In this next page, we're going to be focusing on a special type of condensation called an
intramolecular condensation.
It gets a little bit more complicated.
Dicarbonyl compounds have the ability to self-condensate through cyclization if it will generate a
five or six-membered ring.
Here's the deal.
Any ring below five members – four, three – way so unstable the cyclize on its own.
You’re going to need to put a lot of energy into that system to make it happen.
In order for these reactions to be favorable, we're going to try to make five and six-membered
rings.
That’s actually going to be a theme later on when we talk about other reactions that
condensation products can undergo.
Here I’ve got this diketone.
This could happen to diesters, diketones, whatever, dialdehydes.
The biggest point here is that when you make your enolate, you want to make your enolate
in a place where if you attack the carbonyl, you’ll make either five or a six-membered
ring.
Let me ask you a question.
I’ve actually got four different positions that could make enolates here.
I’ve got position one, position two, position three, and position four.
Now, you guys understand that three and four are the same thing as one and two so I’m
going to erase them because that's just distracting me.
I literally just have a choice between “Do I put the enolate on carbon one or on carbon
two?”
They both have alpha-hydrogens so I could definitely get enolates there.
The real way that you decided this is you decide by the ring that you could make after
you cyclize.
Let’s look at for example enolate 2.
If I were to put my enolate on carbon 2, then it would attack this carbonyl here.
How big would that ensuing ring be?
This being my first, my nucleophile always gets the first carbon into account.
This would be 1, 2, 3, 4.
You always count from nucleophile to electrophile.
Is that big enough?
No way.
Let’s talk about 1.
If I were to make it on 1, it can’t attack that one.
That’s way too close.
It would attack this carbonyl, so then let’s count.
Let’s see if it’s the right size.
Guys, that is a lot better.
That’s a six-carbon chain that is going to connect between the one and the six, making
a six-membered ring.
The NaOEt is going to specifically make an enolate here so that it can cyclize on its
own.
The mechanism is just going to be what you expect for any aldol reaction.
You would kick up the electrons to the O, and then it would protonate.
What I’d like to do is if we already know it’s going to be a six-membered ring, let’s
just draw a six-membered ring and then let’s just add the groups.
From lots of years of experience, I know that it's actually going to be easiest to number
this ring if I start at the top left corner and I work my way around counterclockwise.
Literally guys, this is just me trying to make it look as pretty as possible.
You could have numbered it however you want as long as your numbers are in order, it doesn't
matter.
The only thing you can't do is like 1, 3, 5, 7.
What am I missing?
What groups do I need?
It looks like I'm missing a ketone on 2, so I’m going to put that there.
What else am I missing?
It looks like I'm missing some stuff on 6.
What am I missing on 6?
Could you guys tell me exactly?
Think about it.
Think about exactly what’s not on this ring.
I’m missing an OH that it’s going to protonate.
Am I missing anything else?
I’m also missing a methyl because look, this methyl was not part of the ring.
It's just extra.
There we go.
Now we have it drawn correctly.
This is called what?
What’s the name of this product?
This is our beta-hydroxycarbonyl.
You’re getting professional at this at this point.
But what do we know about beta-hydroxycarbonyls?
They love to dehydrate.
We're going to dehydrate this bad boy and we're going to dehydrate between the alpha
and the beta.
What this is going to make as a final answer is I’m going to get what we call a cyclic
enone.
This is going to be highly favored because that’s really stable.
You’re going to get a cyclic enone that’s what you get when you have basically an intramolecular
reaction that’s an aldol.
This would be specifically an intramolecular aldol.
You might say, “Johnny how did you know it was aldol?”
Because it’s a ketone to our aldehydes, and you’re going to get a cyclic enone.
In the next video, I’m going to show you what happens for intramolecular condensations
of esters.