 So today I'd like to finish our discussion of the reaction of aldehydes and ketones with various nucleophiles and this is basically the content of Chapter 21. So we had started by talking about basic nucleophiles and reviewing how basic nucleophiles including very basic nucleophiles like alcohol nucleophiles and hydride reacted with aldehydes and ketones. Then we went on to other sorts of what I'd say were moderately basic nucleophiles, things like amines where the pK of the conjugate acid was on the order of 10 or 11. And now I'd like to conclude by talking about weakly basic nucleophiles, things that are basic formally like alcohols and water but where the pKa of the conjugate acid is so strong that there it's very acidic and then conversely your nucleophiles are very weakly basic. So just to give you an idea of the theme that I'm talking about if you have some generic aldehyde or ketone like so and maybe I'll draw in my lone pairs just to complete our structure you can think of this aldehyde or ketone as kind of moderately electrophilic. In other words nucleophiles like to attack the carbonyl group but they need to be moderately good or very good nucleophiles. So I'll kind of label this as moderately electrophilic. Now just as water just as an alcohol is a base, a weak base, so is the oxygen of a carbonyl group and you can think of the oxygen of a carbonyl as being able to be protonated. And if you protonate the oxygen of this ketone or aldehyde then the situation changes because at this point you have a full positive charge in the molecule. In other words when we were thinking about aldehydes and ketones we thought about it and recognized that oxygen's more electronegative than carbon and so you could say well there's a very minor resonance structure with a full negative charge on oxygen and a full positive charge on carbon but it's only a minor contributor. As we go to a protonated carbonyl now we come up with a different situation because now you've got a full positive charge. The oxygen, the carbon atom becomes very electrophilic. One way we can think about this is to recognize that although often we'll just draw one resonance structure there's a second resonance structure and in this case our second resonance structure now doesn't have any negative charge in it unlike we had when we had to separate charges on the ketone so it's even more of a contributor and now you have the positive charge on oxygen, on carbon. It's still a minor resonance structure but it's a bigger minor resonance structure, a greater contributor to the reactivity. So while we will maybe just draw one picture very often in the back of our head we'll be keeping in mind that there is this component of reactivity embodied in the second resonance structure that makes this extra electrophilic. And often we will refer to this species, I don't think your textbook uses this name but sometimes I will refer to it as an oxo-carbenium ion. Carbenium ion is a fancy name or a more proper name for carbocation which you learned about in previous quarters and oxo-carbenium ion just tells us there's an oxygen substituent on there because the oxygen can donate in you really don't have very much of the carbocation character if I had to put numbers on it I'd say it's sort of 80-20 in terms of the components but nevertheless you can think of there being a little bit of this resonance structure in with the other and so collectively those two structures will make up a picture of the molecule. All right, this structure here is oxo-carbenium ion really helps embody a lot of the reactivity that we're going to see today and in fact we're going to see reaction upon reaction in which this theme is continuously the same. So let's start by talking about a reaction that occurs and for the most part we don't even pay attention to it. If you have an aldehyde or a ketone again I'm just going to write this sort of generically as R-C-O-R and you have water present let's say you dissolve it in water dissolving acetone or dissolving acid aldehyde in water. You have an equilibrium that in many ways you don't see in many ways it's invisible because if you distill your product out you never end up seeing this you just get back your acetone or your acetaldehyde but you have an equilibrium with this species here we're now instead of having a carbonyl group you have two hydroxy groups. Water has added to the carbonyl group. This species is called a hydrate kind of an obvious thing to call something to which you've added water. Sometimes you'll hear it referred to as a geminal diol. Gemini you might recall is the zodiac character for the twins so geminal just means that there are two hydroxy groups on the same carbon, twinned hydroxy groups if you will and I'll just emphasize that this is not a stable molecule with certain exceptions in other words it's not something that you can put into a bottle in pure form. So you have this equilibrium in the presence of water the equilibrium will lie somewhere to the right or to the middle or to the left depending upon the exact nature of the carbonyl compound. I find it very instructive to take three examples. Your textbook uses the same examples and I think they really embody the idea here. So if you have acetone or indeed just about any ketone unless you have something truly exceptional, oops I should be drawing methyl groups here, if you have acetone or just about any ketone unless you have something truly exceptional would say like a trifluoromethyl group that equilibrium lies far to the left in other words when you dissolve acetone in water you have mostly, mostly, mostly acetone and just a little bit of the geminal diol just a little bit of the hydrate. The ratio at equilibrium is 98 to .2. For acid aldehyde and indeed for aldehydes in general that equilibrium lies more in the middle and so for acid aldehyde the equilibrium lies at 42 to 58 in other words almost one to one. Now I look at these two numbers and I find this really instructive to keep in my head, not the details in the numbers but this gist that the acetone that equilibrium's way to the left acid aldehyde is way in the middle and in a way what that's telling me the way it's influencing my thinking is saying whoa that means that the carbonyl and acid aldehyde is like a thousand times like 10 to the third more electrophilic than the carbonyl and acetone. The carbonyl and, the carbonyl and acid aldehyde is much less happy being a carbonyl. It's much more willing to accept water as a nucleophile and the main reason for this is electronic. The main reason we talked about electron donation into the carbonyl group, the methyl groups donate electrons. You can think of this either just thinking of methyl groups or electron donating or thinking about hyper conjugation thinking back to carbocations and remembering that resonance structure where you have the charges separated in your ketone and saying wait a second this means here we have methyl donating in you can also think of it in a molecular orbital approach thinking about the orbitals of the CH groups overlapping with the pi star molecular orbital. Anyway it's about a thousand fold difference if we take away this remaining methyl group and there's only one aldehyde in this category. The aldehyde is formaldehyde. Formaldehyde really, really, really is unhappy in the aldehyde form. It prefers to exist as the geminal diol form and again the number from your textbook or indeed from anywhere else is 0.1 to 99.9. In other words now we've gone another three orders of magnitude in the direction of favoring the hydrate. And if you want to put this in energetic terms remember for every order of magnitude in equilibrium at room temperature you have about 1.36 kilocalories per mole. So you can sort of think of this in an energy diagram where an aldehyde is like 5 kilocalories per mole more high in energy or more reactive than a ketone and formaldehyde is like another 5 kilocalories per mole just to give you sort of a ballpark on the numbers. All right, at this point I want us to be able to look at how we get at the mechanism because this is going to become a big theme here at the mechanism of how we get from the aldehyde or from the ketone to the geminal diol. So what we're going to do and this is going to be an archetype for all the other mechanisms that we see today. What I'm going to do is write out the acid catalyzed mechanism of addition of water. The main point here is that acetone is not on its own super electrophilic. In other words, acetone and water is not on its own super nucleophilic. So what really speeds up this reaction, what really makes it go is some acid. You can add some acid to the water. If you do this in neutral water, of course, you should remember that even neutral water has a little bit of acid present because neutral water has 10 to the negative 7th molar hydronium ion. So you always have, unless you're in basic solution, you always have some small concentration of H3O plus. And we'll just imagine adding a few drops of aqueous acid, say aqueous HCl to the water. So the first step of this reaction is an equilibrium, Bronsted acid base equilibrium. And for the moment at least, I'm going to be good about writing curved arrows. Later on, we may get a little faster or a little bit sloppier. But we can envision electrons flowing from the lone pair on the acetone to one of the protons on the hydronium ion. They're all the same even though I may write one or the other, and electrons flowing back on to the oxygen atom to give us protonated acetone plus H2O. So if you want to get a feeling for where this equilibrium lies, we know what the pKa of hydronium ion is. The pKa of hydronium ion is the strongest acid you can have in water. pKa of hydronium ion is negative 1.7. Acetone's a bit less basic than water is. This equilibrium is going to lie a bit to the left. You're going to have protons going on and off. The pKa of protonated acetone is about negative 7. And in general, my rule of thumb is if within about 10 orders of magnitude you have a difference of pKa, protons can transfer back and forth. Our equilibrium is going to lie way to the left, but you'll have a significant and sufficiently appreciable concentration of protonated acetone for it to react. All right, let's move on to the next step because what happens in this step is we've bumped up the electrophilicity of the carbonyl by protonating it to make the oxacarbonium ion. Now, at this point, water is good enough nucleophile to attack. So here's our protonated acetone. Here's our water. Electrons flow from the thing that has electrons to the thing that wants electrons from the nucleophile to the electrophile. They flow onto oxygen. Again, we have an equilibrium like so. And finally, protons come off, protons go on. You have 55 molar water. Mass action basically dictates most of the time the protons are going to be on the water instead of on that hydroxy group. And so our final step in this mechanism, in this three-step mechanism, is a proton transfer to water. And again, it doesn't matter which proton I ride or which lone pair I use over here. And so our final step of the equilibrium is that we have our geminal diol and our hydronium ion. And if you look collectively at these three steps, you really see the embodiment of catalysis. In other words, the catalyst is taking intimate part in the reaction. In the first step of the reaction, the hydronium ion that's catalyzing this reaction is being consumed. And we know a catalyst isn't consumed or destroyed in a reaction, which means in the last step of the reaction, it doesn't have to be the last step, but in this case it is the last step. In the last step, we're getting it back. In other words, the entire balanced equation has been embodied in the addition of one molecule of water to the acetone, but that reaction has gone easier in the presence of a catalyst. I could also write a mechanism for this reaction that was base-catalyzed, and your textbook does it. Hydroxide is a good enough nucleophile to add to the carbonyl. You would envision hydroxide adding to the carbonyl to give an oxyanion, the oxyanion putting up a proton from water to give you back hydroxide. Now, technically, I could write an uncatalyzed mechanism. In the case of acetone, that mechanism probably doesn't occur to a heck of a lot. In the case of formaldehyde, which is remember about 10 kilocalories per mole, about, let's say, about six orders of magnitude, more electrophilic, formaldehyde maybe you can have an uncatalyzed mechanism where water attacks the formaldehyde. But we're going to skip over that because basically, that sort of is the exception, and we really set up the theme for the class. The theme is basically a good nucleophile like hydroxide or amine or methyl group, methylithium or hydride. Can attack an unactivated carbonyl, a weak nucleophile like water or alcohol generally requires protonation of the carbonyl first. If you add too much acid, would you protonate the OH group and dehydrate? Now, we are going to see an interesting mix of chemistry coming up. And so, right now we've learned two species that are, can be present. We've seen our acetone and we've seen our geminal diol that's not stable, and we've seen a protonated species along the way that's a, two protonated species that are very high in energy. And we said in acetone, we have 0.2% when you dissolve acetone in water, you have 0.2% as the geminal diol. Later on, we're going to see that there are other species present also in small amounts. So, what you will see, I think it's in the coming chapter, it may be in the one following that, will be this species here, the acetone enol. That will enter into our consideration. This, too, is unstable and higher in energy like the geminal diol. Now, the practical implications of this. This is the stuff, including a little bit of acetone enol. This is the stuff that's going to be present in a water solution of acetone, or even with a little bit of acid. It won't take a lot of acid. You actually, even just with the 10 to the negative 7 molar H3O plus in water are going to have all three of these present in equilibrium. Later on, we're going to start to learn about new reactions like the aldol reaction. And you're going to see that if you really cook the acetone or indeed if you cook other sorts of ketones and aldehydes together, that they will start to react in carbon-carbon bond forming reactions that can be promoted either by base or by acid called the aldol reaction. It'll be one of the biggies that we're learning about soon. And that will go either through the enol or in the case of base catalysis, the conjugate base, the enolate. In practice, if you take acetone and expose it to certain types of acid long enough, you will find species where two molecules of acetone have come together to form the aldol product or even under special conditions, three molecules to form a benzene product, 1-3-5-trimethylbenzene called mesitaline. And you can catch your TAs in their office hours to discuss how three molecules of acetone can come to mesitaline. All right. Well, there's something else that's very, very cool about this reaction. And we know it's an equilibrium. So we know that there's a reverse reaction. Equilibria have forward or reverse reactions. And I want to give you a big, big overarching principle here. And I'll give it a fancy name in a second. The reverse reaction must go through the exact same pathway that the forward reaction went through under these conditions but in reverse. In other words, in acid, the reverse reaction, which we're about to see in a second, must go through the same mechanism. And there's a fancy name for this idea. It's the principle of microscopic reversibility. But like so many things that have fancy names, they really come down to very, very simple ideas. So if we look at the reaction, I'll take the reaction we had here but indeed this is universal so we can take any reaction. Imagine a reaction energy diagram. You've seen these in earlier chapters of your textbook. In other words, something where we have energy on the y-axis and we have some type of reaction coordinate, you can think of it as the progress of the reaction on the x-axis. And imagine a scenario in which we go from reactants to intermediate and on to product. And in this particular case, I'll give us an example of a slightly uphill reaction just as our acetone forming the hydrate was slightly uphill. Now when we make a diagram for a reaction like this, what we're really doing is diagramming out the lowest energy pathway for the reaction. In other words, there may be other pathways for this reaction. There may be a pathway that's way, way, way up here but it's going to be higher in energy. And the basic premise for a reaction is that as we proceed from reactants to intermediates to products over our first transition state, over our second transition state, if we have multiple transition states, we go on over a series of energy barriers and we're going to take the lowest energy path. Well, if this is the lowest energy path in the forward direction, then it's also going to be the lowest energy path in the reverse direction. If there were a hidden lower energy path, they would have gone through that pathway because it would have taken less activation energy to go over those barriers. Your molecules would have gone over that pathway in the forward reaction. And this principle, this idea of microscopic reversibility, then tells us exactly what must be happening in the reverse reaction for even a reaction you haven't seen before. Now, we're going to start with something pretty simple and it should be easy to intuit your way through this reaction even without any high-fancy, high-falutin-fancy principles of microscopic reversibility because all of this chemistry is just about protons going on, protons coming off, and the formation of oxycarbenium ions. So let's imagine the reverse reaction at this point. And here's our geminal diol and we're going to get ourselves back to acetone. And the idea has to be the same as the forward reaction. In other words, the last step of the formation of the geminal diol involved a loss of a proton from the geminal diol. So the first step of the reverse reaction has to be putting a proton onto the geminal diol. So here's our H3O plus and again it's going to be acid catalyzed. So we put a proton onto the geminal diol, electrons flow from the lone pair to the proton, from the hydrogen, the bond to the hydrogen, back on to the oxygen. And for now at least I'm going to be a good person and write everything out step by step. Later on I'll get a little faster, a little bit quicker and expect you to follow along. But in our next step of the mechanism now you can just envision the water getting pushed out by a lone pair of electrons. And so we push in, push out, and we get protonated acetone. It doesn't particularly matter whether I have the proton pointing off to the left or to the right. In the final step of the mechanism we're in water. There's lots of water protons go on. Protons come off just about everything with a lone pair. I'll take the long way around here, pull off my proton and we get our hydronium ion back. And so the mechanism of the reverse reaction looks identical to the mechanism of the forward reaction but in reverse. Some students might wonder why don't we just draw water leaving here and we could draw a carbocation and the answer is there's no difference. I've written one resonance form. I've written the more important resonance form. If you were to write it one way in the forward reaction, I'd probably write it the same way in the reverse reaction but in the back of our head just as I said at the start of the lecture, in the back of our head is the idea that embodied in this structure are two different structures, two different species, two different resonance forms I should say, two different pictures of the same thing. Thoughts or questions at this point? Great question. Does the reversibility, it's applying to the same mechanism in the forward reaction and the reverse apply to other reactions as well? Absolutely. And so we saw cyanohydrin formation on Tuesday a week ago and we didn't talk about cyanohydrin decomposition but the base catalyzed formation, decomposition of cyanohydrin back to aldehydes and cyanide goes by the exact same mechanism in reverse and in discussion section this week you'll be seeing that reaction. We're also going to be seeing some examples in the homework of acetal decomposition and other sorts of reactions and yep, that idea applies and it doesn't apply just because I said it, it applies from this fundamental principle that if the lowest energy path in one direction is indeed the lowest energy, if a path in one direction is indeed the lowest energy path, that path is of course the same in the reverse direction so yeah, it's universal. If you were to change the conditions and say all right, now imagine I threw in a handful of sodium hydroxide into my acetone solution and asked how does my geminal diol decompose now? We've just changed the conditions, then it's going to be a base catalyzed mechanism, then it's going to be pulling a proton off, getting an oxyanion, kicking out hydroxide but as long as the conditions are the same as long as you have the same species present, the same conditions of pH, yep, same forward and reverse. Other questions, good question. You know, this is an interesting reaction. Heat is, so this reaction will go at room temperature, it'll go at higher temperature, you're not going to shift that equilibrium, there's essentially nothing you can do to isolate a bottle of acetone geminal diol or a bottle of acetaldehyde geminal diol. They are that equilibrium, essentially give or take a little bit, stays right where it is regardless of temperature, so yep, no way you can drive that. Now I'm going to now introduce us to something where instead of forming an unstable species, a species that you can't isolate or put in a bottle, we're going to see the formation of a stable species, an acetal, and so that's where we're going to go next. And we will invoke Le Chatelier's principle to drive that reaction. So alcohols in a way are the cousin of water. They're cousin in the sense that they are related. They both are hydroxylic compounds, they both have similar pKa's and pKa's of the conjugate acids and they both can give up a proton. The big difference with water is it has a second proton to give up, okay, so that means there are going to be many reactions that look a heck of a lot alike. So let's take our friend acetone here and let's imagine instead of treating our acetone with water, basically dissolving your acetone in water, we treat it with, I'll write this as H plus for now. And by H plus, I mean a strong acid catalyst and some methanol, we specifically don't want H3O plus because we're going to be seeing a reaction in which water is the product of reaction and we want to drive that reaction by getting rid of the water so you don't want to do this reaction in aqueous acid. Product of the reaction is the, and what's called an acetal. This particular one is called acetone dimethylacetal and the acetal, unlike the geminal diol, is stable. The other product of reaction is water. So you want to go ahead and drive this reaction by getting rid of the water which is why I said we want to use anything but aqueous acid for the reaction. And the reaction is going to proceed. So this in a way is sort of a cousin of our geminal diol, a cousin in the sense that it has a methoxy group and a methoxy group instead of a hydroxy group and a hydroxy group and the reaction is going to go by way of an intermediate that is sort of a sister of the geminal diol and that's by way of a hemiacetal. Hemi means half and so a hemiacetal means a halfacetal. In other words, it's something where you have one group that's an alkoxy group and one group that's a hydroxy group. A hemiacetal in general is unstable and I'm going to say right now that at the end of today's lecture and perhaps even in your backpacks right now, we will see stable hemiacetals, those circumstances can make a special condition and we're going to see how this reaction goes. Now by the way, there's an older system of nomenclature you may see and it may have even popped up in the sapling course. Sometimes you will see the term ketal used. It's fallen out of usage but you'll sometimes see the term ketal used, ketal is an acetal of a ketone so sometimes you'll see this referred to as acetone dimethyl ketal but these days acetal is preferred and similarly sometimes you'll see this referred to as a hemiacetal. I'm going to put both of these terms in parentheses so we don't worry about them but if you hear them either across my lips in the course of our discussions or you see them in any of where you read, that's what you're going to see. Older textbooks tend to use them and as I said they may not have scrubbed it out of sapling. So before we move on to the mechanism, let's take a moment to talk about the acid. So the strongest acid you can have in water is H3O plus. In other words, if I dissolve HCl in water, I don't have HCl in water, it protonates the water. If I dissolve sulfuric acid in water similarly, it's nice to have in mind some idea of what constitutes a strong acid. In general, a strong acid is anything that's got a pKa that's lower than that of the solvent that you're thinking of. So HCl pKa of negative 7, H2SO4 sulfuric acid pKa of I'll just write negative 3 underneath. One popular one that's used. So HCl dissolves in methanol, just great. HCl doesn't dissolve in say benzene so well. It's not so soluble in say a solvent like benzene which is good because we learned that benzene is a good solvent for removing water. We learned about azeotropic distillation which is often important in driving reactions like acetal formation that are equilibria. So another popular acid that organic chemists like to use is paratoluene sulfonic acid. Paratoluene sulfonic acid looks a lot like sulfuric acid, right, sulfuric acid is H2SO4, H-O-S-O, H-O-S-O-O-H. And paratoluene sulfonic acid just has a benzene. The benzene is actually electron withdrawing. So the pKa of tosic acid is about, I'll just write this here, pTSOH is how you usually see it abbreviated which is a fancy name for paratoluene sulfonic acid. Anyway, pKa of negative 7, again a very strong acid. So if we imagine for a moment say taking methanol and dissolving an acid like HCl or H2SO4 or H-O-S-O-O-S-O in it, you have an equilibrium but that equilibrium is going to lie way to the right. You're going to protonate your methanol. Protonated methanol is the strongest species that can exist in methanol, the strongest acidic species that can exist in methanol. And of course, you'll have chloride. This equilibrium lies to the right. You can think of methanol as being very much like water. In other words, the pKa of protonated methanol is just about the same as the pKa of the hydronium ion. It's just a hair lower. It's negative 2 if you want to be exact. In general, you can think of an alcohol or an ether having a pKa of the conjugate acid of about negative 2 or negative 3. The alcohol group's a little bit electron donating. So when we imagine acetal formation and when I imagine this reaction here, it would be an acid like HCl or like H2SO4 dissolved in methanol but I'd think of my acid and the mechanism probably think of it as protonated methanol because that's going to be the strongest species that can occur there. So I want to take us to the mechanism of acetone dimethyl acetal formation and the mechanism is going to be identical to the mechanism we saw for hydrate formation but it's going to occur twice. Now, it's identical in the way that you have to think a little but broadly in that you have to think that well methanol can react analogously to water. Protonated methanol can react analogously to a hydronium ion and then everything is the same. Oxycarbenium ion formation is the same. Everything else is the same except you're going to have methyl groups present. What do I mean? Because you don't actually have to know anything to write this. All you have to do is be able to think your way through. All right, we've got our acetone here and we have our protonated methanol. In acid protons go on and off just about anything with a lone pair of electrons. We flow our electrons from our carbonyl from our weak Lewis base, our very weak I'm sorry, Bronsted base to the Bronsted acid of the protonated methanol just as we had done with protonation from the hydronium ion. Now, you end up with the self-same oxo-carbenium ion. We've seen before the self-same protonated acetone so we know that we can have an attack occur. And I'm going to go a little bit, I think I'll go a little bit faster here. And so, in the next step, and I'll skip curved arrows because by this point you should be able to do this. In the next step, our methanol attacks, and whoops, we still have a proton on it. I don't want to get that fast. So, in the next step, our methanol attacks and then in our next step, a molecule of methanol comes along and pulls off the proton and again I'm going to skip the curved arrows. You can imagine electrons flowing from the lone pair to the hydrogen electrons flowing back onto the oxygen. And so, at that point we have catalyzed the formation of the Hemiacetal intermediate and so far, and I'll go ahead and write this out here in our abbreviated mechanism. And so far, everything is identical to what we wrote in acetal formation or in hydrate formation. All right, I want to take us on at this point to the formation of our acetal from our Hemiacetal to our acetal and everything is going to be identical to what we've seen in acid. Protons go on, protons come off, everything with a lone pair of electrons. So, here we are at our Hemiacetal. Here's our protonated methanol and I'll go through a little bit more slowly, a little bit more carefully now. We'll put a proton onto the hydroxy group. Electrons flow from the lone pair to the proton. Electrons flow from the proton back onto the oxygen atom. Now, in the next step, we're going to kick out a water molecule to form another oxacarbonium ion. Electrons flow from the oxygen, push down, push out the water molecule. Doesn't matter which lone pair of electrons I use, they're both the same. Electrons flow back onto the OH group. We push out a water molecule. We go through the oxacarbonium ion. There's our water byproduct of reaction. And now, another molecule of methanol attacks. Again, electrons flow from the species with electrons to the species that wants electrons. We push up electrons onto the oxygen. Doesn't matter which way I push them to the left or to the right as long as they go from the pi bond onto the oxygen. And in our final step, now, we have lots and lots of methanol protons go on, protons come off, and we recover our protonated methanol catalyst electrons go back onto the oxygen, and we get our protonated methanol. Ah, beautiful question. Okay, one great question. Why doesn't methanol attack this intermediate and kick out water in an SN2 reaction? Anyone? So, okay, so there are two sort of questions here. So you've just mentioned carbocation. I'm going to hold that for a second because I want to come back to that. Two weeks of a nucleophile, okay, we're getting on the right track. What do we know about SN2 reactions? It's a tertiary carbon, and in general, we don't have SN2 displacement at a tertiary carbon, and indeed only very good, very few nucleophiles even do SN2 well at a secondary center. So, yes, it's too weak to be at a secondary center. It's a tertiary center. We essentially never see SN2 at a tertiary, and you mentioned carbocation, and the reaction is indeed an SN1 reaction because if you think about it, remember, the leaving of water as a leaving group to form this species, this is just a carbocation. It's just the other resonance structure. They're one and the same. The oxo-carbenium ion is both the carbocation-like resonance structure and this resonance structure. So, the substitution that we see here is indeed an SN1-type reaction, and indeed not an SN2. Other questions, great questions, exactly what you should be thinking at this point. Another question, exactly, have I planted people in the audience to ask, these are the questions that all of you should be asking. I'm really happy to be hearing these because these are the questions I hear every time from beginning students on this. So, they should be going through your mind, and I am very happy to have people brave enough to ask them. Your question is why doesn't the protonate here, and indeed it does, and that takes us on the path back. So, as we go over a pathway from reactants to intermediates to products in general, we go, and if the reaction is downhill, eventually we end up there. And so indeed that occurs along the reverse pathway. And if you're trying to make an acetal and synthesize it, you drive that reaction as best as possible by kicking out the water. In some cases, it's just by use of a large excess of methanol, Le Chatelier's principle. In some cases, we really push it with Dean Stark distillation to drive the water out. It doesn't work so well with methanol because of the boiling point. Methanol works with other alcohols. In some cases, we add a compound that reacts with the water. For example, methylpropenyl ether or trimethyl orthoformate to react irreversibly with the water and suck it up. And in some cases, one adds a drying agent, a dehydrating agent like molecular sieves. I want to take us on in a couple of directions before we finish up our discussion of acetals, hemiacetals, and hydrates. And one of the things I want to take us on is the reverse reaction. So as I said, all of these reactions are equilibria. All of these reactions can be driven by the use of excess methanol and in turn, all of these reactions can be driven in the reverse reaction. So if we go ahead and carry out hydrolysis with lots and lots of water, we can break down our acetal. In general, that's going to mean just changing our solvent and keeping our acid. So let's take our acetone dimethylacetal and I think at this point I'm going to avoid writing the, all right, I will write the methyl groups one last time here but I expect you to be able to understand them. And so imagine a source of acid in water, aqueous HCl, aqueous sulfuric acid, and lots and lots of water. Then our acetal breaks down, whoops, back to acetone. And if I feel in the mood to balance my equation, two molecules of methanol. Now, at this point I expect you to be able to think your way through the mechanism. It's a good exercise, you should do it yourself. I'm going to give you the gist of it in shorthand. The gist of the mechanism involves protonation. I'm just going to draw out a series of intermediates here, maybe with a few lone pairs to help us keep track of things. The gist of the reverse mechanism is the exact reverse. First step is the same as the last step. First step, protonate on oxygen. Next step, kick out methanol. Next intermediate is our oxacarbene ion. Next step is attack of water. Next step is remove a proton to form our unstable geminal, our unstable hemiasatal. Next step is reprotonate on the methanol. Protons come off, protons go on, all different places, all different lone pairs. Next step is we push out the methanol. And the next step is our penultimate intermediate, our protonated acetone that loses a proton. And so there in a nutshell is the embodiment of the entire mechanism of acetal hydrolysis in reverse. And indeed, this is the only mechanism you're really going to see in all of this acid-catalyzed chemistry with slightly different wrinkles. Intramolecular, intermolecular, but no huge differences, question. Okay, good question. Should we get protonated methanol? We're in a huge sea of water. We have 55 molar H2O, and we've dissolved up a molar, half a molar of our compound. Most of the H3O, most of the protons are going to be on water. Let's take a look at a few more common examples and a little bit of synthetic chemistry. And your book does, your textbook does a nice job with this, so I think I'll conclude by borrowing a few examples that your textbook invoked. So we've already learned that at certain points in synthesis it's necessary to hide a group, to render one group so it can't react. We saw, for example, that in carrying out certain reactions we would want to hide an alcohol so the OH group isn't available by protecting it as a TBDMS ether, protecting it as a silo ether. Acetals are often used to hide a ketone or aldehyde functionality. So for example, and this comes straight from your textbook because I think it's a good relevant example. For example, your textbook poses the question, how could we selectively take this keto ester and reduce it to a keto alcohol? And obviously you see the conundrum. You look at this reaction and immediately you say, well, okay, I know there are plenty of reducing agents that I can use to blast down an ester. Basically we've learned in the course lithium aluminum hydride nukes any carbonyl down to an alcohol. And yet we recognize the problem that if we treat this compound with a powerful reducing agent like lithium aluminum hydride, we're not going, we're also going to reduce the other carbonyl. We're also going to reduce the ketone group. And so the solution to this is to go ahead and to take this compound, this keto ester, and hide the ketone group. And you can do so selectively by making an acetal out of the ketone group. And so if you envision often to make, particularly for acetals of ketones, particularly for the ones I referred to as ketals, because ketones are much less reactive than aldehydes, often you have to work really hard to make that reaction go. And one way to do that, to push the reaction is to make it go intramolecule, to have two alcohol groups in the same molecule. So ethylene glycol, 1, 2 dihydroxyethane and propylene glycol, 1, 3, or 1, 3 propane diol and 1, 2 ethane diol are often used as protecting groups for ketones. And so if one treats with catalytic toxic acid, which as I said is often used in organic solvents, often with Dean Stark distillation to drive out water, often in benzene or toluene, if one treats the ketone and ethylene glycol with catalytic acid, you can selectively protect, you can hide the ketone as the acetal. And now you can envision a three-step process where your first step is to protect. Your second step is to reduce and lithium aluminum hydride followed by workup with water. And I'll say or H3O plus, and I'm going to come back and comment. You know that I like to work up my lithium aluminum hydride reactions with acid. Usually that just means adding a little acid to the reaction. And if you just add a little bit of acid without really cooking things up without heating it, in general your ketal isn't going to break up. And generally you end up to really break down your ketal, to break down your acetals. In general, you need water and acid in a little bit of heat. I'm not at this point you're learning going to get bent out of shape over these differences here. You should just realize if you go into the laboratory, if you pour a little bit of dilute acid in at this point, you're probably not going to hydrolyze your ketal. You can then in your separatory funnel isolate it, dry it over magsulfate. If you then take it in water, boil it up with some aqueous HCl, higher temperatures, longer reaction time, and so forth. You will deprotect, you will remove the protective root. And I'll put in parenthesis of course, ethylene glycol. Although again your organic chemist is probably going to forget about that very quickly. So that's a nice example of where acetals are very useful in synthesis and where they can be used to achieve a goal selectively, which as I said is really one of the big things about organic chemistry. It's that I want to make this molecule, not that molecule. I don't want to make a mixture of molecules. I don't want to make a mixture of diastereomers. And in some cases I don't want to make a mixture of enantiomers, I want a single enantiomer. And this is all what these tools are about. All right, I kind of dropped the hint that intramolecularity ends up giving you a special relationship. Remember water is 55 molar. In other words, when a reaction occurs with water, say the formation of a hydrate like we saw in acid aldehyde where the equilibrium lay right in the middle essentially at 1 to 1 at 42 to 58, that equilibrium lies in the middle because in 55 molar water there's only a certain amount of water present to push in a hydroxy group. But if you have the hydroxy group tied in the molecule, you can have that carbonyl group seeing an even higher concentration of water. In other words, the effective molarity instead of being 50 molar, 55 molar water can be 100 or 1,000 or 10,000 where that hydroxy group is just staring the carbonyl group in the face. And your textbook gives two nice examples that are relevant to the chemistry of carbohydrates which is where I will finally conclude. One of these is a hydroxy pentanol derivative and the other is a hydroxy butanol derivative. And both of these compounds exist in equilibria with their geminal diol with their hemiasatal and those equilibria now lie toward the right. And so I guess I'll make my arrow a little bit longer. In the former case, you have a hydroxy tetrahydropyran type of structure where now the hydroxy group has cyclized into the carbonyl to give a geminal dot, to give a hemiasatal. And in the other case, you have a hydroxy furan structure. And now because that hydroxy group is staring it in the face, your equilibrium, as I said, lies to the right. It's 6 to 94 in the case of the former and it's 11 to 89 in the case of the latter. In other words, if it were just water, if it were just water or just even methanol, you'd have halfway roughly for the equilibrium. You'd have half of the geminal diol, half of the aldehyde or if it were methanol, you'd have half of the hemiasatal, half of the aldehyde. But by the time you have the hydroxy group in the molecule and it can form a 5 or a 6-membered ring, then you have the much higher effective molarity of the hydroxy group staring the carbonyl in the face. The equilibrium lies to the right. 6-membered rings are good because they're unstrained. 5-membered rings are good because they're smaller but you get a little bit of ring strain from eclipsing. Remember the ring strain encyclohexane is effectively zero. The ring strain encyclopentane is about 6 kilocalories per mole. So higher effective molarity of hydroxy group in the second case in the hydroxy butanol but some ring strain there. All right. Now where this really ends up coming into play and why I said you probably have this sitting in your backpack and certainly you have this sitting in your bodies is the sugar molecules. And if we look at a molecule like glucose and I'm going to draw it one way which I think is a little bit closer to what we're seeing here and then I'll draw it another way. So this molecule with all of these hydroxy groups on it is glucose or more specifically what's called beta-D glucose. Beta tells us about the stereochemistry at this position. The OH group is pointing out at us. D tells us about the stereochemistry up at that position but the main thing is this is a sugar that should be familiar to you. If you're drinking unsweet, if you're drinking what do they call high fructose corn syrup you've got glucose in there. If you're eating sucrose it's part of the molecule but if you're eating high fructose corn sugar it's been broken open. If you look at candy where glucose has been added it's there as well. Now if you notice here amidst all of these hydroxy groups is a carbon that is an acetal. This carbon is an aldehyde hiding mask just as our carbon is over here. I haven't drawn in the hydrogen but of course there's a hydrogen here and there's a hydrogen here which means just as our hydroxy pentanol exists in equilibrium with the cyclic form our hydroxy butanol our glucose exists in equilibrium and that equilibrium is way, way, way in favor of the cyclic form even more than 94 to 6. It's like 99 point something, 99.99 something and yet there is a teeny tiny amount of the aldehyde form and I'll draw in the hydrogen here just because to make it a little clear like so and I guess to keep it consistent I will draw it in the CH2OH. And so particularly with the presence of a little bit of acid to catalyze the reaction your equilibrium will go back and forth between the closed form and the open form. In this case there are a couple of reasons why this equilibrium is further toward the cyclic form. One of them is, there is a conformational. The hydroxies just perfectly organized the molecule to form a nice stable chair cyclohexane. You'll often see beta D glucose written in its chair conformation like so and all of the hydroxy groups are equatorial on this ring. All of the hydroxy groups are equatorial. The other reason is electronic. Your hydroxy groups particularly the alpha hydroxy group is inductively electron withdrawing. It pulls electron density away from the carbonyl group and so it makes the carbonyl of glucose even more electrophilic than other aldehyde carbonyls. It's even more prone to react. And I think this is where I'm going to wrap up our chemistry and you'll get a very nice example of some sugar chemistry in the latter homework problems from Smith and I urge you to take a look at them. It's a very nice exercise. All right, we will pick up with our next chapter on Thursday.