Absolutely Small: How Quantum Theory Explains Our Everyday World - Michael D. Fayer (2010)
Chapter 18. Aromatic Molecules
IN CHAPTERS 13 AND 14 we discussed double bonds, and in Chapter 16 we saw that double bonds play a fundamental role in determining the biological properties of fats. Some of the fat molecules discussed, the polyunsaturated fats, have several double bonds, but these double bonds are always separated by a number of single bonds. For example, Figure 16.5 shows a ball-and-stick model of DHA, a polyunsaturated fat with six double bonds. As can be seen, there are two single bonds between each double bond. In this chapter, we will see the wide-ranging impacts of multiple double bonds that are not separated by several single bonds. Quantum theory shows that the nature of the bonding found in the molecule benzene and a vast number of other “aromatic” molecules can explain electrical conductivity in metals, as well as the differences among metals, semiconductors, and insulators that will be discussed in Chapter 19. To understand aromatic molecules and electrical conductivity in metals, it is necessary to discuss the nature of the molecular orbitals that arise when identical atomic orbitals from many atoms interact to form MOs.
BENZENE: THE PROTOTYPICAL AROMATIC MOLECULE
Figure 18.1 shows a diagram of benzene, which is composed of six carbon atoms and six hydrogen atoms. Experiments have determined that benzene is a perfect hexagon with all of the atoms, carbons, and hydrogens in a plane. The angle formed by the bonds from one carbon to its two nearest neighbors is exactly 120°, and the angle formed by the bond of a hydrogen to a carbon and that carbon’s bond to an adjacent carbon is 120°. So the three bonds made by any carbon have a trigonal geometry, which means they are formed by the carbon using three sp2 hybrid atomic orbitals on each carbon. That leaves one unused 2p orbital on each carbon, call them 2pz orbitals, that point in and out of the plane of the page. We know that carbon always forms four bonds. Here, a carbon is only bonded to three other atoms, using up three bonds. The 2pz orbital must be used to make a π double bond, but where is its location in the molecule?
FIGURE 18.1. The geometry of benzene, C6H6. Benzene is planar, and it is a perfect hexagon.
Where Are the Double Bonds?
Figure 18.2 shows two possible double-bonded structures. In both, each carbon makes four bonds. A carbon makes three σ bonds, one to a hydrogen and one to each of the adjacent two carbons. Each carbon also participates in a double bond with one adjacent carbon. The diagrams on the right and left are identical except for the locations of the double bonds.
Two things are wrong with the bonding in benzene depicted in Figure 18.2. In the discussion of double bonds in Chapter 14, Table 14.1 shows that the carbon-carbon double bond length is considerably shorter than the single bond length. In ethylene (double bond) vs. ethane (single bond), the bond lengths are 1.35 Å and 1.54 Å, respectively. So if benzene had alternating double and single bonds, it should have alternating short and long carbon-carbon bonds. However, experiments demonstrate conclusively that benzene is a perfect hexagon; all of the carbon-carbon bond lengths are the same.
If we ignore the diagram’s implied unequal bond lengths, the second problem is the issue of which diagram would be correct, the one on the right or the one on the left? There is no reason to prefer one over the other. An early explanation was that the bonds jumped back and forth between the configuration on the right and the configuration on the left. The jumps were proposed to produce a type of average structure. This idea is in some sense closer to the truth, but the real answer, which then applies to many types of systems, is in the nature of the molecular orbitals that are formed.
FIGURE 18.2. Two possible configurations of double bonds in benzene. In both, each carbon makes four bonds.
Delocalize Pi Bonds
Figure 18.3 is a schematic of the atomic orbitals used to form molecular orbitals in benzene. The top shows the atomic hybrid orbitals used to make the σ bonds. Each carbon uses three sp2 hybrid orbitals to form three σ bonds, one to a hydrogen and one to each of the two adjacent carbon atoms. Forming the three sp2 hybrid atomic orbitals leaves each carbon atom with a leftover p orbital. In the top part of Figure 18.3, take the plane that contains the atoms (the plane of the page) to be the xy plane. Then each of the carbons has an unused pz orbital sticking in and out of the plane of the page. These orbitals are shown in the bottom of the figure. The positive and negative lobes of the pzorbitals are above and below the plane of the ring. In the diagram, the lengths of the bonds between carbon atoms have been exaggerated and the widths of the pz orbitals are reduced so that they can be seen clearly. The pz orbitals actually overlap, as shown more to scale in Figure 14.15.
The six pz atomic orbitals will combine to form molecular orbitals. The six atomic orbitals can contain at most 12 electrons without violating the Pauli Principle. Therefore, the six atomic orbitals will form superpositions to form six molecular orbitals that can also contain at most 12 electrons. The MOs are not associated with a particular atom or even a particular pair of atoms. They are molecular orbitals that span the entire six-carbon system.
In discussing the hydrogen molecule in connection with Figure 12.6, we saw that two atomic orbitals combine to form two molecular orbitals, one bonding MO and one antibonding MO. In Chapter 13, we investigated larger diatomic molecules, such as F2, O2, and N2. In these atoms, three p orbitals on each atom, six atomic orbitals total, combined to form six π MOs, three bonding and three antibonding MOs (see Figure 13.5). Some of these diatomic π MOs were degenerate, that is, they have the same energy.
FIGURE 18.3. Top: Benzene σ bonding. Each carbon makes three bonds using three sp2orbitals that lie in the xy plane. Each carbon has a pzorbital perpendicular to the plane of the benzene ring. Bottom: The carbon pzorbitals have positive and negative lobes that are above and below the plane of the ring. The bond lengths are exaggerated and the pzlobes are made small for clarity of presentation. The lobes of adjacent pzorbitals overlap.
The Bonding and Antibonding Molecular Orbitals
In benzene, six pz atomic orbitals combine to form three bonding MOs and three antibonding MOs, as shown in Figure 18.4. The six carbon 2pz orbitals, one on each carbon atom, have identical energy. This is indicated by the six closely spaced lines on the left-hand side of Figure 18.4. These combine to form six MOs with energy levels shown on the right-hand side of the figure. Three of the MOs have energies lower than the pz atomic orbital energy. These are the bonding MOs. Three of the MOs have energies higher than the atomic orbital energy. These are the antibonding MOs. Figure 18.5 shows the bonding and antibonding energy levels with the six electrons, one from each carbon, placed in the appropriate energy levels. We place the electrons in the lowest energy level consistent with the Pauli Principle. The Pauli Principle (Chapter 11) states that at most two electrons can be in a single orbital, and they must have opposite spin states (one up arrow and one down arrow). The first two electrons go into the lowest energy MO. The next two MOs have the identical energy indicated by two closely spaced lines. Two electrons will go into each of these MOs. The three MOs filled by the six electrons are all π bonding MOs. The π antibonding MOs are empty.
FIGURE 18.4. Left: Benzene has six carbon atoms, each with a 2pzorbital. These have identical energy, which is indicated by the six closely spaced lines. Right: The six pzorbitals combine to form six π molecular orbitals, three bonding (b) and three antibonding (*) MOs.
FIGURE 18.5. Benzene π molecular orbital energy levels with the six electrons placed in the appropriate MOs in the lowest possible energy levels consistent with the Pauli Principle.
The Carbon-Carbon Bond Order is 1.5
Figure 18.5 shows that the six carbon pz electrons occupy three π bonding MOs. Therefore, there are three π bonds shared by six carbons. These three π bonds are in addition to the σ bonds that connect each carbon to its two nearest carbon neighbors. The net result is that each carbon has 1.5 bonds to other carbons. The three π bonds shared by the six carbon atoms contribute a half of a bond joining adjacent carbons. The bonds between carbon atoms are shorter and stronger than a carbon-carbon single bond, but not as short or as strong as a full double bond. The π bonding keeps the molecule rigorously planar. Twisting the ring away from planarity reduces the overlap of the pz orbitals and raises the energy. Figure 18.6 shows a chemical diagram of benzene. A carbon is represented by a vertex. A hydrogen atom is at the end of each line that emanates from a carbon. The circle represents the delocalized π electron system.
FIGURE 18.6. Benzene chemical diagram. A carbon atom is at each vertex, and a hydrogen is at the end of each line from a carbon. The circle represents the delocalized π molecular orbitals.
Many molecules have carbon rings with delocalized π bonding. Another example is naphthalene, which is shown in Figure 18.7. Naphthalene has 10 carbons forming two six-membered rings with eight hydrogens. The two circles represent the delocalized π molecular orbitals. Like benzene, naphthalene is planar and each carbon has 1.5 bonds to the adjacent carbons.
FIGURE 18.7. Naphthalene chemical diagram. Naphthalene has 10 carbons and eight hydrogens. The circles represent the delocalized π molecular orbitals.
THE BENZENE DELOCALIZED PI MOLECULAR ORBITALS
Benzene, naphthalene, and similar molecules are referred to as aromatic molecules. They tend to have a sweet smell. Naphthalene is a mothball, which has a characteristic aromatic smell. Perfumes are more complex aromatic molecules having a number of benzene-like rings, as well as other chemical groups replacing the hydrogen atoms. Small changes in the molecular structure change the aroma, which is what makes one perfume smell different from another.
In Chapter 8, we discussed the particle in a box problem. Figures 8.4 to 8.6 show the particle in a box wavefunctions and energy levels. The wavefunction associated with the lowest energy level has no nodes. The next higher energy state has a wavefunction with one node, the next higher energy state has a wavefunction with two nodes, and so forth. A node is a place where the wavefunction goes to zero, so the probability of finding the electron is zero. The particle in a box is a one-dimensional problem. A node is a point. In Chapter 10, we examined the wavefunctions and the energy levels for a hydrogen atom. Figures 10.2 to 10.6 show representations of the wavefunctions for the 1s, 2s, and 3s hydrogen atom states. The hydrogen atom wavefunctions are three-dimensional. The lowest energy state (1s) has no nodes, the next higher energy state (2s) has a wavefunction with one node. The next higher energy state (3s) has two nodes. These nodes are three dimensional surfaces on which there is zero probability of finding the electron.
The benzene π MOs also have an increasing number of nodes as the energy is increased. Figure 18.8 shows schematics of the benzene π MOs. The shaded areas are the regions of high electron density (high probability of finding electrons) for the π MOs.
FIGURE 18.8. Benzene π molecular orbital energy levels and schematics showing the shapes of the corresponding MOs. As the energy increases, the number of nodes increases. MOs with the same number of nodes have the same energy.
These are three-dimensional electron clouds that extend above and below the plane of the page and do not have sharp boundaries. Also shown are the energy levels with the 6 pz electrons filled in the lowest energy bonding MOs. The lowest energy MO has no nodes. There are two states with the next higher energy. Both of these MOs have one node. The three MOs with no nodes and one node are the bonding MOs. There are also two levels with the next higher energy. These MOs have two nodes. The highest energy MO has three nodes. The three MOs having two and three nodes are the antibonding MOs.
Comparing the schematics of the lowest energy MO with the highest energy MO in Figure 18.8, it can be seen clearly why the former is a bonding MO and the latter is an antibonding MO. The lowest bonding MO has electron density between all of the carbons. The highest antibonding MO has nodes between all of the carbons, so electrons in this MO would not join the carbons together. The pair of bonding MOs, although higher in energy than the lowest energy bonding MO, results in bonding among the carbon atoms. Each of these MOs has one node. The one on the left puts electron density between a pair of carbons on the left and right. The MO on the right puts electron density between three carbon atoms on the top and three on the bottom. In spite of the nodes, these MOs combine with the lowest energy MO to produce the three π bonds that are shared by the six carbons. The two degenerate antibonding MOs have two nodes each. The one on the left clearly does not contribute to bonding because it does not place electron density between any of the carbons. The one on the right puts electron density between two pairs of carbons, but when combined with the one on the right, it does not produce net bonding.
Light Absorption by Aromatic Molecules
Quantum theory can calculate the molecular orbitals of aromatic molecules, as well as their shapes and sizes. There are many ways to test the quantum calculations by comparison to experiments. One of the most useful is to employ optical spectroscopy to measure the wavelengths (colors) of light absorbed by a molecule. We will use naphthalene as an example.
Figure 18.7 shows a diagram of naphthalene, with its 10 carbon atoms. Each carbon atom will contribute one pz orbital with one electron to form the delocalized π electron system. The other three carbon valence electrons are used to form the σ bonds. Ten pz atomic orbitals form the π system, so there will be 10 molecular orbitals, five bonding MOs and five antibonding MOs. In naphthalene, none of the MOs is degenerate; each has a different energy. Figure 18.9 displays a schematic of the energy levels of the π MOs of naphthalene. The left-hand side shows the π energy levels with the 10 π electrons filling the five bonding MOs. The antibonding MOs are empty.
The right side of Figure 18.9 illustrates the effect of absorption of light. Because the energy levels are quantized, only certain energies of light can be absorbed by the molecule. In the figure, ΔE is the lowest energy light that can be absorbed. Consider what happens if we shine light on a sample of naphthalene molecules, beginning with light that is too low in energy to be absorbed by the molecules. The energy of the light is E = hν, where h is Planck’s constant and ν is the frequency. So initially, ΔE > hν; the energy separation between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) is greater than the energy of the photons that are impinging on the sample. As a result, they pass through the sample without being absorbed. Now, we start changing the energy of the light to higher energy (red to blue). When hν = ΔE, light is absorbed, which is detected by a decrease in the amount of light that passes through the sample. An electron is excited from the HOMO to the LUMO. This excitation is shown on the right side of Figure 18.9, which has one electron in the HOMO and one electron in the LUMO. In the left-hand side of Figure 18.9, there are two electrons in the HOMO and none in the LUMO.
FIGURE 18.9. A schematic of the naphthalene π molecular orbital energy levels. There are five bonding MOs and five antibonding MOs. The left side shows the 10 p electrons filling the five bonding MOs. The right side shows the result of absorption of light that raises the energy of a bonding electron into an antibonding MO.
The transition from the HOMO to the LUMO is the lowest energy transition. As drawn, the bonding MOs are closer together than the separation between the HOMO and the LUMO. However, an electron cannot be excited from one filled bonding MO to another filled bonding MO. If we try to take an electron from one bonding MO and put it into another bonding MO, the result would be an MO with three electrons in it. Three electrons in an MO violates the Pauli Exclusion Principle. Then in our optical absorption experiment, as the light color is scanned from red to blue (low energy to high energy), the first color (wavelength) that is absorbed corresponds to the energy ΔE. ΔE can be calculated using quantum mechanics. ΔE depends on the structure of the molecule and the interactions of the atomic orbitals that give rise to the molecular orbitals. The quantum calculations can be tested further by comparison to the wavelengths at which absorption of higher energy light takes place as the light is scanned further and further to the blue. This second absorption occurs because light can take an electron from the HOMO to one energy level above the LUMO. Another absorption at even higher energy will occur by exciting an electron from the HOMO to two levels above the LUMO, and so forth.
NAPHTHALENE TREATED AS A PARTICLE IN A BOX PROBLEM
Using modern quantum theory and computers, the structure of naphthalene can be calculated with great accuracy. The theory will give the bond lengths and bond angles. For example, the bond lengths can be calculated to 0.001 nm, that is, a thousandth of a nanometer. The calculations also yield the frequencies at which light is absorbed with substantial accuracy. The calculations use the masses, number of electrons, and charges of the nuclei. The calculations will include both the σ and π bonding. As we have discussed, the π electrons are not localized on one or two carbon centers, but rather are delocalized over the entire carbon framework of the molecule. The lowest energy absorption of naphthalene, the HOMO to LUMO absorption, occurs at 320 nm, which is in the ultraviolet portion of the optical spectrum.
We can do a very crude calculation by pretending that the π electrons are particles in a box. In Chapter 8, we discussed the particle in a box in great detail. If we take the HOMO to LUMO transition to be a transition of an electron in a box from the n = 1 to n = 2 levels (see Figure 8.7), we can use the formulas derived just below Figure 8.7. For this transition we found that
where h is Planck’s constant, m is the mass of the electron, and L is the length of the box. Here we will take L to be 0.51 nm, the distance across the carbon framework of naphthalene. Then,
Converting this energy to a frequency by dividing by h gives ν = 1.04 × 1015 Hz. Then the wavelength of the light that will be absorbed is λ = 2.87 × 10-7 m = 287 nm. This wavelength is further in the ultraviolet than the true absorption, but not that far from the observed value.
The particle in a box calculation shows that if a particle with the mass of an electron is confined to a box the size of naphthalene, the first absorption will be in the ultraviolet range. The reasonable accuracy of the particle in a box calculation for naphthalene is somewhat fortuitous. Even if we wanted to model naphthalene as a particle in a box, it should be a two- or three-dimensional box, not a one-dimensional box. Such calculations produce results with significant errors. However, an accurate quantum theory calculation will yield the molecular structure and much more accurate frequencies for absorption of light. In addition, if, for example, a hydrogen is replaced by a fluorine, quantum theory will accurately predict how much the frequency of light absorption of fluoronaphthalene is changed from that of naphthalene.