Main content
MCAT
Course: MCAT > Unit 9
Lesson 12: Alpha-carbon chemistry- Alpha-carbon chemistry questions
- Aldol reactions in metabolism
- Keto-enol tautomerization (by Jay)
- Enolate formation from aldehydes
- Enolate formation from ketones
- Kinetic and thermodynamic enolates
- Aldol condensation
- Mixed (crossed) aldol condensation
- Mixed (crossed) aldol condensation using a lithium enolate
- Retro-aldol and retrosynthesis
- Intramolecular aldol condensation
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Mixed (crossed) aldol condensation
How to find the product of a mixed (crossed) aldol condensation. Created by Jay.
Want to join the conversation?
- I understand in this instance why the benzene ring can't form an enolate. What if, for instance, we were working with two different aldehydes that can both form enolates?(9 votes)
- Then you get a mixture of products resulting from all possible combinations.(15 votes)
- Is anything preventing the enolate from attacking the carbonyl carbon(s) of the ethyl 3-oxidanylidenebutanoate instead of the carbon of cyclohexanone? My guess is steric hinderance or is that also a valid product?(7 votes)
- Ethyl 3-oxobutanoate (aka: ethyl 3-oxidanylidenebutanoate, ethyl acetoacetate) has a pKa = ~11, while ethanol's pKa = ~16 (this is the conjugate acid of the ethanoate base used in the reaction).
This means that if you added as much base as ethyl 3-oxobutanoate the latter will be almost entirely deprotonated. I believe that the resulting negative charge and resonance will make the enolate version of ethyl 3-oxobutanoate resistant to nucleophillic attack.(2 votes)
- Is heat the only condition that leads to the double bond?(3 votes)
- In a sense, yes.
There is an activation energy required to form a double bond.
Some elimination reactions have a higher activation energy than others.
But heat always favours elimination, because more of the molecules will be able to get over the energy barrier.(6 votes)
- Isn't hydroxide a poor leaving group? Is it heat and the basic condition that allow it to leave?(2 votes)
- Yes it is a poor leaving group, and in this case it is the ONLY leaving group.... You're correct about heat because it takes energy to break a bond, which allows it to go to a more favored product, which is the other reason why OH leaves.(7 votes)
- at, the subtitles say "sodium methoxide" but should say "sodium ethoxide." 6:29(3 votes)
- I'm confused about hydroxide being a leaving group, doesn't the fact that it's a strong base make it a poor leaving group? Shouldn't it be protonated to water first to make it a good leaving group?(2 votes)
- Hydroxyl's are relatively poor leaving groups. However, the reaction is driven forward with the addition of heat and the eventual formation of the enone. This all has to do with stability and which compound is more (energetically) favorable.(3 votes)
- Around, Jay mentions protonation to form the aldol product. Given that the initial reactant was a ketone, would the product be a ketol rather than an aldol or can aldol be used interchangeable for aldehydes and ketones? 8:00(2 votes)
- The term "aldol condensation" is used as a general term for both aldehydes and ketones.(3 votes)
- I am not getting which hydrogen should migrate and how to know which is the ost acidic hydrogen?(2 votes)
- he didnt put h2o, so how did the molecule got protonated ? at9:23(1 vote)
- It's assumed that you're in an aqueous solution where water is present :)(3 votes)
- @, why do the electrons move to the oxygen? doesn't the alkene carbon without the methyl group NOT have 4 bonds in your product? I do this mechanism and still have the hydroxide group attached along with the double bond. What am I not seeing? 4:10(2 votes)
Video transcript
Lecturer: In the previous
video we looked at aldol condensations
with the same molecule, you call those a simple
aldol condensation. In this video we're going to look at mixed or crossed aldol condensations, so no longer are you starting
with the same molecule. Here we don't have two aldehydes that are the same, we have different aldehydes. We have benzaldehyde on the
left and propanal on the right. We need to figure out what sort of enolate anion that we're going to form. When we add our sodium
hydroxide as our base, what is going to be our enolate anion? To do that we need to
look for alpha carbons. We'll start with propanal. We know the alpha carbon is
the one next to the carbonyl, and so right here this
would be an alpha carbon, and there are two protons
on that alpha carbon, so we have two alpha protons here. Let me go ahead and draw in those. That's a possibility to
form our enolate anion. This aldehyde hydrogen right here is not, we're not going to form
an enolate from that, so we don't have to worry about
the right side of our aldehyde. Let's look at benzaldehyde now. If we think about this carbon
right here it is the carbon next to a carbonyl, however
this carbon already has four bonds to it, so it
doesn't have an alpha proton. Benzaldehyde does not
have an alpha proton, so we don't need to worry about
it forming an enolate anion. We know the enolate anion is going to form from propanal over here. Let me go ahead and
redraw these in a way that makes it easier for me to
see what the product is. I'm going to draw my ring here. I'm going to draw my carbonyl. I like to leave off this
hydrogen because it gets in the way when I'm
thinking about my mechanism. For my propanal over here I prefer to draw a different confirmation of propanal to make it easier to see the product. I like to put my carbonyl
up here like this, and I like to draw this
carbon down like that, and it's going to make it easier to come up with the final result. If I think about deprotonation, if I think about sodium hydroxide, taking one of these alpha protons here, leaving electrons behind
in my alpha carbon, I can think about the structure
of the carbanion enolate anion. There's a negative one
formal charge on this carbon, and that's my nucleophilic enolate anion. Once again I prefer to draw
the carbon as opposed to the oxyanion, it just makes it
easier for me when I'm trying to do a quick mechanism on a
test to figure out a product. Now we have a nucleophile. I'm going to go ahead and put
these electrons in magenta. This is going to be our nucleophile, and our nucleophile is going to attack the electrophilic portion of
our benzaldehyde molecule. If we think about that the
oxygen is partial negative and this carbonyl carbon
here is partially positive, that's the electrophilic
portion of the molecule. That lone pair of electrons here on this carbon is going to attack this carbon, pushing these electrons
off onto the oxygen. That would form an alkoxide intermediate. Let me go ahead and sketch in here. We would form an oxygen with
three lone pairs of electrons, giving it a negative one formal charge, and I'll come back to that. The important carbon-carbon bond
that's formed is right here. Then we have, I'm going to go ahead and draw in everything else. We have our aldehyde, and don't
forget about this down here. In magenta, these electrons in magenta, have formed our new bond right here. Then we would have an
alkoxide intermediate and I'm going to think about protonating it. I'm not concerned with
the exact mechanism here, I'm more concerned about
figuring out the product. We go ahead and proteinate our alkoxide to form our aldol intermediate. In the next step, when
you add heat you're most likely going to form
your conjugated product. Let's think about what would happen next. We still have an alpha carbon. This carbon right here
next to our carbonyl is an alpha carbon, it still
has an alpha proton on it. We can show a proton right here. And we can think about hydroxide
once again acting as a base. Hydroxide acting as a base, coming
along and taking this proton. We can even think about the final product, we can think about these
electrons moving into here and then these electrons
moving out onto here. Again I'm not so concerned with
the exact detailed mechanism, I'm more worried about how to
figure out our product here. When I draw our product, we have our ring, and we now have a double
bond that forms here. We have this going down and
then we have our aldehyde. We form a conjugated product,
we have an enal as our product. Let's follow those electrons. Let's make these electrons in blue here. These electrons in blue moved
in here to form our double bond. We already formed this bond in magenta, the carbon-carbon bond forming part to form our aldol, and
then we lost hydroxide. This would be our final
product, our conjugated product. Again, the goal is just to
figure out how to draw a product from these reactants, and
thinking about where is the alpha carbon and thinking about what
is the enolate that forms. Let's do another one where I walk you through my thought process. Let's look at this reaction. Say you had this on an exam, and your task was to draw the product. We have once again some
different possible alpha carbons. Let's first focus on the cyclohexanone. We know that this could
be an alpha carbon, and we know that this
could be an alpha carbon. Then for this compound over here we have these two carbonyls in this molecule. We know that this could
be an alpha carbon and we know that this could
be an alpha carbon. Now we have to figure
out which one of those alpha carbons is going to be deprotonated when we add sodium
ethoxide as our base here. If you remember one of the earlier videos we talked about an alpha carbon between two carbonyls as having
the most acidic protons. This alpha carbon right
here has two protons on it, and it's easy to deprotonate,
because of the resonance stabilization that we can draw
because of those carbonyls. These protons are the most acidic, much more acidic than the
protons on our ketone here. These are the ones that are
going to be, one of these protons could be deprotonated when
we add sodium ethoxide here. Sodium ethoxide is going
to come along, take one of these protons here, leave
these electrons behind. Let's go ahead and redraw
what we would form here. We're going to form our enolate anions. Let me go ahead and draw that. We would now have a lone pair
of electrons on our carbon, giving that carbon a
negative one formal charge. Let me follow those
electrons, these electrons in magenta are now on our
carbon, forming our carbanion. I'm not going to take the time to draw the resonance structure, the
oxyanions, because I'm just concerned about figuring
out the product here. Now we have a nucleophilic
enolate anion, and we know that's going to attack the
carbonyl of our ketone. Going back over here to
our ketone, the oxygen is partial negative, this
carbon is partially positive, so our nucleophile is going
to attack our electrophile. You can think about these
electrons attacking here, pushing those electrons
off on to your oxygen. If we were to draw the intermediate here we would have our ring. we're going to form an alkoxide, so I'm not going to draw all the lone pairs
on that oxygen right now. I'm more concerned right now with showing the formation of this bond. Let me go ahead and draw in everything and we'll follow some electrons. I put in my carbonyls, then I have these guys over here like that. The electrons in magenta, these electrons formed
our carbon-carbon bond. They formed this bond right here. Then we would form an alkoxide. Then we would go ahead and
protonate to form our aldol. Once again just to save time,
not an exact mechanism but thinking about our intermediate
as being this aldol. Because we have heat
we're probably going to once again form a
conjugated product here and keep going for the complete
aldol condensation. Next we can think about
this alpha carbon right here still having an acidic proton on it. There's still a proton
attached to that alpha carbon. Sodium ethoxide come along. We have a ethoxide anion,
let's go ahead and draw that in here, which could
function as our base. It could take that proton,
and think about these electrons moving into here
to form our double bond, we can think about hydroxide
as a leaving group. That allows us to figure out our product. We have our ring, and then
we now have a double bond, and then we have a carbonyl
over here, then a carbonyl over here, then our oxygen
and our ethyl like that. Let's think about those electrons. The electrons in blue here move
in to form our double bond, and we had already formed a
carbon-carbon bond before, so let's say it's these
electrons right here. Then we have a stable conjugated product. Once again, I'm just thinking about how to draw the product for
a reaction on an exam. Hopefully this helps.