Concept: Concept: Reagents2m
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: Strong oxidizing agents4m
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: Strong oxidizing agents1m
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: Strong oxidizing agents6m
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.
Problem: Provide the major product for the following oxidation reaction.4m
Problem: Determine the major product for the following reaction.6m
Problem: Determine the major product for the following reaction.7m
Concept: Practice 7: Baeyer Villiger Oxidation6m
Hey guys, let's take a look at the following practice question. So here it says provide the necessary reagents for the following transformation. So if we take a look at this what we have initially is a ketone and what we're doing here is we're changing this ketone into an ester and remember an ester is a carbonyl connected to O connected to C so this portion makes it the ester and 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 a ester and in this case what we can do here is we can do the Baeyer Villiger reaction, 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 oxidised, this is one of the rare exceptions where we can. So here we use a peroxyacid, remember a peroxyacid is R C O 3 H. It's kind of like a carboxylic acid except it has one additional oxygen instead of being C O 2 H it's C O 3 H and we're saying here that the most common one that we know of is M C P B A which looks like this. So we're going to say M C P B A stands for meta, which remember means 1, 3 because they are 1, 3 apart, chloro because of the chlorine, peroxy and then remember the peroxy acid portion is this portion, peroxy, benzoic because of the benzene acid. It's like a carboxylic acid so it's enzoic acid acid just like a carboxylic acid. 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 phenol carbon and phenol 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 three oxygens involved and one of these reactive oxygens is going to basically come in and separate one of the carbons out of the way so we can connect directly to the carbonyl carbon.
So if we look here we have a C H 3 and 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's not there then this carbon technically is only connected to one carbon, this carbon, so it's primary. So remember, methyl carbon moves out of the way before, a primary carbon moves out of the way before 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 and here, is 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 and here what we could have done is we could have done a reaction which we call the Haloform test and 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 do know how to do, we'd use an alcohol over acid catalyst and in that way we get that 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.
Show how you could go from cycloheptene to heptanedial by filling in the reagents for A and C and then showing the intermediate in box B.
Determine the correct answer for the question below.
Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. Be sure to indicate the major product if more than one product is formed. Draw all answers in skeletal form.
Give the major product for the following reaction.
Draw the major organic product or reactant in the boxes provided for each reaction. Aqueous work up assumed where necessary.
What is the predominant product from the reaction of 2-hexanol with PCC in CH 2Cl2?