Keto-enol Tautomerism under acidic conditions

Keto-enol Tautomerism under acidic conditions

October 25, 2019 0 By William Morgan


Hello good people. I would like to talk
about keto-enol tautomerism. It is a
conversion of a keto form, here on the
left, into an enol form on the right. A
keto form is any aldehyde or a ketone
that also happens to have an alpha
hydrogen. What do I mean by that? Well,
this carbon here with the yellow dot,
that is the carbonyl carbon. The carbon
next to it is an alpha carbon. Any
hydrogen attached to it is referred to
as an alpha hydrogen. So, generally
speaking you could say that as we go
from a keto form to the enol form, the
alpha hydrogen ends up as a hydrogen on
the oxygen here, giving us the alcohol
functionality. And this bond in green, the
bond between the alpha carbon and the
alpha hydrogen, ends up as a bond between the two carbons. Finally, you could also
say that this bond in orange, so the
pi bond between the carbon and the
oxygen of the carbonyl group, ends up as
a bond between oxygen and hydrogen.
Once again this is just generally speaking,
not necessarily what happens when we
think in terms of reaction mechanisms.
And I will now talk about the reaction
mechanism under acidic conditions. So, let me just scroll down a little bit. Here is
my starting material. I have chosen
ethanal, an aldehyde, but it could have
also been a ketone. I have said that this
happens under acidic conditions so
here’s my protonated water. What is the
first step? The lone pair electrons of
the oxygen will be attracted to the
hydrogen of the protonated water. The
electrons between that particular
hydrogen and oxygen will end up on the
positive oxygen, and we will get an
oxonium ion and the water molecule. What
is the next step? Remember we need to get to an enol form. We need to get to an
alcohol and a double bond. The pi
electrons between the carbonyl carbon
and the oxygen will be attracted to the
positive charge on the oxygen.
The electrons between the alpha carbon…
And let me label that one again. So, that
is my alpha carbon. And I’m also going to
number them. So this is 1, and this is 2.
All right. The electrons between the alpha carbon
and the alpha hydrogen will end up as the
pi electrons, or pi bond electrons,
between carbon 1 and 2. While the oxygen lone pair electrons on
the water will be attracted to the alpha
hydrogen. And so we will end up with an
enol form. So, we have an alcohol that is
also attached to a carbon that is part
of a double bond. So, we have an enol and
we have a protonated water. Remember that
I said that we can go back. So, we can go
from a keto form to an enol form, and we
can go from an enol form to a keto form.
How does that look in terms of the
reaction mechanism. Well, we need to
recover the oxonium ion again. So, the
lone pair electrons of the alcohol will
go back down to form a pi bond between
the oxygen and carbon number 1. The pi
electrons, the pi bond electrons between
carbon 1 and 2 will be attracted to one
of the protons on the protonated water.
And the electrons between that
particular hydrogen and the oxygen on
the water will end up on the oxygen. And
that brings us back to our oxonium ion
and our water. What is the next step?
Recall we are trying to get back to
our aldehyde, so we need to get a
carbonyl group. Or if you like, we need to
get rid of this proton here. This one in
yellow. So, that means that the electrons
that make up the bond between the oxygen and the hydrogen will end up on the
oxygen of the carbonyl. And the lone pair
electrons of the oxygen water will be
attracted to this proton here in yellow. And when
all is said and done, we are back at our
ethanal, and we have also recovered our
protonated water that helped us start
the whole process in the first place. So,
that is all for now, and I will see you
on the flip side.