CHEM 125a: Freshman Organic Chemistry I
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Freshman Organic Chemistry I
CHEM 125a - Lecture 15 - Chemical Reactivity: SOMO, HOMO, and LUMO
Chapter 1. Introduction: “Pathological” Bonding in the BH3 [00:00:00]
Professor Michael McBride: Okay, so now we have a number of different perspectives we can take on understanding what holds molecules together; that is, on bonding. We can look at the real molecule, for example, with X-ray. Or we can do a computer calculation, and if we do a careful enough computer calculation, we can get something that we believe is pretty close to reality, like the total electron density I showed you. Right? And the goal of the computer is to approximate the Schrödinger equation. You can’t do it absolutely, but you can, depending on how much computer power you want to throw at it, you can do it close enough for most purposes. Okay. But you’re neither the molecule — well you are a molecule — but you’re neither the kind of molecule we’re studying, nor are you a computer. You’re a chemist, and your goal is to understand things; let the computer do the heavy lifting mathematically and you understand it. Okay, so let’s see if we can understand some of the things that we didn’t understand when we were looking at the molecule really, with X-ray diffraction. Right? When we were looking, we looked at this molecule and said that the bonding, from the Lewis point of view, was pathological. And you remember how we saw that. We looked at a cross-section through those three atoms, and it looked like that in the electron difference density. And you remember what’s funny about it? What’s pathological? Dana?
Student: The bonds appear to be bent.
Professor Michael McBride: Some of the bonds are bent, and what else?
Student: And there’s another bond that’s missing.
Professor Michael McBride: And one bond is missing. So that bond isn’t there in the electron density, the difference density. And those are bent. They don’t lie on the line between the nuclei. Now, can we understand that, from the point of view of molecular orbitals, of Schrödinger’s equation; why it should look like that in this particular case? So would a computer’s molecular orbitals provide understanding? So I was just showing you, as the class assembled, how you get these molecular orbitals, and I was showing you some of them. There’s the framework of the molecule superimposed on this map, and that triangle on top there, on the right, is what we’re looking at. And so we could look at the thirty-second of the thirty-three occupied orbitals, the next to the highest occupied orbital, and it looks like that. That’s the one we were just looking at, before we went on to the PowerPoint. And you see there is electron density in those — the wavefunction has big value in the region where those bonds are bent. But there are thirty-three of these occupied orbitals, and they put electron density every which way, and it’s very difficult to understand why this is the way it is, why it’s bent. Even if you grant — which is not easy to grant — that this is the proper orbital to look at, and that indeed the electron density of the red and blue, big red and blue lobes, are not on the line between; we don’t know why they are. Right? And there’s so much other electron density in the molecule, with all complicated nodes, that it’s very difficult to know where to look. Right? So the answer is no, it doesn’t provide understanding. It gives far too complicated an answer. It doesn’t tell you why. Even if you believe that it gives you the right numbers, it doesn’t tell you why. Okay? But analysis in terms of pairwise — that is, don’t go over the whole molecule, just look at two atoms adjacent to one another, making a bond, so just a very local view — pairwise bonding overlap of hybrid atomic orbitals explains these two pathologies. So let’s look.
First, I will assert that the best overlap you can get between hybrid orbitals on these carbons — if the nuclei are arranged in an equilateral triangle, so 60º angles — the best hybrids you can get point at the angles shown by the red arrows here. They do it in order to get the best overlap. Now this seems funny, because the angle between the red arrows on the right carbon is greater than 90º. Why do I say it’s funny that it’s greater than — you’re not going to be able to make it 60º because two orbitals that made a 60º angle would not be independent orbitals. Obviously if they were right on top of one another, they’d be the same orbital. But you can’t say they’re not the same orbital if they’re just a little bit away, because still they’re mostly the same orbital. It’s only once they get to 90º that they can be different. What would the orbitals be that would be independent of one another at 90º?
Students: p orbitals.
Professor Michael McBride: p orbitals, right? So 90º is the smallest angle you could get. Why don’t you use 90º? Why do you use greater than 90º? Okay, that’s the question. Why not use p orbitals and make the smallest angle you can? Does anybody see why that is? Okay here, there’s a p orbital that points at 90º. You see its axis, the yellow axis, is a little bit inside the red arrow, and the blue axis, which would be another p orbital, perpendicular to that one, is a little inside the other red arrow; they make an angle of 90º. Got it? Now why don’t the atoms use p orbitals to make the smallest possible angle? Can anybody see why that won’t give the best overlap? How could you change the orbital so as to get better overlap?
[Students speak over one another]
Professor Michael McBride: Claire?
Student: The best overlap is between an s and a p orbital. So if it’s two p orbitals, you don’t have the best overlap.
Professor Michael McBride: Remember, that depends on the distance. And they’re very close to one another. And it turns out it might be that at this angle — if you used the s on one of the carbons, and the p on another — it might be that you would get better overlap. I don’t know that it’s true, but it’s not implausible. But what would be wrong about using this p orbital from that carbon, and on the top one of the triangle using an s, which might give better overlap? Can you see what’s bad about that? If you use all the s of that top carbon there, you can’t use any of it to make any of the other bonds from that carbon. Right? All you have left over on that one is p’s, which give crummy overlap. So even though you might be able make this single overlap better, you couldn’t make all the bonds strong from that top carbon. So you wouldn’t get the best overall overlap. So that’s a good suggestion. But there’s another way to do it, which is to use a hybrid orbital. Now here we’ve put in a little bit less than 20% of s, to give the ratio sp4.1. Right? But notice that it really distorted; it made it much larger in the region where we’re interested in it overlapping. Right? In the process, the axis moved out, to close to where the red arrow is. Right? But even though the bonds didn’t point — let’s see, like this — even though the bonds didn’t point at one another, by extending they overlap more. Got the idea? So the hybridization, by shifting it to one side, gives better overlap, even though the angle might not be quite as good. So you have bent bonds, paradoxically enough, in order to get the best overlap. Isn’t that funny? You’d think the best overlap would be when they’d point at one another. But you get it better when they make an angle because they stretch out.
Okay, so that explains the bent bonds. Now how about the missing bond? Well again we can see that in hybridization. Because what’s left over, after you make those sp4.1’s, three of them, pointing up from the bottom carbon, after you make three sp4.1’s, what you have left over is an sp4.1’s. And it points down, not up. If it pointed up, if the big blue lobe were up and the red down, then it would be part of those other three orbitals. The only way you can make it point up is to make the others point down. Right? And then they would have crummy overlap. So there’s very poor overlap between the tiny lobes of these two. So there’s not enough overlap to get a bond. You don’t stabilize the electrons because you don’t have overlap. Right? And rehybridizing this one, to strengthen this particular bond, by making it point up, would drastically weaken all the others, making them point down. Right? So the best overall overlap, for all the pairs that are involved in these bonds, is to have one bond not there, or to have very, very poor overlap so that the other three are very good. Sam?
Student: So when you hybridized the other two and you get the sp4.1’s, shouldn’t — and I understand that shifts the electrons towards the other atom, but how does that shifting distort the angle?
Professor Michael McBride: Remember the angle between hybrids depends on m and n; spm and spn, we did that last time. If it’s pure p, it’s 90º. If it’s sp3,-sp3, it’s 109.5. If it’s sp2,-sp2it’s 120; and so on. If it’s sp4.1, it makes that certain angle that’s shown there bigger than 90º. Okay? And the only reason for that is so that these hybrids will be different from one another; they won’t be the same. Right? They won’t overlap each other on the same atom, because to the extent they do, they’re the same, and so they can’t at all. They have to be orthogonal, if they’re on the same atom.
Chapter 2. Viewing BH3 via Electron Density and Molecular Orbitals [00:10:47]
Okay, now three views of BH3. First nature; the total electron density. I’m going over the same thing again and again here, with different examples. First look at total electron density; nature. Then we’ll look at molecular orbitals; that’s what a computer does. And then we’ll look at bonds from hybrid atomic orbitals; and that’s what we do, as chemists.
Okay, first we’ll look at the electron cloud of BH3, calculated here, using that same program I just showed you. So we can look at the successive layers of the onion, the high electron density, lower, lower, lower, the various contours. So this lowest one is mostly 1s of boron, but a little bit of the 1s’s of hydrogen too. And one of them doesn’t even show up, because it’s so small. All the hydrogens should be the same, but the graphing capability of the computer, for something that has as few points as this, the way it draws surfaces, doesn’t draw them smooth. Right? It connects points, so it doesn’t even show one of those. Nobody would ever make such a plot. Okay? That’s 0.3 electrons per cubic angstrom. So suppose we go to 0.15 electrons per cubic angstrom, half that density. Now we’re beginning to get smoother surfaces. And it’s on the boron, it’s 1s of the boron, and a little bit of the hydrogens as well. Right? And now we’re going to lower the electron density, 0.05. And now there’s an interesting feature you can see here. You see that dimple? Why is that? Why is there not as much electron density, at this level, around the boron? We’re out in the valence-shell now; not 1s anymore, because the 1s is really down close, high density. It’s because the hydrogen atoms are pulling the electron density away from the valence atoms [correction: orbitals] of Boron.
Often you talk about this in terms of electronegativity. But why is something electronegative? It’s because of what the effective nuclear charge is. The effective nuclear charge of hydrogen, for the valence level, is one. Right? But for boron it should be higher. Boron is the fifth element. But it’s got the 1s electrons shielding the nucleus. So it’s not as high an effective nuclear charge as hydrogen does. So when you put the orbitals together, hydrogen is lower than boron, and the electrons, in those bonds, are mostly on the hydrogen. So you see that dimple in the middle, because there’s not much valence electron density on the boron.
Now we go to 0.02 electrons per cubic angstrom; and finally to 0.002 electrons per cubic angstrom, which is defined as (for purposes of drawing what the van der Waals surface is, that is) how close molecules tend to get to one another at normal pressure in a liquid or in a crystal. Right? And that’s, it’s just found empirically, just by testing and measuring things, that if you draw a surface at 0.002 electrons per cubic Angstrom, you have about how close things would come to one another when you bump these surfaces into one another. Okay, so that’s the van der Waals surface; that’s a definition, right? And we showed once before electrostatic potential, where you show that same surface but color it, to show how the electrons are distributed, and the protons in the molecule. So where would a positive charge on the surface be happy? Where it’s negative. That’s red. And where would it be unhappy, the positive charge, because the molecule is positive there? That’s the blue. So you can see what it is. It’s red where the hydrogens are, because they’re pulling the electrons, and it’s blue where the boron is, because it’s giving up electrons. Okay, so there’s what the real molecule looks like.
Of course I didn’t do it by an experimental measurement from BH3, because you can’t crystallize BH3. I did it by a computer calculation. But I believe that what it gives is what you would see in an experiment, if you could do the experiment. Okay, so that’s the total electron density. Now let’s look at how the computer breaks that up into molecular orbitals in a calculation like this. So it looks at — it makes symmetrical molecular orbitals, like Chladni things, with nodes; we’ve seen lots of those. So here’s the framework, BH3, and we’re going to look at the molecular orbitals the computer calculates. And we’re going to make those molecular orbitals by adding up atomic orbitals, linear combination of atomic orbitals.
And which atomic orbitals will we use? The best would be if we used every atomic orbital. Then we could get perfect flexibility, but the calculation would take too long. So we use just a minimal number of atomic orbitals; namely the 1s orbital on boron, and also the valence orbitals, 2s, 2px, 2py, 2pz, and also the 1s orbitals of hydrogen. So there are one, two, three, four, five, six, seven, eight, there are eight atomic orbitals we’re going to use, to make up our molecular orbitals. How many independent molecular orbitals can we make, if we start with eight atomic orbitals? We can make eight molecular orbitals, the same number you started with. If you tried to make more, they wouldn’t be independent of one another. But you can make — from eight, you make eight. Okay, so these are going to be the energies that’ll turn out of the eight molecular orbitals that we’re going to get, that the computer is going to get, starting from that. If you added more atomic orbitals, if you let it use 3s, 3p, 3d, and so on, then you’d get more molecular orbitals, but they’d be higher in energy, and less relevant. Okay? In fact, how many of these are relevant? How do we know how many we care about, the energy of? Russell?
Student: The total number of electrons.
Professor Michael McBride: How many electrons. We can put two electrons into each. Once we’ve got it filled up, we don’t really care what other energies there could be; at least not for this purpose. Okay, well, it turns out that boron has five electrons, hydrogen has three; eight altogether. So there’ll be eight electrons, four pairs, and that means four occupied orbitals. We don’t care so much about the others, right? So those are the Occupied Molecular Orbitals, OMO. Okay? And then there’ll be, also from this set of atomic orbitals, there’ll be four Unoccupied Molecular Orbitals, UMOs. Okay, now those two are called — nobody talks about OMO and UMO, but they talk about HOMO — that’s very important and we’ll be talking about that — that’s the Highest Occupied Molecular Orbital. And also they’re very interested in the Lowest Unoccupied Molecular Orbital; and you’ll see why very shortly. So the LUMO and the HOMO are very important. In this particular case it turns out there are two HOMOs that have the same energy. Not surprisingly, because it’s such a symmetrical molecule. Okay now, what do you think the very lowest orbital looks like? Katelyn, you’re our expert. What?
Professor Michael McBride: It’s the 1s orbital of boron. So there it is. No surprise there. Not to diminish your contribution, but it’s not surprising. Okay, so there’s 1s. That’s the boron core. Now what’s going to come next do you think? After we’ve got the 1s molecular orbital, what’ll be next?
Professor Michael McBride: 2s. But it’s a molecular orbital, it’ll be made up of everything. So here it is, right? And it’s all the hydrogens and also the boron 2s are contributing to that. Again there’s a dimple in the middle, more of the hydrogens than of the boron. Why? Because the hydrogen 1s is lower than the boron 2s. The boron 1s was lower; that was the first one, right? Okay. And then, so there’s that radial node, the spherical sort of node down the middle, showing it’s a 2s. Right? And then there’s a 2px molecular orbital, and a 2py molecular orbital. And that’s four, and that’s all eight electrons. So that’s what we care about. They’re the occupied molecular orbitals of BH3. Now how about the vacant orbitals? The lowest unoccupied orbital is the last 2p, the 2pz, that’s pointing in and out of the plane of the molecule. And then we’re going to have 3s, which you see has that node in the middle; so two spherical sort of nodes. And then a dx^2-y^2, and finally dxy. Right? So that’s the molecular orbitals that the computer generates for this thing, using what’s called a “minimal basis set”; that is, the lowest number of orbitals you can use to get — that is, through the valence level; you don’t put the higher ones in.
Chapter 3. Assessing Reactivity through HOMO-LUMO Interactions [00:20:26]
Okay, now that’s how the computer does it. Right? And you can get good numbers out of the computer, like the energy of the molecule, or the shape of the molecule, where the nuclei would like to be to minimize the energy. That’s fine, but for understanding it’s not so good. What we want to do, as chemists to understand it, is to partition the total electron density, not into these molecular orbitals, but into atom-pair bonds, and anti-bonds, as you’ll see, and lone pairs; the same stuff Lewis talked about. Okay. That is, we usually do this, but sometimes it doesn’t work. And when doesn’t it work? When it doesn’t work, we must use more sophisticated orbitals, where we say there is a certain phenomenon going on. You know what that phenomenon is? When we can’t use localized bonds?
[Students speak over one another]
Professor Michael McBride: That’s what resonance is, when this approach doesn’t work. Okay, remember that’s where Lewis Theory had to get sort of complicated, Baroque. Okay, so there’s that 2pz orbital, a vacant orbital. Right? Here’s the boron core. But now we’re doing it. We’re happy to use a vacant orbital. That’s part of Lewis’s idea. So 2pz we’ll share with the computer; both of us see that as a lowest vacant orbital. The boron core we understand quite well. Okay? But how are we going to treat those other three pairs of electrons, in the picture Lewis would draw? Where would Lewis put those three pairs of dots? Angela? Can’t hear.
Student: He’d make them the bonds between —
Professor Michael McBride: Those would be the bonds, between B and H. Okay? So what will those look like in our crummy picture? Right? We’re not as good, at least I’m not as good as the computer at drawing these things. But what I’d say, it’s going to be an orbital that’s big on hydrogen, smaller on boron; some hybrid sp2 hybrid of boron is going to overlap with the 1s orbital of hydrogen. It’s going to be bigger on hydrogen than on boron. Why? Why should the lowest, the bonding orbital, be bigger on hydrogen than on boron? Lucas?
Student: The 1s is more shielding.
Professor Michael McBride: Yes. That’s another — that that orbital, the hydrogen orbital, is lower in energy than the boron it’s mixing with. Right? So it’s uneven mixing. Right? Poor energy match. So the one that went down looks mostly like the hydrogen, and the one that went up looks mostly like the boron. So we see one — in fact, we see three of these, one pointing toward each hydrogen. Right? From our point of view they all have the same energy, they’re all B-H bonds, shown as their energy in red there. Okay? But notice, you get the same total energy, as the computer would. The computer divides the pie completely differently. Right? But you get the same answer — right? — and the same geometry that the computer does, from the point of view of what direction do the hybrids point and so on. So we’re getting the same structure, we’re getting the same energy. You get the same total electron density too, if you add all the electrons together. Three bonds add together to give the — the same way, remember, three p orbitals added up together to give a sphere. Right? Our three bonds add up to give the same energy as the computer. So who needs a computer? Right? Our bonds work just as well, if we view them right. Okay? And then there would be the anti-bond, the one that’s — what’s left over, that’s big on boron, small on H, and a node between hydrogen and boron; it’s antibonding, the one that went up in energy when the atoms came together. And again there’d be three of these σ*’s, big on boron, small on H. They would all have the same energy. But again their average energy is the same as the average energy of unoccupied molecular orbitals that the computer gets.
So we’re not doing so bad, following Lewis in this respect, looking at the electron density in terms of bonds rather than in terms of molecular orbitals. Okay? So for many purposes, localized bond orbitals are not bad, and they’re certainly a heck of a lot easier to think about. But, beware of resonance. Right? Because sometimes this isn’t going to work. So we’ll have to understand how that works. So the localized bond orbital picture, pairwise molecular orbitals; that is, two atoms coming together to overlap. And also there’ll be isolated atomic orbitals, like unshared pairs, or isolated vacant orbitals, like the pz orbital of boron that we just saw, that will be — this picture will be our intermediate between hydrogen-like atomic orbitals, which we really understand — you just look up in the table and you’ve got them, right? — and computer molecular orbitals, which are too complicated really to understand in any kind of detail.
It’s handy to sometimes — for some purposes it’s very handy to be able to calculate them and look at them, right? But you don’t do it very often in your head. So when must we think more deeply? When you have these localized orbitals, but they mix, to cause reactivity; that is, a molecular orbital from this molecule mixes with a molecular orbital from this molecule, and one goes down and one goes up. That’s what a reaction is. Or when you have two pairwise orbitals in the same molecule, that we thought of as bonds or vacant orbitals or something like that, and those interact with one another, within the same molecule. We were too simplistic in thinking of them as independent. They interact with one another. That’s what resonance — that’s when resonance occurs is when you have important interactions within a molecule, between orbitals, or between what you thought were orbitals; but were thinking too simply.
So where are we? We’ve gone through atoms, we’ve gone through molecules, and looked at how we can understand the electron density in molecules. And now we’re to the final payoff, which is reactivity. How can we understand reactivity in terms of SOMOs, Singly Occupied Molecular Orbitals, high HOMOs, and low LUMOs? And then how can we recognize functional groups? So even if we never saw them before, we know how reactive they’re going to be, and what with; that’s the payoff. Okay, so now we know that if you have two molecules and you’re interested in their reacting, there’ll be molecular orbitals of the two that come together, interact. And if the molecule gets — if the energy goes down because of that, that’ll be a favorable reaction. If the energy goes up, they’ll bounce off one another. Okay? So, but how many molecular orbitals are there on a molecule?
[Students speak over one another]
Professor Michael McBride: Suppose methane, how many molecular orbitals does it have? It has an infinite number, right? The 13z orbital and so on. Okay? So we have to pick out which orbitals are going to be important in interacting with one another, or else we’re just going to be at a loss, because there’ll be so many. So you have these orbitals on molecule B, say, and those are occupied, and the ones above are vacant, unoccupied. And on molecule A you have a bunch of occupied and a bunch of unoccupied molecular orbitals. And now we’re going to look at orbitals of B mixing with orbitals of A. Right? And which ones? We might look at those two interacting, for example, mix those two, between A and B. Or we could mix those two, or those two, or those two, or those two, or those two, or those two, or those two, or those two — and it goes up infinitely. So this is a problem. Even if we understand how orbitals interact with one another, we’re going to have to narrow our focus, if we want to deal with this. Okay? So there are myriad possible pairwise mixings.
Now there’s one that’s going to be obviously important. Suppose that molecule B has an odd number of electrons, and therefore a singly occupied molecular orbital, and the same is true of molecule A; just suppose that happens. Right? Then when they come together and mix, those two electrons will both go down in energy. So that’s bound to be favorable, substantially favorable, and you’ll get a new bond. That’s like two hydrogen atoms coming together. So if there are SOMOs, on the two molecules — and notice, suppose B had a SOMO but A didn’t. But you’re not just talking about two molecules, you’re talking about two flasks. If you mix — you had to have two B’s that both had SOMOs. So A could be another B. So if you have any SOMOs, you can pair them; B with B, if A doesn’t have any. Right? So if they exist, if there’s a SOMO, then — and that’s true of many atoms, or things called free radicals, like the hydrogen atom, the chlorine atom, the methyl radical — then they’re going to be very reactive. Okay? And for that very reason they’re not very common. You can’t buy a bottle of methyl radicals because they will all have reacted with one another. The same for hydrogen atoms, the same for chlorine atoms. So that’s an easy case to understand, when you have SOMOs. But it’s very rare to encounter it, unless you’re talking about flames, where lots of things are atoms and free radicals, as they’re called. We’ll talk about that a lot more at the end of the semester. Okay, now how about — there are two that could mix. It’s not bad energy match, right? And suppose they have good overlap. Do you care about that one? Why not? Lucas?
Student: It doesn’t exist.
Professor Michael McBride: Oh it exists, it’s an orbital.
Student: Oh, there are no electrons.
Professor Michael McBride: There are no electrons to have that energy. It’s an available energy, but nobody availed themselves of it. So it doesn’t contribute anything. Right? So there’s nothing there. So forget that one. And you can forget most of them for this reason, of the infinite number. Okay, now how about this pair? Now they have — let’s suppose they have good overlap. They’re not bad energy match. Right? So two electrons will go down and two will go up. Do you care? Steven?
Student: No, because since there’s two that are higher energy, they’re antibonding.
Professor Michael McBride: Yes. So they’d more or less cancel. Do they exactly cancel?
Professor Michael McBride: In which direction do they not cancel? Corey?
Student: The lower one is bigger, so it would be even worse.
Professor Michael McBride: The lower one is bonding, but bigger — by bigger, you mean greater than 1/√2 kinds of thing; or no, pardon me, less than 1/√2, it’s smaller, is what you should have said. Okay, it’s smaller. So it doesn’t go down as much as the upper one goes up. So pretty much they cancel. But to the extent they don’t cancel, it’s repulsive, it’s bad to come together, which means the molecules will bounce off. Right? So when you have filled orbitals hitting one another, the molecules repel one another. Okay? So that’s weak net repulsion. That’s not going to explain a reaction. Now how about here? Now we’re mixing a filled orbital with a vacant orbital. So two electrons go down. So you don’t have this canceling. Is this going to be important, this one? Josh?
Student: No, bad energy match.
Professor Michael McBride: Ah, the energy match is so bad that you don’t get very much going down; not enough to make up for the fact that if they’ll come together, there are going to be a lot of other orbitals, molecular orbitals, overlapping, and they’re going to be repulsive, the kind we just looked at. So this weak attraction is not going to be worth much. So it’s negligible mixing because of bad energy match. Now how about here? Now that’s not bad energy match; I mean, it’s not great energy match, but it’s certainly better than any other pair of occupied with vacant. Okay? So that means that pair will go down, right? And you’ll get bonding, if the energy match is good enough that the amount that they go down is significant. Okay? So remember, when we tried the same thing the other direction, from the highest one on the right to the lowest one on the left, the energy match was not good. So that didn’t help. But this one, where it’s high on the left and low on the right, that one will help. Okay, so what it requires, to have a reaction, is that you have a high HOMO, an unusually high HOMO, on one molecule, and an unusually low LUMO on the other, so that you have good energy match, or at least as good as you can get. Right? Then you could get a reaction. And it turns out that there are simple names for things that have unusually high HOMOs and things that have unusually low LUMOs. And you know what those names are?
[Students speak over one another]
Professor Michael McBride: Something that has an unusually high HOMO is called a base, and something that has an unusually low LUMO is called —
Professor Michael McBride: An acid. Right? So that’s a generalization of the idea of acid and base. Most mixing of MOs affects neither the overall energy, nor the overall electron distribution, for one or more of these reasons. Okay? First the electron occupancy can be four; that is, both were occupied, which means two go up and two go down, and the ones that go up, go up a little more. So that’s not a bond. Or zero, and you don’t care about the energy. Or the energy match is poor, so there’s not much mixing all together. Or there’s poor overlap. So even if the energy match were good, you wouldn’t get anything out of it. But, if you have high HOMO and low LUMO mixing, then you get reactivity; unusually high HOMO and unusually low LUMO. And that’s the secret we’ve been going for all semester.
So here’s increasing generality in the concept of acid and base. The names were introduced by Lavoisier, in 1789, and we’ll talk about that. And from his point of view, a base was an element, a substance that could be oxidized, and an acid was something that had been oxidized. Like sulfur was a base; oxidized sulfur, sulfuric acid, was an acid. Phosphorus was a base; oxidized phosphorus was phosphoric acid. Carbon was a base; oxidized carbon was carbonic acid. Okay? Then 100 years later, Arrhenius, in Sweden, said that an acid was something that was — had the idea that there were ions, things that were charged, and that a thing that would give H+ was an acid, and a thing that would give OH- was a base. That’s Arrhenius’s Theory. And that was further generalized, in 1923, by Brønsted and Lowry — Brønsted was a Dane and Lowry an Englishman — who said that an acid was an H+ donor — that’s the same as Arrhenius — but that a base was something that could accept H+; not only OH-, but ammonia, for example, could accept a base [correction: proton]. And Lewis then, in the same year, had the idea that an acid was something that could accept an electron pair, like BH3, and a base was something that could donate an electron pair, like ammonia. Right?
And other names for those acceptors and donors are electrophile, something that loves electrons, will accept them; and a nucleophile, something that loves a nucleus, is something that will give electrons to it. But in the 1960s this was generalized finally to where it is now, that an unusually low vacant orbital is something that can accept electrons; electrons like to be in a low energy orbital. So that is the ultimate definition of what an acid is. And by the same token, an unusually high HOMO has electrons that are eager to avail themselves of some vacant orbital, to lower their energy; and that is a base. Okay? So as so often in science, as the science matures, you get more and more generality to the idea. And HOMO and LUMO, unusually high HOMO, unusually low LUMO, are the ultimate form of acid-base; and that’s what we’ll be talking about. But they have to be unusual. Why do they have to be unusually high and unusually low?
[Students speak over one another]
Professor Michael McBride: Because otherwise you don’t have good energy match, to change the electron energies when they come together. Okay, now here’s what you should have asked. Unusual: compared to what? Okay? So here is what — we’re organic chemists. If we were some other kind of chemists in some other universe, dealing with different elements, we’d have different points of comparison. But what we deal with is things that have a lot of carbon and hydrogen in them. And by far our most common bonds are among — between carbons, or between carbon and hydrogen. So it turns out that the valence orbitals of carbon, the sp hybrid orbitals of carbon, and the 1s orbital of hydrogen, are roughly the same in energy. For our purposes we’ll consider them to be the same. Okay? So you bring two of these together, they overlap, and you get σ and σ*. And we’ve talked about that, the bonding and the antibonding orbital. So most of the occupied orbitals we’re dealing with are σCC or σ CH, and most of the anti-bonding orbitals are σ*CH or σ*CC. So those are our point of comparison. Those are the plain vanilla electrons and holes. Okay?
Chapter 4. Criteria for Assessing Reactivity [00:40:16]
Now let’s look at other things and see whether they’re unusual with respect to this. Now there will be several classes that they fall into. Right? So those are the usual LUMO and the usual HOMO, or the CC analogs. Now what could make things weird? One thing could make things weird is that if you had valence-shell atomic orbitals, that didn’t get mixed with anything. So they’re not σ or σ*, bonding or antibonding; they’re the original orbital, in between these. So that would make things unusual. A second one would be a little bit in between these two. If there’s no overlap, then you have unmixed orbitals. If there’s a lot of overlap, you get bonds. But there could be poor overlap, in which case you don’t go down as far with your bonding orbital, or up as far with your anti-bonding orbitals. So a second case would be poor overlap. A third case would be that the actual atomic energy, the atomic orbital energy that you’re using, the atom you’re using, is not like carbon and hydrogen. So it starts at a completely different place from where the standard orbitals start, before they make the molecular, the bonding orbitals. And fourth can be if there’s charge around. Because obviously if you have a negative charge, it’s not such a good place to put electrons; that means the orbitals are high in energy. If you have a net positive charge, it’s a good place to put electrons; it’s low in energy. So charge can make it. So we’re going to show examples of these four contributors that will allow you to identify when a molecule has an unusual orbital energy. Right? Unusually high HOMO, unusually low LUMO. And then you know that it’ll be a functional group, and you also know what it’ll react with. Because if you have an unusually high HOMO, it’ll react with an unusually low LUMO; and vice versa.
Okay, so first let’s look at the first one, unmixed valence-shell atomic orbitals. So H+, Arrhenius’s acid. Right? That’s a vacant orbital. Right? No electron in it. So that’s obviously unusually low. It’s where this started, the 1s of hydrogen, before it went up or down. So if you had H+, it would be an unusually low vacant orbital. How about NH3? Right? It has an unshared pair, only on the nitrogen. Now, it likes being on nitrogen, compared to carbon or hydrogen. Why? What makes it better to be on nitrogen, in a valence-shell orbital than on — the unshared pair — than to be on an unshared pair on carbon or on hydrogen? Angela?
Student: There’s more protons in nitrogen.
Professor Michael McBride: There are more protons in the nucleus. It’s a lower energy orbital, to start with. Right? But it didn’t mix with anything. There are N-H bonds, in ammonia. Right? And they mixed and went way down in energy, and a vacant orbital went up in energy. But the unshared pair didn’t mix with anything. So it still has that original energy. So it’s unusually high. Compared to what? Compared to what?
[Students speak over one another]
Professor Michael McBride: σCH or σCC. That’s our point of comparison. Okay, or the flip side of that is BH3, which has a vacant orbital. It’s high, to start with, compared with a hydrogen or a carbon, because boron doesn’t have a very big nuclear charge. Right? And its nucleus is screened. Right? So it’s unusually high for an atom. But it’s still atomic, it didn’t mix with anything and go up further. Right? So it’s unusually low for a vacant orbital. So BH3 is an unusually low LUMO. So is it an acid or a base?
[Students speak over one another]
Student: An acid.
Professor Michael McBride: Which?
Professor Michael McBride: I can’t hear you.
Professor Michael McBride: You’ve read through the lecture. You don’t get to vote [laughs]. It’s like H+, right? An unusually low LUMO. So it’s an acid. How about NH3? It’s a base, unusually high HOMO. Okay, how about water? Right? It has unshared pairs of electrons. Yes Lucas?
Student: Sort of a side question. Didn’t we just say that they both had these properties because they had low nuclear charge?
Professor Michael McBride: No, nitrogen has high — which has low nuclear?
Student: Nitrogen is, low nuclear charge compared to carbon.
Professor Michael McBride: Yes. [apparently did not understand the student’s erroneous claim about nitrogen and thought he was speaking of hydrogen vs. boron]
Student: And BH3 is even lower, compared to carbon.
Professor Michael McBride: Yes.
Student: But the energies are completely different because it’s — is it just because of the unshared pair?
Professor Michael McBride: But remember, you have to bear in mind which row of the periodic table you’re in. Right? H, true, it only has one proton, but you’re talking a 1s orbital, low in energy. For carbon, nitrogen, oxygen, boron, you’re talking 2s. Right? So you’re changing the game.
Student: Between BH3 and NH3?
Professor Michael McBride: No, no, not between BH3 and NH3. NH3 is unusually high energy for an occupied orbital. BH3 is unusually low; although it’s high, compared to nitrogen, it’s low for a vacant orbital. It’s an unusually low LUMO. Nitrogen’s an unusually high HOMO. Now oxygen is lower than nitrogen. So water is a base. But it’s not as good a base as ammonia, because its electrons aren’t as high, because it has a bigger nuclear charge, and the same n; it’s in the first complete row in the periodic table. Okay, OH- is even higher. It’s a stronger base than water. Why? Why are its electrons higher in energy than water?
Student: Less protons.
Professor Michael McBride: Got less protons. Right? It’s got a net negative charge, which makes all the electrons high in energy; and specifically the pair we care about, the highest one, unusually high, because it’s got a negative charge. [Technical adjustment] Okay, CH3- is even more of a base, because carbon doesn’t have as many — as big a nuclear charge as oxygen. So when it has a negative charge it’s a really high HOMO. Okay, I got to quit there. We’ll go on to two, three and four next time.
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