Alkyne Oxidative Cleavage - Online Tutor, Practice Problems & Exam Prep

Cleavage of alkynes involves breaking triple bonds, similar to double bonds, using strong oxidizing agents like chromic acid or ozone. This reaction results in carboxylic acids for carbon chains longer than one and carbon dioxide for single carbon chains. For example, a triple bond with three carbons on one side and one on the other yields a three-carbon carboxylic acid and carbon dioxide. Understanding this process is crucial for multi-step synthesis and predicting product outcomes in organic reactions.

General features of alkyne cleavage.

Triple bonds can be cleaved in much the same way that double bonds were, and it turns out that we're going to use really the same exact reagents for these kinds of reactions. We're going to call this cleavage of alkynes. So when you have a cleavage of an alkyne, what that means is that you're basically taking that alkyne and you're splitting it in two. I'm going to use the analogy of scissors, so you're taking these scissors and you're just cutting it right down the middle of that triple bond. The way that we tell what the two products are going to look like, because it's going to split one thing into two, is that we look at how many carbons are on each side. In this case, for this specific triple bond, I would have three carbons on one side and one carbon on the other. So I can expect that my two products are going to be the same thing where I get a three-carbon chain on one side and a one-carbon, lone carbon on the other.

Now it turns out that triple bonds are very sensitive to oxidation, so what that means is that any strong oxidizing agent will work. Both KMnO₄ and ozone are going to really produce the same exact reaction. And what I'm going to wind up getting is a mixture of carboxylic acids for anything that's above one carbon and carbon dioxide for anything that's a single carbon. So in this case, as you can see, I had three carbons on one side, so that means I get a carbon carboxylic acid on that one side. For the other one, I chopped off just a single lone carbon, so that one's going to be fully oxidized to CO₂, carbon dioxide gas. That only happens when you're chopping off one-carbon chains.

What I want to do is really easy. I just wanted to show you guys that, so you know. What I want to do is this multi-step synthesis practice, and I want you guys to try it for yourself. So go ahead and look at this double bond, look at these three reagents, and try to figure out what the end product would look like based on what you know about these reagents. And then I'm going to go ahead and give you the answer. So go for it.

Predict the product of the following reaction.

So I'm starting off with a double bond, and I am halogenating it. Since I am halogenating it, that means I can expect to get vicinal dihalides along that double bond. So what I would expect for the first step is that I would get halogens that look like this: Cl here, Cl here. Remember that they have to face opposite directions because it's actually anti, the position of that. So, that's a reaction that you should have known from the addition chapter. Okay?

Then we have NaH, and we're using it in excess. Now that just means H^- in excess. Do you know what that would do to vicinal dihalides? Actually, what would it do to any dihalides, whether they're geminal or vicinal? It's going to wind up doing a double elimination. So what winds up happening is that one of the H's from here eliminates with one chlorine. An H from here eliminates with the other. So what I wind up getting is a triple bond formed. So, I would wind up getting something that looks like this. Okay? Where I know it looks way different, but if you count my carbons, I started off with 5: 1, 2, 3, 4, 5. I'm ending up still with 5. Okay? It just looks a little funny. And the reason it looks funny is because remember triple bonds have to be in that linear geometry. Okay?

So now I've got a triple bond, and now I can finally well, a question you might be asking is, Johnny, would the excess H^- take this all the way to an alkynide so it has a negative charge? Remember that sometimes if you had excess base, it could take it all the way to a negative. In this case, no. This could never be an alkynide. Why? Because this is an internal alkyne. Do you remember the other type of alkyne that we've talked about? We've talked about terminal alkynes. Terminal alkynes can turn into alkynides because they have an H that can be deprotonated. This alkyne here doesn't have any hydrogens directly on the triple bond. It's got an ethyl on one side, a methyl on the other side. That can't be deprotonated, so it's just going to stay as a triple bond. I hope that makes sense.

So now, we're doing ozone, and this is going to be an ozonolysis reaction. That means I can expect to get carboxylic acids and maybe carbon dioxide. So, I have to go ahead and split this bond. We don't need to know the mechanism, but I'm just going to split that bond in two, and what are we going to get? We're going to get a two-carbon section on one side and a three-carbon section on the other. Does that make sense? So now that I know what my segments look like, I just have to draw the carboxylic acids. So I would get a carboxylic acid that looks like this. Okay. Plus, I would get a carboxylic acid that looks like this. Okay? Notice that one of these is 2 carbons. 1, 2. And one of these is 3 carbons. 1, 2, 3. So overall, it's adding up to 5, which is what I needed it to add up to from my original. Now you might be wondering, Johnny, why didn't we get carbon dioxide like we got in this example here where I had carbon dioxide? Because carbon dioxide only forms when you have a one-carbon segment. And in this case, I didn't. So this would actually be the final answer, just two different carboxylic acids. Alright? So I hope that was good practice for you in multistep synthesis, but also just learning how to do oxidative cleavage of a triple bond. Alright? So let's go ahead and wrap this topic up.

Here’s what students ask on this topic:

What reagents are commonly used for the oxidative cleavage of alkynes?

Common reagents for the oxidative cleavage of alkynes include strong oxidizing agents such as potassium permanganate (KMnO₄) and ozone (O₃). These reagents effectively break the triple bond in alkynes, leading to the formation of carboxylic acids for carbon chains longer than one carbon and carbon dioxide (CO₂) for single carbon chains. The choice of reagent depends on the specific conditions and desired outcomes of the reaction.

What are the products of oxidative cleavage of a terminal alkyne?

When a terminal alkyne undergoes oxidative cleavage, the products are a carboxylic acid and carbon dioxide (CO₂). The terminal carbon, being a single carbon, is fully oxidized to CO₂, while the other part of the molecule forms a carboxylic acid. For example, if the alkyne is ethyne (HC≡CH), the products will be formic acid (HCOOH) and CO₂.

How does the oxidative cleavage of internal alkynes differ from terminal alkynes?

In the oxidative cleavage of internal alkynes, both sides of the triple bond are part of carbon chains, leading to the formation of two carboxylic acids. For example, if the alkyne is 2-butyne (CH₃C≡CCH₃), the products will be two molecules of acetic acid (CH₃COOH). In contrast, terminal alkynes produce a carboxylic acid and carbon dioxide (CO₂).

Why is oxidative cleavage of alkynes important in organic synthesis?

Oxidative cleavage of alkynes is important in organic synthesis because it allows chemists to break down complex molecules into simpler carboxylic acids and carbon dioxide. This reaction is useful for determining the structure of unknown compounds, preparing carboxylic acids, and simplifying multi-step synthesis processes. Understanding this reaction helps predict product outcomes and design efficient synthetic routes.

Can oxidative cleavage of alkynes be used to determine the structure of unknown compounds?

Yes, oxidative cleavage of alkynes can be used to determine the structure of unknown compounds. By breaking the triple bond and analyzing the resulting carboxylic acids and carbon dioxide, chemists can infer the original structure of the alkyne. This method is particularly useful in structural elucidation and confirming the presence of specific alkyne groups within a molecule.