Voiceover: In the last
video we saw a formation of an enolate anion in one of our mechanisms. In this video and the
next one we're going to go in much more detail about how to form enolate anions and what base to use. For example, this base right
here is the ethoxide anion which you could get
from sodium ethoxide, so Na plus OEt minus, and
this could act as a base and take a proton from
an aldehyde or a ketone. In this case we have an
aldehyde, so this is acetaldehyde here, and we need to
find the alpha carbon, so the alpha carbon is the one
next to the carbonyl carbon, so this is the alpha carbon
right here on acetaldehydes. There are three hydrogens
attached to that alpha carbon, we have three alpha protons,
so our base could take any one of those three alpha protons. I'm just going to draw one
in here to simplify things. And so we could show our
base taking this proton and leaving these electrons
behind on our carbon, so we can draw the enolate
anion that would form. So we have our cabonyl
here and then we could show these electrons on this
carbon now, so let me go ahead and follow those
electrons, so in magenta these electrons move off on to
this carbon to form a carbanion. Remember there's also two other hydrogens attached to that carbon there. And we could draw a
resonance structure for this, so we could show these
electrons in magenta moving in here to form a double
bond, push these electrons off on to the oxygen,
and if we do that we'll have a resonance structure showing the negative charge this time on the oxygen. So the oxygen gets a negative
one formal charge now, and then we have a double
bond over here on the right, and so this would be our
other resonance structure. So the electrons in magenta
moved in here to form our double bond, and then let's
make these electrons in here blue, move off on to the
oxygen to make an oxyanion. So this is our enolate anion
here, so we have the carbanion form and then the oxyanion
form of our enolate anion. So your enolate anion here,
which is once again extremely important for reactions
we'll talk about later. So our base has formed our enolate anion. If you think about what happens
to the base, if you protonate ethoxide you're going to form ethanol. So we'll go ahead and show ethanol also formed here in this reaction. So we have an equilibrium
here between our aldehydes and our enolate anion, and
to figure out which direction is favored we need to
know some pKa values. So the pKa value for this
aldehyde is approximately 17, and the pKa value for
ethanol is approximately 16. And so one way to figure out
which direction is favored is to use these equations down here. We could first find the
pKeq by taking the pKa of the acid on the left, so
the acid on the left is our aldehyde, so the pKa is
17, and from that number we subtract the pKa of
the acid on the right. The pKa of the acid on the
right is 16, which is ethanol. So 17 minus 16 gives us one,
and then to find the Keq we can take 10 to the negative
pKeq, so 10 to the negative one is equal to point one,
which is obviously less than one, and so we know
that the equilibrium favors the reactants, the
equilibrium is back in this direction so the equilibrium
favors formation of the aldehyde, so at equilibrium
we're going to have some aldehyde present,
we're also going to have some enolate anion present here. So if you choose sodium
ethoxide as your base you're going to have both the aldehyde and your enolate anion present. Another way to figure out
which direction the equilibrium goes is to think about which
one is the weaker acid. So we have these two acids here. Let me go ahead and change colors. We have a pKa of 17 and a pKa of 16. The lower the pKa the
more acidic something is, so ethanol is more acidic
than R aldehyde and the equilibrium favors formation
of the weaker acid, and so since the aldehyde
is the weaker acid the equilibrium favors
formation of this weaker acid. So that's another way to think about which direction for the equilibrium. What if you wanted to completely
make your enolate anion? So one thing you could
do is add a base like a hydride, so here we have
once again acetaldehyde here. So we have acetaldehyde, and
this time our base is the hydride anion, so we could
get that from something like sodium hydride, Na plus H minus, or potassium hydride, K plus H minus. And so once again we
identify our alpha carbon, which is this one, with
three alpha protons. So we can just go ahead and
draw one of them in there. And we could show our
hydride anion functioning as a base, taking this
proton, leaving these electrons behind on that carbon. So let's go ahead and show the
enolate anion that results. So we have our carbonyl
here, and then we have our electrons on this carbon,
giving it a negative one formal charge, so electrons
in magenta move out on to this carbon, forming the carbanion. And just to save time I
won't draw any oxyanion, but that's the one
that's actually a greater contributor to the resonance hybrid here. So we have our carbanion
and then we would also form hydrogen gas, so if this hydride anion picks up a proton we would form H2. And so let's show those
electrons, so the electrons in red here on our hydride
anion pick up this proton and forming this bond and
so we get hydrogen gas, which would bubble out of
solution, and since this is going to bubble out of
solution we're going to drive the reaction to completion,
so we're going to push the reaction to completion
and so we're going to get our enolate anion so we're
pretty much going to get complete formation of
our enolate anion here. So in the next video
we're going to talk about another base called LDA
which can also give you complete formation of your enolate anion.