Oxidation

Concept: Concept: Oxidizing Agents

2m
Video Transcript

All right guys, so now I want to dive a little deeper into the reagents of oxidation. So 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 gen chem and you had oxidizing agents, reducing agents, oxidation, it's kind of confusing sometimes like the oxidizing agent gets reduced. There's a lot of different stuff you have to memorize or maybe I was just stupid. I don't know. But all I know is that in orgo 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. 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.
So 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.
But that's not what we're going to talk about in this topic. That's it's own separate topic. For right now, I'm trying to deal with these oxidizing reagents that don't break carbon-carbon bonds. 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?

Concept: Example 1: Which compounds can be oxidized?

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Video Transcript

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 two 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 that you have to think about it. You have to think how many bonds to carbon does it already have? It has two. It has one, two bonds to carbon. Is that cool so far? Since it has two bonds to carbon, how many total bonds could it have to oxygen theoretically? Two, because no matter what, carbon can only have four bonds. What that means is that if it has two bonds to carbon, later on I could oxidize it so that it has two bonds to oxygen. So could this be oxidized? Yes. This could be oxidized because I could make it in a form where there's two 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 two carbons, one, two. Is there a way to add a third bond to oxygen? No.
Let's move on to three. Could three 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? Three. How many bonds does it have right now? Only one. So it could actually be oxidized more than once.
Then finally we have compound number four. Three gets a checkmark. Four, 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.
That's the way that it works. All of these could be oxidized except for two which can't because it already has the maximum number of bonds to carbon and oxygen.
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 six reagents. Now it says Cr6+, remember that is the oxidation state of the atom. You're 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 that there's more reagents than this. The Jones reagent is an example of a chromium reagent, where a Jones reagent would use CRO3 and sulfuric acid. All I'm trying to say is that as long as you can 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.
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. I want four 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. 

Concept: Example 2: Oxidizing with strong agent

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Video Transcript

So let's go ahead and look at this first one. What I would get is a carbon that now has two bonds to carbon, so I should draw that six-membered ring, just like before, but now it's going to have two 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 two 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 two bonds to oxygen so that's why you draw two bonds to oxygen and a cyclohexane on the other side.
Go ahead now and try to do the other three. Now you have an example. Try to draw the other three structures. Put them in the box and see if you get the right products. 

Concept: Example 3: Oxidizing with weak agent

6m
Video Transcript

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 the third one is interesting because we talked about how this 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? Three. What that means is that I need to draw a version of this carbon that's going to have three bonds to oxygen.
If you looked at our intro to redox chart, where they talk about things that are getting oxidized and things that are getting reduced, the version of carbon with three 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 it'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 one, two, three bonds to oxygen. Cool.
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 what you'll find is that that carbon once again, only had two bonds to – I mean only had one bond to carbon, but it already had one, two bonds to oxygen. So how many extra bonds to oxygen could it have? Well, we just know that the rule says that you can only have four bonds and one of them has to be a carbon so that means that the last third bond could also be an oxygen.
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 that has one bond to carbon, that means it's going to have three bonds to oxygen. And when you have three bonds to oxygen, you want it to look like a carboxylic acid. When you have two bonds to oxygen, you want it to look like a ketone. When you have one bond to oxygen, you want it to look like an alcohol, which is the one that I have up there. Bueno. Pretty good. Awesome.
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 alcohol. That's really the only difference.
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 two equivalents of oxygen. Let me show you what that looks like.
PCC is the name of this reagent and it is our weak oxidizing agent. Would it be able to oxidize my secondary alcohol? Absolutely. It's going to do the same exact thing. For PCC, I would get the same exact reagent here or the same exact product. Would it be able to oxidize number two? No. Nothing can oxidize number two. It's still no reaction. Would it be able to oxidize number three? 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. 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, “Johnny, 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 one. I'm just going to say one O. At the end of the strong oxidation, how many did we have, bonds to oxygen? We had three. Three O. That means that if we had one oxygen to begin with, and three to end with, we added two equivalents of oxygen.
Well, for PCC, instead of doing two equivalents of oxygen, now we're only going to add one equivalent because now we have two bonds to O instead of one, which is what it started with. If you're starting with one and you ended with two, 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 it, 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 four? 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. If we have an aldehyde already, is it going to do anything to it? No. Once it's an aldehyde, it's not going to oxidize it more. This would be no reaction as well.
Just trying to show you guys the difference between PCC and the other oxidizing agent. It's not that hard. It's just a few details you have to keep in mind. That said, let's go ahead and move on. 

Concept: Concept: Jones Reagent Mechanism

7m
Video Transcript

Now that we understand the process of oxidation pretty well, I want to show you guys at least one of the 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+ reagent and strong acid. 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 H2SO4 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.
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. 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.
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 going to work? Well, let's go ahead and start.
The first step guys is nucleophilic attack because here we have chromic acid. 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-. But we're in the presence of acid, so the O- is going to protonate. 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 once 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. Let's start off with the nucleophilic attack.
This OH here is now right here, attached to the chromium that still has an H on it. Let's go to another easy one. This OH is still here. Nothing changed there. This OH is still here. So they're still in the places that they were before, but there is one new OH that's the one that is in pink because this pink, remember that it grabbed an H, right? That means that this is the OH here. Now hopefully, those colors help you to kind of identify what's going on, where did everything go. All these atoms are still there, so we're doing okay.
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? 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.
What alpha simply means is that instead of eliminating between the alpha and the beta carbon, you're actually going to form a double bond directly on that alpha-carbon to something that's not a carbon. 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, 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. Instead of forming a double bond between two 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 on this O and then finally that O isn't happy, 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. Awesome.
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. 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 two 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.
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. 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. 

Problem: What is the product of the reaction?

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Problem: What is the product of the reaction?

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Problem: What is the product of the reaction?

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Oxidation Additional Practice Problems

Fill in the box with the product(s) that are missing from the chemical reaction equation. Draw only the predominant regioisomer product or products (i.e. Markovnikov or non-Markovnikov products) and please remember that you must draw the structures of all the product stereoisomers using wedges and dashes to indicate stereochemistry. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES.

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Fill in the box with the product(s) that are missing from the chemical reaction equation. Draw only the predominant regioisomer product or products (i.e. Markovnikov or non-Markovnikov products) and please remember that you must draw the structures of all the product stereoisomers using wedges and dashes to indicate stereochemistry. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES.

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Fill in the box with the product(s) that are missing from the chemical reaction equation. Draw only the predominant regioisomer product or products (i.e. Markovnikov or non-Markovnikov products) and please remember that you must draw the structures of all the product stereoisomers using wedges and dashes to indicate stereochemistry. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES.

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What are the full name and the chemical structure of PCC oxidizing reagent?

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What are the chemical components of Jone’s reagent?

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Benzaldehyde (shown below) can be oxidized with Jone’s reagent, but not benzophenone. Why? Please explain it with less than 15 words.

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Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw your answer in skeletal form. You will be graded on the product your draw from the reaction no other information is needed for this question.

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Complete the following reaction by drawing the structure of the principal major product. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.

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Identify the reactants (1˚, 2˚, 3˚ alcohol, aldehyde, etc.) and then draw the organic product of each reaction. If no reaciton occurs, write 'NR'.

 

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Oxidation of a 3° alcohol with chromic acid results in the production of ________.

a. an ester

b. a ketone

c. an aldehyde

d. a carboxylic acid

e. none of these

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Complete the mechanism for the following oxidation of an aldehyde reaction. Draw all the arrows to indicate movement of electrons, write all lone pairs, all formal charges, and all the products for each step. In the dotted box write which mechanistic element is involved in each step.

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Show how the starting material can be converted to the product. Show all the reagents you need and indicate the stereochemistry when appropriate. If a racemic mixture is formed, draw both enantiomers and write “racemic” next to the two structures. You do not need to show arrow pushing like in a mechanism question, only the reactions. All carbon atoms in the products should come from the starting material. Use as many molecules of the starting material as you might need to get to the product.

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Which of the following compounds can be oxidized using CrO3/pyridine?

1) A only

2) B only

3) C only

4) A and B only

5) B and C only

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Draw the structure of the alcohols, which are the direct precursors of the following carbonyl derivatives. Which reagent would you use to achieve these transformations?

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