Keto-enol tautomerization (by Jay) | Alpha Carbon Chemistry | Organic chemistry | Khan Academy

Keto-enol tautomerization (by Jay) | Alpha Carbon Chemistry | Organic chemistry | Khan Academy

August 1, 2019 6 By William Morgan


Voiceover: If you start
with an aldehyde or a ketone
and add a catalytic
amount of acid or base,
you’ll find the aldehyde or ketone
is going to be in equilibrium
with this product over here on the right
which we call an enol.
The name enol comes from the fact
that we have a double
bond in the molecule,
so that’s where the EN part comes in,
and we also have an alcohol.
You can see the OH over here
so that’s where the OL comes in.
This is the enol form
and then over here this is the keto form.
The keto form and the enol form,
and these are different molecules.
They’re isomers of each other
so we call them tautomers
and they’re in equilibrium
with each other.
They’re not different
resonance structures.
Let’s see if we can analyze
our aldehyde or ketone
to see how to form our enol.
If we look at the carbon
that’s next to the carbonyl carbon
we call this the alpha carbon,
and there are two hydrogens
attached to the alpha carbon in this case.
Let me go ahead and draw those in.
Those are called the alpha protons.
If we think about transferring
one of those alpha protons
from the alpha carbon to the oxygen,
even though it’s most
likely not the same proton,
it just helps to think about doing that.
We can also think about
moving the double bond.
Over here on the left,
the double bond is between
the carbon and the oxygen
and we’re moving that
double bond over here
between the two carbons.
Transferring one alpha proton
and shifting your double bond
converts the keto form into the enol form.
Then, we also have a hydrogen, right?
Over here, we still have a
hydrogen left on this carbon,
so let me go ahead and
draw in that hydrogen.
That’s this hydrogen in blue here.
That’s how to think about
converting a keto
tautomer into an enol one.
Let’s look at the acid-catalyzed
mechanism for this.
If we start with our aldehyde or ketone
and add H three O plus,
the first thing that’s gonna happen
is protonation of our carbonyl
and so a lone pair of electrons
picks up this proton like that.
We can go ahead and draw that.
We would protonate our carbonyl
so now our oxygen would have
a plus one formal charge.
Let me just go ahead and
draw in those electrons here.
Let’s say we started with an aldehyde.
We’ll make this an H.
The lone pair of electrons on our oxygen
picked up a proton, like that.
We can draw a resonance
structure for this.
We can move these electrons
off onto our oxygen
so let’s go ahead and show
a resonance structure.
We would have our R group, all right,
and now we would have our oxygen
with two lone pairs of electrons.
Let me go ahead and draw in
those two lone pairs of
electrons on our oxygen.
Then we took a bond
away from carbon, right?
We took a bond away from this carbon
so this carbon right here.
Plus one formal charge on that carbon.
Then we could show the
movement of those electrons.
These electrons right here
I’m saying moving out onto
the oxygen, like that.
This is our intermediate here.
All right.
We know that our alpha
carbon has two protons on it.
Once again, let’s find our alpha carbon.
Here it is right here.
We know we have two
protons attached to it,
two alpha protons, if you will.
In the next step of our mechanism
we’re gonna get a molecule
of water acting as a base.
Let me go ahead and show
a molecule of water here.
The water’s gonna take one
of those alpha protons.
Let’s say once again, it
takes this alpha proton
and leave these electrons behind.
They’re gonna move in here
to form our double bond.
Let’s go ahead and draw our product.
We would have our R group here
and now we would have
a double bond formed
between our two carbons
and then we would have our oxygen,
and then we would have
two lone pairs of electrons on our oxygen.
We would have our hydrogen,
and then we would have
another hydrogen right here.
Let’s go ahead and follow
some of those electrons.
Let’s go ahead and make
these electrons in here blue.
These electrons are
gonna move in, in here.
It doesn’t really matter
which one you say it is,
let’s just say it’s that
one to form our double bond,
and then, the electrons in red
moved off onto this oxygen,
and then we said that these
electrons were in magenta.
You can see that we have
formed our enol here.
This is our enol
and then we started with
our keto form like that.
Keto-enol tautomerization.
Let’s look at the base catalyzed version.
Once again, we start with
our aldehyde or ketone
but this time we’re going to add a base.
Something like hydroxide.
We find our alpha carbon.
Here’s our alpha carbon.
Once again, with two alpha protons.
I’m gonna go ahead and draw
in those two protons here.
The base is gonna take
one of those protons.
Let’s say it takes this
one over here on the right.
That leaves these electrons
behind on this carbon.
Let’s go ahead and draw
the resulting anion here.
We would have our carbonyl like that.
Once again let’s say we
started with an aldehyde
and then we would have
a lone pair of electrons
on this carbon, the carbon in red here.
Let me go ahead and
identify those electrons
so these electrons in here in magenta
have moved off onto this carbon like that,
which gives that carbon a
negative one formal charge.
It’s a carbanion.
There’s still a hydrogen
attached to that carbon in red.
This hydrogen right here
is still attached to it,
I’m just not drawing it in
so we can see a little bit better.
All right.
This is one form of the
anion that we could have.
We could draw a resonance
structure to show the other form,
so if we moved these
electrons in magenta into here
and pushed these electrons
off onto this oxygen.
Let’s draw the resonance structure.
We would have our R group here,
we would have a double bond,
and then our oxygen would have
three lone pairs of electrons
giving it a negative one formal charge,
and then we would have
our hydrogen over here.
The electrons in magenta moved in here
to form our pi bond
and then we can say that
these electrons in here
moved off onto our oxygen.
We could go ahead and show that.
Let me just go ahead and put
the other bracket on here.
We have two forms of this anion.
This is called the enolate anion.
This is the enolate anion.
This is going to be extremely important
in future reactions.
You can see the enolate anion
has two resonance structures.
One where we’re showing the
negative charge on the carbon.
That would be this one right over here.
The negative charge on the carbons.
This is our carbanion form, so carbanion.
Then we also have a resonance structure
where the negative charge is on the oxygen
so we could call this oxyanion.
If you think about which
one contributes more
to the overall hybrid,
oxygen is more electronegative than carbon
and so, it’s better able to have
a negative one formal charge on it.
The oxyanion contributes more
to the resonance hybrids.
All right.
Let’s think about the
last step in our mechanism
to form our enol.
If we think about our oxyanion,
all we’d have to do is
protonate that oxygen here.
We could just go ahead
and draw a water molecule.
We have a water molecule.
This time water’s going
to function as an acid
it’s going to donate a proton.
Let say these electrons in blue
take this proton, leave
these electrons behind,
and so from our oxyanion,
we can go ahead and draw our enol product.
We have our R group here.
We would have our double bond,
we would have our oxygen, all right.
Now protonated like this
to form our enol product.
Let me just go ahead and
show those electrons in blue.
Picked up a proton here to form our enol.
That’s how to get there
using base-catalyze.
Once again, we will talk much more
about the enolate anion
in future videos here.
Let’s look at a situation
where the alpha carbon is a chiral center.
Let’s look at this right here.
Here’s our alpha carbon.
Let’s just say it’s a chiral center.
If R and R double prime are
different from each other
we would have four different things
attached to this carbon.
The alpha carbon here
is SP three hybridized
with tetrahedral geometry.
Let’s say it’s either the
R or the S enantiomer.
It doesn’t really matter which one.
You can see now we have
only one alpha proton.
Only one alpha proton
but because there is an alpha proton
we can form an enol.
In either an acid or
base-catalyzed mechanism
we could think about
the proton here in red,
you could think about
transferring one to this oxygen
and moving your double bond,
and then we form our enol.
Here is our enol.
Now let’s look and see what happened
to the carbon in red right here.
On the left, the alpha carbon was SP three
hybridized with tetrahedral geometry.
Now, this carbon is SP two hybridized
with trigonal planar geometry.
Whatever stereochemical information
we had over here on the left,
whether it was the R or the S enantiomer,
it’s been lost now that
we’ve formed the enol.
The enol is achiral,
it’s flat, it’s planar.
When we reform the keto form,
so one of the possibilities
is to form the enantiomer
that we started with
but the other possibility is
to form the other enantiomer.
You can see that’s what I’ve shown here.
I’ve shown the hydrogen
now going away from us
and our R double prime
group coming out at us.
This is the enantiomer.
Because we formed the enol
we can get a mixture of enantiomers.
Enolization can lead to racemization.
We can get a mixture of enantiomers
and if we wait long enough,
we can get an equal mixture of these guys.
This one and this one
would be in equilibrium
with our enol form.
That’s something to think about
if you have a chiral center
at your alpha carbon.
Let’s look at two quick examples
of keto and enol forms.
Over here on the left
we have cyclohexanone
and on the right would be
the enol version of it.
You could think about one of these
as being your alpha carbon, right,
and you could move these electrons in here
and push those electrons off.
You could see that would
give you this enol form.
It turns out that the
keto form is favored.
The equilibrium is
actually far to the left
favoring formation of the keto form.
Even under just normal conditions,
so not acid or base-catalyzed.
There’s only a trace
amount of the enol presence
however, there are some cases
where the enol is extra-stabilized
and that’s the case for
this example down here.
We have the keto form and
we have the enol form.
Once again, you could think about
these electrons moving in here,
pushing those electrons off
giving you your enol form.
This is a specially-stabilized
enol, right?
This is phenol right here.
We know that phenol has an aromatic ring.
The formation of the enol
form is extra-stabilize
because of this aromatic ring.
This time the equilibrium
is actually to the right
and much more of it is in the enol form
than in the keto form.
In this case, we have some
special stabilization.