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Current time:0:00Total duration:10:18

Voiceover: In order to see
how to form enolate anions, and in this video we're just
gonna look in more detail how to form enolate anions from ketones. And so the ketone we have here is acetone. To find our alpha carbon, we just look at the carbon next to our carbonyl carbon, so this could be an alpha carbon, and this could be an alpha carbon. Each one of those alpha carbons
has three alpha protons, and so there's a total of six. I'm just gonna draw one in here, and this is the one that we're going to show being deprotonated here. So, the base that's going
to deprotonate acetone, we're gonna use LDA, which
is Lithium Diisopropyl Amide And, I could go ahead and
draw in the Lithium here, so Li Plus, and then we see the two isopropyl groups like that, a negative one charge on our nitrogen. So this is a very strong base, it's also very bulky
and sterically hindered. So you can think about
a lone pair of electrons in the nitrogen, taking that proton, leaving these electrons
behind on this carbon, so we can go ahead and draw
the conjugate base here. We would have electrons
on this carbon now, that's a carbanion, so let me go ahead and show those electrons, these electrons in here magenta, are gonna come off onto this carbon. And this carbon is a
[carbanae] because remember there's also two other
hydrogens attached to it. So that's what gives it a
negative one formal charge here. We can draw our resonance structure, we can show these electrons
in magenta moving in here, these electrons coming
off onto our oxygen, so for our resonance
structure we would show the negative charge is now on our oxygen, this would be a negative one
formal charge like that now. So the electrons in magenta moved into here to form our double bond, and then we can show the electrons in here in the blue moving out onto the oxygen. So this is our enolate anion, right, this represents our enolate anion, and for our two resonance structures, we have one with the negative charge on the carbon, so it's our carbanion, and one with our negative charge on our oxygen, that's our oxyanion. Remember the oxyanion
contributes more to the overall hybrid because oxygen is more electronegative than carbon, so that's our enolate
anion that is formed. So, if we form our enolate anion, we're also gonna get another product. So if I think about adding a proton onto our base, we're
going to form an amines. Let me go ahead and draw the
amine that we would make. So we would have, it
would now be this amine. So let's show those electrons. So the electrons over here in red, these electrons right
here pick up this proton forming this bond and we'd form an amine. So this reaction is at equilibrium. So to figure out which
direction this favors, one way to do it would
be to calculate the Keq. And you can see over here on the left, these ways for calculating the Keq, which we talked about in the last video, We first need to calculate the pKeq, which would be the pKa
of the acid on the left, so the acid on the left would be acetone with a pKa of approximately
19 or sometimes you'll see 20, and so from that number
we're going to subtract the pKa of the acid on the
right, which would be our amine. The pKa of this is approximately 36, so 19 minus 36 gives us negative 17. To find the Keq, all we do is take 10 to the negative of that number, so 10 to the negative of negative 17 is the same thing as 10 to the 17th, which is obviously a huge number, much, much greater than one, so we know that the
equilibrium lies to the right. So the equilibrium favors
formation of the enolate anions, and for all practical purposes,
this is a huge number, you're pretty much gonna get complete formation of your enolate anions, so if you add LDA to
acetone you're gonna get again, pretty much all enolate anion, and none of your acetone will remain here. Another way of figuring
that out is to know that the equilibrium favors formation
of the weaker acid. The weaker acid is the one with the higher value for the pKa, remember the lower the pKa the more acidic something is, so acetone is more acidic than this amine, and
so since the amine has the higher value for the pKa, the equilibrium favors
formation of this weaker acid, also to the right like that. Alright, so let's talk
about the pKa of acetone or ketones in general
compared to aldehydes. So this pKa is higher
than that for our aldehyde like we saw in the last video, and we can think about
why by looking at the enolate anion and thinking
about the fact that now we have a methyl group here, and alkyl groups are electron donating so, if it's donating a little
bit of electron density, it already has a negative charge on it, so donating a little
bit of electron density would destabalize this
negative charge a little bit. So if the conjugate base is destabalized, that means that acetone is not quite as likely to donate a proton. So that's the situation
when you have a ketone here. When we had an aldehyde,
we didn't have this electron donating factor, and so that's just one way to think about why an acetone, why a ketone, is not as acidic as an
aldehyde in general. Alright, here's an example where we have a very acidic ketone. So this is a special kind of ketone called a beta-diketone, this
is a beta-diketone here. And if we look for alpha carbons, we look for the carbon
next to our carbonyl, so this is an alpha carbon,
this is an alpha carbon, and this is an alpha carbon, And so the question is which one of those, which one of those
alpha carbons is the one that has the most acidic protons. It turns out to be the
alpha carbon in the center, the one between our two carbonyls, and so there are two alpha
protons on that carbon. And so the pKa for one of those, for one of those acidic
protons is about nine, so that's very acidic, much
more acidic than acetone or acetaldehyde like we talked
about in the last video. And so since a beta diketone
has such acidic protons, we don't need a super
strong base like LDA, we can use something like sodium ethoxide, so negative one formal
charge here on this oxygen, so sodium ethoxide could be used to deprotonate this beta diketone. So we think about one pair of
electrons taking this proton, leaving these electrons
behind on this carbon. So let's go ahead and
draw the conjugate base, so we would have our carbonyls here, and we took a proton away, so now we have our
electrons on this carbon, so a negative one formal charge. So these electrons right in here, magenta, moved onto this carbon, which gives this carbon
negative one formal charge, because there's still a
hydrogen bonded to that carbon, I'm just not drawing it in so
we can see a little bit better. For resonance structures,
we could show these electrons in magenta moving into here, pushing these electrons
off onto the oxygen, so we could draw a resonance
structure for that. So for our resonance structure, we would form a double bond here, and then this oxygen would get a negative one formal charge like that. And then our carbonyl on the
right is still here like that. So that's one of our possible
resonance structures, electrons in magenta moved in
here to form our double bond, and then we could show
these electrons in here moving off onto our oxygen like that. We could have shown our electrons moving over on this side as well, so we can push those electrons off. So let's go ahead and draw another resonance structure here. So let's get some room and let's show the formation of another
resonance structure. This time our carbonyl on
the left is still there. We moved some electrons into here, and then we would form a negative
one charge on this oxygen, so a negative one charge
right here like that. And so let's show those electrons, electrons in red move in
to form our double bond, and then let's show
these electrons in here coming off onto this oxygen like that. So a total of three resonance structures for this enolate anion. And so you could see this negative one formal charge is delocalized, right, it's delocalized over this carbon, it's delocalized on this oxygen, it's delocalized on this oxygen. So the more you delocalize, or spread out a negative charge, the more you stabalize the anion. So this is a very stable
anion because of resonance and also because of conjugation. If you think about, if you think about the conjugation present here,
so here's a double bond, and here's a single bond, and
then here's a double bond. So you have some conjugation
stabalizing it as well. And so we have a very
stable conjugate base for all those reasons
that we just talked about. And since we have a stable conjugate base, this beta diketone is likely
to donate one of these protons, and so that's why it's
pKa value is so low. Alright, so if ethoxide takes
one of these acidic protons, we're gonna form ethanol
as our other product. So we can go ahead and draw
ethanol in here like that, and then we can calculate
the Keq for this reaction. So ethanol's pKa we've already
seen is approximately 16, so if I want to calculate the Keq, first we find the pKeq. So the pKa of the acid on the left, the acid on the left
is this beta diketone, so that'd be nine, from
that number we subtract the pKa of the acid on the
right, which is ethanol, so nine minus 16 gives us negative seven, so that's the pKeq. To find the Keq, we take 10 to
the negative of that number. So 10 to the negative of negative seven, is of course the same
thing as 10 to the seventh, and obviously that's a number
much, much greater than one. And so we know the
equilibrium lies to the right, favoring formation of
the enolate anion here. So favoring formation
of our enolate anion. So for all practical purposes
we're gonna get pretty much nearly complete formation
of our enolate anion. And so, we're gonna get our enolate anion, and that's because even though we use sodium ethoxide, it's not
as strong of a base as LDA, but it's enough,
it's enough to deprotonate our very acidic beta diketone.