2.1A: Another look at the H 2 molecule: bonding and antibonding sigma molecular orbitals

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1 Ashley Robison My Preferences Site Tools Popular pages MindTouch User Guide FAQ Sign Out If you like us, please share us on social media. The latest UCD Hyperlibrary newsletter is now complete, check it out. ChemWiki BioWiki GeoWiki StatWiki PhysWiki MathWiki SolarWiki Periodic Table of the Elements Reference Tables Physical Constants Units & Conversions Lab Techniques ChemWiki: The Dynamic Chemistry E-textbook > Organic Chemistry > Organic Chemistry With a Biological Emphasis > Chapter 2: Introduction to Organic Structure and Bonding II > Section 2.1: Molecular orbital theory: conjugation and aromaticity Section 2.1: Molecular orbital theory: conjugation and aromaticity As we have seen, valence bond theory does a remarkably good job at explaining the bonding geometry of many of the functional groups in organic compounds. There are some areas, however, where the valence bond theory falls short. It fails to adequately account, for example, for some interesting properties of compounds that contain alternating double and single bonds. In order to understand these properties, we need to think about chemical bonding in a new way, using the ideas of molecular orbital (MO) theory. 2.1A: Another look at the H 2 molecule: bonding and antibonding sigma molecular orbitals Let s go back and consider again the simplest possible covalent bond: the one in molecular hydrogen (H 2 ). When we described the hydrogen molecule using valence bond theory, we said that the two 1s orbitals from each atom overlap, allowing the two electrons to be shared and thus forming a covalent bond. In molecular orbital theory, we make a further statement: we say that the two atomic 1s orbitals don t just overlap, they actually combine to form two completely new orbitals. These two new orbitals, instead of describing the likely location of an electron around a single nucleus, describe the location of an electron pair around two or more nuclei. The bonding in H 2, then, is due to the formation of a new molecular orbital (MO), in which a pair of electrons is delocalized around two hydrogen nuclei. An important principle of quantum mechanical theory is that when orbitals combine, the number of orbitals before the combination takes place must equal the number of new orbitals that result orbitals don t just disappear! (we saw this previously when we discussed hybrid orbitals: one s and three p orbitals make four sp 3 hybrids). When two atomic 1s orbitals combine in the formation of H 2, the result is two molecular orbitals called sigma (σ) orbitals. According to MO theory, the first sigma orbital is lower in energy than either of the two isolated atomic 1s orbitals thus this sigma orbital is referred to as a bonding molecular orbital. The second, sigma-star (σ*) orbital is higher in energy than the two atomic 1s orbitals, and is referred to as an antibonding molecular orbital (in MO theory, a star (*) sign always indicates an antibonding orbital). Following the aufbau ('building up') principle, we place the two electrons in the H 2 molecule in the lowest energy orbital, which is the (bonding) sigma orbital. The bonding sigma orbital, which holds both electrons in the ground state of the molecule, is egg-shaped, encompassing the two nuclei, and with the highest likelihood of electrons being in the area between the two nuclei. The high-energy, antibonding sigma-star orbital can be visualized as a pair of droplets, with areas of higher electron density near each nucleus and a node, (area of zero electron density) midway between the two nuclei. Remember that we are thinking here about electron behavior as wave behavior. When two separate waves combine, they can do so with what is called constructive interference, where the two amplitudes reinforce one another, or destructive interference, where the two amplitudes cancel one another out. Bonding MO s are the consequence of constructive interference between two atomic orbitals which results in an attractive interaction and an increase in electron density between the nuclei. Antibonding MO s are the consequence of destructive interference which results in a repulsive interaction and a canceling out of electron density between the nuclei (in other words, a node), 1/8

2 2.1B: MO theory and pi bonds - conjugation The advantage of MO theory becomes more apparent when we think about pi bonds, especially in those situations where two or more pi bonds are able to interact with one another. Let s first consider the pi bond in ethene from an MO theory standpoint (in this example we will be disregarding the various sigma bonds, and thinking only about the pi bond). According to MO theory, the two atomic 2p z orbitals combine to form two pi (π) molecular orbitals, one a low-energy π bonding orbital and one a high-energy π-star (π*) antibonding molecular orbital. These are sometimes denoted, in MO diagrams like the one below, with the Greek letter psi (Ψ) instead of π. In the bonding Ψ 1 orbital, the two shaded lobes of the 2p z orbitals interact constructively with each other, as do the two unshaded lobes (remember, the shading choice represents mathematical (+) and (-) signs for the wavefunction). Therefore, there is increased electron density between the nuclei in the molecular orbital this is why it is a bonding orbital. In the higher-energy antibonding Ψ 2 * orbital, the shaded lobe of one 2p z orbital interacts destructively with the unshaded lobe of the second 2p z orbital, leading to a node between the two nuclei and overall repulsion. By the aufbau principle, the two electrons from the two atomic orbitals will be paired in the lower-energy Ψ 1 orbital when the molecule is in the ground state. Now, consider the 1,3-butadiene molecule. From valence orbital theory we might expect that the C 2 -C 3 bond in this molecule, because it is a sigma bond, would be able to rotate freely. Experimentally, however, it is found that there are significant barriers to rotation about this bond (as well as about the C 1 -C 2 and C 3 -C 4 double bonds), and that the entire molecule is planar. It is also observed that the C 2 -C 3 bond, while longer than the C 1 -C 2 and C 3 -C 4 double bonds, is significantly shorter than a typical carbon-carbon single bond. Molecular orbital theory accounts for these observations with the concept of delocalized π bonds. In this picture, the four 2p z orbitals are all parallel to each other (and perpendicular to the plane of the sigma bonds), and thus there is π-overlap not just between C 1 and C 2 and C 3 and C 4, but between C 2 and C 3 as well. The four atomic (2p z ) orbitals have combined to form four π molecular orbitals. 2/8

3 The lowest energy molecular orbital, Ψ 1, has zero nodes, and is a bonding MO. Slightly higher in energy, but still lower than the isolated p orbitals, is the Ψ 2 orbital. This orbital has one node between C 2 and C 3, but is still a bonding orbital due to the two constructive interactions between C 1 -C 2 and C 3 -C 4. The two higher-energy MO s are denoted Ψ 3 * and Ψ 4 *, and are antibonding. Notice that Ψ 3 * has two nodes and one constructive interaction, while Ψ 4 * has three nodes and zero constructive interactions. The energy of both of these antibonding molecular orbitals is higher than that of the 2p z atomic orbitals of which they are composed. By the aufbau principle, the four electrons from the isolated 2p z atomic orbitals are placed in the bonding Ψ 1 and Ψ 2 MO s. Because Ψ 1 includes constructive interaction between C 2 and C 3, there is a degree, in the 1,3-butadiene molecule, of π-bonding interaction between these two carbons, which accounts for the shorter length and the barrier to rotation. The simple Lewis structure picture of 1,3-butadiene shows the two π bonds as being isolated from one another, with each pair of π electrons stuck in its own π bond. However, molecular orbital theory predicts (accurately) that the four π electrons are to some extent delocalized, or spread out, over the whole π system. 1,3-butadiene is the simplest example of a system of conjugated π bonds. To be considered conjugated, two or more π bonds must be separated by only one single bond in other words, there cannot be an intervening sp 3 -hybridized carbon, because this would break up the overlapping system of parallel 2p z orbitals. In the compound below, for example, the C 1 -C 2 and C 3 -C 4 double bonds are conjugated, while the C 6 -C 7 double bond is considered to be isolated, due to the effect of the sp 3 -hybridized C 5. The presence of C 5 makes it impossible for the π electrons of the C 6 -C 7 pi bond to join the conjugated system on the first four carbons. Conjugated π bond systems can involve oxygens and nitrogen atoms as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups. Later on in chapter 4, we will see that MO theory is very useful in explaining why organic molecules that contain extended systems of conjugated π bonds often have distinctive colors. β-carotene, the compound responsible for the orange color of carrots, has an extended system of 11 conjugated π bonds. 3/8

4 Example Exercise 2.1: Identify isolated and conjugated pi bonds in the compound shown. Solution 2.1C: Aromaticity Molecular orbital theory is especially helpful in explaining the unique properties of a class of compounds called aromatics. Benzene, a common organic solvent, is the simplest example of an aromatic compound. Although it is most often drawn with three double bonds and three single bonds, it is known that all of the carbon-carbon bonds in benzene are exactly the same length Å. This is shorter than a typical carbon-carbon single bond (about 1.54 Å), and slightly longer than a typical carboncarbon double bond (about 1.34 Å). In addition, the π bonds in benzene are significantly less reactive than isolated or conjugated π bonds in most alkenes. To illustrate this unique stability, we will make use of the idea of heat of hydrogenation. The carbon-carbon double bond in an alkene can be converted to a single bond through a process called catalytic hydrogenation essentially adding a molecule of H 2 to the double bond. We will learn more about how this process occurs, both in the laboratory and in living cells, later in the text (section 16.5). For now, what is important to understand is that the hydrogenation process is exothermic: the alkane is lower in energy than the alkene, so hydrogenating the double bond results in the release of energy in the form of heat. Converting one mole of cyclohexene to cyclohexane, for example, releases 28.6 kilocalories. If the benzene molecule is considered to be a six-membered ring with three isolated double bonds, the heat of hydrogenation should theoretically be three times this value, or 85.8 kcal/mol. The actual heat of hydrogenation of benzene, however, is only 49.8 kcal/mol, or 36 kcal/mol less than what we would expect if using the isolated double bond model. Something about the structure of benzene makes these π bonds especially stable. This something has a name: it is called aromaticity. What exactly is this aromatic property that makes the pbonds in benzene so much less reactive than those in alkenes? In a large part, the answer to this question lies in the fact that benzene is a cyclic molecule in which all of the ring atoms are sp 2 -hybridized. This allows the π electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane of the ring. For this to happen, of course, the ring 4/8

5 must be planar otherwise the 2p z orbitals couldn t overlap properly. Benzene is indeed known to be a flat molecule. Do all cyclic molecules with alternating single and double bonds have this same aromatic stability? Quite simply, the answer is no. The eightmembered cyclooctatetraene ring shown below is not flat, and its π bonds are much more reactive than those of benzene. Clearly it takes something more to be aromatic, and this can best be explained with molecular orbital theory. Let s look at an energy diagram for the molecular orbitals containing the π electrons in benzene. Quantum mechanical calculations conclude that the six molecular orbitals in benzene, formed from six atomic 2p z orbitals, occupy four separate energy levels. Ψ 1 and Ψ 6 * have unique energy levels, while the Ψ 2 - Ψ 3 and Ψ 4 *-Ψ 5 * pairs are degenerate (more than one orbital at the same energy level). When we use the aufbau principle to fill up these orbitals with the six π electrons in benzene, we see that the bonding orbitals are completely filled, and the antibonding orbitals are empty. This gives us a good clue to the source of the special stability of benzene: a full set of bonding MO s is similar in many ways to the full shell of electrons possessed by the very stable noble gases like helium, neon, and argon. Now, let s do the same thing for cyclooctatetraene, which we have already learned is not aromatic. The result of molecular orbital calculations tells us that the lowest and highest energy MOs (Ψ 1 and Ψ 8 *) have unique energy levels, while the other six come in degenerate pairs. Notice that y 4 and y 5 are at the same energy level as the isolated 2p z atomic orbitals: these are therefore neither bonding 5/8

6 nor antibonding, rather they are referred to as nonbonding MOs. Filling up the MOs with the eight π electrons in the molecule, we find that the last two electrons are unpaired and fall into the two degenerate nonbonding orbitals. Because we don't have a perfect filled shell of bonding MOs, our molecule is not aromatic. As a consequence, each of the double bonds in cyclooctatetraene acts more like an isolated double bond. Here, then, are the conditions that must be satisfied for a molecule to be considered aromatic: 1. It must have a cyclic structure. 2. The ring must be planar. 3. Each atom in the ring must be sp 2 -hybridized, so that π electrons can be delocalized around the ring. 4. The number of π electrons in the ring must be such that, in the ground state of the molecule, all bonding MOs are completely filled, and all nonbonding and antibonding MOs are completely empty. It turns out that, in order to satisfy condition #4, the ring must contain a specific number of π electrons. The set of possible numbers is quite easy to remember - the rule is simply 4n+2, where n is any positive integer (this is known as the Hückel rule, named after Erich Hückel, a German scientist who studied aromatic compounds in the 1930 s). Thus, if n = 0, the first Hückel number is (4 x 0) + 2, or 2. If n = 1, the Hückel number is (4 x 1) + 2, or 6 (the Hückel number for benzene). The series continues with 10, 14, 18, 22, and so on. Cyclooctatetraene has eight π electrons, which is not a Hückel number. Because 6 is such a common Hückel number, chemists often use the term 'aromatic sextet'. Benzene is best visualized as a planar ring made up of carbon-carbon sbonds, with two donut-like rings of fully delocalized π electron density above and below the plane of the ring (the fact that there is a ring of π electron density on both sides of the molecule stems from the fact that the overlapping p orbitals have two lobes, and the electron density is located in both). This general picture is valid not just for benzene but for all other aromatic structures as well. Let s look at some different aromatic compounds other than benzene. Pyridine and pyrimidine both fulfill all of the criteria for aromaticity. In both of these molecules, the nitrogen atoms are sp 2 hybridized, with the lone pair occupying an sp 2 orbital and therefore not counted among the aromatic sextet. The Hückel number for both pyridine and pyrimidine is six. Rings do not necessarily need to be 6-membered in order to have six π electrons. Pyrrole and imidizole, for example, are both aromatic 5-membered rings with six π electrons. The nitrogen atoms in both of these molecules are sp 2 -hybridized (as they must be for the rings to be aromatic). In pyrrole, the lone pair can be thought of as occupying a 2p z orbital, and thus both of these electrons contribute to the aromatic π system. In imidazole, one lone pair occupies a 2p z orbital and is part of the aromatic sextet, while the second occupies one of the sp 2 orbitals and is not part of the sextet. Molecules with more then one ring can also fulfill the Hückel criteria, and often have many of the same properties as monocyclic aromatic compounds, including a planar structure. Indole (a functional group in the amino acid tryptophan) and purine (a functional group in guanine and adenine DNA/RNA bases) both have a total of ten π electrons delocalized around two rings. The nitrogen in indole and the N 9 nitrogen in purine both contribute a pair of electrons to the π system. The N 1, N 3, and N 7 nitrogens of purine, in contrast, hold their lone pair in sp 2 orbitals, outside of the aromatic system. Example Exercise 2.2: Are the following molecules likely to be aromatic? Explain, using Huckel s criteria. and orbital drawings. Hint: Ions can also be aromatic! 6/8

7 Solution Up to now we have been talking about molecules in which the entire structure makes up an aromatic system. However, in organic chemistry we will more often encounter examples of molecules which have both aromatic and nonaromatic parts. Toluene, a common organic solvent (which is much safer to use than benzene) is simply a benzene ring with a methyl substituent. Benzyaldehyde is benzene with an aldehyde substituent, and phenol is benzene with a hydroxyl substituent. In these substituted benzene compounds, the entire molecule is not aromatic, just the benzene ring part. When a benzene ring is part of a larger molecule, it is called a phenyl group. The amino acid phenylalanine, for example, contains a phenyl group. The amino acids tyrosine, tryptophan, and histidine contain phenol, indole, and imidazole groups, respectively. Pyridoxine, commonly known as vitamin B 6, is a substituted pyridine. The DNA and RNA bases are based on pyrimidine (cytosine, thymine, and uracil) and purine (adenine and guanine). 7/8

8 The flat, aromatic structure of these bases plays a critical role in the overall structure and function of DNA and RNA. Contributors Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) Copyright 2015 Chemwiki Powered by MindTouch Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use 8/8