Carbonyl Condensation Reactions
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HydrocarboxylierungSodium hydroxideCondensation reactionAldolCarbon (fiber)Aldol reactionEnolHydroxideKetoneAldehydeEsterClaisen-UmlagerungMixing (process engineering)Lone pairDehydration reactionAlpha particleBromineClaisen-KondensationWine tasting descriptorsCarbonylverbindungenElectronHydrogenHydroxylOxygenierungBeta sheetNitrogen fixationReaction mechanismAcidBase (chemistry)CobaltoxideProtonationWaterKohlenstoffgruppeElectronegativitySeparation processComputer animation
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ConcentrateMaterials scienceAcidHydrogenReaction mechanismAcetateAldol reactionConjugated systemHydroxideWalkingElimination reactionResonance (chemistry)Carbon (fiber)Process (computing)Lone pairHydrocarboxylierungBeta sheetHydroxylCobaltoxideWaterElectronEnolNitrogen fixationMedical historySeleniteFunctional groupAldehydeVolumetric flow rateExothermieDoppelbindungCondensation reactionAldolTanninSchwermetallLactitolMotion (physics)KetoneWursthülleAldehydeAssetHydrateComputer animation
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AcidHydroxylEnolBeta sheetHydrateReaction mechanismFunctional groupFood additiveWursthülleWaterGesundheitsstörungThermoformingProcess (computing)Aldol reactionBase (chemistry)Dehydration reactionLactitolData conversionAldolConcentrateConjugated systemAldehydeKetoneChemical compoundSetzen <Verfahrenstechnik>PharmacokineticsDoppelbindungLeft-wing politicsHydrocarboxylierungAcid dissociation constantCondensation reactionSense DistrictAlcoholAldehydeAlpha particleAcetoneOxygenierungDeathMoleculeElectronic cigaretteCollisionHydroxideBy-productComputer animation
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ColourantWaterAcetoneHydroxideHydrocarboxylierungWalkingReaction mechanismElectronCobaltoxideFunctional groupHydroxylGesundheitsstörungMoleculeEnolDeuteriumAcidChemical reactionHydrateAldehydeAlpha particleKetoneCondensation reactionAldolBase (chemistry)Sodium hydroxideAlkoxideElimination reactionCarbon (fiber)ProtonationComputer animation
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HydrocarboxylierungAcidCarbonylverbindungenDoppelbindungCarbon (fiber)AldolBase (chemistry)Aldol reactionActivity (UML)AromaticityCondensation reactionChemical compoundEnolHydrogenChemistryMoleculeCHARGE syndromeAlpha particleHydroxylThermoformingProtonationBeta sheetBenzaldehydeEsterReactivity (chemistry)AldehydeWasserstoffionPlant breedingData conversionSodiumPhysical chemistrySeparator (milk)Azo couplingChemical propertyAcetoneKetoneDehydration reactionWaterAldehydeWursthülleComposite materialComputer animation
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Computer animation
Transcript: English(auto-generated)
00:09
Good afternoon we're going to get started, alright. Have you guys noticed that like somewhere in the middle of this class is just like time is just flying by so fast?
00:20
We're starting our last carbonyl chapter and oh my gosh, okay. I made a mistake last time of several people pointed out to me so let's fix that. I did not count carbons.
00:43
So it makes a really good point. Alright I have the right number of carbons here, one, two, three, four, five, but this is one, two, three, four, five, six, I've inadvertently added another carbon, right? Very easy mistake to make, I like to say I did that on purpose to see if you caught
01:01
it but I didn't. Just like that, one, two, three, four, five. Okay very easy to make that mistake so that makes a good point. Alright, we're going to start chapter twenty-four, we're going to learn about the aldol condensation
01:22
and the Claisen condensation.
01:41
We'll start with the, we're going to start with the aldol first but let me just introduce the idea behind this chapter. So we know that enols and enolates from chapter twenty-three are electron rich nucleophiles, they can react with a number of different electrophiles and so we have our carbonyl compound which is an electrophile because we have the partial positive charge on carbon.
02:08
If we remove the beta hydrogen, the alpha hydrogen we get this electron rich nucleophile and then when we add electrophiles it will attack the electrophiles so we incorporate
02:22
electrophile into the alpha position. And so of course our electrophiles are D2O, D3O+, bromine, Br2, and RX. So that's what we talked about in chapter twenty-three.
02:41
So in this chapter we're going to add to those lists of electrophiles and say what if the electrophile was another carbonyl compound? Rather than D2O, D3O+, bromine, RX, what if we had that electron rich carbon group attack another carbonyl and that's the focus of this chapter here.
03:01
So if it does that you have a condensation reaction and we're going to talk about two broad categories of these reactions. If you have an enol or an enolate plus an aldehyde or ketone, it's so the enol or
03:25
enolate of course is the nucleophile and the aldehyde or ketone is the electrophile then you have an aldol addition or an aldol condensation. Depending on what product you get, aldol addition or aldol condensation.
03:41
If you have an enolate plus an ester, so enolate of course being your nucleophile, ester being your electrophile, if you have an enolate plus an ester that's a Claisen condensation.
04:03
So we'll start with aldol addition and aldol condensation and then we'll talk about Claisen and then you'll see once we get through these sections we're going to kind of mix it up because you can have, you can mix the two of these up and get unique products. All right, so we'll start with the aldol, we'll do base catalyze and then we're
04:27
going to do acid catalyze. Your book does not have acid catalyze but I'm going to show you acid catalyze. All right, so here's base catalyze and here's the product that you get and then we're going to go through the mechanism. Very highly likely that this mechanism will be on midterm two.
04:40
Very highly likely you'll see an aldol, very highly likely you'll see a Claisen condensation on midterm two. That's kind of usually how I roll with that material. Okay, that's aldol addition.
05:03
Our product is a beta hydroxy carbonyl so let's see where we get that. Here's alpha position, here's beta position. This is a beta hydroxy carbonyl. So you want beta hydroxy carbonyl to be on your radar. If you see a beta hydroxy carbonyl you can make that potentially by an aldol addition.
05:25
So then what can also happen is this can go further, more sodium hydroxide and typically heat, not necessarily though and you could do a dehydration
05:42
and if you do a dehydration you get an alpha beta unsaturated carbonyl and that's called an aldol condensation product and you'll see that sometimes we stop at the beta hydroxy
06:04
and sometimes we go all the way. But again this is our alpha beta unsaturated carbonyl. So a nice way to make alpha beta unsaturated carbonyls.
06:23
So that's the idea. I'm going to show you conceptually what's happening and then we'll go through the complete mechanism. So here's the what's going on here which is not the complete mechanism. If I ask you for this mechanism on the test then that one I want you to see is the complete mechanism.
06:49
All right. This is where everything we've learned in the carbonyl chapters comes together in this chapter here.
07:00
All right so we have an electron rich nucleophile. It's electron rich nucleophilic carbon. And then over here we of course got the partial positive on the carbonyl carbon, partial negative here.
07:22
So this is electron poor or electrophilic carbon. All right so what we have here is the,
07:41
going to have the lone pairs coming from the electron rich. The lone pairs always come from electron rich and go to electron poor. I'm going to check the carbonyl. We're going to kick electrons up onto oxygen. So here's mixing up everything we learned in chapter 23 with everything we learned in the rest of the chapters.
08:22
All right and then so then you can imagine we do protonation. This is looking like chapter 20 now. We do protonation.
08:46
And then usually heat, although sometimes you don't need heat as we'll see coming up, usually heat you lose water. So that's a dehydration reaction. And that takes you all the way to the aldol condensation product.
09:06
All right so let's go through the complete mechanism. Questions so far conceptually for that reaction? Anybody? Too fast? Okay we'll leave that there.
09:22
All right complete mechanism. All right so hydroxide ion. Now as we've been doing throughout these chapters you can when we make the enolate ion
09:41
you can leave the electrons on carbon or you can push them onto oxygen. I'm going to try to go back and forth here and do it both ways. What I find is that most students like to leave it on carbon in chapter 24. It makes it a little easier to draw products. But let's draw it both ways and I'll show you the alternative way
10:01
right underneath here. So for fair coverage I like to go back and forth here. So and then the alternative is to go onto oxygen. I'll put that one in green.
10:32
So there's our nucleophile. So all we're drawing here is just the resonance structure for the one above.
10:43
And that's what makes it perfectly fine to do this. Either way. So again these guys are resonance structures so it's perfectly fine to draw it that way. And now so now we've made our nucleophilic carbon and now we're going to attack electrophilic carbon.
11:04
So that's going to mean that we're going to attack the aldehyde that has not been deprotonated because once we deprotonate the aldehyde we turn it into a nucleophile. We're not going to have a nucleophile attack a nucleophile. So we want to have, we want to attack carbonyl that has not been deprotonated. And so electrons are going to come
11:21
from the lone pair on carbon. We're going to attack the carbonyl carbon. We're going to kick electrons up onto oxygen. All right. If you draw it the alternative way here then this is what it will look like.
11:49
And again, it's your choice. You do not have to draw both ways but you can, it's going to give you the same product. So let's draw the product here.
12:02
Notice the reversible arrows. This is a reversible reaction.
12:22
And then we're going to protonate. And we're going to protonate with water. And when we protonate with water we're going to regenerate our catalyst.
12:43
That's aldol addition. Beta hydroxy carbonyl is your product if you do just the aldol addition. All right. So this guy right here is going to take you to the same thing. So we'll just do these arrows right here. And I'm just going to put this in a box as the alternative way to draw this.
13:02
So from here on out, sometimes I'm going to push electrons onto oxygen. Sometimes I'm going to push them onto carbon when I make the enolate. All right. So let's keep going here because we want to go all the way to the dehydrated product. So what I'm going to do is scribble this out right here
13:27
and draw one of those hydrogens in because it is a two step elimination. Two step elimination. We make the enolate ion first and then we kick off hydroxyl.
13:41
There is a heavy temptation to just go directly. Let me show you what I'm talking about. We're going to actually remove this acidic hydrogen right here and make an enolate first. We're going to push electrons onto oxygen and then we're going to kick off hydroxyl. There is a heavy temptation to just have the hydroxyl come in
14:02
and remove this hydrogen, push electrons here and kick off the hydroxyl in one step. It has not been shown to be a one step process. So if you do it as a one step process you will miss four points because when you're combining steps into one you miss the points for both steps. So it is a two step elimination. That being said, a bunch of you
14:22
are going to do it as one step on the test and well maybe that won't happen this time. So you make the enolate first and here I like to, when I make the enolate here I like to push electrons onto oxygen because I just like the flow of the whole thing you'll see here. Make the enolate first
14:51
and of course we're going to be kicking off hydroxyl as a leaving group going to something that's really stable. So as we've seen we can have hydroxyl leave if we're going to something that's really stable.
15:01
All right and then the electrons come down from oxygen here and then we move this over and then we kick off hydroxyl. So we're going to do this in the forward direction and in the reverse direction.
15:20
We'll be doing the reverse direction in discussion next week and you can see this will be our product here. All right so let's put a note for ourselves here.
15:42
Important two step elimination. Step one, make enolate. And when you make the enolate you can leave electrons on carbon but you do have to make the enolate first.
16:03
And then in step two, you eliminate hydroxide. Questions on that mechanism, anybody? Yes, oh, that is not irreversible.
16:25
Let me fix that, I don't know why I did that. Thank you, yeah. Okay, more questions, yes. Yes, no it's not.
16:50
Boy I'm having a history of this now. I normally don't, let me fix that. Yes, that is the incorrect number of carbons.
17:05
Okay so now I can't do that anymore for the rest of the quarter, right? Okay, so everybody fix that, I got one too many carbons. I'm doing that a lot lately.
17:20
All right, two step elimination. Everybody have this? So hydroxide's a leaving group, conjugation makes the double bond particularly stable. This allows hydroxide to be removed in a very exothermic reaction. All right, I want to make some points about this aldol condensation here on the next page.
17:41
And that is aldol reactions in equilibrium process as you saw with all my reversible arrows, especially the ones I fixed in the last step. And the concentration of aldol product at equilibrium depends on the substrate use. With aldehydes, the aldol product is favored, but not by a huge amount.
18:11
So 55% at equilibrium. And we're gonna have some ways to tweak this a little to get the product that we do want.
18:20
So 55% at equilibrium is not a good ratio. So that means we still have 45% that is not, that has 45% of starting material. With ketones, starting materials are favored. And so here would be the product. And what you want to do, the first learning curve for aldol is to be able to draw the correct product.
18:42
So you want to go back today and make sure that I'm drawing these products correctly here by going through the mechanism for both of these reactions. And for this one here, it's only 1% at equilibrium. So this kind of goes along with what we talked about
19:00
with hydrates and with acetals. The more electrophilic, the more hydrate, the more acetal you form, and in this case, the more aldol product you form. And so when you try to do this with a ketone, the carbonyl is more hindered and less electrophilic.
19:33
And so that's why we see only 1% of product. And so if you want to drive the equilibrium, we're gonna again talk about some other ways
19:42
to do this also, but if you want to drive the equilibrium, then you want to go all the way to the aldol condensation product. Both reactions can be forced to completion by dehydration, the exothermic dehydration drives the alcohol equilibrium to the right because we're forming such a stable product.
20:00
So all the product wants to be in that form. And so basically if we take this same acetone and we drive the equilibrium usually by heating, again, we're gonna see some examples where we don't even have to heat. Usually you just heat that up
20:22
and you're gonna form the alpha beta unsaturated carbonyl and because this is such a stable product, that's what's gonna drive the equilibrium. So we get 100% conversion.
20:43
Why exothermic? And the answer is right here. Exothermic, the product ketone is conjugated with the double bond which makes it more stable.
21:00
Product ketone is conjugated with the double bond. Conjugation, it makes something more stable, lower energy and that's it. So remember at equilibrium, when we have an equilibrium process, we always favor the most stable compound.
21:24
All right, and that helps us understand a little bit what the next point I want to make about the aldol condensation. And I actually got some students asking me this on Wednesday. Conditions for aldol reaction are the same
21:41
as for hydrate formation, right? Exact same conditions. How do we know we're not gonna form hydrate? What do you think? Well how do we know we're not gonna, let's show the product for hydration. Let's do reversible arrows here.
22:04
Hydration was chapter 20. So here's the hydrate.
22:22
And again, we can form the aldol addition product or we can form the aldol condensation product. In this case, we would heat.
22:47
Why don't we form hydrate instead? Anybody? It's, well, not tautomerization. What do we know about the stability of a hydrate?
23:01
It's not stable, right? So we're in equilibrium conditions. This is unstable. In equilibrium conditions, you always form the more stable product. And this is much more stable. Not that, under equilibrium conditions,
23:40
under equilibrium conditions,
23:44
the most stable product is favored. You absolutely will form hydrate. When you're doing an aldol reaction, you will form hydrate,
24:02
but it's a reversible process. And as soon as you form it, it's gonna be like, okay, that's a dead end. It's gonna go back again. So remember, the molecules don't have brains. They don't know, oh, I have to go all the way to that. That's more stable. They're gonna collide. And if they collide in the right orientation with the right energy, they'll react.
24:20
But if it's a reversible process and it's not something that's stable, it will just go right back again. So they're pretty much gonna try out all the possible things until they hit upon something that's more stable. Once they hit upon something that's more stable, that's low energy, they don't go back again. Okay, so it's just, we like to think of them as being smart, but they're not.
24:47
All right, and I'm supposed to write that down there, but we can just write it up there anyway. All right, so the third point I wanna make. Look at the pKa for enolate ion formation and the above reaction. All right, so certainly this is not favored.
25:04
Forming the enol with hydroxide. Let's look at how much we have. This is where that little thing that I do about how much do we have at equilibrium is actually very helpful. Really comes into play well here. There's the acid-base reaction. We know how to know the direction of equilibrium
25:21
in acid-base. We look for the conjugate acid-base pairs and we look up the pKa for the acid. There's one acid on the right-hand side and there's one acid on the left. So that's approximately 15. And this one is approximately 20. So now we have an idea.
25:41
The reaction is favored to the left, but how much is it favored to the left? At equilibrium, favored to the left. At equilibrium, we have 10 to the third to one.
26:01
No, 10 to the fifth to one. So the equilibrium only contains a small amount of the enolate. Yeah, go ahead.
26:24
Did I draw that wrong? Yes, I did. Oh, what's going on here? Okay. Okay, now really I'm not gonna do that anymore. Okay, I swear. Okay, and here's a good example
26:40
of where we actually want only a small amount of enolate. Because remember, in the aldol, the enolate attacks an aldehyde that hasn't formed an enolate. If we completely deprotonated the aldehyde, it would have nothing to attack. Does that make sense? If we used LDA and we completely formed the enolate,
27:02
then all of the aldehydes converted to the enolate, now we have nothing left to attack. So the fact that we're only forming a tiny amount of the enolate is good because that tiny amount of enolate that we're forming, that one enolate now has 10 to the fifth of these to attack.
27:21
Okay, so it's got a lot of those to attack. And so that's important in this reaction that we have a weak base here. If we used LDA, we wouldn't get that, we wouldn't get, if we used one equivalent of LDA, we would have nothing left to attack. So that's actually very helpful in this reaction. All right, so that's what I say here.
27:41
Even though the equilibrium concentration of enolates small, it still serves as a useful reactive intermediate. Once the enolate reacts with an electrophile, it's removed from the equilibrium, and then now we're gonna form another one that can now attack another aldehyde. So eventually all reacts via low concentration of the enolate ion.
28:03
All right, let's look at the acid-catalyzed mechanism. And notice what I say here. The acid-catalyzed mechanism goes all the way to the dehydrated product, and you can't isolate the addition product.
28:25
Remember that, okay? In base, the dehydration you have to kick off hydroxyl, it's not as favorable. So you can sometimes stop at the beta-hydroxy. In acid, that leaving group's protonated
28:40
and it comes off so much more easily. So let's go through the acid-catalyzed mechanism. You get the same product, side product is water here.
29:02
All right, so we know that in acid-catalyzed mechanism, enol is the nucleophile, not enolate ion. We don't wanna be making enolate ions in an acid-catalyzed mechanism. So let's go through this here. I have put this on exams before also, the acid-catalyzed reaction.
29:29
All right, so just to slow me down a little bit, when I slow down, I make less mistakes. I'm gonna do the changing colors thing, because I'm already over my limit of mistakes for the day. So that slows me down a little bit,
29:43
and that always helps me. So we're gonna make an enol rather than an enolate ion. So notice even though I'm starting with the acetone,
30:00
this is gonna go all the way to the dehydrated product. And let's change this into a two, and I'm gonna remove this. What do I wanna use to deprotonate that with? Water, not hydroxide.
30:20
And we don't want hydroxide ion appearing in an acid-catalyzed mechanism. So I wouldn't use, why do I, okay, you didn't see that. All right, so we're gonna make the enol. That's gonna be our reactive intermediate.
30:59
So there's the enol.
31:02
The enol's going to attack a protonated carbonyl, because that's how we do acid-catalyzed mechanisms, okay? So I already showed how to make a protonated carbonyl. I did that in this step right here. So I'm just gonna have my protonated carbonyl. I don't need to show it twice. There is protonated carbonyl in this reaction mixture.
31:27
So that's ready to go. And then I'm gonna have the electrons on oxygen come down, and we're gonna attack the carbonyl. So big difference with the acid-catalyzed,
31:42
we always attack a protonated carbonyl, and we always attack from the enol, not the enolate ion.
32:06
And notice, we don't have an alkoxide, because we attacked a protonated carbonyl, so that's already protonated for us. We have our protonated carbonyl. We are still gonna have our two-step elimination. Well, actually, it's gonna be three steps, right?
32:21
We're gonna make the enol first. Acid-catalyzed reactions always have more steps. So we're gonna make the enol again.
32:43
Now, what do we know about leaving groups in an acid-catalyzed mechanism? What's something we always do to leaving groups in an acid-catalyzed mechanism? Protonate them first. So I'm gonna protonate that carbonyl.
33:06
I mean, I'm gonna protonate that hydroxyl, make it a better leaving group. So in the acid-catalyzed mechanism, we are not kicking off water.
33:23
We are not kicking off hydroxide, we're kicking off water. So does it look like we're ready to go to kick off our leaving group?
33:46
Electrons on oxygen come down, and we kick off our leaving group. So always a few extra steps in our acid-catalyzed mechanisms.
34:03
And you can see we are one step away. All we need to do is deprotonate the carbonyl now, and we'll be all set with our product. Deprotonate with water, that regenerates our acid catalyst.
34:28
So the kind of mistakes I see on mechanisms like this on the exam are when students mix like half acid-catalyzed, half base-catalyzed. So that would be the biggest mistake. So in other words, like deprotonating with hydroxide
34:42
when you're not gonna have hydroxide appearing in an acid-catalyzed mechanism. So kind of mixing up the two. And that's the thing you wanna get straight before midterm too. Questions on the acid-catalyzed mechanism? Yeah.
35:05
Electrons on the oxygen attack the carbonyl directly. Oh, like you're talking about say, where did I put that? So right here, have water attack the carbonyl.
35:23
The first step, right here? One step below that? Oh, the oxygen attack the carbonyl. So this coming and attacking the carbonyl, not so much.
35:41
Yeah, yes, it absolutely will. But that goes to a hydrate, which is not stable. You are going to form a lot of hydrate in this reaction. You won't isolate hydrate. And how do we know that? How can we know that?
36:01
If we take, if you take labeled water, so O18 water, and you do this reaction, you will have O18 in the carbonyl. And the only way you can do that is if you are attacking the carbonyl. Remember we had the mechanism where we took acetone and then we treated it with hydroxide and water that was labeled and we got labeled acetone.
36:22
So if you did this reaction, if you did this reaction with labeled water, you will see O18 in the oxygen. So we know we're forming hydrate. There's another reaction that we do with these same conditions. Do you remember what that is? From chapter 23?
36:44
Right, so we make the enol. Okay, so we make the enol here and the enol just protonates back again, right? Reverse that reaction, that's also gonna happen. So we don't, the molecules don't necessarily do a straight line like this.
37:01
So we can form the enol and then we can go right back to starting material. How do we know that's actually happening? If we use D2O with deuterium and oxygen labeled, you will start seeing deuteriums incorporated into these alpha positions. That's how we know all of these things are happening.
37:20
But again, at equilibrium, you're gonna form the more stable products. So all these things will happen, but once you form the more stable product, it stays and doesn't go backwards, okay? More questions, that was a really good question, by the way, anybody? All right. Okay, so that's acid catalyzed. Now we wanna talk about crossed aldol.
37:43
So this would be more of a self-condensation where we take our aldehyde, we add base, aldehyde or ketone, we add base and it condenses with itself. We wanted to use two different aldehydes and condense them together.
38:04
So we're gonna have to be really careful if we try to do that. Here's a bad example. We'll call this aldehyde A and we'll call this aldehyde B. And we treat that with sodium hydroxide and H2O.
38:25
So you get four different products. This would be enolate of A adds to A. And if I were you, what I would go home and do today is I would prove to myself
38:40
that these are the four possible products, make sure you know how to do that. Because that's the first learning curve for aldol, okay? This one is enolate of A adds to B. And this one would be enolate of B adds to A
39:08
and enolate of B adds to B.
39:23
All right, so that's not something we would want to do because if there's a specific product out of that four that we want, we wanna make just that product. We don't wanna have to separate those four compounds that would not be a very easy separation. They're gonna have very similar physical properties.
39:41
So we're gonna have a couple ways around this. First way around this is choose one of the reagents so that, choose the reagents so that one of the participants has no alpha hydrogen so that you can only form one enolate. And then you choose the other participant to be more electrophilic and then you can get a clean reaction.
40:02
So here's an example, this is a good example. Carefully choose your reagents. All right, so if you notice that we have benzaldehyde as our first reagent cannot form an enolate. So no alpha hydrogens, therefore no enolate.
40:21
And this is where all of that reactivity comes into play here. So you'll notice here our ketone, we can form the enolates.
40:41
So we're gonna form the enolate, it's symmetrical so there's only one side where we can, it can go on either side, we'll get the same product. So we don't have to worry about kinetic versus thermodynamic enolate. So we're gonna form the enolate here and it will add to benzaldehyde.
41:05
Why will that selectively attack benzaldehyde? Which compounds are the most electrophilic? Benzaldehyde, so this is gonna work really well. When we take two carbonyls,
41:26
one you form the enolate and the other one's more electrophilic, it's gonna be a good reaction. So this is more electrophilic. So we form the enolate here and it will add to benzaldehyde
41:42
because aldehydes are more electrophilic than ketones.
42:02
All right, and so what's the product you get? And in this case you cannot isolate the beta hydroxy carbonyl because the product is so stable it goes directly to this at room temperature.
42:20
Don't have to heat. So formed at room temperature, can't isolate the beta hydroxy carbonyl.
42:42
And that's because the product is so stable. Notice the double bond is not only conjugated with the ketone, it's also conjugated with the aromatic ring. And so that's what we talk about underneath here. You don't even need to heat the reaction to get dehydration to occur.
43:00
Once the enolate of acetone is formed it will more readily add to benzaldehyde than another molecule of acetone. So this is a very good example of an aldol condensation, a mixed aldol where you can get one single product. Questions on this example, anybody? So careful choice of reagents.
43:23
The other thing is maybe you have two compounds that can form enolates but one of them is so much more acidic. So all that acidity we learned about the relative acidity of carbonyls in chapter 23,
43:42
that comes into play here. So here we have active methylene compound.
44:01
A more acidic than a nonactive methylene compound. So when we add base we're gonna exclusively form the enolate on the active methylene compound. We are not gonna form the enolate on the regular carbonyl. And so this is going to be another great reaction here, it's gonna work well. So let's go through that.
44:26
Most acidic hydrogen in the active methylene composition. Which carbonyl is more electrophilic? We have an aldehyde and we have a diester.
44:43
What's more electrophilic? Aldehydes, so we're gonna attack the aldehyde exclusively. When we have one enolate that's gonna exclusively form in one of the reagents and it's gonna attack selectively the other reagent,
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you're gonna have a clean crossed aldol reaction. So that's the most acidic hydrogen. So what you're gonna do is you're gonna form this enolate. Remember with sodium methoxide we completely form this enolate.
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Complete conversion to this enolate. And now we're going to attack the aldehyde. So there isn't any ester left, there isn't any active methylene compound left that's not deprotonated.
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You're gonna protonate then eliminate.
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I'm not gonna show that here, we'll do that in discussion. Protonate then eliminate.
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And once again you can't stop at the beta hydroxy carbonyl because this compound, the product is so stable. That double bond that you just formed is conjugated with both esters. Both carbonyls with both esters.
46:41
Alright so that's gonna work really well. There's a third technique that we're gonna use and that's called a directed aldol and this kind of puts you in the driver's seat. So here's what you do, you prepare the enolate of one carbonyl with LDA. Okay so do chapter 23 chemistry,
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completely form the enolate and then add the second carbonyl to this enolate. So let's see what that looks like.
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LDA gives you the mostly the kinetic enolate. So you form this first and then you add your second carbonyl.
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So we add this next. So this is gonna attack that carbonyl.
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And rather than letting the molecules do their own thing, you're actually in charge here of what's happening. If you know what you're doing
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and this also lets you isolate the beta hydroxy carbonyl compound, depending on what you do if you add water here and you're careful not to add acid, you can isolate the beta hydroxy carbonyl. If you add acid what's gonna happen?
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You're gonna dehydrate aren't you?
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So you're in charge, you direct the reaction. We're gonna stop right there and I hope you guys have a great weekend.