Molecular Modeling Problems

Molecular Modeling Problems

Part A. Structure and Vibrational Spectra 1. Using Vibrational Frequencies to Verify an Equilibrium Structure. Each of the frequencies that make up the vibrational (infrared/Raman) spectrum of a molecule is proportional to the square root of the ratio of the curvature of the potential surface in the vicinity of the structure (the second derivative of the energy) and a mass. As mass is positive, this means that the resulting frequency will be a real number only if the surface curves upward (the second derivative is positive). Therefore, the presence of one or more imaginary frequencies in a calculated spectrum indicates that the structure is not an energy minimum. Use the HF/6-31G* model to obtain equilibrium geometries and vibrational spectra for the chair and boat forms of cyclohexane. Are all frequencies real for both molecules? Which if either molecule is not a minima on the energy surface? Elaborate.

2. Equilibrium Geometry of Disilylene (H2Si=SiH2). As discussed in the previous problem, a structure that gives rise to a vibrational spectrum with an imaginary frequency is not an energy minimum. However, a minimum energy structure can be located from such a structure by “walking along” the geometrical coordinate associated with the imaginary frequency.

Use the B3LYP/6-31G* model to obtain the equilibrium geometry of planar (like ethylene) disilylene, the simplest molecule with a silicon-silicon double bond, and calculate the vibrational spectrum. Are all frequencies real numbers? If they are not, perform the following operations: First, animate any imaginary frequencies to see how disilylene wants to distort to move it toward an energy minimum. Next, distort your structure accordingly. Finally, reoptimize the geometry (of the distorted molecule) and again calculate vibrational frequencies. Is it now an energy minimum? If not, repeat the process until you get a minimum energy structure. Describe the equilibrium geometry of disilylene. 3. Limiting Hartree-Fock Bond Lengths. The Hartree-Fock model almost always yields bond lengths that are shorter than experimental distances. This may be explained by noting that improvements to the Hartree-Fock model involve promotion of electrons from molecular orbitals that are occupied to orbitals that are unoccupied. Occupied orbitals are typically either bonding or non-bonding whereas unoccupied orbitals are typically antibonding. Therefore, electron promotion not only leads to a better description but also presumably to bond lengthening. This in turn suggests that Hartree-Fock bond distances are too short. Compare the experimental bond length for hydrogen fluoride (0.917Ǻ) with that obtained from the Hartree-Fock model using the cc-pVTZ basis set. This is a sufficiently large basis set to approximate the limit of the Hartree-Fock model for equilibrium geometries. Is the calculated distance shorter than the experimental value?

To accompany Physical Chemistry, Ninth Edition by Atkins and De Paula


Repeat your calculations for lithium hydride (the experimental bond length is 1.596Ǻ). Is the Hartree-Fock distance shorter than the experimental length? If it is not, provide an explanation why not. Hint: examine both highest-occupied and lowest-unoccupied molecular orbitals. 4. Bond Lengths in Cyclopropane and Cyclobutane. Use the Hartree-Fock model with the 6-31G* basis set to obtain equilibrium geometries for propane and cyclobutane. Are the carbon-carbon bond lengths in the cycloalkane shorter, longer or about the same as those in propane (the “standard)? Is what you find consistent with the fact that CH bonds on adjacent CH2 groups nearly eclipse each other leading to an increase in non-bonded repulsion? Elaborate. Given your result for cyclobutane, what you expect the carbon-carbon bond lengths in cyclopropane to be? Is your expectation supported by the calculated equilibrium structrure? If it is not, provide a plausible explanation as to why not.



5. Water Dimer. The water dimer exhibits a structure with a single hydrogen bond. While the H --- O distance is not known experimentally, the distance between the two oxygen atoms has been measured as 2.98Ǻ.

O H OHH

H

The Hartree-Fock model with the very large cc-pVQZ basis set shows a similar overall geometry but with an OO distance that is 0.05Ǻ longer than the experimental value. This suggests that electron correlation shortens the hydrogen bond, the opposite of what is normally observed for covalent bonds. To model the effect of correlation on the geometry of water dimer, consider the consequences of promoting electrons from filled to empty molecular orbitals, most simply from the highest-occupied to the lowest-unoccupied molecular orbital. Obtain the geometry of water dimer using the Hartree-Fock/6-31G* model, and display the highest-occupied and lowest-unoccupied molecular orbitals. Is the HOMO bonding, antibonding or non-bonding with respect to the H --- O hydrogen bond? Would loss of an electron lead to shortening or lengthening of the hydrogen bond or would it be expected to have little effect? Is the LUMO bonding, antibonding or non-bonding with respect to the H --- O hydrogen bond? Would gain of an electron lead to shortening or lengthening of the hydrogen bond or would it be expected to have little effect? Overall, would you expect the limiting Hartree-Fock hydrogen bond distance in water dimer to be shorter or longer than or unchanged from the experimental value? 6. Alternative Structure of the Water Dimer. As detailed in the previous problem, water forms a dimer with a single hydrogen bond. However, because water incorporates two electron pairs that may act as hydrogen bond acceptors and two OH bonds that may act as hydrogen-bond donors, it should be possible to construct an alternative dimer with two hydrogen bonds. Does such a structure actually exist?

HO

HO

HH

Attempt to obtain an equilibrium geometry for the doubly hydrogen-bonded structure of water dimer. Use the B3LYP/6-31G* model. If you find such a structure, confirm that it is or is not an energy minimum. If it is an energy minimum, calculate the room-temperature Boltzmann distribution of the two different forms of water dimer. 7. Acetic Acid Dimer. Acetic acid forms a dimer with two equivalent hydrogen-bonds. Has the geometry of acetic acid remained largely unaffected by hydrogen bonding (as suggested in the drawing below), or have significant structural changes occurred?



OCH3C

O H

OC CH3

O

H Use the B3LYP/6-31G* model to calculate equilibrium structures for acetic acid and its dimer. Point out any significant changes in bond lengths between the two. Does the dimer incorporate distinct CO single and double bonds? Have the hydrogen atoms involved in the hydrogen bonds moved to positions halfway between the oxygen atoms, or has each hydrogen atom remained with a single oxygen atom? Do the structural changes (or lack of structural changes) suggest that hydrogen bonds are comparable in strength to normal (covalent) bonds or that they are weaker? Elaborate. Repeat your calculations for trifluoroacetic acid and its dimer. Are geometry changes smaller, larger or about the same as those seen for acetic acid? Is the dimerization energy for trifluoroacetic acid smaller, larger or about the same as that for acetic acid? If the two are different, is the acid with the larger dimerization energy also that with exhibits the larger changes in structure? 8. Infrared Spectrum of Acetic Acid Dimer. Use the B3LYP/6-31G* model to calculate infrared spectra for acetic acid and its dimer. (You already have equilibrium structures if you completed the previous problem.)

OCH3C

O H

OC CH3

O

H Point out and rationalize any significant differences between the frequencies and/or intensities associated with the (two) OH stretching motions in the dimer and the OH stretching frequency in acetic acid. 9. Radical Cation of Diborane. One of the valence molecular orbitals of diborane closely resembles the π orbital in ethylene.

Removal of an electron from this orbital should increase the separation of the boron atoms, just as removal of an electron from the π orbital in ethylene should lengthen the CC bond. Use the B3LYP/6-31G* model to obtain geometries for diborane and its radical cation. Make certain that you start with a distorted (C1 summetry) structure for the radical cation. Also, obtain vibrational frequencies (infrared spectrum) for the radical cation to be certain that it is an energy minimum. Does ionization lead to the “anticipated” increase



in distance between the two boron atoms? If it does not, then explain why not. Hint: Is the molecular orbital in diborane that resembles the π orbital in ethylene the HOMO?

10. Limitations of VSEPR Theory. VSEPR (Valence State Electron Pair Repulsion) theory uses two simple rules to assign geometry. First, the geometry about an atom follows by insisting that electron pairs (bonds or lone pairs) are as far apart as possible. Second, it is more important to separate two lone pairs than it is to separate a lone pair from a bond than it is to separate two bonds. Taken together, these two rules properly account for the observed “see-saw” structure of sulfur tetrafluoride, SF4. Sulfur atom surrounded by five electron pairs (four bonds and a lone pair) assumes a trigonal bipyramidal geometry, and the lone pair prefers an equatorial position. What VSEPR theory does not tell us is whether this is the only structure of SF4, in particular, whether a trigonal pyramid structure (which obeys the first rule but violates the second) might also be an energy minimum.

F

S

F

FF S F

FFF

•• • •

"see saw" trigonal pyramid Use the HF/6-31G* model to obtain geometries for both forms of SF4. Start with C2v symmetry for the see-saw structure and C3v symmetry for the trigonal pyramid structure. Are both structures energy minima or is there only one structure? Detail your reasoning. If there is only one structure, is it the observed structure? If there are two structures, is the observed structure favored? Is the higher-energy structure likely to be observed at room temperature? (Assume that it must make up at least 5% of the mixture to be detected.)

11. Failures of VSEPR Theory. The crystal structure of CaF2 shows that the molecule is bent, at odds with the prediction of VSEPR theory which assigns a linear geometry. Is this a failure of VSEPR theory or is it due to crystal packing? Use the B3LYP/6-31G* model to obtain the equilibrium geometry of CaF2. Start with a bent structure, Does this collapse to linear geometry or does the molecule remain bent?

If "free" CaF2 is linear, calculate the geometry for a molecule with a FCaF bond angle constrained to 140o, and compare its energy to that of linear CaF2. What does the energy difference tell you about the magnitude of the crystal packing energy? If on the other hand, "free" CaF2 is bent, calculate the geometry of "linear" CaF2 and compare its energy to that of the bent molecule.

12. Geometry Changes with Change in the Number of Electrons. The geometry of a molecule depends not only on the constituent atoms, but also on the total number of electrons. Use the HF/6-31G* model to obtain equilibrium geometry for 2-methyl-2-propyl cation (tert-butyl cation), as well as those for the corresponding radical (with one



additional electron) and the anion (with two additional electrons). Describe any changes to the geometry of the central carbon with increasing number of valence electrons, and speculate on the origin of these changes. Hint: examine the lowest-unoccupied molecular orbital (the next orbital to be occupied) in tert-butyl cation. 13. Protonated Ethylene. What is the geometry of protonated ethylene? Is the proton primarily associated with a single carbon does it “bridge” both carbons?

C C HH

H

H

H

H+

C C HHHH

+

Use the B3LYP/6-31G* model to calculate equilibrium geometries for both open and bridged forms of protonated ethylene. Do both structures appear to be minima on the C2H5

+ potential surface or does one of the structures “collapse” to the other? Elaborate. If there is only one energy minimum, is it open or bridged? 14. Protonated Methane. The structure of the ion resulting from protonation of a molecule with a non-bonded electron pair should be similar to the structure of the corresponding isoelectronic neutral molecule, that is, the neutral molecule with the same number of electrons. For example, protonated ammonia should be tetrahedral by analogy with methane. There are, however, no known (characterized) neutral analogues of ions resulting from protonation of alkanes. Here, all the valence electrons are already tied up in σ bonds, and any new bonds must be made at the expense of these existing bonds. Nevertheless, protonated alkanes are observed by mass spectrometry. What do they look like? Use the B3LYP/6-31G* model to explore possible structures for protonated methane (CH5

+). Calculate vibrational frequencies for whatever you uncover to verify it is actually and energy minimum. Describe the bonding in terms of a weak complex or a molecule with a pentavalent carbon. If it is a complex, identify the components and calculate the binding energy. The gas-phase proton affinity of methane is 544 kJ/mol, which is significantly less than the proton affinity of water (691 kJ/mol), but larger than the proton affinity of N2 (494 kJ/mol). What is the calculated proton affinity? (Note that the energy of a proton is 0.) Repeat you calculations and analysis for protonated ethane (C2H7

+). 15. Structure of Diazomethane. Is there a single Lewis structure for diazomethane or is a composite of two structures required for proper description?



N N N– ++ –

N

To decide, compare the geometry of diazomethane obtained from the HF/6-31G* model with those of methylamine, CH3NH2, and methyleneimine, H2C=NH, molecules incorporating normal CN single and double bonds, respectively, and those of trans diimide, HN=NH, and nitrogen, N≡N, molecules incorporating normal NN double and triple bonds, respectively. If two Lewis structures are required, which appears to be the more important?

16. Structure of Ozone. Ozone, O3, leads a proverbial double life. In the upper atmosphere, it protects the earth and its inhabitants from harmful UV radiation, while nearer to the surface it is a serious pollutant contributing to respiratory problems. Draw two different Lewis structures for ozone, one or both of which may require non-zero formal charges. Attempt to obtain equilibrium geometries corresponding to both structures using the B3LYP/6-31G* model. Verify your results by obtaining infrared spectra. If you do obtain two different structures, identify which structure is lower in energy. Is it in accord with the experimentally known equilibrium geometry? If the preferred structure has more than one distinct oxygen atom, which is most positively charged? Most negatively charged? Is your result based on electrostatic charges consistent with that based on formal charges (in the Lewis structure)? 17. Hydrated Hydronuim Cation. Use the B3LYP/6-31G* model to calculate the equilibrium geometry of hydronium cation, H3O+. Next, calculate the structure of a complex between H3O+ and four water molecules. Point out any significant changes that have occurred to the cation a result of interaction with water. In particular, is there evidence for “sharing” the proton? Has the positive charge remained localized on the hydronium cation or has it spread out to the surrounding water molecules? 18. Hydrated Chloride Anion. Use the B3LYP/6-31G* model to calculate the equilibrium geometry for chloride anion surrounded by four water molecules. Has the negative charge remained on the chlorine or has it spread out to the water molecules? 19. Borane Carbonyl. Borane carbonyl, BH3CO, results from interaction of the non-bonded electron pair (on carbon) in carbon monoxide and an empty p-type orbital in borane.

B C OH

HH

Unlike transition-metal carbonyl complexes (see following problem), there is little possibility for significant back bonding, that is, interaction of a high-energy filled molecular orbital on borane with an empty π* orbital on carbon monoxide. This suggests



that neither the CO bond length nor the vibrational frequency of carbon monoxide is likely to change significantly as a result of complexion. Use the B3LYP/6-31G* model to obtain equilibrium geometries and infrared spectra for both carbon monoxide and borane carbonyl. Is there a significant change in CO bond length? Identify the frequency in borane carbonyl corresponding to the CO stretch. Is it smaller, larger or about the same as that in free carbon monoxide?

20. Greenhouse Gases. In order to dissipate the energy that falls on it due to the sun, the earth “radiates” as a so-called “blackbody” into the universe. The “theoretical curve” is a smooth distribution peaking around 900 cm-1 and decaying to nearly zero around 1500 cm-1. This is in the infrared, meaning that some of the radiation will be intercepted by molecules in the earth’s gaseous atmosphere. This in turn means that the earth is actually warmer than it would be were it not to have an atmosphere. This warming is known as the greenhouse effect, to make the analogy between the earth’s atmosphere and the glass of a greenhouse. Both allow energy in and both impede its release. The actual distribution of radiated energy as measured from outside the earth’s atmosphere in the range of 500-1500 cm-1 is given below. The overall profile matches that for a blackbody, but the curve is peppered with holes.

Neither nitrogen nor oxygen, which together comprise 99% of the earth’s atmosphere absorbs in the infrared and causes the “holes”. However, several “minor” atmospheric components, carbon dioxide most important among them, absorb in the infrared and contribute directly to greenhouse warming. Its infrared spectrum shows a strong absorption in the region centering 670 cm-1, the location of the most conspicuous hole in the blackbody radiation profile.



Identify three of the top ten chemicals manufactured worldwide. Use the B3LYP/6-31G* model to calculate the infrared spectra for each and comment whether or not you would expect it to be a significant greenhouse gas.



21. Comparison of Measured and Calculated Frequencies. Frequencies routinely obtained from quantum chemical calculations assume that the potential in the vicinity of the minimum is a quadratic function. This means that they are generally larger than measured frequencies. (The correct potential for say a bond stretching motion goes to zero whereas a quadratic potential goes to ∞. It is possible to correct calculated frequencies for cubic and higher-order terms, but this is generally prohibitive in terms of computer time. It is also possible to correct measured frequencies for anharmonic behavior, leading so-called “harmonic frequencies”. These exist only for diatomic and very small polyatomic molecules, and include: LiF, 914; CO, 2170; N2, 2360 and F2, 923 (all in cm-1). Use the HF/6-311+G** model to calculate equilibrium geometries and vibrational frequencies for LiF (914), CO (2170), N2 (2360) and F2 (923). (Harmonic frequencies in cm-1 are given in parentheses.) What is the average percentage error between calculated and measured (harmonic) frequencies? Are calculated frequencies uniformly too small or too large? Repeat your calculations using the B3LYP/6-311+G** model. What is the average percentage error between calculated and measured (harmonic) frequencies? Is this smaller than the average error for Hartree-Fock calculations? Are calculated frequencies uniformly too small or too large?

22. Methyl Lithium: The structure of methyl lithium appears to be “normal” insofar as it incorporates a carbon with a roughly tetrahedral geometry. This could either mean that the C-Li bond is covalent, or that lithium cation is loosely associated with methyl anion (which also incorporates a tetrahedral carbon). Which of these descriptions is more consistent with atomic charges obtained from the HF/6-31G* model? Elaborate.

Experimentally, methyl lithium exists most simply as a tetramer with the four lithium atoms and four methyl groups at the corners of a cube.

Use the HF/6-31G* model to calculate the geometry of methyl lithium tetramer. Does the conclusion you reached about the nature of the bonding in methyl lithium maintain?

23. Lithium Aluminum Hydride. Lithium aluminum hydride (LiAlH4) is among the most important reducing agents, that is, source of “hydride”. Is it best described as an ion pair between lithium cation and aluminum hydride anion (AlH4

-), a weak complex between lithium hydride (LiH) and aluminum hydride (AlH3), or are one or more hydrogen atoms “shared” by the two metals? To decide, obtain the equilibrium geometry using the B3LYP/6-31G* model. Are the calculated atomic charges consistent with your




geometry? Elaborate. Is there more than one stable structure for lithium aluminum hydride? Start from one or more structures that are different from what you have found and see if you get a different result.

24. Sodium Cyclopentadienide: Is sodium cyclopentadienide accurately represnted in terms of a structure in which sodium cation associates with cyclopentadienyl anion?

Na+

Obtain the equilibrium geometry of sodium cyclopentadienide using the B3LYP/6-31G* model. Is the charge on sodium close to unity? Calculate the equilibrium geometry of cyclopentadienyl anion and compare it with the cyclopentadienyl fragment in sodium cyclopentadide. Is what you find indicative of a weak complex? Elaborate. Calculate the energy of dissociation to cyclopentadienyl anion and sodium cation (you need to obtain the energy for sodium cation). Is the binding energy weak (< 100 kJ/mol), comparable to that of a normal covalent bond (~400 kJ/mol) or somewhere in between? Is your result consistent with the calculated charges and geometries? 25. Proton NMR Spectrum of Cyclohexane. At very low temperatures, the proton NMR spectrum of cyclohexane shows two equal intensity lines at 1.12 and 1.60 ppm. Use the HF/6-31G* model to obtain the equilibrium geometry and the proton chemical shifts. Which line corresponds to the equatorial protons and which corresponds to the axial protons?


Part B. Molecular Stabilities and Reaction Energies 1. Pressure Affects Reaction Enthalpy. The heat (enthalpy) and energy of a chemical reaction are not the same, but are related through a pressure-volume term.

∆H = ∆E+ P∆V at constant pressure However, except for very high pressures, this term is very small and the energy and enthalpy are nearly identical.

Use the B3LYP/6-31G* model to obtain equilibrium geometries for 1,3-butadiene and cyclobutene. Calculate the energy of the ring opening reaction.

Based on space-filling models as a measure, which molecule takes up less volume? Increasing in pressure will drive the reaction toward reactants or toward products? What pressure would be required to make the energy and enthalpy differ by 1 kJ/mol? (Need to supply information on units.) 2. Zero-Point Energies Affect Reaction Energy. The energy resulting from a quantum chemical calculation refers to a molecule resting at the bottom of a potential energy well. On the other hand, a measured enthalpy refers to a molecule in its lowest (“zeroeth”) vibrational state. The difference, known as the zero-point energy, is directly proportional to the sum of all vibrational frequencies, and it is straightforward to correct calculated reaction energies (or experimental reaction enthalpies). Is the correction significant? Use the B3LYP/6-31G* model to calculate equilibrium geometries and vibrational frequencies for acetonitrile, CH3CN, and its isomer, methyl isocyanide, CH3NC. Evaluate the zero-point energy for each. Does this change in the energy of isomerism by more than 10%?

Repeat your calculations and analysis for propene and its isomer cyclopropane.

3. Temperature Affects Reaction Energy. Experimental thermochemical measurements are made at finite temperature (typically 298 K), whereas calculated energies pertain to molecules at 0 K. 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)


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.

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.

4. Mass Affects Reaction Energy. While the Born-Oppenheimer approximation removes nuclear mass from the Schrödinger equation, the energy of a chemical reaction actually depends on mass. This is referred to as an equilibrium isotope effect, and may be attributed to the difference in zero-point energies the between reactants and products (an oversimplification that is adequate for our purpose). The most commonly measured (and largest) isotope effect is that resulting from substitution of deuterium for hydrogen.

Use the B3LYP/6-31G* 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 14N to 15N in nitrogen molecule.

5. Carbon Isotope Effect on CH Bond Dissociation Energy in Methane. How much does the CH bond dissociation energy in methane change as a result of changing the mass of carbon (from 12C to 13C)? Use the B3LYP/6-31G* model to calculate equilibrium geometries and vibrational frequencies for methane and methyl radical and evaluate the change in zero-point energy for bond dissociation. Change the mass of carbon to (13C) for both molecules, rerun the calculation and recalculate the change in zero-point energy for bond dissociation. Assuming an experimental precision of 1%, it this energy change like to be observable?

6. Hydrogen Isocyanide. Under “normal” (laboratory) conditions, hydrogen isocyanide (HNC) is in equilibrium with its more stable isomer, hydrogen cyanide (HCN). According to the B3LYP/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?

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.

7. Ion-Molecule Equilibria. Ions resulting from protonation of neutral molecules in an ion cyclotron resonance spectrometer can be trapped for sufficiently long times to allow ion-molecule equilibria to be established. Measurement of the equilibrium abundance of the ions, together with knowledge of the relative amounts of the neutral molecules, allows determination of the equilibrium constant for the proton transfer reaction and,


therefore, the relative proton affinities of the neutral molecules. For example, measurement of the equilibrium abundance of protonated dimethylamine and aziridine provides the relative proton affinities of the precursors.

dimethylamine-H+ + aziridine → dimethylamine + aziridine-H+

Use the B3LYP/6-31G* model to calculate equilibrium geometries and energies for the four molecules involved in this equilibrium. Which amine 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? If it is not possible, in what ratio would the two amines need to be introduced in order for both ions to be seen?

8. CH Bond Dissociation in Propyne. A bond may dissociate either homolytically, leading to a pair of neutral radicals, or heterolytically, leading to a cation and an anion. Use the B3LYP/6-311+G** to obtain geometries for the two possible radicals resulting from homolytic CH bond dissociation in propyne. Are both radicals stable species? Elaborate. If they are, which is more stable? Rationalize your result. Is the less stable radical likely to be detectable in an equilibrium mixture at room temperature? (Assume that the minor radical needs to be at least 5% of the mixture in order to be seen.)

Heterolytic CH bond dissociation of propyne may either involve loss of a proton, leading to a hydrocarbon anion, or loss of hydride, leading to a hydrocarbon cation. Which is the less unfavorable (less endothermic) process? To decide, obtain geometries for the two hydrocarbon anions and two hydrocarbon cations, as well as for hydride anion. (The energy of proton is 0.) Are both hydrocarbon anions stable species? If they are, which is more stable? Attempt to rationalize your result. Is the less stable anion likely to be detectable in an equilibrium mixture at room temperature? Are both hydrocarbon cations stable species? If they are, which is more stable? Attempt to rationalize your result. Is the less stable cation likely to be detectable in an equilibrium mixture at room temperature?

Is homolytic or heterolytic CH bond dissociation of propyne less unfavorable?

9. Chlorine Nitrate: Chlorine nitrate, ClONO2, possibly plays a role in the balance of ozone in the upper atmosphere, specifically as a “thermodynamic sink” for NO2 and ClO radicals that are known to react with ozone and destroy ozone.

ClO· + NO2· ClONO2

However, were its isomer chlorine peroxynitrite (ClOONO) in equilibrium with chlorine nitrite, it might undergo cleavage of the OO bond. This would lead back to the NO2 and ClO radicals, negating at least in part the scavenging ability of chlorine nitrate.

Use the B3LYP/6-311+G** model to obtain equilibrium geometries and energies for chlorine nitrate and chlorine peroxynitrite and calculate the room-temperature Boltzmann distribution between the two isomers. Is chlorine peroxynitrite likely to be a significant factor? Elaborate

10. 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) has been established experimentally to be approximately 42 kJ/mol higher in energy.

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. Use the corrected energies to assign the ground state for CF2, CCl2 and CBr2. Rationalize any change in preferred ground state (relative to the parent compound) that you uncover.

11. Singlet and Triplet Cyclobutadiene. The singlet state of cyclobutadiene (C4H4) has four π electrons, formally making it an antiaromatic molecule. Does its exhibit a “rectangular” structure with localized single and double bonds or a “square” structure in which the four carbon-carbon bond lengths are all the same? Use the B3LYP/6-31G* model to decide. (Be sure to start with a rectangular structure.) Rationalize your result.

Obtain the geometry of the triplet state of cyclobutadiene. Does this exhibit the same structure (square or rectangular) as singlet cyclobutadiene? Rationalize your result. 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). Is the ground state of cyclobutadiene a singlet or a triplet? Rationalize your result.

12. Singlet and Triplet Cyclopropylidene. Use the B3LYP/6-31G* model to obtain the equilibrium geometry of the singlet state of cyclopropylidene (C3H2).

Is the molecule properly represented by the Lewis structure shown above, that is, does it incorporate “normal” carbon-carbon single and double bonds? If it does not, provide an explanation as well as an alternative structure drawing. It might help you to examine the occupied molecular orbitals, in particular, the π orbitals.

Obtain the geometry of the corresponding triplet state of cyclopropylidene and calculate the singlet-triplet energy difference. Correct your value 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? If it is different, suggest why.

13. Aromaticity of Benzene. The 4n+2 rule suggests that benzene with six π electrons in a ring should be unusually stable (“aromatic”). 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 H2 to 1,3-cyclohexadiene (leading to cyclohexene), or alternatively the energy of adding H2 to cyclohexene (leading to cyclohexane).


Each of these reactions trades a CC π bond and an HH bond for two CH bonds. All three reactions would normally be expected to be exothermic (σ bonds are stronger than π bonds). However, the first reaction also results in loss of aromaticity, which suggests that any difference in energy between the first and second (or third) reactions should reflect the aromatic stabilization of benzene.

Use the B3LYP/6-31G* model to obtain energies for the three reactions. Are the second and third reactions exothermic as expected? Is the first reaction exothermic or endothermic? What is the difference between the energy of the first and the average of second and third reactions? What is your estimate of the aromatic stabilization of benzene?

14. Antiaromaticity of Cyclobutadiene. Use the same reasoning applied in the previous problem to assess loss of stability (“antiaromaticity”) of cyclobutadiene, a molecule with 4n π electrons. Specifically, compare the energy of adding H2 to cyclobutadiene (leading to cyclobutene) with the energy of adding H2 to cyclobutene (leading to cyclobutane). While each reaction trades an HH bond and a CC π bond for two CH bonds, only the first reaction removes interaction of the coplanar double bonds. Use the B3LYP/6-31G* model. You can assume that cyclobutadiene is planar, but do not assume that all four carbon-carbon bonds are the same length. What is the difference in the energies of the two reactions? Is there evidence for significant destabilization of cyclobutadiene? If there is, is the magnitude comparable to the stabilization of benzene (see previous problem)?

15. Aromaticity of 1,3,5-Cyclohexatriene. It is well known that the six carbon-carbon bonds in benzene are identical, midway in length between “normal” single and double bonds. Is this a necessary condition for benzene to be aromatic? To decide, compare energies of benzene and 1,3,5-cyclohexatriene, a hypothetical molecule with alternating single and double bonds. Use the B3LYP/6-31G* model. While you can obtain the equilibrium geometry for benzene, you cannot do this for 1,3,5-cyclohexatriene. This molecule is not an energy minimum and, given the chance, will collapse to benzene. Assume a fixed geometry with alternating single and double bonds set to 1.54Å and 1.32Å. Is the energy difference roughly the same as the aromatic stabilization of benzene (see previous problem), or is it significantly smaller? Do all carbon-carbon bonds in benzene need to be the same in order for the molecule to be aromatic?

16. Aromaticiy of Borazine. Borazine (B3N3H6) is isoelectronic with benzene. Both are planar and both have six π electrons. Is borazine an aromatic molecule?

HN

HB NH

BH

NH

HB

To help you decide, calculate energies for successive addition of one, two and three equivalents of hydrogen. This is analogous to analysis previously carried out for benzene. Use the B3LYP/6-31G* model


HN

HBNH

BH

NH

HB

H2H2N

HBNH

BH

NH

H2B

H2N

H2BNH2

BH

NH

H2B

H2N

H2BNH

BH

NH

H2B

H2 H2

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 previously.

17. Krypton Difluoride. While numerous compounds of xenon are known, the only neutral krypton compound to have been reported to date is krypton difluoride (KrF2). Is its dissociation into krypton atom and fluorine molecule endothermic or exothermic? Obtain equilibrium structures of KrF2, F2 and (the energy of) Kr with the B3LYP/6-31G* model. Also obtain data for XeF2 and atomic xenon. How does the calculated reaction energy compare with that for the corresponding xenon compound? Does thermodynamic stability (with regard to dissociation) a cause for the fact that XeF2 is known whereas KrF2 is not? Elaborate.

18. Energy Content of Hydrazine Fuels. Combustion of which fuel, hydrazine or tetramethylhydrazine, delivers the greater amount of energy on a per gram basis? Use the B3LYP/6-31G* model to obtain energies for the two oxidation reactions. (The products 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?

19. Hydrogenation of Acetylene. Both steps in the hydrogenation of acetylene to ethane are exothermic, each trading an HH bond and a π bond for two CH bonds.

H C C H C CH

H H

H

CH3CH3H2 H2

Which step is the more exothermic? To decide, use the B3LYP/6-31G* model to obtain equilibrium geometries and energies for all reactants and products. What (if anything) does your result say about the relative strengths of the first and second π bonds in acetylene? What have you assumed to reach your conclusion?

20. Hydrogenation of 1,3-Butadiene. Both steps in the hydrogenation of trans-1,3-butadiene to n-butane involve formation of two new CH bonds at the expense of an HH bond and a π bond.

C C

H

H C

H

C

H

H

H

C C

CH2CH3

HH

HH2 H2

CH3CH2CH2CH3

Which step is the more exothermic? To decide, use the B3LYP/6-31G* model to obtain equilibrium geometries and energies for trans-1,3-butadiene, 1-butene and n-butane as well as molecular hydrogen. Does your result support the idea that adjacent (conjugated) double bonds are stronger than isolated double bonds? Elaborate.


21. Hydrogenation 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 B3LYP/6-31G* model to obtain equilibrium geometries for all molecules in the two reactions.

vs. C C

C

HH

HC

H

CH3C CHH2

Which reaction is more exothermic? What, if anything, does your result say about the interaction (conjugation) energy of adjacent double bonds vs. that of a double bond adjacent to a triple bond?

22. Ketene Dimer. There are six “cyclic” structures for the dimer of ketene, H2C=C=O.

C

O C

O

CH2

CH2

C

O O

C

CH2 CH2

C

H2C C

CH2

O

C

H2C CH2

CO

C

H2C C

O

CH2

C

H2C O

CCH2O

O

O

O

Before you do any calculations, guess which structure 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). According to the B3LYP/6-31G* model, which structure is actually preferred? Are two or more structures likely to be seen in an equilibrium mixture at room temperature? Assume 5% as the limit of detection of any structure. Is the dimerization exothermic or endothermic? Is this consistent with changes in bond strengths and the strain of the resulting four-membered ring? Elaborate.

23. Molecular Phosphorous. The most stable arrangement of molecular phosphorus is tetrahedral P4 (white phosphorus). Dissociation into 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 contrary, the stable form of molecular nitrogen is N2. N4 has never been detected.

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.

24. Dimerization of Alkylboranes. Borane (BH3) exists in equilibrium with its dimer, diborane (B2H6), an equilibrium that strongly favors the dimer.


Trimethylborane also coexists with its dimer, although here equilibrium stongly favors the monomer.

What is the reason for the change? Is it that a hydrogen bridge is strongly favored over a methyl bridge? To help you decide obtain geometries and energies for methylborane and the four possible dimers of methylborane: 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. 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)? Summarize your results to come up with an explanation for the difference in the two equilibria.

25. Hydrogen Azide. It is not unreasonable to expect that hydrogen azide might have a high-energy cyclic isomer, cyclotriazine. While such a structure is likely to be highly strained, its Lewis structure does not involve separated positive and negative charges.

Attempt to obtain equilibrium geometries for both hydrogen azide and cyclotriazene 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 energy 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.)

26. 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.

27. Why is Cyclopentadiene a Strong Acid? In the gas phase, cyclopentadiene is a much stronger acid than 1,3-pentadiene, that is, proton-transfer between cyclopentadiene and 1,3-pentadiene is exothermic.


cyclopentadiene + 1,3-pentadiene - H+ → cyclopentadiene-H+ + 1,3-pentadiene

Use the B3LYP/6-31G* model to obtain equilibrium geometries for cyclopentadiene and cyclopentadienyl anion (the deprotonated form of cyclopentadiene). Are there any conspicuous structural changes between the two molecules that might help to rationalize the high acidity? Elaborate.

28. Gas and Aqueous-Phase Basicities of Amines. The relative proton affinities of free amines differ significantly from those in water. For example, in the gas-phase the proton affinity of trimethylamine is 84 kJ/mol greater than that of ammonia, whereas in water the two are nearly the same.

Me3N + NH4+ → Me3NH+ + NH3 ∆Egas= -84 kJ/mol; ∆Eaq = xx kJ/mol

Use the B3LYP/6-31G* model to obtain equilibrium geometries for ammonia, trimethylamine, ammonium ion and trimethylammonium ion, and calculate the energy of proton transfer between the two amines. Is your result in reasonable accord with the experimental value?

You can’t actually perform the calculations in water. However, you can model the energy of the reaction in water by explicitly including water molecules in the calculation. At the minimum it is necessary to attach enough 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 ions. 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. Is there a serious flaw in the model? Elaborate.

29. Boiling Points of Ethanol and Ethylamine. While ethanol and ethylamine have similar molecular weights and molecular structures and while both can participate in up to 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.

30. Cycloalkynes. The fact that the C-C≡C bond angles in acyclic alkynes want to be linear makes it unlikely that triple bonds will easily incorporate into small rings. Identify the smallest cycloalkyne with both C-C≡C bond angles 170° or larger? Start with cyclohexyne and use the B3LYP/6-31G* model to obtain equilibrium geometries.

Obtain geometries of the corresponding cycloalkenes as well as for hydrogen molecule, calculate energies of hydrogenation reactions.


Is there a relationship between C-C≡C bond angle and hydrogenation energy?

31. Tetrahedrane. Tetrahedrane, (CH)4 in the form of a tetrahedron, has yet to be experimentally characterized, although X-ray crystal structures exist for derivatives, for example, tetra-tert-butyltetrahedrane. Somewhat surprising is the observation that the carbon-carbon bond length short for a linkage between sp3 carbons, and nearly identical to that found in cylopropane (1.49Ǻ vs. 1.50Ǻ).

Use the B3LYP/6-31G* model to obtain the geometry of tetrahedrane itself. Does it also incorporate a short carbon-carbon bond? Next, obtain geometries for all remaining molecules involved in hydrogenation of tetrahedrane, first to bicyclo[1.1.0]butane and then to cyclobutane.

H2 H2

Are both steps exothermic? Is the first step more or less favorable (exothermic) than the second step? Rationalize your result.



Part C. Molecular Orbitals and Orbital Energies

1. Triplet State of Methylene. Use the STO-3G model to obtain the equilibrium geometry of singlet methylene, and examine the highest-occupied and lowest-unoccupied molecular orbitals (the HOMO and LUMO, respectively). What effect, if any, would promotion of an electron from HOMO and LUMO be expected to have on the geometry, in particular, the HCH bond angle? Is the equilibrium geometry of triplet methylene obtained from the STO-3G model consistent with your expectations? Elaborate.

2. Mo2. Calculate the energy of dimolybdenum assuming a Mo-Mo bond bond distance of 1.94Ǻ, and display the valence molecular orbitals. Use the STO-3G model. Classify each molecular orbital as bonding, non-bonding or antibonding and as σ, π or δ. How many “bonds” connect the two molybdenum atoms?

3. Chromium-Chromium Multiple Bonds. Multiple bonding between transition metals is quite common. Consider, for example, the compound shown below, in which the two chromium atoms are bridged by four acetate (MeCO2

-) groups and each then coordinated to an axial pyridine ligand.

Perform an STO-3G energy calculation on the experimental crystal structure of this molecule [the .spartan file is available in chromium-chromium multiple bond] and, starting from the HOMO, display all the occupied molecular orbitals until you are satisfied that you have located all those that are bonding between the two chromium atoms. How many have such orbitals (“bonds”) have you located? Characterize each of as a σ, π or δ orbitals and rationalize the ordering of their energies. How would you describe the metal-metal bond in this molecule? 4. Molecular Orbitals of Diborane. Diborane was originally thought to look like ethane, and only later was assigned the correct structure with two hydrogen atoms positioned above and below the plane formed by the remaining six atoms. In fact, diborane is much more closely related to ethylene than it is to ethane. These two molecules have the same number of valence electrons, that is, they are isoelectronic. Do the molecular orbitals of diborane resemble those of ethylene?


Obtain equilibrium geometries for diborane and ethylene using the STO-3G model. Attempt to associate each of the six valence molecular orbitals in diborane with an orbital in ethylene. Point out differences in both ordering and detailed shape. Describe the molecular orbital in diborane that is associated with the π orbital in ethylene. Is it the HOMO? Locate the unoccupied molecular in diborane that resembles the π* orbital in ethylene. Is it the LUMO?

5. Lone Pairs in Water and Hydrogen Sulfide. Describe the two highest-occupied molecular orbitals of water as obtained from an STO-3G calculation. Are they primarily bonding, non-bonding or antibonding? Is the highest-occupied orbital a σ orbital or a π orbital?

Repeat your calculations and analysis for hydrogen sulfide. Point out (and attempt to rationalize) any significant differences between the two molecules.

6. Lithium and Sodium Hydroxide. Water is bent (the HOH angle is ~105o) but both lithium hydroxide and sodium hydroxide are linear. To understand why, compare the highest occupied molecular orbitals for the three molecules. Use the STO-3G model to furnish equilibrium geometries. Characterize the HOMO of water as bonding, antibonding or nonbonding and as σ or π. How (if it all) is it different from the HOMO in lithium hydroxide and sodium hydroxide? How does this relate to the observe change in geometry?

7. Adding and Removing Electrons. Molecular orbitals may show distinct bonding or antibonding character. Loss of an electron from an occupied molecular orbital due to excitation or ionization, or gain of an electron by a previously unoccupied molecular orbital from excitation or electron capture could lead to changes in bonding and accompanying changes in molecular geometry.

Use the STO-3G model to obtain equilibrium geometries for ethylene (H2C=CH2), formaldimine (H2C=NH) and formaldehyde (H2C=O), and examine the HOMO for each. What would you expect to happen to the geometry around carbon (remain planar vs. pyramidalize), to the C=X bond length (lengthen, shorten or remain the same) and (for formaldimine) to the C=NH bond angle (increase, decrease or remain the same) were an electron to be removed from this orbital?

Obtain equilibrium geometries for the ethylene, formaldimine and formaldehyde radical cations using the STO-3G model, and compare with those of the corresponding neutral molecules. Are the changes in geometry in line with what you expect?


Examine the LUMO for each of the three molecules. Guess what would happen to the geometry around carbon, C=X bond length and (for formaldimine) the C=NH bond angle were an electron to be added to this orbital.

Obtain equilibrium geometries for the ethylene, formaldimine and formaldehyde radical anions using the STO-3G model, and compare with those of the corresponding neutral molecules. Are the changes in geometry in line with what you expect?

8. Carbon Monoxide as a Ligand. Carbon monoxide is certainly the most common ligand in transition-metal organometallic compounds. It acts by donating electrons from the σ lone pair on carbon into an empty orbital on the metal and accepting electrons from the metal atom into low-lying π* orbital.

Use the B3LYP/6-31G* model to obtain the geometry for carbon monoxide and examine the HOMO. Is it bonding, antibonding or essentially non-bonding between carbon and oxygen? What, if anything, would you expect to happen to the CO bond as electrons are donated from the HOMO to the metal atom? Display the LUMO. (There are actually two equivalent orbitals, designated LUMO and LUMO+1, and you can base arguments on either one.) Is it bonding, antibonding or essentially non-bonding between carbon and oxygen? What if anything would you expect to happen to the CO bond if electrons were donated from the metal atom into this orbital? Taken together, will donor-acceptor interactions involving carbon monoxide and a metal affect the ligand?

Obtain the geometry the iron pentacarbonyl, Fe(CO)5, using the B3LYP/6-31G* model. It has a trigonal bipyramidal structure with three equatorial and two axial CO ligands. Delete one of the equatorial ligands to make iron tetracarbonyl, Fe(CO)4. Calculate the energy of this fragment but do not optimize the geometry. Display the HOMO. Does it have significant amplitude in the location where the carbon monoxide ligand will attach? If so, does it have the proper symmetry to interact with the LUMO in CO?

9. Binding Ethylene to a Metal. Two limiting structures can be drawn to represent ethylene bonded to a transition metal. The first may be thought of as a weak complex in that it maintains the carbon-carbon double bond, while the second destroys the double bond in order to form two new metal-carbon σ bonds, leading to a three-membered ring (a metallacycle). The bonding is “real” metal-alkene complexes presumably falls in between these two extremes.

M M

Optimize the geometry of ethylene using the B3LYP/6-31G* model, and examine both the HOMO and LUMO. Is the HOMO bonding, antibonding or non-bonding between the two carbons? What if anything should happen to the carbon-carbon bond as electrons are donated from the HOMO to the metal? Do you expect the carbon-carbon bond length to decrease, increase or remain about the same? Elaborate.

The LUMO is where the next (pair of) electrons will go. Is this orbital bonding, antibonding or non-bonding between the two carbons? What, if anything, should happen to the carbon-carbon bond as electrons are donated (from the metal) into the LUMO? Is


the expected change in the carbon-carbon bond due to this interaction in the same direction or in the opposite direction as any change due to interaction of the HOMO with the metal? Elaborate.

Optimize the geometry of ethylene iron tetracarbonyl using the B3LYP/6-31G* model. Compare the length of the carbon-carbon bond in the complex to that in ethylene and, based on this comparison, describe the bonding between the ethylene and the metal.

CO

Fe

COCO

CO

Delete the ethylene ligand (and the ligand attachment point), and calculate the energy of the resulting (iron tetracarbonyl) fragment (do not optimize the geometry). Examine both the HOMO and LUMO of this fragment. Is the LUMO likely to interact with the HOMO of ethylene? Elaborate. Would you expect electron donation from ethylene to the metal to occur? Is the HOMO likely to interact with the LUMO of ethylene? Elaborate. Would you expect electron donation from the metal to ethylene to occur?

10. Anticipating Excited-State Geometries. Excited states may be thought of as arising from electron promotion from filled to empty molecular orbitals. In this view, the lowest-energy excited state should arise from promotion of an electron from the HOMO to the LUMO. Any influence that the bonding or antibonding character of the HOMO may have on geometry is weakened and in return the consequences of any bonding or antibonding interactions present in the LUMO are felt. Therefore, examination of the HOMO and LUMO for a molecule in its ground state should provide insight into the geometry of the excited state.

Obtain the equilibrium geometry for nitrogen using the STO-3G model. Examine the HOMO and LUMO. Would you characterize the lowest-energy electronic transition as n --> π* or π --> π*? Elaborate. Would you expect the NN bond length in the ground state to shorten, lengthen or remain unchanged upon excitation? Elaborate

Obtain the equilibrium geometry for acetone and examine the HOMO and LUMO. Would you characterize the lowest-energy electronic transition n --> π* or π --> π*? Elaborate. Would you expect the CO bond in the ground state to shorten, lengthen or remain unchanged upon excitation? Would you anticipate any other changes in geometry? Elaborate.

11. Anticipating Color Changes. The color of a molecule may reasonably be assumed to be related to λmax, the difference in energy the ground and first excited states. This in turn should be reflected by the difference in energies between the highest-occupied and lowest- unoccupied molecular orbitals, the so-called, HOMO-LUMO gap. The smaller the gap, the lower the energy of the spectral transition and the more the color will tend toward the “red”. The energies of molecular orbitals depend on structure, and it is reasonable to expect that the changes in the HOMO/LUMO gap with changes in molecular structure will anticipate the changes in color.


Use the STO-3G model to obtain equilibrium geometries for azobenzene (R=R´=H), 4-hydroxyazobenzene (R=OH, R´=H) and 4-amino-4´-nitroazobenzene (R=NH2, R´=NO2).

N NR R´

Azobenzene is orange. Use the change in HOMO/LUMO gap to predict whether the color of each of the other molecules will be shifted toward the red or blue end of the spectrum.

12. SN2 Reaction of Methyl Iodide. Attack of cyanide anion, onto the backside of methyl iodide leads to acetonitrile and iodide anion.

N C N C CH3 + ICH3 I– – A simple picture is that the HOMO of cyanide interacts with the LUMO of methyl iodide, leading to formation of a carbon-carbon bond and loss of a carbon-iodine bond. Use the STO-3G model to obtain the geometry of methyl iodide. Examine the LUMO. Is it bonding, antibonding or non-bonding between carbon and iodine? What would you expect to happen to the CI bond were a pair of electrons to be added to this orbital?

13. Ionization Potentials of Amines: Experimental ionization potentials for a series of amines and amides follow: NH3, 10.1; MeNH2, 8.9; Me2NH, 8.2, Me3N, 7.9; PhNH2, 7.7; NH2CHO, 10.2; NF3, 12.9; NCl3, 10.1. Use the 6-31G* model to obtain equilibrium geometries for these compounds, and plot HOMO energy vs. ionization potential. Point out (and try to rationalize) any significant deviations from linear correlation.



Part D. Electron and Spin Densities 1. CPK Models vs. Electron Density Surfaces for Hydrocarbons. Do the sizes and shapes of molecules depicted by simple space-filling (CPK) models reflect those based on electron density surfaces? Obtain electron density surfaces for methane, ethane, propane, n-butane, n-pentane and n-hexane that enclose 99% of the total number of electrons. Use the HF/6-31G* model. Plot the volumes of these surfaces against the volumes of the corresponding space-filling models. Is there a reasonable correlation?

2. Sizes of Alkali Metal Cations. Compare electron density surfaces for lithium, sodium and potassium cations that enclose 99% of the total number of electrons. Use the HF/6-31G* model. Which is smallest and which is largest? Rationalize what you observe. How do the calculated radii compare with the usual van der Waals radii of the three elements? (The radii may be obtained from the calculated volumes.)

3. Sizes of Hydride and Halide Anions. Compare electron density surfaces for hydride, fluoride chloride and bromide anions that enclose 99% of the total number of electrons. Use the HF/6-31G* model. Which is smallest and which is largest? Rationalize what you observe. How do the calculated radii compare with the usual van der Waals radii of the three elements? (The radii may be obtained from the calculated volumes.)

4. Methyl Anion, Ammonia and Hydronium Cation. Methyl anion (CH3-), ammonia

and hydronium cation (H3O+) are isoelectronic, in that all three have 8 valence electrons. Does this mean that they are the same size? To tell, compare electron density surfaces that enclose 99% of the total number of electrons. Base these on equilibrium geometries obtained from the HF/6-31G* model. If the three sizes are different, which is smallest and which is largest? What is the percentage change in volume from the smallest to largest? Rationalize what you observe.

Repeat your calculations and analysis for the corresponding second-row systems: silyl anion (SiH3

-), phosphine (NH3) and sulfonium cation (H3S+). Is the (percentage) change in volume noted here smaller, larger or about the same as noted for the first-row compounds? 5. Bonding in Diborane. Which Lewis structure, that lacking or including a boron-boron bond, provides the better description of diborane?

HB

HB

HB

HB

H

HH

H

To decide, examine the electron density surface for diborane in between the two boron atoms. Is it convex (suggesting buildup of electron density) or concave (suggesting depletion of density)? Use the B3LYP/6-31G* model and choose a surface that encloses 40-50% of the total number of electrons.


6. Hydrocarbon Radicals. Loss of hydrogen atom from a hydrocarbon leads to a neutral radical. The common wisdom is that the more delocalized the unpaired electron, the less likely the radical is to react. This is the idea behind the clinical use of so-called antioxidants, molecules that can donate a hydrogen atom to a reactive (and potentially destructive) radical leading to a stable (and presumably less destructive) radical.

Use the B3LYP/6-31G* model to obtain equilibrium geometries for ethyl radical (from ethane), vinyl radical (from ethylene), ethynyl radical (from acetylene) and phenyl radical (from benzene), and compare spin density surfaces. For which, does the unpaired appear to be most localized? For which does it appear to be most delocalized? Obtain equilibrium geometries for ethane, ethylene, acetylene and benzene (as well as hydrogen atom) and calculate CH bond energies. Do the bond dissociation energies parallel the extent to which the spin is delocalized in the corresponding radicals?

7. Benzene Radical Cation. Ionization of benzene results in removal of an electron from the highest occupied molecular orbital which, in this case, is a pair of degenerate (equal energy) molecular orbitals.

This breaks the degeneracy and as a consequence necessarily leads to distortion away from six-fold symmetry. Use the B3LYP/6-31G* model to obtain the geometry of benzene radical cation. You need to start from a structure that has been distorted from six-fold symmetry. Does the resulting structure show delocalized bonds (as benzene) or is there significant bond alternation? Elaborate. Examine the spin density for benzene radical cation. What if anything does it tell you about where the electron was removed from?

8. Triplet Methylene: The ground state of methylene (CH2) is known to be a triplet. If triplet methylene is linear, the 2p orbitals on carbon not involved in CH bonding (2px and 2py were the molecule oriented on the z axis), would be equivalent and each could hold one of the unpaired electrons.

Use the B3LYP/6-31G* model to obtain the geometry for linear triplet methylene. Does the spin density reflect this simple picture? Distort and display the spin density.

Distort your structure away from a linear geometry and redetermine the geometry? Does it return to a linear structure? If not, compare the spin density for bent triplet methylene with that for the linear molecule. How has it changed?



Part E. Atomic Charges and Electrostatic Potential Maps 1. Atomic Charges and Atomic Electronegativities. Are charges anticipated by atomic electonegativities? Use the HF/6-31G* model to calculate equilibrium geometries and charges for first-row hydrides: LiH, BH3, CH4, NH3, H2O and HF, second-row hydrides: NaH, AlH3, SiH4, PH3, H2S and HCl as well as HBr. Plot the charge on hydrogen against the electronegativity of the atom to which it is bonded. (You can add H2 to you plot without having to do any calculations.) Is there a simple relationship between the two quantities? Elaborate. 2. Dipole Moments for Ammonia and Trifluoroamine. Why is the dipole moment in ammonia (1.47 debyes) much larger than that in trifluoroamine (0.23 debyes)? Is it because the HNH bond angle in ammonia is smaller than the FNF bond angle in trifluoroamine? Is it because NH bonds are polarized with nitrogen at the negative end in the same direction as the nitrogen lone pair, whereas NF bonds are polarized with nitrogen at the positive end opposite to the direction of the lone pair? To help you decide, calculate equilibrium for ammonia and trifluoroamine using the B3LYP/6-31G* model. Is the HNH bond angle smaller than the FNF bond angle? Compare electrostatic potential maps for the two molecules. Is there a marked difference in polarity of the two molecules?

3. Ordering the Acidities of Ammonia, Water and Hydrogen Fluoride. In the gas phase, hydrogen fluoride is a stronger acid than water, which in turn is a stronger acid than ammonia. Is the trend in acidities reflected in the value of the electrostatic potential (at hydrogen) in the three molecules? To decide, obtain geometries and electrostatic potential maps for the three molecules using the HF/6-31G* model. Display the maps side-by-side on the same color scale with hydrogen clearly exposed.

4. Weak vs. Strong Acids. Ethanol is an example of a very weak acid, acetic acid of a moderately strong acid and sulfuric and nitric acids of very strong acids. Obtain geometries and electrostatic potential maps for these molecules using the HF/6-31G* model. Display the maps side-by-side on the same color scale with the acidic hydrogen clearly exposed. Do the maps reveal the known ordering of acid strengths in these compounds? Use electrostatic potential maps to try to identify compounds that are stronger acids than the nitric or sulfuric acid.

5. Acetic Acid Dimer. Acetic acid forms a symmetrical hydrogen-bonded dimer.

OCH3C

O H

OC CH3

O

H Obtain equilibrium geometries for acetic acid and its dimer using the HF/6-31G* model. Display electrostatic potential maps for the two side-by-side and on the same color scale.


Is there evidence for charge transfer away from the acidic hydrogen in acetic acid onto to oxygen to which it hydrogen bonded? 6. Inside Buckyball. Buckmeister fullerene, more commonly known as or buckyball or C60, is a fascinating molecule on several grounds. For one, it is a form of elementary carbon, distinct from the other two known forms (graphite and diamond). It should be noted, however, that C60 is just one of several ball-shaped molecules that have come to light in the past two decades. There are perhaps many elementary forms of carbon!

Second, buckyball has been shown to be able to encapsulate a number of atoms and small molecules which are then unable to diffuse out. The obvious question is how the molecules get inside. The answer is that they don’t “get in”. Rather, buckyball needs to be formed from around them. The size of the cavity (and by inference the maximum size of a molecule that can be encapsulated) can be estimated using the known geometry of buckyball and the van der Waals radius of carbon. However, there is no way to assess the environment presented to a molecule held inside the cavity, and to know whether it will be more “hospitable” to neutral molecules or to cations or anions. An electrostatic potential map should provide insight. A file containing the electrostatic potential map for C60 is available in buckyball electrostatic potential map. The map has been sliced in half to allow the electrostatic potential for not only the outside but also the inside of buckyball to be seen. For comparison, the electrostatic potential map for benzene is provided alongside. The two maps are on the same scale (-100 to 100 kJ/mol). Relative to benzene, does the map for the “outside” of C60 show more or less charge variation than the map for benzene? Would you expect it to be more or less soluble than benzene in non-polar solvents? In polar solvents? Elaborate. According to the electrostatic potential map, is the “inside” environment of C60 significantly different from the “outside” environment? Would cations, anions or neutral atoms and molecules be best accommodated inside of buckyball?

7. SN2 Reaction of Chloride and Methyl Bromide. Electrostatic potential maps are not limited to stable molecules but can also be applied to transition states and more generally to the sequence of structures connecting reactants and products. Consider, for example, SN2 reaction of chloride anion with methyl bromide leading to methyl chloride and bromide anion

Cl– + CH3Br → CH3Cl + Br–


Use the HF/6-31G* model to obtain an energy profile. Start with chloride 3.5Å from the carbon in methyl bromide and move it in steps of 0.1Å to 1.7Å. Plot both the energy and the energy corrected for the effect of aqueous environment vs. the CCl distance (the “reaction coordinate”). Is the reaction endothermic or exothermic in the gas phase? How do the solvent corrections affect overall endo or exothermicity? Does this preference maintain as the energies are corrected for the effect of aqueous solvent? Are your results consistent with the observation that bromide is a better leaving group than chloride? Elaborate.

Examine electrostatic potential maps for the structures along the reaction profile, with particular focus on the reactants and products and on the transition state. For which (reactant, product or transition state) is the charge most localized? For which is it most delocalized? What does this say about the relative abilities of Cl and Br to act as leaving groups? Does the energy along the reaction pathway appear to correlate with the extent to which charge is delocalized? Elaborate. Does the change in energy due to the solvent correction correlate with the extent to which charge is delocalized? Elaborate.



Part F. Transition States and Activation Energies

1. Vinyl Alcohol to Acetaldehyde. Even though vinyl alcohol (H2C=C(H)OH) is less stable than its isomer acetaldehyde (CH3C(H)=O), it can be stored almost indefinitely. This suggests that the isomerization reaction is likely to have a sizable activation barrier..

Use the B3LYP/6-31G* model to obtain equilibrium geometries for acetaldehyde and vinyl alcohol. What is the room-temperature Boltzmann ratio of acetaldehyde and vinyl alcohol according to this model? Is vinyl alcohol likely to be detectable in an equilibrium mixture? Obtain the transition-state geometry for the isomerization and calculate the activation energy. Is it large enough that once formed, vinyl alcohol is likely to persist for long times? Elaborate.

2. Dichlorocarbene Addition to Ethylene. Singlet dichlorocarbene adds to ethylene to yield 1,1-dichlorocyclopropane.

CCl2 + H2C CH2

Cl Cl

Obtain an equilibrium geometry for singlet dichlorocarbene using the B3LYP/6-31G* model and display an electrostatic potential map. This shows the distribution of charge on the accessible surface. Where is the electrophilic site, in the ClCCl plane or out of the plane? How would you expect dichlorocarbene to approach ethylene? Does this lead to a product with the proper three-dimensional geometry? Elaborate.

Next, obtain the transition state for the addition. Is the calculated structure consistent with the conclusions reached above regarding the electrophilic character of dichlorocarbene? Elaborate.

3. Hydroxycarbene. The energy of singlet hydroxycarbene (HCÖH) is estimated to be ~220 kJ/mol higher than that of its isomer, formaldehyde. This means that in order for hydroxycarbene to be detected, the energy barrier leading to formaldehyde needs to be sizable (>100 kJ/mol). Additionally, the barrier to dissociation to hydrogen and carbon monoxide molecules also needs to be sizable.

HCÖH H2CO

HCÖH H2 + CO

Use the B3LYP/6-311+G** model to calculate equilibrium geometries of all reactants and products of the two reactions above. Are the calculations in reasonable accord with the estimated difference in energy between the two isomers? Is the barrier to isomerization to formaldehyde sufficiently high to permit hydroxycarbene to be detected? Is the barrier to dissociation to hydrogen and carbon monoxide sufficiently high to permit hydroxycarbene to be detected?

4. Dimerization of Borane. Borane (BH3) dimerizes to diborane.

BH3 + BH3 B2H6


Is there an activation energy associated with dimerization? To see if there is, first obtain the equilibrium geometry of diborane. Use the B3LYP/6-31G* model. Starting from this structure, repeat the geometry calculation constraining each of the two bridging hydrogen atoms to different boron atoms to maintain a distance of 1.3Ǻ (close to the equilibrium value), to 1.4Ǻ, 1.5Ǻ, etc., all the way to 2.5Ǻ. Plot energy vs. (constrained) BH distance. Does your plot contain an energy maximum? If it does, obtain the transition state and calculate the barrier to activation.

5. Isomerization of Methyl and Ethyl Isocyanide. The isomerization of methyl isocyanide (CH3N+≡C-) to acetonitrile (CH3C≡N) is one of the most thoroughly examined chemical reactions. The process presumably involves migration of CH3 as an intact unit, the transition state being a 3-member ring with partial single bonds between the methyl group and both nitrogen and carbon. Use the B3LYP/6-31G* model to obtain structures for reactant (methyl isocyanide) and transition state, and calculate the activation energy for the isomerization. Is your result similar to the value of 159 kJ/mol estimated from experimental rate data? Use the same model to calculate the activation energy for the analogous reaction of ethyl isocyanide. Assuming that the error is the same as for the methyl isocyanide isomerization, what is your best guess for activation energy for reaction of ethyl isocyanide? Which reaction appears to be faster? Offer an explanation of your result.

6. Hydroboration of Alkenes vs. Alkynes. Addition of an alkylboranes to a carbon-carbon double bond (“hydroboration”) followed by oxidation is a common way to alcohols, for example, addition of dimethylborane to 2,3-dimethyl-2-butene yields 2,3-dimethyl-2-butanol.

Me2BH + Me2C=CMe2 →→ Me2CH-CMe2OH

The reaction is more general as borane also adds to carbon-carbon triple bonds. Use the B3LYP/6-31G* model to obtain structures and energies for reactants and transition states for addition of dimethylborane (Me2BH) to both ethylene and acetylene. Which reaction has the lower activation energy? Offer an explanation for your result.

7. Kinetic Isotope Effects and Hydrogen Abstraction. Reaction of chlorine atom with d1-dichloromethane can either involve hydrogen abstraction (leading to HCl) or deuterium abstraction (leading to DCl).

CCl

Cl

H

D

Cl•

Cl•

C•Cl

ClD + HCl

C•Cl

ClH + DCl

hydrogenabstraction

deuteriumabstraction

The two reactions follow the same pathway, pass through the same transition state and lead to the same products. The only difference is whether a hydrogen or deuterium atom is abstracted. This affects the zero-point energies of the transition states and products, and leads to kinetic and equilibrium isotope effects, respectively.


Obtain equilibrium geometries and infrared spectra for dichloromethyl radical and hydrogen chloride using the HF/6-31G* model. Use calculated zero-point energies to evaluate the equilibrium isotope effect, that is, the energy of the reaction.

Cl2CH + DCl Cl2CD + HCl••

Use the Boltzmann equation to calculate the equilibrium distribution of products at 25°C. Obtain the structure of the transition state for hydrogen abstraction from dichloromethane by chlorine atom. Again use the HF/6-31G* model. Calculate zero-point energies for the two possible monodeuterated transition-states. Use the Boltzmann equation (with transition state zero-point energies instead of reaction product zero-point energies) to calculate the kinetic distribution of products at 25°C. Do equilibrium and kinetic isotope effects favor the same product? If they do not, suggest conditions that would favor abstraction of hydrogen and conditions that would favor abstraction of deuterium.



Part G. Flexible Molecules 1. Conformers for n-Butane. Obtain equilibrium geometries for both anti and gauche conformers of n-butane using the HF/6-31G* model. At what temperature does the minor conformer make up less than 1% of an equilibrium mixture? At what temperature does it make up more than 40%? Don’t forget that 360o rotation about the central CC bond leads to two equivalent gauche conformers.

2. Dipole Moment of n-Butane. Using the results from the previous problem, calculate the value of the dipole moment for a sample of n-butane at room temperature. To what value does the dipole moment go as the temperature is raised? Elaborate.

3. Infrared Spectra of Acrolein Conformers. Acrolein (propenal, H2C=CC(H)=O) exists as a mixture of syn and anti conformers.

C CH

H H

C HO

C CH

H H

C OH

Could these be distinguished using infrared spectroscopy, in particular, from the infrared frequency corresponding to the CO stretch? This requires both that the minor conformer contribute at least 5% to an equilibrium mixture and that the CO stretching frequencies of the conformers differ by at least 5 cm-1. Use the B3LYP/6-31G* model to obtain geometries and infrared spectra for both conformers. Do not start with planar structures. Which conformer is lower in energy? What percentage of a room-temperature equilibrium mixture corresponds to the higher-energy conformer? If it less than 5%, what temperature would be needed to bring it to 5%? Compare the infrared spectra for the two conformers. Do the CO stretching frequencies differ by 5 cm-1 or more? Is there anything else that would help to distinguish the spectra? Elaborate.

4. Dipole Moment of Formic Acid. Formic acid (HCO2H) incorporates one rotatable bond and may exist in either cis (HOC=O = 0o) or trans (HOC=O = 180o) conformers. Use the B3LYP/6-31G* model to obtain geometries for both conformers. Which is more stable and by how much? Rationalize the preference. What is the dipole moment of a sample at room temperature? Is it dominated by one conformer or do both conformers contribute significantly?

5. Carbonic Acid. Three “planar”conformers can be drawn for carbonic acid.

C O

HO

OH

C O

HO

OH

OH

H OC O

According to the HF/6-31G* model, which conformer is preferred? What is % abundance of the two higher-energy conformers room temperature? Does the ordering of dipole


moments for the three conformers parallel the ordering of energies? If it does, provide an explanation as to why.

6. Hydrazine and Tetrafluorohydrazine. Obtain an energy profile for rotation about the NN bond in hydrazine and obtain a Fourier fit. Use the HF/6-31G* model and step (:NN: dihedral angle) from 0 to 180o in 20o increments. (It is not necessary to step all the way to 360o to identify the unique energy minima and to obtain the connecting barriers.) Which term(s) dominate the Fourier fit? Rationalize your result in terms of what you know about the relative “sizes” of bonds and lone pairs.

Repeat your calculations and analysis for tetrafluorohydrazine. Is this profile qualitatively similar to that for hydrazine insofar as the location of the energy minimum and the locations and heights of the rotational barriers? Which term(s) dominate the Fourier fit? Point out any significant differences between the two and provide a rationale.

7. Dinitrogen Tetraoxide. What is the preferred conformation of dinitrogen tetraoxide (O2NNO2) which results from oxidation of hydrazine? Are the nitro groups coplanar or perpendicular (or somewhere in between)? Use the B3LYP/6-31G* model to obtain the equilibrium geometry starting with a twisted geometry. Rationalize your result.

8. Hydrogen Peroxide and Hydrogen Disulfide. Obtain an energy profile for rotation about the OO bond in hydrogen peroxide and obtain a Fourier fit. Use the HF/6-31G* model and step the HOOH dihedral angle from 0 to 180o in 20o increments. (It is not necessary to step all the way to 360o to identify the unique energy minima and to obtain the connecting barriers.) Which term(s) dominate the Fourier fit? Rationalize your result in terms of what you know about the relative “sizes” of bonds and lone pairs.

Repeat your calculations and analysis for rotation about the SS bond in hydrogen disulfide. Is the profile qualitatively similar to that for hydrogen peroxide insofar as the location of the energy minimum and the locations and heights of the rotational barriers? Which term(s) dominate the Fourier fit? Point out any significant differences between the two and provide a rationale.

9. 1-Butene. Use the HF/6-31G* model to obtain an energy profile for rotation about the central CC single bond in 1-butene (CH3CH2─CH=CH2). Step from 0 to 180o in 20o increments. (It is not necessary to step all the way to 360o to identify the unique energy minima and to obtain the connecting barriers.) How many distinct energy minima are there? If more than one, which is favored? Which term(s) dominate the Fourier fit?

10. 1,3-butadiene. Use the HF/6-31G* model to obtain an energy profile for rotation about the central CC single bond in 1,3-butadiene. Step from 0 to 180o in 20o increments. (It is not necessary to step all the way to 360o to identify the unique energy minima and to obtain the connecting barriers.) Is this profile qualitatively similar to that for 1,3-butadiene insofar as the location of the energy minima? Specifically, is there a trans coplanar minimum? Is there a cis minimum that is actually slightly distorted from planarity? Is the trans minimum lower in energy than the cis minimum? Which term(s) dominate the Fourier fit?


11. Acrolein and Glyoxal. Use the HF/6-31G* model to obtain an energy profile for rotation about the central CC single bond in acrolein (H2C=C(H)-C(H)=O). Step from 0 to 180o in 20o increments. (It is not necessary to step all the way to 360o to identify the unique energy minima and to obtain the connecting barriers.) Is this profile qualitatively similar to that for 1,3-butadiene insofar as the location of the energy minima (see previous problem)? Specifically, are the CC and CO double bond in the “cis” structure of acrolein coplanar or (as in the case of 1,3-butadiene) and the locations and heights of the rotational barriers? Which term(s) dominate the Fourier fit? Point out any significant differences between the two and provide a rationale.

Repeat your calculations and analysis for glyoxal (O=(H)C-C(H)=O) and answer the analogous questions.

12. λmax vs. Diene Conformation. A very simple way to model the energy of an electronic transition from ground to excited state (λmax in the UV/vis spectrum) is to assume that it parallels the difference in energy between the highest-occupied and lowest-unoccupied molecular orbitals (the HOMO-LUMO gap). To what extent does this gap (and λmax) for a diene depend conformation? To what extent does the change in the gap parallel the change in energy of the ground-state molecule with change in conformation?

Use the energy profile for 1,3-butadiene obtained in a previous problem (varying the CCCC dihedral angle in each from 0° to 180° in 20° steps) to plot both the energy and the HOMO/LUMO gap as a function of dihedral angle. At what dihedral angle is the HOMO/LUMO gap the largest? At what dihedral angle is it the smallest? Is there much difference in the HOMO/LUMO gap between cis and trans-planar diene conformers? Does the variation in total energy closely follow the HOMO-LUMO gap or are the two uncorrelated?

13. Twist-Boat Cyclohexane. In addition to the familiar chair structure, there is a second stable form of cyclohexane. This structure is usually described as “twist boat”, insofar as the CH2 groups on the opposite side of the six-member ring point in the same direction. Obtain equilibrium geometries for both chair and twist boat conformers of cyclohexane using the HF/6-31G* model, and calculate the room-temperature equilibrium distribution of the two. Is the higher-energy conformer sufficiently abundant (>5% of the total) such that it is likely to be seen? Elaborate.

14. Inversion in Phosphine and Trifluorophosphine. Phosphine (PH3) is pyramidal and inverts via a planar transition state. Use the HF/6-31G* model to obtain geometries for both pyramidal and planar forms. Also perform analogous calculations on the two structures of ammonia (NH3). Is this barrier in phosphine smaller, larger or about the same as that for ammonia? Provide a rationale if it is markedly different.

Repeat your calculations for trifluorophosphine (PF3). Is there a significant increase or decrease in inversion barrier relative to that for phosphine. Obtain planar and pyramidal structures for trifluoroamine (NF3). Is any change in inversion barrier from ammonia similar to the change found in the phorphorus compounds?


15. Inversion in Aziridine. Is pyramidal inversion of a nitrogen that is part of a three-member ring expected to be more or less difficult that inversion of an acyclic amine. To decide, use the HF/6-31G* model to obtain geometries for dimethylamine (to act as a reference), aziridine and their respective inversion transition states.

Calculate inversion barriers (difference between pyramidal and planar forms) for both molecules. Is the barrier in aziridine significantly larger than that in dimethylamine? Provide a rational for your results. 16. Pseudorotation in Phosphorus Pentafluoride. Pseudorotation exchanges equatorial and axial positions of a five-coordinate, trigonal-bipyramidal center, for example, the phosphorus in phosphorus pentafluoride (PF5).

F P

F

F

F

F

axial

equatorial

axial

equatorial

The process, known as pseudorotation, simultaneously decreases the FPF angle involving two axial fluorines and increases the FPF angle involving two of the equatorial fluorines. The transition state is a structure in which (what were) the two axial fluorines and two of the equatorial fluorines form the base of a square-based pyramid. Continuing the motion returns to the stable trigonal-bipyramidal geometry but with axial and equatorial fluorines exchanged. Repeated pseudorotation moves fully scramble the fluorines.

F P

F

F

F

FF P

F

F

F

F

Use the HF/6-31G* model to obtain geometries for both trigonal bipyramidal (D3h symmetry) and square-based pyramidal (C4v symmetry) forms of PF5 and calculate the barrier to pseudorotation. Even though the latter is presumed to be a transition state, you can obtain its geometry by starting from a structure with C4v symmetry. Is your result consistent with failure of 19F NMR spectroscopy to distinguish between equatorial and axial sites even at temperatures as low as -100o C? Elaborate/ 17. Pseudorotation in Iron Pentacarbonyl. Iron pentacarbonyl, Fe(CO)5, adopts a trigonal bipyramidal (D3h symmetry) geometry with distinct equatorial and axial positions. However, equatorial and axial positions these are assumed to rapidly exchange by way of a square-based pyramid (C4v symmetry) structure, in process known as pseudorotation. Use the B3LYP/6-31G* density functional model to obtain geometries for both forms. (Even though the square-based pyramid is presumed to be a transition



state, if you start with a C4v symmetry structure, this will be maintained in the geometry optimization.) Calculate the activation energy for pseudorotation. Is it significantly higher than that for PF5 (xx kJ/mol)? Is it likely that the 13C NMR spectrum likely to exhibit one or two resonances? Elaborate.

18. Nitroamide. Is the amino group in nitroamide (O2N-NH2) planar or pyramidal? While the inherent preference is for a pyramidal geometry, delocalization of the lone pair on the amino group into the nitro group is best accommodated by a planar structure. Use the B3LYP/6-31G* model to determine the equilibrium geometry of nitroamide, starting from a non-planar structure. If you find that the molecule is non-planar, calculate the barrier to inversion.

Molecular Modeling Problems

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