so for this practice problem we have got acetone, a ketone, and then we've got
a two carbon aldehyde and so this product mixture in the presence of sodium
hydroxide we're gonna get a number of products, as a hint we've got strong base
with aldehydes and ketones, probably gonna be doing some kind of enolate
chemistry so if you want to go back and check out my tutorial on aldol
condensation that's gonna give you a tip and as it happens we're gonna
want to be able to find four possible products with the mechanisms
that lead to them, so see if you can find four products for this reaction mixture
so what we want to recall about aldehydes and ketones in the presence of
a strong base like hydroxide is that they have alpha protons that can be extracted
to generate enolate anions, so either of them are enolizable so let's take a
look first at the aldehyde, all we can do is, so this is the carbonyl carbon this
is the position that is alpha to the carbonyl, and alpha protons are going to
be slightly acidic, at least acidic enough for a strong base like hydroxide
to extract them, around pKa 19 and so what we're gonna get is the enolate, so
here's our enolate, and then enolates as we know our are good sources of
nucleophilic carbon and so then enolate is swimming around, we have this aldehyde
that we formed the enolate from let's say it bumps into an asset a molecule of
acetone, well it easily can react with a carbonyl containing compound
like a ketone and so actually let's have this go directly to the proton, save a
step, and so what we've done is we've generated this new carbon-carbon bond
and this is aldol, this is what we're doing with aldol, so the enolate has reform the
carbonyl, this pi bond acted as a nucleophile, attacked that carbon and now
we have the hydroxyl right there and then these two carbons are
still there so we want to make sure that those are still there so
that's what those are and then what can happen after that is we can go ahead and
enolize again but instead of drawing a structure that reflects the electron
density going back up what we can do is actually just eliminate here and get the
pi bond there so there is our first aldol condensation product, this is the
aldol addition product, but it's usually tough to isolate there, probably
especially with heat gonna go ahead and get all the way to the aldol
condensation product so there is our first product there. but here's the thing
when this, if this enolate from the aldehyde is formed and is swimming
around in solution it might bump into acetone first but it also might bump
into an aldehyde first, it might bump into a molecule of aldehyde that has
not yet been enolized because in solution we've got billions and trillions
and trillions of these molecules and its who knows what is going to bump into
what, so this aldehyde might enolize again by the same steps, let's just go ahead
and draw it, so we're gonna grab that alpha proton, we're gonna generate the enolate
anion and then it might bump into a molecule of the aldehyde, one of itself
that hasn't yet been enolized and so once again let's just grab this instead
so here's our aldol addition product now once again we can go ahead and grab an
alpha proton and we can kick off this hydroxyl and that is going to be our
second aldol condensation product so this is an example of crossed alcohol
products because we have two compounds
both the aldehyde and ketone both of which can enolize and be the nucleophile
but they can also both be the electrophile because the carbonyl containing
compound is enolized to become the nucleophile but it's also what is
attacked by the nucleophile, that is what we call crossed alcohol products so
we've shown both products that result from an aldehyde enolate but the thing
is acetone as we just said is also enolizable so let's see what happens if we
make the enolate from acetone, so there is our enolate that results from acetone, if we
extracted a proton from acetone we can have that go ahead and attack either a
molecule of aldehyde, let's grab that proton, there is our aldol addition product, once again
this is the bond we formed, in fact if we wanna go ahead and number it, just at
least for one of these, 1, 2, 3, 4, 5 we can see that there has been a new bond
between carbons three and four because when this carbonyl reforms
this kicks out and we've got that new bond there, so there's our aldol
addition product and then again we can go ahead and do this and we will get this
condensation product. note that these are structurally all different, maybe
slightly but they are different molecules. and then lastly once again we
can generate the enolate from acetone and we'll get this and instead of
attacking the aldehyde maybe it attacks another molecule of acetone, let's take that proton
instead, and so there is our aldol addition product. once again we've got a
new carbon-carbon bond, let's number, 1, 2, 3, 4, 5, 6 and that's going to be
1, 2, 3, 4, 5, 6 we've got a new bond between carbons three and five and so there that
is right there, that's the new sigma bond that was generated in the aldol addition
and now lastly we can
go ahead and take this proton and kick off the hydroxyl and we get our fourth
and final aldol condensation product. so this once again is an example of crossed
aldol products because we had two compounds both an aldehyde and ketone
that can act both as the enolate and the thing that the enolate is attacking so that ends up
with four products which are shown here
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