 Okay, when we were talking about this second topic about molecular orbitals, there was a slide on there that I forgot, total oversight. I had shifted my notes around. And so I wanted us to cover that before we go on back to this lecture on energies because it's an important part of our discussion of the interaction between filled orbitals and unfilled orbitals. And I want to bring you back to our explanation of the energetic consequences of having orbitals interact, the orbitals that are filled interacting with orbitals that are empty. And I gave you this equation that said that you can approximate the energy that you get when filled orbitals interact with unfilled orbitals with the simple equation that says orbital overlap in the numerator and the difference between filled and empty orbital energies in the denominator. And we said in great detail about what happens as you raise the energy of filled orbitals, the nucleophiles, and you drop the energy of unfilled orbitals. And that's the electrophiles. But I didn't say anything about this orbital overlap. And so I want to stop and talk about that. And the most important concept for us to consider is the importance of symmetry when we talk about this idea of orbital overlap. So let's talk about the interaction of canonical atomic orbitals or hybrid orbitals. Let's imagine some example where we have two S orbitals interacting with each other. For example, to make a hydrogen molecule, two Hs. As long as I have two S orbitals, these are one S orbitals interacting with each other, I can imagine some combination where I have like phases. And that's the kind of interaction that would lead to a hydrogen-hydrogen bond. Or you can imagine the interaction of two P orbitals. And what we have to imagine here is that there's two possible ways that this can interact. One where we have like phases. And what we expect is this overlap area should be even bigger in the products. So the product molecule, the hydrogen molecule has a lower energy orbital that looks like this. That's constructive interaction between the two orbitals. Here I can imagine an interaction where I make a pi bond. And the symmetry of the pi bond looks very big in the middle where we have constructive interaction. And so the end result looks like this. And I can even imagine the interaction of a P orbital with an S orbital as long as I tilt this in this way so that everything is some sort of ends up with some sort of a positive constructive, I guess, in terms of wave functions constructive interference. If you call it inter, well that's not a good. Here's what I intend to draw, something that looks like this. Okay, so that's, those are examples of forming bonds. And in this case I would end up with some sort of a sigma bond again. So graphically when we graph out the energies, as long as orbitals have the right symmetry, I can produce two new orbitals, a bonding combination and an anti-bonding combination. And the bonding combination will be lower in energy because these interactions are productive. So these are examples of things with the correct symmetry interacting. But the symmetry doesn't have to be correct. Let's talk about an alternative situation where I try to interact an S orbital with a P orbital but I've got them aligned incorrectly. So what I'll expect out of this interaction, if I look at the way I've chosen to phase these, as I'll get a constructive interaction here, but I'll get an equally destructive bonding interaction here, an anti-bonding interaction on the bottom part. Every k-cal per mole of energy benefit or electron volt of energy benefit I get here, I'll get an equally destructive interaction there. So the end result would be no net change. And that would be an example of having two orbitals interact. I can depict the energetic consequences of the wrong symmetry here simply by saying that there would be no interaction. That the two new resulting orbitals I would get would have no essential difference. So if you want to see some energy benefit, you have to match the symmetry of the orbitals that are interacting so that the hashed part interacts only with hashed part or at least more hashed part than unhashed part. So I might get bonding interactions here but anti-bonding interactions there. There's no net energetic benefit between mixing those orbitals. Okay, so symmetry has to match in order to receive an energy benefit. And you'd get an equally useless interaction if I tried to make a p orbital interact edge on like this. And there it's equally destructive and equally constructive interactions. There's a simple rule that relates to this in organic chemistry that has to do with bicyclic molecules and it's called Brett's rule. And let's talk about Brett's rule. It's basically just a vocalization or some sort of a statement of this idea of symmetry. So I'm going to draw out a bicyclic molecule. This is a bicyclo 2, 2, 2 ring system. And so there's all kinds of molecules and natural products and base reagents that have this kind of bicyclic structure. And what this rule is saying is that if I look at this bridgehead here, that's this point of intersection between the rings, what it says is I'm going to have a hard time creating a pi, let me make that look a little better, a pi bond there. Then any reaction where I try to generate that, it's going to be incredibly unstable and potentially impossible to isolate any molecule that looks like this. This is what we would refer to as a bridgehead olefin. And here's that bridgehead right there with that dot. Why is that? Let me go ahead and redraw this pi bond in terms of the p orbitals that are combining together or that it would have to combine together to make a pi bond like that. So I'm going to redraw this bicyclic molecule. This time I'm going to draw that hydrogen atom at the two position of this bicyclo octane system. And what I want to do is now that I can see all three of the things that are attached to this carbon atom, I'm going to draw the p orbital that would have to interact or the way the p orbital would be aligned. That set of atoms with that geometry. So there's one p orbital and if I want to combine that with a p orbital over here in order to make a bond, the problem is I look at the three things that are attached to this dotted carbon. Here's one, here's one, and here's one. And they're almost in the plane of the board. And what that means is that the p orbital that's on this carbon atom has to be sticking almost straight out of the board. And I'll phase that back into that orbital like this. And what you can see when I do that is those p orbitals are orthogonal. It doesn't matter that they're close together. It doesn't matter that they're both p orbitals and have the potential in theory to form a pi bond. But they have the wrong geometry and will interact with the wrong symmetry. You'll get no significant pi bonding. In fact, a better resonance structure for this molecule right here would be to have two radicals at those two carbons. And that's the way you would expect it to react. It'll end up undergoing radical reactions very easily. Okay, so Brett's rule is an expression of this importance of symmetry when two orbitals interact. You won't get an energetic benefit unless you have significant overlap, and that was part of this equation, orbital overlap over here. Okay, one last thing related to the importance of orbital overlap. So I'm going to go ahead and draw just an interaction of two p orbitals on carbon atoms. And I want you to try to imagine what difference doesn't make if those two carbon atoms are close together or far apart. Which you have to imagine, what happens as I bring these two carbon atoms closer together? As I bring these two carbon atoms closer together, I'll expect to see more and more overlap between those empty p orbitals. And you only get that with second row atoms where the p orbitals are approximately the same size. You only get stable pi bonds with second row atoms. Boron, carbon, nitrogen, oxygen, and fluorine. Once you talk about transition metals, all that pi bonding stuff ceases to have this phenomenal strength of pi bonds among second row atoms. Let me go ahead and draw out some bond lengths for you. If you go back and you look at bunches and bunches and bunches of crystal structures, for simple, straightforward, carbon-based compounds, what you'll find is that typically, bond lengths for alkenes are about 1.46 angstroms is a good number. There's not a huge degree of variation unless you strain the system in some way. If I look at a simple alkyne, there's a more S character for those SP-hybridized carbon atoms in an alkyne and they're substantially shorter bonds. What that means is that the p orbitals in an alkyne are closer together and overlap more effectively. So what are the chemical consequences of that shorter bond length, of the better overlap? The consequences are that pi bonds, alkyne pi bonds, are stronger and more stable. So let's go ahead and take an example of that. I'll go ahead and take a simple substrate molecule and we'll imagine treating this with an epoxidizing agent, metachloroperoxybenzoic acid. I expect you to know that peroxy acids make epoxides and there's two different pi systems that you could make an epoxide with. The product in this reaction, the major product in this reaction results from attack on the alkene and it has nothing to do with the fact that the oxirane, the epoxide you would make from the alkyne would be unstable. I mean sure it would be unstable but you would not get an 86% yield of this product if this were reacting faster. What it tells you is that at least 86 out of 100 molecules was epoxidizing faster on the alkene than on the alkyne. And you can't argue that that's happening because of steric hindrance because this is less hindered. You can't argue that it's easier to approach just one or two faces here because the electron density in alkyne is symmetrical all around. You can approach the alkyne from any orientation you want. There's no possible rationale for why this is true, why this is the more nucleophilic of these two double bonds except the fact that this bond is weaker and more energetic. So if you look at a C-C pi bond it's got an energy that's up here and if you look at a C-C pi bond in alkyne it's more stable, it's lower in energy. This is supposed to be a triple bonded C here. It's lower in energy and it's less nucleophilic. Even though there's two pi bonds in the alkyne, it's still attacking less than the alkene. Okay, so that's all the importance of bond length and the more effective orbital overlap in this alkyne because those two carbon atoms are closer together and they make the bond more stable but also less nucleophilic. Okay, so that's that missing slide that I had from lecture two so you'll have to sort of in your mind paste that back in. It was there on the lecture notes that I posted online but now let's return to this topic of energy that we were discussing when we left on Friday. Okay, I want to start off with a very simple equation. I want to talk about the reactivity of methane. If you go over to Saudi Arabia or to Africa where they're drilling for crude oil, at night the skies are lit up because when you pump crude oil out of the ground there's also methane gas coming out with it and it's too expensive to package up that methane gas and to ship it to other continents like our continent even though we need that methane gas. That's natural gas, so they burn it off. It's called flare off and so why is it so expensive to ship a gas? It just takes a lot of infrastructure to ship gas to other continents. One thing that people are trying very hard to do right now is to convert methane gas, find ways to convert methane gas into liquids that can still serve as fuels that are easy to ship. It's easy to ship liquids, it's hard to ship gas and you can't do this at present. You can't mix simple oxidizing agents with methane and make liquids like methanol out of them even though methanol is a fantastic fuel for things like fuel cells. And so why is this such a problem? What's the problem with this equation here? And what I'd like you to know how to do is at least assess in a simple way is the problem free energy. Is it that this is not thermodynamically possible for us to do? Any time I give you some reaction I'd like you to have some idea about whether that's thermodynamically reasonable. Is that an exothermic process? If you're talking about a reaction that's not exothermic then there's no real point talking about it actually. So our criterion that you know from general chemistry is that the free energy change for that transformation needs to be negative. It needs to be less than zero. Otherwise you'll be trying to go uphill. And unfortunately it's very, very hard to know free energies when I draw out reactions like this. I don't have a good, well maybe I have some intuition. And the problem is this, that free energy is composed of an enthalpy term that means things like bond strength, strain, issues like that. But there's this other part of free energy that's entropic in nature and that's very hard to assess. Now fortunately if you're clever when you make comparisons of things, comparisons of starting materials to products, transition states to transition states as long as we keep things similar we can ignore that entropic term T delta S. If I compare two transition states that look sort of similar except you're just attacking different bonds. If I have reactions where I have two molecules reacting and I get two molecules out as products, that's kind of similar in terms of molecularity. So as long as I'm really careful to compare similar things I can ignore this entropic part of free energy and we can just focus on solving this delta H part, the enthalpy. There is a simple way for you to estimate the enthalpy change in reactions. And I'm going to show you how to do that. It's going to require some upfront effort on your part and I'll make sure you put in that effort. So Hess's law is actually very general but I'm going to show you how to apply Hess's law to estimate enthalpy change in reactions based on bond dissociation energies, B, D, E's. And so let's take an example and I'll try to explain what Hess's law means in terms that relate to this methanol, this methane-methanol equation. So let's go back to this methanol equation where we'd really love to know whether we're asking too much of this oxidation reaction. And so I don't really know what this enthalpy change here is but I have a pretty good idea for the energetic cost of breaking this oxygen-oxygen bond. And so I want to imagine that. I don't know this number here but I can know this number. What does it cost to break this hydroxyl-hydroxyl bond to give two hydroxyl radicals? In other words, to homolytically dissociate that. I have a reasonable idea about what it costs to break a C-H bond to give two radicals. And I also have a reasonable idea about what it takes to break this water-H bond to give two radicals. And I'm just going to continue on with this idea here. And I hope I'm getting my pieces right. And so what you can see here is I can imagine some equation here where I've broken this bond to give these pieces. And now I just have to imagine recombining these pieces and I've redone my pieces here. So let me just say equals here. That's supposed to be an equal sign. What you can see is I get the same pieces this way as if I break the bonds over the other way. And we can construct a thermodynamic cycle from this. In other words, let's imagine that I take this piece over here, this hydroxyl, and make a bond. You can see I'm going to get those, the methanol piece and the water piece on the other side. So let me ignore this equivalent sign. And I just want to imagine that in two steps, breaking bonds, I'll just write break, and forming bonds, I can imagine making those same products there. Even if that has nothing whatsoever to do with the actual reaction. Maybe I use a platinum catalyst for this. Or maybe I just simply heat it hot enough to happen. It doesn't matter. As long as I know the enthalpy change here and the enthalpy change there, I can simply add them together and I will get the exactly right enthalpy change. This is what Hess's law says. That the mechanism doesn't matter. As long as you can create some sort of a thermodynamic cycle, some equivalent way to get to your products, it doesn't matter whether the mechanisms are the same, the enthalpy change will be correct. So as long as we just memorize a few bond association energies like this one and this one and this one and this one, we can simply add and subtract them and we'll come out with the right enthalpy change. So let's take a look at some bond energies. And what we're really looking for here, if we want this to be a favorable process, what we'd like to do, if we want to get more energy out of forming bonds than the energy we have to put into break bonds, we want to look for processes where we either break weak bonds, that's cheap or processes where we form very stable bonds because we get a lot out of that. That's what we want to look for. So let's review bond association energies. I'll give you some numbers that I want you to memorize and then you can do this at your leisure for any reaction. And let me just summarize. So reactions tend to be favorable if you break weak bonds or form stable bonds. So let's review some bond association energies. I'm going to give you a definition. Bonds association energy is defined to be bond strength. And both of those are related to this equation. I'll give you one example for an Rh bond. Could be CH3H or in ethane, propane, doesn't matter. But by definition, a bond association energy or if I use the word bond strength refers to a homolytic process that gives you two radicals. By definition. I didn't invent this definition. I'm stuck with it just like you. But this is the definition of bond strength. And that's super important. So by definition, the enthalpy change associated with this for this reaction right here and that's a standard state free energy or free enthalpy change, that's defined to be the bond association energy. So why am I harping on this idea of bond strength? Because I'm going to refer to strong bonds and weak bonds. And one of the unfortunate consequences of that is that I'm going to define an OH bond as a strong bond. Now you know as well as I do that protons exchange very rapidly in water or in proton transfer reactions. But the word strength implies that I'm not talking about proton transfers. When I say the word bond strength, I'm talking about homolytic dissociation to give radicals. And this is a very costly process. The bond association energy for water is about 110 kcals per mole. That's considered a strong bond. And I'll give you some more numbers to help you put that into perspective. So again, as soon as I say the word bond strength, I'm not talking about anions or bases or acids. I'm talking about radical reactions. Okay, so let's take a look at some sample numbers. We're going to start off by looking at CH bonds because those are very common. And then we'll look at bonds between carbon and other second group atoms. So start off by thinking about alkanes like ethane. The bond association energy, there is nothing special. They're typically around 100 kcals per mole, high 90s. It varies a little bit. The larger the alkanes get, the weaker those bonds get by a little bit, but they're around 100 for a typical CH bond in an alkane. But look what happens when I change the hybridization and I allow this carbon to contribute more S character to that sigma bond. It becomes more stable. That becomes a stronger bond by 12 kilocalories per mole. How many times more stable is the CH bond in this alkene? This is in kcals per mole. That means there's a 12 kilocalorie per mole difference. And if I divide that by 1.4, how many factors of 10 is that? Instability, about 8, about 10 to the 8? That's about 100 million times more stable because of that extra S character. And you can see what's going to happen as I contribute even more S character to that CH bond. I should expect it to get even stronger. I've never seen any reaction where something plucks an H off the end of an alkan and H atom and leaves you with a carbon radical at the end of an alkan. Those bonds are too strong. Nothing does that. That's not to say you couldn't prepare some alkan or radical some other way, but you're not going to do it by plucking off this H. So you can see all CH bonds are not created equal. Pay attention to hybridization. That's essential. Now last, another thing that will have an obvious effect and I think you know this on CH bonds is nearby pi systems. So probably sometime back in organic chemistry and undergraduate organic chemistry you learned that you could brominate the benzylic position of molecules. And that's because this CH bond is weak. And to put some numbers on that it's 85 K-cals per mole. And it's very similar for an allylic CH as it is for a benzylic CH. So resonance can affect bond strengths. Okay, let's take a look at a few more other common bonds and bond association energies. Let's take a look at some bonds, not to hydrogen but between second row atoms, carbon to carbon, carbon to oxygen. There's nothing special about a carbon-carbon bond. The first thing I want you to note is that it's weaker than a CH bond. A process that leaves H dot sucks. That's why those bonds are higher in energy. But if you generate two carbon-based radicals, at least there's bonds that can help other orbitals that can help stabilize carbon-based radicals. So a C-C bond is weaker than a CH bond. If I look at a C-C pi bond, of course, now there's two bonds in there, you should expect that to be a much more stable pi bond because to dissociate that to give two carbines, I guess, it doesn't cost exactly twice the energy. But it's very expensive to simultaneously break both a sigma and a pi bond. And if I look at an alkyne, then of course it becomes very expensive. Now I want to compare this to carbon-oxygen bonds. Those are the main ones you'll think about or compare these with in organic chemistry. And we're going to start off by noting that there's nothing really significantly different between the bond energy, between the strength of the carbon-carbon bond and the strength of the carbon-oxygen bond. What's important is when we look at the pi bond. It is, of course, more expensive to break both of those bonds at once. And I don't really want to focus on these energies individually. There's a set of numbers I want to extract out of this that I expect you to use over and over again. And it's the individual energy of just the pi bond here. There's a sigma bond here. There's a sigma bond here. Let's forget about that and just look at what's different. What's different is there's that extra pi bond. And I can create a number to help me think about that. The energy of a C-C pi bond is the difference between those two numbers. And it's about 64 k cals per mole. The obvious point here is that a pi bond between two carbon atoms, this second bond, is substantially weaker than a sigma bond. Carbon-carbon pi bonds are weak. You want to look for reactions that break carbon-carbon pi bonds and trade them for carbon-carbon sigma bonds. Reactions that break carbon-carbon pi bonds and trade them for carbon-carbon sigma bonds will tend to be favorable. Okay, let's look at carbon-oxygen pi bonds. The strength of this pi bond between carbon and oxygen is the difference here. And now we're talking about 94 k cals per mole. That's stronger than a single bond. The carbon-oxygen pi bond, that second bond, is stronger than the first one that was there. This is considered a strong bond. You want to look for reactions that form carbon-oxygen pi bonds. Reactions that form carbonyl groups will tend to be thermodynamically favorable. So when I look at reactions like that methane reaction, I'm looking for reactions that break pi bonds between carbon. And I'm looking, that wasn't an example of one. But in general, when I look for reactions that I want to be thermodynamically driven, I'm looking for things that trade C-C pi bonds for C-O pi bonds. And we'll talk about some of those examples later in the course. Okay, let's, so those are examples of weak and strong bonds. Let's take a look at some single bonds that I would consider to be strong. And I've already talked about one of those. And that's an OH bond, is strong, 110 k-cals per mole for a typical alcohol OH bond. That would be very unusual for you to have any reaction that plucks that hydrogen atom off. And that's not to say that there aren't reactions that generate RO dot, but they don't proceed by plucking off an H and H atom. So I already mentioned another example of a strong single bond. And that was this insanely strong CH bond at the end of an alkyne. Again, I've never heard of any reaction where anything can pluck this H off of the end of an alkyne. Okay, things I have not yet talked about. Silicon fluorine. Oh my God, crazy stability here. 141 k-cals per mole for the stability of a silicon fluorine bond. This is why you use tetrabutylammonium fluoride to deprotect silal ethers because you get this huge thermodynamic driving force. You also get kinetic speed associated with that process. But it's hugely thermodynamically driven to form silicon fluorine bonds. And then the last common example of a super stable bond, and I'm not even sure how to draw this, I can draw it like this. And it doesn't matter whether I draw this, I can have a fifth bond of phosphorus. There's another resonance structure. If you want, you can draw it like this as a way to help you rationalize the stability. But the importance is this bond is super stable. If you have a reaction where you put in triphenylphosphine, there's a massive thermodynamic driving force for somewhere in that process for you to generate triphenylphosphine oxide as the byproduct. And so the energy there, again, is 136 k-cals per mole. Hugely favorable, so reactions like Mitsunobu reactions, Appel reactions, we'll talk about these later, are driven by the formation of PO bonds. Okay, so those are examples of very strong bonds. You want to look for reactions that form those types of bonds. They'll tend to be enthalpically favorable, and they'll tend to be thermodynamically favorable with negative delta Gs. Okay, let's take a look at some weak single bonds. Let's keep our focus on single bonds right now. Okay, so carbon-silicon bonds, single bonds tend to be weak. If you have a chance, you'd like to trade a carbon, and I don't have the number here, a carbon-silicon bond for an oxygen-silicon bond, or even better, for a fluorine-silicon bond. That'll be thermodynamically favorable. Look at all these bonds to halogens. If you take a simple alkyl iodide and you put it under a TLC lamp, you'll slowly see your spots start to turn brown because you're homilizing that weak bond. The carbon-iodine bond will start to homilize and generate I dot and cyclohexyl dot. Carbon-halogen bonds tend to be weak. Bonds to halogens tend to be weak, especially for halogens that are not second-row atoms. So those are examples of weak bonds. You can expect those kinds of bonds to break in thermodynamically favorable reactions. Okay, now let's look at bonds where the two atoms are the same. Carbon-carbon bonds are sort of unique in this respect because, I mean, they're not super weak, but they're not strong. They're just sort of right in between. But if you stay in the second row where atoms are about the same size, carbon, nitrogen, oxygen, fluorine, yeah, 81, I'm using, this is specifically ethane, the 81's an average. So as you, you know, as you add substituents, those numbers will change. So let's go ahead and take this specific set of molecules where you have the same atoms. Look what happens is I increase the electronegativity to nitrogen-nitrogen. Now it's a weaker bond. And then I make it even more electronegative and it gets even worse. So you can see what's happening with this hydrogen peroxide methane reaction. I looked at that. In half a second I had judged that that ought to be thermodynamically favorable, it's thermodynamically favorable for hydrogen peroxide to react chemically with methane. That's not to say that there's a kinetic pathway to do that without a catalyst. But any reaction that breaks a hydroxyl-hydroxyl bond is going to tend to be thermodynamically favorable. That's a weak, crummy bond. And then finally we get down to fluorine, fluorine, and that's very weak. So there's this sort of trend in getting weaker and weaker. So the reason why they flame off methane and they don't convert it to methanol using hydrogen peroxide is not a thermodynamic problem, it's just that you can't stop at methanol. Any oxidant that can oxidize methane will continue to oxidize it further. They can't figure out how to stop the oxidation of methane at methanol. So let me continue to draw some of these diatomic molecules you already know. And the reason why you already know these is because in sophomore organic chemistry you learn that you can homilize these with light or heat in order to generate radicals. To do free radical allylic halogenation or free radical benzylic halogenation. Okay, so that's Hess's law. Let's go ahead and use these numbers and we'll apply this to the, to this methanol equation. So you can see how we use these numbers and how we add up those bond association energies. And I think I erased my equation so I'll rewrite it here. And what I want to do is I want to add up all the energies of the bonds broken and add up all the energies of the bonds formed. So what's the energy of this oxygen-oxygen bond? I'm wondering if I've written down the wrong number because sometimes I've written down numbers for specific molecules and then other times I've written down numbers for just generic bond types. And it's 98 for a CH bond which is relatively strong. So we add up all the energy it costs us to break these bonds. We'll come up with a number of 149 k-cals per mole. And then we come over here and we look at the bonds that we form. Let me not put that line there. So the OH bond in water or just a generic OH bond. So about 110 and then for a CO bond. And so when we add these together we get 189. You can, you can tell already that you're generating more energy by forming bonds than you had to put in to break these bonds. And so when we do our final calculation we're going to say delta H is equal to bonds formed. Well actually if we want this to turn out to have the right sign, bonds broken. And that's going to be equal to 149 minus 189 which is equal to minus 40. Minus 40 k-cals per mole down. That means downhill. Enthalpically this is a favorable process. That's enthalpically favorable for you to do this reaction. We don't know anything about how fast this is. And in fact it's not fast. You could store hydrogen peroxide in methane as long as there's no catalyst there. But enthalpically that's favorable. So just some simple powerful numbers you can use anytime you want to judge whether reactions look like they ought to be thermodynamically favorable. And I've put some problems on the problem set for you to add those together. Okay so let's go ahead and talk about energies from a different source that's becoming more and more common and it's computers. More and more often nowadays you're going to hear people refer to electronic structure calculations. They'll say stuff like DFT, B3 lip, Hartree, Fock. These are all numbers that are basically enthalpies. And let's talk about what those numbers mean. You can use these, this Hess's law and bond energy calculation anywhere. Sitting in a pub drinking beers while you're listening to a research seminar. But energies from calculations you're going to have to sit in front of a computer. And it's not hard. Nowadays you just draw the structure in, hit go, and it'll pop out some energy. So what do those energies mean? And let me replace this. So when you hear people use terms like density functional theory. Or you see very strange looking things like B3 lip. I mean somebody did some density functional theory calculation using Beck's third iteration of the Lee-Yang Par using Los Alamos National Labs. But it doesn't matter what the letters are. The general rule is the more letters you see here, the more accurate the calculation is. So you see something like this. And it's like that's not as good. And it's just generally true. What do these numbers mean? The calculations are doing this. They're, if you wanted to calculate the enthalpy change. And what you get out of these, you get out of these an energy. And they're not going to say delta H or delta G. They're just going to tell you it's an energy from these calculations. That's approximately the same as an enthalpy value. That means they did not take entropy into account. They're just giving you like an enthalpy number. And so if I went to a computer and I drew these four molecules in, each one of these molecules has an energy associated with it. Water has an energy associated with it. It doesn't matter which bond. That's not what the computer tells you. It just says you draw that structure in and I will tell you what the energies are. And it would tell you something like this. Like the energy of a methane molecule is minus 25.222.572382 k cal per mole. It used to be, you needed very large, that's not for the bond. That's for the whole methane molecule. It's like what do you do with that? It doesn't mean anything by itself. It's only when you add the energy of this and the energy of that, the two starting materials, and you compare it to the energies of the two products added together. That's the only way that those numbers can be useful to you and this is another example of Hess's law. So what are they doing? What they're doing, what the computer is doing is it's taking the energies you get from all these nuclei. I'll draw out these, the nuclei of all these carbon and hydrogen atoms. Probably I've got too many and it's assessing how much energy do I get by taking a bunch of nuclei plus a bunch of electrons and combining them together to make this molecule. And of course you'll come up with a different number here because you've got a different number of nuclei and electrons to make methanol. And then you can ask the computer to tell me how much energy you get by making these types of things and then you can apply Hess's law. That's why you can add the energy of this methanol, energy of water and subtract from it the energy of the two starting materials added together and you'll come out with phenomenally accurate numbers for that. Now again it doesn't take entropy into account or sorry entropy into account but that's okay. Okay, so use numbers from electronic structure calculations in the same kind of way that you use this entropy, in the same way that you use enthalpy. It doesn't take entropy into account. And I want to talk about entropy before we leave today because it's a huge bugaboo for us. This term right here, it's really going to place everything we talk about on insecure footing. Okay, so I want to draw a very simple interaction and what I'm trying to draw is an idealized hydrogen bond. You can use electronic structure calculations, phenomenally accurate electronic structure calculations to calculate out the ideal hydrogen bond. And if we compare that and then I can draw a water molecule in a computer and get the energy out of that and say well what's the energy of just two water molecules added together, energy of this and then I add that to the energy of that and what do I find out about what do I get out of making a hydrogen bond between water? What I find here is that the energy for this by in and you can measure this accurately and get directly a delta H is that that's thermodynamically favorable. 4.6 k-cals per mole, this hydrogen bond is enthalpy favorable relative to just two water molecules floating in infinite distance in space, that's no good. The problem here is this, I'm not so good at math. What's the difference between those numbers? Like 1.7 or something like that? What this tells you is that a hydrogen bond between two water molecules is thermodynamically unfavorable. It is thermodynamically unfavorable for us to form this hydrogen bond. This is the problem with entropy and this is why you have to exercise immense caution when you use enthalpy values. What is this equation really saying here? What this equation is really saying, it should be minus T double S. So here's what this equation is really saying and where did I get these numbers? I wasn't really asking about the energy difference here. I was asking about the energy difference between this idealized hydrogen bond and all other possibilities and there are many, many, many ways to make a non-ideal hydrogen bond. Maybe I could have some sort of distorted bond where the angle here doesn't look so good. Or maybe I can have these two water molecules poking at each other with the oxygens. Or maybe I can have some hydrogen bond where this molecule is bent at some not so good angle. The number of ways to make the not perfect hydrogen bond, the non-ideal hydrogen bond is so immense. There are so many ways to make it something other than the perfect hydrogen bond that this becomes thermodynamically unfavorable. What you find is if you take a box full of zillions of water molecules that only one out, here's what this number tells you, that only one out of 16 pairs of water molecules will be in this ideal hydrogen bond state. All the others will be in some other paired organization state. So if you go to a computer and you draw this in and you conclude that, oh, that's what water looks like, you've come up with the wrong result because you didn't take into account entropy. This is the problem with entropy. It is very hard to draw every other possible configuration into a computer and get the energy out. You don't have the time to do that. Nobody has the time to do that. And that's what makes entropy hard. Okay, one last, I'm going to give you a number to keep in mind for entropy. And this is a useful number. It tells you how much uncertainty you should have whenever you ignore entropy. There's a very famous calculation that was published by Bill Jenks, a famous physical organic chemist. And he just took some numbers and a piece of paper and sat down and did some rough calculations. And what he was able to demonstrate got him one of the most highly cited papers in the proceedings of National Academy of Sciences. So, and I did not intend to do that. I intended to draw the Diels-Alder product here. So this is a 4 plus 2 cycle addition reaction. The point here is not, who cares about the fact that you form carbon-carbon bonds. The point is if you have any reaction where two molecules have to get close and align and orient correctly, like ballroom dancers, not forming any bonds, just getting close to forming bonds, pairing up one mole of A plus one mole of B so that everything is paired up costs a huge amount of energy and has nothing to do with enthalpy. What you find here is that this is a favorable reaction in terms of enthalpy. You could have guessed that by looking at the fact that we just traded two pi bonds, C-C pi bonds for single bonds. So just that bond analysis would have told you that an enthalpy it's good. The problem is that bringing two cyclopentadines close together, not forming bonds, no bonding yet, and just getting them together and getting 6.02 times 10 to the 23rd of these pairs ready to form bonds with just the right distances, just the right orientation costs. This tells you about how much it can cost you just to bring a molecule of A and a molecule of B together. It can cost you up to 40 kilocalories per mole. How many factors of 10 and disfavorability of that? It's 10 to the 10th unfavorable just to bring pairs of molecules together. So every time I see a reaction like methane plus hydrogen peroxide where two things have to come together, I'm taking this into account. Now it can be less than that. Sometimes this is a reaction that has a high entropic component because this carbon has to get close to this carbon at the same time that this carbon has to get close to that one. That has an higher entropical requirement than two things that just simply have to collide with each other. So entropy can cost you as little as zero but as much as this. So there's this huge uncertainty somewhere between zero and 14 k-cals per mole if we ignore entropy when we think about two things coming together. The entropic costs for intramolecular molecules where two things don't have to collide, that's a lot cheaper. So entropically. Okay, so energies. Use Hess's law, memorize those bond association energies. I want you to know how to calculate enthalpy changes for reactions. If you get energies from electronic structure calculations, remember that ignores entropy because you didn't draw every other possible alternative in. And then finally, don't forget this entropy component because entropy can really kill you when you bring two things together.