Forming enolates from ketones using LDA and sodium ethoxide. Created by Jay.
Want to join the conversation?
- @9:06aren't the two isomers drawn below the first the same structure? are the two molecules not mirror images of each other?(5 votes)
- As drawn, the two resonance structures on the bottom are identical; I believe the point was just to show that the electrons could move around. It would come into play if the side chains of the ketones were different. If the side chains are the same, or there are no side chains, you'd just need one of those structures on the bottom, along with the top one.(2 votes)
- @3:58Why does equilibrium favor the formation of the weaker acid ?(3 votes)
- From what i remember from earlier chem, the weaker the acid (lewis acid) the less likely to accept electrons; therefore, if it is the reactant, it will not react as well as a stronger acid, or if its the product, it will not be as likely to "backwards-react" as seen in the equilibrium, and just stay as it is.(1 vote)
- after the aldehyde is deprotonated by a base, what would make it stay as a nucleophilic carbanion rather than converting into an enol?(2 votes)
- So essentially you're asking: How can we obtain the conjugate base of the keto-enol tautomer? Use a base strong enough to deprotonate the proton...in slight excess. The most common base in practice is LDA.(2 votes)
- So if I understand correctly from the video, a strong base is not necessarily needed to deprotonate a strong acid? Is it because a strong acid and strong base will full react to form salt and water and that is not what is needed here?(2 votes)
Is the molecule with the negative charge on the oxygen the major product (compared to the one with the negative charge on the carbon)? And does it happen because of the higher stability of the molecule due to the presence of the electrons on the most electronegative atom (oxygen)?
(I know it is a resonance structure but it should be possible for a structure to tend more toward one of the limit molecules, the major one. Right?)(2 votes)
- Can someone help me with what I'm missing at3:58: the reason LDAH is a weaker acid (~36) is mainly because of the bulky isopropyl groups of LDA right? Otherwise, is the main reason because the enolate anion seems to be a more stable base than LDA?(1 vote)
- The bulky isopropyl groups of LDA is more for limiting its ability of being both a strong base and a strong nucleophile. As a strong base, it'll simply deprotonate acidic protons. As a bulky and sterically hindered base, it will not exhibit (SN2) nucleophilic tendencies.
As for which is more acidic/stable... the pKa would indicate that the dissociation constant (Ka) for the enol is higher and thus more readily deprotonates (forming the enolate) when compared to LDA/LDA-H. This can be due to a number of reasons which include thermodynamics (Gibbs free E), the enolate anions stability and inductive effects.
I would just caution that the wording "more stable base" isn't always true. Thermodynamics plays a role (especially with LDA). For example, when using LDA to deprotonate an alpha carbon of a ketone with two differing alpha carbons (secondary vs tertiary), the deprotonation of the less substituted (subsequently forming of a less stable enolate) is favored at low temperatures. Conversely, at higher temperatures the deprotonation of the tertiary carbon is favored (and a more stable enolate is formed).(2 votes)
- For both enolate formation from aldehydes and ketones, the use of a base is because the alpha hydrogen from the carbon is acidic, correct? So formation of an enolate should always use a base? What is the difference between an enolate and an enol? I noticed that in the keto-enol tatuomerization, the use of a base or an acid could be done to create an enol...why is that?(1 vote)
- Question 1: Yes, the proton of the alpha carbon is more acidic than the conjugate acid of base used (pka alpha H < pka conjugate acid), thus the enolate is highly favored.
Question 2: When attempting to form the enolate (strong nucleophile), you should always use a base. In literature, the most commonly used base is LDA. Realize that the enolate has a negative charge...the presence of certain acids will protonate the enolate forming the enol (less nucleophilic). This may or may not work for the researcher.
Question 3: The enolate is the charged (more nucleophilic) deprotonated version of the enol. The ending 'ate' would suggest that its the conjugate base of some acid. The acid here would be the enol. Thus, the enolate is simply the conjugate base of the enol.
Question 4: Keto-enol tautomerization is simply a matter of equilibria. Take any ketone (such as acetone) and you'll find that some minute amount of enol exist in solution (99.999% ketone vs 0.001% enol). This suggests that the MORE STABLE tautomer is the ketone and not the enol. However, the stability of the enol can be improved with the introduction of either an acid or base, thus, the percentages might change from 99.999/0.001 to 75/25.
It is important to note that the formation of either the enol or enolate is heavily dependent upon the purpose of synthesis. Sometimes a researcher might want to trap the enol as an intermediate (with the use of silane groups such TMS/TBS/TBDPS etc.)
Hope that helps!(2 votes)
- How can you calculate the number of resonance structures for a given compound?(1 vote)
- You can refer to bond order but there is no such all in one formula for calculating resonating structures. To find out you have to draw possible resonating structures of the species.(1 vote)
- 4:50Why would donating electron density destabilize negative charge?(1 vote)
- The negative charge is not donating the electron density.
It's the methyl group which donates the electron density which leads to destabilization of the negative charge.(1 vote)
Voiceover: In order to see how to form enolate anions, and in this video we're just gonna look in more detail how to form enolate anions from ketones. And so the ketone we have here is acetone. To find our alpha carbon, we just look at the carbon next to our carbonyl carbon, so this could be an alpha carbon, and this could be an alpha carbon. Each one of those alpha carbons has three alpha protons, and so there's a total of six. I'm just gonna draw one in here, and this is the one that we're going to show being deprotonated here. So, the base that's going to deprotonate acetone, we're gonna use LDA, which is Lithium Diisopropyl Amide And, I could go ahead and draw in the Lithium here, so Li Plus, and then we see the two isopropyl groups like that, a negative one charge on our nitrogen. So this is a very strong base, it's also very bulky and sterically hindered. So you can think about a lone pair of electrons in the nitrogen, taking that proton, leaving these electrons behind on this carbon, so we can go ahead and draw the conjugate base here. We would have electrons on this carbon now, that's a carbanion, so let me go ahead and show those electrons, these electrons in here magenta, are gonna come off onto this carbon. And this carbon is a [carbanae] because remember there's also two other hydrogens attached to it. So that's what gives it a negative one formal charge here. We can draw our resonance structure, we can show these electrons in magenta moving in here, these electrons coming off onto our oxygen, so for our resonance structure we would show the negative charge is now on our oxygen, this would be a negative one formal charge like that now. So the electrons in magenta moved into here to form our double bond, and then we can show the electrons in here in the blue moving out onto the oxygen. So this is our enolate anion, right, this represents our enolate anion, and for our two resonance structures, we have one with the negative charge on the carbon, so it's our carbanion, and one with our negative charge on our oxygen, that's our oxyanion. Remember the oxyanion contributes more to the overall hybrid because oxygen is more electronegative than carbon, so that's our enolate anion that is formed. So, if we form our enolate anion, we're also gonna get another product. So if I think about adding a proton onto our base, we're going to form an amines. Let me go ahead and draw the amine that we would make. So we would have, it would now be this amine. So let's show those electrons. So the electrons over here in red, these electrons right here pick up this proton forming this bond and we'd form an amine. So this reaction is at equilibrium. So to figure out which direction this favors, one way to do it would be to calculate the Keq. And you can see over here on the left, these ways for calculating the Keq, which we talked about in the last video, We first need to calculate the pKeq, which would be the pKa of the acid on the left, so the acid on the left would be acetone with a pKa of approximately 19 or sometimes you'll see 20, and so from that number we're going to subtract the pKa of the acid on the right, which would be our amine. The pKa of this is approximately 36, so 19 minus 36 gives us negative 17. To find the Keq, all we do is take 10 to the negative of that number, so 10 to the negative of negative 17 is the same thing as 10 to the 17th, which is obviously a huge number, much, much greater than one, so we know that the equilibrium lies to the right. So the equilibrium favors formation of the enolate anions, and for all practical purposes, this is a huge number, you're pretty much gonna get complete formation of your enolate anions, so if you add LDA to acetone you're gonna get again, pretty much all enolate anion, and none of your acetone will remain here. Another way of figuring that out is to know that the equilibrium favors formation of the weaker acid. The weaker acid is the one with the higher value for the pKa, remember the lower the pKa the more acidic something is, so acetone is more acidic than this amine, and so since the amine has the higher value for the pKa, the equilibrium favors formation of this weaker acid, also to the right like that. Alright, so let's talk about the pKa of acetone or ketones in general compared to aldehydes. So this pKa is higher than that for our aldehyde like we saw in the last video, and we can think about why by looking at the enolate anion and thinking about the fact that now we have a methyl group here, and alkyl groups are electron donating so, if it's donating a little bit of electron density, it already has a negative charge on it, so donating a little bit of electron density would destabalize this negative charge a little bit. So if the conjugate base is destabalized, that means that acetone is not quite as likely to donate a proton. So that's the situation when you have a ketone here. When we had an aldehyde, we didn't have this electron donating factor, and so that's just one way to think about why an acetone, why a ketone, is not as acidic as an aldehyde in general. Alright, here's an example where we have a very acidic ketone. So this is a special kind of ketone called a beta-diketone, this is a beta-diketone here. And if we look for alpha carbons, we look for the carbon next to our carbonyl, so this is an alpha carbon, this is an alpha carbon, and this is an alpha carbon, And so the question is which one of those, which one of those alpha carbons is the one that has the most acidic protons. It turns out to be the alpha carbon in the center, the one between our two carbonyls, and so there are two alpha protons on that carbon. And so the pKa for one of those, for one of those acidic protons is about nine, so that's very acidic, much more acidic than acetone or acetaldehyde like we talked about in the last video. And so since a beta diketone has such acidic protons, we don't need a super strong base like LDA, we can use something like sodium ethoxide, so negative one formal charge here on this oxygen, so sodium ethoxide could be used to deprotonate this beta diketone. So we think about one pair of electrons taking this proton, leaving these electrons behind on this carbon. So let's go ahead and draw the conjugate base, so we would have our carbonyls here, and we took a proton away, so now we have our electrons on this carbon, so a negative one formal charge. So these electrons right in here, magenta, moved onto this carbon, which gives this carbon negative one formal charge, because there's still a hydrogen bonded to that carbon, I'm just not drawing it in so we can see a little bit better. For resonance structures, we could show these electrons in magenta moving into here, pushing these electrons off onto the oxygen, so we could draw a resonance structure for that. So for our resonance structure, we would form a double bond here, and then this oxygen would get a negative one formal charge like that. And then our carbonyl on the right is still here like that. So that's one of our possible resonance structures, electrons in magenta moved in here to form our double bond, and then we could show these electrons in here moving off onto our oxygen like that. We could have shown our electrons moving over on this side as well, so we can push those electrons off. So let's go ahead and draw another resonance structure here. So let's get some room and let's show the formation of another resonance structure. This time our carbonyl on the left is still there. We moved some electrons into here, and then we would form a negative one charge on this oxygen, so a negative one charge right here like that. And so let's show those electrons, electrons in red move in to form our double bond, and then let's show these electrons in here coming off onto this oxygen like that. So a total of three resonance structures for this enolate anion. And so you could see this negative one formal charge is delocalized, right, it's delocalized over this carbon, it's delocalized on this oxygen, it's delocalized on this oxygen. So the more you delocalize, or spread out a negative charge, the more you stabalize the anion. So this is a very stable anion because of resonance and also because of conjugation. If you think about, if you think about the conjugation present here, so here's a double bond, and here's a single bond, and then here's a double bond. So you have some conjugation stabalizing it as well. And so we have a very stable conjugate base for all those reasons that we just talked about. And since we have a stable conjugate base, this beta diketone is likely to donate one of these protons, and so that's why it's pKa value is so low. Alright, so if ethoxide takes one of these acidic protons, we're gonna form ethanol as our other product. So we can go ahead and draw ethanol in here like that, and then we can calculate the Keq for this reaction. So ethanol's pKa we've already seen is approximately 16, so if I want to calculate the Keq, first we find the pKeq. So the pKa of the acid on the left, the acid on the left is this beta diketone, so that'd be nine, from that number we subtract the pKa of the acid on the right, which is ethanol, so nine minus 16 gives us negative seven, so that's the pKeq. To find the Keq, we take 10 to the negative of that number. So 10 to the negative of negative seven, is of course the same thing as 10 to the seventh, and obviously that's a number much, much greater than one. And so we know the equilibrium lies to the right, favoring formation of the enolate anion here. So favoring formation of our enolate anion. So for all practical purposes we're gonna get pretty much nearly complete formation of our enolate anion. And so, we're gonna get our enolate anion, and that's because even though we use sodium ethoxide, it's not as strong of a base as LDA, but it's enough, it's enough to deprotonate our very acidic beta diketone.