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Unformatted text preview: C h ap ter P 1 : M olecul ar O rbitals , Electron a nd S pin D en si ty and El ectrostati c P oten ti al Map
Introduction The Hartree-Fock approximation together with the LCAO approximation and the notion of a basis set allows us to turn in this chapter to an entirely different way of thinking about quantum mechanics. Here, “pictures” replace mathematical functions. The most familiar pictures are the solutions themselves (the molecular orbitals). These provide an alternative to Lewis structures for discussion of chemical bonding and a conceptual framework for the treatment of chemical reactivity. The electron density also provides a model for bonding and allows characterization of overall molecular size and shape. The spin density locates unpaired electrons allowing characterization of the properties and reactivities of radicals. Finally, the electrostatic potential shows the distribution of charge in a molecule. Mapped onto the electron density, it provides a means to identify likely acidic or basis sites in a molecule as well as to provide a hint to its solubility. This chapter explores how to represent these and other quantities as pictures, and how to use the pictures to reveal fundamental information about molecular structure and chemical reactivity. 1 Isosurfaces The quantities that we wish to represent in terms of pictures are functions of the Cartesian coordinates in three dimensions. The most familiar way to show them on a two-dimensional computer screen (or on a printed page) it to “cut a 2D plane or slice” through the data and connect locations of equal value as contour lines. While this may provide useful insight, defining the location of the plane is likely to be problematic for any but planar molecules. An alternative and unambiguous approach is to define a surface for which the value of the function is a constant.
f(x,y,z) = constant The constant will be referred to as an isovalue and the surface will be referred to as an isovalue surface or isosurface. We will be concerned with several different isosurfaces. The two most important depict the molecular orbitals and the distribution of electrons or electron density. While it is not obvious what physical meaning may be attached to a particular molecular orbital isosurface, different electron density isosurfaces easily lend themselves to different interpretations, most conspicuously an isosurface that depicts a molecule’s overall size and shape. We will use such an electron density isosurface both for this purpose as well as for the foundation for so-called property maps, in particular, electrostatic potential maps. Here, the value of the electrostatic potential, depicting distribution of charge, is displayed on the “exposed” surface of a molecule. The result is a view of molecular charge distribution from the perspective of a something approaching and “sitting on” the surface. 2 Molecular Orbitals from Atomic Orbitals Molecular orbitals from both Hartree-Fock molecular orbital and density functional models are solutions of approximate forms of the Schrödinger equation. The underlying assumption is that the electrons move independently and that their motions do not correlate. According to the LCAO approximation, molecular orbitals ψi are written as linear combinations of atom-centered basis functions φμ (loosely referred to as atomic orbitals). ψi = Σμ cμi φμ The sum is carried out over all atomic orbitals and cμi are the linear coefficients of the molecular orbitals (often incorrectly referred to as the molecular orbitals themselves). The number of molecular orbitals is the same as the number of atomic orbitals, and the ψ form an orthonormal set of functions. ∫ ψi ψj dτ = δij δ, the Kronecker delta, is 1 for i=j and 0 otherwise Most of the molecules that we will encounter are closed-shell molecules. Here, the molecular orbitals are each occupied with two electrons, starting from the lowest-energy molecular orbital. Since the number of linear combinations will generally be larger than the half the total number of electrons, not all the molecular orbitals will be occupied. For example, within the framework of a minimal basis set, only one of the two molecular orbitals for hydrogen molecule are occupied, and only five of the seven molecular orbitals for water molecule is occupied. The remaining orbitals are unoccupied. While the number of occupied molecular orbitals does not depend on the size of the basis set, the number of unoccupied molecular orbitals grows with increase in basis set size.
The Hartree-Fock method with the minimal STO-3G basis set will be employed for all examples and problems in this chapter dealing with the sizes and shapes of molecular orbitals. A minimal basis set for hydrogen and helium atoms comprises one atomic orbital (1s), for lithium and beryllium, two atomic orbitals (1s and 2s) and for boron through neon, five atomic orbitals (1s,2s,2px,2py,2pz). Except for hydrogen and helium, the 1s orbital is referred to as a core orbital. 1s for hydrogen and helium and the 2s and 2p orbitals for the lithium through neon are referred to as valence orbitals. Core and valence orbitals for sodium and magnesium are 1s,2s,2px,2py,2pz and 3s, respectively and for aluminum through argon are 1s,2s,2px,2py,2pz and 3s,3px,3py,3pz, respectively. For first-row transition metals scandium to zinc, the core is 1s,2s,2px,2py,2pz ,3s,3px,3py,3pz and the valence is 3dzz,3dxx-yy,3dxy,3dxz,3dyx,4s,4px,4py,4pz. 3 If the orbital surface (or set of surfaces) is confined to a single atom or more generally to atoms that are not close together, the orbital is said to be nonbonding. Molecular orbitals that are made up primarily of core atomic orbitals are also classified as non bonding even though they may involve neighboring atoms. Removing electrons from an occupied non-bonding orbital, or adding electrons to an unoccupied molecular orbital should have little overall effect on structure and bonding. If the orbital contains a surface that extends continuously over two neighboring atoms, the orbital is regarded as bonding with respect to these atoms. Removing electrons from an occupied bonding orbital should weaken the bond between these atoms and cause them to move further apart, while adding electrons to an unoccupied bonding orbital should strengthen the bond and cause the atoms to draw closer together. Two different kinds of bonding orbitals are depicted below. The drawing and corresponding isosurface on the left correspond to a σ bonding orbital while the drawing and isosurface on the right correspond to a π bonding orbital. Note that the π-bonding orbital is divided into two regions separated by a node. These regions correspond to isosurfaces of equal magnitude but opposite sign. In between the value of the molecular orbital must go to zero (a change of sign) and the “zero” isosurface is referred to as a node. It is important to recognize that, while the absolute sign of a molecular orbital is unimportant, the sign of one part of an orbital relative to another part of the same orbital is important. A
A B B σ bonding π bonding It is also possible for an orbital to contain a node that divides the region between two neighboring atoms into separate “atomic” regions. Such an orbital is regarded as antibonding with respect to these atoms. Adding electrons to an occupied antibonding orbital should weaken the bond and push the atoms apart, while removing electrons from an unoccupied antibonding orbital should strengthen the bond and draw the atoms closer together. Two different kinds of antibonding orbitals are depicted below. The drawing and corresponding isosurface on the left correspond to a σ 4 antibonding orbital while the drawing and isosurface on the right correspond to a π antibonding orbital. node A B node A B σ antibonding π antibonding Most commonly, σ and π bonding orbitals are referred to as simply σ and π orbitals, while σ and π antibonding orbitals are referred to as σ∗ and π* orbitals (pronounced “σ star” and “π star”, respectively). In summary, bonds can be strengthened either by adding electrons to bonding molecular orbitals or by removing electrons from antibonding molecular orbitals. Conversely, bonds can be weakened either by removing electrons from bonding molecular orbitals or by adding electrons to antibonding molecular orbitals. Occupancy changes to non-bonding molecular orbitals would be expected to have little overall effect on bond srengths. We start by considering only the occupied molecular orbitals for a few simple molecules and later turn to the unoccupied molecular orbitals. 5 Occupied Molecular Orbitals While molecular orbitals actually have no physical meaning, they are commonly related to the bonds and lone pairs in conventional Lewis structures that are familiar to generations of chemists. A few simple examples serve to make the connection. Hydrogen Molecule There is a single occupied molecular orbital, being made up of the sum of the 1s atomic orbitals on the two hydrogen atoms. ψHHσ= 0.551sH + 0.551sH This is a σ bonding orbital, and corresponds to the single line between the hydrogen atoms in the Lewis structure. H-H Removal of an electron (by excitation or ionization) will lead to bond weakening and lengthening. According to Hartree-Fock STO-3G calculations, the HH bond distance in H2 is 0.712Ǻ, while that in H2+ is 1.061Ǻ (an increase of 49%). Carbon Monoxide The seven occupied molecular orbitals of carbon monoxide divide into two core orbitals and five valence orbitals. The two core orbitals are not very interesting and are not shown. Suffice it to say that the lower-energy core orbital is localized almost entirely on oxygen (using primarily the 1s atomic orbital), while the higher-energy core orbital is localized almost entirely on carbon (using primarily the 1s atomic orbital). They do not contribute significantly either to the properties of chemical behavior of carbon monoxide. All five valence molecular orbitals are delocalized over both centers but to different extents. 6 Molecular orbital coefficients for the five valence molecular orbitals are given as the columns in Table P1-1, with the atomic orbitals as the rows. They assume that the molecule is oriented along the Z axis. For example, the highest-energy occupied molecular orbital (the HOMO) is written in terms of 1s, 2s and 2pz atomic orbitals on both carbon and oxygen. ψHOMO = -0.165 1sC + 0.748 2sC + 0.575 2pzC + 0.001 1sO + 0.040 2sO 0.445 2pzO
The occupied molecular orbitals for carbon monoxide are provided in carbon monoxide occupied orbitals in the Chapter P1 directory. Three of the five valence orbitals (the two lowest energy orbitals and the HOMO) are σ orbitals. The two orbitals in between have the same energy (they are said to be degenerate) and are π orbitals. The lowest-energy σ orbital corresponds roughly to the “single bond” part of the CO triple bond in a conventional Lewis structure. The other two σ orbitals may loosely be described as lone pairs. The lower-energy orbital extends more from oxygen than from carbon, while the higher-energy σ orbital (the HOMO) extends more from the carbon end. This is consistent with the fact that carbon monoxide almost always prefers to bind to a metal center from carbon. While the HOMO in carbon monoxide is primarily a nonbonding orbital, it is σ bonding between carbon and oxygen. Removal of an electron (by excitation or ionization) should lead to modest bond weakening and lengthening. According to Hartree-Fock STO-3G calculations, the CO bond distance in carbon monoxide is 1.145Ǻ, while 7 Table P1-1: Valence Molecular Orbitals of Carbon Monoxide σ C 1s 2s 2px 2py 2pz O 1s 2s 2px 2py 2pz -.124 .244 0 0 -.166 -.223 .771 0 0 .210 σ -.170 .559 0 0 -.065 .132 -.643 0 0 .615 π 0 0 .446 0 0 0 0 .794 0 0 π 0 0 0 .446 0 0 0 0 .794 0 σ -.165 .748 0 0 -.575 -.001 .049 0 0 -.445 8 that in the corresponding radical cation is 1.206Ǻ (an increase of only 5%). Note that the Lewis structure requires a formal charge negative charge on carbon and a formal positive charge on oxygen.
- :C O:+ This helps to explain why the dipole moment of CO is nearly zero (with the carbon at the negative end).
Molecular orbitals not only provide insight into molecular geometry and changes in geometry with changing number of electrons, but they may also be used to suggest why a particular chemical reaction occurs as it does, or why a chemical reaction does not occur as expected. For example, the fact that the HOMO in cyanide anion is more concentrated on carbon (on the right) than on nitrogen, suggests, as is observed, that cyanide will act as a carbon nucleophile and not a nitrogen nucleophile. While this might seem counterintuitive as nitrogen is more electronegative than carbon, and thus more likely to hold the negative charge, it is entirely reasonable. Because nitrogen is more electronegative than carbon, it holds on to its electrons better than does carbon, meaning that it will be the poorer nucleophile. 9 Singlet Methylene The four occupied molecular orbitals of singlet methylene divide into one core orbital, which is localized on carbon, and three valence orbitals, which are delocalized onto all three centers. Molecular orbital coefficients for the three valence molecular orbitals are given as the columns in Table P1-2, with the atomic orbitals as the rows. They assume that the molecule is oriented in the YZ plane. For example, the highest-energy occupied molecular orbital (the HOMO) is written in terms of 1s, 2s and 2pz atomic orbitals on both carbon and oxygen. ψHOMO = -0.141 1sC + 0.643 2sC - 0.645 2pzC - 0.234 1sH - 0.234 1sH The occupied molecular orbitals for singlet methylene are provided in methylene occupied orbitals in the Chapter P1 directory. The two lowest-energy orbitals “replace” the two CH bonds in a Lewis structure.
H C H Note that, unlike the bonds in the Lewis structure, both molecular orbitals reflect the fact that the two hydrogens (and the two CH bonds) 10 Table P1-2: Valence Molecular Orbitals of Singlet Methylene σ C 1s 2s 2px 2py 2pz H1 1s H2 1s .215 -.666 0 0 -.173 -.270 -.270 σ 0 0 0 -.539 0 .467 -.467 σ -.141 .643 0 0 -.645 -.234 -.234 11 in methylene are equivalent. More generally, all molecular orbitals must show any equivalences (“symmetries”) that exist in the geometry of the molecule. This means that the orbital surface covering one section of a molecule must either be the same or the negative of the surface covering an equivalent section. The reason that the surface can either be the same (symmetric) or the negative (antisymmetric) is a consequence of the fact that the observable (the electron density) does not depend on the orbital but rather on the square of the orbital. An exception to this rule occurs where the energies of two or more molecular orbitals are exactly the same. This will be discussed when we consider the π molecular orbitals of benzene. The HOMO of singlet methylene, like the higher-energy CH bonding orbital, is also described by two surfaces. The larger surface is directed from carbon away from the two hydrogens and may be described as non bonding. The smaller surface encompasses both CH bonding regions. Although it is hard to track the path of the node, it actually passes almost exactly through the carbon. This means that this surface possesses little if any CH bonding or antibonding character. The non-bonding character of the HOMO dominates, consistent with the fact that singlet methylene behaves as an electronpair donor (a nucleophile). Note, however, that the HOMO is H---H bonding. While removal of an electron (by excitation or ionization) should not lead to significant changes in CH bond distances or bond strengths, it should affect an increase in HCH bond angle. According to Hartree-Fock STO-3G calculations, CH bond distances in methylene and its radical cation are nearly the same (1.123Ǻ vs. 1.132Ǻ) while the bond angle has increased from 101o to 136o. 12 Ethylene The eight occupied molecular orbitals of ethylene divide into two core orbitals and six valence orbitals. The two core orbitals are made up primarily of sums and differences of the 1s atomic orbitals on the two carbon atoms. These are non-bonding molecular orbitals and have little effect on the structure and properties of ethylene. Five of the six valence molecular orbitals reside in the plane of the molecule (they are σ orbitals). They replace the carbon-carbon σ bond and four carbon-hydrogen σ bonds (single lines between C and H) in the Lewis structure. However, unlike the CC and CH bonds in the Lewis structure, they are delocalized onto all six centers and (their squares) reflect the symmetry of ethylene. The remaining molecular orbital rises out of the plane (it is the π orbital and also the HOMO), and involves only the two carbons. It corresponds to the “second” line between the carbons in the Lewis structure. Molecular orbital coefficients for the six valence molecular orbitals are given as the columns in Table P1-3, with the atomic orbitals as the rows. They assume that the molecule is oriented in the XZ plane and that the atoms are numbered as in the diagram below.
H1 C1 H2 C2 H4 H3 13 Table P1-3: Valence Molecular Orbitals of Ethylene σ C 1 1s 2s 2px 2py 2pz C 2 1s 2s 2px 2py 2pz H1 1s H2 1s H3 1s H4 1s -.180 .468 0 0 -.120 .468 -.666 0 0 .120 .112 .112 .112 .112 σ -.136 .410 0 0 .203 -.410 0 0 0 .203 .224 .224 -.224 -.224 σ 0 0 .397 0 0 0 .643 .397 0 0 .256 -.256 -.256 .256 σ -.015 .017 0 0 -.500 -.015 .017 0 0 -.500 -.217 -.217 -.217 -.217 σ 0 0 0 0 0 0 0 0 π 0 0 .632 0 0 0 .632 0 .393 0 -.393 0 .351 0 -.351 0 .351 0 -.351 0 14 The occupied molecular orbitals for ethylene are provided in ethylene occupied orbitals in the Chapter P1 directory. Alternatively, one may view of the five valence molecular orbitals of ethylene as arising from the interaction of the valence orbitals of two singlet methylene fragments (see previous discussion). Addition and subtraction of the lower-energy CH bonding orbital in methylene
15 leads to the two lowest energy CH bonding combinations in ethylene. Addition of the higher-energy CH bonding orbital in methylene leads to the next higher energy CH bonding combination, while subtraction leads to the highest-energy CH bonding combination. The latter is the highest-occupied σ orbital in ethylene (the HOMO is the π orbital). Underneath this is a CH bonding combination that results from addition of the two highest-occupied molecular orbitals of methylene. 16 Benzene Benzene has 21 occupied molecular orbitals divided into six core orbitals and 15 valence orbitals. The valence orbitals are further divided into 12 in-plane σ orbitals (accounting for the six CC and six CH σ bonds and three out-of-plane π orbitals. The “chemistry” of benzene is dictated by its π orbitals. These are shown below. The occupied π orbitals for benzene are provided in benzene pi orbitals in the Chapter P1 directory. The lowest-energy π orbital is bonding between all pairs of carbon atoms, and reflects the six-fold symmetry of benzene. Neither of the two higher energy π orbitals reflects the full symmetry of benzene. (There remains only one linear combination that does. This is a high-energy unoccupied molecular orbital that is antibonding between all pairs of carbon atoms.) This does not violate “the rule” that the square of the wavefunction must exhibit the full symmetry of the molecule, because the two molecular orbitals in question have the same energy (they are degenerate). As discussed in Chapter Y, any linear combination of degenerate orbitals is an acceptable solution, whether or not it reflects the underlying symmetry of the molecule. There is a corollary. The fact that fewer molecular orbitals with the appropriate symmetry are available to contain the valence electrons means that there must be degeneracies among the π orbitals. 17 Unoccupied Molecular Orbitals Occupied molecular orbitals are commonly (and often incorrectly) related to the bonds and lone pairs in the Lewis structure for the ground state. However, such an association will generally not be possible for unoccupied molecular orbitals, which may be viewed as “left over” linear combinations of atomic orbitals. That is to say, there is no Lewis structure “equivalent” of a σ or π antibonding molecular orbital.
Interpreting unoccupied molecular orbitals for any but small basis sets may be problematic especially where diffuse functions are involved. Here, the lowest-energy unoccupied molecular orbitals may not be associated with specific atoms or bonds but rather be dispersed far from far away from them. Examination of the few lowest-energy unoccupied molecular orbitals may be informative in two ways. The first is to anticipate the likely changes to bonding that will occur in moving beyond the Hartree-Fock model. Recall from our discussion of full configuration interaction (Chapter X) that a better wavefunction may be constructed by adding to the ground state (Hartree-Fock) wavefunction contributions from all possible excited-state wavefunctions. These in turn may be seen as arising from promotion of one or more electrons from filled molecular orbitals (in the Hartree-Fock wavefunction) to unfilled molecular orbitals. For example, promotion from the highest-filled to lowest-unfilled molecular orbitals in ethylene (π to π*) removes an electron from an orbital that is bonding with respect to the two carbons and puts it into an orbital that is antibonding. This implies that the CC bond length in the ππ* excited state of ethylene will be longer than that in the ground state. It also implies that the CC bond length obtained from the Hartree-Fock model will be too short. We will explore this in greater detail in Chapter P2. The second use of unoccupied molecular orbitals is to anticipate the reactivity of molecules with nucleophiles in the same way that that examination of occupied molecular orbital may be used to anticipate the reactivity with electrophiles. In particular, a molecule in which the lowest18 unoccupied molecular orbital (the LUMO) is accessible may be an ideal candidate for attack by a nucleophile, and the location of the LUMO may reveal where the attack is likely to occur. For example, the LUMO in acetone is very low in energy and primarily localized on the carbonyl carbon. This anticipates the dominant chemical reaction of carbonyl compounds. The LUMO in methylene is localized entirely on carbon (it looks like a 2p atomic orbital) and is therefore nonbonding in character. Were this orbital to accept electrons, the CH bonds would not be affected. Triplet Methylene Methylene has received a great deal of attention among chemists, both experimental and theoretical. The singlet state that we have described with the two highest energy electrons in an in-plane σ orbital is significantly lower in energy than the singlet that would arise were one of the electrons to be promoted to the LUMO (an out-of-plane π orbital). However, the ground state of methylene is not a singlet but rather a triplet. Here, four different valence molecular orbitals are occupied, not three as in the singlet. Two are doubly occupied and closely correspond to two the CH bonding orbitals in the singlet. The remaining two are singly occupied, one corresponding to the HOMO of the singlet and the other to the LUMO. 19 Triplet State of Methylene: On the basis of the HOMO and LUMO of singlet methylene, predict whether the HCH bond angle in the corresponding triplet will be smaller, larger or about the same. Elaborate. Obtain equilibrium geometries for both states of methylene using the STO-3G model to test your prediction. Mo2. Display the valence molecular orbitals of dimolybdenum as obtained from an STO3G calculation assuming a Mo-Mo bond distance of 1.94Ǻ. Classify each as bonding or antibonding and as σ, π or δ. How many bonds connect the two molybdenum atoms? Chromium-Chromium Quadruple Bonds: Numerous compounds with quadruple bonds between transition metals have been reported. Among them are compounds in which two chromium atoms a bridged by four acetate (MeCO2-) groups with or without axial ligands, for example, pyridine. Using the experimental crystal structure [chromium-chromium quadruple bond] perform an STO-3G energy calculation. Starting from the highest energy, 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? Would you characterize this molecule a incorporating a Cr-Cr quadruple bond? Characterize each of as a σ, π or δ orbitals and rationalize the ordering of their energies. Diborane and Ethylene: Diborane and ethylene are isoelectronic, that is, they have the same number of electrons. Are the molecular orbitals that accommodate these electrons are similar? Obtain equilibrium geometries for diborane and ethylene using the STO-3G model. Associate each of the six (occupied) valence molecular orbitals for diborane with an orbital in ethylene. 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? Borazine and Boraphosphazine: Both borazine and boraphosphazine have a total of six π electrons (the same as benzene), three from each nitrogen (phosphorous) lone pair. As with benzene, both molecules are planar, and all six bonds that make up the ring are the same length as benzene. 20 HN HB H B N H NH BH Obtain equilibrium geometries for borazine and boraphosphazine, and compare the three occupied π molecular orbitals of each with those presented earlier for benzene. Use the STO-3G model. Point out similarities and differences. Specifically, is there evidence for localization of the π orbitals onto the electronegative nitrogen (phosphorus) atoms? Lithium and Sodium Hydroxide: Water is bent (the HOH angle is ~105o) but both lithium hydroxide and sodium hydroxide are linear. A clue to the cause behind the change in geometry follows by comparing their highest occupied molecular orbitals. Use the STO-3G model to obtain equilibrium geometries for the three molecules. 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? Adding and Removing Electrons: Molecular orbitals are delocalized throughout the molecule and 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). Examine the HOMO for each, and guess what would happen to the geometry around carbon (remain planar vs. pyramidalize), to the C=X bond length and (for formaldimine) to the C=NH bond angle, were an electron to be removed from this orbital. Choose from the possibilities listed below. remove electron add electron from HOMO to LUMO H2C=CH2 geometry around carbon C=C remain planar pyramidalize remain planar pyramidalize lengthen lengthen shorten shorten remain the same remain the same remain planar pyramidalize remain planar pyramidalize H2C=NH geometry around carbon C=N lengthen lengthen shorten shorten remain the same remain the same increase increase decrease decrease remain the same remain the same
21 <CNH H2C=O geometry around carbon C=O remain planar pyramidalize remain planar pyramidalize lengthen lengthen shorten shorten remain the same remain the same 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. Choose from the possibilities listed above. 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? The first excited state of formaldehyde (the so-called n−>π* state) can be thought of as arising from the promotion of an electron from the HOMO (in the ground-state of formaldehyde) to the LUMO. The experimental equilibrium geometry of the molecule shows that the carbon has pyramidalized and that the CO bond has lengthened. Rationalize the experimental result using what you know about the HOMO and LUMO of formaldehyde and your experience with calculations on the radical cation and radical anion of formaldehyde. SN2 Reaction of Methyl Iodide: Attack of a nucleophile, for example cyanide anion, onto the backside of methyl iodide leads to acetonitrile and iodide anion.
N C– CH3 I N C CH3 + 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 Hartree-Fock 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? Dimer of Nitric Oxide: Nitric oxide (NO) only reluctantly forms a dimer, the structure of which is believed to in the form of a four-membered ring rather than open chain.
O N O N O N N O Use the Hartree-Fock STO-3G model to obtain the equilibrium geometry of nitric oxide. Display the highest-energy occupied molecular orbitals for both spin manifolds. Where is the unpaired electron? What is the total NO bond order? Draw a correlation 22 diagram for two NO molecules leading to the cyclic dimer and predict the form of the highest-energy occupied molecular orbital (in the dimer). Use the STO-3G model to obtain the geometry of the cyclic dimer to confirm or contest your prediction. An alternative dimer in which the nitrogen on one nitric oxide interacts with the oxygen on the other (and vice versa) is not observed. Use the same molecular orbital displays and correlation diagram to rationalize why. 23 Orbital Energies A common misconception is that the total (electronic) energy of a molecule is given by the sum of the energies of the occupied molecular orbitals. In fact, the correct expression for the total energy is much more complicated, but can be expressed in a form that involves the orbital energies. expression of total energy involving orbital energies This, aside, the energies of the HOMO and LUMO (the frontier molecular orbitals) and the difference in energy between them are important quantities to many chemists in that they can be related to quantities that can actually be measured. The HOMO is related to the ionization potential, the energy required to completely remove an electron from an atom or molecule. The LUMO is related to the electron affinity, the gain in energy following capture of an electron. The difference in energy between the HOMO and LUMO (the so-called HOMO-LUMO gap) is related to the energy of a UV/visible spectral transition.
There is a serious problem with interpreting the LUMO energy in terms of an electron affinity and the HOMO-LUMO gap in terms of a spectral transition. The LUMO energy is almost always a positive quantity implying that the electron affinities are almost always negative (an electron will not add) and meaning that the HOMO-LUMO gap is almost always larger than the energy required for ionization. These are both false. 24 Electron Donor and Acceptor Groups: The extent to which filled and empty molecular orbital interacts and the stabilization that results from interaction, depends both on the extent to which the orbitals overlap and on their difference in energy. A good electrondonor group is one with a high energy HOMO, and a good electron-acceptor group is one with a low energy LUMO. Use the HF/6-31G* model to calculate equilibrium geometries for methylamine, methanol, methane thiol and methyl chloride (methyl attached to electron-donor groups) and acetonitrile, methyl formate, nitromethane and 1,1,1-trifluoroethane (methyl attached to electron-acceptor groups). Rank the four donors and the four acceptors. 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) and significant deviations from linear correlation. Rates of Diels-Alder Reactions: The rates of Diels-Alder reactions generally increase with the π-donor ability of electron-donor groups (EDGs) on the diene, and the with πacceptor ability of electron-withdrawing groups (EWGs) on the dienophile.
+ EDG Y EDG = R, OR EWG = CN, CHO, CO2 H EWG X Donors “push up” the energy of the HOMO on the diene and acceptors “pull down” the energy of the LUMO on the dienophile, and the resulting decrease in HOMO-LUMO gap, leads to stronger interaction between diene and dienophile and to a decrease in energy barrier.
LUMO diene Orbital Energy HOMO dienophile Experimental rates for Diels-Alder cycloadditions of cyclopentadiene and a series of cyano-substituted alkene dienophiles are provided below (expressed in log units relative to the rate of cyclopentadiene and acrylonitrile). 25 acrylonitrile (cyanoethylene) 0 trans-1,2-dicyanoethylene 1.89 cis-1,2-dicyanoethylene 1.94 1,1-dicyanoethylene 4.64 tricyanoethylene 5.66 tetracyanoethylene 7.61 Obtain geometries for these compounds using the STO-3G model with the basis set. Plot LUMO energy vs. the log of the relative rate. Which dienophile has the smallest LUMO energy (leads to the smallest HOMO-LUMO gap)? Is this the dienophile that reacts most rapidly with cyclopentadiene? Which has the largest LUMO energy? Is this the compound that reacts most slowly? Does the HOMO-LUMO gap correlate with relative reaction rate? HOMO/LUMO Gap and UV/visible Spectra: A simple model for UV/visible spectroscopy is that light absorption causes an electron to jump from the HOMO to the LUMO. The smaller the HOMO-LUMO gap the lower the energy of the spectral transition. As orbital energies depend on structure, it is reasonable to expect that the change in the HOMO/LUMO gap with change in molecular structure will anticipate the change in color. Use the STO-3G model to obtain equilibrium geometries for azobenzene (R=R´=H), 4hydroxyazobenzene (R=OH, R´=H) and 4-amino-4´-nitroazobenzene (R=NH2, R´=NO2).
R N N 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. 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 first (lowest-energy) excited state arises 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 may offer clues about the geometry of its first 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 using the STO-3G model 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. 26 Symmetry of Orbital Interaction. The Woodward-Hoffmann Rules Woodward and Hoffmann [R.B. Woodward and R. Hoffmann, The Conservation of Orbital Symmetry, Verlag-Chemie, Weinhein, 1970] introduced organic chemists to the idea that interactions high-energy occupied and low-energy unoccupied molecular orbitals may be key to understanding why some chemical reactions proceed easily whereas others do not. For example, the fact that 1,3-butadiene and ethylene undergo (Diels-Alder) reaction to form cyclohexene may be seen as a consequence of the fact that the HOMO of the former may interact constructively with the LUMO of the latter. + On the other hand, interaction between the HOMO on one ethylene and the LUMO on another ethylene is not favorable, and reaction to form cyclobutane should not easily occur. + The success of the Woodward-Hoffmann rules and more generally the use of molecular orbitals for reagents to anticipate chemical reactivity and product selectivity may be traced to the Hammond Postulate. This states that the transition state for an exothermic reaction will more closely resemble reactants than products. Aromaticity of Diels-Alder Transition State: Another way to phrase the WoodwardHoffmann rules is in terms of special stabilization or “aromaticity” of its transition state. In the case of the Diels-Alder reaction this suggests that its occupied molecular orbitals will resemble the three occupied π orbitals of benzene. Use the Hartree-Fock 3-21G
27 model to obtain both the transition state geometry for the Diels-Alder reaction of ethylene and 1,3-butadiene, and the equilibrium geometry of benzene. Examine the several highest-occupied molecular orbitals in the transition state. Are there three that correspond closely to the three π orbitals of benzene? Elaborate. 28 Electron Density The electron density (or more simply density) for a closed-shell molecule at a particular point in space, ρ(x,y,z), gives the number of electrons at this location. It results from summing the product of all pairs of atomic orbitals, φ, each pair multiplied by an element of the density matrix, P. ρ(x,y,z) = ΣμΣυPμυφμ(x,y,z) φυ(x,y,z) The latter is made up of a sum over all occupied molecular orbitals, where the c are the molecular orbital coefficients. The factor of two accounts for the fact that (in a closed-shell molecule) each molecular orbital is doubly occupied. Pμυ= 2Σkcμkcυk
This expression may be generalized to molecules with one or more unpaired electrons. In the so-called unrestricted formalism that is most commonly employed, the two spin systems are treated independently leading to two different density matrices. Further details are provided later in the chapter. The electron density provides information about the positions of the nuclei (in the center of regions of highest density), about the bonding in molecules (which atoms are connected to which atoms) and about overall molecular size and shape (the shapes of regions needed to contain a large fraction of the total number of electrons). As illustrated below for benzene, these different interpretations follow from different values of the electron density. Electron density surfaces for benzene are provided in benzene electron densities in the Chapter P1 directory. The three density surfaces (left to right) correspond to isovalues that lead to enclosure of approximately 25%, 50% and 98% of the total number of electrons. 29 Specific isovalues which lead to these enclosures for Hartree-Fock 6-31G* calculation are 0.35, 0.1 and 0.002 electrons2/au3, respectively (au stands for the atomic unit of length; 1 au = 1.8898Å). The image to the extreme right is a conventional space-filling model.
The Hartree-Fock method with the 6-31G* basis set will be used for all examples and all problems for this and the remaining sections in this chapter. For a first-row element, this comprises a core made up of 6 s-type Gaussians and a valence that is split into only two parts made up of 3 and 1 s–type Gaussians and 3 and 1 sets of p-type Gaussians (sharing the same exponents), supplemented by a set of d-type Gaussians. Hydrogens are represented by two sets of s-type Gaussians, the first comprising 3 Gaussians and the second a single Gaussian. Regions of highest density (left surface) surround the carbons and identify their locations. While this does not provide “new” information (atomic locations are part of the input for the calculation), the connection between electron density and atomic positions is the basis behind the use of X-ray diffraction as a means of establishing molecular structure. The fact that hydrogens are not associated with the highest electron densities and as a consequence are “not seen”, is consistent with the difficulty of assigning hydrogen positions from X-ray diffraction. Regions of slightly lower density (center surface) do reveal hydrogen locations as well as the “bonds” between atoms. The latter is essentially the same information conveyed by a conventional skeletal model. Note that this is “new” information as bonds are not part of the input for a calculation.
Actually, the center of electron density “on hydrogen” is shifted toward the atom to which it is bonded. This results in bonds to hydrogen from X-ray diffraction that are too short by a tenth of an Å or more. An even lower value of the density (right surface) may be used to portray overall molecular size and shape. This is essentially the same information provided by a conventional space-filling (CPK) model. Both convey how closely molecules can “pack” in a liquid or solid and, therefore, “how close” a reagent can get to a molecule in advance of reacting with it. Thus, it is natural to use this density surfaces as the basis of property maps, in particular, electrostatic potential maps. 30 CPK Models vs. Electron Density Surfaces for Hydrocarbons: To what extent do the sizes and shapes of molecules depicted by simple space-filling (CPK) models reflect those based on electron density surfaces? Determine density surfaces for methane, ethane, propane, n-butane, n-pentane and n-hexane that enclose 98% of the total number of electrons. Use the Hartree-Fock model with the 6-31G* basis set. Plot the resulting enclosure volumes against the volumes of the corresponding space-filling models. Sizes of Alkali Metal Cations: Determine density surfaces for lithium, sodium and potassium cations that enclose 98% of the total number of electrons. Use the HartreeFock model with the 6-31G* basis set. Compare the three surfaces. 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.) Sizes of Hydride and Halide Anions: Determine density surfaces for hydride, fluoride chloride and bromide anions that enclose 98% of the total number of electrons. Use the Hartree-Fock model with the 6-31G* basis set. Compare the three surfaces. Which is smallest and which is largest? Rational 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.) Methyl Anion, Ammonia and Hydronium Cation: Obtain equilibrium geometries for the methyl anion (CH3-), ammonia and hydronium cation (H3O+) using the Hartree-Fock model with the 6-31G* basis set. All three have 8 valence electrons (they are isoelectronic). For each, calculate and display an electron density surface selecting the isovalue such that it encloses 98% of the total electron density. Compare the three surfaces. Which depicts smallest molecule and which depicts the 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? 31 Spin Density An electron can have one of two possible spin states, α or β. Most molecules have an even number of electrons, half of which are α or “spinup” ( ) and half of which are β or “spin-down” ( ). That is to say, the electrons occur in pairs, one α and the other β. Molecules in which all electrons are paired are referred to as closed-shell molecules. Molecules with one unpaired electron are known as free radicals or simply radicals. While common as reactive intermediates, radicals are rarely sufficiently stable to allow them to be isolated. Rather, if given the chance they combine to make closed-shell molecules, for example, two methyl radicals combine to make ethane. H3C + CH3 H3C-CH3 Molecules with two unpaired electrons are referred to as triplets. The most familiar example of a molecule with a triplet ground state is the oxygen molecule. Triplets have an even number of electrons, meaning that there is a corresponding singlet state (all electrons paired). Singlet oxygen is significantly higher in energy than triplet oxygen. Radicals and other molecules with unpaired electrons may be detected by Electron Spin Resonance (ESR) spectroscopy. Details of the experiment. The spin density at a particular point in space, ρspin(x,y,z), gives the difference in density formed by electrons of α spin and the density formed by electrons of β spin. It results from the summing the product of all pairs of atomic orbitals, φ, each pair multiplied by the difference between an element of the α and β density matrices, Pα and Pβ, respectively. ρspin(x,y,z) = ΣμΣυ(Pαμυ-Pβμυ)φμ(x,y,z) φυ(x,y,z) The latter are made up of sums over all occupied α and β molecular orbitals, where the c’s are molecular orbital coefficients. Pαμυ= 2Σkcαμkcαυk Pβμυ= 2Σkcβμkcβυk The spin density is zero everywhere for a molecule in which all electrons are paired. For a molecule with one or more unpaired electrons, regions of nonzero spin density “locate” the unpaired electrons. For example, the spin density surface for the allyl radical shows that the unpaired electron resides on the terminal carbons and not on the central carbon, in accord with Lewis
32 structures. Note, however, that unlike the Lewis structure, the calculated spin density shows deviation from pairing at the hydrogen bonded to the central carbon.
• • The spin density surface for allyl radical is provided allyl radical spin density in the Chapter P1 directory. The usual notion is that a radical in which the unpaired electron is delocalized will be less reactive than a radical in which the unpaired electron is localized, that is, resides on a single atom. Thus, allyl radical would be expected to be less reactive than 2-propyl radical. An important practical example is provided by the role which vitamin E is believed to play in scavenging free radicals that might otherwise cause damage through their reaction with unsaturated fatty acids found in cellular membranes. Vitamin E transfers a hydrogen atom to the radical to give stable products that can then be safely excreted. Such compounds are referred to as antioxidants.
R• + vitamin E
O HO vitamin E RH + vitamin E• The spin density surface clearly shows that the unpaired is spread over the three of the carbon in the aromatic ring. 33 The spin density surface for the radical formed by hydrogen abstraction from vitamin E is provided vitamin E radical spin density in the Chapter P1 directory. 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 it is to react. 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? Most delocalized? Obtain equilibrium geometries for ethane, ethylene, acetylene and benzene (as well as hydrogen atom) and calculate CH bond energies. Do these parallel the extent of spin delocalization seen in the corresponding radicals? 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, Obtain the geometry of benzene radical cation using the STO3G model. 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? 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 Hartree-Fock model and the 6-31G* basis set 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? BHT: 3,5-di-tert-butyl-4-hydroxytoluene (butylated hydroxytoluene or BHT) is one of a class of synthetic antioxidants. 34 Use the Hartree-Fock model with the 6-31G* basis set to obtain an equilibrium geometry for the radical formed from BHT by removal of the hydrogen on the hydroxyl group. Compare the spin density to that for the radical formed from Vitamin E (see text). Is the radical formed from BHT likely to be more or less reactive than that formed from vitamin E? Elaborate. 35 Electrostatic Potential The electrostatic potential is the energy of interaction of a positive point charge located in the vicinity of a molecule with the nuclei and electrons of the molecule. ε(p) = ΣAZA/rAp - ΣμΣυ Pμυ∫(φμ(r)φυ(r))/rpdr The first summation is over all nuclei with nuclear charge (atomic number) ZA. rAp is the distance between the nucleus A and the point charge. The second summation is over all pairs of basis functions. Pμυ is an element of the density matrix. Integration is over all space and rp is the distance between the point charge and the integration variable. Note that this expression is not exact and corresponds only to the first term in a power series expansion. Higher-order terms in the expansion account for polarization of electron density. Regions of negative electrostatic potential are associated with electrons that are loosely bound, for example, the π electrons in ethylene. Aside from the sign, the electrostatic potential surface qualitatively resembles the HOMO of ethylene. negative potential HOMO For a molecule like ammonia with a single non-bonded pair of electrons, a surface of negative potential is also similar to the HOMO, although only the latter shows a significant component on the hydrogens. negative potential HOMO 36 Where there are several non-bonded pairs, for example, two pairs in water and three pairs in hydrogen fluoride, parallels between negative electrostatic potential surfaces and HOMO’s are less appropriate. negative potential HOMO negative potential HOMO The reason is that potential surfaces weight all occupied molecular orbitals according to their availability, the one, two and three non-bonded electron pairs in ammonia, water and hydrogen fluoride, respectively, receiving the greatest weight. Finally, note that the electrostatic potential surfaces for these three molecules correlate closely with conventional Lewis structures. The Lewis structure for ammonia has a single lone pair pointing away from the cone defined by the nitrogen and three hydrogens, while that for water has two lone pairs pointing in the remaining two tetrahedral directions and the Lewis structure for hydrogen fluoride has three lone pairs arranged tetrahedrally around fluorine.
N H H H O H H F H Negative electrostatic potential surfaces may be used to distinguish among molecules that appear to be structurally very similar. A good example is provided by comparison of isosurfaces for benzene and pyridine for which of electrostatic potential is -40 kJ/mol. Whereas space-filling models and electron density surfaces for the two molecules are nearly identical, 37 the electrostatic potential surfaces are completely different. As expected, the negative potential surface for benzene shows that it is the π electrons that are loosely bound. These extend equally above and below the plane of the ring. Quite to the contrary, the loosely-bound electrons of pyridine are in the plane of the ring. The “chemistry” of the two molecules would therefore be expected to be entirely different.
Electron density and electrostatic potential surfaces for benzene and pyridine are provided in benzene and pyridine electron densities and electrostatic potentials in the Chapter P1 directory. Surfaces of constant positive electrostatic potential are almost always less informative, as they are dominated by regions close by the nuclei (and not available “to do chemistry”). However, as we shall see in a later section, the “exposed” area of positive potential may be a useful indicator of acid strength.
Electrostatic Potential Surfaces for Hydrocarbons: Obtain equilibrium geometries for ethylene, acetylene, allene, cyclopropane and cyclobutane using the Hartree-Fock model with the 6-31G* basis set, and for each calculate and display an electrostatic potential surface with an isovalue of -80 kJ/mol. For which molecule, ethylene or acetylene, does the surface extend further from the molecule? Which molecule would you expect to be the more reactive toward electrophiles? Does the electrostatic potential surface for allene look like a composite of two ethylene surfaces of are there differences. Which molecule, cyclopropane or cyclobutane, might be expected to more behave like ethylene toward electrophiles? Explain. Electrostatic Potential Surfaces for Carbon Monoxide: Obtain the equilibrium geometry of carbon monoxide using the Hartree-Fock model with the 6-31G* basis set and display an electrostatic potential surface with an isovalue of -80 kJ/mol. Question or explanation that it incorrectly suggests that the oxygen end should be more reactive than the carbon end. 38 Using the Electrostatic Potential to Obtain Atomic Charges Charges are part of the everyday language of chemistry and among the most common quantities demanded from quantum chemical calculations. Despite this, they are neither measurable nor may they be determined uniquely from calculation. The reason is easy to understand. While nuclei may be treated as point charges, electrons need be treated in terms of a distribution that extends throughout all space (even though it is primarily concentrated in regions around the individual nuclei and between nuclei that are close together). For example, while a surface for hydrogen fluoride that encloses a large fraction of the total number of electrons reflects the fact that more electrons are associated with fluorine than with hydrogen (consistent with F ), it does not reveal how the direction of its dipole moment, H many electrons belong to each nucleus.
H F There is no unique way to proportion electrons, although there are a number of well-defined recipes that seem to lead to reasonable atomic charges. One of the more popular of these is based on fitting the electrostatic potential. This involves three steps starting with a wavefunction. 1. Select a set of (several thousand) points located outside the van der Waals surface. In quantum chemical terms, this is a surface that encloses 98% or 99% of the total number of electrons (see discussion earlier in this chapter). 2. Calculate the electrostatic potential at each of these points. 3. Fit the set of energies to an approximate potential in which atomcentered charges have replaced nuclei and electrons. The charges are parameters in the fitting procedure, subject to the requirement that they sum to the total charge on the molecule. While it is not possible to say if charges obtained by fitting the electrostatic potential (or by any other means) are “right”, simply because there is no “right”. The best that can be done is to conclude that they are “reasonable” based on what we know (or believe we know) about a molecule’s overall charge distribution. 39 A major shortcoming of atomic charges based on fits to the electrostatic potential is the need to be able to “see” all the atoms (not only the one of interest). Even if only a single is hidden from view, for example, the iron in ferrocene, the charges on the remaining atoms are ill defined as the total charge on the atoms that are visible is ill defined. In these situations an alternative method for charge assignment is needed. One such alternative is due to Mulliken. This starts with the idea the number of electrons at any point in space, ρ(r), may be written as a summation over the product of all combinations of atomic basis functions, φμ(r) and φυ(r), each multiplied by an element of the density matrix, Pμυ. ρ(r) = ∑μ ∑υ Pμυ φμ(r)φυ(r) Integrating over all space, leads to a number which is equal to the total number of electrons in the molecule. This may be expressed as a sum of products of two numbers, the first an element of the density matrix and the second an element of the overlap matrix, S, which reflects the extent to which two basis functions interpenetrate. ∫ρ(r) = ∑μ ∑υ Pμυ ∫φμ(r)φυ(r)dr = ∑μ ∑υ Pμυ Sμυ = ∑μPμμ + 2 ∑μ ∑υ Pμυ Sμυ = n It is reasonable to assign to assign the electrons associated with a diagonal element to the atom on which the basis function resides. It is also reasonable to assign the electrons associated with an off-diagonal element where both functions are on a single atom to that atom. What is not apparent is how to accommodate electrons that are associated with two different basis functions, that is, products Pμυ Sμυ where μ and υ are on different atoms. The Mulliken recipe gives half of the total to each atom (well defined but completely arbitrary) and leads to a set of electron populations associated with the basis functions, qμ and to a set of atomic electron populations, qA. Corrected for atomic number, the latter give rise to Mulliken charges, QA. qμ = Pμμ + 2 ∑μ ∑≠υ Pμυ Sμυ qA = Pμμ + 2 ∑μ qμ QA = ZA - qA 40 Electrostatic Charges: Which step in the recipe for electrostatic charges outlined above leads to ambiguity? 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 reasonable relationship? 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 electrostatic 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? Dimethylsulfoxide and Dimethylsulfone: Bonding in dimethylsulfoxide, (H3C)2SO, and dimethylsulfone, (H3C)2SO2, can either be described as hypervalent, meaning that sulfur is surrounded by more than the normal complement of eight valence electrons, or as zwitterionic, meaning that sulfur maintains its octet but bears unit positive charge in dimethylsulfoxide and dipositive charge in dimethylsulfone. Calculated charges for the two molecules relative to that for dimethylsulfide, (CH3)2S, should be able to suggest which of the two descriptions is more appropriate, or whether neither is appropriate. Compare sulfur charges for dimethylsufoxide, dimethylsulfone and dimethylsulfide. Use equilibrium geometries from the HF/6-31G* model. Is the sulfur in dimethylsulfoxide significantly more positive than that in dimethylsulfide? Do the calculations suggest that it bears a full positive charge as required by a zwitterionic bonding model? Is the sulfur in dimethylsulfone even more positively charged than that in dimethylsulfoxide? Would you conclude that dimethylsulfoxide and dimethylsulfone are best represented as hypervalent molecules, as zwitterions, or something in between? Anticipating Electrophilic Reactivity of Alkenes: Bromination of alkenes involves a two-step reaction in which the first is addition of Br+ (an electrophile) and the second step 41 is trans addition of Br- (a nucleophile) to the intermediate cyclic bromonium ion, for example bromination of cyclopentene.
Br2 H Br+ H Br– Br H H Br Calculate equilibrium geometry for cyclopentene bromonium ion using the HF/6-31G* model and examine the electrostatic charge on bromine. Is it significantly positive (as implied by the drawing above)? 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? Lithium Aluminum Hydride and Sodium Borohydride: Lithium aluminum hydride (LiAlH4) commonly known as LAH is one of the most important reducing agents available to organic chemists. Use the B3LYP/6-31G* model to investigate its geometry. Do calculated charges suggest that it is better described as an ion pair, that is, a weak complex between lithium cation and aluminum hydride anion (AlH4-), or as covalently bound? Repeat your calculations and analysis for sodium borohydride (NaBH4). 42 Dipole Moments and Atomic Charges The dipole moment of a molecule may be considered as the imbalance of nuclear and electmay be written in terms We
Dipole Moments for Ammonia and Trifluoroamine: Whereas ammonia has a sizable dipole moment (1.47 debyes) that for trifluoroamine is very small (0.23 debyes). One possible explanation is that the ammonia is more “puckered” than trifluoroamine (HNH bond angle is smaller than FNF bond angle). Another is that NH bonds are polarized with nitrogen at the negative end, “adding to” the nitrogen lone pair whereas NF bond are polarized with nitrogen at the positive end, “subtracting from” the lone pair. To decide if the first explanation is plausible, calculate equilibrium for ammonia and trifluoroamine using the B3LYP/6-31G* model. Is the HNH bond angle smaller than the FNF bond angle? To decide if the second explanation is plausible, compare electrostatic potential maps for the two molecules. Is there a marked difference in polarity of the two molecules? cis and trans-Diazene and Difluorodiazene: Calculate equilibrium geometries for both cis and trans forms of diazene, HN=NH. Which is more stable? Provide a rationale for your result. Repeat your calculations for difluorodiazene. Does this show the same preference? If not, provide an explanation as to why not. 43 Electrostatic Potential Map We have seen that an electron density surface may be constructed to portray overall molecular size and shape. Therefore, it may be used to identify how close a molecule can get to its environment or to an incoming reagent. Were the electrostatic potential to be calculated at all accessible locations on this density surface and then displayed in the same obvious manner “on top of” the surface, it would provide a powerful “visual” means to convey molecular charge distribution. "electron density" This composite display of electron density and electrostatic potential is known as an electrostatic potential map. In practice, it is constructed by “coloring” each location on the density surface according to the value of the potential evaluated at that location. By convention, colors toward the “red” end of the visible spectrum represent negative values of the potential, colors toward the “blue” end represent positive values of the potential and colors in the “green” represent values of the potential that are close to zero.
The origin of this arbitrary convention is that red light is lower in energy (more negative) that blue light. For example, an electrostatic potential map for benzene is constructed by first specifying an electron density surface that depicts the molecule’s overall size and shape, that is a surface which encloses ~98% of the total number of electrons. 44 The electron density surface for benzene is provided in benzene electron density in the Chapter P1 directory. Next, the value of the electrostatic potential is evaluated for “all points” on this surface and the regions surrounding these points are colored according to value of the potential. In the case of benzene, regions above and below the π system will be negative and colored red, whereas regions around the periphery of the ring will be positive and colored blue. Note that, except for coloring, the electron density surface is unchanged (insofar as depicting the size and shape of benzene). The monochrome image, conveying only structural information, has been replaced by a color image, conveying the value of the electrostatic potential in addition to structure). The electrostatic potential map for benzene is provided in benzene electrostatic potential map in the Chapter P1 directory. The observed “perpendicular” geometry of the dimer of benzene is easily rationalized from the electrostatic potential map of the monomer, which clearly shows opposite charge distribution for the σ and π systems. In order to take advantage of favorable electrostatic interactions instead of “suffering” from them, benzene dimer prefers to adopt a perpendicular instead of a parallel geometry. 45 Electrostatic potential maps for perpendicular and parallel structures of benzene dimer are provided in benzene dimer electrostatic potential map in the Chapter P1 directory. Carrying the argument one step further, electrostatics can explain why benzene does not crystallize in a stack, but instead prefers a perpendicular arrangement, albeit a perpendicular arrangement in three dimensions. A space-filling model for a small chunk of crystalline benzene is provided in benzene crystal in the Chapter P1 directory. There is a very important lesson in this example. While in one dimension a stack is a very compact arrangement, it is actually a rather akward packing arrangement in three dimensions. Electrostatic potential maps have found a myriad of uses. A particularly important application is to characterize regions in a molecule as electron rich or electron poor and to provide indication of the extent to which a molecule presents a hydrophilic or hydrophobic environment. This can then be correlated with solubility or transport properties, quantities that are very difficult to calculate directly. For example, the electrostatic potential map for icotidine (left) shows a much more polar exposed surface than that for amitripyline (right), suggesting that latter would be less soluble in water (blood) and more likely to move through a cellular membrane. 46 icotidine amitripyline The plot below shows that the area on an electrostatic potential map for which the absolute value of the electrostatic potential is >100 kJ/mol (the socalled polar area) correlates reasonably well with the log of the rate of the transport between blood and brain. Electrostatic potential maps for the compounds shown in this plot are provided in bloodbrain transport vs. polar area in the Chapter P1 directory. 47 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 this reflected in the relative values of the electrostatic potential (at hydrogen) in the three molecules? Use the HF/6-31G* model to obtain geometries and electrostatic potential maps for the three hydrides. Display the three maps side-by-side and on the same color scale. Does the potential at hydrogen show the expected trend? Elaborate. 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. Strengthening a Carboxylic Acid: Electrostatic potential maps may be used to anticipate the effects of substituents on acid strength, for example on the strength of formic acid. This provides an alternative to experiment for “engineering” molecules with particularly high (or low) acidities. Use the HF/6-31G* model to obtain equilibrium geometries for a series of substituted formic acids (XCO2H, X=H, CH3, CF3, phenyl). Rank the acids in order of increasing strength. Examine one or more additional X groups in an attempt to find an even stronger acid. Rationalize your result. Is Azulene Composed of Two Aromatic Ions? Azulene is commonly represented as the fusion of two aromatic ions, both with 6π electrons, the seven-membered ring cycloheptatrienyl (tropylium) cation and the five-membered ring cyclopentadienyl anion. If this picture is at all realistic, then the “cycloheptatrienyl side” of azulene should be positively charged while the “cyclopentadienyl side” should be negatively charged. A suitable “neutral” reference compound is naphthalene (a more stable C10H8 compound). Use the Hartree-Fock 6-31G* model to calculate equilibrium geometries and electrostatic potential maps for naphthalene and azulene. Display the two maps side-by-side and on the same scale. Do you see evidence of charge separation in azulene? Is it in the expected direction? What effect, if any, would you expect charge separation to have on the energy of azulene relative to that of a molecule in which there is no significant charge separation? Elaborate. Acetic Acid Dimer: Acetic acids forms a symmetrical hydrogen-bonded dimer.
O H3C C O H O HO C CH3 48 Obtain equilibrium geometries for both acetic acid and acetic acid dimer using the Hartree-Fock 6-31G* model. Display electrostatic potential maps for the two side-by-side and on the same scale. Is there evidence for charge transfer away from the acidic hydrogen in acetic acid onto to oxygen to which it hydrogen bonded? Do electrostatic charges support your conclusion? Lewis Acid Catalysis of Diels-Alder Reactions Lewis acids such as BF3 are known to accelerate Diels-Alder reactions. This may be rationalized by assuming that they complex to the dienophile, for example, to the nitrogen in acrylonitrile, and withdraw electrons to make it more electron deficient. This should lead to a lowering of lowest-energy molecular orbital (LUMO) and decrease in the HOMO-LUMO gap (see discussion in the previous problem). Use the B3LYP/6-31G* model to obtain equilibrium geometries for acrylonitrile with and without BF3 complexed. Does the LUMO energy decrease as a result of Lewis-acid complexation? Display electrostatic potential maps for the two molecules. (Be careful to use the same scale.) Is there a noticeable decrease in the potential above the carboncarbon bond? Buckyball. Inside and Out: Buckyball (Buckermeisterfullerene or fullerene or C60) has received a great deal of attention since its discovery in 1985. Aside from being a third stable form of elemental carbon (the other two are graphite and diamond), provides an interesting environment both outside as well as inside. Obtain equilibrium geometries of C60 and (for comparison) benzene using the HartreeFock model with the 3-21G basis set. Display electrostatic potential maps for the two side-by–side and on the same scale (-100 to 100 kJ/mol). Relative to benzene, is the surface of fullerene show more or less charge variation? Would you expect fullerene to be more or less soluble than benzene in non-polar solvents? In polar solvents? Elaborate. Fullerene incorporates a small cavity. Is this likely to be “hospitable” to non-polar or polar molecules? To tell, examine a contour (slice) plot of the electrostatic potential, focusing on the cavity. Polar Area as a Measure of Aqueous Solubility: 49 ...
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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 University of California, Berkeley.
- Spring '09