Keto-enol tautomerization

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.