Aldol reaction | Alpha Carbon Chemistry | Organic chemistry | Khan Academy

Aldol reaction | Alpha Carbon Chemistry | Organic chemistry | Khan Academy

January 1, 2020 66 By Kody Olson


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.