How to think about the aldol condensation using retrosynthesis. Created by Jay.
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- AT7:26why does it need to be a directed aldol synthesis instead of the more simple version? is it because we do not end up with the conjugated form?(8 votes)
- You are doing a "crossed aldol" condensation with two different carbonyl compounds.
Let's call them A and B.
If you just mix them together with a base, you will get a mixture of four different aldols: A-A, A-B,B-A, and B-B.
Your maximum yield would be about 25 %.
It is much better to form the enolate of one compound first and then add it to the second carbonyl compound.(8 votes)
- At7:41, can we use NaOH to carry out this reaction?(4 votes)
- The pka of alpha-protons on ketones are about 20, while that of water is about 16. You can use these to predict where the equilibrium would lie. If you reacted 3-pentanone (your ketone) with NaOH, you would form water. Using the pka's from earlier, take the pKa of the reactant minus the pKa of the product (20-16=4).so Keq is equal to 10^-4. since this answer is less than one, the equilibrium will favor the reactants. This means that you will have plenty of hydroxide and ketone (but very little of your desired enolate) floating around in solution when you add in butanal in your second step. Aldehydes have a lower pKa for alpha protons than ketones, so will be more reactive with the residual hydroxide. This will form a different enolate that can go on to react with other carbonyl groups. Basically, using NaOH will lead to the formation of a mixture of aldol products, which is not very useful synthetically. We have to convert the ketone to our desired enolate (using a strong base like LDA) then add in our electrophile.(6 votes)
- At5:09, he includes EtOH. What is the EtOH required for in the aldol condensation? Donating proton to the oxyanion intermediate product?(3 votes)
- why not just do an ozonolysis reaction?(1 vote)
- what is cannizzaro's reaction ?(2 votes)
- The Cannizzaro reaction can take place with aldehydes that lack a hydrogen on the alpha carbon. Under basic conditions, half of the aldehyde molecules are oxidised to the corresponding carboxylic acid and the other half are reduced to the corresponding alcohol. I don't think it is covered by any videos in this series.(1 vote)
- Hey guys just a quick one, how do you know where to assign the alpha carbon? On the video, it is assigned as the carbon before the carbonyl group however what if there is no carbonyl group in the molecule? Thanks guys!(1 vote)
- in the ethanol solution will the Na in NaOH form sodium ethoxide?(1 vote)
- It would! The hydroxide anion (OH-) could deprotonate the H-OEt in ethanol leaving EtO- or an ethoxide which would go with the Na+ to make sodium ethoxide. But, it is important to notice that the pKa of water is 14 and the pKa of ethanol is 16 which means that water is 100 times more acidic than ethanol! As such our newly formed sodium ethoxide would immediately deprotonate a nearby water molecule to reform NaOH and ethanol. In solution this process of deprotoantion and reprotonation forms an equilibrium solution of ~100:1 NaOH to NaOEt(2 votes)
Voiceover: We've done a lot of aldol condensations, and in particular, we've looked at the mechanism in great detail. In this video, we're going to think about doing things in reverse. So we're going to start with a retro aldol reaction and see how that way of thinking can apply to retro synthesis. And so let's start with cinnamaldehyde right here. And if we do a retro aldol reaction, the mechanism is pretty much the exact reverse of an aldol condensation. And so if we had a base like sodium carbonate, we're gonna form benzaldehyde and acid aldehyde. A cinnamaldehyde is the molecule that gives cinnamon its smell, and benzaldehyde smells like almonds, and so this is a pretty cool experiment to do. You can stat with the molecule that smells like cinnamon and end with a molecule that smells like almonds. Let's analyze our cinnamaldehyde starting compound here to see how we would have formed those products. We know the carbon next to our carbonyl is our alpha carbon. And we know the carbon next to that is the beta carbon. And so looking at the structure, all right, we know that there's a hydrogen attached to our alpha carbon like that. So we think about breaking this double bonds, right? We can see this this two carbon setup over here on the right for acid aldehyde, right, so those are the two carbons, and this is the hydrogen that's bonded to that alpha carbon there. Looking at the structure again on the right, we know that there are two more hydrogens bonded to that alpha carbon, and so therefore, if we're doing things in reverse, we can think about adding two hydrogens to this alpha carbon, and then go ahead and draw two hydrogens right here, so I'm just going to write H2. And then we can think about adding an oxygen to the beta carbons, and we can think about adding H2O. I know there's already a hydrogen on that beta carbon, so I could think about this being the hydrogen on that carbon. All right and then if I add an oxygen to that carbon, that would be this oxygen right here. So just a, a way of thinking in reverse, right? Adding water, right, breaking your double bond and adding an oxygen and two hydrogens will, is one way of thinking about forming your product here. So or you can think about the reverse. You can think about losing water from this portion and sticking these two fragments together to give you your aldol condensation product. So once again, this way of thinking can be can be very useful if you're trying to retro synthesize something, so if you're thinking in reverse. So for example, let's say a question on a test was show how you could synthesize this enone here, so once again we're gonna do the same kind of analysis. We're gonna find our alpha carbon right here; this is our alpha carbon, and this is beta carbon. All right, so attached to our alpha carbon, all right, we know that there's a hydrogen right here, so we're gonna think about breaking this double bond, right? So let's go ahead and draw a retro synthesis arrow here. So we're gonna break that double bond, so on the left, all right, we're gonna have our benzine ring. We're gonna have our carbonyl here. And then we're gonna have a hydrogen attached to this carbon, attached to our alpha carbon. And remember, we're gonna add two hydrogens to our alpha carbon. So we're gonna add two hydrogens to our alpha carbon, and that would give us the ketone that we would need. Right, for our beta carbon, we know there's already a hydrogen attached right here, and so we're going to add an oxygen to this beta carbon on the left, so this is our way of thinking about it. So over here on the right we're gonna have we're gonna have our carbonyl. All right, and then we're adding an oxygen, right, we're adding a oxygen right here. And then we had that hydrogen, so I'm gonna go ahead and draw that hydrogen in blue. And then we can go ahead and draw our ring. So we have our ring right here. And then we have our nitro group coming off over here. So just to check ourselves and make sure that this is the right way of thinking about it in reverse, if we take water from this portion, right? And stick those two fragments together. Then that would give us our conjugated enone product over here on the left. And so that's, again, one way of thinking about how to do these sorts of problems. And so if they if they wanted you to give some some reaction conditions here, you can go ahead and, and draw what you would need, right? So you would say, I need to start with this ketone, right? Like this and then I would also need an aldehyde, so I'm redrawing the 4-Nitrobenzaldehyde product, right? So we have 4-Nitrobenzaldehyde like this. All right, so this compound, this ketone, of course, is this compound right here. And then 4-Nitrobenzaldehyde here is the same one that we drew right here. So if you take those two, right, and you, and you put them together, you should form your conjugated enone. Because remember, thinking about your alpha protons, right, there are no alpha protons on our 4-Nitrobenzaldehyde. Right, so this,this carbon right here doesn't have any alpha protons, so the only source for alpha protons would be this carbon right here. This is your only alpha carbon with alpha protons on it. So in terms of in terms of what else we would need, we would need a base, all right, to do our alpha condensation, so we could add a something like sodium hydroxide as our base. And we could we could create a solution of sodium hydroxide and water, and then also ethanol so if you took these ketone right in this aldehyde and use these reaction conditions, you should form the conjugate enone as your major product. And so that how to how to synthesize it, so once again thinking in reverse. Let's do another one. This one's a little bit different because we don't have an enone for an enol as our target compound. We have an aldol as our target compound. Well, we can use the same kind of thinking. We can find our carbonyl and identify the carbon next to that as being our alpha carbon and then thinking about this as being our beta carbon, right? And we know that this is the bond that forms. So we can think about breaking this bond. And and we know right, we know that we have a hydrogen on this carbon right here, as I'm just gonna see if I can stick it into there. So when I think about retrosynthesis, all right I go ahead and draw my retrosynthesis arrow in here, so on the left, I mean I can see on the left that I'm going to have this carbon two, three, four and five, right? So I have these five carbons right here so, we go ahead and draw those in. So we have our carbonyl, and then we have our five carbons, and then I already have this hydrogen in magenta, and we're only going to add on hydrogen this time. So let me go ahead and draw in one hydrogen in green because this isn't the congregated product. We stopped at the aldol if you're thinking about it in reverse. So we're not going to add two hydrogens, we're only going to add one here, and so what else would we need; so over here on the right, let's think about the fact that there is hydrogen attached to this carbon right here. And then how many carbons will we have for our other compound, one two three and four, so we need a 4-carbon carbonyl. Right, and there's a hydrogen attached to it, so a 4-carbon, carbonyl. Let me go ahead and draw that in, one, two, three four, and then we have this hydrogen ion glued to it. So now, we've identified and thinking about it in reverse, what compounds we would need to form this aldol product. We would need this ketone and this aldehyde. But, in terms of in terms of actually synthesizing the aldol compound on the left, we have to be a little bit careful about how to do it. We need to do a directed aldol addition here, so let's think about starting with this ketone, all right? So think about starting with this ketone here, so let me go ahead and redraw it. I'm going to draw it in a slightly different way. All right, so that's this ketone right here. And to that ketone, we're going to add LDA. So we're going to form a lithium enolate here, and once we've formed our lithium enolate, then we can add our aldehyde. So in our second step, we can add our aldehyde in here, so we talked about this in earlier video. So we have our aldehyde as a 4-carbon aldehyde, all right, so we're going to add butanal. And then that's going to give us a, a lithium alkoxide, right? So we need to we need to protonate the alkoxide, so in the workup, we can add some water here. And that would give us our aldol product, so that would give us this, so that is the way to synthesize our aldol products. Let me go ahead and draw it in here, right, so we would form this. Okay, so a directed aldol addition. Let's do, let's do one more of this retrosynthesis approach. And let's think about how to make this molecule. So this is a little bit different than before. Okay, but we can do our same stuff. So we can identify our alpha carbon, right, the one next to our carbonyl. And then our beta carbon is the one next to that, right? So we can think about breaking this double bond, all right, and adding two hydrogens to our alpha carbon, right? So let’s go ahead and do that, so we’re going to break that double bond and add two hydrogens to the alpha carbons, so thinking about this in terms of retrosynthesis, we have a ring here. All right and then let me, let me go ahead and draw this over here. We would have two hydrogens over here. Let me go ahead and get some more room. All right, so probably easier to think about adding your oxygen to your beta carbon first, so let's do that. All right, we know we're gonna add an oxygen to our beta carbon. So let's go ahead and put in our oxygen on our beta carbon; I'll do that in green. Right, and then we're gonna think about two hydrogens to our alpha carbons, so we're gonna draw in our alpha carbon like that, and we're gonna add two hydrogens; I'm gonna put those in green like this. And this is a little bit more room, and then we would have our carbonyl like that, and then we would have to make sure count your carbons when we're doing these. All right? So let's go ahead and show those carbons. So this carbon right here is this carbon. All right. And then, this carbon is this one right here. And then this carbon, our carbonyl one is right here. And so you can see this is actually an intramolecular aldol condensation. So if you think about losing water right here, right, and sticking those two fragments together, you would form this compound on the left, and so this is a pretty cool reaction. Right, we could we could redraw this, right, so on an exam, right, you might, I mean, you could draw it like that, but most of the time, you would see it in a different conformation. All right, so you would see the compound looking like this. So let me go ahead and draw that. So identify your carbons. Carbon in magenta is the one right here. Carbon in blue is the one right here. And the carbon in red is the one right here, which has your carbonyl. And then we are going to go ahead and make our alpha carbon right here green, so this carbon right here, so this must have been, this must have been your starting compound, right, and so if you add a base, right, you can convert, you can convert this molecule into your target compound. So once again an intramolecular aldol condensation. And we'll talk in more detail about those in the next video.