Keto Enol Tautomerism Acid and Base Reaction and Mechanism

Keto Enol Tautomerism Acid and Base Reaction and Mechanism

July 22, 2019 62 By William Morgan


Leah here from leah4sci.com and in this video
I’ll take you through the mechanism for Keto
– Enol Tautamerization in both acidic and
basic solutions. But first, what is Keto – Enol
Tautomerism When you see the word Keto, I
want you to think of the ketone functional
group and that’s where you have a carbonyl,
carbon double bound to oxygen with R groups
on either side. In this case we have a Methyl
and another methyl. Keep in mind though that
with KET Keto – Enol Tautomerization, the
carbonyl on the ketone or carbonyl on the
aldehyde both qualify as the Keto form in
this reaction. The Enol comes from two words
together. When you have a pi bond in a molecule,
that is an alkene which gives ua the ene.
And if you have an OH on one of the same carbons,
meaning the carbon holding the pi bond also
has an alcohol, that’s where we get the ol
portion drop the e and we have the Enol, a
molecule with a pi bond between carbons and
an alcohol on the same group. I like to think
of Tautomerization as a fancy for Isomerization
in equilibrium. Specifically, Tautomer refer
to the Keto and Enol form being in equilibrium
with each other. So for example if I start
out with Enol with an alcohol on pi bond or
both on Carbon 2 and then re-draw the skeleton
as an Isomer, meaning I have the same exact
atoms but maybe they’re positioned differently
on the molecule. We’ll still have the Oxygen
on the same position but now I want you to
imagine that we’re taking the green hydrogen
and moving it to the bottom of the molecule
and we’re taking that pi bond and moving it
up between the carbon and the Oxygen. Now
I’m showing it as the same green Hydrogen
but keep in mind in the mechanism, you have
a different hydrogen being added and removed.
But the net atoms on the molecule are the
same and these two are going to be in equilibrium
with each other.
But even though they’re in equilibrium, they’re
not 50-50, the average pair of tautomerism
will favor the keto form. This is more stable
with the concentration of electrons sitting
between carbon and Oxygen rather than having
the concentration of electrons between the
two carbon atoms. Every now and then you’ll
find a molecule where the Enol is more favorite
and this has to do with resonance and stability
more than anything else. For example, if I
have Phenol which is benzene with an alcohol,
I can take any of the pi bonds, call that
the Enol and then try to show to Tautomerization
as follows; We’ll swap out this Hydrogen and
show it in green to show what’s moving, but
now I remove that pi bond, put it between
carbon and Oxygen, put an extra hydrogen here.
The problem with this specific setup tautomers
is that the first one is aromatic and it has
resonance within the benzene ring. That makes
it very stable.
The second structure lost resonance on this
part of the molecule making it slightly less
stable. So this is one example where the Keto
form is less stable than the Enol form. But
in a simple regular molecule, the Keto form
is going to be more stable. Now this will
come up in different Organic Chemistry reactions
but the first time you’ll probably see this
is when you’re studying alkyne hydration reactions.
So you’ll have your Alkyne reacting to form
an Enol and it’ll automatically go to the
Keto form. You’ll have Oxymercuration in acidic
conditions and hydroboration in basic conditions.
So let’s start with KET under acidic conditions.
Remember that in acid you have lots and lots
of protons floating around in solution but
protons don’t float freely in solution. Instead,
you’ll see them as hydronium being carried
by a water molecule where the Oxygen has a
lone pair and a positive charge. Oxygen pulls
some of the positivity from the nearby Hydrogen
atoms making them partially positive. And
so the Oxygen on the ketone will use one of
its partially negative electrons to reach
for a partially positive hydrogen atom and
form a new bond. Since Hydrogen can only have
one bond, the electrons between hydrogen and
the green oxygen collapse onto the Oxygen
and this is the first step that you’ll see
in every carbonyl reaction under acidic conditions.
This is deprotonation or activation step.
Let’s show our product. We have the Oxygen
with one lone pair that hasn’t change. The
black electrons are now sitting on Oxygen
as a bond between itself and hydrogen. And
on the side we have a green water molecule
in solution with one lone pair that it had
initially and the red lone pair that it got
from the collapsing bond between itself and
hydrogen. Don’t forget your formal charges,
the blue oxygen has three bonds and one lone
pair with the formal charge of plus one. This
makes Oxygen positive. But Oxygen is unhappy
with the positive charge because it’s an electronegative
atom so it’ll start pulling on the bond between
itself and carbon making that carbon very
partially positive. But even more importantly,
if we have a Hydrogen nearby, the electron
density from the bond between carbon and hydrogen
will also be influenced towards that positive
carbon, and that’s the key. Because another
water molecule in solution can come along
and use its lone pair to grab that Hydrogen
atom leaving the electrons behind and leaving
them to collapse towards that partially positive
carbon which in turn causes the electrons
between carbon and Oxygen to collapse onto
the Oxygen. So this is a chain reaction where
all the electrons quickly move to the next
place, or the next place where they want to
be ultimately being drawn to that positive
charge on Oxygen.
So what is that give us as a result? The Oxygen
now has just the one bond between itself and
carbon. It still has the blue lone pair, it
still has the black bond to the green hydrogen
atom but it now also has a red lone pair where
the pi bond collapse onto the Oxygen giving
me that second set of electrons. We also have
another pair of red electrons that used to
hold carbon to hydrogen now sitting as a bond
between the two carbon atoms. What happened
to that purple hydrogen atom? It’s now floating
around in solution bound to a water molecule
as a new hydronium.
And last but not the least, a catalyst is
something that helps the reaction but doesn’t
get used up. We started with Hydronium, we
carried out the reaction and look at that,
we reformed hydronium. The hydrogen that was
added is the green hydrogen. The one that
was removed is a purple hydrogen. So they
are different but the net number of atoms
stay the same and that is why these two are
considered Tautomers or Isomers but specific
Isomers where we started out with the Keto
form for the ketone and ended with the Enol
form. Keep in mind that this is a reversible
reaction. And when you think about your alkyne
reactions you started with an Enol and you
have to trace this path backwards to get to
the ketone so let’s see how to do that in
an acidic solution.
Here we have the Enol and the first we do
is that last thing we do in the previous reaction.
We need to put a Hydrogen back. How do we
do that? We find a partially positive hydrogen
in solution, one sitting on hydronium where
Oxygen has a lone and a positive charge. But
we can’t simply show the electrons coming
on to grab it because that will leave this
carbon here with an incomplete octet very
unhappy. Instead, remember how we triggered
the previous reaction? It had to do with the
lone pair on Oxygen. If the lone pair on Oxygen
comes down creating a bond between itself
and carbon, carbon now has too many bonds
causing this bond to get kicked out and that
is how you grab the hydrogen. Break the bond
between hydrogen and oxygen and let’s see
what we have. Oxygen now has just one lone
pair where the second pair the purple electrons
are sitting as a bond between itself and carbon.
The red electrons that used to be a pi bond
are now sitting as a bond between itself and
a hydrogen atom and we have a free floating
water molecule in solution. Don’t forget your
formal charges, Oxygen has one lone pair,
three bonds and a positive charge.
When nearly done, we now have to get rid of
that hydrogen atom and how do we do that?
well we find the most negative specie in solution,
a water molecule with lone pairs on Oxygen.
Oxygen reaches for Hydrogen causes these electrons
to collapse back onto the carbonyl Oxygen
and that gives us our final product. Carbon
is till double bound to Oxygen, we had green
lone pair of electrons we just got a second
lone pair as the blue electrons and let’s
not forget our purple hydrogen sitting on
that lower carbon.
Now let’s take a look at Keto-Enol Tautomerization
under basic conditions. In acidic condition
we have many protons in solution and the most
negative species was the partially negative
Oxygen. Under basic conditions, we have a
lot of Hydroxide where the most negative is
the OH minus and the most positive is a neutral
water. You do not want to show positive charges
in a basic solution. In base we have negatives
and I like to think of it as aggressive atoms
which act faster giving us fewer steps of
the mechanism. We have hydrogen sitting over
here that we ultimately want to remove. So
we’ll bring in our base, our OH minus which
we’ll use one of its lone pair of electrons
to grab that hydrogen atom. When it grabs
hydrogen the electrons binding it to carbon
will collapse towards that partially positive
carbonyl carbon which in turn will cause the
electrons between carbon and oxygen to collapse
onto Oxygen. Again a chain reaction where
the electrons just keep moving and moving
towards the Oxygen atom. What we get is Oxygen
with a single bond to carbon. We have the
two initial purple lone pairs with a third
lone pair that we just added from the collapsing
pi bond. That third lone pair gives me a negative
charge but don’t forget the pi bond that we
just formed by the moving electrons between
the two carbon atoms. What happens to the
OH minus in solution? Will since it’s now
bound to the green Nitrogen atom and has two
lone pairs, it’s a neutral water molecule.
But what about that negative Oxygen? We don’t
wanna leave it negative so we’ll find another
water molecule on solution. Remember the most
positive species in a negative solution is
the partially positive hydrogen. And so Oxygen
will use one of its negative electron pairs
to grab that hydrogen, breaking this bond
and collapsing the electrons onto Oxygen.
What is that give us? the skeleton stays the
same, Oxygen has the two lone pairs, what
happened to the third lone pair is now sitting
as a bond between itself and a hydrogen atom
from water. The pibond hasen’t changed so
we’ll leave it as it were it was and in addition
we also have another OH minus formed in solution.
The OH minus act as a catalyst in this reaction.
We’ve started with an OH minus turned it
into water but we got another OH minus which
means the net OH minus came back at the end
and that’s the base catalyzed reaction. But
just like with the acidic form, in your alkyne
reactions you’re going to see the Enol before
the keto form, so let’s show to convert on
your basic conditions. Here we have an Enol
under basic conditions which means we have
lots of OH minus present. We’ll show an OH
minus here as the aggressive attacking molecule
and we’ll also show a water molecule because
that will be our source of protons in this
reaction. We can show hydroxide using one
of its lone pair of electrons to grab Hydrogen
off of the Oxygen. This causes the bond between
Oxygen and Hydrogen to collapse downward towards
the carbon. But carbon has gonna have too
many bonds so instead of getting kicked on
to the nearby carbon, think of it as getting
kicked on to the nearby carbon not being happy
where it is and the electrons quickly seek
out something else. To show that in one arrow.
We just show the pi electrons reaching towards
the hydrogen atom but keep in mind that this
carbon that is forming the bond not this one
even though the arrow appears to come from
between the two. Hydrogen can only have one
bond so we collapse this bond onto Oxygen
as a lone pair and then one step we got our
product. The skeleton stay the same, Oxygen
have the two green lone pairs so we show that
first, the black bond is now a pi bond between
carbon and oxygen, the red pi bound to electrons
are now sitting as single bond between itself
in the purple hydrogen atom. The former hydroxide
is now sitting with an extra hydrogen as a
water molecule in solution and the former
water molecule is now sitting as hydroxide
with that extra pair of electrons giving it
the negative charge. And there you have it,
acid and base catalyzed keto-enol tautomerization.
If you found this helpful, please let me know
in the comments below and be sure to share
this with your classmates so they too can
finally understand this reaction. And of course,
for even more help with Alkyne Reactions,
be sure to download my alkyne reaction cheat
sheet breaking down every reaction, reagent
and product to help you understand what’s
going on.