Lecture 27. CH4 Molecular Orbitals and Delocalized Bonding
Shaka, Athan J.
 This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. 
 Right, let's pick up where we left off. We had just gone through Linus Pauling's brilliant exposition on hybrid orbitals, which are still used to this very day to rationalize a lot of things to do with structure and reactivity. But let's take a closer look, let's compare the hybrid orbital approach, which really says let's make pairs of electrons make the bonds with the molecular orbital approach, which says look, the way you have to do it is the way you did it with an atom. You have to put all the positive charges where they're supposed to be, and then you have to solve it. Then you have to get the solution, and then you can put in your electrons into the orbitals, and they go into the lowest energy once. This is a more delocalized approach to bonding because if you're including all the nuclei at once, then there's no reason why you should be drawing lines. That makes it a little harder to understand though when you're drawing structures and you're pushing the lines around, chemists love those lines because it allows them to write and rationalize reactions. But occasionally a reaction goes that doesn't seem to behave, and that could be an indication that maybe the theory of the lines is falling apart a little bit. Let's talk then in that vein about more delocalized bonding. In the MO picture, we don't make hybrid orbitals, so the orbital is used, but it's completely different. They must not confuse the two. In the molecular orbital picture, what we said we had to do was we could combine together atomic orbitals that had the same symmetry, similar energy, and overlapped in space. For tetrahedral molecule with a central atom like methane, we've got four carbon valence atomic orbitals, the S and the three P's. We have to know the structure. Why? Well, otherwise we'd have to move the things around and calculate the energy as a function of R and R2 and R3. We already know the structure. It's tetrahedral. What we want to understand is not the structure so much. We want to understand the bonding. How is it working? What's going on? We assume we know the structure. It's the same structure as Pauling's, but the description of the bonding is going to be different, and that's the thing that we're going to focus on. In the MO approach, you have to know the structure first. If you don't, you have to search through all possible structures. If you're doing that, that is a very, very, very long-winded and intensive calculation, although sometimes people do it just to see what might be the most stable structure if there's no experiment or it hasn't been made. Let's put our hydrogen atoms at the corners of the same cube that we had before, HA, HB, HC, and HD, and let's see what we're going to do with these four. The carbon is at the center, and with our four hydrogen orbitals and our four carbon orbitals, we get eight molecular orbitals because recall in the linear combination of atomic orbitals, rubric, we have the same number of orbitals at the end in terms of molecular orbitals as what we started with. Okay, now question is, which orbitals can we combine? The answer is rule two said they have to have the same symmetry for this molecule. What that means is that we cannot do what Pauling did because we cannot combine the carbon 2S and the carbon 2PZ or the other two P's because those have different symmetry. Therefore if we want to consider the 2S on carbon, we cannot combine it with any of the other atomic orbitals on carbon. We have to instead combine it with combinations of the hydrogen orbitals at the four positions of the cube. And that's completely different then because now we won't necessarily end up with four identical lines in terms of four localized bonds. We will end up with a tetrahedral structure, that's for sure, because we started with that, but our description of the bonding will be different. Now the game is this. We have the carbon orbital. Each of the four of them has different symmetry. We have to combine the four hydrogens so that they have the same symmetry as the carbon orbital. And that's pretty easy to do because if the carbon orbital is a big fat round S, then that means that these guys have to be as round as possible so they should just all be red and that should be one of them. And if the carbon orbital changes sign, like it's plus here and minus here, well we've got two hydrogens on the top of the cube, they should be plus. Two hydrogens on the bottom of the cube, they should be minus and so forth. And so there's a combination of hydrogens that goes perfectly with each of the four combination on the carbon and that's how we're going to do it. So for 2S we use 1SA plus 1SB plus 1SC plus 1SD. All of them as round as can be. And this then, if they're all red, the 2S is red and all these guys are red, this makes a gigantic orbital with no nodes that goes around all the nuclei and has roughly spherical symmetry and it's called A1 which is a symmetry label that lets the spectroscopists know what the symmetry of the molecular orbital is. And that name A1 has to do with its symmetry under the tetrahedral point group TD that you'll learn about later in the course. For now we're just going to treat A1 as a label that tells us which orbital we're talking about. Remember that less nodes is lower energy and so we've got this big red thing with this big fluffy thing sort of looks like a teddy bear with no nodes therefore A1 is the lowest symmetry. That's what we predict. Above it then are three degenerate because they're related by symmetry orbitals and there's one for each of the p orbitals on the carbon atom. And those three are called T2, T means three, a name that has to do again with their symmetry under the tetrahedral point group. The one and two have meanings but I don't want to go into the exact meaning now. These are the four bonding orbitals and they are not the same but there's four of them and they still predict an exactly tetrahedral structure not surprisingly because we started out with an exactly tetrahedral structure. The other four combinations have more nodes. I could take the S as red and I have to have something the same symmetry but I can pick them all blue. Now that's a disaster because now there's nodes in between the two nuclei so that one's no good and likewise I could pick the PZ with red on the top and blue on the bottom and I could perversely pick these two guys to be blue rather than red and then there would be destructive interference and again that would be a very high energy solution. Therefore we can easily see that there are four bonding orbitals so there's going to be four bonds, good, and there's four anti-bonding orbitals and since there's only eight electrons the four bonding orbitals are filled and the four anti-bonding orbitals are empty and that's almost always the way it works out when you do things correctly because not surprisingly nature finds its way into making the more stable configurations. Here then is an MO diagram. I have four, it's a little bit harder to draw the tennis tournament when you don't have just two players but I have four hydrogen 1s orbitals on one side and then I have the carbon orbitals and I have the 2s of the carbon lower than the 1s of the hydrogen, that's good and then I have the 2p higher. Well I'd have to know that but I can certainly figure that out by looking up the ionization energy of the atoms which is well known. And so I can order them like that and I have one very good combination, the teddy bear which is down at the bottom, the A1 and then I have the three that are identical, the T2 and they're at slightly higher energy. The key is there's no way that the A1 and the T2 have the same energy. They have to have different energies and that is an experimentally testable fact that we can look at. So we fill up the bottom with eight electrons, four from the hydrogen, four from the carbon, the same way we always do with molecular orbital diagrams. Once we calculate the diagram we fill it up from the bottom. And here is an actual calculation of these orbital contours from the excellent website of Dr. Stefan Imel. Here is the A1 symmetry between the carbon 1s. Remember that when we did lithium, when we drew those contours, there was hardly any overlap. What you're seeing here with the carbon 1s is exactly the same thing. The green contour of the carbon 1s is nowhere near the white protons that are sticking out like tinker toy spokes. And so the 1s is doing no bonding at all with the hydrogens. On the other hand, when we take the 2s, we get this big spherical look, it can't be exactly spherical because of the underlying tetrahedral shape, but we get this big green contour. And what's contoured here, the surface is drawn at the 90% probability that an electron in there is inside that region. And you can see that it goes around all the nuclei, the carbon and all four hydrogens. So these two electrons in this orbital have this whole big space to go around it. And it's a very good overlap with the 1s atomic orbitals of hydrogen. Then the next one up is the 2p and I've shown here one horizontal and the 2p, one side is, I apologize, I didn't pick the same colors that Dr. Imel picked, but let's say green is positive and blue is negative rather than red and blue. Same difference. Here there is a node, but the important thing is, is that the node is right at the carbon atom. So although there's a node, it is not in between the two atomic nuclei. What was bad with H2, with the anti-bonding, is that the node was between them so that there was no possibility of the electron being in there to glue the positive charges together. Here we've got one side building up density to hold these two hydrogens in and the other side building up density to hold these two hydrogens in and the node is right at the carbon where it doesn't do any harm because it's not between the bonds. If we look at the other two, they're exactly the same as this one, but they're just flipped. They're all degenerate. They have exactly the same energy because of symmetry. So there's another one that's in and out. Again, it has a node right at the carbon atom and then the final one, the third one, is just up and down. It's nice that he's drawn them with these different perspectives because as you draw them slightly differently and look at them, you get a much better idea of what these surfaces actually look like than if you just have one perspective. These are the three then that we said had T2 symmetry and these are the three other bonding orbitals. They have a node but it's not a bad one to make an anti-bonding orbital. Question is then, on one hand we have the approach where we first monkey with the carbon orbitals and we keep the hydrogens and then we make a bond to each hydrogen in turn. That's nice because that's how we might draw it on a piece of paper if we were in organic chemistry for example or just regular Lewis structure. On the other, we have the MO approach and we can calculate the energy of these orbitals so I just drew them qualitatively but we can calculate them and under certain approximations and we can make the approximation better and better if we want to do more work. We don't need to do any more work to say look, in the MO there are two different kinds of orbitals. There's one that's T2 and there's a lower one that's A1 and therefore what we do is we go to somebody who does photoelectron spectroscopy and we say can you leak in some natural gas and ionize it and tell us what the binding energy of the electrons are in this region and how many orbitals there are and if we do this experiment there is a difference because if there are four equivalent hybrid orbitals each with two electrons then they're all the same. We get one band in the photoelectron spectrum, pick an electron out of any of those and boom and you never ionize again. It's so unlikely that you ionize period that you don't ionize twice so we don't have to worry about that and then the carbon core electrons which are the only ones left are much, much higher energy so we don't have to worry about them either because we can pick the photon so that they aren't going to come up. We are going to put in x-rays for example and therefore if there's one band in the photoelectron spectrum, Pauling is correct and MO theory is washed up and if there's two bands in the photoelectron spectrum then MO theory even though it's a little more complicated and we don't have to stick like lines is more correct. And let's then have a look and this experiment just like the paramagnetism of O2 is definitive and of course it comes down in favor of MO theory otherwise we wouldn't have gone to all this trouble. Here on slide 666 is the photoelectron spectrum of methane and there are two bands and they are assigned to T2 and A1 and they even have about the right ratio of size. You can see that one looks like it's maybe three times bigger than the other. There are a lot of factors that go into the cross sections of photo ejection so you can't just say if there's three orbitals you get three times the area or anything as simple as that but nevertheless it does seem to make some sense. And if we look at it the vibrational progression because methane is a much more complex molecule and can really vibrate in a lot of different ways which is of course one reason why it's a very bad greenhouse gas because you put it up in the atmosphere and it can absorb infrared in a lot of different ways and then radiate it back down to earth. We can't resolve all the little things like we could in N2. They just appear as kind of a ragged like a porcupine envelope but nevertheless we can say it's the T2 really that's doing the lion's share of the bonding in methane and the A1. The big teddy bear is rather less important because of the less vibrational progression when we eject an electron from it. This then says look we have to prefer the MO theory to the theory of localized bonds even clever localized bonds with hybrid orbitals because now we have an experiment that's clearly indicating that one approach is better and it disagrees with the other one and that's what science is all about. It's easy to talk but you can talk all you want if somebody does an experiment and shows the ocean is rising then it's rising and it doesn't matter whether you want to say it's not or if it's measured you can quibble with the measurement but you first have to understand how the measurement works and in fact the measurements along those lines are extremely reliable even though the system itself is quite complicated. Alright let's talk about some bigger molecules. We don't want to just leave off with methane although to a physical chemist methane is a big molecule in a lot of ways and I think you can see why because they may want to know a lot of things about it that an organic chemist may not be interested in and an engineer just wants to burn it and get some power. So we can look at some conjugated hydrocarbons and so-called aromatic systems. I've never smelled an aromatic molecule that was aromatic to me like a rose but they're called aromatic and they certainly do have smells but they usually smell like gas or mothballs or something like that. So let's look at benzene, naphthalene and a few others butadiene so-called conjugated hydrocarbons and this will be kind of an interesting, another interesting way to simplify what could be an extremely complex if we did it from scratch it would just be terrible calculation and make it simple and leave enough meat on the bone that we can still come to some interesting conclusions about the stability of the system. First let's just consider ethylene, C2H4. Ethylene is an extremely important molecule in the fine chemical industry. It's the feedstock for a lot of things and a lot of people are spending time and energy to figure out how to make ethylene efficiently when we run out of other ways that we have been making it. Well we've got the 2S orbitals on the two carbons and the two P orbitals and we've got the 4 1S orbitals on the hydrogens so there are going to be 12 molecular orbitals and because there's 12 atomic orbitals and 12 is a lot but we can still figure out what they should look like. Just like methane, even though this is bigger we're going to have to start by putting the nuclei at their preferred position, 120 degree bond angles between the hydrogen and the carbon and then we're going to have to solve for what the molecular orbital energy should be by putting in electrons that are allowed to go over the entire nuclear framework and only if they don't want to go over here for their own reasons are they not going to. It's not that we're capturing them between two nuclei. Here then on 668 is the molecular orbital diagram for ethylene. We've got our four hydrogens again but now we've got two carbons and we've got all these states labeled from the bottom, sigma G, 2AG with symmetry labels, 2B, 1U and so forth and so on. We don't have to understand what all of them mean but many of them are overlapped between the S orbitals on carbon and the S orbitals on the protons and then there's two right in the middle there marked pi G and pi U that are just coming from the carbon and that's because those two are orthogonal to the plane in which the protons are so those two have different symmetry than what the S can be and therefore they only come from the carbon and they can either be this way, the two carbons or that way and of course only one of them is filled because there's one double bond in ethylene. There's a sigma bond and then there's a pi bond and so one of those is filled right up to where it's stable. They're slightly below the 2P level. We slot in all our electrons all the way up. What we find is that it's filled right up to that level that involves just the P orbitals of carbon and here is the photoelectron spectrum of ethylene. We can also get that and now the ethylene is quite complicated so these peaks that all have vibrational progressions are very, very hard to resolve and they may not be resolvable at all. It depends how carefully and how clever the experimenter is but at least they aren't so wide that they overlap with each other and we can't see them at all and they've all been assigned as you can see 3B, 3U and 3AG and so forth and they're all assigned and they all correspond to ejecting electrons from lower and lower in the molecular orbital diagram all the way to 2AG after that. Maybe you need too much energy and they didn't have such energetic photons. The symmetry labels again treat them just as labels for now. Don't dwell on them too much. Once you study group theory you'll know exactly what they mean. For us though it's the highest occupied molecular orbital, the so-called HOMO and the lowest occupied, excuse me, the lowest unoccupied molecular orbital, the LUMO that are going to be the most important thing. Why? Because if I'm going to be making a bond then it's the electrons right at the top of the cake that are going to fly off and go somewhere and make the bond. The ones that are held down further down in energy are not going to be likely to be the ones going first and likewise if some other atom is coming up, rimming with electrons and says hey take some electrons from me, where are they going to go? Well they're going to go into the lowest unoccupied molecular orbital because they're going to come in and they're going to go down to the most stable state and therefore the highest occupied molecular orbital, what it is and the lowest unoccupied molecular orbital, what it looks like are very, very, very important to understanding chemical reactions and therefore we can focus mostly on them and forget about most of the other ones and that's a key simplification. The highest occupied one I've labeled as 1B3U and the lowest unoccupied is 1B2G and they both involve the carbon P orbitals and so these two, the highest occupied and the lowest unoccupied only involve pi electrons, it's only the pi electrons that are involved and that is a key simplification when you consider the reactivity and structure of things with double bonds like ethylene. We can treat all the sigma orbitals as just like core electrons and since they aren't important, they aren't going to be participating, we can rationalize whatever structure they have in any way we want and usually the prescription is those aren't so important, we just say they're sp2 hybrids, it doesn't matter what we call them because they're never going to do anything, they're just going to be filled and we might as well take a simple approach like Pauling's approach and just say well they're making 120 degree and there's a sigma framework that's holding the atoms in place and then there's these delicate pi orbitals and for those we reserve the molecular orbital treatment because those are going to do something and so we want to treat them more accurately and because there's only two orbitals, this and like that, it's basically down to the same thing as back to H2. The fact that they're P rather than S doesn't matter, the math is exactly the same and we've already done it so we can leverage that now and just say we've got this perpendicular pi system with two P orbitals and they can either be in phase or out of phase and we can calculate what we want to do so it's just like H2 in this case. Our pi wave function then while we suppose the molecular orbital for the pi is C1, 2PZ1 plus C2, 2PZ2 where 1 and 2 are referring to the two carbons and C1 and C2 are the two coefficients yet to be determined but we know they're going to come out to be equal because the carbons are the same bisymmetry. We have to make a secular determinant because this is molecular orbital, we've got to calculate the orbital energies, remember we have the Coulomb integral and the exchange integral and so here it is H11 minus ES11, we have the overlap integral H12 minus ES12, we have the exchange integral and so forth. The determinant so this times this is equal to zero and since the two carbon atoms are equivalent then H11, that integral and H22, that integral are exactly the same so we can make those the same and H12 is already the same. If we wanted to make quantitative progress here and we wanted to proceed the way we have in a lot of other cases helium and hydride, we'd have to get out our integrals, our Thetas and sine Thetas and all that and go to work but we don't want to do that, none of that's too easy to do and we're only after some qualitative description of what's going on, we don't expect it to be quantitative, it'd be very hard to make it quantitative even if we calculate these pi orbitals so well by just leaving out everything else and saying forget it about the other stuff, we could be making some big mistakes, some big errors there. Let's simplify things and not do any integrals at all and that will be a very, very good method and that's called the Huckel approximation. In integral whatever it is is just a number by the time you integrate it, it's a number with units but it's a number and we're going to assume that we're going to just replace integrals with these symbols, numbers, alpha and beta and we're going to calculate things very quickly then. The Huckel approximation makes three assumptions and at first they seem laughable but in fact they're very nicely, physically motivated. The first thing is the overlap integral, remember how much trouble that was to calculate s and how we had to do all those things. Well here what we're going to say is this, we're going to say that the overlap integral is zero unless you're talking about the overlap to the same place. So the overlap even between two neighbors is zero and that seems very counterintuitive why because you're claiming they're making a bond but when you look at what came out of our analysis before s was just a player in the denominator that just changed things slightly in the denominator. It never dictated whether a bond formed or not. It was okay so it doesn't really matter if we set it to zero and if we set it to zero then we're getting rid of a lot of math because all those things go away because they're multiplied by s12. So those are gone and then when it's the same we set it to one because we're saying it's normalized and then all the Coulomb integrals, the H integrals are the same for equivalent carbons and sometimes we assume they're the same even if the carbons aren't quite equivalent because we're just too lazy to figure out if they're that much different or it's difficult to figure out what they would be. And then the exchange or they're also called resonance integrals in the business. Those integrals vanish except for nearest neighbors and the rationale for that is that if I'm here and you've got a p orbital then that integral is going to have some value but once I get out to here and I've got something intervening in between it's too far away. So I could calculate it but it'd be small and I don't want to waste all my time calculating something that is 0.1% of the answer but is extremely difficult to calculate. That's a poor use of my time. So those three approximations let us simplify our secular determinant and get rid of all the calculations. Conventionally you write alpha for one of the integrals in beta for the exchange and our secular determinant just becomes alpha minus e beta beta alpha minus e equals to 0. And if I, that's a quadratic equation and the solution for the energies is alpha plus or minus beta. Not surprisingly it's very similar to what we had with H2. There's a good combination and a bad combination. Now in fact there's two electrons going in and we didn't calculate any electron, electron repulsion or anything like that. So all we're going to do is double the energy when we say what it actually is. The advantage of this however is we can do something that's a little bit more intimidating, much more intimidating than helium for example. Let's try benzene. Benzene is C6H6. We would have a 6 by 6 determinant and it would have entries everywhere. So just expanding that out as this times the determinant of 5 by 5 will take forever. And that's going to be a major problem to try to calculate something like that. So let's figure out how to get our 6 PZ orbitals in the pi system to yield 6 MOs without doing that. Let's set up the determinant then which I've done here on slide 676 in the Huckel approximation. The nice thing is except for terms right near the diagonal in the array each of which has beta when they're off because they all have the same value. And the 6 carbons all have the same integrals for the H11, H22 and so forth. So those are all alpha. And then I have my energy, the overlap is gone so there's no energy off the diagonal, that's very convenient. Everything's zero everywhere except for the top and bottom and that's because as I go around the structure 6 is next to 1. So you always end up with 1 up there. But all the rest are zero so most of the thing goes away. And we can even make it simpler by since we've got zero and we know beta's not zero we can divide everything by beta so that the certain things are 1 if they're off the diagonal. And then we can just redefine alpha minus E divided by beta to be some variable. So let's let X equals to alpha minus E divided by beta. And then all the other terms are 1 then we end up with the following polynomial to root. If you expand out the secular determinant you end up with a sixth order polynomial. It's X to the 6 minus 6X to the 4th plus 9X squared minus 4 equals zero. So that's why you took a course in algebra in high school because now this should be a duck soup problem. In fact it may not be so easy if you just stare at it but there's a trick. I'm going to let you explore the trick on the homework. Because it's only even powers we can redefine Y is equal to X squared and then we've only got a cubic. And if we're lucky we might be able to factorize a cubic and figure out how to actually get the answer. Or we can plot it even for lazy in Mathematica and we can see where it crosses zero and if it crosses zero at some convenient place that is easy to tell because it's an integer then we can try factorizing it with X minus that integer X naught or N naught and see if it factorizes. And if you do that and you're persistent you find that you have repeated roots because of the Y equals X squared and then some of them are repeated again just because of the structure of what you end up with at the end in the cubic and you end up with X is equal to plus or minus one or plus or minus one or plus or minus two. Those are the six roots of that equation. Well X was alpha minus E over beta and therefore there are the corresponding energies E1 which is the best one alpha plus two beta recall that these are usually negative. E2 is equal to E3 is alpha plus beta. E4 is equal to E5 that's alpha minus beta and then E6 is alpha minus two beta. So there's nice symmetry about how the energies come out. Now how do we get what the molecular orbitals are? All what we had to do before we go back we actually put in the energy to the original equation that we had and then we find the orbital that we get that corresponds to that energy. The degenerate ones that have E2 and E3 the same and E4 and E5 the same require a little bit of thought to how to sort those two out. But whatever orbitals we get recall that they have to be orthonormal. The molecular orbitals have to be orthonormal as well. So that's an important thing that lets you simplify it. But I'll let you do that and here I've given you the answer. The first one not surprisingly the lowest energy molecular orbital for benzene is when all the six recall benzene as a hexagon planar is when all the six p orbitals have the same phase. So I get a ring of positive electron density around the top holding all the carbons together and I get a ring around the bottom of the opposite sign but when I square it I find I build up density in between the carbons even though they're off axis and that holds them all together. And then the next two have some nodes but the nodes aren't bad. So those are between at certain carbons and the orbitals are just like methane they hold the other parts together. So there's one that has a node this way that holds these parts together and there's one the other way that holds these parts together. And then there's the bad three that have the opposite combinations basically and the highest one is when this one's up and this one's down and this one's up and this one's down and so forth and that has nodes in between everything and tons of nodes and you can see it's a very crinkled wave function that's going to have very high energy and that's the least favorable one. That's a strongly anti-bonding orbital. And in fact by drawing simple figures just based on the cyclic structure you can arrive at some simple rules to predict aromatic stability. You draw the structure point down and so for hexagon you draw point down and then at each vertex you put a line which is an MO and you have one on the bottom then you have two that's E2 and E3 then you have these two that's E4 and E5 and then you have one at the top that's E6 and you've only got six electrons because you only had one electron in each p orbital to start with so you fill up the bottom to orbital with two this one with two this one with two game over. So you pick the three good ones the three winners and you left all the anti-bonding ones unoccupied and all the electrons are paired and all these orbitals are delocalized so we don't predict any difference although in the Lewis structure of course for benzene we draw a single bond and a double bond because we have to because of the Lewis structure and then we say well there's resonance we move this here and move them around like a mouse trap and the molecular orbital we don't have to do that they're already set up to be delocalized. On the other hand if we pick cyclobutadiene and we stand a square on its points then we have a lowest one E1 then we have E2 and E3 and then we have E4 but we have four carbons in cyclobutadiene and therefore we've only got four electrons so we put two in the bottom but then we've got two equal ones here well that's like O2 again so one goes here and one goes there. Well that's not aromatic stability that's predicting a di-radical I wouldn't take this kind of very qualitative thing too seriously I do some spectroscopy to figure out what you actually get but whenever it comes out like that where the two aren't paired and there's some left the conclusion is it's not aromatic and what happens then is the ones that have 4n plus 2 pi electrons become stable and the ones with 4n like 4 rather than 6 are not especially stable and not stable at all in some cases so only the 4n plus 2 pi systems are predicted to have aromatic stability and that comes out of a more detailed calculation to not just playing games with shapes but it's interesting to just play games with shapes because it's very quick and often it's good enough. Alright let's do a practice problem let's consider the following molecules and let's consider them and see which one is an organic chemist would predict to be aromatic so they had the special stability benzene was used as a solvent for a long time why because it doesn't react with anything so it's a perfect solvent if things won't dissolve in water or the reaction slow in water you dissolve it in benzene and the reaction goes very quickly problem was of course benzene is quite a potent carcinogen like a lot of these systems and so if you're breathing a lot of it and boiling it up and breathing it on a daily basis it's not the greatest and that's why they went away from it. Let's consider then benzene we know is aromatic let's consider naphthalene which is in moth balls that's that smell azuline which you probably haven't seen nor smelled and then cycloactatetraene let's take these three and see what they are well first thing is the names if you don't know what the structures are referring to the names are useless and we have to know they aren't going to be aromatic unless in the Lewis structure they would have alternating single and double bonds and we could do resonance and move the things around so that they would all be equivalent half the time single half the time double if they aren't even like that then forget it it's not going to be an aromatic system or the part that can't do that trick with resonance is not going to be an aromatic system and they have to have a cyclic structure as well so that the snake bites its tail because that's important in how those two last terms come in if those don't come in a lot of things change it turns out and then let's draw the structures and count them so for naphthalene we've got two benzene rings fused together now if we were actually going to do the MO treatment we'd have to be careful because these two carbons without any hydrogens on could have a different coulomb integral than the others and the others would all be the same and so we should take that into account we should have alpha 1 and alpha 2 for lazy we just say well they're about the same column alpha and I've drawn the Lewis structure here one of them and we could play a game with resonance then and draw one the other way we count the electrons in the pi system we forget about the sigma we forget about the hydrogen and we see there's 1 2 3 4 5 6 7 8 9 10 that's 4n plus 2 for n equals 2 and so it's planar and it is in fact aromatic for azuline which is a structural isomer of naphthalene there's a 7 membered ring glued to a 5 membered ring and I can still draw the double bonds the same way it's kind of amazing that it works like that and it's a beautiful blue color and again I count up it's a cyclic system I count up there's 10 pi electrons and the delocalized MO's go over the whole thing and it's planar and it's very nice example of an aromatic system that's not so trivial like a benzene one and then here what I've shown you is a mushroom that is blue it's absolutely amazing but in fact this mushroom makes an azuline derivative for some reason probably extremely interesting chemistry usually when plants make interesting molecules it's because they're keeping bugs off or keeping other things away because too bad for plants they can't move and if you're stuck and bugs are crawling all over you you need chemical warfare to keep them at bay usually then if you see a brightly colored mushroom like this you admire its beauty and its beautiful blue color but let someone else figure out if it's safe to eat there are people who do that for a living and it's very interesting to see how courageous they are they usually eat a very tiny piece and then wait and see if what happens and then they eat a little bit more and I'll let them do that but don't eat the mushrooms that you find around that are this beautiful blue color or the ones with the orange gills either because that might be the last thing you do. For cycloactatetraene I've drawn it like a stop sign with four double bonds but it only has eight pi electrons so that's not 4n plus 2 and in fact the system is not aromatic and it's not planar it has a 3d structure and looks much more like alternating single and double bonds that are localized so that one not surprisingly if it's not aromatic it doesn't usually have a trivial name. Okay we're going to leave it there and for our very last lecture what I want to do is go through where we started, where we got to and all the things we covered because we've covered a lot of ground from electrons and photons to atoms to molecules and we've done it in a certain semi-systematic way that I hope has made certain things that you wondered about much clearer and maybe caused you to want to learn more about some other things that you've never heard of so we'll leave it there and then sum up in the next lecture.
 Right, let's pick up where we left off.  We had just gone through Linus Pauling's brilliant exposition on hybrid orbitals, which are still  used to this very day to rationalize a lot of things to do with structure and reactivity.  But let's take a closer look, let's compare the hybrid orbital approach, which really  says let's make pairs of electrons make the bonds with the molecular orbital approach,  which says look, the way you have to do it is the way you did it with an atom.  You have to put all the positive charges where they're supposed to be, and then you  have to solve it.  Then you have to get the solution, and then you can put in your electrons into the orbitals,  and they go into the lowest energy once.  This is a more delocalized approach to bonding because if you're including all the nuclei  at once, then there's no reason why you should be drawing lines.  That makes it a little harder to understand though when you're drawing structures and  you're pushing the lines around, chemists love those lines because it allows them to  write and rationalize reactions.  But occasionally a reaction goes that doesn't seem to behave, and that could be an indication  that maybe the theory of the lines is falling apart a little bit.  Let's talk then in that vein about more delocalized bonding.  In the MO picture, we don't make hybrid orbitals, so the orbital is used, but it's completely  different.  They must not confuse the two.  In the molecular orbital picture, what we said we had to do was we could combine together  atomic orbitals that had the same symmetry, similar energy, and overlapped in space.  For tetrahedral molecule with a central atom like methane, we've got four carbon valence  atomic orbitals, the S and the three P's.  We have to know the structure.  Why?  Well, otherwise we'd have to move the things around and calculate the energy as a function  of R and R2 and R3.  We already know the structure.  It's tetrahedral.  What we want to understand is not the structure so much.  We want to understand the bonding.  How is it working?  What's going on?  We assume we know the structure.  It's the same structure as Pauling's, but the description of the bonding is going to  be different, and that's the thing that we're going to focus on.  In the MO approach, you have to know the structure first.  If you don't, you have to search through all possible structures.  If you're doing that, that is a very, very, very long-winded and intensive calculation,  although sometimes people do it just to see what might be the most stable structure if  there's no experiment or it hasn't been made.  Let's put our hydrogen atoms at the corners of the same cube that we had before, HA, HB,  HC, and HD, and let's see what we're going to do with these four.  The carbon is at the center, and with our four hydrogen orbitals and our four carbon  orbitals, we get eight molecular orbitals because recall in the linear combination of  atomic orbitals, rubric, we have the same number of orbitals at the end in terms of molecular  orbitals as what we started with.  Okay, now question is, which orbitals can we combine?  The answer is rule two said they have to have the same symmetry for this molecule.  What that means is that we cannot do what Pauling did because we cannot combine the  carbon 2S and the carbon 2PZ or the other two P's because those have different symmetry.  Therefore if we want to consider the 2S on carbon, we cannot combine it with any of the  other atomic orbitals on carbon.  We have to instead combine it with combinations of the hydrogen orbitals at the four positions  of the cube.  And that's completely different then because now we won't necessarily end up with four  identical lines in terms of four localized bonds.  We will end up with a tetrahedral structure, that's for sure, because we started with that,  but our description of the bonding will be different.  Now the game is this.  We have the carbon orbital.  Each of the four of them has different symmetry.  We have to combine the four hydrogens so that they have the same symmetry as the carbon  orbital.  And that's pretty easy to do because if the carbon orbital is a big fat round S, then  that means that these guys have to be as round as possible so they should just all be red  and that should be one of them.  And if the carbon orbital changes sign, like it's plus here and minus here, well we've  got two hydrogens on the top of the cube, they should be plus.  Two hydrogens on the bottom of the cube, they should be minus and so forth.  And so there's a combination of hydrogens that goes perfectly with each of the four  combination on the carbon and that's how we're going to do it.  So for 2S we use 1SA plus 1SB plus 1SC plus 1SD.  All of them as round as can be.  And this then, if they're all red, the 2S is red and all these guys are red, this makes  a gigantic orbital with no nodes that goes around all the nuclei and has roughly spherical  symmetry and it's called A1 which is a symmetry label that lets the spectroscopists know what  the symmetry of the molecular orbital is.  And that name A1 has to do with its symmetry under the tetrahedral point group TD that  you'll learn about later in the course.  For now we're just going to treat A1 as a label that tells us which orbital we're talking  about.  Remember that less nodes is lower energy and so we've got this big red thing with this  big fluffy thing sort of looks like a teddy bear with no nodes therefore A1 is the lowest  symmetry.  That's what we predict.  Above it then are three degenerate because they're related by symmetry orbitals and there's  one for each of the p orbitals on the carbon atom.  And those three are called T2, T means three, a name that has to do again with their symmetry  under the tetrahedral point group.  The one and two have meanings but I don't want to go into the exact meaning now.  These are the four bonding orbitals and they are not the same but there's four of them  and they still predict an exactly tetrahedral structure not surprisingly because we started  out with an exactly tetrahedral structure.  The other four combinations have more nodes.  I could take the S as red and I have to have something the same symmetry but I can pick  them all blue.  Now that's a disaster because now there's nodes in between the two nuclei so that one's  no good and likewise I could pick the PZ with red on the top and blue on the bottom and  I could perversely pick these two guys to be blue rather than red and then there would  be destructive interference and again that would be a very high energy solution.  Therefore we can easily see that there are four bonding orbitals so there's going to  be four bonds, good, and there's four anti-bonding orbitals and since there's only eight electrons  the four bonding orbitals are filled and the four anti-bonding orbitals are empty and that's  almost always the way it works out when you do things correctly because not surprisingly  nature finds its way into making the more stable configurations.  Here then is an MO diagram.  I have four, it's a little bit harder to draw the tennis tournament when you don't have  just two players but I have four hydrogen 1s orbitals on one side and then I have the  carbon orbitals and I have the 2s of the carbon lower than the 1s of the hydrogen, that's  good and then I have the 2p higher.  Well I'd have to know that but I can certainly figure that out by looking up the ionization  energy of the atoms which is well known.  And so I can order them like that and I have one very good combination, the teddy bear  which is down at the bottom, the A1 and then I have the three that are identical, the T2  and they're at slightly higher energy.  The key is there's no way that the A1 and the T2 have the same energy.  They have to have different energies and that is an experimentally testable fact that we  can look at.  So we fill up the bottom with eight electrons, four from the hydrogen, four from the carbon,  the same way we always do with molecular orbital diagrams.  Once we calculate the diagram we fill it up from the bottom.  And here is an actual calculation of these orbital contours from the excellent website  of Dr. Stefan Imel.  Here is the A1 symmetry between the carbon 1s.  Remember that when we did lithium, when we drew those contours, there was hardly any overlap.  What you're seeing here with the carbon 1s is exactly the same thing.  The green contour of the carbon 1s is nowhere near the white protons that are sticking out  like tinker toy spokes.  And so the 1s is doing no bonding at all with the hydrogens.  On the other hand, when we take the 2s, we get this big spherical look, it can't be exactly  spherical because of the underlying tetrahedral shape, but we get this big green contour.  And what's contoured here, the surface is drawn at the 90% probability that an electron  in there is inside that region.  And you can see that it goes around all the nuclei, the carbon and all four hydrogens.  So these two electrons in this orbital have this whole big space to go around it.  And it's a very good overlap with the 1s atomic orbitals of hydrogen.  Then the next one up is the 2p and I've shown here one horizontal and the 2p, one side is,  I apologize, I didn't pick the same colors that Dr. Imel picked, but let's say green  is positive and blue is negative rather than red and blue.  Same difference.  Here there is a node, but the important thing is, is that the node is right at the carbon  atom.  So although there's a node, it is not in between the two atomic nuclei.  What was bad with H2, with the anti-bonding, is that the node was between them so that  there was no possibility of the electron being in there to glue the positive charges together.  Here we've got one side building up density to hold these two hydrogens in and the other  side building up density to hold these two hydrogens in and the node is right at the  carbon where it doesn't do any harm because it's not between the bonds.  If we look at the other two, they're exactly the same as this one, but they're just flipped.  They're all degenerate.  They have exactly the same energy because of symmetry.  So there's another one that's in and out.  Again, it has a node right at the carbon atom and then the final one, the third one, is  just up and down.  It's nice that he's drawn them with these different perspectives because as you draw  them slightly differently and look at them, you get a much better idea of what these surfaces  actually look like than if you just have one perspective.  These are the three then that we said had T2 symmetry and these are the three other bonding  orbitals.  They have a node but it's not a bad one to make an anti-bonding orbital.  Question is then, on one hand we have the approach where we first monkey with the carbon  orbitals and we keep the hydrogens and then we make a bond to each hydrogen in turn.  That's nice because that's how we might draw it on a piece of paper if we were in organic  chemistry for example or just regular Lewis structure.  On the other, we have the MO approach and we can calculate the energy of these orbitals  so I just drew them qualitatively but we can calculate them and under certain approximations  and we can make the approximation better and better if we want to do more work.  We don't need to do any more work to say look, in the MO there are two different kinds of  orbitals.  There's one that's T2 and there's a lower one that's A1 and therefore what we do is  we go to somebody who does photoelectron spectroscopy and we say can you leak in some natural gas  and ionize it and tell us what the binding energy of the electrons are in this region  and how many orbitals there are and if we do this experiment there is a difference because  if there are four equivalent hybrid orbitals each with two electrons then they're all the  same.  We get one band in the photoelectron spectrum, pick an electron out of any of those and boom  and you never ionize again.  It's so unlikely that you ionize period that you don't ionize twice so we don't have to  worry about that and then the carbon core electrons which are the only ones left are  much, much higher energy so we don't have to worry about them either because we can pick  the photon so that they aren't going to come up.  We are going to put in x-rays for example and therefore if there's one band in the photoelectron  spectrum, Pauling is correct and MO theory is washed up and if there's two bands in the  photoelectron spectrum then MO theory even though it's a little more complicated and  we don't have to stick like lines is more correct.  And let's then have a look and this experiment just like the paramagnetism of O2 is definitive  and of course it comes down in favor of MO theory otherwise we wouldn't have gone to  all this trouble.  Here on slide 666 is the photoelectron spectrum of methane and there are two bands and they  are assigned to T2 and A1 and they even have about the right ratio of size.  You can see that one looks like it's maybe three times bigger than the other.  There are a lot of factors that go into the cross sections of photo ejection so you can't  just say if there's three orbitals you get three times the area or anything as simple  as that but nevertheless it does seem to make some sense.  And if we look at it the vibrational progression because methane is a much more complex molecule  and can really vibrate in a lot of different ways which is of course one reason why it's  a very bad greenhouse gas because you put it up in the atmosphere and it can absorb  infrared in a lot of different ways and then radiate it back down to earth.  We can't resolve all the little things like we could in N2.  They just appear as kind of a ragged like a porcupine envelope but nevertheless we can  say it's the T2 really that's doing the lion's share of the bonding in methane and the A1.  The big teddy bear is rather less important because of the less vibrational progression  when we eject an electron from it.  This then says look we have to prefer the MO theory to the theory of localized bonds  even clever localized bonds with hybrid orbitals because now we have an experiment that's clearly  indicating that one approach is better and it disagrees with the other one and that's  what science is all about.  It's easy to talk but you can talk all you want if somebody does an experiment and shows  the ocean is rising then it's rising and it doesn't matter whether you want to say it's  not or if it's measured you can quibble with the measurement but you first have to understand  how the measurement works and in fact the measurements along those lines are extremely  reliable even though the system itself is quite complicated.  Alright let's talk about some bigger molecules.  We don't want to just leave off with methane although to a physical chemist methane is  a big molecule in a lot of ways and I think you can see why because they may want to know  a lot of things about it that an organic chemist may not be interested in and an engineer just  wants to burn it and get some power.  So we can look at some conjugated hydrocarbons and so-called aromatic systems.  I've never smelled an aromatic molecule that was aromatic to me like a rose but they're  called aromatic and they certainly do have smells but they usually smell like gas or  mothballs or something like that.  So let's look at benzene, naphthalene and a few others butadiene so-called conjugated  hydrocarbons and this will be kind of an interesting, another interesting way to simplify what  could be an extremely complex if we did it from scratch it would just be terrible calculation  and make it simple and leave enough meat on the bone that we can still come to some interesting  conclusions about the stability of the system.  First let's just consider ethylene, C2H4.  Ethylene is an extremely important molecule in the fine chemical industry.  It's the feedstock for a lot of things and a lot of people are spending time and energy  to figure out how to make ethylene efficiently when we run out of other ways that we have  been making it.  Well we've got the 2S orbitals on the two carbons and the two P orbitals and we've  got the 4 1S orbitals on the hydrogens so there are going to be 12 molecular orbitals  and because there's 12 atomic orbitals and 12 is a lot but we can still figure out what  they should look like.  Just like methane, even though this is bigger we're going to have to start by putting the  nuclei at their preferred position, 120 degree bond angles between the hydrogen and the carbon  and then we're going to have to solve for what the molecular orbital energy should be  by putting in electrons that are allowed to go over the entire nuclear framework and only  if they don't want to go over here for their own reasons are they not going to.  It's not that we're capturing them between two nuclei.  Here then on 668 is the molecular orbital diagram for ethylene.  We've got our four hydrogens again but now we've got two carbons and we've got all these  states labeled from the bottom, sigma G, 2AG with symmetry labels, 2B, 1U and so forth  and so on.  We don't have to understand what all of them mean but many of them are overlapped between  the S orbitals on carbon and the S orbitals on the protons and then there's two right  in the middle there marked pi G and pi U that are just coming from the carbon and that's  because those two are orthogonal to the plane in which the protons are so those two have  different symmetry than what the S can be and therefore they only come from the carbon  and they can either be this way, the two carbons or that way and of course only one of them  is filled because there's one double bond in ethylene.  There's a sigma bond and then there's a pi bond and so one of those is filled right up  to where it's stable.  They're slightly below the 2P level.  We slot in all our electrons all the way up.  What we find is that it's filled right up to that level that involves just the P orbitals  of carbon and here is the photoelectron spectrum of ethylene.  We can also get that and now the ethylene is quite complicated so these peaks that all  have vibrational progressions are very, very hard to resolve and they may not be resolvable  at all.  It depends how carefully and how clever the experimenter is but at least they aren't  so wide that they overlap with each other and we can't see them at all and they've  all been assigned as you can see 3B, 3U and 3AG and so forth and they're all assigned  and they all correspond to ejecting electrons from lower and lower in the molecular orbital  diagram all the way to 2AG after that.  Maybe you need too much energy and they didn't have such energetic photons.  The symmetry labels again treat them just as labels for now.  Don't dwell on them too much.  Once you study group theory you'll know exactly what they mean.  For us though it's the highest occupied molecular orbital, the so-called HOMO and the lowest  occupied, excuse me, the lowest unoccupied molecular orbital, the LUMO that are going  to be the most important thing.  Why?  Because if I'm going to be making a bond then it's the electrons right at the top of the  cake that are going to fly off and go somewhere and make the bond.  The ones that are held down further down in energy are not going to be likely to be the  ones going first and likewise if some other atom is coming up, rimming with electrons  and says hey take some electrons from me, where are they going to go?  Well they're going to go into the lowest unoccupied molecular orbital because they're going to  come in and they're going to go down to the most stable state and therefore the highest  occupied molecular orbital, what it is and the lowest unoccupied molecular orbital, what  it looks like are very, very, very important to understanding chemical reactions and therefore  we can focus mostly on them and forget about most of the other ones and that's a key simplification.  The highest occupied one I've labeled as 1B3U and the lowest unoccupied is 1B2G and they  both involve the carbon P orbitals and so these two, the highest occupied and the lowest  unoccupied only involve pi electrons, it's only the pi electrons that are involved and  that is a key simplification when you consider the reactivity and structure of things with  double bonds like ethylene.  We can treat all the sigma orbitals as just like core electrons and since they aren't  important, they aren't going to be participating, we can rationalize whatever structure they  have in any way we want and usually the prescription is those aren't so important, we just say  they're sp2 hybrids, it doesn't matter what we call them because they're never going to  do anything, they're just going to be filled and we might as well take a simple approach  like Pauling's approach and just say well they're making 120 degree and there's a sigma  framework that's holding the atoms in place and then there's these delicate pi orbitals  and for those we reserve the molecular orbital treatment because those are going to do something  and so we want to treat them more accurately and because there's only two orbitals, this  and like that, it's basically down to the same thing as back to H2.  The fact that they're P rather than S doesn't matter, the math is exactly the same and we've  already done it so we can leverage that now and just say we've got this perpendicular  pi system with two P orbitals and they can either be in phase or out of phase and we  can calculate what we want to do so it's just like H2 in this case.  Our pi wave function then while we suppose the molecular orbital for the pi is C1, 2PZ1  plus C2, 2PZ2 where 1 and 2 are referring to the two carbons and C1 and C2 are the two  coefficients yet to be determined but we know they're going to come out to be equal because  the carbons are the same bisymmetry.  We have to make a secular determinant because this is molecular orbital, we've got to calculate  the orbital energies, remember we have the Coulomb integral and the exchange integral  and so here it is H11 minus ES11, we have the overlap integral H12 minus ES12, we have  the exchange integral and so forth.  The determinant so this times this is equal to zero and since the two carbon atoms are  equivalent then H11, that integral and H22, that integral are exactly the same so we can  make those the same and H12 is already the same.  If we wanted to make quantitative progress here and we wanted to proceed the way we have  in a lot of other cases helium and hydride, we'd have to get out our integrals, our Thetas  and sine Thetas and all that and go to work but we don't want to do that, none of that's  too easy to do and we're only after some qualitative description of what's going on,  we don't expect it to be quantitative, it'd be very hard to make it quantitative even if  we calculate these pi orbitals so well by just leaving out everything else and saying  forget it about the other stuff, we could be making some big mistakes, some big errors  there.  Let's simplify things and not do any integrals at all and that will be a very, very good  method and that's called the Huckel approximation.  In integral whatever it is is just a number by the time you integrate it, it's a number  with units but it's a number and we're going to assume that we're going to just replace  integrals with these symbols, numbers, alpha and beta and we're going to calculate things  very quickly then.  The Huckel approximation makes three assumptions and at first they seem laughable but in fact  they're very nicely, physically motivated.  The first thing is the overlap integral, remember how much trouble that was to calculate s and  how we had to do all those things.  Well here what we're going to say is this, we're going to say that the overlap integral  is zero unless you're talking about the overlap to the same place.  So the overlap even between two neighbors is zero and that seems very counterintuitive  why because you're claiming they're making a bond but when you look at what came out  of our analysis before s was just a player in the denominator that just changed things  slightly in the denominator.  It never dictated whether a bond formed or not.  It was okay so it doesn't really matter if we set it to zero and if we set it to zero  then we're getting rid of a lot of math because all those things go away because they're multiplied  by s12.  So those are gone and then when it's the same we set it to one because we're saying it's  normalized and then all the Coulomb integrals, the H integrals are the same for equivalent  carbons and sometimes we assume they're the same even if the carbons aren't quite equivalent  because we're just too lazy to figure out if they're that much different or it's difficult  to figure out what they would be.  And then the exchange or they're also called resonance integrals in the business.  Those integrals vanish except for nearest neighbors and the rationale for that is that  if I'm here and you've got a p orbital then that integral is going to have some value  but once I get out to here and I've got something intervening in between it's too far away.  So I could calculate it but it'd be small and I don't want to waste all my time calculating  something that is 0.1% of the answer but is extremely difficult to calculate.  That's a poor use of my time.  So those three approximations let us simplify our secular determinant and get rid of all  the calculations.  Conventionally you write alpha for one of the integrals in beta for the exchange and our  secular determinant just becomes alpha minus e beta beta alpha minus e equals to 0.  And if I, that's a quadratic equation and the solution for the energies is alpha plus  or minus beta.  Not surprisingly it's very similar to what we had with H2.  There's a good combination and a bad combination.  Now in fact there's two electrons going in and we didn't calculate any electron, electron  repulsion or anything like that.  So all we're going to do is double the energy when we say what it actually is.  The advantage of this however is we can do something that's a little bit more intimidating,  much more intimidating than helium for example.  Let's try benzene.  Benzene is C6H6.  We would have a 6 by 6 determinant and it would have entries everywhere.  So just expanding that out as this times the determinant of 5 by 5 will take forever.  And that's going to be a major problem to try to calculate something like that.  So let's figure out how to get our 6 PZ orbitals in the pi system to yield 6 MOs without doing  that.  Let's set up the determinant then which I've done here on slide 676 in the Huckel approximation.  The nice thing is except for terms right near the diagonal in the array each of which has  beta when they're off because they all have the same value.  And the 6 carbons all have the same integrals for the H11, H22 and so forth.  So those are all alpha.  And then I have my energy, the overlap is gone so there's no energy off the diagonal,  that's very convenient.  Everything's zero everywhere except for the top and bottom and that's because as I go  around the structure 6 is next to 1.  So you always end up with 1 up there.  But all the rest are zero so most of the thing goes away.  And we can even make it simpler by since we've got zero and we know beta's not zero we can  divide everything by beta so that the certain things are 1 if they're off the diagonal.  And then we can just redefine alpha minus E divided by beta to be some variable.  So let's let X equals to alpha minus E divided by beta.  And then all the other terms are 1 then we end up with the following polynomial to root.  If you expand out the secular determinant you end up with a sixth order polynomial.  It's X to the 6 minus 6X to the 4th plus 9X squared minus 4 equals zero.  So that's why you took a course in algebra in high school because now this should be  a duck soup problem.  In fact it may not be so easy if you just stare at it but there's a trick.  I'm going to let you explore the trick on the homework.  Because it's only even powers we can redefine Y is equal to X squared and then we've only  got a cubic.  And if we're lucky we might be able to factorize a cubic and figure out how to actually get  the answer.  Or we can plot it even for lazy in Mathematica and we can see where it crosses zero and if  it crosses zero at some convenient place that is easy to tell because it's an integer then  we can try factorizing it with X minus that integer X naught or N naught and see if it  factorizes.  And if you do that and you're persistent you find that you have repeated roots because  of the Y equals X squared and then some of them are repeated again just because of the  structure of what you end up with at the end in the cubic and you end up with X is equal  to plus or minus one or plus or minus one or plus or minus two.  Those are the six roots of that equation.  Well X was alpha minus E over beta and therefore there are the corresponding energies E1 which  is the best one alpha plus two beta recall that these are usually negative.  E2 is equal to E3 is alpha plus beta.  E4 is equal to E5 that's alpha minus beta and then E6 is alpha minus two beta.  So there's nice symmetry about how the energies come out.  Now how do we get what the molecular orbitals are?  All what we had to do before we go back we actually put in the energy to the original  equation that we had and then we find the orbital that we get that corresponds to that  energy.  The degenerate ones that have E2 and E3 the same and E4 and E5 the same require a little  bit of thought to how to sort those two out.  But whatever orbitals we get recall that they have to be orthonormal.  The molecular orbitals have to be orthonormal as well.  So that's an important thing that lets you simplify it.  But I'll let you do that and here I've given you the answer.  The first one not surprisingly the lowest energy molecular orbital for benzene is when  all the six recall benzene as a hexagon planar is when all the six p orbitals have the same  phase.  So I get a ring of positive electron density around the top holding all the carbons together  and I get a ring around the bottom of the opposite sign but when I square it I find  I build up density in between the carbons even though they're off axis and that holds  them all together.  And then the next two have some nodes but the nodes aren't bad.  So those are between at certain carbons and the orbitals are just like methane they hold  the other parts together.  So there's one that has a node this way that holds these parts together and there's one  the other way that holds these parts together.  And then there's the bad three that have the opposite combinations basically and the highest  one is when this one's up and this one's down and this one's up and this one's down  and so forth and that has nodes in between everything and tons of nodes and you can see  it's a very crinkled wave function that's going to have very high energy and that's  the least favorable one.  That's a strongly anti-bonding orbital.  And in fact by drawing simple figures just based on the cyclic structure you can arrive  at some simple rules to predict aromatic stability.  You draw the structure point down and so for hexagon you draw point down and then at each  vertex you put a line which is an MO and you have one on the bottom then you have two  that's E2 and E3 then you have these two that's E4 and E5 and then you have one at  the top that's E6 and you've only got six electrons because you only had one electron  in each p orbital to start with so you fill up the bottom to orbital with two this one  with two this one with two game over.  So you pick the three good ones the three winners and you left all the anti-bonding ones unoccupied  and all the electrons are paired and all these orbitals are delocalized so we don't predict  any difference although in the Lewis structure of course for benzene we draw a single bond  and a double bond because we have to because of the Lewis structure and then we say well  there's resonance we move this here and move them around like a mouse trap and the molecular  orbital we don't have to do that they're already set up to be delocalized.  On the other hand if we pick cyclobutadiene and we stand a square on its points then we  have a lowest one E1 then we have E2 and E3 and then we have E4 but we have four carbons  in cyclobutadiene and therefore we've only got four electrons so we put two in the bottom  but then we've got two equal ones here well that's like O2 again so one goes here and  one goes there.  Well that's not aromatic stability that's predicting a di-radical I wouldn't take this  kind of very qualitative thing too seriously I do some spectroscopy to figure out what  you actually get but whenever it comes out like that where the two aren't paired and  there's some left the conclusion is it's not aromatic and what happens then is the ones  that have 4n plus 2 pi electrons become stable and the ones with 4n like 4 rather than 6  are not especially stable and not stable at all in some cases so only the 4n plus 2 pi  systems are predicted to have aromatic stability and that comes out of a more detailed calculation  to not just playing games with shapes but it's interesting to just play games with shapes  because it's very quick and often it's good enough.  Alright let's do a practice problem let's consider the following molecules and let's  consider them and see which one is an organic chemist would predict to be aromatic so they  had the special stability benzene was used as a solvent for a long time why because it  doesn't react with anything so it's a perfect solvent if things won't dissolve in water  or the reaction slow in water you dissolve it in benzene and the reaction goes very quickly  problem was of course benzene is quite a potent carcinogen like a lot of these systems and  so if you're breathing a lot of it and boiling it up and breathing it on a daily basis it's  not the greatest and that's why they went away from it.  Let's consider then benzene we know is aromatic let's consider naphthalene which is in moth  balls that's that smell azuline which you probably haven't seen nor smelled and then  cycloactatetraene let's take these three and see what they are well first thing is  the names if you don't know what the structures are referring to the names are useless and  we have to know they aren't going to be aromatic unless in the Lewis structure they would have  alternating single and double bonds and we could do resonance and move the things around  so that they would all be equivalent half the time single half the time double if they  aren't even like that then forget it it's not going to be an aromatic system or the  part that can't do that trick with resonance is not going to be an aromatic system and  they have to have a cyclic structure as well so that the snake bites its tail because that's  important in how those two last terms come in if those don't come in a lot of things  change it turns out and then let's draw the structures and count them so for naphthalene  we've got two benzene rings fused together now if we were actually going to do the MO  treatment we'd have to be careful because these two carbons without any hydrogens on  could have a different coulomb integral than the others and the others would all be the  same and so we should take that into account we should have alpha 1 and alpha 2 for lazy  we just say well they're about the same column alpha and I've drawn the Lewis structure here  one of them and we could play a game with resonance then and draw one the other way  we count the electrons in the pi system we forget about the sigma we forget about the  hydrogen and we see there's 1 2 3 4 5 6 7 8 9 10 that's 4n plus 2 for n equals 2 and  so it's planar and it is in fact aromatic for azuline which is a structural isomer  of naphthalene there's a 7 membered ring glued to a 5 membered ring and I can still  draw the double bonds the same way it's kind of amazing that it works like that and it's  a beautiful blue color and again I count up it's a cyclic system I count up there's 10  pi electrons and the delocalized MO's go over the whole thing and it's planar and it's  very nice example of an aromatic system that's not so trivial like a benzene one and then  here what I've shown you is a mushroom that is blue it's absolutely amazing but in  fact this mushroom makes an azuline derivative for some reason probably extremely interesting  chemistry usually when plants make interesting molecules it's because they're keeping bugs  off or keeping other things away because too bad for plants they can't move and if you're  stuck and bugs are crawling all over you you need chemical warfare to keep them at bay  usually then if you see a brightly colored mushroom like this you admire its beauty  and its beautiful blue color but let someone else figure out if it's safe to eat there  are people who do that for a living and it's very interesting to see how courageous they  are they usually eat a very tiny piece and then wait and see if what happens and then  they eat a little bit more and I'll let them do that but don't eat the mushrooms that you  find around that are this beautiful blue color or the ones with the orange gills either because  that might be the last thing you do.  For cycloactatetraene I've drawn it like a stop sign with four double bonds but it only  has eight pi electrons so that's not 4n plus 2 and in fact the system is not aromatic and  it's not planar it has a 3d structure and looks much more like alternating single and  double bonds that are localized so that one not surprisingly if it's not aromatic it doesn't  usually have a trivial name. Okay we're going to leave it there and for our very last lecture  what I want to do is go through where we started, where we got to and all the things we covered  because we've covered a lot of ground from electrons and photons to atoms to molecules  and we've done it in a certain semi-systematic way that I hope has made certain things that  you wondered about much clearer and maybe caused you to want to learn more about some  other things that you've never heard of so we'll leave it there and then sum up in the  next lecture.
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Right, let's pick up where we left off. We had just gone through Linus Pauling's brilliant exposition on hybrid orbitals, which are still used to this very day to rationalize a lot of things to do with structure and reactivity. But let's take a closer look, let's compare the hybrid orbital approach, which really says let's make pairs of electrons make the bonds with the molecular orbital approach, which says look, the way you have to do it is the way you did it with an atom. You have to put all the positive charges where they're supposed to be, and then you have to solve it. Then you have to get the solution, and then you can put in your electrons into the orbitals, and they go into the lowest energy once. This is a more delocalized approach to bonding because if you're including all the nuclei at once, then there's no reason why you should be drawing lines. That makes it a little harder to understand though when you're drawing structures and you're pushing the lines around, chemists love those lines because it allows them to write and rationalize reactions. But occasionally a reaction goes that doesn't seem to behave, and that could be an indication that maybe the theory of the lines is falling apart a little bit. Let's talk then in that vein about more delocalized bonding. In the MO picture, we don't make hybrid orbitals, so the orbital is used, but it's completely different. They must not confuse the two. In the molecular orbital picture, what we said we had to do was we could combine together atomic orbitals that had the same symmetry, similar energy, and overlapped in space. For tetrahedral molecule with a central atom like methane, we've got four carbon valence atomic orbitals, the S and the three P's. We have to know the structure. Why? Well, otherwise we'd have to move the things around and calculate the energy as a function of R and R2 and R3. We already know the structure. It's tetrahedral. What we want to understand is not the structure so much. We want to understand the bonding. How is it working? What's going on? We assume we know the structure. It's the same structure as Pauling's, but the description of the bonding is going to be different, and that's the thing that we're going to focus on. In the MO approach, you have to know the structure first. If you don't, you have to search through all possible structures. If you're doing that, that is a very, very, very long-winded and intensive calculation, although sometimes people do it just to see what might be the most stable structure if there's no experiment or it hasn't been made. Let's put our hydrogen atoms at the corners of the same cube that we had before, HA, HB, HC, and HD, and let's see what we're going to do with these four. The carbon is at the center, and with our four hydrogen orbitals and our four carbon orbitals, we get eight molecular orbitals because recall in the linear combination of atomic orbitals, rubric, we have the same number of orbitals at the end in terms of molecular orbitals as what we started with. Okay, now question is, which orbitals can we combine? The answer is rule two said they have to have the same symmetry for this molecule. What that means is that we cannot do what Pauling did because we cannot combine the carbon 2S and the carbon 2PZ or the other two P's because those have different symmetry. Therefore if we want to consider the 2S on carbon, we cannot combine it with any of the other atomic orbitals on carbon. We have to instead combine it with combinations of the hydrogen orbitals at the four positions of the cube. And that's completely different then because now we won't necessarily end up with four identical lines in terms of four localized bonds. We will end up with a tetrahedral structure, that's for sure, because we started with that, but our description of the bonding will be different. Now the game is this. We have the carbon orbital. Each of the four of them has different symmetry. We have to combine the four hydrogens so that they have the same symmetry as the carbon orbital. And that's pretty easy to do because if the carbon orbital is a big fat round S, then that means that these guys have to be as round as possible so they should just all be red and that should be one of them. And if the carbon orbital changes sign, like it's plus here and minus here, well we've got two hydrogens on the top of the cube, they should be plus. Two hydrogens on the bottom of the cube, they should be minus and so forth. And so there's a combination of hydrogens that goes perfectly with each of the four combination on the carbon and that's how we're going to do it. So for 2S we use 1SA plus 1SB plus 1SC plus 1SD. All of them as round as can be. And this then, if they're all red, the 2S is red and all these guys are red, this makes a gigantic orbital with no nodes that goes around all the nuclei and has roughly spherical symmetry and it's called A1 which is a symmetry label that lets the spectroscopists know what the symmetry of the molecular orbital is. And that name A1 has to do with its symmetry under the tetrahedral point group TD that you'll learn about later in the course. For now we're just going to treat A1 as a label that tells us which orbital we're talking about. Remember that less nodes is lower energy and so we've got this big red thing with this big fluffy thing sort of looks like a teddy bear with no nodes therefore A1 is the lowest symmetry. That's what we predict. Above it then are three degenerate because they're related by symmetry orbitals and there's one for each of the p orbitals on the carbon atom. And those three are called T2, T means three, a name that has to do again with their symmetry under the tetrahedral point group. The one and two have meanings but I don't want to go into the exact meaning now. These are the four bonding orbitals and they are not the same but there's four of them and they still predict an exactly tetrahedral structure not surprisingly because we started out with an exactly tetrahedral structure. The other four combinations have more nodes. I could take the S as red and I have to have something the same symmetry but I can pick them all blue. Now that's a disaster because now there's nodes in between the two nuclei so that one's no good and likewise I could pick the PZ with red on the top and blue on the bottom and I could perversely pick these two guys to be blue rather than red and then there would be destructive interference and again that would be a very high energy solution. Therefore we can easily see that there are four bonding orbitals so there's going to be four bonds, good, and there's four anti-bonding orbitals and since there's only eight electrons the four bonding orbitals are filled and the four anti-bonding orbitals are empty and that's almost always the way it works out when you do things correctly because not surprisingly nature finds its way into making the more stable configurations. Here then is an MO diagram. I have four, it's a little bit harder to draw the tennis tournament when you don't have just two players but I have four hydrogen 1s orbitals on one side and then I have the carbon orbitals and I have the 2s of the carbon lower than the 1s of the hydrogen, that's good and then I have the 2p higher. Well I'd have to know that but I can certainly figure that out by looking up the ionization energy of the atoms which is well known. And so I can order them like that and I have one very good combination, the teddy bear which is down at the bottom, the A1 and then I have the three that are identical, the T2 and they're at slightly higher energy. The key is there's no way that the A1 and the T2 have the same energy. They have to have different energies and that is an experimentally testable fact that we can look at. So we fill up the bottom with eight electrons, four from the hydrogen, four from the carbon, the same way we always do with molecular orbital diagrams. Once we calculate the diagram we fill it up from the bottom. And here is an actual calculation of these orbital contours from the excellent website of Dr. Stefan Imel. Here is the A1 symmetry between the carbon 1s. Remember that when we did lithium, when we drew those contours, there was hardly any overlap. What you're seeing here with the carbon 1s is exactly the same thing. The green contour of the carbon 1s is nowhere near the white protons that are sticking out like tinker toy spokes. And so the 1s is doing no bonding at all with the hydrogens. On the other hand, when we take the 2s, we get this big spherical look, it can't be exactly spherical because of the underlying tetrahedral shape, but we get this big green contour. And what's contoured here, the surface is drawn at the 90% probability that an electron in there is inside that region. And you can see that it goes around all the nuclei, the carbon and all four hydrogens. So these two electrons in this orbital have this whole big space to go around it. And it's a very good overlap with the 1s atomic orbitals of hydrogen. Then the next one up is the 2p and I've shown here one horizontal and the 2p, one side is, I apologize, I didn't pick the same colors that Dr. Imel picked, but let's say green is positive and blue is negative rather than red and blue. Same difference. Here there is a node, but the important thing is, is that the node is right at the carbon atom. So although there's a node, it is not in between the two atomic nuclei. What was bad with H2, with the anti-bonding, is that the node was between them so that there was no possibility of the electron being in there to glue the positive charges together. Here we've got one side building up density to hold these two hydrogens in and the other side building up density to hold these two hydrogens in and the node is right at the carbon where it doesn't do any harm because it's not between the bonds. If we look at the other two, they're exactly the same as this one, but they're just flipped. They're all degenerate. They have exactly the same energy because of symmetry. So there's another one that's in and out. Again, it has a node right at the carbon atom and then the final one, the third one, is just up and down. It's nice that he's drawn them with these different perspectives because as you draw them slightly differently and look at them, you get a much better idea of what these surfaces actually look like than if you just have one perspective. These are the three then that we said had T2 symmetry and these are the three other bonding orbitals. They have a node but it's not a bad one to make an anti-bonding orbital. Question is then, on one hand we have the approach where we first monkey with the carbon orbitals and we keep the hydrogens and then we make a bond to each hydrogen in turn. That's nice because that's how we might draw it on a piece of paper if we were in organic chemistry for example or just regular Lewis structure. On the other, we have the MO approach and we can calculate the energy of these orbitals so I just drew them qualitatively but we can calculate them and under certain approximations and we can make the approximation better and better if we want to do more work. We don't need to do any more work to say look, in the MO there are two different kinds of orbitals. There's one that's T2 and there's a lower one that's A1 and therefore what we do is we go to somebody who does photoelectron spectroscopy and we say can you leak in some natural gas and ionize it and tell us what the binding energy of the electrons are in this region and how many orbitals there are and if we do this experiment there is a difference because if there are four equivalent hybrid orbitals each with two electrons then they're all the same. We get one band in the photoelectron spectrum, pick an electron out of any of those and boom and you never ionize again. It's so unlikely that you ionize period that you don't ionize twice so we don't have to worry about that and then the carbon core electrons which are the only ones left are much, much higher energy so we don't have to worry about them either because we can pick the photon so that they aren't going to come up. We are going to put in x-rays for example and therefore if there's one band in the photoelectron spectrum, Pauling is correct and MO theory is washed up and if there's two bands in the photoelectron spectrum then MO theory even though it's a little more complicated and we don't have to stick like lines is more correct. And let's then have a look and this experiment just like the paramagnetism of O2 is definitive and of course it comes down in favor of MO theory otherwise we wouldn't have gone to all this trouble. Here on slide 666 is the photoelectron spectrum of methane and there are two bands and they are assigned to T2 and A1 and they even have about the right ratio of size. You can see that one looks like it's maybe three times bigger than the other. There are a lot of factors that go into the cross sections of photo ejection so you can't just say if there's three orbitals you get three times the area or anything as simple as that but nevertheless it does seem to make some sense. And if we look at it the vibrational progression because methane is a much more complex molecule and can really vibrate in a lot of different ways which is of course one reason why it's a very bad greenhouse gas because you put it up in the atmosphere and it can absorb infrared in a lot of different ways and then radiate it back down to earth. We can't resolve all the little things like we could in N2. They just appear as kind of a ragged like a porcupine envelope but nevertheless we can say it's the T2 really that's doing the lion's share of the bonding in methane and the A1. The big teddy bear is rather less important because of the less vibrational progression when we eject an electron from it. This then says look we have to prefer the MO theory to the theory of localized bonds even clever localized bonds with hybrid orbitals because now we have an experiment that's clearly indicating that one approach is better and it disagrees with the other one and that's what science is all about. It's easy to talk but you can talk all you want if somebody does an experiment and shows the ocean is rising then it's rising and it doesn't matter whether you want to say it's not or if it's measured you can quibble with the measurement but you first have to understand how the measurement works and in fact the measurements along those lines are extremely reliable even though the system itself is quite complicated. Alright let's talk about some bigger molecules. We don't want to just leave off with methane although to a physical chemist methane is a big molecule in a lot of ways and I think you can see why because they may want to know a lot of things about it that an organic chemist may not be interested in and an engineer just wants to burn it and get some power. So we can look at some conjugated hydrocarbons and so-called aromatic systems. I've never smelled an aromatic molecule that was aromatic to me like a rose but they're called aromatic and they certainly do have smells but they usually smell like gas or mothballs or something like that. So let's look at benzene, naphthalene and a few others butadiene so-called conjugated hydrocarbons and this will be kind of an interesting, another interesting way to simplify what could be an extremely complex if we did it from scratch it would just be terrible calculation and make it simple and leave enough meat on the bone that we can still come to some interesting conclusions about the stability of the system. First let's just consider ethylene, C2H4. Ethylene is an extremely important molecule in the fine chemical industry. It's the feedstock for a lot of things and a lot of people are spending time and energy to figure out how to make ethylene efficiently when we run out of other ways that we have been making it. Well we've got the 2S orbitals on the two carbons and the two P orbitals and we've got the 4 1S orbitals on the hydrogens so there are going to be 12 molecular orbitals and because there's 12 atomic orbitals and 12 is a lot but we can still figure out what they should look like. Just like methane, even though this is bigger we're going to have to start by putting the nuclei at their preferred position, 120 degree bond angles between the hydrogen and the carbon and then we're going to have to solve for what the molecular orbital energy should be by putting in electrons that are allowed to go over the entire nuclear framework and only if they don't want to go over here for their own reasons are they not going to. It's not that we're capturing them between two nuclei. Here then on 668 is the molecular orbital diagram for ethylene. We've got our four hydrogens again but now we've got two carbons and we've got all these states labeled from the bottom, sigma G, 2AG with symmetry labels, 2B, 1U and so forth and so on. We don't have to understand what all of them mean but many of them are overlapped between the S orbitals on carbon and the S orbitals on the protons and then there's two right in the middle there marked pi G and pi U that are just coming from the carbon and that's because those two are orthogonal to the plane in which the protons are so those two have different symmetry than what the S can be and therefore they only come from the carbon and they can either be this way, the two carbons or that way and of course only one of them is filled because there's one double bond in ethylene. There's a sigma bond and then there's a pi bond and so one of those is filled right up to where it's stable. They're slightly below the 2P level. We slot in all our electrons all the way up. What we find is that it's filled right up to that level that involves just the P orbitals of carbon and here is the photoelectron spectrum of ethylene. We can also get that and now the ethylene is quite complicated so these peaks that all have vibrational progressions are very, very hard to resolve and they may not be resolvable at all. It depends how carefully and how clever the experimenter is but at least they aren't so wide that they overlap with each other and we can't see them at all and they've all been assigned as you can see 3B, 3U and 3AG and so forth and they're all assigned and they all correspond to ejecting electrons from lower and lower in the molecular orbital diagram all the way to 2AG after that. Maybe you need too much energy and they didn't have such energetic photons. The symmetry labels again treat them just as labels for now. Don't dwell on them too much. Once you study group theory you'll know exactly what they mean. For us though it's the highest occupied molecular orbital, the so-called HOMO and the lowest occupied, excuse me, the lowest unoccupied molecular orbital, the LUMO that are going to be the most important thing. Why? Because if I'm going to be making a bond then it's the electrons right at the top of the cake that are going to fly off and go somewhere and make the bond. The ones that are held down further down in energy are not going to be likely to be the ones going first and likewise if some other atom is coming up, rimming with electrons and says hey take some electrons from me, where are they going to go? Well they're going to go into the lowest unoccupied molecular orbital because they're going to come in and they're going to go down to the most stable state and therefore the highest occupied molecular orbital, what it is and the lowest unoccupied molecular orbital, what it looks like are very, very, very important to understanding chemical reactions and therefore we can focus mostly on them and forget about most of the other ones and that's a key simplification. The highest occupied one I've labeled as 1B3U and the lowest unoccupied is 1B2G and they both involve the carbon P orbitals and so these two, the highest occupied and the lowest unoccupied only involve pi electrons, it's only the pi electrons that are involved and that is a key simplification when you consider the reactivity and structure of things with double bonds like ethylene. We can treat all the sigma orbitals as just like core electrons and since they aren't important, they aren't going to be participating, we can rationalize whatever structure they have in any way we want and usually the prescription is those aren't so important, we just say they're sp2 hybrids, it doesn't matter what we call them because they're never going to do anything, they're just going to be filled and we might as well take a simple approach like Pauling's approach and just say well they're making 120 degree and there's a sigma framework that's holding the atoms in place and then there's these delicate pi orbitals and for those we reserve the molecular orbital treatment because those are going to do something and so we want to treat them more accurately and because there's only two orbitals, this and like that, it's basically down to the same thing as back to H2. The fact that they're P rather than S doesn't matter, the math is exactly the same and we've already done it so we can leverage that now and just say we've got this perpendicular pi system with two P orbitals and they can either be in phase or out of phase and we can calculate what we want to do so it's just like H2 in this case. Our pi wave function then while we suppose the molecular orbital for the pi is C1, 2PZ1 plus C2, 2PZ2 where 1 and 2 are referring to the two carbons and C1 and C2 are the two coefficients yet to be determined but we know they're going to come out to be equal because the carbons are the same bisymmetry. We have to make a secular determinant because this is molecular orbital, we've got to calculate the orbital energies, remember we have the Coulomb integral and the exchange integral and so here it is H11 minus ES11, we have the overlap integral H12 minus ES12, we have the exchange integral and so forth. The determinant so this times this is equal to zero and since the two carbon atoms are equivalent then H11, that integral and H22, that integral are exactly the same so we can make those the same and H12 is already the same. If we wanted to make quantitative progress here and we wanted to proceed the way we have in a lot of other cases helium and hydride, we'd have to get out our integrals, our Thetas and sine Thetas and all that and go to work but we don't want to do that, none of that's too easy to do and we're only after some qualitative description of what's going on, we don't expect it to be quantitative, it'd be very hard to make it quantitative even if we calculate these pi orbitals so well by just leaving out everything else and saying forget it about the other stuff, we could be making some big mistakes, some big errors there. Let's simplify things and not do any integrals at all and that will be a very, very good method and that's called the Huckel approximation. In integral whatever it is is just a number by the time you integrate it, it's a number with units but it's a number and we're going to assume that we're going to just replace integrals with these symbols, numbers, alpha and beta and we're going to calculate things very quickly then. The Huckel approximation makes three assumptions and at first they seem laughable but in fact they're very nicely, physically motivated. The first thing is the overlap integral, remember how much trouble that was to calculate s and how we had to do all those things. Well here what we're going to say is this, we're going to say that the overlap integral is zero unless you're talking about the overlap to the same place. So the overlap even between two neighbors is zero and that seems very counterintuitive why because you're claiming they're making a bond but when you look at what came out of our analysis before s was just a player in the denominator that just changed things slightly in the denominator. It never dictated whether a bond formed or not. It was okay so it doesn't really matter if we set it to zero and if we set it to zero then we're getting rid of a lot of math because all those things go away because they're multiplied by s12. So those are gone and then when it's the same we set it to one because we're saying it's normalized and then all the Coulomb integrals, the H integrals are the same for equivalent carbons and sometimes we assume they're the same even if the carbons aren't quite equivalent because we're just too lazy to figure out if they're that much different or it's difficult to figure out what they would be. And then the exchange or they're also called resonance integrals in the business. Those integrals vanish except for nearest neighbors and the rationale for that is that if I'm here and you've got a p orbital then that integral is going to have some value but once I get out to here and I've got something intervening in between it's too far away. So I could calculate it but it'd be small and I don't want to waste all my time calculating something that is 0.1% of the answer but is extremely difficult to calculate. That's a poor use of my time. So those three approximations let us simplify our secular determinant and get rid of all the calculations. Conventionally you write alpha for one of the integrals in beta for the exchange and our secular determinant just becomes alpha minus e beta beta alpha minus e equals to 0. And if I, that's a quadratic equation and the solution for the energies is alpha plus or minus beta. Not surprisingly it's very similar to what we had with H2. There's a good combination and a bad combination. Now in fact there's two electrons going in and we didn't calculate any electron, electron repulsion or anything like that. So all we're going to do is double the energy when we say what it actually is. The advantage of this however is we can do something that's a little bit more intimidating, much more intimidating than helium for example. Let's try benzene. Benzene is C6H6. We would have a 6 by 6 determinant and it would have entries everywhere. So just expanding that out as this times the determinant of 5 by 5 will take forever. And that's going to be a major problem to try to calculate something like that. So let's figure out how to get our 6 PZ orbitals in the pi system to yield 6 MOs without doing that. Let's set up the determinant then which I've done here on slide 676 in the Huckel approximation. The nice thing is except for terms right near the diagonal in the array each of which has beta when they're off because they all have the same value. And the 6 carbons all have the same integrals for the H11, H22 and so forth. So those are all alpha. And then I have my energy, the overlap is gone so there's no energy off the diagonal, that's very convenient. Everything's zero everywhere except for the top and bottom and that's because as I go around the structure 6 is next to 1. So you always end up with 1 up there. But all the rest are zero so most of the thing goes away. And we can even make it simpler by since we've got zero and we know beta's not zero we can divide everything by beta so that the certain things are 1 if they're off the diagonal. And then we can just redefine alpha minus E divided by beta to be some variable. So let's let X equals to alpha minus E divided by beta. And then all the other terms are 1 then we end up with the following polynomial to root. If you expand out the secular determinant you end up with a sixth order polynomial. It's X to the 6 minus 6X to the 4th plus 9X squared minus 4 equals zero. So that's why you took a course in algebra in high school because now this should be a duck soup problem. In fact it may not be so easy if you just stare at it but there's a trick. I'm going to let you explore the trick on the homework. Because it's only even powers we can redefine Y is equal to X squared and then we've only got a cubic. And if we're lucky we might be able to factorize a cubic and figure out how to actually get the answer. Or we can plot it even for lazy in Mathematica and we can see where it crosses zero and if it crosses zero at some convenient place that is easy to tell because it's an integer then we can try factorizing it with X minus that integer X naught or N naught and see if it factorizes. And if you do that and you're persistent you find that you have repeated roots because of the Y equals X squared and then some of them are repeated again just because of the structure of what you end up with at the end in the cubic and you end up with X is equal to plus or minus one or plus or minus one or plus or minus two. Those are the six roots of that equation. Well X was alpha minus E over beta and therefore there are the corresponding energies E1 which is the best one alpha plus two beta recall that these are usually negative. E2 is equal to E3 is alpha plus beta. E4 is equal to E5 that's alpha minus beta and then E6 is alpha minus two beta. So there's nice symmetry about how the energies come out. Now how do we get what the molecular orbitals are? All what we had to do before we go back we actually put in the energy to the original equation that we had and then we find the orbital that we get that corresponds to that energy. The degenerate ones that have E2 and E3 the same and E4 and E5 the same require a little bit of thought to how to sort those two out. But whatever orbitals we get recall that they have to be orthonormal. The molecular orbitals have to be orthonormal as well. So that's an important thing that lets you simplify it. But I'll let you do that and here I've given you the answer. The first one not surprisingly the lowest energy molecular orbital for benzene is when all the six recall benzene as a hexagon planar is when all the six p orbitals have the same phase. So I get a ring of positive electron density around the top holding all the carbons together and I get a ring around the bottom of the opposite sign but when I square it I find I build up density in between the carbons even though they're off axis and that holds them all together. And then the next two have some nodes but the nodes aren't bad. So those are between at certain carbons and the orbitals are just like methane they hold the other parts together. So there's one that has a node this way that holds these parts together and there's one the other way that holds these parts together. And then there's the bad three that have the opposite combinations basically and the highest one is when this one's up and this one's down and this one's up and this one's down and so forth and that has nodes in between everything and tons of nodes and you can see it's a very crinkled wave function that's going to have very high energy and that's the least favorable one. That's a strongly anti-bonding orbital. And in fact by drawing simple figures just based on the cyclic structure you can arrive at some simple rules to predict aromatic stability. You draw the structure point down and so for hexagon you draw point down and then at each vertex you put a line which is an MO and you have one on the bottom then you have two that's E2 and E3 then you have these two that's E4 and E5 and then you have one at the top that's E6 and you've only got six electrons because you only had one electron in each p orbital to start with so you fill up the bottom to orbital with two this one with two this one with two game over. So you pick the three good ones the three winners and you left all the anti-bonding ones unoccupied and all the electrons are paired and all these orbitals are delocalized so we don't predict any difference although in the Lewis structure of course for benzene we draw a single bond and a double bond because we have to because of the Lewis structure and then we say well there's resonance we move this here and move them around like a mouse trap and the molecular orbital we don't have to do that they're already set up to be delocalized. On the other hand if we pick cyclobutadiene and we stand a square on its points then we have a lowest one E1 then we have E2 and E3 and then we have E4 but we have four carbons in cyclobutadiene and therefore we've only got four electrons so we put two in the bottom but then we've got two equal ones here well that's like O2 again so one goes here and one goes there. Well that's not aromatic stability that's predicting a di-radical I wouldn't take this kind of very qualitative thing too seriously I do some spectroscopy to figure out what you actually get but whenever it comes out like that where the two aren't paired and there's some left the conclusion is it's not aromatic and what happens then is the ones that have 4n plus 2 pi electrons become stable and the ones with 4n like 4 rather than 6 are not especially stable and not stable at all in some cases so only the 4n plus 2 pi systems are predicted to have aromatic stability and that comes out of a more detailed calculation to not just playing games with shapes but it's interesting to just play games with shapes because it's very quick and often it's good enough. Alright let's do a practice problem let's consider the following molecules and let's consider them and see which one is an organic chemist would predict to be aromatic so they had the special stability benzene was used as a solvent for a long time why because it doesn't react with anything so it's a perfect solvent if things won't dissolve in water or the reaction slow in water you dissolve it in benzene and the reaction goes very quickly problem was of course benzene is quite a potent carcinogen like a lot of these systems and so if you're breathing a lot of it and boiling it up and breathing it on a daily basis it's not the greatest and that's why they went away from it. Let's consider then benzene we know is aromatic let's consider naphthalene which is in moth balls that's that smell azuline which you probably haven't seen nor smelled and then cycloactatetraene let's take these three and see what they are well first thing is the names if you don't know what the structures are referring to the names are useless and we have to know they aren't going to be aromatic unless in the Lewis structure they would have alternating single and double bonds and we could do resonance and move the things around so that they would all be equivalent half the time single half the time double if they aren't even like that then forget it it's not going to be an aromatic system or the part that can't do that trick with resonance is not going to be an aromatic system and they have to have a cyclic structure as well so that the snake bites its tail because that's important in how those two last terms come in if those don't come in a lot of things change it turns out and then let's draw the structures and count them so for naphthalene we've got two benzene rings fused together now if we were actually going to do the MO treatment we'd have to be careful because these two carbons without any hydrogens on could have a different coulomb integral than the others and the others would all be the same and so we should take that into account we should have alpha 1 and alpha 2 for lazy we just say well they're about the same column alpha and I've drawn the Lewis structure here one of them and we could play a game with resonance then and draw one the other way we count the electrons in the pi system we forget about the sigma we forget about the hydrogen and we see there's 1 2 3 4 5 6 7 8 9 10 that's 4n plus 2 for n equals 2 and so it's planar and it is in fact aromatic for azuline which is a structural isomer of naphthalene there's a 7 membered ring glued to a 5 membered ring and I can still draw the double bonds the same way it's kind of amazing that it works like that and it's a beautiful blue color and again I count up it's a cyclic system I count up there's 10 pi electrons and the delocalized MO's go over the whole thing and it's planar and it's very nice example of an aromatic system that's not so trivial like a benzene one and then here what I've shown you is a mushroom that is blue it's absolutely amazing but in fact this mushroom makes an azuline derivative for some reason probably extremely interesting chemistry usually when plants make interesting molecules it's because they're keeping bugs off or keeping other things away because too bad for plants they can't move and if you're stuck and bugs are crawling all over you you need chemical warfare to keep them at bay usually then if you see a brightly colored mushroom like this you admire its beauty and its beautiful blue color but let someone else figure out if it's safe to eat there are people who do that for a living and it's very interesting to see how courageous they are they usually eat a very tiny piece and then wait and see if what happens and then they eat a little bit more and I'll let them do that but don't eat the mushrooms that you find around that are this beautiful blue color or the ones with the orange gills either because that might be the last thing you do. For cycloactatetraene I've drawn it like a stop sign with four double bonds but it only has eight pi electrons so that's not 4n plus 2 and in fact the system is not aromatic and it's not planar it has a 3d structure and looks much more like alternating single and double bonds that are localized so that one not surprisingly if it's not aromatic it doesn't usually have a trivial name. Okay we're going to leave it there and for our very last lecture what I want to do is go through where we started, where we got to and all the things we covered because we've covered a lot of ground from electrons and photons to atoms to molecules and we've done it in a certain semi-systematic way that I hope has made certain things that you wondered about much clearer and maybe caused you to want to learn more about some other things that you've never heard of so we'll leave it there and then sum up in the next lecture.
an atom. You have to put all the positive charges Pauling's, but the description of the bonding is with combinations of the hydrogen orbitals at the all. It depends how carefully and how clever the 0. And if I, that's a quadratic equation and the although in the Lewis structure of course for di-radical I wouldn't take this kind of very photons to atoms to molecules and we've done it