 Okay, I wanted to start off by just considering a very simple reaction here where you have some sort of alkoxide and they have some sort of chloro-alkane and you try to make an ether out of that. This is one of the first reactions you learn in sophomore organic chemistry. It's making ethers using the Williamson ether synthesis. So I've got two possible products here and I want you to, I'll call them A and B and I'd like you to try to assess which of these you think forms faster and which of these things you think forms more slowly. Or what do you think is going to be the major product in this reaction? So the other byproduct is chloride in this case. It's the anion chloride for each of these cases. This is definitely an irreversible reaction. The chloride is not going to displace the ethoxide anion to regenerate the starting materials. And in cases where reactions are irreversible or they don't go backwards, you get the product that forms fastest. The product that forms fastest is the major product in the reaction. So in this case, what we expect, I hope it is your intuition that you'll attack here on the allylic carbon faster than you'll attack here on the sp2 substituted or the sp2 hybridized carbon atom. And I want us to be able to analyze reactions in terms of reaction coordinate energy diagrams. So we're going to start incorporating a new type of diagram, yes, in rare cases, yes, generally the conditions are very hard to sway one way or the other. But I'll give you an example where that's possible. Okay, so in irreversible reactions, you get the product that forms fastest. Let's try to graphically depict this idea. And I want to start off by pointing out that I'm going to show you a diagram with a completely different type of y-axis. So so far in this course, we've been talking about molecular orbital energy diagrams where I have some sort of an energy axis, E, and I look at the energies of molecular orbitals, and I've got these little energy lines. This is completely different. I'm not graphing molecular orbital energies here. I'm graphing free energies of starting materials or free energies of products. You measure free energies in K-cals per mole, and that means you can convert those into numerical ratios. 1.4 K-cals per mole on this axis equals a factor of 10. Those other molecular orbital energy diagrams, those were in electron volts. You don't convert those into factors of 10. So when we write reaction coordinate diagrams, there's always this mystical sort of x-axis called the reaction coordinate. Normally that means something like a bond distance. As oxygen gets closer and closer and closer and closer to carbon, you could consider this to be a carbon oxygen bond distance if you wanted. But we generally don't worry about putting units on there or assigning it. So let's go ahead and diagram this reaction. I'll draw the starting materials right here sort of in the middle, and I want to show that this has two possible directions that it can go. So in this case, we can have it going in one direction over here to generate some products. But there's also an alternative reaction pathway. And so one of these directions will lead to product A, and one of these directions will lead to product B. Which of these sides here goes to, which product am I depicting over here? Yeah, this is product A. We said that we like this because we're attacking to do a regular SN2 reaction on an SP3 hybridized center. Our intuition, I hope your intuition is that this should be faster and this should be slower. And the reaction that we believe should be faster should have the lower energy transition state. So I'll put A here and I'll put B here. And so let me go ahead and put a little energy line there. Now if I take an NMR, a crude NMR for this reaction, probably what I would see is I would see this product in there. Who knows what the yield would be? 88%, 94% doesn't matter. The important point is if I don't see any of this in my NMR, just based on the limits of detection by NMR, I can probably say that the ratio is greater than 99 to 1. Regardless of what the total yield is, if I can't detect any of this, I can normally detect about 1% or maybe sometimes less depending on how much signal to noise you have. If I see a 99 to 1 or some, all I really know is greater than 99 to 1, if I can't detect that minor isomer, it's at least a 99 to 1 ratio. What does that allow me to say about the free energies of these two transition states? Just by looking at an NMR, a crude NMR. It's at least two factors of 10 in favor of this. It's at least 2.8 k-cals per mole greater or lower in energy to go to product A. And we give this difference in free energies between the transition states, this designation, delta, delta G double dagger. Here's where that comes from. We label the energy to go to this difficult transition state as delta G double dagger. Double dagger refers to a transition state. That's the transition state energy to go to product B. And I'll put a little B subscript there. And then we would label this transition state energy as the difference in energy between starting materials and product A. And there's that little double dagger to tell me it's the transition state. And the difference between those two delta Gs, this is a delta, this is a difference in free energies, the difference between these two differences we label as this. This is a very common phrase. And all we know is this is greater than 2.8 k-cals per mole. We don't know if it's 4, we don't know if it's 7, we don't know if it's probably not 100 k-cals per mole. But this was what correlates to a greater than 99 to 1 ratio, a greater than 100 to 1 ratio. So we can construct a pretty scholarly diagram just by looking at a crude NMR. And I would expect you to be able to do that. So remember, as long as we're drawing reaction coordinate diagrams and measuring things in k-cals per mole, I want you to be able to convert these types of free energy differences into ratios. And I want you to be able to convert ratios into free energy differences and draw reaction coordinate diagrams. Which of these is more stable? Let me draw, I've got so many things drawn here. So let me just sort of focus your attention on that. Which of these two products is the more stable product? B is more stable. It's more stable by a lot. What's the important point here? I'm going to write it out so we don't forget. And what I ought to do is I ought to tell you more boldly about half the time less stable products form faster. So if you came away from sophomore organic chemistry thinking that more stable things form faster, you completely missed the point. You completely lost it somewhere. When they were giving you some special rule. About half the time less stable things form faster. So we really need to separate the stability of the products from the rates of chemical reactions because they don't say the same things. So just to come down here, we can now modify our diagram. I'm just going to pull this board down a little bit. What we're saying here is that product B, which forms more slowly, is actually more stable than product A. And we don't know how I didn't give you a value for that, but I'm showing that with my diagram. It has to do with resonance. There's resonance donation between these lone pairs and pi star, and you don't have any of that resonance in the allillic ether. Okay. So rates of chemical reactions are super important. So I'm going to draw another reaction coordinate diagram for a hypothetical reaction. And I'm going to label this for a reaction where product or starting material, this is different from what I showed over there. Now I'm talking about some hypothetical reaction where A goes to B. And what I want to do is I want to draw this thing called a transition state and just ponder on the fact that there isn't really something, there is no species called the transition state in the reaction. The transition state is some arrangement of atoms where you've got halfway bonding, half of a bond that's half formed and half broken or a set of bonds that are half formed and half broken. You could never isolate anything like that where bonds are half formed or half broken. So this is a total fiction. There isn't really any species called the transition state at any point in time in your reaction. It's just this fiction. But you can pretend that there is. Any time I can measure a difference here in K-Cal's per mole, I can convert that into numerical ratios. Even though there isn't anything called a transition state, if I pretend there is, I can now measure an energy difference here and we'll call this delta G double dagger going to the transition state from A. As long as I can measure an energy difference in K-Cal's per mole, I can convert that into factors of 10, into numerical ratios. Let's do that. Let's draw out the equations that are necessary to make that work. So this would be the rate equation for that chemical reaction. We'd have rate, reaction rate is equal to something called a rate constant times the concentration of A. If you have 10 times more A, the reaction is 10 times faster. If you have 1 tenth as much as A or if you had 10 times more solvent, the reaction is 10 times slower. Concentration matters. If the rate constant is 10 times bigger, your reaction is 10 times faster. The rate constant is usually correlated with all this arrow pushing that we do. All this arrow pushing, all this stuff that we're doing in Chem 201, arrow pushing is related to rate constants. When you get to chemistry 202, you'll talk about reaction kinetics and start incorporating concentration into those ideas because they obviously influence the rates of chemical reactions. Okay, so how do you convert the free energies I have over there to numerical ratios? Here's the equation that you use. It's equal to a numerical ratio of rate constants. And the numerical ratio of rate constants has to do the rate constant for the reaction versus this other mystical thing. So what is this telling you? It's telling you if you have some sort of a delta G value here, and I'm just going to arbitrarily assign a number here to 23 k cals per mole for that kind of an energy barrier. So if you have an energy barrier here of 23 k cals per mole, there's two different pathways. You can go this direction, which is not very fast, or you can go the opposite direction where there's no energy barrier. And that's what k double daggers. It's when there's no energy barrier, when you're just falling off of this energy hill, how fast is that? So this is equal to the fastest possible reaction when there's no energy barrier. And we can put numbers on that, and I'd like you to know what those numbers are. What is the fastest possible rate constant? There's some limits on that. We'll get to that in just a second. But one thing that I want to do is note that there's temperature in this equation and that you can use that in the laboratory. Temperature is a part of this equation for every reaction you run. And you don't have to know anything about A or B here. And you can make some good rule of thumb estimates for how temperature is going to affect your reaction. And here's the rule of thumb that I'd like you to use in the lab. Excuse me. So for most reactions that you run in the lab, if you turn your thermometer, if you turn your heating bath up by 10 degrees, you should expect your reaction to get twice as fast. If you go up by 20 degrees, 4 times faster. If you go up by 30 degrees, 8 times faster. Now you're starting, wouldn't you like to get your PhD 8 times faster? You are not, when you read some procedure in the literature and you're following that procedure, you don't have to sit there and be a victim of the way the procedure was written. You have the choice to raise the temperature on your reaction and make your reactions go faster. And it roughly follows this rule. It's actually a pretty good rule for most of the reactions that occur between minus 78 and 100 degrees, which is most of the reactions that you all run. So it's a very good rule of thumb. Now the downside is your side reactions will also go faster. And sometimes they go a little bit faster than the one you want. But this is just a simple, powerful rule that you can apply. If you're worried that, gee, I don't want to have to wake up at 2 a.m. to do my work up, well, turn the heat up. And then you can work the reaction up at 5 p.m. and then go home for dinner. So a simple rule of thumb and it comes from this equation right here. Okay, so let's talk about this fastest possible rate. What is the fastest possible rate constant when there's no energy barrier if you're just falling off this energy hill? We're going to put some limits on that. I'm very frequently in this class going to mention things related to rate constants. And I want you to know the range of values that rate constants can have. Now, unfortunately, and maybe this is good for us, just to keep our lives simple and during this quarter when we're just doing a lot of arrow pushing, you're almost never going to know anything about K. The absolute value of K. I'm very frequently going to give you relative rate constants. This reaction is 10 times faster than that one. I'm not telling you how fast either reaction is. I'm just telling you that one of them is 10 times faster. This reaction is 10 times slower. That reaction is twice as fast. So I'll give you a lot of information about relative rate constants, but you're rarely going to know the absolute value for the rate constant for any elementary reaction. Let me go ahead and put some limits. If you see anybody spout out some rate constant that doesn't fall within this range, you need to question the validity of that. So let's take three different cases of elementary reaction processes. So for unimolecular reactions, and that could be something like an electrocyclic ring opening where two molecules don't have to collide with each other, it's just one molecule that sits there and does something suddenly. That's a unimolecular reaction. You'll find that the slowest possible rate is somewhere, and I'm going to be a little bit fuzzy on this, and we use per seconds. It's like a frequency. 10 to the minus 8 times a second. That's how fast this bond might break if it was very unfavorable. So that's about the slowest possible rate. And it correlates with the year long time scale. If you run through the calculations, we typically don't care very much about reactions that take several years, right? You're not going to write anything in your thesis about reactions that take several years. So if somebody says my reaction occurs with a rate constant, a unimolecular rate constant, a 10 to the minus 10, you might want to ask them how they measured that on the time scale of an NIH or an NSF grant. Okay, so the rate constants for unimolecular reactions, things where two things don't have to collide with each other are usually greater than that. And they are always in every single case slower than this value. This is the fastest possible rate constant, and that's that K double dagger. That's the fastest possible rate constant for any unimolecular process. And what is it that sets that limit? No, this can be moving all, you're getting close to it, but this can diffuse around faster, slower. It's the vibration of bonds. No bond can break faster than it vibrates. If it's vibrating this slowly, it can't suddenly be out here faster than it was vibrating. The two atoms can move away from each other only as fast as they are vibrating. So this is the vibrational frequency. That's the frequency of things like CC and CH stretches in your IR spectrum. If you convert in wave numbers or whatever you're using in your IR spectrum, it correlates with this. That's typical CC, CH bond vibration frequencies measured in Hertz. Okay, so let's talk about bimolecular reactions like that SN2 reaction that I showed you before with the thoxide attacking an allylic chloride. So here we have different units now. Notice in the units we have a molar measurement number. That's sort of reminding you that concentration matters. If I add 10 times more reagent, my reaction will go 10 times faster. If I add 10 times more solvent, it's more dilute. My reaction will go 10 times slower. So this is just your reminder that concentration matters when you require two things to collide with each other. Two molecules to collide. When you require an athoxide to collide with an allylic chloride. So in bimolecular reactions, your rate constants will always be greater than this. And again, this correlates with sort of year-long time scales. You don't care about bimolecular reactions that take years to operate. What's the fastest possible bimolecular rate constant? It's this, something on the order of 10 to the 9th per molar per second. What is it that sets fundamentally the limit on the fastest possible rate of a bimolecular reaction? It's diffusion. Things can't react faster than they diffuse towards each other. And this number is a little bit soft because different solvents have different viscosities, right? If you did your reaction in molasses, then the diffusion would be very slow. Whereas if you do a reaction in ether that is very low viscosity, things diffuse around in the solution very quickly. So for a bimolecular elementary reaction, the fastest possible rate is limited by diffusion. So when you look at these numbers here, these transition state energies, it matters whether you're thinking about unimolecular processes or bimolecular processes, this sort of rate ratios that you get here. And so this is diffusion and this is bond vibration. Okay, and if somebody gives you some number that's outside that range, question them. Say, oh, wait a second. I thought that diffusion was limiting the rates of bimolecular reactions. How can you come up with 10 to the 11th for a bimolecular process? Okay, we don't have any examples of termolecular reactions. All of our elementary reactions are either unimolecular or bimolecular in organic chemistry. Let's suppose you promised your advisor that you were going to have that alkyl bromide ready by tomorrow. I got to strip that silo group off. This is a tert-butyl dimethyl silo group. I got to strip that silo group off and then run on an appell reaction with triphenylphosphine and carbon tetra bromide, convert the alcohol to the bromide. I got to get that silo ether off. And then I can convert that into the bromide. Pretty typical reagent would be pyridine hydrofluoric acid. I mean, you can buy that sort of a salt complex. It's a source of fluoride. Let's suppose just before you started class today, you set up your reaction. And let's go ahead and draw out our TLCs for this reaction here. That's pyridine. It's a salt you make by mixing pyridine with hydrofluoric acid. It's not really important for the, it's milder than just plain HF because you have pyridine. Okay, so let's suppose at 8 a.m. you set up your reaction and here's your silo ether. I'll just write starting material. And then down here, just before you came in to class today, it's 10 a.m., let's suppose you TLC'd it again. You ran in and got your TLC and now you see something that looks like this. So that's 10 a.m. What's your reaction going to look like at 12 o'clock at noon? This is the fundamental question. And the important point is that rate constants affect how you organize your time. Do I have time to get that thing worked up by this afternoon and get the next reaction started? What if I won't have time? What can I do to make it faster or slower or better organize my time? Oh no, I can't go anywhere this afternoon and I have to stay and keep TLC'ing my reaction to find out what's going to happen. I look at this and I can make some pretty strong guesses about what's going to happen with this reaction. And what I want to do is I want to just graph out the reaction progress as I measure the amount of product that's forming in this reaction. And so let's just go ahead and start off by drawing out 8 a.m. at the start of the reaction. I'll just write a.m., p.m. And what I want to try to do, so here's 10 a.m. and we're interested in what's going to happen around noon time. So I've already showed you what's happening at 8 a.m., it's all starting material. You just set up the reaction. Here's 100% product and 0% product is down here at the bottom. And so at the beginning of your reaction you have 0% product. How much product do you have at 10 a.m.? About 50%. I'll just roughly graph that here. Here's my 100% mark. And so the question we're asking is how much product are you going to have at 12 o'clock? So here's what I would love to see and that's totally wishful thinking, right? If this is what you're thinking is going to happen, then you don't understand rate constants and the relationship to half-lives. The way to think about this is not in terms of rate constants, you need to start thinking in half-lives. What this tells you, what that TLC is telling you is that the half-life of the reaction is about two hours. If you go another two hours you will use up another half of the leftover starting material. So 50% of my starting material is left over, I'm going to consume half of that in the next two hours. So how much product will I have at 12 o'clock? You'll have 75%. And then at 2 o'clock in the afternoon, now I have 25% starting material left over. How much am I going to consume in the next two hours? Yeah, now it's going to be another 12% so now I'm not very good at math, it's going to be about 88%. And then I will go another two hours, another half-life and I've used up another half of the remaining stuff. And then I keep going out here. What you'll find is that it takes about eight half-lives and I'm going to abbreviate that with T1 half. It takes about eight half-lives to reach completion and I mean really complete to reach greater than that. Now usually you could work it up at 99% completion and be perfectly happy with the yield but, you know, we try to be even better than that. It takes about eight half-lives for reaction to reach completion. In other words, reactions don't go like this because most reactions are first order in starting materials and that red pin is not doing it. Not every reaction but most reactions follow a first order dependence on starting material. That means the rate of the reaction gets slower and slower and slower as I have less and less starting material there. So again when you set up a reaction you can predict what state the reaction is going to be in at three o'clock in the morning. Do I really need to come in and work that up at three in the morning? No, there's no chance that that's going to be ready because after eight hours it was only half done. If your reaction is half done after eight hours, how long will it take to reach completion? Days, don't come in and TLC it every hour. You'll know very, very quickly what the course of that reaction is going to be. You're not doing work by TLCing reactions that are obviously going to take days to run. So use this information. There are a few reactions that follow zero with order dependence on starting materials. Those reactions are rare. There are some organic catalytic reactions that obey those rates. There are a few that involve second order dependence on starting material but most reactions have a first order dependence on starting material concentration. Okay, so super important idea, this idea that half lives are really what govern our day-to-day activities in the laboratory and you need to think in terms of half lives. So don't think in rate constants. Even though I'm going to tell you, oh this reaction is 10 times faster, that one is 10 times slower, when you're organizing your time think half lives because you look at a clock in the laboratory and that's measured in minutes and hours. So for typical unimolecular reactions where one thing goes to one thing, there's a pretty useful way to convert half lives and I should say is approximately, well, I'll give you the exact number. There we go. So here's the way to convert rate constants, those things measured in per second into half lives. In seconds, it's this simple equation, natural log of 2 over K. Now I'm too stupid to know what natural log of 2 is, I just know it's something like one point, I don't even remember what it is, I won't even try. So here's what I do. There we go and that's close enough every single time for me trying to get things done and estimate how long things are going to take. So the half life is approximately the inverse of the rate constant and that's a good enough. For bimolecular reactions where there's two different reagents, that's only true if both of your reagents are present at one molar. Now usually you run reactions with things at about half molar so you'll be a little bit off for bimolecular reactions. Okay, so it's important to think in terms of half lives when you're in the lab doing stuff and organizing your time. Let's change our gears here. We've been talking about kinetic reactions where the rate of the reaction determines the product selectivity and I want to switch around now and talk about equilibrium reactions. And I'm going to give you an example of and the mechanism for this is a little bit complex because it involves a lot of proton transfers. I'm not going to draw out the mechanism for you. This is peritoluene sulfonic acid. It's frequently used as a catalyst for forming and hydrolyzing acitals. And if you do a hydrolysis reaction in water at zero degrees, there's two possible products that you'll get or that you can see and it depends on the conditions for the reaction. So in this product both the methoxy and the ethoxy groups have been hydrolyzed off. So you can see there's a mixed acetal here and you can sort of imagine that being a carbonyl group. And so if you hydrolyze those off you get this lactone, this cyclic ester but there's another product that you can get which is this ethyl ester and you can get either one of these depending on the reaction conditions and that comes from hydrolyzing off this methoxy group or sorry, hydrolyzing or cleaving this carbon oxygen bond in the ring and then hydrolyzing off the methoxy group. Now the product ratio depends on the conditions of the reaction. If you stop this reaction after 20 minutes and I always advise you to start TLCing your reactions as soon as they start because if you always wait until 10 hours because the procedure said 10 hours before you take your first TLC you have no idea what happened in the reaction. Maybe it was done in three minutes and you should have worked it up and then it just started decomposing after that. Here's what you see after just 20 minutes. You see that the main product for the reaction has cleaved the ring and the methoxy group has been hydrolyzed off but if you wait 12 hours, if you just let it sit you get a totally different result. You can't even detect that ethyl ester in there. It's all lactone. What has happened in this reaction? All we've done is we've changed the time and the conditions are exactly the same. This is a case where this minor product that you see over here is the result of an equilibration of the starting materials of the products under the condition of the reaction. It turns out that the products equilibrate. There's a huge entropic advantage to having this alcohol in the presence of an acid catalyst form an ester through a fissurous stratification reaction and I'll draw the acid catalyst there but we don't need to talk about the mechanism. There's six steps in the arrow pushing mechanism. The point is that under the conditions of the reaction this is occurring and if you wait long enough it will go to the thermodynamically more stable ring product here. So this is what we would call a reaction that's under thermodynamic control. If you stop the reaction very quickly this identifies for you the kinetic product, the product that forms fastest which tells you something about transition state energies. It tells you this is forming initially much faster than this other product but then we know that these can equilibrate. So immediately we can draw some sort of an energy diagram based on these ideas and what we know from this final result here, see this greater than 99 less than 1, what's the difference in free energies between those two products? I'll call one of them the closed ring over there I'll call the lactone. And then this other one we'll call the hydroxyester. We don't know the difference in free energies but it's at least a 100 to 1 ratio therefore it's at least 2.8 kcals per mole and initially the hydroxyester was forming at least 20 times faster like 595 is like 20 to 1, right? This was forming at least 20 times faster. So we know that and maybe none of this was forming, maybe all of this occurred through some equilibration reaction. So we don't know maybe the difference in free energies here was 1,000 kcals per mole but we know it was at least not a factor of 10 but a factor of 20 which is greater than 1.4 plus 0.4. That's a factor of 2 and this is a factor of 10, 1.4 kcals per mole is a factor of 10. So we know that the difference in transition state energies going that favored this was at least 1.8 kcals per mole, lower in energy. Maybe none of this formed directly, maybe it all goes to this first and then it equilibrates over to that. Okay, so it's important to look at numerical ratios. You need to always be willing to and quick to translate those into free energies that you can put in a diagram. Just from looking at a TLC plate you can start to draw out free energy diagrams. By estimating is it 10 to 1, 1 to 1, equal amounts? That's very powerful stuff. Okay, so how can you learn about just reaction rates? How can you learn about rate constants for reactions? Which reactions are fast and which reactions are slow? I think you guys recognize this probably as a Deals-Alder reaction. It is amazing how pathetic cyclohexenone is as a Deals-Alder partner. It's one of the worst Deals-Alder reactants imaginable. So here's two closely related Deals-Alder reactions. Here's another one where they had some extra substituents in there and they only reported this one in the presence of silica gel. So for some reason adding silica gel accelerated that. And I can make some guesses but that's not the point of this. So in this reaction they get a 29% yield and this reaction with silica gel you get a 64% yield. And so which of these is the faster overall reaction? This is usually what you see when you read papers. It's stuff like this. And the answer is you can't possibly know which is the faster reaction. You can't know which is the better reaction. Here's the information they didn't give you. Because if they were heating that first one at 200 degrees to make it work because it's so crummy. It looks good until you realize that I mean how often have you run a reaction at 200 degrees? That's obviously a slow and crummy reaction. And you don't know that until you look at the temperature. That's what tells you it's slow and crummy. If we look at this one over here they told you the temperature actually it's 25 degrees Celsius. But until you read the fine print you don't realize what's going on when you read the experimental they let this thing sit for five months. The point is that you will not learn about reactivity. If you spent your entire time in sophomore organic chemistry or some introductory chemistry course hoping to learn about which reactions are good and which reactions are bad. The only way you can say about which reactions are good and which reactions are bad is not by looking at yields it's by looking at rates and times. Or sorry temperatures and times. So I'm going to give you mechanisms, complex ones. And you're going to be asking yourself gee which way should I start pushing the arrows? Is this way good? Does this look like something that ought to be fast? Well that depends have you seen something similar in the literature that occurs quickly? And the only way to know which things are quick are things that occur at low temperatures and rapid reaction times. So again it's not sufficient to look at yields in fact those tell you very little about which reactions are fast and which reactions are slow. So there's very few tabulated listings of actual reaction rates. You need to pay attention to those other details. Let me show you how to extract some information. This is called a Mukayama Aldol reaction, enol silolithors, this pi bond is very nucleophilic and attacks the carbonyl. And the mechanism is not what I'm getting at here. I don't think that's important for this for the point that I want to make. The point that I want to make is if you're reading the paper, you need to home in on this. They cooled the reaction to minus 78 degrees when they did it. And what that tells you immediately is that every single arrow pushing step in this mechanism must be fast. If there were any slow steps in the reaction mechanism, they couldn't possibly have cooled this down to minus 78 and had it work. So when you draw the mechanism, you know every step in that mechanism is fast and every time you draw a similar mechanism on your paper, those steps will be fast if they look like the steps in this mechanism. This is immediately giving you a wealth of information about which kinds of reactions are fast and which kinds of reactions are slow. Okay, so for reactions you cool, that means every single elementary step in the mechanism must have been fast. Conversely, for reactions you heat, nobody heats reactions because it's easier than letting them sit at room temperature. You heat reactions to make slow, crummy reactions, fast reactions. And so if you're heating up a reaction, something was slow. So for example, when you do a fissurist verification reaction and we look at the temperature and the time, they had to boil this. I'm sure they would have boiled it hotter if ethanol had a higher boiling point. But that's the boiling point of the solvent. And it's still slow, 12 hours. So these are conditions for making an ester. I'm not going to draw the mechanism for this fissurist verification. You all should have learned that back in sophomore organic chemistry. It involves six elementary steps. What this tells you, the fact that they had to heat this is that at least one of the steps, the high energy step in the mechanism was slow. At least one of the steps in the mechanism was slow. And you don't know which one, but at least one was slow. And usually you can make a guess as to which of the elementary steps was the slow step in the reaction. Okay, so the important point is from now on, if you haven't been already, you cannot be satisfied simply with yields. I expect you from now on to look at temperatures and times. And I actually have a time here, it was one hour. But usually you don't cool reactions to minus 78 and let them go for four days. That's very inconvenient. When you cool reactions to minus 78, they take less than a day usually. So from now on, I want you to pay attention to temperatures and times because they're telling you about whether the elementary steps in the mechanism are fast or slow. And you need to keep that information in the back of your mind. Okay, so finally on when we come back on Friday, we're going to start doing some more arrow pushing. Yes, two reasons. So there may be side reactions that occur and maybe temperature can affect the ratios of side reactions. It's exothermic and if things start to overheat, at least in this case, things will go very badly. So titanium tetrachloride can cause problems if you allow them to generate heat at their own pace. So it usually means they're too fast and usually you're trying to get selectivity. Okay, so Friday, finally we're going to start talking about some reactions and enough of the background stuff and we'll begin with some carbocation stuff and some arrow pushing.