Now that we understand the process of oxidation pretty well, I want to show you guys at least
one of the example mechanism of oxidation so you'll understand kind of the electron
movement that creates this oxidative environment.
So guys, one of the most common reagents that's used for oxidation is Jones reagent.
And Jones reagent is the name that we give to the combination of a chromium 6+ reagent
and strong acid.
This takes a lot of different forms and you'll see it written a lot of different ways, but
essentially this would be the combination of CrO3 with something like H2SO4.
This is very, very common.
But you will see it other ways because once the H2SO4 reacts with the CrO3, it's going
to change and it's going to look a little bit different.
But as long as you see a chromium reagent with a strong acid, that's pretty much a Jones
reagent.
What I want to do is I want to take you guys through this mechanism step-by-step and what
we're going to start off with is an alcohol.
As you can notice, we're starting off with a secondary alcohol.
What would we expect a secondary alcohol to become after it reacts with a strong oxidizing
agent?
Well, remember that you can only oxidize as much as you can without breaking carbon-carbon
bonds.
So what that means is that if it's a secondary alcohol and I need to keep both of those R
groups there, the most that this can become is a ketone.
So actually that's what we're trying to do.
What we're trying to do is we're trying to get rid of this alcohol, get rid of this H
and make a ketone.
How is that going to work?
Well, let's go ahead and start.
The first step guys is nucleophilic attack because here we have chromic acid.
Chromic acid which is formed by the chromium and the acid reacting together.
And chromic acid is kind of unique because it's got all these crazy dipoles pulling away
from the chromium.
What kind of reactivity do you think the chromium is going to display?
Guys, it's going to be a super strong electrophile, right?
Because it's got basically no electrons.
So in our first step, we would expect the alcohol that's a decent nucleophile, it's
got electrons on it, to attack the chromium.
Now the chromium already has enough bonds.
It doesn't want more.
So if we make a bond we have to break a bond.
So one of these double bonds is going to become an anion.
So now I would expect that I would get an O-.
But we're in the presence of acid, so the O- is going to protonate.
Then that's what I'm going to get.
I'm going to get an OH there, instead of the double bond.
So now basically once step forward, we get this huge molecule.
And I know that it can be difficult to keep track of where things are, so I'm going to
use colors to show you where everything is.
Let's start off with the nucleophilic attack.
This OH here is now right here, attached to the chromium that still has an H on it.
Let's go to another easy one.
This OH is still here.
Nothing changed there.
This OH is still here.
So they're still in the places that they were before, but there is one new OH that's the
one that is in pink because this pink, remember that it grabbed an H, right?
That means that this is the OH here.
Now hopefully, those colors help you to kind of identify what's going on, where did everything
go.
All these atoms are still there, so we're doing okay.
Now we did our nucleophilic attack.
The next step is a very kind of interesting and rare step called alpha-elimination.
Guys, you've heard of elimination reactions already.
They make double bonds, right?
We've always dealt with beta-elimination.
Beta-elimination is kind of just the go-to elimination that we use.
But in some specific mechanisms in organic chemistry, we actually see alpha-eliminations
take place and oxidation is one of them.
Oxidation mechanisms have a lot of alpha-elimination.
What alpha simply means is that instead of eliminating between the alpha and the beta
carbon, you're actually going to form a double bond directly on that alpha-carbon to something
that's not a carbon.
In the next step, I'm going to eliminate.
This is my alpha carbon because it's the one that has the oxygen on it.
If I were to say, this is an alpha alcohol.
This is where my alcohol is.
My alpha carbon is the one that's attached to the alcohol.
Well, you deprotonate with a conjugate base of whatever your acid was, water in this case.
Instead of forming a double bond between two carbons, I just form a double bond directly
on that carbon and a non-carbon atom like O.
Now the O needs to break a bond because the O isn't happy having that many bonds.
So the O is going to break a bond and actually it's going to do another alpha-elimination.
So this double bond is going to break and make a double bond on this O and then finally
that O isn't happy, so we're going to basically get rid of that H eventually with water.
It's not going to happen all in this step.
In fact, maybe I just won't draw that for right now because that would happen in the
second step, but eventually you have an OH and then it gets deprotonated.
Awesome.
The most important part here guys is not what happens to the chromic acid after because
I don't really care.
That was just my oxidizing agent.
What I'm really concerned about is what's happening to this carbon.
Well, it's looking a lot more like a ketone, isn't it?
Because what I did was I made a double bond between the C and the O. And those two R groups
are still there and I got rid of the H. So now what I'm going to have is this O needs
to be deprotonated and I can deprotonate it with water or whatever other conjugate you
want.
What you're going to get is a ketone.
Cool.
You're going to get a ketone plus you'll get your chromic acid that eventually reforms.
So you'll get something that looks like this.
But we don't really care about that because that's just the oxidizing agent.
What we really need to be able to draw here is how to get to the ketone.
Cool?
Awesome guys.
Hopefully, that helps you understand the process of oxidation a little bit better and also
it introduces the concept of an alpha-elimination, which will come up again in other reactions.