Oxidative Cleavage: Videos & Practice Problems

Strong oxidative cleavage of double bonds occurs with hot KMnO₄, resulting in the formation of ketones, carboxylic acids, and carbon dioxide. Internal alkenes yield ketones, while terminal alkenes produce carboxylic acids due to the presence of hydrogen. A one-carbon fragment fully oxidizes to carbon dioxide. Understanding the geometry of the molecule is crucial; cutting a ring produces a straight chain, while multiple cuts yield distinct fragments. This reaction emphasizes the addition of oxygen, illustrating the concept of oxidation in organic chemistry.

General properties of strong oxidative cleavage.

In this video, we're going to discuss a really important reaction that happens to double bonds and that's called strong oxidative cleavage. Strong oxidative cleavage is the reaction of double bonds with hot KMnO₄. Now it's really important to know because KMnO₄ can do several things depending on the temperature. Now if you guys recall, KMnO₄ in a cold environment is actually going to do a 1,2-syn diol reaction. It's going to add vicinal diols to the double bond. But if I increase the heat, it's not adding alcohols anymore. It's actually just going to slice right through. You can think of that spaceship I talked about for potassium permanganate. Just think that if you rev up the heat, it's just going to crash right through the double bond and split it in half. And that's exactly what strong oxidative cleavage does.

What we're going to do is take this example. Notice that I have this 7-carbon chain. I'm going to split it in 2 areas and we're going to kind of inspect each part of it and see what happened. Now it's important to realize that you're going to get 3 different types of products when you react with strong oxidative cleavage. You're going to get ketones — that's specifically what you're going to see exactly why you get each one: Ketones, carboxylic acids, and carbon dioxide. So, when would you get each one? Well, you get ketones for what I call internal double bonds, for internal alkenes. You have carboxylic acid for terminal alkenes. And you get carbon dioxide for 1-carbon fragments. If you're just splitting off 1 carbon, that is going to become a carbon dioxide, fully oxidized. And that's kind of the theme here. I know that we haven't really gone over a rigorous oxidation and reduction chapter yet. But oxidation simply can be simplified as adding oxygens to things. And the prevailing theme here is that we're just going to add as many oxygens as possible to these cuts. So as you can see, if I go ahead and I slice down those double bonds, cleavage means that they're going to be completely sliced off and I'm going to get completely different chunks of carbons. Well, I would wind up getting a 1-carbon chunk here, so keep that in mind. I would get a 4-carbon fragment here and then I would get a 2-carbon fragment here. The question is, what does each of these become? Well, this is an example of an internal double bond. Notice that it's surrounded by R groups on both sides. Well, that's not going to change. When it's internal, it becomes a ketone because those R groups aren't going anywhere. As a general rule, we're not breaking off R groups. We're just breaking the double bond and that's it. R groups are in place.

Now over here, we have examples of terminal. So let me show you what terminal is. Terminal would be like this and like this. Notice how they both are at the end of a chain and they have an H on them. The biggest thing here is that it has at least one H coming off. It's not surrounded by R groups on both sides. It has an H. So the fact that it has an H means that that H can be oxidized to OH. So that's when we get our carboxylic acids because we had one hydrogen that could be broken off into 2 different carboxylic acids. We had 1 hydrogen each. Basically, we saw what happens when you have 0 hydrogens. You get a ketone. Let's also write that down. Internal is 0 hydrogens. Terminal is when you have 1 hydrogen. So guess what the last one is. When you have 2 Hs, right? So when you have 2 hydrogens, that means that your one carbon fragment, that's all you have left. And that's when you get fully oxidized to carbon dioxide. Carbon dioxide is considered the most oxidized form of carbon because all it is is carbon and oxygen. In fact, it's inorganic carbon because it doesn't have hydrogens anymore. It just goes into the atmosphere. That's called inorganic carbon. So that's basically the way we do it.

Also, you have to think geometrically. If I start off with a ring instead of a straight chain or whatever, a branched chain, and I cut it in one place, what would I expect that ring to become if I just cut it in one place? Just think of it like anything around the house. Let's say you have a rubber band and you cut it in one place, you stretch it out, it's going to become a straight chain. It's not going to break into different pieces. It's just going to be one piece that's a different shape. So think of it like that. Now if you took a rubber band and you sliced it right through the middle, both sides, now you would get 2 pieces. But that's if you have 2 cuts. That's what I want you guys to be thinking about. Does this make sense? If this is a rubber band or some kind of household item and I took it and I cut it in 2 different places there, would it actually make 3 fragments? In this case, yes, definitely. That being said, let's move down to this molecule. I want you guys to analyze how many different hydrogens are coming off of the ends of the whole bonds to determine the products and also the shape. Tell me how many different fragments you would get.

Predict the product of the following reaction.

• A.
• B.
• C.
• D.

Answer: (D)

Here’s what students ask on this topic:

What is oxidative cleavage in organic chemistry?

Oxidative cleavage in organic chemistry refers to the reaction where a double bond in an alkene is broken, resulting in the formation of two separate carbonyl-containing fragments. This reaction typically involves strong oxidizing agents like hot potassium permanganate (KMnO₄). The products can include ketones, carboxylic acids, and carbon dioxide, depending on the structure of the original alkene. Internal alkenes yield ketones, terminal alkenes produce carboxylic acids, and a one-carbon fragment fully oxidizes to carbon dioxide.

What are the products of oxidative cleavage with hot KMnO₄?

When oxidative cleavage is performed with hot KMnO₄, the products depend on the type of alkene. Internal alkenes (those with R groups on both sides of the double bond) yield ketones. Terminal alkenes (those with at least one hydrogen attached to the double-bonded carbon) produce carboxylic acids. If the cleavage results in a one-carbon fragment, it is fully oxidized to carbon dioxide (CO₂).

How does the geometry of a molecule affect the products of oxidative cleavage?

The geometry of a molecule significantly affects the products of oxidative cleavage. For example, if you start with a ring structure and make a single cut, the ring will open up into a straight chain. Multiple cuts in a ring or chain can yield distinct fragments. The position of the double bonds and the presence of hydrogen atoms on the carbons involved in the double bond determine whether the products will be ketones, carboxylic acids, or carbon dioxide.

Why does hot KMnO₄ produce different products compared to cold KMnO₄?

Hot KMnO₄ produces different products compared to cold KMnO₄ due to the increased reactivity at higher temperatures. Cold KMnO₄ typically adds vicinal diols (syndials) to the double bond, resulting in the formation of alcohols. In contrast, hot KMnO₄ causes oxidative cleavage, breaking the double bond completely and forming carbonyl-containing products such as ketones, carboxylic acids, and carbon dioxide, depending on the structure of the original alkene.

What is the role of hydrogen atoms in determining the products of oxidative cleavage?

The presence of hydrogen atoms on the carbons involved in the double bond is crucial in determining the products of oxidative cleavage. Internal alkenes, which have no hydrogen atoms on the double-bonded carbons, yield ketones. Terminal alkenes, which have at least one hydrogen atom, produce carboxylic acids. If the cleavage results in a one-carbon fragment with two hydrogen atoms, it is fully oxidized to carbon dioxide (CO₂).