 My name is Dave Van Branken and I'd like to welcome you to OpenCourseWare's version of Chemistry 201, Organic Reaction Mechanisms. Over the next 30 lectures or so, I'll be taking you through how we at UC Irvine push arrows in Organic Reaction Mechanisms. And so let's get started. Okay, welcome to lecture one for Organic Reaction Mechanisms 1. Our first lecture is to introduce arrow pushing and I want to invite you to read in the required textbook for the class. It's by Ian Fleming, Molecular Orbitals and Organic Chemical Reactions. I suggest that you read the preface and then skim chapter 1. There's important concepts in there that you will use, but you don't need them to have all of those concepts memorized for this first lecture. The other pair of papers that I would recommend for you to read is one by O'Hagan and Lloyd that introduces the history of the curved arrow in Organic Reaction Mechanisms. And you might want to go back and take a look at the original paper in 1922 where arrow pushing mechanisms were first introduced. Let's go ahead and start off by just reminding ourselves of what this course is all about. Okay, so what is Chemistry 201? It's part of a two-quarter graduate level chemistry sequence at UC Irvine to teach basic Organic Reaction Mechanisms. About 15 years ago, we realized that students were finishing their first quarter of graduate level Organic Chemistry courses knowing a lot of things but not being really good at arrow pushing mechanisms. And we decided to correct that by reorganizing the course content. So in the first quarter of Organic Reaction Mechanisms at UC Irvine, we cover arrow pushing mechanisms, molecular orbital theory, structure, bonding, and energetics. And that's what my course is going to be about. We also teach a second quarter course that covers equally important concepts, pericyclic reactions, radical reactions, confirmations, stereoelectronics, and mechanistic tools. So let's go ahead and get on with the business of arrow pushing and how we see that fitting into a broader picture of Organic Chemistry. And I want to start off with a basic question and that is what happens when you see a mechanism like this and you're asked or a reaction like this and you're asked to write a mechanism for that reaction. Frequently, you'll draw something on a piece of paper and wonder if what you have drawn is correct and so you'll go to find answers to what is the mechanism for this reaction. You can look in the backs of books, you can look on the web, you have so many resources available to you. And unfortunately, if you go to those resources and look for what is the mechanism for this reaction, you won't get a clear and consistent answer. This is one of the most important reactions in biology. It's formation of an emine. It's one of the, of an extremely important reaction in Organic Chemistry in the field of organocatalysis. And if we look at the answer that's present out there, you'll find various different mechanisms. So here's an example of a mechanism that I found and it starts off by showing that this nitrogen forms a bond with this carbon atom. You can see that bond over here. And then in the next step, you can see a hydrogen atom moves from this nitrogen over here to the oxygen. And then there's the interaction of a proton here. But if I look at some other resource, I won't find exactly the same mechanism. Here's another mechanism that is out there for this reaction. And it starts off showing arrows. And that's always valuable, these curved arrows. This one starts off by showing you that you break a CO double bond. And then an electron pair on nitrogen attacks the carbon atom. And then this proton gets lost or this hydrogen atom gets lost. It's not the same as the mechanism that I just showed you. The arrows are in different places. The steps in bonding are different. Here's another mechanism. This one doesn't show the pi bond, the double bond breaking. It shows the lone pair on oxygen attacking the positive charge on a proton. And then it shows you that this CO double bond breaks to give the electrons to oxygen and satisfy that positive charge. And then in the next step, it shows you that this oxygen atom is grabbing a hydrogen. This is clearly different from the mechanism that I just showed you. And if you were looking for the right answer that was going to be graded, you'd want everything to look the same. Here's another mechanism. This one is in color, so it must be correct. So here the nitrogen atom is attacking a carbon and that kicks the electrons from this double bond up to oxygen. And then this negative charge bends over and grabs a hydrogen atom. This is a different mechanism from the one that I just showed you and from the other three mechanisms. So why is it that organic chemists can't seem to agree on the reaction mechanisms for this very important reaction? This is a common question that you'll face among a series of common questions. Why can't organic chemists agree? We all agree this reaction is important even for this simple and fundamental reaction. Worse yet, you might be trying to answer some question on a problem set that you'll be graded on or on an exam and you'll wonder when you try to draw out an answer, why doesn't my mechanism match the answer in the textbook? Is the textbook wrong? Is the answer you just wrote wrong? And more perniciously, if you try to look in the literature, the research literature and you write out some mechanism and you'd like to know whether the mechanism you wrote as plausible is correct, unfortunately, you'll find in many cases that your mechanism doesn't match that printed in the journal article. And the journal article, because it's peer reviewed, has a level of gravitas that makes you feel like, well, my mechanism must be wrong. Whatever I see printed in the literature or in the journal article must be correct. And the way to resolve all these questions is to reduce it down to a fundamental problem. They're all related in a sense. And that is that organic chemists have never agreed on rules for arrow pushing mechanisms. Not since 1922 when they were first introduced, organic chemists have never agreed what they were going to use arrows to represent. And so that's the fundamental problem that leads to all of these questions on the part of people who start in the field of organic chemistry. And so we're going to attempt to resolve this problem. So I want to start off by showing you what are the key features of an arrow pushing mechanism that I expect when somebody says that they're drawing a plausible arrow pushing mechanism. And the first and most important feature of an arrow pushing mechanism is that you need to break a reaction, a chemical transformation into a series of elementary reactions. So I've taken that imine forming mechanism here and I've broken it down into a series of elementary reactions. How do you know that this is a series of elementary reactions? One, two, three, four, five, six. Well, let's go ahead and give a definition to elementary reaction so we're all on the same page. So by definition, an elementary reaction has two properties, an elementary reaction has a single transition state. So if this truly is an elementary reaction step, it has to have one and only one transition state. And it has to match an energy barrier and the reaction coordinate free energy diagram. If I draw out a mechanism that has six steps, that correlates with a reaction coordinate energy diagram that has six energy barriers on the way to products. So elementary reactions have a single transition state. There's another property of elementary reactions. And that is that they're microscopically reversible. And what that means is that if I say that in this step I'm going to use this lone pair to pull off a hydrogen atom, to pull off a proton from this acid, then if I talk about the reverse reaction, I have to use this A minus to pull the proton back off and go backwards. I can't use a different mechanism or a different type of reaction to do that. It has to involve the same transition state. Now we're not going to use this microscopic reversibility property so much in this quarter, but it's still an important property of elementary reactions to keep in mind. So what's the second feature of an arrow pushing mechanism, number two? The second feature, of course, is to add curved arrows, curly arrows to your reaction mechanism. So first we need to know how many steps are in the mechanism and then for each of those elementary reaction steps we need to add curved arrows to show us which bonds are forming and which bonds are breaking. So how do you decide whether the curved arrows that you've drawn are any good? Whether they represent reactions that are likely to happen or not likely to happen. And I want to try to get you to think about chemical interactions when two molecules approach each other, when two molecules are in proximity, how do we assess whether that's a good interaction or a bad interaction? So if you think about the factors that influence non-bonding interactions, this is the kind of equation that organic chemists and inorganic chemists and all chemists will tend to use to think about non-bonding interactions. If you want to assess whether the interaction between two entities is good, there's three terms in this equation and they are all distance dependent in some way. So the first term, the first factor that we think about when two molecules interact is governed by this equation and you'll recognize this as a form of Coulomb's law. You can see there's a distance dependence, so the farther away the weaker these interactions and this is the interaction between two charges. And so we frequently use this to think about interactions between charge groups. For example, you might find these types of two functional groups in an enzyme active site where some aspartate carboxylate side chain with a negative charge interacts with some quaternary ammonium ion or maybe negatively interacts with some sort of sodium ion. These are really informed by this Coulomb's law equation. There's another two terms in this non-bonding interaction equation that are also heavily distance dependent. One of those terms is repulsive. It leads to positive energy and the other term is negative and it leads to some sort of an attractive force. And these are what we refer to as the Van der Waals interaction. So these two types of terms tell us why or help us to explain why and to what extent helium wants to form a liquid. It also tells us why methane wants to be a liquid on the surface of the moon titan of Saturn or why methane might want to have some positive interaction of methane. These are good for quantifying these types of interactions. Unfortunately, this equation is not very useful to organic chemists. So organic chemists and throughout this course and throughout your career you should think about a different equation to quantify even in an empirical way the way molecules interact. So if you're really interested in chemical reactions that form covalent bonds then we use a different type of equation that has three different terms that are also distance dependent. And so the first term in that equation you'll immediately recognize as this Coulomb's law term that we had in the previous equation. So if we want to think about the interaction of this alkoxide with T-butyl bromide, this is ready to form a covalent bond. And you could use this equation to judge how good the interaction is with this negatively charged oxygen in this carbon atom here with a partial positive charge. And it would explain to you why oxygen doesn't want to attack bromine because that has a partial negative charge. It would also tell you because of this term right here that there are repulsive interactions to consider. So when we draw these out, each of these, each of these out, it's easy to see that the negative charge on oxygen might be attracted to this carbon that has a partial positive charge. It's easy to see that as this ethoxide anion gets closer and closer maybe this carbon atom here might start bumping into these methyl groups. And then finally if we really want to understand the outcome of this reaction it turns out that this reaction is going to generate a double bond. It's called an E2 elimination reaction. We have to somehow turn to this third term. This third term tells us why you form covalent bonds. This third term tells us that the oxygen, maybe even if it wants to form a bond to this carbon atom will instead choose to pull a proton off of this carbon atom and generate a C-C double bond. So we have to think about this third term here. And we have to think about what arrow pushing is going to mean to us and how we're going to use it. So let's go ahead and come up with some sort of an agreement. We're going to use arrow pushing to represent something. And we have to decide whether we're going to use arrow pushing to represent charge interactions, whether we're going to use arrow pushing to represent repulsive interactions or whether we're going to use arrow pushing to represent attractive interactions like this. Now I think we can all agree that charge is an obvious concept. We reinforce this concept with notions like, oh, opposites attract or familiarity breeds contempt. I would argue that we don't need to use arrow pushing to represent the interactions of charge. I think this is intuitive even to non-scientists, this idea of opposites attracting and like charges repelling. And I would say similarly that sterics is also an obvious concept. So these repulsive interactions, if you've ever used a hammer to drive a nail into a board or set a book upon a table, you know that the hammer doesn't fly through the nail even though it's mostly empty space. You know that a book doesn't pass through a table. And these are all concepts that are completely obvious. We don't need arrow pushing to demonstrate those concepts. We need arrow pushing to help us understand the things that are hard to see. And so there I would argue that it's the third term that's not so obvious. And if we look in a little bit more detail at this third term we can see these abbreviations here that stands for occupied orbital interacting with unoccupied orbital. We can see some terms here that reflect the difference in energy of those orbitals. Coefficient on the atoms, overlap of orbitals. This third term is mysterious and may not be familiar to you. And it involves things that we don't see and that is molecular orbitals. When I drew those molecules for you for a thawkside interacting with t-butyl bromide, nowhere in there did I draw the molecular orbitals. And so let's use curved arrows to represent this. This is the part that's not obvious. And this is where we really need help. And ultimately it's this term right here that governs the formation of covalent bonds. And that's really what we want to explain with our arrow pushing mechanisms. I want to take you back to a book that was published in the 1970s and it really guided my generation of chemists in thinking about reaction mechanisms and arrow pushing. It's called Frontiers, Orbitals and Organic Chemical Reactions by Ian Fleming. Not the Ian Fleming who wrote the James Bond series. This is Ian Fleming the chemist. And midway through this book here on page 49 is a discussion of electrophilic aromatic substitution. You may have drawn out mechanisms that involve these types of arenium-ion intermediates. And buried within this discussion is an incandescent statement and I want to draw your attention to that. And it starts off with curly arrows. Really a justification for the use of curly arrows. And so let's just look at that in more detail. The statement says curly arrows when used with a molecular orbital description of bonding work as well as they do simply because they illustrate the electron distribution in the frontier orbital. And let me stop there. They work as well as they do. I want my curved arrows to work for me. I want my curved arrows to predict for me which reactions are good and which reactions are bad. And Fleming continues by saying, and for reaction kinetics it is the frontier orbital that is most important. I want my arrow pushing to tell me which reactions are fast, which reactions are slow, which reactions are plausible and which reactions are implausible. And Fleming is telling us how to do that. He's saying that your curved arrows should illustrate the electron distribution in the frontier orbital. And so we're going to turn that into our mandate for arrow pushing. And our mandate for this class, this is going to be our central cannon for mechanistic arrow pushing. So let me re-craft that statement into a general principle that we're going to use for the rest of this quarter. And that is that we're going to use curved arrows to depict the interaction of filled orbitals with unfilled orbitals. Every single curved arrow you draw will represent that. So for example, this set of curved arrows that we've drawn here, you may have drawn mechanisms like this many, many times. And when it's worked well for you, it has worked well because it represents the interaction of filled orbitals with unfilled orbitals. So when you draw this SN2 displacement reaction, what you're really saying is that the electrons in this non-bonding lone pair, this frontier orbital, are interacting with the sigma star orbital for this carbon bromine bond, this anti-bonding orbital. And of course, when you put electrons into an anti-bonding orbital, you break the bond. So you've been doing this already and didn't know it. You've been using this principle and maybe you didn't know why your arrow pushing worked so well for you. Now it's important for me to say what arrow pushing is supposed to mean, but let me give you something that's just as important and that is what does arrow pushing not mean? So let's rephrase this. Do not use curly arrows to depict the motion of atoms. So it is true that this nitrogen gets closer to carbon. That's not what this arrow is meant to represent. And it is true that this bromine atom is moving farther and farther and farther away from carbon, but that's not what this arrow pushing is meant to represent. Just as important as saying what curved arrows are meant to represent, it's important for me to say what curved arrows do not represent. Curved arrows are not meant to represent the motion of atoms. It is certainly true that the nitrogen is getting closer to this carbon atom and it is certainly true that this bromine is moving farther away from carbon in this reaction, but that's not what these arrows represent. Likewise, we don't want to use arrow pushing to represent the interaction of charges. And this is one of the fatal flaws that you'll see often in the use of arrow pushing to depict reaction mechanisms. So this, we now have our mandate. We have a central cannon that we can follow when we draw reaction mechanisms. So now we're ready to take complex chemical transformations and reduce them down to a series of elementary reactions and use arrow pushing to depict what's going on in each of those steps. Now I want to start this course in a way by giving you seven basic rules for arrow pushing mechanisms. We have one central cannon that says that arrow, my name is Dave Van Branken and I'd like the interaction of filled orbitals with unfilled orbitals, but we need to have some guidelines for how to apply that simple cannon, that simple rule for arrow pushing. So we're going to break this down into seven independent rules that we need to follow in order to make sure that we're using arrow pushing in the most effective way possible without misguiding ourselves. So let me start off by giving you rule number one and that is draw correct Lewis structures. When you draw organic molecules, they need to be correct. So let's start off with the first subpart of this rule and that is when you draw correct Lewis structures, you need to obey the octet rule. This is fundamental and it's a major driving force in many, many organic reactions. And let me just remind you about the application of the octet rule and let's start off down here by looking for this at the structure for methane sulfonic acid. If we were to draw the correct Lewis structure for methane sulfonic acid, you'd immediately see that there's six bonds to this sulfur atom. This sulfur atom has more than eight electrons and so it might look like it violates the octet rule. But we need to remind ourselves that the octet rule only applies to second row atoms, lithium, beryllium, boron, carbon, nitrogen and oxygen. And 90 to 95 percent of our reactions will mainly be carbon, nitrogen and oxygen. So we won't think about beryllium, sometimes we'll think about lithium. But these are the atoms that obey the octet rule. Sulfur is a third row atom. It's below oxygen and so we don't have to worry about the octet rule for sulfur or for selenium or for transition metals or for phosphorous or for silicon or anything below that second row. Okay, part two of our rule is draw every substituent on every atom except carbon. Now we need to make sure that our Lewis structures are correct. Oftentimes you'll see organic compounds drawn out using a keyboard typed out like this. Now when you try to translate this into a Lewis structure, even though it looks like this carbon is bound to that carbon and that's bound to the oxygen, this would be completely incorrect. It doesn't obey the octet rule and it doesn't show all the attachments of the atoms. In fact, this carbon atom here needs to be attached to that carbon atom. And as soon as you draw the Lewis structure like this you'll realize that there's something wrong. This is the chemical formula for acetone and we need to have drawn a carbonyl group on top of this carbon. We need to show every single charge on our atoms. So here's an example of a stable molecule called trimethylamine and oxide. These types of compounds are important in Polanovsky rearrangements, cornblum oxidations. Now you may be tempted, infinitely tempted to want to form a bond between this oxygen and nitrogen. To somehow use this negative charge to satisfy that positive charge. But if you choose to take this electron pair and move it down to nitrogen, you've committed a fatal mistake. You've just violated the octet rule and implied that nitrogen now has 10 electrons and that's absolutely impossible. So remember to obey the octet rule and remember to include the charges on your structures. Those charges can be, can serve as an important guide for you in predicting reactivity. So here's a molecule called tert-butyl isonitrile. Isonitrile's tend to smell very bad, horribly bad in fact. But nitrile's have the interesting property that they can be nucleophiles or electrophiles. And if you don't draw the charges for this isonitrile group with the bonding, it may not be completely obvious to you that that carbon atom is nucleophilic. And it may not be completely obvious to you in what way this acts as an electrophile. And so including those charges can help you get started when you start doing arrow pushing. Even if you don't start your arrows with the charges, it'll remind you that this carbon atom is the nucleophilic component in this reaction. And then finally, I've drawn it in bold here and in capital letters. This is the most important part of this. And that is draw your lone pairs. Draw the lone pairs. 95% of all organic arrow pushing starts with a lone pair. So when you're confronted with some reaction like this where this alkene reacts with the bromine-bromine double bond, the reaction doesn't start with formation of an oxygen-bromine bond. So why should you draw the lone pairs on that oxygen atom? Well, the lone pairs on the oxygen atom explain why this double bond is so reactive. So if you don't draw this lone pair donating into this pi star orbital making this alkene very nucleophilic, you'll wonder why this reaction is so much faster than all other bromination reactions of pi bonds that you have seen. And when you draw the lone pair there, it reminds you that this is going to be an extremely fast reaction driven by the nucleophilicity of those oxygen lone pairs. Okay, so it's important for us to follow these basic rules about drawing correct Lewis structures. Now let me go back and step back for a second because you're going to be tempted to misapply this simple rule. And that is you'll be tempted to use dative bonds for arrow pushing. Let me remind you of what dative bonds are. The term dative means donor, donor bond. And this is frequently a representation system that's used in organometallic chemistry. So when you see an organometallic complex like this nickel-tryphenylphosphine complex, these complexes use dative bonds. This is the tradition in organometallic chemistry. You cannot use this nickel-phosphorous bond for arrow pushing. It's a dative bond. This is the structure of diborane. When you buy boron as a chemical it actually comes as a dimer and there's a hydrogen atom that is bonded to two different boron atoms in this dimer of borane. These are dative bonds. If this were a regular Lewis structure we'd have to have a positive charge somewhere on this structure to show that the proton has too many bonds. And if I don't see that charge there I know this must be a dative bond, each of these bonds. We cannot use these bonds for arrow pushing. Here's a donor bond in a pi complex between bromine and an olefin and you get a sense that something is wrong here by the fact that this is dashed. If you try to use this for arrow pushing you're going to come out with the wrong charges. Let me show you what happens if you try to use dative bonds for arrow pushing. Let's go back to this nickel-tryphenylphosphine complex and I've highlighted this dative bond here in red. According to this structure, if this were a correct Lewis structure and this phosphorous really were neutral, if I take these electrons away from nickel and give them to phosphorous, that neutral phosphorous has to pick up a minus charge. And conversely the nickel has to become positive. These are the wrong charges. If you use dative bonds for arrow pushing you'll end up with the wrong charges on your products. If you want to use arrow pushing with molecules that have data bonds, dative bonds, you have to convert the dative bond into a charge separated illage structure. So the correct charges and a correct Lewis structure would have a positive charge on phosphorous because phosphorous has four bonds. Three phenyls, one nickel. The nickel has one bond too many. It has to have a negative charge. And now when we put the correct charges for this correct Lewis structure, when we take the electrons away from nickel and give them to the positively charged phosphonium phosphorous, that will now have the correct neutral charge. So be very careful when you see structures that have dative bonds because you can't use those dative bonds for arrow pushing. Okay, let's look at rule number two of our seven rules. Make arrows start with bonds or lone pairs. It's a very simple rule. What we're really saying is that every arrow, these curved arrows that I've drawn in red here, every single arrow must start with a filled orbital. And these three things, lone pairs, pi bonds and sigma bonds are simply representations of filled orbitals. Arrow pushing is supposed to represent the interaction of filled orbitals with unfilled orbitals. So here the correct arrow pushing for this interaction of a hydroxide anion with acid aldehyde with this carbonyl carbon would have my arrow start with a lone pair. That's a representation of a non-bonding orbital. And here's a representation of a pi bonding orbital donating into an antibonding orbital for an HO bond. And here's an arrow that starts with a bond. This represents a sigma bond or a sigma bonding orbital. So arrow pushing needs to start with one of these, your arrows need to start with one of these three types of features of a correctly drawn Lewis structure. So just to come back to this corollary to this, just remember that arrows do not start on charges. They start with lone pairs, pi bonds or sigma bonds. We already said that the interaction of charges are important in organic chemistry, but we're not going to use arrow pushing to represent that. Iodide anions are nucleophilic, and we all understand that the negative charge is important for that. But if you want to draw the correct structure for this, you need to draw the lone pair on iodine attacking this carbon atom. Likewise, we come over to this hydroxide case. It's not the negative charge that forms a bond with that carbon atom. It's the electrons in this one of these three lone pairs that forms the bond. And it might look very compelling to you to take borohydride anion and add the negative charge to this partially positively charged carbon atom. If you do that, if you choose to do that, you'll end up with a structure that is wrong at every single level of detail, violating the octet rule, being incorrect, not having the correct charges. So you have to avoid this temptation to use arrow pushing to represent the interaction of charges. Never start arrows on charges. Charges don't form bonds. Electrons form bonds. And there are electrons in lone pairs, in pi bonds and in sigma bonds. Okay, finally, arrows do not start on atoms. And I'm going to use language that will make you think that they do. I'm going to say, oh, that bromide pops off. That's not to say that your arrow pushing should show that bromide popping off. That's not what arrow pushing is supposed to represent. I might say, oh, you add a proton to that carbon atom there. But that's not what your arrow pushing is supposed to represent. Your arrow pushing is not supposed to represent the thing that I say. It's supposed to represent the interaction of filled orbitals with unfilled orbitals. You should start your arrow pushing here with this double bond and use it to attack the proton. Even if I say you add the proton to that carbon atom, don't use the arrow pushing to represent things that I say. Okay, rule number three. I just told you where arrows start. Now let's talk about where arrows end. Arrows end on atoms or bonds. And this is sort of a troubling rule because arrows are supposed to represent the interaction of filled orbitals with unfilled orbitals. And unfortunately, even when you correctly draw a Lewis structure, we don't represent unfilled orbitals with Lewis structures. And so when I make a rule that says arrows should end on atoms and bonds, it's because these correlate vest with the three canonical types of unfilled orbitals. And there are three. There's empty P orbitals like carbocations. There's pi star orbitals when you attack a double bond. And there are sigma star orbitals when you break sigma bonds as in an SN2 reaction. So let's take a look at the correct ways to end an arrow. Here I've got an arrow ending on an atom. Here I've got another arrow ending on an atom. So here what I'm representing is that this lone pair is donating into an antibonding orbital, a sigma star orbital for this H chlorine bond. Here I'm starting my arrow with a lone pair. And I'm ending on an empty, on an atom. But that atom is representing an empty P orbital on this carbocation. Here I have my arrow ending on an atom. And it's very common in organic chemistry in these kinds of reactions to show the formation of a double bond. And to have this arrow terminate in the middle of that double bond. It's a way to show that we're about to form a CC double bond. But there's an alternative representation that I would encourage you to use that's much more rarely used but is more correct. And that is to show that when you attack this hydrogen atom here, this proton, that those electrons donate into this carbon atom empty P orbital. If you make this arrow end on that carbon atom and not in the middle of the bond, it's more obvious that you understand that there's an empty P orbital on that carbon atom. And that that's what's driving that reaction. Okay. As we said before, we have to mention this idea not to use arrows to represent the interaction of charges. So arrows do not end on charges. They never end on charges. Fifty percent of the time you will end up with the wrong result. If you're wondering what will happen when you take a methyl anion with this very unhappy protonated hydrofluoric acid, you might be tempted to make this minus charge attack the positive charge. If you do that, you're going to take a fluorine that has too many bonds and give it even more bonds than it wants to have. In fact, the correct answer here is that you want to have the lone pair on this carbon atom attack the proton and give the electrons to fluorine. So don't make arrows attack charges. Charges don't form bonds, electrons form bonds. So once again, we don't have arrows terminating an empty space that we're using our arrows so that they should end on atoms or bonds. So even though the bromide pops off and floats away, we're not trying to show that the bond goes nowhere. We need to show that bond ending on bromine. So it should look like this and not simply flying away into empty space. Every arrow should end on an atom or on a bond. Okay, let's look at our fourth rule. And this is really a guideline. It's not governed by any fundamental principles of quantum mechanics or thermodynamics. It's really a guideline to keep you from making mistakes. And the three arrow rule, obey the three arrow rule, says don't draw elementary steps, elementary reaction steps with more than three curved or curly arrows. If you did that, it means you probably didn't break the mechanism into elementary steps. And you might think that this will never happen to you, but let's take a simple example where you do a transamination reaction, an aminolysis of a cyanomethylester. These are very easy to form amide bonds with cyanomethylesters because these are easy to attack. So you probably can see that ultimately you have to form a bond between nitrogen and carbon. You know that nitrogen is a strong nucleophile. And so it's very reasonable to expect that the lone pair on this nitrogen might want to attack that carbonyl carbon. And when it does, you're going to kick the electrons up to the oxygen. And now it's getting exciting because we can see that that's going to help pop out this very good leaving group, this cyanide stabilized methoxide leaving group. And when we use those electrons to push back down and kick out that leaving group, we know that we have to pick up a proton, here it is right below here, in order to generate this alcohol product that I showed. Now the problem with this is that this is wrong in so many different ways. First of all, we're implying that this carbon-oxygen bond is plucking off this proton. And that's not what's plucking off this proton. It's a lone pair on oxygen. More importantly, we're implying that all of these things are happening simultaneously. When we have all of these arrow pushing, all of these curved arrows in a single step, it implies that they're all concerted and simultaneous. And that's not what happens in this reaction. More likely, this reaction involves six elementary reaction steps. Now frequently, you'll have cases where you don't need to show all of the steps. As you gain more experience in organic chemistry, and you want to show mechanisms to other organic chemists where the mechanisms are well understood, then you can take a shortcut. And the shortcut can be used like this. So if we want to show a tautomerization reaction, to show how in the presence of an amine base that this ketone might tautomerize to this enol, it's a two-step mechanism. And it's very well known by organic chemists. And if you wanted to show somebody what's going on, maybe you don't need to show what's going on in both steps. And if you feel like it's totally obvious to the listener or to the reader, then we can do this. We can simply stack arrows on top of each other to tell the reader there's two steps and only show the arrow pushing for the first step. Don't layer on top of that arrow pushing for the second step because that implies it's concerted. But if you have a five-step mechanism and you don't want to show all the steps, simply stack the other five arrows here to show that you've left out five of those reaction steps. But to convey to the reader that you know that there were five steps involved and you simply chose not to show them. Okay, so what's another issue with the three-arrow rule? It's not a rule that's governed by quantum mechanical principles. Once you leave this class, once you leave this course, Chemistry 201, you are free to violate the three-arrow rule. It's just a safety device to prevent you from trying to overlay too many elementary reaction steps on top of each other. Here's an example of a reaction that I would show using four arrows. But there is always a way. And what this reaction, what this series of arrows is showing you is the reason this parole is so nucleophilic is because this lone pair on nitrogen is donating into the pi star orbital here. That pi bond is nucleophilic. It's donating into this pi star orbital and making this double bond nucleophilic. And that's why it so quickly attacks this aceliomion. Now, you can always obey the three-arrow rule. The way that you would show the arrow pushing for this and still satisfy the three-arrow rule is to start by drawing a resonance structure. We can draw a resonance structure for this parole that shows why it's so nucleophilic at this carbon atom. And we can do that with three arrows. Now we can use this resonance structure for our key carbon-carbon bond forming step to show how we do this electrophilic addition process. So you can always manage to follow the three-arrow rule as long as you use resonance structures as an intermediary. Number five, don't draw term molecular reactions. Don't draw term molecular elementary reactions. In a term molecular elementary reaction, three molecules simultaneously react in an elementary step. So if we look at the, oh, I was supposed to, oh, yeah, okay. Rule number five, don't draw term molecular reactions. And more importantly, don't draw term molecular elementary reactions. So here's a very common reaction that's used to add methyl groups to phenolic OH groups. And the most typical conditions involve use of potassium carbonate as a base in DMF. And you'll be tempted, because you know the potassium carbonate is a base, you'll be tempted to draw a reaction in which you deprotonate this phenolic group to generate an alkoxide, and then you use those electrons that are being released to form a bond to this methyl iodide. And if you draw it like this, you're implying that all three of these molecules are colliding simultaneously with perfect alignment such that the lone pair on this base is perfectly aligned with the antibonding orbital on this HO bond, and that at exactly that point in time, this is also colliding with this carbon atom of methyl iodide in order to make a new oxygen carbon bond and displace the carbon iodine bond. Most collisions between reagent molecules do not lead to a chemical reaction. Most collisions between organic molecules do not lead to formation of bonds and cleavage of bonds. Most collisions are unproductive. So the chances that any three molecules will collide simultaneously with the right energy with the right alignment is inconceivable. There are no elementary reactions with a single transition state where three molecules simultaneously collide. That's not plausible. The better way to think about this reaction is that the base pulls this proton off. Proton transfers are fast, extremely fast, and extremely facile. Don't be afraid to draw this pre-equilibrium where you pull the proton off and generate a phenoxide anion. Manfred Eigen won the Nobel Prize for showing how fast these kinds of reactions are, and now this phenoxide anion in a second elementary step can now collide with this methyl iodide in order to make a new oxygen carbon bond. So when you draw a term molecular elementary reactions, you're missing the fact that this reaction has two different transition states. Rule number six, H is always attached to something. You'll hear me discuss mechanisms and talk about proton transfers, hydride. You'll hear me talk about hydrogen atom transfers. When I say that, I'm referring to functional groups, not to free species. There is no reaction that involves a free proton H plus. There is no reaction that involves a hydride anion. There is no reaction that involves a hydrogen atom. These species have no role in solution phase chemistry. Maybe if you fly into the center of the sun where the energies are incredible, that's where you might find species like this, free protons with no electrons associated with them. H dot or a hydride anion. But in a typical organic reaction or in a biochemical system, there's no role for these kinds of species. What's the way you're supposed to represent this? Well, instead of having a lone pair attacking H plus, there is no such thing. The H was attached to something before the oxygen attacked. The H is attached to something after the oxygen attacked. And in the transition state, the H is attached to two things. But at no point in time is H plus a proton just dangling in free space. When you look at hydrogen atom transfers, it is true that this key teal radical will pick up a hydrogen atom, but it doesn't pick up a free hydrogen atom. It will pluck a hydrogen atom off of another molecule. And we represent that with these kinds of fish of caros, and we'll discuss the use of those more in chemistry 202. And then finally, we're going to talk about many, many hydride reduction reactions. But there is no such species as free hydride. Hydride is a functional group that's attached to reagents such as this hydridoborate species. So remember to always make sure that your hydrogen atoms are attached to something. Don't leave a simple H atoms behind. So in other words, when you draw the deprotonation of this oxonium ion to give a neutral carbonyl species, don't just leave that H plus dangling off in space. Use something else to pluck that proton off of the oxygen. You will very frequently see in the literature and in textbooks that people violate this rule flagrantly because they don't want to commit to what is acting as the acid in their system. They'll frequently draw H plus. And so let me give you an alternative to that that's more correct, because we all agree that there is no such reagent as H plus. And the day that you can go to my lab or your lab or any laboratory and find a bottle that contains just H plus, then I will allow you to draw this reagent. Okay, instead of doing that, let me invite you to use the following alternatives that are much more sophisticated than this. So instead of writing H plus, which is just a symbol, why not write HA? It does not take significantly more time to draw an A than it takes to draw a plus here. And in doing this, you're showing the reader that you understand that the H is attached to something else, and then you can use that HA bond in your arrow pushing. If you feel like you have to draw H plus, you're compelled for some reason, put quotation marks around it to show that you understand there is no such thing as H plus, and you're just using that as a symbol. That'll prevent people from thinking that you don't know just basic principles of chemistry. Okay, we have our last rule to consider that we're going to use for arrow pushing, and that is avoid proton transfers through four-membered transition states. And you will very frequently be tempted to violate this rule. Don't draw intramolecular proton transfers if they involve four-membered transition states. So let's take a look. Here's an intermediate that you might end up with if you're cyclizing an amino ester onto itself to make a new carbon-nitrogen bond. So it's very easy for amino esters to cyclize to make lactam rings. And as soon as you make a new carbon-nitrogen bond, you need to explain various different aspects to this proton transfer reaction. Ultimately, you'll want to transfer this proton either to this O minus or to the oxygen atom over here on this methoxy group. And I want you to avoid the temptation to reach over with this lone pair and pluck off that proton. The transition state for that would involve four atoms, oxygen, proton, nitrogen, carbon, and would be square shaped and would be absolutely implausible relative to the alternatives. Likewise, if you draw the tautomerization of this enol into a ketone where this proton ends up on this carbon atom, I want you to avoid the temptation to use this carbon atom to pluck off that proton. That violates fundamental principles of orbital symmetry if we claim that this is happening. And it wouldn't be just slow, but for typical reaction geometries, this would be utterly impossible. So what do you do instead to avoid this temptation? So when you do acid-catalyzed proton transfers, what I'd like you to first do instead of using four-member transition states is I'd like you to first decide if your reaction conditions are acidic or basic. And then once you've decided that, you can use symbols to show how protons can be transferred back and forth to these symbolic reagents. If you think that this tautomerization of an enol to a ketone is occurring under acidic conditions, even slightly acidic conditions, then symbolize an acid as HA. Instead of moving this proton directly to that carbon atom, why don't you simply use this carbon atom to pluck off a proton from your acid? Those reactions will be very fast, faster than any transition state that involves a four-membered ring. And then in your second step, you can use your A minus that's left over the conjugate base from your acid to pluck the proton off this oxonium ion. This two-step transformation is the real mechanism for tautomerization, and you don't have to go through a four-membered transition state. If you think the reaction conditions are basic, so under basic conditions, you have species like B to act as your base with a lone pair, and the protonated conjugate acid of B, BH plus, to act as your acid. So if you want to explain how a proton gets from this oxygen, for example, over to this nitrogen atom, do it in two steps. Use your symbolic base to pull a proton off of this, and then use the lone pair to pull the proton back off onto the nitrogen. You pluck the proton off of here, you re-deliver it to the nitrogen, and you do it in two steps, and you avoid a four-membered transition state. So it means that you'll be drawing more structures when you draw mechanisms, but it means that you'll be drawing the correct mechanisms. Okay, so that's the seven rules that we need to obey when we do mechanistic arrow pushing. So rule number one, draw correct Lewis structures. This is important for second row atoms, carbon, nitrogen, oxygen, boron, even lithium. Rule number two, make arrows start with bonds or lone pairs. These are representations of filled orbitals. Sorry, that should be, yeah, sorry, bonds or lone pairs. These are representations of filled orbitals, of non-bonding lone pair orbitals, of sigma bonding orbitals, or pi bonding orbitals. Rule number three, make arrows end on atoms or bonds. These are our best representation. They're not perfect, but they are our best representation of unfilled orbitals, of empty p orbitals, of pi bonds, anti-bonding orbitals, or of sigma anti-bonding orbitals. Rule number four, obey the three-arrow rule. Don't have more than three curved arrows in an elementary reaction step. Rule number five, don't draw termolecular elementary reactions. Most collisions are not successful at forming or breaking bonds. It's very unlikely when two things collide that they will form a bond, it is virtually impossible to have three species simultaneously collide and form a bond. Rule number six, H is always attached to something. And this is particularly important to remember for proton transfer steps. So use symbols like HA and BH to represent species that are transferring protons. And then finally, avoid proton transfer through four-membered transition states. It's always faster to pluck a proton off and re-deliver it or vice versa than to transfer a proton directly through a four-membered transition state. All right, that's it for the first lecture. And I'll see you during the next lecture when we introduce the idea of molecular orbitals, atomic orbitals, and hybrid atomic orbitals, and how they can be predictive in allowing us to apply these various rules.