In this video, I want to introduce you to a mechanism
called the aldol reaction.
And it's easily one of the most important mechanisms and
reactions in all of organic chemistry because it's a
powerful way to actually create carbon-carbon bonds.
And it'll actually be a little bit of a review of what we saw
with enol and the enolate ions and the keto-enol tautomerism.
I have trouble saying that.
Anyway, let's start with a couple of aldehydes.
And just for convenience I'll make them identical.
So let's say that this is one of aldehydes, it just has a
carbon chain there.
That's what the r is, it could be of any length.
Who knows what's there.
And then I have another carbon here.
And then this is bound to the carbonyl group and we're going
to make it an aldehyde, although you could do this
reaction with a ketone as well.
And just to make things clear, this carbon right here-- which
is going to be involved in a lot of the business here-- let
me draw it's hydrogens.
Normally we don't have to draw its hydrogens.
And just as a bit of review, the carbon that is next to the
carbonyl carbon is called an alpha carbon.
If this was a ketone, this would have also been an alpha
carbon if this was a carbon.
And we're going to see in this reaction, besides just
exploring the reaction, is that these hydrogens are
actually much more acidic then traditional hydrogens attached
to carbons on the rest of the chain.
And it comes from the fact that this proton can be given
to something else, the electron can go to that
carbon, and then it'll be resonance stabilized.
And we're going to see that in a second.
Now I said I would draw two molecules of that, because we
need two molecules.
We're actually going to be, to some degree,
joining the two molecules.
So let me draw another aldehyde right over here.
And I'm going to draw it symmetric to this, because
it'll make it, I think, a little bit easier to visualize
the two molecules.
Actually, let me just draw it the same way.
But I'll draw it in a different color.
So you have the r group and then you have the
oxygen right there.
And I won't draw all of the hydrogens on this guy, but
this and this are the exact same molecule.
It's just the hydrogens are implicit here.
Now, the aldol reaction I'll show you will be in a basic
environment.
So you could imagine that it'll be catalyzed by a base.
And so, imagine we have some hydroxide laying around.
Some of the hydroxide anion.
Let me do that in a different color.
So let's say we have some hydroxide anion floating
around-- negative charge, just like that.
I just told you that these hydrogens are much more acidic
then hydrogens anywhere else on a carbon chain-- these
alpha hydrogens.
So you could imagine a situation where an electron
from the hydroxide is given to one of these hydrogen protons
and then the electron that was associated with that hydrogen
is now given back to this alpha carbon.
And so if that were to happen, the next step in our reaction
would look like this.
And I'll draw it in equilibrium.
Actually, let me draw it this way.
So the products of that step would be in equilibrium, with,
you have your carbon chain or the rest of your molecule
right there.
And that's just to show that it could be anything.
It's attached here to the alpha carbon, which is now
going to be negative-- I'll show that in a second-- which
is attached to the carbonyl group, which is
attached to a hydrogen.
And actually, I'll stop drawing that
hydrogen for now, too.
Just, we know it's there.
But I'll keep drawing this hydrogen right over here.
The other hydrogen was taken away and this alpha carbon now
has a negative charge because it got the
electron from that proton.
And of course, we have the hydroxide.
It grabbed this hydrogen and it is now water.
Now the reason why this was acidic to begin with is
because this is resonance stabilized.
And I'll show you that it's resonance
stabilized right now.
This alpha carbon right here can give its electron to the
carbonyl carbon.
And if the carbonyl carbon gets an electron, it can give
an electron to this oxygen up here.
It'll break the double bond.
So this configuration is resonance
stabilized with this.
So I could draw it like this.
You have your r and then you have a single
bond to this oxygen.
It now gained an electron.
It is now negative.
And you now have a double bond, just like that.
And I could draw this hydrogen if I like, or I don't have to.
It's implicitly over there now.
And you might be familiar with this.
This is the Enolate anion.
This right here is the enolate ion.
If we had a hydrogen right here, it would be enol, and we
would say hey, this is the keto form,
this is the enol form.
We've seen this before.
Now, what's interesting about the enolate ion is it can act
as a nucleophile.
It can do a nucleophilic attack on the other aldehyde's
carbonyl group.
But it does it in kind of a non-conventional way.
And I'll show you how it does it right now.
So it does the attack like this.
So let me draw this guy over here.
So you have the carbonyl group and then you have its alpha
carbon and then you have an r group right over there.
There's actually a hydrogen right over here, as well.
I just flipped it over.
This and this are the same molecule.
And let me make it clear-- these two guys right here are
residence forms. And, once again, this is the reason why
it's easier to take this hydrogen than other hydrogens
on a traditional carbon chain.
Easier to take an alpha hydrogen to a carbonyl group
because you have this resonance structure.
But this enolate ion, especially this configuration
of it, you can imagine it doing something like this.
You can imagine this oxygen giving back the electron to
the carbonyl carbon-- to this carbon right here.
And when that happens, then this guy is going to be giving
up an electron.
And that electron that he gives up-- let me do it in a
new color-- this electron that he gives up could go and do a
nucleophilic attack on this carbonyl group.
And so if that carbonyl carbon gets-- let me do this in a new
color-- if this carbonyl carbon gets an electron, then
it could give away an electron to that oxygen right up there.
So the next step after this, we would have
something like this.
And once again, I'll show it as happening in equilibrium.
So from here we go right over there, and what we have is a
situation-- let me draw this guy on the left first. So we
have a double bond to this oxygen now.
Actually, let me draw the second.
So this is this oxygen.
We now have a double bond.
And let me do it in this same purple color right over here.
And then we have the rest of what was an aldehyde.
Where you have-- let me do it in that same color-- and then
you have your r group right over there.
But now this electron gets in an attack
on this other aldehyde.
So this guy right here, this alpha carbon is that same
alpha carbon we've been dealing with, is now bonded to
this carbonyl carbon.
So it is now bound to this carbonyl
carbon right over here.
And so it will look like this.
Let me draw it with the right colors.
Get the orange out.
So that carbonyl carbon, it now has a single
bond to this oxygen.
This electron was taken back by it.
So this oxygen now has a negative charge.
And it is bound to its alpha carbon.
And then that is bonded to another group, probably a
carbon chain or something that contains a carbon chain or
another function or group.
Whatever you want to call it.
And then the final step.
This anion can get rid of its negative charge by essentially
grabbing a hydrogen maybe from this water
that was formed before.
Obviously, not going to be the same molecule, but it could
grab it from this in a previous step.
This water molecule that was formed in a previous step.
And of course, this is all in a basic environment.
So it can give an electron to this hydrogen, and then the
hydrogen proton would lose an electron to the hydroxide and
the hydroxide will become negative again.
And so what will be the final product?
The final product will be-- and I'm just going to try my
best to redraw this thing right over here.
You have this part of the molecule, so you have this
carbonyl group right over here.
It is attached to this radical group right over there.
So that is this part.
And I can even do the same colors.
This bond right over here is this bond right over here.
And then this carbon is attached to a carbon that's
attached to a hydroxyl group now.
So it'll look like this.
And let me draw it.
So this oxygen is now this oxygen, and it just captured
this hydrogen.
So it is now a hydroxyl group.
It's now an -OH group.
And then, finally, this guy is bound to what
was an alpha carbon.
It's not anymore.
What was an alpha carbon, which is then bound to a
radical group.
And if we want, we can remember that there was
always, from the get-go, there was always a
hydrogen over here.
So why is this called the aldol reaction
and why does it matter?
Well, it's called the aldol reaction because what we
formed is both an aldehyde-- notice this is an aldehyde--
and it's an alcohol.
So that's where the word aldol comes from.
But the more important thing about this-- and I don't want
to mislead you-- could have also done this with a ketone
You could have had a methyl group or a ethyl group.
You could have had a big carbon chain here.
It still would've worked.
So the aldol reaction doesn't only form things that are
aldehydes and alcohols.
It could have formed something that's both a
ketone and an alcohol.
But that's why it's called the aldol reaction.
But the more important thing about the aldol reaction is,
one, it shows you how the enolate ion can be a
nucleophile.
It shows you why the alpha hydrogens are more acidic than
hydrogens on other parts of carbon chains.
But the most useful aspect of it is it's a useful way to
actually join two carbon chains together.
Notice, we were able to join this alpha carbon right here
to this carbonyl carbon over here to form this aldol.
Or sometimes this will be called-- because this is still
an alpha carbon right here, this is an alpha carbon, this
is a beta carbon-- and so sometimes this will be
referred to as a beta hydroxy.
And we've probably used things from the pharmacy that has
this word in it.
This is also called a beta hydroxy.
This is alpha, this is beta.
It has a hydroxyl group on the beta carbon.
Beta hydroxy aldehyde.
Anyway, hopefully you found that entertaining.