Oxidation of Alcohols - Video Tutorials & Practice Problems

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.

Provide the major product for the following oxidation reaction. 

• A.
• B.
• C.
• D.

Provide the major product for the following oxidation reaction. 

• A.
• B.
• C.
• D.

Provide the major product for the following oxidation reaction. 

• A.
• B.
• C.
• D.

 Provide the major product for the following oxidation reaction.

• A.
• B.
• C.
• D.

Determine the major product for the following reaction. 

• A.
• B.
• C.
• D.