 Good morning. So I'm excited today. We begin the second half of the course today, which of course means yesterday's lecture won't be on the midterm. So okay, the usual question. But no seriously, I'm excited because now we get on to the topic that I hinted back way in lecture two. I said that basically the bulk of this course is covering carbonyl chemistry and even beyond lectures the six weeks of the class we continue with the theme of carbonyls in the very end of the class where we talk about carbohydrates and about peptides and proteins. But we've been talking for the past essentially five weeks about the electrophilicity of the carbonyl group. And now we're getting into this second half of carbonyl compounds, the fact that they're nucleophilic at the alpha carbon under the right conditions. So back in our first week I drew a diagram sort of like this. I said okay, here's some generic carbonyl compound, not necessarily a ketone or ester, or aldehyde, we could mean an ester. And the general gist of it is that the carbonyl compound is electrophilic. It can react with nucleophiles. We come to this second half and so I'll say now we bring in this notion that the alpha carbon under the right conditions can react as a nucleophile. In other words, it can react with electrophiles. And we're going to be exploring this idea extensively over the next four lectures. We're going to be spanning this idea from chapter 23 in your Smith textbook to chapter 24 and discovering the richness of the chemistry of enols and enolates, which are the intermediates by which we have this reaction. Okay, so we talked before about various aspects of carbonyl compound structure and we said we talked about the reactivity of a carbonyl compound. We talked a little bit about in water equilibrium with a hydrate species and we said that for example in acetone there's very little of the hydrate and typically it's not a big factor in the behavior. In other words, if I took an NMR spectrum of acetone in water I wouldn't see the hydrate because it's just a minuscule amount of the compound. Well, there's another minuscule amount that can be present that's very important in reactivity and that is the enol totomer. So I'll write an equilibrium where we have and let's just take this as acetone right now. We'll be going through with various examples but let's just pretend you have acetone or some ketone or aldehyde let's say. And so we have this equilibrium where we have another species and now the species is obviously very different in other words you have the carbonyl compound and then you have a species that has a double bond and an alcohol and that's why it's called an enol because en means a double bond and ol means alcohol. And so we refer to this as the ketone totomer and this as the enol totomer. So totomer means an isomer where you've moved hydrogen around. And so you notice of course if this is acetone here we have three hydrogens attached to this carbon and then in this particular enol of acetone we would have two hydrogens attached and one here. So it's the same formula, it's the same molecular formula and yet a different arrangement of atoms. Totomers are a type of constitutional isomer. Remember we've learned about stereoisomers and then we've learned about constitutional isomers. And by constitutional isomer I mean we have a different connectivity, stereoisomers have the same connectivity. Now as I said the enol totomer is in equilibrium with the ketone totomer and that equilibrium lies way, way to the left. I'll tell you just how far to the left in just a moment but I wanted to introduce our players. Now if we imagine pulling off the hydrogen of the enol totomer in other words deprotonating, removing that hydrogen with a base we get what we call an enolate anion and I'll be sort of semi-generic here. In other words we could be acetone, we could be something else but I want to show you that there are two substituents attached to the carbon of the double bond there to the, to what we call the alpha carbon. And the enolate anion can be thought of as having two different resonance structures so we can envision one resonance structure where we have our negative charge on the oxygen and a second resonance structure where we simply move these electrons down and move the electrons onto the alpha carbon. Remember we use a double headed arrow, not a reaction arrow, not an equilibrium arrow but a double headed arrow to indicate resonance structures and I will very explicitly draw my lone pairs and draw my negative charges like so and just to be a good person I'm going to put this in brackets to remind us, to help remind us that these are two pictures of the same thing. It's not one, it's not the other unlike the Ketol and Enol totomer they are not two distinct species that you can identify distinctly, they are one and the same at the same time. And so I will write resonance structures here to remind us of this and I will also remind us that since this is the anion of an Enol we call it an enolate anion. All right and these two sets of variants on the carbonyl, the Enol totomer and the Enolate anion are going to be occupying our thinking for the next couple of weeks, for the next four lectures as we explore the reactivity of carbonyl compounds and specifically their ability to react as nucleophiles at the alpha carbon. In terms of understanding reactivity and this reactivity as nucleophiles I think the Enolate is a very obvious place to start. So if I draw the structure of an Enolate like so and again I'm keeping it kind of generic you could think of this as acetone Enolate or you could think of it as some other Enolate but in fact this is pretty much universal behavior of Enolates and we imagine some electrophile, now again in the spirit of keeping things generic I'm going to write it as E plus. That doesn't necessarily mean we have a positively charged species. You've seen positively charged electrophiles, you've seen carbocations in the past, protons or H3O plus of course are cationic electrophiles. But by the end of today's lecture you'll also be seeing methyl iodide and SN2 type alkylating agents acting as electrophiles. So here in very cartoony, very loose fashion I'll write E plus in quotation marks and the reactivity at the alpha carbon as a nucleophile can be thought of as bringing our electrons down from the oxygen, bringing our electrons over from the double bond over to the atom bearing the positive charge. I guess technically if I want to do this right my arrow should end up at the atom and again this is sort of an abstraction of an atom. And so if I want to complete my drawing that constitutes our reaction. Now if you're not comfortable with moving all of these curved arrows around realize that this is just the representation that this first part of moving electrons down and moving these electrons is just the embodiment of how we get from the resonance structure on the left to the resonance structure on the right. In other words when I'm interconverting these to resonance structures in my mind's eye I'm bringing these electrons down and bringing these electrons onto the alpha carbon but I don't necessarily need to draw that out because by this point in my course and by this point in your thinking you should be comfortable enough with that operation in your mind. Okay so that constitutes the reactivity of enolates and I just want to take a moment before I take questions to talk about the reactivity of enols again over viewed in this very, very generic fashion. So in the case of an enol this minor species that's present in equilibria with ketones and aldehydes and to a lesser extent with other carbonyl compounds. We can imagine here's our enol and again I'll have my generic electrophile E plus and we can imagine electrons flowing down from the oxygen into the carbon-oxygen bond giving rise to a carbon-oxygen double bond concurrently with electrons flowing from the carbon-carbon double bond. To the electrophile and if we continue with our grammar of electron pushing our grammar of the curved arrow now that's going to leave us with a protonated carbonyl and a bond to our electrophile and now we can imagine losing a proton again I'm talking very much in abstractions here so I'm not talking about a proton just popping off but a proton being taken off by base and so at this point we can imagine going ahead and getting to and I'll write that proton in quotes because again one would envision something taking it off so we can envision getting to the carbonyl compound that's reacted at the alpha position. Question can the negatively charged carbon absolutely it is one and the same whether I write this reaction as a flow of electrons like such or whether I write the reaction as I guess again technically to be correct since it's a resonance structure I'm going to write it in the exact same geometry whoops the exact same geometry because the two resonance structures share a geometry okay if I want to think about it like this this is the self same thing it is the exact same way of writing this and so these are not really different they're just different ways of representing the same thing and as one moves in their understanding of organic chemistry you go from sort of needing to explicitly write that ladder one out to recognizing that the one image embodied in the former which happens to be the major resonance structure the one image embodied in the former ends up being enough for us to think our way through. Other questions let's move on to some specifics so this is this provides an overview this provides a very generic introduction to the chemistry of enols and enolates and now let's talk a little bit about some specifics about the prevalence of all the carbonyl compounds with the exception of di carbonyls which we're going to talk about very soon in a lecture or so with all the carbonyl compounds aldehydes and ketones are the most in all like with aldehydes being a little bit more in all like than ketones your textbook just gives sort of a generic answer that there's less than 1% enol present for a typical carbonyl compound but it's a lot less than 1% there's very little enol so acetone is an equilibrium if you spray acetone on your glassware you're spraying mostly acetone but you're spraying a little bit of acetone enol it's an equilibrium and the position of that equilibrium K is equal to the equilibrium constant is equal to 1.5 times 10 to the negative 7 in other words there's less than 1 part in a million of acetone enol present at equilibrium cyclohexanone in general more substitution makes for a more stable double bond we learn this in the chemistry of alkenes that in general when you put alkyl groups on a double bond it's more stable so cyclohexanone exists in equilibrium with the enol and it's a little bit more enolic than acetone the equilibrium constant is 5 times 10 to the negative 5 again just a hair more than 1 part in a million whereas before I guess we had a hair less than 1 part in a million as I said in general aldehydes are a little bit more enolic than ketones that's not necessarily surprising remember the aldehyde carbonyl group is pretty darn unhappy as a carbonyl group it's only got one alkyl group donating in we saw this when we talked about the chemistry of hydrates and we said that acetaldehyde exists about 50 percent about half hydrate and half as the aldehyde form in water whereas acetone is vastly predominantly the ketone form not the geminal diol and so similarly we see that acetaldehyde is a little bit more enolic than say acetone which would probably be the best comparison so in acetaldehyde you again have an equilibrium again that equilibrium lies way way to the left but now you have just a hair more of the acetaldehyde enol the equilibrium is at 2 times 10 to the negative 5 for your equilibrium constant so compared to acetone a couple orders more orders of magnitude more in spite of being a minor equilibrium component the reactivity of enols is very important in the reactivity of carbonyl compounds and that makes sense even if you have a little bit of something when it reacts more gets generated and so a reaction can proceed and proceed and proceed being small doesn't mean you're unimportant and so we can take a look at how enols and ketones keto and enol totemers interconvert and look at the mechanisms of formation another question who wants to answer this question someone enol form of formaldehyde can't no alpha carbon on formaldehyde sorry acid aldehyde so that's a very good way of thinking about it whenever you have an equilibrium you're comparing the energy of a ground state and a product and so the bigger the difference in energy the bigger the equilibrium constant the bigger whether it's positive or negative you know depending on whether it's greater than one or less than one depending on how the energy difference goes so anything that either lowers the energy of the reactant or raises the energy of the product makes that equilibrium constant greater in magnitude whether it's you know in one direction or the other direction so when we talked about acetone versus cyclohexanone they're both ketones but we're lowering the energy of the enol by making the double bond more stable so here's the acetone enol here's the cyclohexanone enol in the case of acid aldehyde versus acetone now you can think of it as we're raising the energy of the reactant because acid aldehyde is higher in energy it's less stabilized relative and it's always a question of reference frame but that's a good way to think about it so later on we're going to see things that provide a lot of stabilization of the enol specifically conjugation all right so that's talking about reactants and about products about keto and enol tautomers let's now look at how we get from the keto tautomers to the enol tautomers and this reaction can be casually catalyzed by either acid or base I want to start off by talking about acid catalyzed enol formation and I'm going to write out the mechanism very slowly and carefully very meticulously because I think this really is important to think your way through we've already seen on the quiz how important mechanism is and clarity and expression and thinking on mechanism the mechanism on the quiz was kind of long and complicated involving many steps the mechanism of acid catalyzed enol formation is relatively simple and easy if you have an acid hydronium ion for example protons go on and off everything with lone pairs of electrons the carbonyl of acetone or the carbonyl of a ketone or the carbonyl of an ester is weakly Lewis basic weakly Bronsted basic and so you have an equilibrium where you can protonate the carbonyl or if we're looking at hydronium ion as catalyst hydronium ion has been consumed in the first step of this reaction and now we're going to recreate it in the second step of the reaction I'm going to again draw our protonated acetone and now I will explicitly draw in my alpha hydrogens because we're going to use them very often when you're writing a mechanism or thinking about a mechanism you will write different parts of a molecule at one time so here's our water here's our alpha proton and now water is weakly basic very weakly basic water can pull off the alpha proton we just push our electrons up on to the oxygen and again we have an equilibrium and I will be very, very explicit and try to draw in my all of my relevant hydrogens here to help keep us on track and I will try to remember all of my lone pairs and all of my charges so I'm not a bad person and here we go that's an example of how we go from the ketone potomer to the enol potomer. Thoughts or questions at this point is water able to pull off the hydrogen so you mentioned the term addition elimination and we've seen addition elimination in acid chlorides that's where a nucleophile adds to a double bond we kick up a lone pair of electrons it comes back and pushes something out nucleophile like an amine adds to an acid chloride we kick electrons up on to the oxygen they kick back down kick out the chloride so this reaction mechanistically is an acid base reaction and the only thing that's maybe surprising or confusing to you is the fact that we're continuing to push our electrons all the way up now technically technically I could go ahead and stop at this point but of course if I stop at this point I recognize in my mind's eye well this is a really, really funny resonance structure a really unimportant bad resonance structure of that and so we just keep pushing and of course protons come on and off all different acidic positions in acid base equilibria and so the proton whoops I forgot my positive charge so much for being a good person over here so the protons can come on and off the oxygen if then we head in the reverse pathway and guess what? Most of the time your equilibrium partitions back in other words when you have an equilibrium like this so here's our energy here's our reaction coordinate when you have a reaction like this where we're going from reactant from our acetone plus H3O plus to our protonated acetone and then to the enol and technically if I'm writing this out correctly this is acetone protonated acetone plus H2O what's happening in our mechanism is first we're going uphill to an intermediate and then we're coming down to our product to our enol that's higher in energy so when you're at this point at the intermediate you can partition back you can partition forward it goes both ways most of the time it actually ends up going back down and ultimately of course it's the difference in energy between our product and our reactant that determines the position of the equilibrium. There was another question it does indeed take the H from the oxygen and the mechanistic reverse of this step is taking the H from there most of the time it takes the H from there some of the time we go forward and it goes in the other direction or should I say well it's not technically the partitioning it's not going to be most of the time it's going to be half and half because most of the time we're not forming this intermediate so technically it can be a little more of one or a little bit more of the other but the point is that heads us back sometimes we pull off this proton and it heads forward this proton is acidic. Other question can the water also perform a nucleophilic attack great question this is what's so profound this is why people can find organic chemistry confusing because yes it can and yes it does and we saw that we saw we talked about acid catalyzed hydrate formation acid catalyzed geminal diol formation so in your acetone in water with a little bit of acid honestly even without a little bit of acid because water has hydronium ion 10 to the negative seventh molar in your acetone and water there is this gamish this mixture of acetone the main component and then a little bit of the hydrate and even less of the enol so absolutely and so keeping this in one's mind oh yeah this is the pathway we're thinking about now because this leads to the good stuff we're going to see over the next four lectures is often confusing to initial to beginning students great question really important because this is the part of understanding it all that's okay and you look at this mechanism here this manifestation of catalysis and realize what this means for the uncatalyzed mechanism it means that in the uncatalyzed mechanism you're not actually changing the difference in energy I'm trying to draw this curve for the uncatalyzed mechanism between acetone and acetone enol at the exact same level you're not actually changing the difference in energy between acetone and acetone enol what we're doing is providing a low energy pathway that's not accessible without acid in other words there is hypothetically a one step mechanism that converts acetone to acetone enol but the energy barrier is much higher so when any acid is present that includes that 10 to the negative seventh molar H3O plus hydronium ion in pure water acid can catalyze this reaction now the other thing and I've been beating on this beating on this like a dead horse and the other thing that's important is reactions proceed through the reverse reaction proceeds through the exact same mechanism when cyanohydrin's form by adding cyanide to a carbonyl to give an oxyanion and then protonating the oxyanion to get the alcohol we learn that cyanohydrin's break down under the same conditions by deprotonating the oxygen to give an oxyanion and then kicking out cyanide when enols form by first protonating the carbonyl to give a protonated carbonyl compound and then deprotonating the alpha carbon to give the enol that means that the reverse pathway proceeds backwards by the same mechanism and so we can write and you should be able to write in your sleep the reverse mechanism we simply go ahead and start with the enol and hydronium ion protons flow the proton goes to the alpha carbon electrons flow down from the oxygen and now in the second step that's the step in going from our intermediate water back to our acetone now water just acts as a base and pulls off the proton on the carbonyl electrons from flow from the lone pair on water to the protons electrons flow from the bond back on to the oxygen lots of questions all right let's see where this gets us in terms of some reactivity and then we'll get to enolates which get really funny. I'm going to take probably the simplest reaction that I can think of and that reaction is going to be deuteration so let's take cyclohexanone as a variation here and we'll envision dissolving it in D2O I'll put parenthesis as solvent D2O is just heavy water it's just deuterium oxide it's just the isotopamer of water with deuterium every glass of water you drink contains zillions of molecules of deuterium or zillions of deuterium atoms one out of every 7,000 hydrogen atoms is deuterium at natural abundance meaning in your proteins in your lipids in your carbohydrates there is a miniscule amount of deuterium and one can by electrolysis or distillation concentrate the deuterium to get pure D2O if we treat our cyclohexanone with D2O and a little bit of catalytic acid I'm going to get away from H3O plus and D3O plus for a second to remind us one actually has to go into the laboratory and get a real compound I've written DCL DCL is just hydrochloric acid made with deuterium so it's a strong acid it dissociates in D2O what will happen is you will replace all of your alpha protons with deuterium and this is kind of cool you can see this for yourself if you ask your lab TA to give you some cyclohexanone and some D2O and a little bit of DCL and you dissolve your cyclohexanone in D2O you would be able to take an NMR spectrum and you would see the alpha protons the hydrogens next to the carbonyl at about two and a half parts per million and as the reaction proceeded the peak in the NMR from those hydrogens the peak at two and a half parts per million would disappear as those hydrogens were replaced by deuterium so how does this occur well it's the exact same mechanism that we saw before just repeated over and over again I'll write this in abbreviated fashion we can envision that in D2O we protonate the D2O with DCL remember when you pour HCl into water you have H3O plus and Cl minus the D3O plus is just like H3O plus in its reactivity you can put a deuterium onto the carbonyl to give a protonated or in this case deuterated cation and now in the second step of this reaction of course from balancing my equation in my second step of my other product is D2O in the second step of the equation just as we have pulled the protons off the alpha position of acetone with H2O for protonated acetone now the D2O can pull off our alpha protons and so again we have an equilibrium now we take ourselves to the enol plus if I want to balance my equation D2O H plus I think I won't continue on the sideboard here so one thing I should point out is we have lots and lots of D2O present water is 55 molar deuterium is practically the same as water D2O is going to be essentially 55 molar. Protons as I've been saying again and again go on and off every protonated species in other words when we have D2O H plus and lots and lots of water so I will write plus excess D2O you have an equilibrium and of course mass action drives this equilibrium and so you'll get a little bit of H2O HOD plus D3O plus in other words basically you've got tons and tons of D2O your water just spreads out your protons just spread out to make little bits of HOD. If I do a reaction with 55 molar D2O and I have one molar solution or a tenth molar solution of ketone as my protons go in I get this little bit of HOD in this mass of D2O and D3O plus. All right let's continue our mechanism here so we have our enol with a deuterium on it and we have lots and lots of D3O D2O and a little bit of D3O plus we can protonate our enol. I'm just writing the same mechanism as I wrote before without the curved arrows and without filling in all of my lone pairs of electrons and now we can have further proton, further deuteron transfer and you can envision this mechanism just continuing we form the enol, we protonate the enol with deuterium, we form the enol some more, we protonate with deuterium and eventually we've washed out all of those enolizable alpha positions and now we have the fully deuterated molecule. Organic chemists are terribly bad at balancing equations but if I want to write a balanced equation I would write that our ketone plus 4 D2O goes with catalytic DCL in D2O, solvent goes to the fully deuterated ketone plus 4 HOD in lots and lots of D2O. So one of the other properties of carbonyl compounds is that their alpha proton is acidic. So for example in acetone the alpha proton has a pKa of about 19 and that's interesting because if you think about it, if you think about say a regular methyl compound or a regular alcohol compound the pKa is 20 is about 50. In other words the pKa of an alkane is while nominally acidic, very weakly acidic, it is very, very, very weakly acidic. It's only acidic in the sense that you can think of its conjugate base as being a very, very, very strong base and yet by the time you get over to acetone putting that carbonyl there shifts by 31 orders of magnitude the acidity of that proton. It shifts the equilibrium massively because when I think of the conjugate base of acetone I don't just think of this, I think about that special resonance stabilization that we get as the enolate. And you can see how picturing these two resonance structures together, the structure on the left is really very inadequate to explain this stability. The structure on the right explains it beautifully. We have a negative charge on the oxygen. We have oxygen as electronegative. We know alkoxide has or alcohols are reasonably happy to lose a proton. They're weak acids but they're not very, very, very weak acids. PK of an alcohol is 16 or 17. The PKA of an alkane is 50. That oxygen does a heck of a lot and with acetone with our PKA of 19 we're getting a lot of stabilization from this resonance structure. In other words, this is the major resonance structure. This is the one that's important in determining the reactivity and this one is the stability and this one is the minor resonance structure. Now, more generally the carbonyl group provides stabilization and your textbook gives a number of compounds and I'll give you these compounds and then I want to give you a generalization that I keep in my own head. So your textbook gives you acid aldehyde with a PKA of 17 and it gives you ethyl acetate with a PKA of 25 for the methyl group and acetonitrile with a PKA of 25 and dimethylacetamide with a PKA of 30 for the alpha. And really the numbers that I keep in my head, these are a lot of details, but the numbers that I keep in my head are as follows. I keep in my head two numbers for all of this, that the alpha protons for the ketone and aldehyde family are about PKA 20. Twenty's good enough. Yeah, I'll keep in the back of my head, aldehydes are a little bit more acidic. I'll keep in my mind that 20 is typical for say cyclohexanone, that acetone is 19, but those details are small. Those are unimportant. The other big picture while aldehydes and ketones are about PKA 20, in general the carboxylic acid family, esters, nitriels, even amids, even carboxylic acids under the right conditions, you first remove the acidic proton, so it's a carboxylate, and then you remove a second proton. In general, a good number to keep in mind is about 25. Yeah, there's some variations, but it's not as super important as keeping these pictures in mind. And one of the pictures from this is in general ketones and acid and aldehydes are less acidic than water. In other words, hydroxide will take off only a small amount of the protons, only in equilibrium concentration. Ethoxide will pull off only in equilibrium concentration, but stronger bases, those can pull off protons quantitatively. Since the alpha protons, the ketones and aldehydes are acidic, it shouldn't surprise you that base can also catalyze enol formation. So let's come back to our acetone molecule as our sort of archetypal ketone, and now let's envision hydroxide as a base, I could do this with alkoxide too, but some sort of moderately strong base, not super, super strong. We'll talk about super, super strong in a moment, but something like hydroxide or an amine, like triethylamine. We can pull off the alpha proton. The alpha proton is weakly acidic. Remember pKa of 20 is the number you want to keep in your head, so we have an equilibrium here with the enolate, but that equilibrium lies slightly away from the enolate, and so we can protonate. In other words, you generate a little bit of enolate, but that you don't generate the enolate quantitatively, and so we can protonate. If we protonate back on carbon, we're headed back in the same position, we're headed back to the start of the reaction, if you protonate on oxygen on the other hand, now we've proceeded on to the enol. And just as in our acid-catalyzed enol formation, hydroxide here, although it's being consumed in the first step, is being recreated in the second step, and so it is not being taken up or destroyed or created in the reaction overall, it is acting as a catalyst. Similarly, we're not changing the position of the ketoenol equilibrium. We're only creating a lower energy pathway to allow the ketone and enol forms to interconvert. Okay, good question. If you're working your equilibria, technically yes. In this particular example, that is an astute question. In this particular example, with these particular pKa's, technically you're right. There will be more enolate than enol present. If you want to come to an example where now you have very little enolate and the equilibrium still goes, substitute in this reaction as base triethylamine, because now, remember we said that for acetone we're going to have 1.5 times 10 to the negative 7 is our equilibrium constant. So there's not a lot of enol. And we're having our pKa for acetone of 19, which means you actually do have, you know, one in 1,000 or one in 10,000 enolate, one in 1,000. So if you substitute triethylamine pKa about 10 or 11 for the conjugate acid, you'll still catalyze this reaction, but now you'll have very, very little of the enolate. Do you use a Dean Stark apparatus? No need here. We're not actually trying to change the position. And, okay, could one. Now you're asking a smart question. Can one drive this reaction? And the answer is you can't drive it by removing water because the reaction is unimolecular overall. There's no net change in the molecule. It's an isomerization. So the way to drive the reaction to the enolate is to go ahead and to use a very strong base. And unfortunately, there's very little one can do to try to generate the enol as an isolated species, although there's an Israeli chemist who actually is focused on stable enols because another thing about organic chemists, in addition to being very bad at balancing equations, is if someone says it can't be done, enols are unstable. You can't make a stable enol. You can bet you an organic chemist will go out there and say, how do I make a stable enol? We'll see some stable enols that are very special enols, but for a regular one. All right, so where is all of this leading? Let's go and do some enolate chemistry. So just like with DCL and D2O, if I take cyclohexanone with catalytic sodium deuteroxide, just the deuterium analog of sodium hydroxide in D2O as solvent and I let it sit, again you can do this for real in an NMR tube. Johnny or Kim could go back to my laboratory and do this right now and show you those alpha protons would wash out and be replaced by a deuterium. Similarly, if I imagine for a moment that I had something that had chirality at the alpha position, let's imagine for a moment that I have this ketone where I have a methyl group and let's say that I have just the synantiomer and I treat this ketone with let's say sodium hydroxide since we're talking enolate chemistry, sodium hydroxide in water. Then what will happen will be I lose my enantiomeric purity. I will get a mixture of the two enantiomers. I will get the racemate. I will get an equal mixture of the R and the S. What's happening here in this latter example? We're forming little bits of the enolate repeatedly. The enolate is flat. The enolate loses its chirality. I will say that we go via this. In other words, you can picture in your mind's eye. The base pulls off the alpha proton. We get the enolate. The enolate is flat. We have no chirality to the enolate. If a proton comes from the back, we get the original enantiomer. If it comes from the front, we get the new enantiomer and so our compound racemizes in aqueous base. So in addition to being bad at balancing equations, in addition to wanting to do what can't be done, organic chemists are control freaks. We want to control the molecules and make them do stuff. And often we want to do stuff to make stuff that's useful like new medicines. So whereas weak base doesn't make a lot of enolate, a strong base does and we can use that. If you have something like cyclohexanone in a base, I'll write it as B minus, let's say, a very strong base. In other words, pKa of the conjugate acid much greater than 20, then we have an equilibrium that lies way, way, way to the right, often so far to the right that we can just ignore the left arrow to give us the enolate anion and our protonated acid. And the base that's extremely valuable to organic chemists is diisopropyl amide anion or more specifically the lithium salt, lithium diisopropyl amide for those of us who are control freaks, for those of us who want to make molecules do stuff. This base is really good because it's a strong base. pKa of the conjugate acid, the pKa for diisopropyl amine is about 40. So I'll write pKa, IPR2 and H is a nice shorthand and I'll say tilde approximately 40. Remember we're comparing, I said if you want to keep one number in your head, keep 20 for the pKa of cyclohexan, for the pKa of a ketone, that's actually the pKa of cyclohexanone to within a unit but again just keep that number in your mind. The other thing that's really important about the diisopropyl amide anion is it's big. Those diisopropyl groups and there are two of them are bulky, they get in the way. In other words, lithium diisopropyl amide is a good base but not a good nucleophile and again I'm going to come down to this confusion in organic chemistry students when you're starting out. Oh my God, there's all this stuff to know. How do I know that this compound like cyanide is acting as a nucleophile? How do I know that diisopropyl amide is acting as a base while you start to learn these patterns? The very bulky ones are better at pulling off protons, the little bitty ones and cyanide is just a little bullet, is better as a nucleophile. So the reaction here lies so far to the right that our lithium enolate, I won't even bother to write the reverse arrow, our lithium enolate is essentially formed stoichiometrically and of course, the other component in this reaction is diisopropyl amine. So we have an acid base reaction, stronger acid plus stronger base reacts to give weaker acid plus weaker base. How do you know where that equilibrium lies? You just look at the pKa's of the acid. Okay, we've got the weak acid pKa, 40 on the left, we have a strong acid on the right and remember, the other number I like to keep in mind is about 10 orders of magnitude. In other words, if you've got things within a pKa range of about 10 units, I'd say yeah, you've got a little bit of an equilibrium but you go to 20 units apart and it's like bam, that reaction goes essentially all the way. Often when organic chemists are writing out synthetic reactions, we'll write in a little bit of shorthand, for example, I will say LDA, it's widely used in tetrahydrofuran as a solvent in order to minimize its reaction as a nucleophile and minimize equilibration processes, often you'll do these reactions at negative 78 degrees Celsius, that's the temperature of a dry ice bath which makes it particularly popular. Remember early on I said that E plus is sort of an abstraction, I said you'd see methyl iodide. So you can envision a reaction like this and the first step, we add cyclohexanone to LDA at negative 78 degrees. In the next step, we drip in methyl iodide and allow the reaction to warm up, maybe perform an aqueous workup on it with a little bit of acid and we've methylated our cyclohexanone. What's happening here? Same thing we saw with protons. We've generated our enolate stoichiometrically, here's our methyl iodide, whether it's a proton, whether it's methyl iodide, we have a good electrophile, electrons flow from the alpha carbon to the methyl and we have the electrophilic methyl carbon and we do an SN2 displacement to give the methylated cyclohexanone an iodide anion. Most of the time organic chemists don't go ahead and buy lithium diisopropyl amide. We typically make it by acid-base chemistry and butyl lithium is widely available. It's the most common alcohol lithium, widely used. You can mix it with diisopropyl amine in THF if you like before you add your cyclohexanone and you pull off a proton, you get butane plus lithium diisopropyl amide and again I'm now getting into writing things in shorthand so I'm not writing as many lone pairs of electrons here and this too is an acid-base reaction and butyl lithium is very, very strongly basic. The pK of an alkane is about 50, pK of diisopropyl amine, we already said is 40, that equilibrium lies way, way to the right. You might ask why don't we use butyl lithium as a base comes back to what I was saying about sterics. If I use butyl lithium the main reaction, the only reaction that you would detect that you would write on say an exam if you want to get down to the brass tacks of being an O-chem student would be addition of the butyl lithium to the carbonyl. It's not sterically hindered enough to not add. We go to the lithium diisopropyl amide with those big isopropyl groups and now it's much more basic than nucleophilic and butyl lithium toward a carbonyl is much more nucleophilic than it is basic. Let's have some fun now with variations among enolates. So imagine we have our methyl cyclohexanone and we treat it with base. Now we can get two different enolates. We've got two different types of alpha protons. We've got alpha protons that are opposite the methyl and alpha protons that are next to the methyl. In other words, we can get this enolate or we can get this enolate the less substituted enolate or the more substituted enolate. Now, in general, we know that more substituted double bonds are more thermodynamically stable. The more substituted enolate is what we call the thermodynamic enolate, slower in energy among an equilibrium between the two enolates, it would form. But we also know that size matters. We have a bulky side of the molecule. We have a non-bulky side of the molecule. If a base, particularly a sterically hindered base, needs to get in to pull off a proton, it's going to grab the easier one if it's bulky. So the enolate that is less substituted is often referred to as the kinetic enolate, the one that forms first, the one that is easier to form. So we can go ahead and think of it as these protons are more accessible to base. This proton is technically a little bit more acidic. I will write here, just for comparison, more substituted enolate, more thermodynamically stable. So let's take a moment to play with this notion because as I said, organic chemists love to control reactions, love to control reactivity. And it's important because when you're making molecules for medicines or for drugs, you need the molecule that you want. And I'll take a couple of examples from your textbook here because they're good examples. If we take our methyl cyclohexanone and we treat it with LDA and THF at negative 78, you may see it written just as LDA, you may see it written as LDA, THF, you may see it written as LDA, THF, negative 78. And we treat it with methyl iodide. We deprotonate to form the kinetic enolate and we end up with the 2,6-dimethyl cyclohexanone. So I'll write via the kinetic enolate. Borrowing an example from your textbook, which if you want to ask questions about after class, I will tell you some of the subtleties of, but if you treat with a weaker base that can equilibrate, your textbook uses sodium ethoxide in ethanol. And basically the way I will write this is to mix sodium ethoxide, methyl iodide in ethanol, then because we're under conditions where we're thermodynamically equilibrating our emulates, your predominant product, your main product is going to be the 2,2-dimethyl cyclohexanone via the thermodynamic enolate. And so this is a taste of the type of control that organic chemists can get. Now I'll write this as major product. Technically here, if I do this reaction with LDA, this is the major product. In other words, I'm going to get 99 parts of this in one part, this. Technically this is the major product. I'll probably get about 80 parts of this and 20 parts of that. I know there's one question, but I want to wrap up with a couple of last points about stereochemistry and stereoisomers at this point. Because I think it's pretty and I think it's cool and I think it's intellectually deep and I think that a lot of this is what organic chemistry is all about. So technically I'm lying in that structure, maybe to be more precise, I'm telling you half a story. And the other half of the story is when I draw this structure, I'm living in flatlands. Each of those carbons with the methyl group is a stereogenic center and so we have different types of species that we can get. We have the species, the stereoisomer to be more specific in which the two methyls are on the same side of the ring. We have the cis stereoisomer. We have the stereoisomer in which the two methyl groups are on the opposite side of the ring. The transstereoisomer and so when we form our enolate and now I will specify my methyl stereochemistry if our methyl group adds from the back, we get the transstereoisomer. If our methyl group adds from the front, we get the cis stereoisomer. And there's one more level of subtlety and I'll show you that and then that's going to wrap up today's lecture. So imagine for a moment I take this compound and almost invariably it's going to be racemic if I haven't specified otherwise. And so I treat with LDA in THF at negative 78 degrees and then I treat with methyl iodide just as I've done. And you'll get a mixture of the cis and trans compound but of course the trans compound is going to be the racemic. In other words, we generate our enolate and the methyl group if it adds opposite the methyl group depending on whether this methyl group is out or back we're going to get the racemic compound. In the case of the cis, the cis compound if it adds from the same face, the cis compound is a mesocompound. So in other words, whether I draw it like this or I draw it like this, this is just the self-same thing. All right, well that gives us a taste of the richness of stereochemistry as well as a beginning on this notion of control and you can see even though we've talked at this point about control of the regiochemistry of addition, we've only seen that there is still a depth and richness to the stereochemistry of addition and that will go largely beyond the scope of this class. Thank you.