 Okay, let me take you back to where we left off on Wednesday. On Wednesday, we started talking about orbitals. We're going to spend the rest of this quarter using arrow pushing as a way to describe orbital interaction. So we really need to be comfortable with these three types of orbitals, atomic orbitals, hybrid atomic orbitals, and molecular orbitals. And really, the rest of this quarter is going to be molecular orbitals. So I introduced you to this idea of a molecular orbital interaction diagram that allows us to graphically depict something we don't see when we draw Lewis structures. What we don't see when we draw Lewis structures is the energies of the electrons that are attacking. That's what we don't see. And so the molecular orbital interaction diagram is designed to show you graphically the energetic consequences of an interaction like this. Let's just imagine that water does an SN2 attack on chloromethane. The way to think about that is that my arrows represent the interaction of this filled orbital here, this non-bonding lone pair on oxygen. We can represent that with an N with a subscript O. Attacking an antibonding orbital, donating its electrons into this antibonding orbital, and therefore breaking the bond. And I would depict that interaction like this. A filled orbital like this, non-bonding orbital on oxygen, is interacting with an unfilled orbital, this sigma star antibonding orbital for the carbon chlorine bond. And if I mix two orbitals, I have to get two orbitals out. I get a filled orbital that's lower in energy and an unfilled orbital. I create a new bond, a new oxygen carbon bond. I'm representing it like this. I create a new antibonding orbital with that that goes along with that new bond that we create. Okay, so the important aspects of this diagram that we said is we said that whenever the energy difference here is small, whenever nuclear files are high in energy and energetic, whenever electrophiles are low in energy and easy to attack and easy to nomate into, that results in a small interaction, a small difference in the orbitals that are interacting, and that results in a very big energy benefit. You get a very big drop in energy for the new orbital that you create, and consequently your reactions will be exothermic and fast, and that's what we care about. We care about reactions that are fast. Okay, so that's the MO interaction diagram. Now, you may not be experienced using those, but sometimes I'm going to fall back on those to try to explain the energetic consequences of our arrow pushing. So let's continue talking about this idea that there's a way to depict things with MO diagrams. So we're going to talk about frontier orbitals, and I want you to consider another SN2 reaction. I'm kind of getting tired of these very simple reactions, but we can't, but we have to start with things that are simple. And I want you to think about this idea that maybe ammonia could come close to bromomethane. Maybe something would happen as they interact with each other. I mean, if you want to think about the interaction of an ammonia molecule with bromomethane, we have a rule that we have to satisfy, and that is that when two reactants approach, all of the filled orbital MOs in one molecule will interact with all the unfilled MOs in the other molecule, and vice versa. It's going to be this crazy, crazy way to think about things. If we really want to be accurate about what's happening in this these two things approach, we have to think not about just this, but we have to think about every orbital interacting. Let me show you what that means in terms of equations. Remember on that first day I said I'm going to give you an equation with three terms that involves electrostatics, that involves sterics, and some sort of mysterious MO interaction mathematical term that looked like this, and there's a summation symbol in there that says you have to add up all the interactions for all the MOs in order to accurately quantify this. What that means is I can't just think about this interaction with sigma star here. I also have to add up the energy I get by doing a deprotonation there and attacking that on antibonding orbital. Simultaneously, as this lone pair gets closer and closer, it's also overlapping with that empty orbital. I also have to think about the electrons in the CH bond coming back and attacking and doing a deprotonation. There's an antibonding orbital that's empty, there's a filled orbital, and you have to think about those interactions, and then you have to add up every single interaction. It just gets crazy. If you want to be accurate about the interactions between filled orbitals and unfilled orbitals, you have to add all this stuff together. And so let me use an MO interaction diagram to depict this. What I want to do is I want to take an axis for energy, and I want to depict one of the reactants, the ammonia, and it's a whole series of filled and unfilled orbitals. I've got filled orbitals for the ammonia and then unfilled orbitals, and I'm not going to draw all of them. We don't need to draw all of them. You could add up the electrons in the Lewis structure there and figure out how many filled orbitals I need. So I'm just going to draw some of them. And let me draw the MOs for the other reactant, the bromomethane. It's also going to have orbitals that are filled and unfilled. I see bonds in bromomethane, therefore I must have some sort of bonding type orbitals and some empty orbitals. So I'm not going to draw all of them. I just want to draw some of them so you can see how crazy this interaction diagram would be. I have to think about the overlap of this, for example, non-bonding orbital on nitrogen interacting with sigma star for the carbon bromine bond, but I also have to think about this interacting with every single other orbital simultaneously. And so the number of interactions is massive and we're not going to be any good at this. Now I want to come back to the form of this equation because there's something in here that greatly simplifies our lives. And that is the fact that the interactions here are dependent on the difference in energies. In other words, when I look at this, I can tell that one of these orbitals is higher in energy and more nucleophilic than the others. And because this energy difference is in the numerator, I will get less interaction energy if I take one of these lower energy orbitals, like this one down here, and try to interact it with one of these higher energy orbitals. None of these are as energetic and as good as this. And as a consequence of that, we can ignore this. And when I come over here and I look at the empty orbitals, none of these orbitals are as easy to add into as this sigma star, the lowest energy unoccupied orbital, so we can ignore those. And usually, we're going to be good at identifying which ones of these is most nucleophilic and which ones is not, which ones are not, and so we're going to ignore these as well. We're going to end up, because this is in the numerator and is only, our overall quantity is only big when the two orbitals interacting are close in energy. We're going to be able to reduce every single interaction like this into the interaction of one filled orbital and one unfilled orbital. And that's like this, right here. It greatly simplifies our lives. Kenichi Fukui won the Nobel Prize in 1981 for recognizing that when the denominator is small, this term gets big. That's the Nobel Prize, right there, for Frontier Orbital there. You could have won that if you had recognized that the mathematical consequences of that. What that means is that if we're interested in the interaction of filled orbitals with unfilled orbitals, then we can approximate and say, we don't have to be exact. We just have to look at the orbital interaction with just the homo and the lumo, and we don't need to think about all these other arrows, just the arrows that matter most. So that's what Frontier Orbital Molecular Theory is all about. Frontier Molecular Orbital Theory says, in simple terms, just look at the homo and the lumo. And so let me just write that down to remind you. So I'm going to stop using the term homo and lumo for the rest of the quarter. I may throw in every once in a while. From now on, I'm going to refer to those as the Frontier Orbital. So the energy frontiers, the lowest energy empty orbital, the highest energy filled orbitals. We're going to refer to as Frontier Orbital. And so we're going to spend the rest of our time using arrow pushing to indicate these interactions that matter most. Very powerful stuff, even though it seems very simple. Okay, so let's go ahead and take a look at the canonical Frontier Orbitals. We care about, I'm trying to convince you to care about the energies of these orbitals. Which orbitals are high in energy? Which orbitals are low in energy? That's our surrogate for asking which electrons are nucleophilic, which electrons are not so nucleophilic? Which things are easy to attack and which things are not easy to attack? So for the rest of this quarter, we're going to talk about six canonical Frontier Molecular Orbital. Molecular orbitals can have bizarre shapes. Bizarre three-dimensional shapes spread out over entire molecules. But when you look at the Frontier Orbital and the shapes that they adopt, there's only six basic shapes to those Frontier Orbital. And I'm going to, let me try to give you the energetic consequences here. I'm going to start off by drawing a P orbital. For example, on carbon. P orbitals are empty. That's like a carbocation or an empty P orbital on a borane. If I take two P orbitals on carbon, I can mix them together to get pi bonds. One will be lower in energy. And one will be exactly the same amount higher in energy. So pi orbitals are filled. That's a pi bond. Relative to that P orbital, pi should be equally lower and pi star, sorry, this should be an anti-bonding orbital, should be equally higher. Now, if I take another orbital and I mix it in with these, in order to get a sigma bond, those are the lowest energy orbitals of all, the lowest energy filled orbitals of all. I can also mix in some S-character and get a non-bonding orbital. If you mix in S-character, that's definitely lower in energy than a P orbital. Those are things like SP3 lone pairs, SP2 lone pairs. Lone pairs are the highest energy filled orbitals that we will ever talk about. Ninety-five percent of the time when you push arrows, your first arrow will start with a lone pair, a non-bonding lone pair with a pair of electrons in one of these n-type orbitals. And then the highest energy unfilled orbital that we can talk about will be a sigma star orbital. Yes, you have a question? Because the other orbitals have to hybridize in order to bond. So in other words, if I draw ammonia, this has a certain hybridization because the nitrogen needed to use a certain type of hybridization to form bonds with the other Hs. If it didn't form any bonds with anything else and was just an individual nitrogen atom, your view of hybridization is totally arbitrary. It's only once things start to form bonds with particular geometries, that's when you can decide on the hybridization. Okay, so what do we have? We've got a total of six types of orbitals. We've got three filled orbitals, three types of unfilled orbitals. These are our canonical frontier orbitals. And we are going to spend the rest of the quarter talking about the interaction of these three filled orbitals with these three unfilled orbitals. There's only nine possible combinations. I basically covered the entire class for you right here with this diagram. So how hard could this be? And what we'll find out is it's immensely difficult and very, very interesting to tease apart reactions into these combinations. So like I said, we're really interested in how high an energy our orbitals are. How reactive this pair of electrons is in this orbital or how low an energy this antibonding orbital is. We care about what things make things high in energy or low in energy, make orbitals higher in energy and lower in energy. So what I want to do is I want to spend the rest of this section on frontier molecular orbitals talking about how does electronegativity perturb the energies of these orbitals? How does bond length perturb the energies of those orbitals? How does conjugation perturb the energies of those orbitals? Because if we can decide which molecular orbitals are high in energy, we know where to start our arrow pushing from. So let's go ahead and start to talk about those trends. I'm going to start off by talking about electronegativity. And we've already covered this. I talked about it in the context of atomic orbitals. So let me go and explain what I mean by this. Molecular orbitals centered on more electronegative atoms are lower in energy. So for example, if I try to imagine a lone pair on carbon, and let's suppose it's a carb anion, I'll put a little minus there just to remind myself. I'm going to compare a bunch of anions. If you wanted to draw that out, you might draw something like this. It doesn't matter whether it's butylithy or some butyl anion or we're not going to worry about what the R groups are that are attached to the carbon. What would happen if I replaced that carbon atom with a nitrogen atom with something more electronegative? As soon as I switch over to a nitrogen atom, I now have one more proton in the nucleus. Nitrogen has one more proton in the nucleus. If you were an electron, you would want to be on the nitrogen orbital. What you can expect is every single orbital that involves that nitrogen atom will now be lower in energy, and that includes the non-bonding orbital on nitrogen, the lone pair of nitrogen. You kind of already knew that nitrogen lone pairs are less nucleophilic and less basic than carb anion lone pairs. And if I switch to oxygen and have hydroxide or alkoxide, again, let me be generic there, I should expect that to be less nucleophilic and less energetic than the lone pair on nitrogen. As I keep going more and more electronegative, I'll continue to see every single orbital that involves that atom drop. And so finally, I'll get down to fluorine, which has no bond or fluoride anion that has nothing bonded to it. And I should expect that to be the least nucleophilic and least energetic of all. So I'm really telling you stuff you already knew. You had an intuition, I'm sure, before you walked into this class, that fluoride anions are less nucleophilic than carb anions. You already knew this effect of electronegativity. I'm just trying to graph it for you in terms of molecular orbitals. So let's go ahead and compare, talk about the impact on anti-bonding orbitals. Let's think about this for a moment. Imagine a nucleophile attacking ethane in order to pop out some sort of a leaving group. How many of you like this arrow pushing? Probably very few. Well, the arrow pushing itself is fine, but as a reaction, that doesn't look very plausible to you. And why is that? Let's go ahead and sketch out what happens as we take a sigma star orbital, like this carbon-carbon one, and we graph that. What happens if I replace this atom here with something else? I'm interested in the anti-bonding orbital here versus some other atom. Your intuition already tells you what's going to happen to this sigma star orbital energy. If I replace this carbon atom right here with something that's more electronegative, like a nitrogen atom, you might think that that empty orbital is easier to add into. If I replace that carbon atom with an oxygen atom and become even more electronegative, then that orbital energy is lower, and it's easier to donate into. And then finally, we get to something like a fluorine anion or a fluorine group. And now you're starting to get to the point where maybe you can imagine popping out a fluoride anion. You may have had all kinds of reasonings or rationales for why it's better with fluorine. Like, oh, fluoride anion is more stable. Then, no, that's completely wrong. The reason why you could potentially push out a fluoride but never a methyl is because the sigma star orbitals get lower in energy when you replace carbon with something that's more electronegative. It has nothing to do with the leaving group, and that thermodynamic aspect to the reaction. So that's why it's this effect of electronegativity on sigma star. That's why you can't do this, but you can do this when this is something, when X is something that's electronegative. Okay, so that's the effect of electronegativity. You knew this stuff already. You just didn't have some sort of MO diagram picture for why that is true. Let's look at the effect of bond lengths on orbital energies. On the energies of frontier orbitals. Let's go ahead and take a look at the effect of making bonds either longer or shorter. In this case, I'll look at the effect of making a bond longer. So I want to sketch out the frontier orbitals that I might associate with a carbon fluorine bond. And I'll draw a sigma star orbital for that carbon fluorine bond, really high in energy. And I'll draw the sigma bond really low in energy. Whoops, sigma orbital. That's filled. I mean, sigma star is always really high. And what I want to do is I want to imagine, if I look at the periodic table, let me sketch out the periodic table here. As I drop down this row from fluorine to chlorine to bromine to iodine, what's happening as I drop from the second row with fluorine to the third row with chlorine is that atoms get bigger and bonds get longer. Orbital interact less effectively when they're farther away. When I go to a carbon chlorine bond, what will happen? And I'll try to distort this and make that look really long because that's the way I'd like you to remember that bond. When I go to a longer bond, the overlap between this carbon and chlorine is poorer and so I get a weaker bond. And because that bond is weaker, it's easier to break. This is the way to represent this graphically. Look at the effect of things being farther away. Poor overlap makes weaker bonds and those weaker bonds we represent like this. That means that that bond is more nucleophilic. If you ever have any reason to use the CF bond to attack something, you can assume that the carbon chlorine bond will be faster to attack as a nucleophile. I'm not saying you'll ever do that. That's totally irrational. But if you had reason to, you could make this comparison. And as I keep dropping down the periodic table to larger and larger atoms like carbon bromine, the bonds get weaker, the electrons in that sigma bond become higher in energy. And the bond becomes easier to break. And finally, I'll get to a carbon iodine bond. This is sigma star, sigma, I didn't intend to have stars here. Sorry, I'm going star happy. Those are the filled orbitals. Yeah, so these are all bonding orbitals. It's the antibonding orbitals that get the star. Now I'm going to make that look really long so you get a sense. I'm exaggerating how much of a difference there is in bond length. So this is the effect of bond length. As you go to longer and longer bonds, the bond itself becomes more nucleophilic, but it also becomes easier to break the bond. So if you ever wondered why it's so popular in organic chemistry to attack alkyl iodides, it has nothing to do with leaving group ability. If you learn some rationale that iodides are better, you totally learned something that was completely fictitious. This is why it's easier to attack alkyl iodides because the sigma star orbital for that carbon iodine bond is lower in energy because the bond is long. And you can take this bond length and apply it to anything you want. Longer bonds are more nucleophilic and less stable. Longer bonds are also easier to break. So let's extend this. Boron is a second row atom like carbon. Boron, carbon, nitrogen, oxygen, fluorine. Aluminum is a third row atom. Which of these do you think is more nucleophilic? These bonds are longer. These bonds are higher in energy. The aluminum, hydride bonds are more nucleophilic. These bonds are way, way, way, way, way, way more nucleophilic. You could have guessed that just by looking at the fact that aluminum is in the third row and boron is the second row. So again, for the same reason that alkyl iodides are more reactive than alkyl fluorides, these are better nucleophiles. This is a better and more reactive nucleophile. And I'll just write EL plus for my electrophile. It could be a carbonyl. If you've ever worked with sodium borohydride or lithium aluminum hydride, you already know that tetrahydridoaluminate can burst into flames when you expose it to alcohols or to water. And borohydride will just dissolve and react more slowly. There are no non-bonding electrons on these two, in a correctly drawn Lewis structure. If there were, those would be the most reactive part. Okay, so let's go ahead and look at the effect of conjugation on our canonical frontier orbitals. I'd like for you, I mean, what you're going to find is very, very quickly, I'm not going to be drawing these MO diagrams anymore. Probably within a couple of lectures, you're not going to see them anymore. But I expect you to be using them to think about structures that you draw. We're going to be drawing large, hairy, complex molecules. And this is the kind of stuff you have to fall back on. Bond length, electronegativity, issues like that. So you notice it was this, when I drew it out, it was this pinching effect of the orbitals as I went to longer bonds. Lumo's got lower in energy, homo's got higher in energy. So we're going to see the same effect as we add conjugation to molecules. And what I wanted to do is I wanted to try to sketch out the pi orbitals for ethylene. There I go with the star again. Okay, so the filled orbital for this pi bond in ethylene, there's a filled orbital. That's the pi bond that you can see, that second bond. And then there's the anti-bonding pi orbital. And now you don't see it. We don't draw those when we draw Lewis structures. And so what would happen if I conjugate that double bond to another C-C pi system? What would happen is I would watch the filled orbital get higher in energy and the unfilled orbital get lower in energy. And as I keep increasing the conjugation, I'll keep seeing this effect increase. Pi will get higher in energy, pi star will get lower in energy. And this is all relative to a P orbital, the energy of just a plain P orbital. And you can continue this on ad infinitum, et cetera. And you'll see this pinching effect. So what's the implication of that? It's pretty profound. This is just like making bonds longer, this idea of pi conjugation. Let's go ahead and compare two molecules. Here's pentadiene. One of them is what, 1-3 pentadiene and one of them is 1-4 pentadiene. And what I want to do is I want to think about these acting as nucleophiles. Maybe if I had some sort of a carbocation or an oxocarbenium ion, I could imagine this pi bond attacking to make a new carbon-carbon bond. We'll talk about those types of prins reactions or electrophilic aromatic substitution later. Which of these is faster? Conjugated or non-conjugated? Now the conjugated one is faster. And the question is why? How could you have known that? I think in sophomore organic chemistry, we tell you something stupid like, oh, it makes an allyl cation and that's more stable. That's not the reason why it's faster. The reason why it's faster is over here. When you add conjugation, you raise the energy of the pi electrons. The HOMO in this, the pi electrons are higher in energy than these pi electrons. That's why this is faster. Okay, now let's look at, I don't know if I'd call this the reverse, but it's almost the reverse. Let's think about something different. Let's think about a nucleophile attacking the pi system here. So here the conjugated system reacts faster with electrophiles. Which of these should react faster with a nucleophile? The conjugated or the non-conjugated? The conjugated should react faster. They both react faster. Nucleophiles and electrophiles react faster with conjugated systems. And again, in your introductory organic chemistry class, you probably made up some idea like, well, it makes a more stable alyl carbanion or something that's just totally wrong. That's not why nucleophiles add faster. Nucleophiles add faster because pi star is lower in energy when you have conjugation. That's the effect of conjugation. It's just like longer bonds. And what's bizarre? And it's not bizarre, but I don't want you to get stuck in this conundrum here. Here's two molecules. They have the same molecular formula. These two molecules, the conjugated one is more stable. Conjugated molecules, molecules with conjugated pi systems are more stable than comparable molecules that have no conjugation. How can it be that the more stable of these two molecules reacts faster with nucleophiles and faster with electrophiles? And it has to do with this diagram and what I'm not showing you. When I conjugate two pi bonds together, what I did not show you is that there's a bunch of other orbitals. There's other, there's actually two different pi orbitals. There's other sigma orbitals. Those get way lower in energy when you conjugate. Overall, this whole molecule is more stable when it's conjugated than this one. But when I just look at the most reactive pair of electrons, and that's all I care about with this arrow pushing, the most reactive pair of electrons, I guess up here, is higher in energy, even though the molecule is overall stabilized. So it's possible to raise the homo in energy even while you're lowering all the other orbitals really lower. Don't confuse this pair of electrons that I'm showing with my arrow with the overall stability of the molecule. Those are two completely different things. These two electrons can become more reactive even though the overall molecule is more stable. And you need to separate those ideas. Okay, so that's all we have to say about orbitals for today. And we still have some time, so we're going to go on to the next topic. But again, I handed out a summary sheet for you that summarizes the effects of things like electronegativity on orbital energies. If those sheets are standing around, you can pass them around. Things like p-character on orbital energies. Things like bond length on orbital energies. Effects of conjugation. And I expect you to be using these when you draw out structures. If you draw out a steroid or an alkaloid or some lithium enolate, you can extend these very simple concepts to very complex molecules. Okay, in the remaining time, we're going to start on the next topic. And that's going to be a discussion of energy. Yeah, so I was worried about that. Yeah, use the black pen. There we go. Okay, so the Gibbs equation. I think all of you are familiar with this kind of an equation. You learned it in freshman chemistry, G-chem. You learned it in organic chemistry. And then you never use it again. And I want to boil this idea down to something very simple. I want you to imagine owing somebody some money. How much of these would you rather owe to somebody? Not obvious. It's totally not obvious. I don't feel like it's obvious to me. I'm a pretty clever person. Let's imagine putting some gas in your tank. How many liters of gas does it take to fill your tank? Or even if I have gallons. Does it matter whether I ask gallons? It doesn't matter whether I ask gallons or liters. You have no idea. You could, if you sat long enough, you could think about it. The point I'm making is that logarithms and exponents are not intuitive. They're not intuitive. They're not intuitive to you. They're not intuitive to me. They're not intuitive to anybody, but maybe a math genius like Rimonogen. It's not helpful for us to use numbers like this. And this is the problem with this equation. That's fundamentally the problem. So I'm going to ask you to not think like this because it's not intuitive. So let's talk about what is intuitive. And I'm going to encourage you to convert free energies into things that are intuitive. For the rest of your time as an organic chemist, I want to ask you to do this. Ratios are intuitive. And more importantly, base 10 ratios are intuitive to you. They're intuitive to you, to me. They are intuitive to my 4-year-old daughter. She can tell the difference between 1 and 10 and which is more. So I want to draw a table for you. I want to think about this. It doesn't matter what the actual molecules are. Any set of reactants and products where we think about an equilibrium. And I want to think about the impact of putting numbers, energies, into this equation over here that involves either natural logs or exponents. And what I want to do is I want to try to equate energies with numerical ratios. So let me make the table. So what I'll do is I'll draw some sort of a delta G value. And then I'll go ahead and draw the ratio of A to B. Whenever you have a free energy difference of some number between starting materials and products or between two transition states, doesn't matter. So I'm going to start off in the middle of this table. When delta G is 0 and the free energy difference is 0, what's the ratio of those two species at equilibrium when the free energy difference is 0? 1 to 1. And I have a very clear idea what 1 to 1 means. My daughter has a clear idea what 1 to 1 means. That's very easy to grasp. Now what's important is what happens is I increase this A to B ratio from 1 to 10. I want you to know what that free energy difference is. And it turns out that that free energy difference is 1.4 kcals per mole. It favors the products in this case, so I'm going to put a negative sign there. I have no idea what, how to think about 1.4 kcals per mole. But I know how to think about 10 to 1. I know 10 is bigger than 1. And I know 10 is a lot bigger than 1. What's powerful here is that if the free energy difference is minus 2.8 kcals per mole, that the energy difference is now 1 to 100. And you get the trend here. I now expect you to know the trend. What happens if the equilibrium difference here is 1 to 1,000, what's the free energy difference? Minus 4.2. I usually can't even remember the point 2. I'm like, oh, minus 4. But we'll put the 4.2 here. And I'm going to stop using kcals per mole. In this class, I always use sort of American kcals per mole instead of kilojoules per mole. Okay, so every time from now on that I give you an energy difference, whether it's an energy difference between starting materials and reactants, whether it's an energy difference between two transition states, I expect you to convert that into a set of ratios in your mind because ratios are intuitive. Oh, this transition state is 1.4 kcals per mole over an energy than this other one. That means it's 10 times faster. That's intuitive. If it's 4 kcals per mole, it's about 1,000 times faster. If the energy difference between starting materials and products is 1,000 to 1, you have a chance of getting over 99.9% yield in that reaction because the equilibrium will favor products over reactions. And we can go the other direction as well. If the energy difference is positive, 1.4 kcals per mole, now it goes the other way, 10 to 1 in favor of starting materials, you don't have a prayer of getting a good yield out of that reaction. Nothing you do can change that thermodynamic difference. And if I add another 4.4 kcals or 1.4 kcals per mole to that penalty, and so now that's really a disfavored reaction by 100 to 1, and you can keep filling. So I expect you to know that 1.4 kcals per mole is a factor of 10, and I expect you to apply that. So free energies are only useful to the extent that you can convert them to ratios. I expect you to be very fast with this 1.4 kcals per mole bit. Okay, let me give you one more energy or one more sort of rule to add on to that. Let me first by writing out the 1.4 rule. So 1.4 kcals per mole is equal to a factor of 10, either in equilibrium constant or in a rate constant. Yes, that's at 298, so I think on the problems that I ask you to do some conversion for a different temperature. Okay, I'm not sure how to describe this. I'll call this the expert rule. Once you're really good at 1.4, you're really good at applying this and doing these factors of 10, I'll give you another rule. And that is at 0.4 kcals per mole is a factor of 2. These are the only ones I really remember. So if I see an energy difference in transition states that's only 0.4 kcals per mole, oh, this transition state is 0.4 kcals per mole in energy lower. That means it's twice as fast. And I'm not really good with 0.4, so when I see 0.5, it's like, oh, that's about a factor of 2. That's kind of the level at which I think. So you'll have some chance to practice with that on the problem set. Okay, so let's go ahead and let's talk about some real examples of this. If you're drawing this on a piece of paper, I advise you just to draw the cyclic ester, the lactone moiety there. That's all we really care about for right now. This turns out to be a clinically used drug. And there are some substituents on there, but that doesn't matter. And what I want you to think about is the reaction of this with water. So if you wanted to hydrolyze this open in order to give this hydroxy acid, then what kind of yield can you expect for this reaction? I can tell you right now that the free energy change is this. And why would you want to hydrolyze that open? It turns out that if you don't, it's not soluble. It has very poor solubility. You can't administer this particular derivative of a Rhino-T-Can as a drug. What's the yield? Very low. It's very low. At equilibrium, the ratio is going to be 1,000 to 1. That's three factors of 1.4. If it's uphill by 4.2 K-cals per mole, you have no chance of doing this. In fact, there is a way to hydrolyze this. What you have to do is not use acid. You do it in base so that this side is not the OH but the carboxylate and then it becomes thermodynamically favorable. That's what they administer as the drug in order to get solubility out of this. So you can't get better than a 0.1% yield if you're trying to isolate the free acid there. So let me go ahead and extend this idea that K-cals per mole give us useful ratios to transition states. Let's suppose you do a lithium aluminum hydride reduction on this and you get a 95% yield of the axial alcohol and in case it's not obvious, there's the hydride group that I just added. This is what you get after workup. There's two possible transition states for this and in one transition state, the hydride adds from the top and in the other transition state, the hydride adds from the bottom to give you this product right here. And so what's the difference in energy for those two transition states for axial attack versus equatorial attack? You can imagine the starting material going through one transition state or the other transition state and the only thing you have perhaps is a TLC plate or some sort of isolated yield, how do you convert this into transition state free energies? What's the maximum amount of the other isomer that could have been obtained in this reaction? 5%. You already know two numbers in a ratio. I've got two numbers. I can make a ratio out of that. That means the ratio is at least 20 to 1, right? I don't know how much of the minor isomer. Maybe it was 1%, but I know it was at least 20 to 1. It might be greater. And so what does that mean? It's a factor of 2 and it's a factor of 10 to 1. How much energy for a factor of 10? 1.4 k cals per mole. How many, how much energy for a factor of 2? 0.4 k cals per mole. That means I know that the energy difference between this unfavorite axial attack and the favorite equatorial attack is at least, I don't know how much greater it is than 1.8, but at least I can give people a meaningful number. So every single time you look at a TLC plate and say, well, that spot looks about twice as big. Or you get a yield and you say, well, I got 60% of this isomer. I didn't get 40% of the other. You can create ratios and when you can create ratios, you can create free energy differences. And it doesn't matter whether it's transition states or equilibrium values between starting material and product and I encourage you to do that. Okay, we're going to come back and finish up these ideas of how to use energy. But I want you to practice this 1.4 k cals per mole. It's immensely useful, right? When you're discussing things with other chemists and you use numbers like this, it's not as powerful. As when you use numbers like that.