Oxidizing Agent - Online Tutor, Practice Problems & Exam Prep

Oxidation involves the addition of oxygen to a molecule, typically using oxidizing agents like Jones' reagent, which consists of chromium (Cr) and a strong acid. A secondary alcohol can be oxidized to a ketone without breaking carbon-carbon bonds through a mechanism involving nucleophilic attack and alpha elimination. Understanding the distinction between strong and weak oxidizing agents is crucial for predicting reaction outcomes. Remember, the key rule is that oxidizing agents add oxygen while preserving carbon-carbon bonds.

Reagents

All right guys. Now I want to dive a little deeper into the reagents of oxidation. Just so you guys know, all of these reagents that we're going to talk about in this topic can be generalized as oxidizing agents. Now I remember when I was in general chemistry and you had oxidizing agents, reducing agents, oxidation. It was kind of confusing sometimes like how the oxidizing agent gets reduced. There's a lot of different stuff you have to memorize. Or maybe I was just confused. I don't know. But all I know is that in organic chemistry, it's really not complicated at all. All you have to think about is that the oxidizing agent is the thing that oxidizes your molecule. So if you're trying to oxidize a molecule, make more bonds to oxygen, you're going to use an oxidizing agent. It's that easy. There's a general rule that you really need to follow with all these reactions. Even if you don't know the mechanism, there's just a rule that you can use and that's that oxidizing agents are going to add as much oxygen as possible while not breaking any carbon-carbon bonds. Now this is a little bit of a lie. There are some oxidizing reagents that can break carbon-carbon bonds. Some examples that you might already know would be, for example, ozonolysis. That would be an example of an oxidation that can. That's not what we're going to talk about in this topic. That's its own separate topic. For right now, I'm trying to deal with these oxidizing agents that don't break carbon-carbon bonds. So I'll get to what they are in a second, but even before we know what the reagents are, we could already jump into a practice problem. What I'm wondering is out of these four molecules here, which of them actually could be oxidized? What I'm basically saying is how many of them could you add bonds to oxygen without breaking a carbon-carbon bond. So I'm going to go ahead and let you guys figure that out. Get back to me. Which of these could be oxidized?

Strong oxidizing agents

All right. So I hope that you didn't say all of them because there are some that can and some that can't. Let me give you an example. The first one, there's a carbon here that has a bond to an oxygen. It has 2 bonds to carbon and then it has one bond to H. Would you guys agree with that? So now my question is, is there a way that I could turn this carbon into a carbon that has more bonds to oxygen? Here's the way you have to think about it. You have to think well, how many bonds to carbon does it already have? It has 2. It has 1, 2 bonds to carbon. Is that cool so far? Since it has 2 bonds to carbon, how many total bonds could it have to oxygen theoretically? 2. Because no matter what, carbon can only have 4 bonds. So what that means is that if it has 2 bonds to carbon, later on, I could oxidize it so that it has 2 bonds to oxygen. So could this be oxidized? Yes. This could be oxidized because I could make it in the form where there's 2 bonds to oxygen. So let's move on to the next one.

The next one, this carbon, could it be oxidized? No. This one is not going to be able to be oxidized because notice that it already has its maximum number of bonds to oxygen because it has 2 carbons. 1, 2. Is there a way to add a 3rd bond to oxygen? No. Let's move on to 3. Could 3 be oxidized? Yes. Because it only has one bond to carbon. So that means if it only has one bond to carbon, then it could have how many bonds to oxygen? 3. How many bonds does it have right now? Only 1. So it could actually be oxidized more than once. Then finally, we have compound number 4. 3 gets a check mark. 4, could this carbon be oxidized? Yes. Once again, because it only has one bond to carbon, so that means that we could take away that H and we could make another bond to oxygen there. So that's the way that it works. All of these could be oxidized except for 2, which can't because it already has the maximum number of bonds to carbon and oxygen.

All right. So what reagents are going to do this? Well, strong oxidizing agents are agents that are going to add the maximum number of oxygens possible while following the rule of not breaking any carbons. These reagents are going to be KMnO4. KMnO4 is a reagent that you've probably already seen, but in case you haven't, potassium permanganate, very strong oxidizing agent. Also, your chromium 6 reagents. Now it's a Cr⁶⁺. Remember that is the oxidation state of the atom. You are not going to have to calculate oxidation states in organic chemistry. What you should know is that if you see chromium present in any of these weird molecules, these are all examples of strong oxidizing agents. It turns out there's more reagents than this. The Jones reagent is an example of a chromium reagent where Jones' reagent would use CrO₃ and sulfuric acid. So all I'm trying to say is that as long as you see some kind of chromium in the reagent, think this is a strong oxidizing agent. You don't have to actually calculate out the oxidation state. So what I want you guys to do for this next practice problem is go ahead and draw the new oxidation products of each of these molecules. So I want 4 different things in these boxes. If it's not going to react, put no reaction. But I want to see all the different oxidation products. So go ahead and try to do the first one.

Strong oxidizing agents

So let's go ahead and look at this first one. What I would get is a carbon that now has 2 bonds to carbon. So I should draw that 6 membered ring just like before. But now it's going to have 2 bonds to oxygen. So what I'm expecting to get here is a ketone. Why a ketone? Because a ketone would be the version of that carbon that has 2 bonds to oxygen. Now you might be like Johnny, how would I know if it's a ketone? How about if it's another functional group? I don't want you to think about the functional group. Honestly, all I want you to think is how many bonds to oxygen is this thing able to make. In this case, that carbon could make 2 bonds to oxygen, so that's why you draw 2 bonds to oxygen and a cyclohexane on the other side. Go ahead now and try to do the other 3. Now that you have an example, try to draw the other 3 structures, put them in the box and see if you get the right products.

Strong oxidizing agents

All right. So this first one would be no reaction. We know that because we said that the second molecule can't be oxidized at all. So there's no point in even drawing a product. Now carbon right here has only one bond to another carbon. So that means if it has one bond to carbon, how many oxygens can it possibly have? 3. So what that means is that I need to draw a version of this carbon that's going to have 3 bonds to oxygen. If you looked at our intro to redox chart where I talk about things that are getting oxidized and things that are getting reduced, the version of carbon with 3 oxygens would be a carboxylic acid. So what I would do is I would draw that carbon with a double bond O and with an OH. What that's going to do is now that's going to keep my carbon-carbon bond, so I'm not breaking the rule. I'm not breaking that bond. But now I have 1, 2, 3 bonds to oxygen. All right. Cool.

So now how about this last one? Well, this last one, I'll move out of the way so you guys can see it. This last one you will find is that that carbon once again only had one bond to carbon, but it already had 1, 2 bonds to oxygen. So how many extra bonds to oxygen could it have? Well, we already know that the rule says you can only have 4 bonds and one of them has to be to carbon. That means that the last third bond could also be an oxygen. So what I'm going to do here is I'm going to try to move out of the way here. I'm going to draw this molecule once again also as a carboxylic acid. Why? Because basically, when you're oxidizing something with 2 bonds to oxygen, you want it to look like a ketone, and then when we have one bond to oxygen, you want it to look like an alcohol, which is the one that I have up there. Okay. Good. Awesome.

So now what I want to do is I want to show you guys another reagent. It turns out that even though we deal with strong oxidizing agents a lot, there's also a reagent that's called a weak oxidizing agent. Now a weak oxidizing agent would simply be one that doesn't oxidize multiple times. The way that we're going to define it in particular is that it can only add one equivalent of oxygen to primary alcohols. That's really the only difference. So what that means is that it's going to do the same thing as all the other reagents, KMnO₄, chromium 6, same thing, except in one situation. In a primary alcohol, instead of going all the way to a carboxylic acid, it's going to go one equivalent instead of 2 equivalents of oxygen. So let me show you what that looks like. Pcc is the name of this reagent and it is our weak oxidizing agent. So would it be able to oxidize my secondary alcohol? Absolutely. It's going to do the same exact thing. So for PCC, I would get the same exact reagent here or the same exact product. Would it be able to oxidize number 2? No. Nothing can oxidize number 2. It's still no reaction. Would it be able to oxidize number 3? Yes, it would. But this is our special situation. Notice that I have a primary alcohol. Whenever you have a primary alcohol, what that means is that for a strong oxidizing agent, I would have taken it to a carboxylic acid like this. But for a weak oxidizing agent like PCC, I'm going to go to an aldehyde instead. So that means that I'm actually going to draw this thing like this with an H instead of an OH. That's the biggest difference here. That's actually the only major difference that we have with PCC is that instead of getting carboxylic acid, we get an aldehyde.

Now you might be wondering, what do you mean by one equivalent of oxygen? All I mean is that notice that at the beginning how many bonds to oxygen did we have? We had 1. I'm just going to say 1 O. At the end of the strong oxidation, how many did we have bonds to oxygen? We had 3. 3 O. So that means that if we had 1 oxygen to begin with and 3 to end with, we added 2 equivalents of oxygen. Well, for PCC, instead of using doing 2 equivalents of oxygen, now we're only going to add one equivalent because now we have 2 bonds to O instead of 1, which is what it started with. So if you're starting with 1 and you're with 2, that means you only added one equivalent of oxygen and that's what this definition has to do with. But if you want to remember, just say that primary alcohols go to aldehydes. That's another way of saying it. That's maybe less complicated and that's always right. You could just say it like that the rest of your life if you want to and that's fine. Let's get down to our last structure. Would it be able to oxidize my 4? And the answer is no. This would be no reaction. Why is that? Well, because it's already an aldehyde. Notice that aldehyde is the product of PCC. PCC is going to make an aldehyde. So if we have an aldehyde already, is it going to do anything to it? No. Okay. So once it's in aldehyde, it's not going to oxidize it more. This would be no reaction as well. Okay. So just trying to show you guys the difference between PCC and the other oxidizing agents. It's not that hard. It's just a few details you have to keep in mind. All right. So that said, let's go ahead and move on.

Jones Reagent Mechanism

Now that we understand the process of oxidation pretty well, I want to show you guys at least one 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 plus reagent and strong acid. So 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 HClO₄ 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, okay?

So 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. So 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. Okay. 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 gonna work? Well, let's go ahead and start. The first step guys is nucleophilic attack because here we have 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 negative but we're in the presence of acid so the O negative is going to protonate, right? And 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, one 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. Okay? Let's start off with the nucleophilic attack. This O here is now right here attached to the chromium. It still has an H on it. Let's go to another easy one. This O is still here. So nothing changed there. This O is still here. Okay. So they're still in the places that they were before. But there is one new O that's the one that is in pink because this pink, remember that it grabbed an H, right? So that means that this is the O here. Okay? So now, hopefully, those colors help you to kind of identify what's going on. Where did everything go? All the atoms are still there. So we're doing okay. All right. So 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? And 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. So what alpha simply means is that instead of eliminating between the alpha and the beta carbon, right, you're actually going to form a double bond directly on that alpha carbon to something that's not a carbon. So 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, you know, 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. And instead of forming a double bond between 2 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 in this O. And then finally, that O isn't happy, so we're going to take 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. Okay? Awesome. So 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. Okay. 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 2 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. And 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. So 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.

Intro

All right, guys. Now we're going to do some practice problems involving oxidation. The two big things you have to remember are that you can't break carbon-carbon bonds and also how to tell the difference between a strong and a weak oxidizing agent. So now go ahead and try to do the first problem.

What is the product of the reaction?

• A.
• B.
• C.
• D.

What is the product of the reaction?

• A.
• B.
• C.
• D.

What is the product of the reaction?

• A.
• B.
• C.
• D. No reaction

Here’s what students ask on this topic:

What is an oxidizing agent in organic chemistry?

An oxidizing agent in organic chemistry is a substance that facilitates the oxidation of another molecule by accepting electrons from it. This process typically involves the addition of oxygen or the removal of hydrogen. Common oxidizing agents include potassium permanganate (KMnO₄), chromium trioxide (CrO₃), and Jones' reagent, which is a combination of CrO₃ and a strong acid like H₂SO₄. These agents are used to convert alcohols to ketones or carboxylic acids, among other transformations.

How does Jones' reagent work in oxidation reactions?

Jones' reagent, a mixture of chromium trioxide (CrO₃) and sulfuric acid (H₂SO₄), is a strong oxidizing agent used to oxidize primary alcohols to carboxylic acids and secondary alcohols to ketones. The mechanism involves the formation of chromic acid (H₂CrO₄), which acts as a strong electrophile. The alcohol nucleophilically attacks the chromium, leading to a series of steps including nucleophilic attack, alpha elimination, and deprotonation, ultimately forming the oxidized product.

What is the difference between strong and weak oxidizing agents?

Strong oxidizing agents, like Jones' reagent (CrO₃/H₂SO₄) and potassium permanganate (KMnO₄), can fully oxidize primary alcohols to carboxylic acids and secondary alcohols to ketones. Weak oxidizing agents, such as PCC (Pyridinium chlorochromate), typically oxidize primary alcohols to aldehydes without further oxidation to carboxylic acids. The choice of oxidizing agent depends on the desired level of oxidation and the sensitivity of the substrate.

Can you explain the mechanism of oxidation of a secondary alcohol using Jones' reagent?

The oxidation of a secondary alcohol using Jones' reagent involves several steps. First, the alcohol nucleophilically attacks the chromium in chromic acid (H₂CrO₄), forming a chromate ester. This is followed by an alpha elimination, where a proton is removed, forming a double bond between the carbon and oxygen. Finally, the intermediate is deprotonated, resulting in the formation of a ketone. Throughout the process, the carbon-carbon bonds remain intact, and the secondary alcohol is converted to a ketone.

What are some common examples of oxidizing agents used in organic chemistry?

Common oxidizing agents in organic chemistry include:

• Potassium permanganate (KMnO₄)
• Chromium trioxide (CrO₃)
• Jones' reagent (CrO₃/H₂SO₄)
• Pyridinium chlorochromate (PCC)
• Osmium tetroxide (OsO₄)
• Hydrogen peroxide (H₂O₂)

These agents are used to achieve various oxidation states in organic molecules, such as converting alcohols to aldehydes, ketones, or carboxylic acids.