Baeyer-Villiger Oxidation - Online Tutor, Practice Problems & Exam Prep

The Baeyer-Villiger oxidation reaction transforms aldehydes and ketones into carboxylic acids and esters, respectively, using peroxyacids like mCPBA. This reaction is regioselective, influenced by the migratory aptitude of substituents, with hydrogen having the highest aptitude, followed by tertiary and secondary carbons. The mechanism involves three main steps: protonation of the carbonyl oxygen, nucleophilic addition of the peroxyacid, and migration of the substituent to form the final product. The stereochemistry remains unchanged throughout the reaction, retaining the original configuration.

General Reaction:

What's up everybody? In this section, we're going to take a look at an oxidation reaction that is very unique but also very useful. Okay. It's the Baeyer-Villiger oxidation reaction. And the Baeyer-Villiger oxidation reaction reacts aldehydes or ketones with peroxyacids. Alright? And it turns the ketones into esters and aldehydes into carboxylic acids. Alright? And like I said, it's reacting with peroxy acids. Where have we seen peroxy acids before? Yeah. Peroxy acids are our epoxidation reagent when we're making an epoxide from an alkene. Okay. So remember, we can have some sort of alkene plus our peroxy acid, and we'll get an epoxide like that. Right? That little three-membered ring with oxygen. Okay? When we use the same reagents with an aldehyde or a ketone, we get this oxidation reaction where we increase our number of bonds to oxygen. Okay? This reaction is unique in that it differs from our other oxidation reactions in that we can oxidize a ketone, which is usually the most oxidized we can make something with, like our Jones oxidation, which is our strong oxidizing agent. We can actually further oxidize a ketone. Okay? So that is really useful, especially when we get into the carboxylic acid derivatives chapter where it's really important to be able to go from aldehydes and ketones to carboxylic acid derivatives. Okay? So this oxidation is regioselective based on the migratory aptitude of the groups that are attached to a carbonyl carbon. Alright. So migratory aptitude is probably a new word for all of you, and it's something we'll define once we get down to the general reaction area. I even know it's basically just how well does one group or how likely is one group to move compared to another group. Okay. That migratory aptitude trend is similar to the cation stability trend. Alright. There's really not a great way to memorize this, but hydrogen has the highest migratory aptitude. Alright. Followed by tertiary then secondary carbons. Alright. After that we have aryls. So that’s like our benzene rings are aromatics. And the worst is going to be our primary carbon. Okay? So this is similar to our carbocation stability trend. And I should say that this is relatively controversial still. What I have listed here is the most up-to-date and the most agreed upon trend, but you may see some variation from your professor or your textbook. Alright. And just go with whatever your professor expects you to know, but it's most likely going to be this. Okay? So now to get into the general reaction. Alright. We start off with an aldehyde or a ketone, and we react it with our peroxyacid. Alright. In both of these reactions, we use MCPBA, alright, which is metachloroperoxybenzoic acid. And what this looks like is we have a peroxy acid. Right? This here is our peroxy acid, and attached to it is a benzene with a metachlorine. Okay. So that's our MCPBA. But really all we care about is this peroxyacid portion of it. Alright. And what the reaction does is it basically inserts an oxygen between our carbonyl carbon and our group with the highest migratory aptitude. Alright. So looking at the ketone on the left here, we have a primary carbon and we have a tertiary carbon directly attached to our carbonyl carbon. Alright. Looking back to our trend, tertiary has a higher migratory aptitude than primary. Alright. So that will be our group that migrates. And what we mean by migrating, right, or by our migratory aptitude is we have an oxygen atom from our peroxy acid that is essentially just going to insert between the carbonyl carbon and the group with the highest migratory aptitude. Okay? So that group has to shift. It has to migrate away to make that new bond to oxygen and break the bond with the carbonyl carbon. Alright. You'll see exactly why this is called migration when we get to the mechanism. Alright. But that is what our migration our migratory aptitude is describing is which group can best separate from the carbonyl carbon and make that new bond to the oxygen. Okay. So here it's the tertiary group, and that's what we end up with in our product. We have that oxygen inserted between our carbonyl carbon and our tertiary carbon. Alright. Looking at the next example, we have an aldehyde this time. Alright. So we have a hydrogen compared to a tertiary carbon. Which group here has the highest migratory aptitude? Yeah. Here, it’s the hydrogen. Okay. So now our oxygen needs to insert on that right side. So that hydrogen is our group that's going to migrate. Alright? And that's what we see here. We inserted our oxygen between our carbonyl carbon and our migrating group. Okay. Before we move on to the mechanism, we need to talk about stereochemistry. Alright. And the stereochemistry for these reactions is really easy. Alright. It just stays exactly the same. So looking at this example, we have a ketone. We're reacting with MCPBA again. Alright. So we know we're doing our Baeyer-Villiger oxidation, and we have a secondary carbon on the left side and a tertiary carbon attached to that carbonyl carbon on the right side. Which of those has the higher migratory aptitude? Yeah. The tertiary does. Okay. So that's the side we're going to insert oxygen on. Alright. That tertiary carbon will migrate away, and we'll insert our oxygen here. And all we need to know about our stereochemistry is that it stays exactly the same. Okay. So this was a wedge in our starting material. Alright. It is going to be a wedge in our product here. Alright? And if we looked at our absolute configuration here, our R and S, it's going to be retained as well. Okay? So now, in the next video, we'll get into the mechanism and really see where the word migration comes from. Okay?

Mechanism:

Alright, everybody. Let's go ahead and take a look at the mechanism for the Baeyer-Villiger oxidation reaction. This mechanism takes place in 3 major steps. 2 of those 3 steps, you should be very familiar with. Alright? You've been doing them the whole time, more or less that you've been in organic chemistry. Okay. And then there's just one new step which is where we get the term migration from. And we'll discuss that when we get there. Okay. So starting off with the first step, this is one that you've done a 1000 times. It is our pronation step. Alright. So in our pronation step, we just protonate the oxygen in our carbonyl. Okay. So we have a ketone. We're reacting it with a peroxy acid. Right? That acid has an acidic proton. Those are gonna react together. And what we'll have is this oxygen. We'll use a lone pair to make a new bond to that hydrogen. We made a bond, we have to break a bond. So we kick those electrons up to that oxygen. Alright. That's all our pronation step is. And that gets us to our first set of intermediates. Okay. We're missing some atoms and some charges on these intermediates. Do you know what we're missing? Yeah. Well, we just protonated our carbonyl, so we need a hydrogen there. Right? And what charge will that give it? Yeah. That will give it a positive charge. Okay. What about the oxygen that just lost that hydrogen? Well, now that has an extra lone pair, it's gonna have a negative charge. Okay? So now that we have our complete intermediates here, we're gonna do our second step. Alright. Which is a nucleophilic addition step. It's our nucleophilic addition. Alright. And this is just like any other nucleophilic addition you've done before where the negative charge of whatever our nucleophile is, in this case, it's our what was our peroxy acid, it's that conjugate base, that attacks the partial positive charge of our carbonyl carbon. Alright. So we just have this arrow here. We do our nucleophilic addition right there. We made a bond, so we have to break a bond. So that's what we'll do right there. Okay. That will get us to our next intermediate. And again, we're missing a couple atoms or charges here. Right. Where are we missing something? Well, yeah, remember, we added this hydrogen in our first step. Okay. This oxygen right here, that is the same oxygen from our carbonyl carbon we've had the whole time. Okay? So we need to have a hydrogen there. Will it have a charge at all? No. It won't. Right? Okay. And if we wanted to layout all of the atoms from our peroxy acid, we could say that these 2 oxygens, the carbonyl and this r group are gonna be those 2 oxygens, the carbonyl and the r group there. Okay? So this is the new step in the mechanism that you haven't seen before, and this is our migration step. Okay? So this is how we go from that molecule to our final product, and it's our migration. Okay? So that oxygen of our carbonyl that came from our carbonyl, okay, the yellow one, it really wants to reform that carbon oxygen double bond. Alright. So it's gonna do that. It's gonna use a lone pair, and it's gonna reform that carbon oxygen double bond. Okay? Now here you would probably expect that the bond between the carbon and the green oxygen that came from our peroxy acid, you'd expect that to break, but that's not actually what happens. This is where our migration occurs. Okay? So if we look at the 2 groups attached there, we have a primary carbon and we have a tertiary carbon. Which of those has a higher migratory aptitude? Yeah. The tertiary does. Okay. So what we wanna do is we migrate that group. And the migration, what that looks like is this bond here that I'm coloring in green will actually break that bond and will use the 2 electrons in that bond to move the whole group. Okay? So this group, the tertiary group with the higher migratory aptitude will migrate to the oxygen from our peroxy acid. Okay? So that one arrow right there is where we get the word migration. Alright. From there, we made a new bond, so we need to break a bond. Okay. We're gonna break the oxygen oxygen bond, and that is just gonna swing over and form another carbon oxygen double bond. Alright? And again, we made a bond, so we have to break another bond. So this carbon oxygen double bond gets kicked up to the oxygen. Alright? So in total, our migration step just has 4 arrows. Alright? And they all move in the same general direction. Okay? If we have it drawn how we have here, they're all moving from the oxygen from our original carbonyl all the way out to the oxygen from our carbonyl in our peroxyacid. Okay? So 4 arrows there, and that gets us almost to our final product. Alright. We still have this hydrogen right there. Okay. So we're still gonna have a hydrogen in this product right here. Alright. So what is the last thing we need to do to make our final neutral product? Yeah. We just need to deprotonate. Alright. You may remember that I said there's 3 major steps. Alright. This is not a major step. Alright. This is just a deprotonation. We're gonna have everything highlighted in green over here come in. So we're gonna have our 2 oxygens, our carbonyl, and our r group. Right? That has an it's gonna be 1 oxygen. That has a negative charge on it. Alright. That is just gonna come in and deprotonate our carbonyl here. Just like that. Okay. So we're gonna get a neutral final product, plus we're gonna get a byproduct of a carboxylic acid. With whatever r group was part of our peroxy acid that we started with. Okay? So this is the mechanism for the Baeyer-Villiger oxidation. Remember, it's the 3 major steps, protonation, nucleophilic addition, migration. Alright. And then the last thing we have to do is just deprotonate to get our major neutral product. Okay. Go ahead and attempt the practice problems, that are following this and then we'll solve them in the next video.

Predict The product:

Okay. This first practice problem was asking us to predict the major product formed from this reaction. Okay. So first things first, we just need to identify everything involved. Alright? So our reagent here that we have or our starting material is a ketone. Alright. And we're reacting that with this molecule here, CF₃CO₃H. What type of molecule is that? Is it a peroxyacid? Well, yes. Alright. And how do I know that? Well, anytime we see this general structure here, the CO₃H, attached to whatever on the left of that carbon. Alright. This can be anything on the left there. Whenever we see that CO₃H, we know that it's going to be a peroxyacid. Alright. So anytime you see a molecule written out with CO₃H, you know it's peroxyacid. Anytime you see something that just has CO₃H, right, and again, this that R group there, that can be anything. Whenever we see either of those, we know it's a peroxy acid. Okay? So, we have a peroxy acid and we have a ketone. What type of reaction is it? Yeah. It's going to be a Baeyer-Villiger oxidation reaction. Okay. So now that we've identified what type of reaction it is, we've identified our reagents. What's the first thing we should do? Yeah. Well, we should just look at our ketone and look at each side of the ketone and identify which has the higher migratory aptitude. Okay. So looking at this ketone, on the right side, we have this carbon here. What type of carbon is that? Yeah. It's going to be secondary. Okay. Then on the left here, we have this carbon there attached to our carbonyl carbon. What type of carbon is that? Yeah. That is a tertiary carbon. Okay. So now we need to think back to our trend for migratory aptitude. And what has a higher migratory aptitude? A tertiary carbon or a secondary carbon? Yeah. Tertiary does. Okay. So we know that that is going to be the migrating group. Alright. And remember that we can just kind of shortcut this reaction by thinking of our oxygen from our peroxy acid just inserting itself between the carbonyl carbon and the migrating group. So we can think about that oxygen just getting pushed in there. Alright. And pushing that tertiary carbon down and the carbonyl carbon up. Alright. And the last thing we need to think about is the stereochemistry here. What's going to happen to that wedged methyl? Well, yeah, our stereochemistry is retained. Okay. So it's just going to stay as a wedge. It's literally just going to get pushed down. Okay? So when we draw our product for this, we have our carbonyl. Alright. And if we were to number our carbons, we would have 1, 2, 3, 4, 5, 6. So we're going to expand that from a 6-membered ring to a 7-membered ring where we inserted our oxygen right here. Alright. And then we're going to have the rest of our carbons. Let's draw that in black. Oops. Alright. So if we number these, we would have 1, 2, 3, 4, 5, 6, 7. Alright. So where does our wedged methyl go? Yeah. Well, we numbered it the same. So it's still going to be on carbon 6. Alright? It's just going to be one further away from our carbonyl. Alright. So this is what our product will look like. We expanded our ring, we retained our stereochemistry, and we migrated the group, the tertiary carbon with the highest migratory aptitude. Okay? So go ahead and give this next question a try, drawing out the mechanism, and then we'll solve that in the next video.

Show the mechanism:

Alright. This practice question was asking us to show the mechanism and predict the product for this reaction. Okay. So again, we first just want to identify our reaction and classify it. Okay. So we are reacting a ketone there on the left, and on the right here, we have this molecule. Do you know what it's called? Yes. This is MCPBA. Okay? And MCPBA is a peroxy acid. We know that because of this section right here, this is our CO₃H. And that is what we have in every peroxy acid. And whenever we have that with a ketone or even with an aldehyde, we're going to do our Baer Villager oxidation. Okay. So now that we've figured all that out, the first thing we should do for a question like this is just go ahead and predict the product. Baer Villager oxidation reactions are pretty easy to predict the product because we just have to insert an oxygen. So we need to look at our ketone. We have our carbonyl carbon here, a secondary carbon there, and then we have this group over here, which is our aryl. Does an aryl or a secondary carbon have a higher migratory aptitude? Yes, the secondary carbon does. So that is the group that we want to migrate. So remember, we're just going to think about inserting an oxygen between that carbonyl carbon and that secondary carbon. So what that would look like as a product is that we would still have this aromatic ring, still have our carbonyl, and now just insert our oxygen. So we'll have a new bond from our carbonyl carbon to our oxygen, and then a new bond from that oxygen to our secondary carbon. If we were to number these carbons, that would be 1 and 2. This is carbon 12. We just inserted the oxygen there. We know that is our product.

Now, for the mechanism, we just need to recall that it has three major steps. Do you remember what those three major steps are? Yes. We have protonation, nucleophilic addition, then our migration step. In our protonation step, what do we do? In our protonation step, we just protonate our carbonyl. I'm going to redraw our MCPBA over here. We use the hydrogen of our CO₃H to protonate. Our oxygen makes a bond to that hydrogen. We made a bond, we have to break a bond. So what we're left with is our original group here, and now we just have an oxygen with a positive charge. That is our protonation step.

In the second step, what do we do? What's the second step called? This is our nucleophilic addition. Here we add our peroxy acid that now has a negative charge to our carbonyl. This has a negative charge; it still has this benzene with the meta-chlorine. That negative charge is attracted to the partial positive charge of our carbonyl carbon. We make a bond there, and we break a bond. And that's our nucleophilic addition. So just those two arrows. And what we get as a product is we have all these black carbons here are coming from our starting material. So now we don't have a carbonyl, we have just an OH, and we have our ethyl attached there. Now we have this new bond, which I'm going to show in green, to our red oxygens. This is our next intermediate. This is where the new step comes in for the Baer Villager oxidation. This is where our migration step comes in. In this step, we start at the oxygen that originated on our carbonyl. If we were to trace this oxygen back through our intermediates, it starts out as our carbonyl oxygen. And what do we do? We reform that carbonyl, the carbon oxygen double bond. Then we have our migration step occur right here. We already decided that that ethyl, that secondary carbon, has a higher migratory aptitude compared to the aryl group, the benzene on the left. So we're going to migrate that ethyl. This bond I'm going to draw in green here is the bond that will migrate. And what we'll do is we'll just migrate that bond, those two carbons to the oxygen from our peroxyacid. We made a bond, so we need to break a bond. So this oxygen oxygen bond breaks and swings over, making another carbon oxygen double bond. That breaks this carbon oxygen double bond and kicks the electrons up there. This step, our migration step, has four arrows total, and they all move in the same direction. Here we're going from left to right. That gets us something very close to our final product. Let's draw this in black. So we still have this benzene. Right? We reformed our carbonyl. What's attached to that oxygen? It still has this hydrogen attached. So that tells us it's going to have a positive charge. We just broke this bond here and moved it over to the oxygen. So that carbonyl just has this green bond to that oxygen, which in turn has a new bond to our two carbons of the ethyl. And again, if we wanted to number those, we could say this is carbon 1 and this is carbon 2. This would be carbon 1, this would be carbon 2. So that's basically our final product. The last thing we have to do is just neutralize it. So we need to deprotonate that proton. What do we need to use for that? We just use the carboxylate that we just formed. So all of this right here just ends up looking like this. It's a carboxylate, the conjugate base of a carboxylic acid, which is a carboxylate. And we just use that negative charge of the carboxylate to deprotonate the proton on our carbonyl. That is the mechanism for this Baer Villager oxidation reaction. And remember the three major steps, protonation, nucleophilic addition, and migration. And remember that the migration step has four arrows. It will always have four arrows, and they'll always go in that same direction. Let's go ahead and move on to the next section.

Determine the major product for the following reaction. 

• A.
• B.
• C.
• D.

Provide the necessary reagents for the following transformation.

Hey guys. Let's take a look at the following practice question. Here it says, provide the necessary reagents for the following transformation. If we take a look at this, what we have initially is a ketone. What we're doing here is we're changing this ketone into an ester. Remember, an ester is a carbonyl connected to O connected to C. So, this portion makes it the ester. Remember, a ketone is a carbonyl with a carbon on either side. Now, there are a couple of ways that we can change a ketone into an ester. In this case, what we can do here is we can perform the Baeyer-Villiger reaction. Now, remember the Baeyer-Villiger reaction uses a peroxyacid to change a ketone into an ester. So, ketones normally don't get oxidized. This is one of the rare exceptions where we can. So here, we use a peroxyacid. Remember, peroxyacid is RCO3H. It's kind of like a carboxylic acid except it has one additional oxygen. We'll say here that the most common one that we know of is mCPBA, which looks like this. We're going to say mCPBA stands for meta, which remember means "position 3" because they are 3 positions apart. Chloro because of the chlorine. Peroxy and remember the peroxyacid portion is this portion. Peroxybenzoic because of the benzene. Acid, it's like a carboxylic acid so it's benzoic acid just like a carboxylic acid would. Now, here, remember the Baeyer-Villiger reaction follows what we call a migratory aptitude. Basically, the migratory aptitude means the groups that are more likely to leave the carbonyl carbon so that oxygen can come in its place. So the migratory aptitude is: H would move out of the way first, then a tertiary carbon, then a secondary carbon which is equal to a phenyl carbon, and phenyl means benzene. Then a primary carbon and finally a methyl carbon will be the last to move out of the way. So, if we take a look here, what's going to happen here is we have this peroxyacid which has 3 oxygens involved. One of these reactive oxygens is going to basically come in and separate one of the carbons out of the way so it can connect directly to the carbonyl carbon. If we look here, we have a CH3 (Methyl), and here we pretend that this bond doesn't exist because one of them is going to have to move out of the way. So, if we pretend this bond is not there, then this carbon technically is only connected to 1 carbon, this carbon, so it's primary. So, remember, a primary carbon moves out of the way before a methyl carbon. So, the extra oxygen here would basically shoot in, move that carbon out of the way, and implant itself right here so it can connect directly to the carbonyl carbon. But don't worry, that primary carbon over here would still be around. Here, let's see one carbon, you'd still have the same chain, 1, 2, 3. So that’s what we have to say here. So, Baeyer-Villiger would be the way to change this ketone into an ester. Now, remember, I told you that there was another way that we could have changed a ketone into an ester because I said there were a couple of ways. Here, it wouldn't be possible to make this particular ester in any other way. But we could have gone from this ketone to this ester, this particular ester, by a different method. Here what we could have done is we could have done a reaction which we call the haloform test. Once we made the haloform test, that would change my ketone into a carboxylic acid. We would then change our carboxylic acid into an ester which we know how to do. We’d use an alcohol over an acid catalyst. And in that way, we get the ester that we need. Now, we'll come upon a question like this later on as we talk about alpha substitution reactions. But just realize these are the two major ways to make a ketone into an ester, either doing Baeyer-Villiger or doing this method where we use the haloform test followed by changing the carboxylic acid that we form into an ester. So, these are the two major ways to do these types of transformations.

Here’s what students ask on this topic:

What is the Baeyer-Villiger oxidation reaction?

The Baeyer-Villiger oxidation reaction is a chemical reaction that transforms aldehydes and ketones into carboxylic acids and esters, respectively, using peroxyacids like mCPBA (meta-chloroperoxybenzoic acid). This reaction is unique because it can further oxidize ketones, which are typically the most oxidized form in many other oxidation reactions. The reaction is regioselective, meaning the position where the oxygen is inserted depends on the migratory aptitude of the substituents attached to the carbonyl carbon.

What reagents are used in the Baeyer-Villiger oxidation?

The primary reagent used in the Baeyer-Villiger oxidation is a peroxyacid, such as mCPBA (meta-chloroperoxybenzoic acid). The peroxyacid reacts with the carbonyl compound (either an aldehyde or a ketone) to insert an oxygen atom between the carbonyl carbon and the substituent with the highest migratory aptitude. This transforms aldehydes into carboxylic acids and ketones into esters.

What is migratory aptitude in the Baeyer-Villiger oxidation?

Migratory aptitude refers to the tendency of a substituent attached to the carbonyl carbon to migrate to the oxygen atom during the Baeyer-Villiger oxidation. The trend in migratory aptitude is similar to carbocation stability: hydrogen has the highest migratory aptitude, followed by tertiary, secondary, and primary carbons, with aryl groups (like benzene rings) having lower aptitude. This determines which substituent will migrate and where the oxygen will be inserted in the final product.

What is the mechanism of the Baeyer-Villiger oxidation?

The mechanism of the Baeyer-Villiger oxidation involves three main steps: protonation, nucleophilic addition, and migration. First, the carbonyl oxygen is protonated by the peroxyacid. Next, the peroxyacid's conjugate base performs a nucleophilic addition to the carbonyl carbon. Finally, the substituent with the highest migratory aptitude migrates to the oxygen from the peroxyacid, forming the final product. The stereochemistry of the starting material is retained throughout the reaction.

How does stereochemistry affect the Baeyer-Villiger oxidation?

In the Baeyer-Villiger oxidation, the stereochemistry of the starting material is retained in the final product. This means that if the starting carbonyl compound has a specific configuration (e.g., R or S), this configuration will be preserved after the reaction. The reaction does not alter the stereochemistry of the migrating group or the remaining substituents on the carbonyl carbon.