimental values. While care has been taken to ensure the integrity of this collection, because the data derive from a variety of experimental techniques and has been assembled over many decades, heats of formation for individual entries vary widely in quality. The most egregious source of error is that the structure is incorrect, meaning that the reported heat does not correspond to the reported structure. Because most of the compounds in the NIST database are fairly simple and readily available commercially, it is likely that only a very few structures are incorrectly assigned. More likely sources of error include incomplete combustion and poorly characterized combustion products. Hydrocarbons and oxycarbons present fewest problems as their combustion leads only to carbon dioxide and water, the amounts of which may easily be determined. However, combustion of molecules with other elements may give rise to a complex mixture of products and greater uncertainty. Despite their importance, heats of formation are not routinely determined (or at least are not routinely reported) for new compounds. While combustion experiments are straightforward and the results easily interpreted, accurate measurements require considerable diligence and may need to be repeated several times to establish useful error limits. More relevant, combustion experiments may require (and destroy) significant quantities of compound. Very few synthetic chemists are willing to part with hundreds of mg (a huge amount by modern standards) of a compound that they have just spent weeks or months preparing. One alternative source of experimental thermochemical data follows from measurement of equilibrium constants. The best examples of this are for ionmolecule reactions carried out in the gas phase using ion cyclotron resonance spectroscopy and related techniques. Most important among these 6 are protonation/deprotonation reactions, leading to gas-phase basicities and acidities. For example, it is possible to accurately establish the energy of protonation of a base, B, relative to that of a standard base, Bstandard, by determining the amounts of the two protonated bases present at equilibrium. BH+ + Bstandard B + BstandardH+ Equilibrium measurements require that the proton affinities of the two bases be similar (within ~10 kJ/mol). In practice, a proton affinity “ladder” of base strengths (involving hundreds of individual compounds) has been constructed, allowing an appropriate standard to be selected over a range of several hundred kJ/mol. The advantages to this kind of approach are that it is quite accurate and requires only miniscule amounts of materials. The obvious disadvantage is that it applies to ion-molecule reactions only. The NIST database contains a large collection of heats of formation for both negative and positive ions based on such measurements. Another source of experimental thermochemical data on positive ions derives from measurements of ionization potentials and electron affinities. M + e- M+ + 2eM + e- MThe former are widely available whereas the latter are less common. 7 Reaction Energies and Boltzmann Distributions As commented previously, the energy (heat) of a balanced chemical reaction may be obtained using either heats of formation or total energies. ∆E(reaction) = Eproduct 1 + Eproduct 2 + … - Ereactant 1 - Ereactant 2 - … A negative ∆E indicates an exothermic (energetically favorable) reaction, while a positive ∆E indicates an endothermic (unfavorable) reaction. An important special case is where there is only one reactant molecule and one product molecule. Here, the reactants and products are isomers, and the reaction energy accounts directly for their difference in their energies. ∆E(reaction) = ∆E(isomer) = Eisomer 2 – Eisomer 1 A negative ∆E(reaction) means that isomer 2 is more stable than isomer 1, and that the reaction will proceed as written (it is exothermic). The equilibrium composition of a mixture of isomers is given by the Boltzmann equation. Isomer 1 Isomer 2 % Isomer i = Isomer 3 ... 100 exp (-EIsomer i /kT) ! exp (-1060 E Isomer j ) j ΔEisomer i is the energy of isomer i in hartrees relative to the energy of the lowest-energy isomer. T is the temperature (in K) and k is the Boltzmann constant. In the case where the equilibrium is between two isomers, Isomer 1 Isomer 2 and assuming room temperature (298 K) the expression becomes. [ Isomer 1 ] = exp [-1060 (Eisomer1 – Eisomer2 )] [ Isomer 2 ] HCN vs. HNC: Under “normal” (laboratory) conditions, hydrogen isocyanide (HNC) is in equilibrium with its more stable isomer, hydrogen cyanide (HCN). According to the HF/6-31G* model, what is the room-temperature Boltzmann distribution of isomers? Assuming that 5% as the lower bound for detection, what is the lowest temperature that would be needed to see the minor isomer? 8 Radioastronomy confirms that both hydrogen cyanide and hydrogen isocyanide are present in interstellar clouds. The interesting observation is that they occur in similar amounts. Speculate why. Protonation of Dimethylamine and Aziridine: Molecules contained the high vacuum environment of an ion cyclotron resonance spectrometer can be protonated and ionmolecule equilibria established. The relative abundance of the ions involved in the equilibrium can be measured which, together with knowledge of the relative amounts of the neutral molecules, allows highly accurate determination of relative proton affinities. Use the HF/6-31G* model to calculate the equilibrium geometries of the four molecules involved in the proton-transfer equilibrium between dimethylamine and aziridine. dimethylamine-H+ + aziridine = dimethylamine + aziridine-H+ Which molecule has the higher proton affinity and by how much? Assuming a 1:1 mixture of the two amines, and that the limit of detection of the “minor” ion is 5%, is it possible to directly establish the relative proton affinities of dimethylamine and aziridine? 9 Classifying Chemical Reactions Chemical reactions may be conveniently be divided into one of three categories depending on the extent that overall bonding is maintained. Reactions that Change the Total Number of Electron Pairs The most common reactions of this type are homolytic bond dissociation reactions. Hartree-Fock models are expected to provide bond energies that are too large. To see why this is so, consider bond dissociation in H2. H–H → H• +H• While the Hartree-Fock energy for the product (two hydrogen atoms) is exact (each contains only a single electron), that for the reactant (hydrogen molecule) is too high. This means that the bond dissociation reaction will not be sufficiently endothermic. In reality, homolytic bond dissociation reactions are not very important as they are only infrequently encountered. However, they serve as “absolute standards” with which to judge the performance of different theoretical models. An important possible “exception” involves comparisons between reactants and transition states (activation energies), whic may also lead to a change in the number of electron pairs. These will be discussed in Chapter P4. Reactions that Conserve the Total Number of Electron Pairs This category separates reaction into two classes, depending on whether or not total bond count in addition to total electron-pair count is maintained. Protonation of trimethylamine (defining its absolute proton affinity) and association of trifluoroborane and carbon monoxide (leading to trifluoroborane carbonyl) are examples of the latter. (CH3)3N: + H+ (CH3)3NH+ BF3 + :CO BF3CO Here the product contains one more bond than the reactants but the same number of electron pairs. Examples of reactions that conserve bonds as well as electron pairs are more common. Among the most important are comparisons of structural isomers, for example, comparison of allene and propyne. Here, the reactant incorporates two double bonds and the product one single and one triple bond. CH2=C=CH2 → CH3CH≡CH2 10 Some structural isomer comparisons, for example, that between 1-butyne and 2butyne, not only maintain overall bond count, but also maintain the numbers of individual bond types involving (in this case, two single bonds and one triple bond). HC CCH2CH3 → CH3C≡CCH3 In this case, only the nature of the “atomic hybrids” (environments) are altered. One of the carbon-carbon single bonds in 1-butyne made from sp and sp3 hybrids and the other from two sp3 hybrids, whereas both of the carbon-carbon bonds in 2-butyne are made from sp and sp3 hybrids. This and other reactions like it will be put into a third class (see next section). Addition reactions such as cycloaddition of 1,3-butadiene and ethylene to form cyclohexene (a Diels-Alder reaction) and of addition of molecular bromine to cyclohexene to yield trans-1,2-dibromocyclohexane, also conserve total bond count (but not individual bond counts). In the first case, there is a loss of two π bonds but a gain of two new carbon-carbon σ bonds, while in the second case a π bond and a Br-Br bond are traded for two C-Br bonds. + Reactions that conserve the total number of electron pairs would be expected to benefit from error cancellation to a greater extent than reactions that do not. Because of this, it might be anticipated that their energies will be described even by quantum chemical models that do not properly take electron correlation into account, specifically Hartree-Fock models. We will explore this later in the chapter. Reactions that Conserve Individual Bond Counts Processes that conserve not only the total number of bonds and electron pairs but also the numbers of each kind of chemical bond (and each kind of non-bonded lone pair) represent another very important and very diverse class of chemical reactions. Some structural isomer comparisons, for example, that between 1-butyne and 2-butyne mentioned above, formally fit this category, as do all comparisons of regio and stereoisomers. For example, comparison of isobutene and cis and trans2-butene. The latter two are stereoisomers and both are regioisomers of isobutene. 11 H3 C H C H3 C H3C C CH3 C H H H3 C C H C H H C CH3 Also of note are reactions that compare the energies of protonation (proton affinities) of closely-related molecules, for example, that compare the proton affinities of ammonia and trimethylamine. This is a quantity that can be accurately measured although not necessarily directly (see previous discussion). (CH3)3NH+ + NH3 → (CH3)3N + NH4+ In all of these cases, comparisons are between molecules with the same number of each kind of chemical bond and of each kind of electron pair. Only the local environment changes. It might be expected that these kinds of reaction would benefit from error cancellation to a greater extent than the previous types of reactions. The obvious suspicion is that it will be “easier” to calculate relative quantites, for example, the relative proton affinities of ammonia and methylamine, than it is to calculate absolute quantities, for example, the absolute proton affinity of ammonia. This will be tested in the next section. At the outset it should be pointed out that while absolute energies will certainly be required in some instances, in many others relative energies will suffice. 12 Limiting Behavior of Hartree-Fock, Density Functional and MP2 Models for Reaction Energies We first set out to establish or at least to estimate the limiting behavior of Hartree-Fock, B3LYP and MP2 models with regard to reaction energies using examples of each of the types processes discussed in the previous section. This will allow us to separate the effects of the LCAO approximation from effects arising from replacement of the exact manyelectron wavefunction by an approximate Hartree-Fock, density functional or MP2 wavefunction. As commented in the introduction to this chapter and in Chapter P2, it is not possible to actually reach the limit for either Hartree-Fock and B3LYP models, nor is it possible to establish the MP2 limit. However, it should be possible to use a sufficiently large basis set such that the addition of further functions to the basis set will have only a small effect on calculated reaction energy. As for geometry comparisons, the ccpVQZ basis set will be employed. This should be sufficiently large to be able to reflect the properties of the limit, but small enough to be applied to the molecules in the comparisons provided. Of more practical concern, ccpVQZ is about as large a basis set that can be applied to the molecules used in the comparisons that follow. The results provided in this section based on use of the cc-pVQZ basis set will be paralleled by those obtained using the slightly smaller cc-pVTZ basis set. This will allow us to quantify the extent to which reactions energies based on the cc-pVQZ basis set actually approach limiting values. The first comparison involves AH bond energies in molecular hydrogen and in one-heavy-atom hydrides and AB bond energies in two-heavy-atom hydrides. AH A +H AB A +B Reference bond energies have been obtained from the G3(MP2) recipe (see discussion later in this chapter) rather than from experiment. This provides a more uniform baseline and allows comparisons to be extended to molecules where experimental bond energies are not accurately known (if known at all). G3(MP2) heats of formation have already been corrected for zero-point vibrational energy and for finite temperature (298 K). That is, they correspond to the experimental data most commonly found in the chemical literature (including that in the NIST database). 13 These corrections are based on vibrational frequencies obtained from the HF/6-31G* model scaled by 0.9 to take account of a known systematic error in these frequencies. This means that energies from the Hartree-Fock, B3LYP and MP2 calculations also need to be corrected. Signed differences between B3LYP/cc-pVQZ and MP2/cc-pVQZ and G3(MP2) AH bond energies are shown in Figure P3-1 and between G3(MP2) AB bond energies are shown in Figure P3-2. Excel spreadsheets containing AH and AB bond dissociation energies discussed in this section (including Hartree-Fock bond energies) are provided on the CD-ROM accompanying this text (limiting AH bond energies and limiting AB bond energies, respectively). The Hartree-Fock cc-pVQZ model provides a very poor account of both AH and AB bond energies and individual errors have not been included in the figures. As expected, Hartree-Fock bond energies are much larger than G3(MP2) values, with mean absolute deviations of 125 and 160 kJ/mol for AH and AB bond energies, respectively. Bond energies calculated using the smaller cc-pVTZ basis set (not provided in the figures but available in the Excel spreadsheet) are nearly identical with those, confirming that the problem is the Hartree-Fock approximation and not the use of a finite basis set. The B3LYP/cc-pVQZ model provides a far better account. Mean absolute errors are 8 and 17 kJ/mol for AH and AB bond energies, respectively. Individual bond energies are always smaller than G3(MP2) values, and usually within 10 kJ/mol of G3(MP2) values for AH bond energies and 20 kJ/mol for AB bond energies. The B3LYP/cc-pVTZ model yields bond energies that differ from the corresponding cc-pVQZ values by only 1-2 kJ/mol, although the difference is 6 kJ/mol for the OH bond energy in water and 3-4 kJ/mol for some bonds involving two heteroatoms. This suggests that B3LYP models are more sensitive to basis set than Hartree-Fock models. Still, most of the discrepancy between B3LYP and G3(MP2) bond energies does not appear to be due to limitations in the basis set. Individual bond energies from the MP2/cc-pVQZ model show much wider variation from G3(MP2) values than those from the corresponding B3LYP model, although mean absolute errors are virtually identical (9 kJ/mol for AH bond energies and 17 kJ/mol for AB bond energies). These range from 25 kJ/mol too large for the bond energy in hydrogen, to 35 kJ/mol too small 14 Figure P3-1: Signed Deviations Between B3LYP/cc-pVQZ (left) and MP2/cc-pvQZ (right) and G3(MP2) AH Bond Energies (kJ/mol) Figure P3-2: Signed Deviations Between B3LYP/cc-pVQZ (left) and MP2/cc-pvQZ (right) and G3(MP2) AB Bond Energies (kJ/mol) 15 for the OO bond energy in hydrogen peroxide. In part, the problem may reside with the basis set. Bond energies from the MP2/cc-pVQZ model differ by as much as 18 kJ/mol (in Cl2) from those obtained from the corresponding cc-pVTZ model, and differences on the order of 10 kJ/mol are common. This result, while disturbing, is not unexpected. MP2 models (unlike Hartree-Fock and density functional models) directly use “excited-state” wavefunctions which involve both higher-order and more diffuse component than required in the ground state. In summary, “limiting” Hartree-Fock models fail to provide an acceptable account of bond dissociation energies. The corresponding B3LYP and MP2 models perform much better, although sizable deviations are noted for individual molecules, in particular, for MP2 models. Bond energies obtained from both Hartree-Fock and B3LYP models do not change significantly in moving from the cc-pVQZ to the smaller cc-pVTZ basis set. On the other hand, large changes are noted for some molecules between the corresponding MP2 models. The second comparison (Table P3-1) involves energy differences among structural isomers. Here, the total number of electron pairs is conserved but the numbers of individual bond types are not maintained. Both experimental data and data from the G3(MP2) thermochemical recipe have been used as references. “Limiting” Hartree-Fock, B3LYP and MP2 isomer energies have been corrected for zero-point energy and finite temperature in the same way as G3(MP2) energy differences. Both the mean absolute error of Hartree-Fock isomer energies from experimental enthalpies and the mean absolute deviation from G3(MP2) values are 11 kJ/mol, much less than the errors seen previously for bond energy comparisons. Note, however, that differences in isomer energies are much smaller than bond energies. The largest individual error from experiment is 33 kJ/mol, and involves comparison of 1,3-butadiene with its highly strained isomer, bicyclo[1.1.0]butane. The corresponding deviation from G3(MP2) is 21 kJ/mol. Other large errors (from experiment) involve comparisons of propyne and cyclopropene and 1,3-butadiene and cyclobutene (both 20 kJ/mol). Other large deviations (from G(MP2)) involve comparisons of acetaldehyde and oxirane (17 kJ/mol) and acetonitrile and methyl isocyanide (16/kJ/mol) and methyl ethyl ketone and 2,3-dihydrofuran (16 kJ/mol). As with bond dissociation energies, isomer energy differences 16 from Hartree-Fock calculations with the cc-pVTZ basis set (not provided in the table) are nearly identical to those with the cc-pVQZ basis set. Table P3-1: Errors in “Limiting” Hartree-Fock, B3LYP and MP2 Energies of Structural Isomers (kJ/mol) HF reference propyne B3LYP MP2 G3(MP2) Expt. isomer allene 7 -9 18 1 7 cyclopropene 113 99 99 100 93 cyclopropane 43 40 19 38 29 2-butyne 32 37 21 39 36 cyclobutene 68 67 40 56 48 bicyclo[1.1.0]butane 141 131 90 120 108 cyclobutane 54 53 35 47 46 1,4-pentadiene 55 56 83 67 70 methyl isocyanide 84 99 113 100 88 vinyl alcohol 52 43 44 41 43 oxirane 132 121 110 115 118 ethanol dimethyl ether 44 44 54 50 51 acetic acid methyl formate 70 68 75 70 75 methyl vinyl ketone cyclobutanone 32 27 11 22 23 2-hydroxy-1,3-butadiene 52 41 38 38 2,3-dihydrofuran 51 45 35 43 divinyl ether 104 91 105 95 102 mean absolute error 11 9 (10) 5 - mean absolute deviation from G3(MP2) 11 5 (12) - 5 propene 1,3-butadiene isobutene cyclopentene acetonitrile acetaldehyde 17 An Excel spreadsheet containing structural isomer energies for Hartree-Fock, B3LYP and MP2 models with both cc-pVQZ and cc-pVTZ basis sets is provided on the CD-ROM accompanying this text (limiting structural isomer energies). The B3LYP model offers some improvement over the corresponding Hartree-Fock model. The mean absolute error from experiment is reduced to 9 kJ/mol and the mean absolute deviation from G3(MP2) is reduced to 5 kJ/mol. The largest individual deviation error from experiment is reduced to 9 kJ/mol (for comparison of 1,3-butadiene and bicyclo[1.1.0]butane). The largest individual deviations from G3(MP2) are 11 kJ/mol (for ). Consistent with previous results for bond energies, isomer energies change only slightly in moving to the smaller cc-pVTZ basis set (data are not provided in the table). The “limiting” MP2 model shows a mean absolute error with experimental results of 10 kJ/mol, similar to that for the corresponding B3LYP and MP2 models. However, the largest individual error (comparison of 1,3-butadiene and bicycle[1.1.0] butane) is larger (30 kJ/mol). The mean absolute deviation with G3(MP2) is 12 kJ/mol, somewhat greater than that for the other two models. The final comparison (Table P3-2) involves regio and stereoisomers and exemplifes reactions in which both the total number of electron pairs and the numbers of individual bond types are conserved. As with the previous comparisons, the reference is to G3(MP2) energy differences, and the “limiting” Hartree-Fock, B3LYP and MP2 isomer energies have been corrected for zero-point energy and finite temperature in the same way as G3(MP2) energy differences. An Excel spreadsheet containing region and stereoisomer energies for Hartree-Fock, B3LYP and MP2 models with both cc-pVQZ and cc-pVTZ basis sets is provided on the CD-ROM accompanying this text (limiting regio and stereoisomer energies). The “limiting” Hartree-Fock model provides a good account of differences in isomer energies. The mean deviation from G3(MP2) energy differences is 5 kJ/mol, and with a single exception (comparison of cyclobutene and methylenecyclopropane) individual deviations are 7 kJ/mol or less. This is not unexpected. Regio and stereoisomers are more closely related to each other than structural isomers, and energy comparisons should benefit from 18 and cancellation of errors. Results from Hartree-Fock calculations with the smaller cc-pVTZ basis set (not provided in the table) are nearly the same. Table P3-2: Errors in “Limiting” Hartree-Fock, B3LYP and MP2 Energies of Regio and Stereoisomers (kJ/mol) HF reference 1,3-butadiene B3LYP MP2 G3(MP2) Expt. isomer 1,2-butadiene 55 44 51 51 53 1-butyne 24 27 21 19 20 methylenecyclopropane 26 19 33 40 44 trans-2-butene 2 1 6 6 7 cis-2-butene 9 7 11 11 10 1-butene 13 15 18 16 17 methylenecyclobutane 87 80 90 87 86 1,4-pentadiene 21 29 31 30 30 1,1-dimethylallene 53 44 52 53 53 1,3-dimethylallene 55 45 58 57 1,2-pentadiene 62 55 66 65 65 2-methyl-2-propenal 7 4 1 3 9 trans-2-butenal 13 10 13 6 5 mean absolute error 4 8 (3) 2 - mean absolute deviation from G3(MP2) 4 7 (3) - 2 2-butyne cyclobutene isobutene cyclopentene 2-methyl-1,3-butadiene methyl vinyl ketone 19 In terms of mean absolute deviation from G3, the B3LYP/cc-pVQZ model is slightly inferior to the corresponding Hartree-Fock model for region and stereoisomer energy comparisons, although the difference is not large. This suggests that errors inherent to Hartree-Fock theory in large part cancel where reactants and products have the same number of bonds and electron pairs, and differ only in detailed environment. MP2 20 Practical Hartree-Fock and Density Functional Models for Reaction Energy Calculations Except for very small molecules, Hartree-Fock, B3LYP and especially MP2 models with very large basis sets such as cc-pVTZ and cc-pVQZ are not practical for calculation of the energies for any but reactions involving very small molecules. While this will slowly change with improvements in computer speed and storage capabilities, at the present time these basis sets are primarily of value in judging the limits and ultimately the quality of the underlying (Hartree-Fock and B3LYP and MP2) models. Here we examine the performance of two smaller basis sets, 6-31G* and 6-311+G**, both of which are presently easily applicable to much larger molecules (with molecular weights approaching 500 amu). The primary focus will be to establish the ability of the models to reproduce G3(MP2) relative energies to “chemical accuracy” that is, within 8 kJ/mol. A summary of mean absolute errors for collections of AH and AB bond dissociation energies are provided in Table P3-3. These include HartreeFock, B3LYP and MP2 models with the 6-31G* and 6-311+G** basis sets as well as with the cc-pVTZ and cc-pVQZ basis sets previously examined. An Excel spreadsheet containing AH and AB bond dissociation energies for HartreeFock, B3LYP and MP2 models with 6-31G*, 6-311+G**, cc-pVQZ and cc-pVTZ basis sets is provided on the CD-ROM accompanying this text (AH and AB bond dissociation energies). Hartree-Fock models lead to unacceptable results for AH and AB bond energies, irrespective of basis set. They should not be employed for this purpose. The B3LYP/6-311+G** model yields similar AH bond energies (and a similar mean absolute error) to that of the corresponding model with the ccpVQZ basis set. The MP2 model is more sensitive to basis set. The mean absolute error in AH bond energies obtained from the MP2/6-311+G** model is three times larger than that from the corresponding model with the cc-pVQZ basis set. This result could have been anticipated by previously noted differences between MP2 models with the cc-pVQZ and cc-pVTZ basis sets (see Table P3-1). The 6-31G* basis set provides a much poorer account of AH bond energies for both B3LYP and MP2 models. In terms of mean absolute error, bond energies from the B3LYP model with the 6-31G* and cc-pVQZ basis sets are comparable, while those obtained using the 6-311+G** basis set are not as good. 21 Table P3-3: Summary of Deviations from G3(MP2) of Hartree-Fock, B3LYP and MP2 AH and AB Bond Dissociation Energies (kJ/mol) Hartree-Fock Basis Set AH AB B3LYP MP2 AH AB AH AB 6-31G* 20 15 54 20 6-311+G** 10 27 24 21 8 17 cc-pVTZ cc-pVQZ 125 160 22 9 17 There are situations where the 6-311+G** basis set may not be adequate to provide an accurate account of isomer energies. A particularly simple example is the identity of the molecule resulting from dimerization of chlorine oxide, ClO. It is an important example because the observed (high) concentration of ClO in the stratosphere above Antarctica as a function of time of year corrlelates with the observed (low) concentrations of ozone, O3. Current thinking involves a catalytic photochemical mechanism that starts with (ClO) dimer formation and eventually loss of atomic chlorine. This in turn reacts with ozone leading to molecular oxygen and chlorine oxide. ClO + ClO (ClO)2 (ClO)2 + hυ Cl + ClOO ClOO Cl + O2 Cl + O3 ClO + O2 The obvious choice for the structure of ClO dimer is chlorine peroxide, ClOOCl, but “limiting” (cc-pVQZ basis set) B3LYP calculations show that chloryl chloride, ClClO2, is only 19 kJ/mol higher in energy. The fact that analogous calculations with smaller basis sets show progressively larger isomer differences (47 kJ/mol from the cc-pVTZ basis set and 132 kJ/mol from the 6-311+G** basis set) strongly suggest that the separation is even smaller (and perhaps actually favoring chlorine perchlorate. This would have serious impact on the proposed ozone destruction mechanism. Relative CH Bond Dissociation Energies in Hydrocarbons: CH bond energies in acetylene, benzene and ethylene are known to be 113, 25 and 17 kJ/mol smaller than that in ethane, whereas that in ethane is known to be 25 kJ/mol larger. Given this knowledge, which bond in propyne is more likely to break, that on C1 (the alkyne) or on C3 (the methyl group)? Which bond in propene is most likely to break, those on C1 or C2 (the alkene) or on C3 (the methyl group)? Use calculations from the B3LYP/6-31G* model to back up your answers. Use the B3LYP/6-31G* model to obtain equilibrium geometries for allene, H2C=C=CH2, and the radical resulting from CH bond dissociation. Calculate the bond dissociation energy relative to that of methane (you will need to do calculations on methane and methyl radical). Is this the result you expect based on the experimental result for bond dissociation in ethylene? (You can confirm this result by performing calculations on ethylene and vinyl radical.) If it is not, provide an explanation as to why not. Chlorine Nitrate: Chlorine nitrate, ClONO2, may play an active role in the balance ozone in the upper atmosphere. Specifically, the molecule may serve as a sink for NO2 and ClO free radicals that are known to react with ozone and destroy ozone. ClO + NO2 ClONO2 23 Use the B3LYP/6-311+G** model to obtain equilibrium geometries for the two free radicals as well as chlorine nitrate. Is radical recombination sufficiently exothermic to make chlorine nitrate an effective radical sink? Elaborate. Chlorine peroxynitrite (ClOONO) is a plausible isomer of chlorine nitrate. Were it in equilibrium with chlorine nitrite, it might undergo cleavage of the OO bond leading back to the initial radicals. Obtain the equilibrium geometry of chlorine peroxynitrite using the B3LYP/6-311+G** model, and calculate the room-temperature Boltzmann distribution between the two isomers. Is chlorine peroxynitrite likely to be involved? Elaborate. Carbon-Fluorine Bond Stengths in Fluoromethanes: Carbon-fluorine bond lengths in fluoromethanes, CFnH4-n (n=1-4), decrease with increasing number of fluorines, from 1.38Ǻ in fluoromethane to 1.32Ǻ in tetrafluoromethane. Does increase in the number of fluorines also lead to increase in CF bond energies? Use the B3LYP/6-31G* model to evaluate energies for each of the of the reactions below. CH3F2 → CH3. + F. CH2F2 → CH2F. + F. CF3H → CHF2. + F. CF4 → CF3. + F. Do CF bond energies in the fluoromethanes parallel CF bond distances? Repeat your calculations and analysis for the analogous fluorosilanes, SiFnH4-n (n=1-4). Repeat your calculations and analyses on both carbon and silicon compounds substituted with chlorine instead of fluorine, CClnH4-n and SiClnH4-n (n=1-4). Singlet and Triplet Carbenes: Molecules incorporating divalent carbon are referred to as carbenes or methylenes. The parent compound, CH2 (methylene), is known to possess a triplet ground state, with one unpaired electron residing in an in-plane σ orbital and the other in an out-of-plane π orbital. The lowest-energy singlet state (with both electrons in the σ orbital) is known experimentally to be approximately 42 kJ/mol higher in energy. Because triplet methylene and other triplet carbenes have one fewer electron pair than the corresponding singlets, Hartree-Fock models will always bias in favor the former. Thus, the (estimated) Hartree-Fock limiting (cc-pVQZ basis set) energy difference for triplet and singlet methylene (triplet favored) is 118 kJ/mol. B3LYP and MP2 models show no such bias. Limiting (cc-pVQZ basis set) singlet-triplet energy splittings are 48 and 60 kJ/mol (in favor of the triplet). These two models appear to correctly assign the ground state, and (in the very few cases for which experimental data exist) provide a good account of singlet-triplet energy differences. Use the B3LYP/6-31G* model to obtain equilibrium geometries of both singlet and triplet states of methylene, difluoromethylene (CF2), dichloromethylene (CCl2) and dibromomethylene (CBr2). Adjust the calculated singlet-triplet energy differences for CF2, CCl2 and CBr2 to account for the error in the singlet-triplet difference for CH2 and assign the ground state for each. Rationalize any change in preferred ground state (relative to the parent compound) that you uncover. 24 Singlet and Triplet Cyclobutadiene: While cyclobutadiene, C4H4, is a very short-lived molecule, the cyclobutadienyl ligand is common throughout organometallic chemistry. Here, the coordinated metal may change the number of π electrons from 4 (meaning that cyclobutadiene is formally an antiaromatic molecule) to a lesser or greater number. Does cyclobutadiene itself possess a singlet or triplet ground state? To tell, obtain equilibrium geometries for both singlet and triplet cyclobutadiene using the B3LYP/6-311+G** model, starting from structures that are not square. Describe the geometries. Is either or both square? If not, provide a rational as to why not? Calculate the singlet-triplet energy difference in cyclobutadiene, and correct it to account for the error in the corresponding difference in methylene (experimentally the triplet is favored by 42 kJ/mol). Which state is favored and by how much? Cyclopropylidene and Tropylidene Use the B3LYP/6-31G* model to obtain equilibrium geometries for both singlet and triplet state of cyclopropylidene. Correct the calculated singlet-triplet energy difference to account for the error in the corresponding difference in methylene (experimentally the triplet is favored by 42 kJ/mol). Is the ground state different from that for methylene? Examine both the carbon-carbon bond distances in singlet cyclopropylidene as well as its highest-occupied molecular orbital (use cyclopropene as a reference). Speculate on the cause behind the state preference Repeat the calculations and analysis for tropylidene. Use cycloheptatriene as a reference and locate the molecular orbitals in the carbene that correspond to the three occupied π orbitals in cycloheptatriene. Dissociation of Krypton Difluoride: While numerous compounds of xenon are known, the only neutral krypton compound to be reported is the difluoride, XeF2. Is such a species thermodynamically stable with regard to dissociation into xenon atom and fluorine molecule? To tell, use the B3LYP/6-31G* model to obtain equilibrium structures of XeF2, F2 and (the energy of) Xe. Sulfur-Sulfur Linkages in Proteins: Nearby cysteine residues in proteins may form sulfur-sulfur linkages. In so doing they impose geometrical constraints on the protein which in turn affects its secondary structure. How strong are sulfur-sulfur bonds? To tell, consider the model reaction. H3CS-SCH3 2 CH3S Use the B3LYP/6-311+G* model to calculate equilibrium geometries for dimethyl disulfide and thiomethoxy radical, and calculate the energy of SS bond cleavage. How does it compare with other bond energies (see Excel spreadsheet AH and AB bond dissociation energies for typical examples)? 25 The next comparison involves energy differences among structural isomers. Here, the total number of electron pairs is conserved but the numbers of individual bond types are not maintained. G3(MP2) serves as a reference. As with the previous bond energy comparisons, only a summary of mean absolute errors is provided (Table P3-4). An Excel spreadsheet containing structural isomer energies for Hartree-Fock, B3LYP and MP2 models with 6-31G*, 6-311+G**, cc-pVTZ and cc-pVQZ basis sets is provided on the CD-ROM accompanying this text (structural isomer energies). Hartree-Fock models with the 6-31G* and 6-311+G** basis sets actually yield slightly lower overall errors for this particular set of comparisons than the corresponding models with larger “limiting” basis sets. Overall errors for B3LYP models do not change significantly with basis set, while those for MP2 models with the two smaller basis sets are larger than those for the larger sets. Individual errors show wider deviations (see Excel spreadsheet). The B3LYP/6-31G* and B3LYP/6-311+G** models show the best overall performance, although those of the corresponding Hartree-Fock models are quite similar. 26 Table P3-4: Basis Set Summary of Deviations from G3 of Hartree-Fock, B3LYP and MP2 Energies of Structural Isomers (kJ/mol) Hartree-Fock B3LYP MP2 6-31G* 9 8 17 6-311+G** 10 7 12 16 7 cc-pVTZ cc-pVQZ 27 Acidity of Propyne: Loss of hydride anion from propyne can either occur from the sp hybridized carbon or from the sp3 hybridized carbon. The former might be expected as acetylene is a much stronger acid than ethane, while the latter might be expected as it would lead to a delocalized anion. Which deprotonation is thermodynamically favored? H3 C C C– H3C C CH H2C C C H2 – Use the B3LYP/6-311+G** model to obtain equilibrium geometries for the two anions. Which is lower in energy? Is the less stable anion likely to be detectable in an equilibrium mixture at room temperature? (Assume a threshold of 5%.) Isomers of Carboranes: Carboranes are compounds of carbon, boron and hydrogen. Depending on stoichiometry, they exhibit caged (closo), partially caged (nido from the Latin word for nest) or open (arachno from the Greek word for spider) structures. are B4C2H6 and B10C2H12 simple examples of closo structures. The former can exist in one of two isomers, and the latter in one of three isomers. H C HB H B BH HB BH C H HB CH HB BH C H ChemDraw B10C2H12for isomers Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of B4C2H6 and the three isomers of B10C2H12. Examine carbon-carbon bond lengths in the appropriate isomer of each. Do these appear to be “normal” single bonds or are the significantly shorter or longer? If the latter provide a rationalization? Identify the lowestenergy isomer for B4C2H6 and B10C2H12 and provide a rationale for the observed preferences. Performance of Practical Models for Hydrogenation Reactions: A hydrogenation reaction maintains overall bond count but does not maintain individual bond counts. For example, the products of hydrogenation of ethane have the same number (eight) of σ bonds as the reactants but two new CH bonds have replaced a C-C bond and H-H bond. CH3-CH3 + H-H → 2CH4 Calculate hydrogenation energies for ethane, hydrazine, hydrogen peroxide and fluorine using the Hartree-Fock, B3LYP and MP2 models with both the 6-31G* and 6-311+G** 28 basis sets (six models in total). Correct these for zero-point energy and finite temperature using the data in Appendix X. Compare these to the results from the G3(MP2) recipe. CH3-CH3 H2N-NH2 HO-OH F- F -60 kJ/mol -185 kJ/mol -347 kJ/mol -548 kJ/mol Which (if any) models yield hydrogenation energies that are within +/-12 kJ/mol of the G3(MP2) values? Which model provides the best results overall? Energy Content of Hydrazine Fuels: According to the B3LYP/6-31G* model, which fuel delivers the greater energy on a per gram basis, hydrazine or tetramethylhydrazine? The products of oxidation are N2 and water for hydrazine and N2, water and CO2 for tetramethylhydrazine. Make certain to include the mass of the oxidizer (O2) in your calculations. How does the better of the two fuels compare with molecular hydrogen on a per gram basis? Combustion of Hydrocarbons and Fluorocarbons: Hydrocarbons are commonly used as fuels but fluorocarbons are not. Of course, fluorocarbons cannot be mined or drilled from the earth, but is there also a fundamental reason? Use the B3LYP/6-31G* model to calculate the energy of complete combustion of methane, ethane and propane (to CO2 and water) and tetrafluoromethane, hexafluoroethane and decafluoropropane (to CO2 and OF2). Point out differences between energies of combustion for the hydrocarbon and analogous fluorocarbon and relate this to their use as fuels. “Combustion” of Silanes: Silicon-oxygen polymers (“sand”) result from combustion (in oxygen) of silanes and other silicon-containing compounds. This makes it difficult to assign heats of formation. One clever solution is to “burn” such compounds in fluorine (F2) rather than in oxygen. This leads only to gaseous products (SiF4 and HF) the amounts of which may easily be determined, for example, combustion of silane. SiH4 + 4F2 SiF4 + 4HF Use the B3LYP/6-31G* model to determine energies of complete “combustion” (in F2) of silane, disilane, trisilane, 2-silyltrisilane and 2,2-disilyltrisilane (resulting in only SiF4 and HF as products). On a per gram basis, combustion of which of these produces the greates amount of heat? H SiH4 H3 S i SiH3 H3 S i Si S i H2 H3 S i S i H3 SiH3 H3 S i SiH3 Si SiH3 H3 S i SiH3 Compare silane combustion energies with those of the corresponding hydrocarbons (see previous problem; you need to perform B3LYP/6-31G* calculations on 2-methylpropane and 2,2-dimethylpropane). On a per gram basis, are alkanes or silanes better fuels? 2,4-Cyclohexadiene vs. Phenol: As a rule, enols (unsaturated alcohols) are less stable than their keto isomers (aldehydes and ketones). For example, vinyl alcohol is estimated to be 43 kJ/mol less stable than its “keto” tautomer, acetaldehyde. CH3C(H)=O H2C=C(H)OH 29 ∆E ≈ 43 kJ/mol Use the B3LYP/6-31G* model to obtain the equilibrium geometry and energy of 2,4cyclohexadienone and its enol tautomer, phenol. O OH Which is more stable? If this is an exception to the “rule”, provide a rationale as to why. What temperature would be needed in order for the higher-energy structure to be present as 10% of the equilibrium mixture? Addition vs. Substitution: Alkenes typically undergo addition reactions whereas aromatic compounds typically undergo substitution reactions. For example, reaction of bromine and cyclohexane yields trans-1,2-dibromocyclohexane not 1-bromocyclohexene, whereas bromination of benzene yields bromobenzene not trans-5,6-dibromo-1,3cyclohexadiene. Br + Br2 Br vs. + HBr Br Br + Br2 Br vs. + HBr Br What is the reason for the change in preferred reaction in moving from the alkene to the arene? Use the HF/6-31G* model to obtain equilibrium geometries and energies for reactants and the products of both addition and substitution reactions of both cyclohexene and benzene (four reactions in total). Is your result consistent with what is actually observed? Are all four reactions exothermic? If one or more are not, provide a rationale as to why. Bromonium Ions: Addition of Br2 to an alkene, for example, cyclopentene, is usually represented as a two-step process. In the first step, the electrophile Br+ adds to the double – bond giving rise to a “bromonium ion” intermediate. In the second step, Br reacts with the intermediate to give a trans brominated product. Br2 Br– Br H H Br The 13C NMR spectrum of the intermediate cyclopentenebromonium ion has been recorded and shows resonances at 19, 32 and 115 ppm (relative to tetramethylsilane). This is consistent with one of two possible structures. Either the bromine is bonded to both carbons, leaving it with the (formal) positive charge, or it bonded to only a single 30 carbon, leaving the positive charge on the other carbon. In the latter case, the bromine needs to transfer between the two carbons with a very low energy barrier. Br+ + H H H Br + H H H Br Use the B3LYP/6-31G* model to obtain equilibrium structures, relative energies and 13C NMR chemical shifts for the two alternative structures. Is the cyclic structure better represented as a three-membered ring (like cyclopropane or oxacyclopropane) or as a weak complex between cyclopentene and bromine cation? Examine the carbon-carbon bond distance to tell. Which structure is lower in energy? Which better fits the observed 13 C NMR spectrum, the bridged structure or an equal mixture of the two open structures? Would you expect the NMR spectrum to change were the measurement carried out at 160K? Elaborate. Are the energetic and NMR results consistent with each other? Hydrogenation of Acetylene: Both steps in the hydrogenation of acetylene to ethane involve formation of two new CH bonds from the loss of an HH bond and a π bond. Use the HF/6-31G* model to obtain equilibrium geometries and energies for all molecules involved in the hydrogenation reaction. Which step is the more exothermic? Rationalize your result. H C C H H2 H H C H2 C H CH3CH3 H Ketene Dimer: There are six possible structures for the dimer of ketene, H2C=C=O. CH2 CH2 C O O C CH2 O O C C C O O H2 C CH2 C CH2 O C H2 C C CH2 O O O C H2 C CH2 C C C H2 C O O CH2 Before you do any calculations, predict which of these is likely to be the most stable based on what you know about the relative stabilities of CC and CO π bonds (lost in the dimerization) and CC and CC σ bonds (gained in the dimerization). Then, obtain equilibrium geometries for all six using the HF/6-31G* model. Which structure is actually preferred. Assuming thermodynamic control, are two or more structures likely to be seen (assume 5% as the limit of detection of any structure). Is the preferred structure in line with your prediction? Is the dimerization exothermic or endothermic? Is this consistent with changes in bond strengths and the strain of the resulting four-membered ring? Elaborate. Molecular Phosphorous: The most stable arrangement of molecular phosphorus (white phosphorus) is tetrahedral P4. Dissociation to two molecules of P2 is estimated to be endothermic by >200 kJ/mol, and can only be detected upon heating to 1000 K. To the 31 contrary, the stable form of molecular nitrogen is N2, and N4 has never been detected let alone isolated. Use the B3LYP/6-31G* model to obtain equilibrium geometries and energies for tetrahedral P4 and N4 (as well as P2 and N2) and calculate energies for the two dissociation reactions. Are your results consistent with what is known (or unknown) about the two systems? Is tetrahedral N4 actually a stable molecule? Explain your reasoning. Is tetrahedral As4 a stable molecule? Use the B3LYP/6-31G* model to obtain its equilibrium geometry and vibrational frequencies. Is its dissociation to As2 endothermic (like P4) or exothermic? What is the room temperature equilibrium distribution of As4 and As2? Tetramethylbutane: The central carbon-carbon bond in 2,2,3,3-tetramethylethane (tetramethylbutane) is known to be significantly weaker than the CC bond in ethane. One explanation is that bond cleavage relieves the crowding of the methyl groups. Another is that is tert-butyl radical (the product of bond dissociation of tetramethylbutane) is much more stable than the methyl radical (the product of dissociation of ethane). To find out which explanation is correct or if both contribute, calculate energies for the following four reactions. Use the B3LYP 6-31G* model. (Do not to start with a symmetrical staggered structure for tetramethylbutane.) Me3C-CMe3 + H3C· → Me3C· + H3C-CH3 Me3C-CMe3 + H2 → 2Me3CH H3C-CH3 + H2 → 2CH4 Me3CH + H3C· → Me3C· + CH4 The first reaction directly compares carbon-carbon bond energies of tetramethylbutane and ethane. The next two reactions provide the energies of hydrogenation of tetramethylbutane and ethane (both giving rise to uncrowded products). The last reaction compares bond energies of these (uncrowded) products (providing a measure of the relative stabilities of the methyl and tert-butyl radicals). Is the bond energy in tetramethylbutane less than that in ethane? If so, by how much? Is there evidence for crowding in the geometry of tetramethylbutane? Is the central carboncarbon bond similar in length to the bond in ethane? Are all single bonds staggered? Does the difference between the hydrogenation energies of tetramethylbutane and ethane partially or fully account for the difference in bond energies? Does the difference in difficulty of formation of tert-butyl and methyl radicals partially or fully account for the difference in bond energies? Dimerization of Alkylboranes: Borane (BH3) exists in an equilibrium with its dimer, diborane (B2H6) that strongly favors the dimer (from B3LYP/6-31G* calculations). ΔE = -164 kJ/mol Trimethylborane also dimerizes but the equilibrium favors the monomer 32 ΔE = 113 kJ/mol What is the reason for the change? Is a hydrogen bridge strongly favored over a methyl bridge? To tell, consider the dimerization of methylborane which may lead to four possible products: cis and trans isomers with both methyl groups terminal, with both methyl groups bridged and with one methyl group terminal and the other bridged. Use the B3LYP/6-31G* model to obtain equilibrium geometries for all four isomers in addition to methylborane. Which dimer is favored? Is dimerization to one or both of the isomers with the two methyl groups in terminal positions exothermic (as with dimerization of borane)? Is dimerization to the isomer with both methyl groups in bridged positions endothermic (as with dimerzation of hexamethyldiborane)? Speculate on the reason for the difference in dimerization energies between borane and trimethylborane. Repeat the calculations for dimerization of both fluoroborane and chloroborane, each leading to four different products. Is bridging by hydrogen of fluorine (chlorine) favored? Obtain dimerization energies for alane (AlH3) to dialane (Al2H6) and trimethylaluminum (AlMe3) to hexamethyldialane (Al2Me6). Continue to use the B3LYP/6-31G* model. Point out any significant differences from those of the corresponding boron compounds. Hydrogen Azide: Hydrogen azide is purported to have a high energy cyclic isomer, cyclotriazine. This is not unreasonable. While the cyclic isomer is likely to be highly strained, its Lewis structure doe not involve separated positive and negative charges. Attempt to obtain equilibrium geometries for both using the B3LYP/6-311+G** model. (Start from a non-planar geometry for the cyclic isomer.) Are both acyclic and cyclic structures energy minima? Justify you answer. If both structures are minima, which is more stable? What temperature would be required in order for both to be observed in an equilibrium mixture? (Assume that the minor isomer must be at least 5% of the mixture in order to be detected.) Repeat your calculations and analysis for methyl azide and methyl substituted cyclotriazene. Isomers of Adamantyl Cation: Tertiary carbocations, such as tert-butyl cation are more stable than secondary carbocations, such as isopropyl cation. It is also important that the 33 carbocation center is planar or nearly planar. Which is more important when only one of the two can be realized? The isomers of adamantly cation provide the opportunity to tell. 2-adamantyl cation incorporates a non-planar tertiary carbocation center, whereas the 1adamantyl cation incorporates a planar secondary carbocation center. Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of adamantyl cation. Are the CCC bond angles at the cation center in 2-adamantyl cation significantly distorted from ideal (120o)? Which cation is lower in energy? Does it appear that the need to accommodate a planar cation is more important that the benefit achieved in going from a secondary to tertiary center? Elaborate. The final comparisons (Table P3-5) exemplify reactions in which both the total number of electron pairs and the numbers of individual bond types are conserved, specifically comparisons of regio and stereoisomers. As with previous comparisons, data from G3(MP2) calculations serve as a reference. 34 Table P3-5: reference Deviations from G3(MP2) of Hartree-Fock, B3LYP and MP2 Energies of Regio and Stereoisomers (kJ/mol) isomer Hartree-Fock B3LYP MP2 6-31G* 6-311+G** 6-31G* 6-311+G** 6-31G* 6-311+G** G3 1,3-butadiene 1,2-butadiene 50 53 39 43 47 49 49 2-butyne 1-butyne 28 24 33 27 23 18 22 cyclobutene methylenecyclopropane 29 26 25 20 32 31 44 isobutene trans-2-butene cis-2-butene 1-butene 1 8 13 2 9 13 1 7 18 1 7 15 4 11 18 5 11 16 5 11 17 cyclopentene methylenecyclobutene 86 85 79 79 89 89 85 2-methyl-1, 3-butadiene 23 48 58 21 51 60 34 40 53 29 42 53 30 45 60 27 49 62 29 55 62 methyl vinyl trans-2-butenal ketone 2-methyl-2-propanal 7 9 14 9 3 5 10 5 7 4 13 1 7 4 mean absolute error 5 5 8 8 4 4 – 1,4-pentadiene 1,1-dimethylallene 1,2-pentadiene 35 Isomers of Pentavalent Phosphorus Halides, PFnCl5-n: While pentavalent phosphorus halides adopt trigonal bipyramidal geometries with distinct axial and equatorial positions, these positions rapidly interconvert via pseudorotation (see discussion in Chapter P5). Therefore, observed properties, for example, the dipole moment, will be those of an equilbrium mixture of all possible isomers and depend on temperature. Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of SF4Cl (with Cl in either an equatorial or axial position). Which isomer is lower in energy? Why? Calculate the dipole moment of a sample at room temperature. Is it different at 50 K? At 1000 K? Repeat the calculations for SF3Cl2 (3 isomers), SF2Cl3 (3 isomers) and SFCl4 (2 isomers). Are the results consistent with those for SF4Cl? Elaborate. Compute dipole moments at 50 K, room temperature and 1000 K. Oxy Acids of Phosphorus: The oxy acids of phosphorus can exist in one of two isomeric (tautomeric) forms, one in which the phosphorus is trivalent and the other in which it is pentavalent. O R' P R OH R' P R H The experimental evidence points strongly to the pentavalent species as the dominant form. Specifically, the infrared spectra of phosphorus oxy acids contain lines characteristic of P=O and PH bond stretches and no evidence of OH stretches (except where R or R’ is OH). The one exception appears to be the bis(trifluoromethyl) compound (R=R’=CF3). Despite the apparent preference for pentavalent forms, the known “chemistry” of the phosphorus oxy acids demands involvement of the trivalent structure. This suggests that the two must be in equilibrium, meaning that they two are close in energy. Use the B3LYP/6-31G* model to obtain equilibrium geometries for both trivalent and pentavalent forms of dimethylphosphonate (R=R’=OMe). Precede your calculations by a search of possible conformers using molecular mechanics (discussion of conformational searching will be provided in Chapter P5). Which form is lower in energy? Is your result consistent with the experiment? Elaborate. What temperature would be required for the higher energy structure to be present as 5% of an equilibrium mixture? Repeat your calculations with the bis(trifluoromethyl) compound (R=R’=CF3). If, as the experimental data suggest, there is a change in preferred structure, propose a reason why. Adamantene: The six atoms involved in a carbon-carbon double bond prefer to lie in the same plane, for example, the two carbons and four hydrogens in ethylene lie in the same plane. This does not appear to be possible for adamantene, the olefin formed from lss of hydrogen from the stable hydrocarbon, adamantane. - H2 36 Use the B3LYP/6-31G* model to obtain the equilibrium geometry for adamantene. Does the molecule actually incorporate a double bond, that is, with a carbon-carbon distance in the usual range of 1.30 to 1.34Ǻ? Is one or both of the carbons involved in the bond puckered? Are they twisted relative to each other? Display the HOMO of adamantene, and describe how it differs from the HOMO of a “normal” alkene. Relate the hydrogenation energy of adamantene to that of 2-methyl-2-butene, a molecule with similar “substituents” on the double bond. (Use the B3LYP/6-31G* model to obtain geometries for 2-methylbutane and 2-methyl-2-butene.) adamantene + 2-methylbutane adamantane + 2-methyl-2-butene What does this tell you about the strengths of the two π bonds? Bond Separation Reactions and Interaction of Substituents: A bond-separation reaction relates any molecule that can be written in terms of a Lewis structure to the set of “simplest molecules” that contain the same bonds. For example, the bond separation reaction for acrolein, relates it to ethane, ethylene and formaldehyde, the simplest molecules with carbon-carbon single and double bonds and a carbon-oxygen double bond. Two molecules of methane need to be added to the left hand side to achieve stoichiometric balance. H2C=C-C(H)=O + 2CH4 → H2C=CH2 + H3C-CH3 + H2C=O Aside from the conformation of the reactant, the energy of a bond separation reaction is well-defined and unique as long as the Lewis structure is well defined and unique. Because it maintains both total and individual bond counts, it might be expected to be reasonably well described even by models that provide poor account of electron correlation. Bond separation reactions may be used to determine whether two substituents bonded to carbon interact constructively (bond separation reaction is endothermic), destructively (bond separation reaction is exothermic) or not at all. XCH2Y + CH4 → CH3X + CH3Y Use the HF/6-31G* model to obtain equilibrium geometries and energies for all molecules involved in bond separation reactions of molecules, CH2X2 (X=CH3, CMe3, CN, F, SiH3). Interpret the calculated reaction energies both in terms of steric interactions and the σ and π donor/acceptor properties of the individual substituents: σ donor, π donor σ acceptor, π acceptor σ acceptor, π donor σ donor, π acceptor Me, CMe3 CN F SiH3 Which of the substituents interact favorably? Which interact unfavorably? Provide a rationale for each. Compare bond distances in CH3X and CH2X2 compounds. Do the changes parallel the interaction energies? Elaborate. 37 Bond Stengths vs. Bond Separation Energies in Fluoromethanes: In an earlier problem (see Chapter P2), you found that CF bond energies in fluoromethanes, CFnH4-n (n=1-4), increase with increasing number of fluorines. Is this trend reflected in the bond separation energies? Use the HF/6-31G* model to calculate energies for all reactants and products for the three bond separation reactions below. CH2F2 + CH4 → 2 CH3F CF3H +2CH4 → 3CH3F CF4 + 3CH4 → 4CH3F Do the bond energies of these reactions (normalized for the number of CF bonds) parallel the bond dissociation energies calculated previously? Repeat your calculations and analysis for bond separation reactions of fluorosilanes, SiFnH4-n (n=1-4). Repeat your calculations and analysis for bond separation reactions on both carbon and silicon compounds substituted by chlorine instead of fluorine, CClnH4-n and SiClnH4-n, (n=1-4). 38 Ion Molecule Reactions Quantum chemical calculations are not restricted to uncharged molecules, but may also be applied to cations and anions. While there are virtually no experimental data relating to the geometries of in the gas phase, there is a wealth of data relating to their energies. The two most common sources are proton transfer reactions, that is, acidities and basicities, and electron transfer reactions, that is, ionization potentials and electron affinities. There are an enormous number of X-ray crystal structures of ions (together with their counterions). Discussion has already been provided in Chapter P2. As discussed earlier in this chapter, relative acidities and basicities are most commonly determined by ion-cyclotron-resonance (ICR) spectroscopy. What is actually measured is the equilibrium abundance of the ions involved in a proton transfer reactions between two bases (or two acids) of similar strength, for example, that between methylamine and ethylamine (bases) or between acetic acid and propanoic acid (acids). The neutral molecules are not seen. MeNH2 + EtNH3+ MeNH3+ + EtNH2 MeCO2H + EtCO2- MeCO2- + EtCO2H Measurements are all finally related to a single standard base (ammonia) and standard acid (???), giving rise to absolute basicities and acidities. Relative Acidities of Propene and Propyne: Is propene or propyne the stronger acid in the gas phase? Use the B3LYP/6-311+G** model to calculate the geometries for the reactants and product of the reaction. Be certain to consider all possible anions resulting from deprotonation of propene and propyne. propene + propyne-H+ propene-H+ + propyne Rationalize your result. Is it anticipated by comparison of electrostatic potential maps for propene and propyne? Elaborate. Why is Cyclopentadiene a Strong Acid? Cyclopentadiene is a much stronger acid than 1,3-pentadiene in the gas phase, that is, the following reaction is highly exothermic. cyclopentadiene + 1,3-pentadiene - H+ 39 cyclopentadiene-H+ + 1,3-pentadiene Use the B3LYP/6-31G* model to obtain equilibrium geometries for cyclopentadiene and its deprotonated form (cyclopentadienyl anion). Note any structural changes that have occurred to cyclopentadiene as a result of deprotonation, and use these to rationalize its high acidity. Gas and Aqueous-Phase Basicities of Amines: The relative base strengths (proton affinities) of free amines are known to differ significantly from those in water. For example, in the gas-phase, the proton affinity trimethylamine is 84 kJ/mol greater than ammonia, whereas in water, the two are of nearly equal. Use the B3LYP/6-31G* model to obtain equilibrium geometries for the ammonia and trimethylamine and their respective protonated forms (ammonium and trimethylammonium ions). Is the energy that you calculate for the transfer of a proton between the two in accord with their relative gasphase proton affinities? Me3N + NH4+ Me3NH+ + NH3 For practical reasons, you can’t perform the equivalent B3LYP calculations in water. You can, however, model the energy of the reaction in water by explicitly including a specific number of water molecules in your calculation. The obvious thing to do is to attach the minimum number of water molecules to account for all possible amine-water hydrogen bonds. This number is four for ammonia and ammonium ion, but only one for trimethylamine and trimethylammonium ion. Obtain equilibrium geometries for the four amine/ammonium ion complexes involved in the reaction below and calculate the relative “aqueous-phase” proton affinities of trimethylamine and ammonia. Me3N…H2O + NH4+ …(H2O)4 Me3NH+…H2O + NH3…(H2O)4 Do the results show the observed trend in proton affinities in moving from the gas phase into water? Elaborate. Redo your calculations with trimethylamine and trimethylammonium ion “attached to” four water molecules instead of only one, and calculate the energy of the reaction below. Me3N…(H2O)4 + NH4+ …(H2O)4 Me3NH+…(H2O)4 + NH3…(H2O)4 Does this appear to be a better model? Elaborate. Try to identify a serious flaw in both models. Boiling Points of Ethanol and Ethylamine: While ethanol and ethylamine have similar molecular weights and molecular structures and while both can participate in three hydrogen bonds, the two molecules have very different boiling points. Ethanol is a liquid at STP whereas ethylamine is a gas. Is this because ethanol forms stronger intermolecular hydrogen bonds than ethylamine? To tell, obtain equilibrium geometries for ethanol and ethylamine and their respective hydrogen-bonded dimers, and compare hydrogen-bond energies. Use the B3LYP/6-31G* model. ethanol + ethanol ethanol … ethanol ethylamine + ethylamine ethylamine … ethylamine Which dimer is more tightly bound? Is your result consistent with the ordering of boiling points? Elaborate. 40 Exploring Chemical Concepts with Quantum Chemical Models The availability of reliable and practical quantum chemical models offers the opportunity to critically examine a number of the qualitative models that chemists have advanced over many decades. Among the most important are conjugation, aromaticity and ring strain, the first leading to molecules that are stablized and the second to molecules that are destabilized. Conjugation Molecules incorporating adjacent double bonds that are coplanar or nearly coplanar are known to be preferred over non-coplanar arrangements or arrangements in which the double bonds are separated by one or more sp3 “spacers”. Thus, the double bonds in both trans-1,3-pentadiene and in trans2-butenal are coplanar and each is more stable than its non-conjugated isomer, 1,4-pentadiene and 3-butenal, respectively. Hydrogenation of 1,3-Butadiene: Both steps in the hydrogenation of trans-1,3butadiene to 1-butene and then to n-butane involve formation of two new CH bonds resulting from the loss of an HH bond and a π bond. Use the HF/6-31G* model to obtain equilibrium geometries and energies for all molecules involved in both steps of the hydrogenation. Which step is the more exothermic? Does this support the idea that conjugated double bonds are stronger than isolated double bonds? Elaborate. H H C C H H C C H2 H C H H CH2 CH3 H2 C H CH3 CH2 CH 2 CH3 H Hydrogenation Energy of 1,3-butadiene vs. But-1-yne-3-ene: Compare the energy of the first step in the hydrogenation of 1,3-butadiene (see previous problem) with that of hydrogenation of the double bond in but-1-yne-3-ene. Use the HF/6-31G* model to obtain equilibrium geometries for all molecules in the two reactions. H C H C C vs. H C H2 CH3C CH H Which reaction is more exothermic? What does this say about the conjugation energies of two double bonds vs. a double bond and a triple bond? 41 Hydrogenation of 1,2-butadiene: Both steps in the hydrogenation of 1,2-butadiene first to 1-butene and then to n-butane involve formation of two new CH bonds resulting from the loss of an HH bond and a π bond. Use the Hartree-Fock 6-31G* model to obtain equilibrium geometries and energies for all molecules involved in both steps of the hydrogenation. Does the double bond in 1,2-butadiene appear to be stronger or weaker than that in 1-butene? Is the ordering of bond strengths consistent with the ordering of double bond lengths in the two molecules? Based on hydrogenation energies, would you conclude that the double bond in 1,2-butadiene is stronger or weaker than that in 1,3butadiene (see first problem in this section)? Would you characterize 1,2-butadiene as a conjugated molecule? Elaborate. CH3 H C H C C CH2CH3 H H2 C H H C H Conjugated vs. Non-Conjugated Cyclic Dienes: 1,4-cyclohexadiene is known to be slightly (4 kJ/mol) more stable than its isomer, 1,3-cyclohexadiene. This is a surprising result as the former is a non-conjugated diene whereas the latter is a conjugated diene. Use the HF/6-31G* model to obtain equilibrium geometries for the conjugated dienes, 1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene and their nonconjugated isomers, 1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene. Assign the preferred isomer and calculate the energy difference for each pair. Do the calculations reproduce the experimental result for cyclohexadiene? Do they show like preferences for cycloheptadiene and cyclooctadiene? Aromaticity Whereas coplanar arrangements of two or more connected π bonds show modest energy stabilization, much larger effects are noted for some cyclic arrangements of π bonds. Specifically, molecules with 4n+2 π electrons in a planar or nearly planar arrangement are known to be unusually stable or aromatic, and are likely to exhibit chemistry that is quite distinct from that of unsaturated molecules (even conjugated unsaturated molecules). The archetypical example is benzene with three carbon-carbon double bonds (involving six π electrons) confined in a planar six-membered ring. One way to “measure” the aromatic stabilization of benzene is to compare the energy of adding H2 (leading to 1,3-cyclohexadiene) with the energy of adding hydrogen to 1,3-cyclohexadiene (leading to cyclohexene), or the energy of adding hydrogen to cyclohexene (leading to cyclohexane). H2 H2 H2 2 4 k J /m o l -110 kJ/mol - 1 2 0 k J /m o l 42 The three steps are similar insofar as each trades a CC π bond and an HH bond for two CH bonds. Normally, one would tend to think of such a trade as energetically favorable as σ bonds are stronger than π bonds. However, the first hydrogenation step is actually slightly endothermic, while the remaining two steps are both strongly exothermic. The difference between the first and second (or third) steps (~140 kJ/mol) reflects the additional (aromatic) stabilization resulting from arrangement of three π bonds in a planar six-membered ring. The individual hydrogenation reactions do not conserve the numbers of each kind of chemical bond (each reaction does maintain overall bond count), and would not be expected to benefit fully from cancellation of errors. However, the difference between the first and second reactions (or between the first and third reactions) which actually relates to aromatic stabilization can be expressed as a reaction that does maintain individual bond counts. Previous experience with reactions of this type suggests that its energy should be reasonably well described by quantum chemical methods that do not take electron correlation into account. + 2 1,3,5-Cyclohexatriene: The six carbon-carbon bonds in benzene are all the same, intermediate in length between “normal” single and double bonds. The hypothetical molecule 1,3,5-cyclohexatriene is identical to benzene except that the CC bonds alternate between single and double bonds. 1,3,5-cyclohexatriene is not an energy minimum and, if given the chance, will collapse to benzene. Is the energy difference between benzene and 1,3,5-cyclohexatriene roughly the same as the aromatic stabilization of benzene (~ 140 kJ/mol; see discussion above), or is it significantly smaller? To decide, use the HF/631G* model to obtain an equilibrium geometry for benzene and an energy for 1,3,5cyclohexatriene, assuming a fixed geometry with alternating single and double CC bonds set to 1.54Å and 1.32Å. Aromaticity of Thiophene: Thiophene is a likely candidate for an aromatic molecule with two double bonds and an out-of plane lone pair on sulfur making a total of six π electrons. S Compare reaction energies for the individual steps in the hydrogenation of thiophene to tetrahydrothiophene. As was the case for benzene, addition of the first equivalent of hydrogen breaks one of the double bonds and leading to loss of aromaticity. Addition of the second equivalent only breaks a double bond. Therefore, the difference in energy between the two provides a measure of the aromatic stabilization. 43 X X = S thiophene X = NH pyrrole X = O furan X H2 X H2 X = S tetrahydrothiophene X = NH tetrahydropyrrole X = O tetrahydrofuran Use the HF/6-31G* model to obtain geometries and energies for thiophene, dihydrothiophene and tetrahydrothiophene as well as hydrogen, and calculate hydrogenation energies for the two steps. Is the difference in hydrogenation energies comparable to that between the corresponding first and second hydrogenation energies for benzene (see discussion above) or is it significantly smaller or greater? Comment on the aromaticity of thiophene. Perform analogous calculations and analyses on pyrrole (X=NH) and furan (X=O). Do either or both of these molecules show aromaticity? Elaborate. Order the aromaticity of thiophene, pyrrole and furan. Aromaticiy of Borazine and Boraphosphazine: Borazine, B3N3H6, is isoelectronic with benzene and like benzene is known to be a planar molecule with six π electrons (originating from the three out-of-plane nitrogen lone pairs). HN HB H B N H NH BH Does this mean that borazine is aromatic? Use the B3LYP/6-31G* model to calculate energies for successive addition of one, two and three equivalents of hydrogen (analogous to what has previously been done for benzene). HN HB H B N H NH H2 BH H2N HB H2 B N H NH H 2 H2 N BH H2B H2 B N H2 NH H2 H2N BH H2 B H2 B N H NH BH Is the first hydrogenation step significantly more difficult (more endothermic) than the second and third steps? If it is, estimate the “aromaticity” of borazine and compare this to the value for benzene obtained from the same analysis. Use the B3LYP/6-31G* model to calculate energies for the individual steps in the hydrogenation of the phosphorous analogue of borazine (boraphosphazine). HP HB H B P H PH BH H2 H2P HB H2 B P H PH BH H2 H2 P H2B H2 B P H2 P H H2 H2P BH H2 B H2 B P H PH BH Rank the aromatic stabilization of benzene, borazine, and boraphosphazine. Cyanuric Acid: Heating urea leads to loss of ammonia and production of hydrogen cyanate. This in turn trimerizes to cyanuric acid, which may either take on a keto or enol structure. Both are aromatic molecules insofar as both involve six π electrons in a planar six-membered ring. Which is preferred and why? 44 O HN O • H2N NH3 + HN C O NH 2 O HO NH N H O N N OH N OH Use the B3LYP/6-31G* model to obtain equilibrium geometries for urea, ammonia, hydrogen cyanate and the keto and enol forms of hydrogen cyanate trimer. Is loss of ammonia from urea an endothermic or exothermic process? Is it is endothermic, what do you think drives the reaction? Is trimerization to the more stable of the two forms of cyanuric acid endothermic or exothermic? Is the overall reaction endothermic or exothermic? What is the preferred structure of cyanuric acid? Cyclobutadiene: Whereas a molecule with 4n+2 π electrons in a cycle is thought to be unusually stable (aromatic), a cyclic arrangement of 4n electron has the potential of being unusually unstable (antiaromatic). A good example of this is singlet cyclobutadiene. Compare the energy of hydrogenation of cyclobutadiene (to cyclobutene) with that of hydrogenation of cyclobutene (to cyclobutane). Use the B3LYP/6-311+G** model. The two reactions are similar in that each trades an HH bond and a CC π bond for two CH bonds. However, only the first reaction removes interaction of the coplanar double bonds. Aromaticity of Cyclopentadienyl and Cycloheptatrienyl Radicals: Cyclopentadienyl radical assumes a planar geometry with a total of five π electrons. Cycloheptatrienyl radical is also planar but has seven π electrons. Do these radicals benefit from aromatic stabilization? To tell, use the same test previously employed for benzene and related systems, that is, compare energies for the first step in an overall hydrogenation reaction with that of subsequent steps. Use the B3LYP/6-31G* model to obtain equilibrium geometries for all the molecules in the above two reactions. For each, calculate the sequence of hydrogenation energies. Would you conclude that either cyclopentadienyl and/or cycloheptatrienyl radical is aromatic? Elaborate. Aromaticity of Cyclopentadienyl Anion and Cycloheptatrienyl (Tropylium) Cation: The cyclopentadienyl anion is among the most common ligands found in transition-metal organometallic compounds. Here it maintains a planar (or nearly planar) geometry and, is generally thought to contribute six electrons to the valence shell of the metal to which it is bonded. Is cyclopentadienyl anion aromatic? Certainly it fits the obvious criteria of possessing a planar cyclic arrangement of 4n+2 π electrons (as does benzene). However, these electrons are spread over only five carbons (and not six as in benzene). 45 The opposite situation exists for cycloheptatrienyl cation (commonly known as tropylium). It is also planar with six π electrons, but these are spread over seven carbons. Do these ions benefit from aromatic stabilization? To tell, use the same test previously employed for benzene and related systems, that is, compare energies for the first step in an overall hydrogenation reaction with that of subsequent steps. – + H2 H2 – + H2 – H2 + + Use the B3LYP/6-31G* model to obtain equilibrium geometries for all the molecules in the above two reactions. For each, calculate the sequence of hydrogenation energies. Would you conclude that either cyclopentadienyl anion and/or cycloheptatrienyl cation is aromatic? Elaborate. Ring Strain A molecule with an sp3 center incorporated into three-membered rings would be expected to be energetically disfavored. The main reason is that the centers cannot adopt their ideal tetrahedral values (bond angles of 109.5o). Another reason is that planarity of the skeleton leads to eclipsing interaction involving hydrogens on adjacent centers. This combination of factors is commonly referred to as ring strain, and the associated “cost” in energy is commonly referred to as the strain energy. For example, cyclopropane, is known to be ~60 kJ/mol less stable than its isomer, propene, while cyclohexane is ~80 kJ/mol more stable than 1-hexene. The latter reflects the fact that, all other things being equal, σ bonds are stronger than π bonds. The difference (~140 kJ/mol) might be considered as the strain energy of cyclopropane. cyclopropane propene cyclohexane 1-hexene -60 kJ/mol 80 kJ/mol Isomerization energy is but one way to assess strain energy. Hydrogenation reactions may also be used to distinguish strained from unstrained molecules. For example, hydrogenation of cyclopropane to propane is exothermic by ~144 kJ/mol, while hydrogenation of cyclohexane to hexane is exothermic by only ~44 kJ/mol, yielding a strain energy of ~100 kJ/mol. 46 Ring Strain in Oxirane and Aziridine: Which molecule is more strained oxirane or cyclopropane? Use the HF/6-31G* model to evaluate the hydrogenation energy of oxirane (leading to dimethyl ether) relative to that of cyclopropane (leading to propane). (Hydrogen molecule appears on both sides of the equation and its energy is not needed.) Offer an explanation for your result. O H2 C CH2 + CH2 H3 C O CH3 H3 C + CH3 CH2 H2 C CH2 Repeat the calculation for hydrogenation of aziridine (leading to dimethylamine) relative to hydrogenation of cyclopropane. NH H2 C CH2 + CH2 H3 C NH CH3 H3 C CH3 + CH2 H2 C CH2 Which is more strained, aziridine or oxirane? Suggest an explanation for your result. Strain Energies of Small-Ring Cycloalkanes: As discussed above, cyclopropane is a highly strained molecule whereas cyclohexane is not. Is the strain energy of cyclobutane closer to that of cyclopropane or cyclohexane? Of cyclopentane? Is cycloheptane more strained than cyclohexane? To assess the strain in cyclopropane, cyclobutane, cyclopentane and cycloheptane relative to that in cyclohexane, obtain energies of reactions with n = 1, 2, 3 and 5. Use the HF/6-31G* model. (Energies for cyclopropane and propane are available if you completed the previous problem.) H2C CH2 CH3 CH3 CH3 H2C CH3 + CH2 + (CH2 )4 (CH2 )n (CH2 )n (CH2 )4 Cycloalkynes: Because alkynes exhibit a linear or nearly linear C-C≡C-C unit, it is unlikely that they will be able to incorporate into small rings. Identify the smallest cycloalkyne where both C-C≡C bond angles are within 10° of being linear (Start with cyclohexyne and minimize with molecular mechanics. Use the B3LYP/6-31G* model to calculate the equilibrium geometry of this cycloalkyne and cycloalkynes with one and two fewer carbons. Also obtain geometries of the corresponding cycloalkenes as well as hydrogen molecule. Calculate energies of the hydrogenation reaction. (CH2 )n C (CH2 )n H2 C C H n = 4–9 C H Is there a relationship between C-C≡C bond angle and hydrogenation energy for the three cycloalkynes? Strain Energies in Polycyclic Alkanes: Polycyclic cycloalkanes are likely to be even more strained than cycloalkanes. An extreme example of a strained molecule is tetrahedrane. While the parent compound, C4H4, has yet to be experimentally characterized, X-ray crystal structures exist for several simple derivatives, and appear to 47 be quite “normal”. For example, tetra-tert-butyltetrahedrane, shows carbon-carbon bond lengths of 1.49Ǻ, virtually identical with those found in cylopropane (1.50Ǻ). Use the B3LYP/6-31G* model to calculate equilibrium geometries for all molecules involved in hydrogenation of tetrahedrane, first to bicyclo[1.1.0]butane and then to cyclobutane. H2 H2 Cubane: It may come as a surprise that despite what appears to be a highly-strained carbon skeleton, derivatives of cubane are rather common (the crystal structures of well over a hundred of them have been reported). Use the B3LYP/6-31G* model to obtain equilibrium geometries for cubane and the first and second hydrogenation products (the former is unambiguous (all twelve CC bonds are the same), but the second is not. Assume the structure shown below. H2 H2 Inter and Intramolecular Hydrogen Bonding Discussion to be written The Enol of 1,3-Malonaldehyde: The enols of most carbonyl compounds are higher in energy than the keto forms. A possible exception arises in dicarbonyl compounds such as 1,3-malonaldehyde where there is the possibility of intramolecular hydrogen bonding. O O O H O H O O Use the B3LYP/6-31G* model to obtain equilibrium geometries for malonaldehyde and for cis-3-hydroxyacrolein. Start with a conformation of the latter that allows a hydrogen bond to be formed. Also obtain the geometry of trans-3-hydroxyacrolein (a molecule that cannot form an intramolecular hydrogen bond). Is cis-3-hydroxyacrolein lower in energy than malonaldehyde? If so, does this appear to be due to the hydrogen bond (compare the energies of cis and trans-3-hydroxyacrolein to tell)? Problem (using canned files) on hydrogen-bonding in base pairs and related systems. 48 H3C CH3 N O O N H H3C H N O N H O H2N N N H N N N CH3 O H N N CH3 N H N CH3 49 N N H H H HN O N H N N H O N N N CH 3 N NH H N N O H H HN N CH3 NH2 N N N N H CH3 N O N N CH3 N N N H HN O N H3 C O C H3 C H3 N N O Metal-Ligand Binding in Transition-Metal Organometallics Discussion to be written Metal-Ligand Binding Energies in Iron Carbonyls: Both ethylene and acetylene bind to iron tetracarbonyl to form stable organometallics. Obtain equilibrium geometries for these two compounds as well as for their component fragments (ethylene, acetylene and iron tetracarbonyl), and calculate metal-ligand binding energies. Use the B3LYP/6-31G* model. Which is more tightly bound? Is your result consistent with changes to the C=C and C≡C bonds in ethylene and acetylene that result from binding to the metal carbonyl? Elaborate. Perform analogous calculations on butadiene iron tricarbonyl and cyclobutadiene iron tricarbonyl as well as 1,3-butadiene, cyclobutadiene and iron tricarbonyl. Which ligand is more tightly bond to iron tricarbonyl, butadiene iron or cyclobutadiene? Is this result in line with the relative bond length changes in the two dienes? Elaborate. Metal-Ligand Binding Energy in Benzene Chromium Tricarbonyl: Stable πcomplexes between chromium tricarbonyl and arenes are common and are sufficiently stable to allow “chemistry” to be performed on the ligand. Cr OC CO CO Use the B3LYP/6-31G* model to obtain equilibrium geometries for benzene chromium tricarbonyl as well as for benzene and chromium tricarbonyl. Is the metal-ligand binding energy of the same order of magnitude as a normal covalent bond (250-400 kJ/mol), or of the same order of magnitude as a hydrogen bond (20-30 kJ/mol) or somewhere in between? Compare the binding energy in benzene chromium tricarbonyl to those in complexes considered in the previous problem. Does benzene appear to be a better (more tightly bound) ligand than ethylene or acetylene? Than butadiene or cyclobutadiene? Rationalize you results. Tebbe Reagent: Early transition metal carbenes, LnM=CRR’, catalyze a variety of related processes involving olefins, including metathesis and polymerization. The Tebbe reagent, Cp2Ti=CH2 Me2AlCl, exhibits similar behavior suggesting that it is nothing more than a weak complex between a titanium methylidene and a Lewis acid. 50 Use the B3LYP/6-31G* model to establish the equilibrium geometry for the Tebbe reagent as well as for its methylidene and Lewis acid components, and calculate the energy of complexation. is both an excellent olefin metatesis catalyst for olefin 51 Using Approximate Equilibrium Geometries Is it actually necessary to obtain geometries with the same theoretical model required to accurately calculate reaction energies, or are geometries from a simpler (lower computation cost) model sufficient? This is an important question because determining equilibrium geometry may easily require one or two orders of magnitude more computation than calculating energy at a fixed geometry, due to the need of multiple calculations at different geometries. It is also straightforward to answer. Table P3-6 compares energies for structural isomers obtained from B3LYP/6-311+G** calculations using four different sets of geometries: 3-21G, 6-31G*, B3LYP/6-31G* and “exact” (B3LYP/6-311+G**). Also provided are reference isomer energies obtained from G3(MP2) calculations. The conclusion is clear. The magnitude of the error introduced as a result of using approximate geometries is insignificant in comparison with that inherent to the underlying B3LYP/6-311+G** model. Even the HF/3-21G model provides satisfactory geometries. Some properties will be more sensitive to small changes in geometry, for example, the dipole moment. Also, the infrared spectrum of a molecule (discussed in Chapter P2) requires that the geometry be calculated using the same theoretical model. However, for reaction energy calculations, the savings in computation cost achieved by the use of approximate geometries more than offsets any small changes in results. 52 Table P3-6: Effect of Use of Approximate Geometries on the Energies of Structural Isomers from B3LYP/6-311+G** (kJ/mol) 3-21G reference propyne 6-31G* B3LYP/6-31G* B3LYP/6-311+G** G3(MP2) isomer allene -7 -7 -7 -7 3 cyclopropene 102 101 101 101 102 cyclopropane 38 38 38 38 38 2-butyne 36 37 37 37 3 bicyclo[1.1.0]butane 132 130 130 130 119 cyclobutane 47 46 47 47 48 1,4-pentadiene 63 63 63 63 65 methyl isocyanide 100 100 100 102 102 vinyl alcohol 45 43 43 43 43 oxirane 124 122 121 121 116 ethanol dimethyl ether 48 48 47 47 49 acetic acid methyl formate 69 69 68 68 70 methyl vinyl ketone cyclobutanone 24 23 24 24 23 divinyl ether 101 98 98 98 102 1 0 0 - - propene 1,3-butadiene isobutene cyclopentene acetonitrile acetaldehyde mean absolute deviation from “exact” mean absolute error from G3(MP2) - 53 Silaolefins: Only a very few compounds incorporating a carbon-silicon double bond (“silaolefins”) have been characterized, among them compounds 1-4. On the contrary, stable compounds with silicon-carbon single bonds (“silanes”) are common. There is indirect evidence that suggests that the lack of silaolefins is due at least in part to the high reactivity of silicon-carbon double bonds. Specifically, all of the compounds that do exist have large bulky groups shielding the double bond, thereby making it difficult for reagents to approach. Me SiMe3 SiMe(Cme3)2 Si M e3Si M e3Si C Me OSiMe3 Si (1-adamantyl) 1 2 CMe3 (CMe3)Me2Si Si C (CMe3)Me2Si C C M e3Si Si C CMe3 3 C (CMe3)Me2Si 4 We examine the kinetic reactivity of silaolefins elsewhere in this text. For now we, probe their thermochemical stability, specifically the thermochemistry of hydrogenation relative to that of normal olefins. Obtain equilibrium geometries for 1,1-dimethylsilaethylene (Me2Si=CH2) and isobutene, as well as their hydrogenation products, trimethylsilane and isobutene. Use the B3LYP/6-31G* model. Evaluate the energy of the reaction comparing hydrogenation energies for 1,1-dimethylsilaethylene and isobutene. Me2Si=CH2 + Me3CH → Me3SiH + Me2C=CH2 What does it tell you about the strength of the SiC double bond in the silaolefin relative to the CC double-bond strength in isobutene? Repeat your calculations for the germanium analogue of isobutene, that is, evaluate the energy of the reaction. Me2Ge=CH2 + Me3CH → Me3GeH + Me2C=CH2 Do you conclude silicon or germanium forms the stronger π bond to carbon? Elaborate any assumptions that you have made to reach this conclusion. Finally, repeat both sets of calculations using geometries obtained from the HF/3-21G model (in lieu of the B3LYP/6-31G* model), following these with energy calculations with the B3LYP/model. Have the results changed significantly. Estimate the computation cost of this approach relative to that employing “exact” geometries. Incorporating Atoms and Molecules into BuckyBall: The cavity inside Buckyball, aka, Buckmeister fullerene, fullerene of C60), is not large enough to incorporate anything but atoms and very small molecules (atomic inclusion compounds are common but only a few examples of molecular complexes have been reported). 54 Host-guest complexes need to be formed by “assembling” the fullerene in the presence the atoms/molecules that are to be incorporated, for example, a helium complex needs to be formed by assembling it helium gas. The fullerene “mesh” is very tight, and entrapped atoms and molecules cannot escape. Therefore, it is not clear if complex formation is thermodynamically driven force or whether complexes exist because fullerene does not provide an exit for its guest. Quantum chemical calculations may be employed to decide. These are large molecules and “full” treatments with models capable of producing reliable binding energies may be prohibitive in terms of computer time. An alternative is to obtain equilibrium geometries with a simple model and follow these with energy calculations with a better (and computationally more expensive) model. In this example, the HF/3-21G model serves the first role and the B3LYP/6-31G* model the second role. Even so, calculations of this magnitude are not suitable as “homework exercises”, and the results have been provided to examine and ponder. No quantum chemical calculations are required. Results for C60 and for helium, neon, hydrogen molecule, methane and water complexes of C60 obtained form the B3LYP/6-31G* model based on HF/3-21G equilibrium geometries are provided in C60 and C70 complexes. Also provided results for C70 and for the C70-hydrogen molecule complex. Calculate binding energies for the five C60 complexes. fullerene (host) + guest host-guest complex Which if any of the complexation reactions are thermodynamically driven? Are your results consistent with electrostatic potential maps for the inside surface of fullerene examined in Chapter P1? Elaborate. Calculate the binding energy of hydrogen molecule into C70. Is the guest more or less inclined thermodynamically to associate with the inside of the larger C70 host than it is with the smaller C60 host? Finally, calculate binding energies for hydrogen molecule inside both C60 and C70 cages obtained from B3LYP/6-311+G** energies again assuming HF/3-21G equilibrium geometries. Do you see a marked change either in the absolute numbers and/or in the difference between binding energies for the smaller and larger hosts from you previous results? Elaborate. Results for both C60 and C70 and together with their complexes with hydrogen molecule, obtained form the B3LYP/6-311+G** model based on HF/3-21G equilibrium geometries are provided in C60 and C70 hydrogen molecule complexes. 55 56 Thermochemical Recipes This chapter has focused on the calculation of energies of diverse chemical reactions. One lesson we have learned is that reactions that involve explicit bond making or breaking require “better” quantum chemical models than reactions that maintain bonding and only change local environment. Another lesson is that even the “best” of the models that are presently routinely applicable to molecules of moderate size may lead to errors of unacceptable magnitude. It would certainly be desirable to have a single practical model that would be able to provide energies for any reaction to within 4-8 kJ/mol and be applicable to molecules with molecular weights up to 400 amu. One possibility is the G3(MP2) method that we have used throughout this chapter to supply reference data (in lieu of experimental data). In practice, it is applicable only to very small molecules (with molecular weights below 150 amu). G3(MP2) heats of formation for several hundred small molecules have been compared with experimental values contained in the NIST thermochemical database. This leads to a mean absolute error of 6.2 kJ/mol and an RMS error of 8.3 kJ/mol. G3(MP2) actually involve a combination of several different quantum chemical models, and perhaps is best referred to as a recipe. The first step is to obtain an equilibrium geometry using the MP2/6-31G* model. Next, two energy calculations are performed at this geometry, one an MP2 calculation with the G3MP2large basis set an the other a QCISD(T) calculation with the 6-31G* basis set. The energy differences between the MP2 calculation with the two basis sets and the QCISD(T) and MP2 calculations with the same basis set are summed. This assumes that the changes in energy due to increase in the size of the basis set from 6-31G* to the G3MP2large and improvement in treatment of electron correlation from the MP2 to the QCISD(T) are independent. This assumption is the reason that G3(MP2) is as widely applicable as it is. Were it necessary to perform a QCISD(T) calculation (which scales as the seventh power of the number of basis functions) with the G3MP2large basis set, the procedure would be much more limited. The fact that G3(MP2) heats of formation obtained are in reasonable accord with experimental values validates the assumption. The final step in the recipe involves a HF/6-31G* frequency calculation (preceded by a HF/6-31G* geometry). This allows for calculation of the zero-point energy and for correction for finite temperature. 57 A new recipe, designated T1, has been formulated that requires significantly less computation than G3(MP2), but yields heats of formation that are nearly identical. Both the QCISD(T) and Hartree-Fock frequency calculations have been eliminated and the MP2/G3MP2large calculation has been replaced by a RI-MP2 calculation in which the G3MP2large basis set is approximated using so-called dual basis set techniques. A HF/6-31G* geometry replaces the MP2/6-31G* geometry. Without further modification, these changes to G3(MP2) result in heats of formation that are not sufficiently accurate to be useful in thermochemical calculations. A successful recipe follows by introducing a total of 66 linear regression parameters based on Mulliken bond orders calculated from the Hartree-Fock wavefunction. These have been determined using a training set of more than 1100 G3(MP2) heats of formation. A plot of T1 vs. G3(MP2) heats of formation for this set is provided in Figure P3-3. Table P3-7 compares structural isomer energies obtained from T1 heats of formation for a variety of simple systems with those obtained both from the G3(MP2) recipe and with experimental heats. The mean absolute and rms errors is 6.6 and 9.0 kJ/mol, roughly the same as those for the G3(MP2) recipe. As discussed earlier in this chapter, while the energy of a molecule is one of its most fundamental properties, it is only rarely determined. This means that quantum chemical models (or combinations of quantum chemical models) that were able to provide routine and reliable energies (heats of formation) are of significant value. 58 Figure P3-3: Comparison of Heats of Formation Obtained from the T1 and G3(MP2) Recipes (kJ/mol) 59 Table P3-7: Comparison of Energies of Structural Isomers from T1 and G3(MP2) Recipes with Experimental Values (kJ/mol) formula (reference) isomer T1 G3(MP2) expt. C2H3 N (acetonitrile) methyl isocyanide 102 100 88 C2H4 O (acetaldehyde) vinyl alcohol oxirane 41 113 41 115 43 118 C2H4 O2 (acetic acid) methyl formate 68 70 75 C2H6 O (ethanol) dimethyl ether 49 50 51 C3H4 (propyne) allene cyclopropene -3 91 1 100 7 93 C3H6 (propene) cyclopropane 36 38 29 C4H6 (1,3-butadiene) 2-butyne cyclobutene 1,2-butadiene 1-butyne methylenecyclopropane bicyclo[1.1.0]butane 35 50 43 56 80 113 37 56 51 58 85 120 36 48 53 56 92 108 2-methyl-2-propenal cyclobutanone 2-hydroxy-1,3-butadiene 2,3-dihydrofuran divinyl ether 7 4 25 48 35 96 6 3 23 38 35 97 5 9 23 38 43 102 C4H8 (2-methylpropene) trans-2-butene 1-butene cyclobutane 4 15 46 6 16 47 7 17 46 C5H8 (cyclopentene) 2-methyl-1,3-butadiene methylenecyclobutane 1,4-pentadiene 1,1-dimethylallene 1,2-pentadiene 42 89 69 86 98 37 87 67 90 101 40 86 70 93 105 5 4 - C4H6 O (methyl vinyl ketone) trans-2-butenal mean absolute error 60 The version of Spartan provded with this textbook (Spartan Student) does not provide for calculations using the T1 thermochemical recipe. However, the associated Spartan Molecular Database includes T1 data for all the molecules required for the problems that follow. Double-Bond Disproportionation Reactions: A double-bond disproportionation reaction relates the energy of a molecule incorporating a double bond with the average of the energies of molecules incorporating the corresponding single and triple bonds. For example, the double-bond disproportionation reaction for ethylene relates its energy to the average of the energies of ethane and acetylene. Experimentally, this reaction is endothermic by 38 kJ/mol, meaning that a CC double bond is weaker than the average of carbon-carbon single and triple bonds. 2 H2C=CH2 → H3C-CH3 + HC≡CH Use the T1 recipe to calculate the energies of double-bond disproportionation reactions for ethylene, methyleneimine (H2C=NH), formaldehyde (H2C=O) and thioformaldehyde (H2C=S). (Build the molecules as you normally would and simply retrieve from the Spartan Molecular Database.) Adjust your results for the last three to account for the error in the reaction of ethylene. Note any significant differences among the four disproportionation energies and try to rationalize them. 61 Calculation of Entropy and Gibbs Energy The change in the Gibbs energy (ΔG) for a chemical reaction follows from the change enthalpy (ΔH) by way of the usual thermodynamic relationship. ΔG = ΔH - TΔS T is the temperature in K and ΔS is the change in entropy. The entropy is a sum of translational, rotational and vibrational parts. S = Str + Srot + Svib The translational part, which depends on the total molecular mass, M, and pressure, P (n is the number of moles, and R and k are the gas and Boltzmann constants), cancels in a mass-balanced equation. 3 Str = nR 2 2!MkT + ln 3/ 2 2 nRT P The rotational part depends on the principal moments of inertia, IA, IB and IC, which in turn depend on the geometry. s is termed the symmetry number which is usually unity. 3 Srot = nR 2 ( !vA vBvC )1/2 + ln s vA = h2/8!IA kT, vB = h2 /8!IB kT, vC = h2/8!IC kT The vibrational part depends on the frequencies. It is based on the linear harmonic oscillator approximation, and incorrectly goes to ∞ and not ½ RT as the frequency goes to 0. It needs to be adjusted to approach the proper limit. Svib = nR ! i (uieui – 1)–1 – ln (1 – e –ui ) µi = h!i/kT 62 ...
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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 Berkeley.

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