Alcohols, Epoxides and Ethers
By James Ashenhurst
Demystifying Alcohol Oxidations
Last updated: March 26th, 2019
Today’s post in a nutshell: many oxidations of alcohols might seem a little mysterious, but essentially proceed through a glorified E2 mechanism. Many oxidants work by attaching a good leaving group to oxygen, which is followed by elimination.
The Familiar Key Step At The Heart Of (Almost) All Alcohol Oxidation Reactions
When I was learning organic chemistry I remember the reagents for oxidation reactions completely coming out of left field.
KMnO4, K2Cr2O7, PCC, CrO3, Swern, Dess—Martin ? Hold on. Where did these reagents come from? How do they work? Why chromium? What’s the mechanism?
In my course, the details of these reactions were completely glossed over. “ Don’t worry about the mechanism! No time to go through this! “ the instructor said. I was left with the impression that there was something deeply mysterious about alcohol oxidation.
Only later did I learn that it’s not mysterious at all. In fact the key mechanism is very familiar.
Let me show you what I mean.
Here’s a reaction we’ve seen before. Elimination of alkyl halides to give alkenes through an E2 mechanism. Base removes hydrogen, we break C-H, form C-C (π) and break C-LG. The result is an alkene.
Now imagine a slightly different E2 reaction, except one where the good leaving group is on oxygen. We’ll leave it vague, as “LG” for now.
See how we break C-H, form C-O (π), and break O-LG, forming a new C-O π bond in the process. Since we’ve formed a new C-O bond at the expense of a C-H bond, an oxidation has occurred.
Believe it or not, most oxidation reactions of alcohols proceed exactly this way!
Oxidants Are Essentially Just Fancy Reagents For Attaching Good “Leaving Groups” Directly To Oxygen
Hold on, you might say. It can’t possibly be that simple. Why do we have so many different types of oxidizing agents? And why do the mechanisms (like the Jones oxidation here for example) seem so complicated?
Yes, there are a lot of steps in a typical oxidation reaction. However, most of these steps consist of:
– activating the oxidant (such as in the Swern oxidation, where oxalyl chloride converts DMSO to an electrophilic sulfur species, or in chromate oxidations, where strong acid converts chromate (such as K2Cr2O7) to the active oxidant [H2CrO4]
– coordination of the alcohol to the oxidant, followed by proton transfer(s) (seen in the mechanisms of most chromium oxidants, and Dess-Martin periodinane).
These steps are important, of course, but only in a supporting role. If you’ll excuse the analogy, they’re just foreplay that precedes the main event.
The effect of these beginning steps is simply to install a good leaving group on oxygen. That “good leaving group” can take many forms. It’s illustrated here with each oxidant, in green. There are, of course, many, many more oxidizing agents for alcohols than those depicted, but almost all of them essentially work the same way.
Treatment of each of these substrates with base then results in breakage of C-H, formation of C-O (π) and breakage of O-LG.
Each of these “leaving groups” accepts the pair of electrons from the bond to oxygen, reducing its oxidation state by 2 in the process. [remember – the oxidant is reduced, the substrate is oxidized]
What About Oxidation Of Aldehydes To Carboxylic Acids?
So if oxidation of alcohols to aldehydes and ketones is essentially an E2 reaction, how do we explain oxidation of aldehydes to carboxylic acids?
See, given what we’ve just shown, you might initially think it works something like this:
That’s actually not what happens. [Why not? the aldehyde carbon is a good electrophile, and any species basic enough to remove the C-H is more likely to add to the aldehyde C ]
It actually follows the same type of process as with alcohols! However, there’s a trick.
There’s a missing ingredient not mentioned in the diagram above. Water.
What happens is that water adds to the aldehyde, forming a hydrate. [If this looks unfamiliar, you’ll see MANY variations of this type of mechanism in your upcoming chapter on aldehydes and ketones. This is a sneak preview]
NOW, the oxidant attaches to one of the hydroxyl groups of the hydrate. The E2 from here is much easier to visualize.
This also helps to explain one key observation I made tangentially in the last post. The reagent CrO3/pyridine (Collins’ reagent) will oxidize primary alcohols to aldehydes and stop there. However, if water is present, this oxidation will go all the way to carboxylic acids. That’s because the water will form a hydrate with the aldehyde, allowing for further oxidation.
No hydrate, no further oxidation.
Why Don’t Ketones Oxidize Further?
This also explains why ketones don’t oxidize further. There’s no hydrogen that can be removed in an E2-type process that will lead to a new double bond!
It’s similar to the old question of why this alkyl halide (below) doesn’t undergo elimination. There’s no hydrogen on the “beta” carbon (i.e. on the carbon adjacent to the carbon bearing the good leaving group) that can be removed, so no elimination occurs.
The same could be said for why tertiary alcohols don’t oxidize.
The Bottom Line
So the bottom line for alcohol oxidation is the following.
- Pretty much every alcohol oxidation reaction you’ll encounter has the same key step: an E2-like deprotonation of C-H that results in formation of a new C-O pi bond and breakage of a transient leaving group.
- Aldehydes oxidize to carboxylic acids after formation of a hydrate.
- Ketones don’t oxidize further because there’s no C-H bond that can be broken that would result in a new C-O pi bond.
In the next post we’ll move to something completely different: intramolecular reactions of alcohols, a perennial subject of organic chemistry exams.
Next Post – Intramolecular Reactions Of Alcohols And Ethers
The main exception you’ll encounter is KMnO4, which likely proceeds through a C-H abstraction/internal return type mechanism followed by collapse of the hydrate to give the new carbonyl. That mechanism is mentioned in exactly zero introductory textbooks, so you likely don’t “need” to know this unless you are exceptionally curious about organic chemistry. [back to post]