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Organic chemistry
Course: Organic chemistry > Unit 12
Lesson 2: Aldol condensationsRetro-aldol and retrosynthesis
How to think about the aldol condensation using retrosynthesis. Created by Jay.
Want to join the conversation?
- ATwhy 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? 7:26(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)
- At, can we use NaOH to carry out this reaction? 7:41(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)
- At, he includes EtOH. What is the EtOH required for in the aldol condensation? Donating proton to the oxyanion intermediate product? 5:09(3 votes)
- The ethanol serves both as a solvent for the reaction and as a proton donor.(3 votes)
- Will there be retrosynthesis on the MCAT?(3 votes)
- at, how do you know that LDA is what is used. How would you know the following steps?? 7:42(3 votes)
- why not just do an ozonolysis reaction?(1 vote)
- Wouldn't ozonolysis create a new carbonyl on the alpha-carbon?(3 votes)
- is a lecture present on cannizzaro's reaction ?(2 votes)
- 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)
- The α carbon is the first carbon that attaches to a functional group.(2 votes)
- 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)
Video transcript
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.