Keto-enol Tautomerism under basic conditions – Part 1

Keto-enol Tautomerism under basic conditions – Part 1

October 24, 2019 0 By William Morgan


Hello good people. I would like to talk
about keto-enol tautomerism under the
basic conditions. So, it is a conversion
of a keto form into an enol form and
back again. A keto form is any aldehyde
or a ketone that also has an alpha
hydrogen. What do I mean by that? Well,
this carbon here with the yellow dot,
that is the carbonyl carbon. The one next
to it is an alpha carbon. Any hydrogen
attached to it is referred to as an
alpha hydrogen. So, generally speaking as
we go from a keto form to an enol
form, the alpha hydrogen will end up over
here attached to the oxygen. The bonds
that were between the alpha hydrogen and
the alpha carbon, pardon me, not the bonds. The bond, or the electrons of the bond,
that were between the alpha hydrogen and the alpha carbon will end up as the pi
bond between the two carbons. And the pi bond between the carbonyl carbon and the
oxygen will end up as a bond between
oxygen and hydrogen. Alright. Let’s look
at a more specific example. So, here’s my
starting material. I have chosen a ketone
this time, 2-butanol. And let me number
the carbons 1, 2, 3, 4. And this is my
carbonyl carbon, which makes this carbon, the alpha carbon. But also this carbon,
the alpha carbon. Alright. I said this happens in the basic
conditions. So, base comes in and the
electrons, the lone pair electrons of my
base will be attracted to one of the
alpha hydrogens. As that happens the
bonds, not the bonds, the electrons that
were the bond between the alpha carbon
and the alpha hydrogen will end up on
the alpha carbon. And so I will end up
with a carbanion and a water molecule.
So, this carbanion, the carbon that
carries the negative charge, has a lone
pair of electrons on it. I am going to
number the carbons again. So, carbon
number 3 has a lone pair.
What is the next step? Remember we are
trying to get to an enol form. So,
that means we need to get a double bond,
and we need to get an alcohol
functionality. So, the lone pair electrons
on carbon number 3… Let’s do that in
the right colour. They will end up forming
a pi bond between carbon 3 and carbon
2. The lone pairs of the electrons from
the carbonyl will be attracted to the
hydrogen on the water molecule. And the
electrons of a pI bond between carbon
and oxygen will move up onto the oxygen.
As a result we’ll end up with a double
bond between carbon 3 and 2 and an
alcohol functionality on carbon 2. And we
have also recovered our base. So, how do
we go back? How do we go from our enol
form back to a keto form? Well, we need to recover the carbanion again. So, how do
we do that? Well, the pi electrons between
carbon 2 and 3 will end up back on carbon 3.
The lone pair electrons on the oxygen
will end up between carbon number 2 and
the oxygen to form a pi bond. And at the
same time the lone pairs of our base
will grab the proton on the alcohol, and
the electrons that make up the bond
between the oxygen and the hydrogen of
the alcohol will end up on the oxygen. So,
we’ll get a carbanion back, and we’ll
get water back. What is the final step?
Remember we are trying to get back to
our ketone.
So, the final step will be that these
lone pair electrons on our carbanion
will be attracted to a proton on our
base. And the electrons that were between
the hydrogen and the oxygen on our base
will end up on the oxygen. And so we are
back at our ketone, and we have also
recovered our base again. That is all for
now, and I will see you on the flip side.