Ozonolysis Full Mechanism - Online Tutor, Practice Problems & Exam Prep

The ozonolysis reaction involves oxidative cleavage of alkenes, forming unstable ozonide intermediates. The mechanism consists of three steps, each with three arrows, leading to the formation of either aldehydes and ketones through reductive workup (using DMS or zinc and acetic acid) or carboxylic acids and ketones via oxidative workup (using H₂O₂). The reductive pathway preserves carbonyl groups, while the oxidative pathway converts aldehydes to carboxylic acids, emphasizing the importance of understanding reaction mechanisms in organic synthesis.

General Mechanism:

What's up, everybody? In this section, we're going to take a look at the full mechanism for the oxidative cleavage reaction, ozonolysis. The mechanism for the ozonolysis reaction proceeds through intermediates called ozonides. Alright. And what ozonides are, are there they are, cyclic molecules that were formed by the addition of ozone. Okay. So it was going to have 3 oxygens, and they're always going to be really unstable. Alright. That instability in these intermediates is what drives our reaction forward. Okay? And ozonolysis is always combined with a reductive or oxidative workup. Alright? The reductive workup forms aldehydes and ketones, while the oxidative state of the workup forms carboxylic acids and ketones. Okay? Our reagents for the reductive workup are typically either DMS, which is, dimethyl sulfide that we see right here, or zinc and acetic acid. Alright. For our oxidative workup, we typically use H₂O₂, which is just hydrogen peroxide. Okay? Now with this analysis, the mechanism proceeds the same through the first 3 steps regardless of which workup we're going to use. Okay. So we're going to break this up and first look at the first three steps that are common to either oxidative or reductive analysis. And then we'll look at the reductive workup mechanism and then the oxidative workup mechanism after that. Okay? So the general mechanism is 3 steps and each step has 3 arrows. Okay? So all this analysis mechanism has 3 steps and 3 arrows per step. Okay? In our first step, we react our alkene with our ozone molecule. Alright. What is our ozone molecule missing here? Yeah. It's missing its charges. Okay? Remember, we have a positive charge on this middle oxygen and a negative charge on that oxygen there. Okay? That negative charge is going to act as a nucleophile, and it's going to add to our double bond. Okay? So this negative charge is going to come in and add to our alkene there. Alright. We made a bond, so we have to break a bond. So we break the pi bond in that alkene and attack our oxygen at the far end of the ozone molecule. Okay? That breaks the oxygen-oxygen pi bond and puts a lone pair on oxygen. Okay. So we had our 3 arrows there. They're all moving in the same direction. That gets us to our first ozonide intermediate. Alright? And this intermediate is called molozonide. Alright. Remember, our ozonide intermediates are unstable. Alright. And that instability is going to drive this next step. Alright. But first, you may be wondering, where is this hydrogen right here? Where did that end up? Well, remember, this first step is going to be a syn addition. Alright. Where our ozone adds either from the top of that alkene or the bottom and the stereochemistry of that alkene is retained. So what I mean by that is that the hydrogen is cis to the ethyl right there. Okay. So it's going to be cis to the ethyl in our molozonide intermediate here. So that hydrogen is going to end up onto this wedge right there. Alright? So from here, this second step, it again has 3 arrows. Alright. We want to start at one of the oxygens that's making a bond to carbon. Alright. So either the one on the right or the left, and it doesn't matter here. Alright. So we can just choose one. We'll choose the one on the right here and we'll make a carbon-oxygen double bond. Alright. That's our first arrow. We made a bond. We have to break a bond and we're going to break the carbon-carbon bond that connects the two oxygens. Alright. So that is this bond right here in green. We're going to break that bond and we're going to make another carbon-oxygen double bond. Alright. Again, we made a bond, so we have to break another bond. So we're going to break this oxygen-oxygen bond and give that top oxygen a lone pair. Okay? So that gives us 2 molecules again. We have an aldehyde on the left. Alright? And then we have that other intermediate on the right. What are we missing on that molecule? You're missing our charges again. Okay. So we need a negative charge on that oxygen and a positive charge on this oxygen. Okay. And what we do next isn't actually a mechanistic step. It just makes it easier to draw the next mechanistic step. Okay. Okay. What we want to do is scoop up this molecule with the charges and flip it back towards ourselves. Okay? And when we do that, we end up putting that negative charge on the bottom here and the positive charge right next to it. Okay? The reason we do that is because these two molecules need to react together. Alright. And if you remember, the dipoles on carbonyls always look like this, where we have a dipole pulling electrons away from the carbonyl carbon towards that oxygen. Okay. So both of these carbonyl carbons have partial positive charges. So they're electrophilic. Alright. And then we have an oxygen with a negative charge. That's going to be nucleophilic. Okay. So we're going to do a nucleophilic addition here. And this step again has three arrows, and this negative charge will draw an arrow to the carbonyl carbon there. Alright. We made a bond. We have to break a bond. So we're going to break a carbon-oxygen pie bond, and that is going to attack the other carbonyl carbon. Alright. Our third and last arrow for this step is breaking the other carbon-oxygen pi bond

Reductive Workup Mechanism:

Alright. Now we're going to take a look at the mechanism for the reductive workup. Okay. So we're going to start where we left off from the general ozonolysis mechanism where we formed ozonide. And we're going to add in one of our reductive workup reagents, which is dimethyl sulfide or DMS. Okay? What happens here is our DMS attacks our ozonide intermediate. So we'll draw this again down there. Alright. Our DMS attacks one of the oxygens that's already making a bond to another oxygen. Okay? So one of either this oxygen or that oxygen. It doesn't matter which, but it has to be one of them. Alright? So it'll make a bond to one of those oxygens. We made a bond, so we have to break a bond. Alright. We're going to break the oxygen-oxygen sigma bond here, and that is going to swing over and make a carbon-oxygen pi bond. Okay? We made a bond, so, again, we have to break a bond. So we're going to break the carbon-oxygen bond there, and we're going to make a carbon-oxygen pi bond on the other side. Alright. Again, we made a bond, so we need to break a bond, and we're going to break the carbon-oxygen bond there on the other side and give those electrons back to our sulfur there. Okay? So overall, we're doing 4 arrows in this step, and we're making 2 carbonyls. Okay? So if we were to track the electrons from this oxygen-oxygen double bond, where do they end up in our products? Yeah. Well, they end up making the pi bond from our aldehyde. Right? We have a methyl and a hydrogen there. Okay. We have a methyl and a hydrogen here. So that bond is going to be the pi bond over here. Okay. What about the electrons in this green bond here? Where do those end up? Yeah. Well, those end up being the pi bond on our other carbonyl. Okay. We have the methyl and the ethyl on either side. Alright. We have our ethyl and our methyl on either side, and the pi bond here came from that carbon-oxygen double bond. Okay? If we were to look at the other carbon-oxygen bond that broke, we know that that bond lets highlight it in pink, that bond there ends up being the oxygen-sulfur pi bond in our product there. Okay? So our product here is DMSO. That's what the DMS turns into. Alright. And since it was the reductive workup, we form an aldehyde and a ketone. Alright. And with our reductive workup, we can form aldehydes, we can form ketones, we can't form carboxylic acids. Okay? And in the next video, we will take a look at the oxidative workup mechanism.

Oxidative Workup Mechanism:

What's up, guys? We're finally getting to the oxidative workout mechanism. Alright. And sorry to let you down, but there's not actually a mechanism here. Okay? This is unknown. Alright. I know you're all super bummed about that. You don't get to learn another mechanism, but it is important to know what an oxidative workup does in an ozonolysis reaction. Okay. When we react our ozonide intermediate with our oxidative workup reagents here, H₂O₂, which is hydrogen peroxide. Anything that would have turned into an aldehyde in our reductive workup will turn into a carboxylic acid. Okay. So here we form a carboxylic acid and a ketone. Okay. So if we look up to our reductive workup of the same ozonide intermediate, we formed a ketone and an aldehyde.

Alright. We basically just oxidize that aldehyde up to a carboxylic acid. Okay. So in your ozonide intermediate, anywhere that you see a hydrogen attached to one of these carbons making 2 oxygen sigma bonds, that hydrogen will become an OH in your carboxylic acid. Okay? So super important to know, super useful reaction to use. Okay? Go ahead and try the practice problem below, and then in our next video, we'll come in and solve it.

Predict the products and show the mechanism:

Alright, guys. This question is asking us to predict the products and show the full mechanism for this reaction. The first thing we need to do is just identify what reaction is occurring. And to do that, we need to identify what our reagents are. So we start off, and we have an alkene. In our first step, we're reacting that with O₃. This is a huge clue. Anytime we see O₃, which is ozone, we should automatically be thinking of our ozonolysis reaction. And in our second step, we're reacting with DMS. Remember, DMS is dimethylsulfide, which just looks like that. A sulfur atom with 2 CH threes attached. So we're reacting an alkene with ozone followed by DMS. What would you call this reaction if you had to classify it really specifically? Yes, this is going to be ozonolysis with a reductive workup.

Whenever we're doing ozonolysis reactions and we're asked to predict products and show mechanisms, the easiest thing to do first is just to predict the products. Remember Johnny taught you to use the scissors to just cut this bond and separate it? Well, yes, we can still do that here even though we have to show the mechanism. So we want to think about using our scissors to cut our alkene right down the middle. We're going to pull those two pieces apart. On the left, we're going to have that 5-membered ring. It still has half of that double bond. What do we put on the other half of that double bond? Yes, we put our oxygen there. The other side, the other half will just have that one or two carbons there and then half of that pie bond. What goes on the other half of the pie bond? Yes, the other oxygen. I'm going to fill in the hydrogen there, but remember anytime we see a hydrogen coming off our alkene with a reductive workup, it'll be a hydrogen of an aldehyde in our product. So those are our two products.

Now let's just show how we make them with our mechanism. What is the first step of our mechanism? Yes, we just need to react our alkene with ozone. So ozone O₃, remember, is three oxygens bound together. We have an oxygen. Let's draw this in blue, actually. We have an oxygen-an oxygen-and an oxygen. We need one oxygen-oxygen double bond. It doesn't matter which side we do it on. I'm just going to put it on the left. That to our alkene. This negative charge will come to our alkene. This negative charge will come in, make a bond to a carbon of our alkene. And remember, we could have drawn our ozone the other way where the negative charges on the other oxygen. Then we just would have attacked that left carbon of our alkene. It doesn't make a difference in the mechanism or in our final product. But we'll make that bond there. We made a bond, so we have to break a bond. So we'll break this carbon-carbon double bond and make a bond to oxygen. Again, we made a bond, so we have to break a bond. So we're going to break this oxygen-oxygen pi bond right there. There are three arrows for this step in the mechanism. We're ready to draw the product of that step. So here we'll have our 5-member ring. It has that methyl and this hydrogen, and then we have our two new bonds to oxygen just like that. You may be wondering why didn't I show stereochemistry here? And really, the stereochemistry of the intermediates doesn't matter because we end up making these carbonyls, that are SP₂ hybridized. So SP₂ hybridized atoms don't have any stereochemistry. We don't have to worry about our wedges or our dashes. So we don't really care about it in the mechanism unless we're asked to show the stereochemistry of our intermediates. So this is our first ozonide intermediate. And remember, it's called molozonide. And to get to our next intermediate, we again need to react with three arrows. And, again, they're all going to react in a cyclic way. Our first arrow starts at one of the oxygens making an oxygen-carbon bond. So we'll choose this one on the right, but, again, it doesn't matter which one. And we'll make that carbon-oxygen pi bond. We made a bond, so we have to break a bond. Which bond are we going to break? Yes, we're going to break the bond that connects the two oxygens. So this green bond right here, we're going to break, and we're going to use those electrons to make another carbon-oxygen pi bond. Again, we made a bond, so we have to break another bond. So we're going to break this oxygen-oxygen bond right there. So if we want to draw the product from all of these atoms on the right, what would that look like? Yes, it would look pretty similar to an aldehyde. So we'll have the methyl and the hydrogen there. We'll have an oxygen-carbon double bond attached. And then we'll have an oxygen attached to that oxygen. What will our charges be here? Yes, we'll have a negative charge on this oxygen and a positive charge on this oxygen. What will the other half look like? This half highlighted in blue. Yes, that will just be a ketone. We'll have that, and then a ketone just like that. What is our next step? Well, yes. Remember, first, we have to flip this molecule over. So I'm just going to go ahead and redraw them. We can flip over either molecule, but I think it's easiest to flip over the one with the charges. So remember, we want to scoop it up and flip it towards ourselves. What that will look like is we'll have all of this flipped over. We'll have our oxygen, our other oxygen. We'll still have our charges, negative charge, a positive charge. And we'll keep the other side exactly the same. Just that 5-member ring with the carbonyl with the ketone. What happens here? Yes, well, again, just like every step here, we have three arrows. And they all move in the same direction. What we want to do is use that negative charge to attack our carbonyl carbon of our ketone, right, and do a nucleophilic addition. We made a bond, so we have to break a bond. These electrons will come all the way over to the other carbonyl carbon of the molecule that originally attacked. We made a bond, so we have to break a bond. This is our last arrow. So we break this carbon-oxygen pi bond right there. So we're at three arrows there. We're done with this step, and we formed our next ozonide intermediate, which has an oxygen right here. A carbon attached to that, that is our 5-membered ring. On the other side, we have our hydrogen and our methyl. And then attached to those, we have our two oxygens. I know that can be kind of hard to follow, so I'm going to map everything out here. These two oxygens right there that are attached to each other are going to be the two oxygens that are attached to each other in this ozonide intermediate. This carbon attached to those two oxygens is going to be that carbon there on the left. And then this carbon and that oxygen are going to be the other carbon and oxygen of our ring in this intermediate. So this is our next intermediate. What was this one called? Yes, this one is called just ozonide. And this is where our ozonolysis stops, and our workup begins. So now we have to do our reductive workup with DMS. So we need to draw on our DMS, which is just a sulfur with two methyls. What is the first arrow that we need to draw here? Yes. So remember, we have four total. The first one goes from the sulfur to one of the oxygens making an oxygen oxygen bond. It doesn't matter which one. We'll just go ahead and go to this one on the left. So we just made a sulfur oxygen bond. Once we make that bond, we have to break a bond. We're always going to break the oxygen oxygen bond. That oxygen oxygen bond will swing over and make a carbon oxygen double bond. And, actually, I drew that bond to the wrong place. This oxygen goes right is bonded to the same carbon as the other oxygen. But it doesn't change the mechanistic step here. So that oxygen-oxygen bond goes to form a carbon-oxygen bond right there. We made a bond, so we need to break a bond. So we're going to break the other carbon-oxygen bond of that carbon and use those electrons to make another carbon-oxygen pi bond. We made a bond, so we have to break another bond and draw our fourth arrow of our reduction step. And we're going to break this carbon-oxygen bond right there and give those electrons back to sulfur. So we have our four arrows of our reductive step, and that gets us to our final product. So our aldehyde let's get rid of this highlighting here and show that our aldehyde here in yellow comes from this oxygen, that methyl, and that hydrogen, while our ketone in blue comes from this oxygen and all of those carbons of that 5-membered ring. What is our one byproduct that we didn't show? Yes, it's DMSO. So we also have DMSO, which is just dimethylsulfoxide. So that is our osmosis mechanism there. Let's go ahead and move on to the next section.

Here’s what students ask on this topic:

What is the full mechanism of ozonolysis?

The full mechanism of ozonolysis involves three main steps, each with three arrows. First, the alkene reacts with ozone (O₃), forming a molozonide intermediate. This intermediate is unstable and rearranges to form a more stable ozonide. The ozonide can then undergo either a reductive or oxidative workup. In the reductive workup, reagents like dimethyl sulfide (DMS) or zinc and acetic acid are used, leading to the formation of aldehydes and ketones. In the oxidative workup, hydrogen peroxide (H₂O₂) is used, converting any aldehydes into carboxylic acids while ketones remain unchanged.

What are the reagents used in the reductive workup of ozonolysis?

In the reductive workup of ozonolysis, the common reagents used are dimethyl sulfide (DMS) and zinc with acetic acid. These reagents help in reducing the ozonide intermediate to form aldehydes and ketones. DMS attacks the ozonide, breaking the oxygen-oxygen sigma bond and forming carbon-oxygen pi bonds, resulting in the formation of aldehydes and ketones. Zinc and acetic acid work similarly by reducing the ozonide intermediate.

What products are formed in the oxidative workup of ozonolysis?

In the oxidative workup of ozonolysis, hydrogen peroxide (H₂O₂) is used as the reagent. This converts any aldehydes formed in the ozonide intermediate into carboxylic acids. Therefore, the products of the oxidative workup are carboxylic acids and ketones. The key difference from the reductive workup is the oxidation of aldehydes to carboxylic acids.

What is the role of ozonides in the ozonolysis mechanism?

Ozonides are crucial intermediates in the ozonolysis mechanism. They are formed by the addition of ozone to the alkene, resulting in a cyclic structure. These intermediates are highly unstable, which drives the reaction forward. The ozonide can then undergo either a reductive or oxidative workup to form the final products. In the reductive workup, ozonides are reduced to aldehydes and ketones, while in the oxidative workup, they are converted to carboxylic acids and ketones.

How does the stereochemistry of the alkene affect the ozonolysis mechanism?

The stereochemistry of the alkene is retained during the initial step of the ozonolysis mechanism, which involves a syn addition of ozone. This means that the relative positions of substituents on the alkene are preserved in the molozonide intermediate. For example, if a hydrogen and an ethyl group are cis on the alkene, they will remain cis in the molozonide intermediate. This retention of stereochemistry is important for predicting the final products of the reaction.