In this video, I want to discuss the specific tautomers of dicarbonyl compounds.
I’ve already introduced the fact that tautomers happen to carbonyls whether you like it or
not.
This is just a natural process that occurs at equilibrium.
Just because it happens, doesn't mean that it happens at very appreciable amounts.
In fact, most carbonyl compounds are going to favor the keto tautomer.
Why?
Thermodynamic.
It’s just more stable.
The keto tautomer is more stable than a vinyl alcohol.
But beta-dicarbonyls are special because beta-dicarbonyls are shaped in such a way that they’re actually
going to favor the enol tautomer.
How do we explain that?
Let's look at the beta-dicarbonyl for a second.
First of all, let’s define beta-dicarbonyl that you have one carbonyl and another one
beta to the one that you currently have.
We remember that the alpha-carbon of a carbonyl has a pKa of around 20.
These are pKas that we’re talking about here.
Does anyone want to take a stab at what the pKa is between a beta-dicarbonyl?
What’s the pKa of this hydrogen here?
Notice that now it has two carbonyls on both sides.
This is going to be a ridiculous pKa of 10, depending – 9, 11, 13 – around that area.
How do you explain that?
That’s a crazy acidity.
That’s as acidic as ammonium, NH4+, with a positive charge.
How do you get something that acidic that’s just an alkane?
Because of the stability of the enol tautomer.
Look at what the enol tautomer is going to look like.
The enol tautomer, first of all you’re going to have two tautomers that are possible.
What you're going to get first of all is a bunch of overlapping atoms that can resonate.
This is called conjugation.
Conjugation makes molecule stable.
Conjugation is going to stabilize these.
On top of that, we’ve even got hydrogen bonding because the enol from one side is
going to be able to hydrogen-bond with the keto from the other.
This is the enol side and this is the keto side.
It can yield to hydrogen-bond with itself.
What winds up happening is that if you take away a proton from the beta-dicarbonyl, the
methylene carbon in the middle, you’re actually going to get 75% enol in equilibrium depending
on what the R groups are obviously.
If these R groups are different things, the number will change a little bit.
I'm just letting you know this is an extremely favored tautomerization.
Beta-dicarbonyl compounds are exceptionally acidic due to the high stability of the conjugate
base.
One thing to keep in mind tis that since this is going to happen so frequently, if you happen
to have a chiral center at the alpha-carbon, let’s say that this had been a chiral center.
It's not now because I’ve got two Hs.
But if it was a chiral center, that chiral center will always be racemized.
Wat the heck does that mean?
It just means that you can't isolate one enantiomer.
You’re going to wind up getting a mixture of both, 50-50.
Why?
Because when you form the enol, you’re going to lose your stereochemistry.
You’re going to turn trigonal planar.
You’re going to lose the wedge and dash structure.
Then when you add, you could either add the proton from the front or from the back.
You're going to get a racemization of any alpha-carbon.
That's pretty much as a rule.
You’re going to racemize all of your alpha-carbons.
This being said, beta-dicarbonyls are very important for organic synthesis because they're
so good at doing tautomerization.
Let’s do this problem here.
I want you to look at the problem and evaluate which of these beta-dicarbonyls is going to
be the most acidic one of all.
Think about the one that can form the most tautomers.
Do that and then I’ll answer.