Now that we have epoxides and know how to make them, it's important to know what can
we do with them.
It turns out that ethers are highly unreactive.
Ethers barely react with anything.
But because of the fact that epoxides are three-membered rings, they're very, very highly
strained.
What that means is that even though ethers as a function group are not very reactive,
epoxides are because they have so much potential energy locked up in those bonds.
They're going to want to break open first chance that they get.
So it turns out that we have two different ways that we can open the epoxide ring and
they have to do with different reagents that we use.
These are called the acid-catalyzed ring opening reaction and the base-catalyzed ring opening
reaction.
Let's start off with acid and then we'll work on to base.
So in an acid-catalyzed reaction – notice first of all that I'm using the words acid-catalyzed.
That actually already gives us a hint about the mechanism.
Remember when I said whenever you have an acid catalyzed mechanism, what's the first
step going to be?
Protonation.
You always need to protonate first.
If it says acid-catalyzed, you always need to protonate first.
So what that means is that the very first step is if I'm using an acid with an epoxide,
the O, the nucleophilic O, is going to go ahead and grab the H from the acid because
that H has a positive charge.
So now what I'm going to get is a molecule that looks like this, where everything is
still in the same exact place, except that I've got a formal charge on this O because
the O has too many bonds.
So now what happens?
I've got a Cl- and the Cl- wants to break open the ring, but it's trying to decide which
side is it going to break.
Is it going to break the most substituted, the tertiary, or the least substituted, the
secondary?
That's supposed to be a three by the way.
Let me draw that again.
The tertiary or the secondary?
The answer to this is that the chlorine or whatever anion we're using, even if it was
just a neutral nucleophile, is going to be the most attracted to the side of the ring
with the most positive character.
So what we're looking at is the side of the ring that could stabilize that carbocation
the best because that – I'm sorry, I just said carbocation.
It's not a carbocation.
It's just a cation.
It's just a positive charge.
But that positive charge can delocalize a little bit into those two different atoms.
So the question is which side is going to be the one that has the most positive character,
the secondary or the tertiary.
The answer is the tertiary.
So I'm going to go ahead and attack here.
If I make that bond, I have to break a bond.
So I'm going to break the bond to the O. What this is going to give me is a new compound
that looks like this.
Let's say that this bond right here is the red bond that I'm drawing here.
Oops.
That red is the same as this.
What is it going to have on it?
Well, let's say that this chlorine attached from the front, right there.
What else would it have?
Well, it would have two methyl groups, so I'm going to draw methyl, methyl.
And then on the other side of the red bond what am I going to have?
Well, if the chlorine attached from the back, what that means is that the alcohol is going
to go – I mean from the front, the alcohol is going to go to the back.
By the way, this alcohol that I'm drawing here is just that right there.
Notice that it's still attached in this position.
So it's still going to be attached there.
On top of that, I'm still going to have two other groups.
I'm going to have a propyl and I'm going to have an H. So that H would just stay where
it is in the front.
Obviously, you don't need to draw H's.
At the end of the day, I could have just drawn this without the H. I could just erase that
if I wanted to.
Does that make sense?
So notice what my product is here.
Well, my product is that I'm going to get – what my product is is that I'm going to
get a nucleophile attaching to the more substituted position and I'm going to get an alcohol at
the least substituted position.
Just so you guys know, this is unique to acid-catalyzed ring opening.
Base-catalyze ring opening is actually going to be different.