Kumada Coupling Reaction - Video Tutorials & Practice Problems

Kumada Coupling Reaction

In this video, we're going to take a look at the Kumada coupling reaction. Now, the Kumada coupling reaction involves the coupling of a carbon halide and a reagent that we've used numerous times in the past, Grignard's reagent. The reaction uses a palladium or nickel catalyst in the formation of predominantly biaryl or β vinyl products. Other types of products can be created, but those are usually the two that we're aiming for. Here the use of a palladium or nickel catalyst allows for stereoselectivity with the Grignard reagent. We've used the Grignard reagent in other reactions before, but with this reaction, the catalyst that we employ allows us greater control over the configuration of our final product. If we have an alkene, we can use certain types of Grignards to give it either an E or a Z configuration. The Kumada coupling reaction can mirror the generic form of a cross-coupling reaction.

In a generic cross-coupling reaction, we have R₁X, which represents our carbon halide and R₂MgX, which represents our coupling agent, where M represents our transition metal, and L represents a ligand. The C-X would just be our byproduct formed within this generic coupling reaction. Transitioning to the Kumada coupling reaction, we see that we still have a carbon halide involved. The R₁ of our carbon halide will either be a vinyl or an aryl group. Then, as our coupling agent, we just have our Grignard reagent. Here, the R₂ group of the Grignard reagent is represented by vinyl, aryl, or an alkyl group. The C portion of our coupling agent is represented by MgX, where X is Cl, Br, I, or some triflate group. As usual, X represents chlorine, bromine, iodine, or some triflate group.

Through the use of palladium and nickel catalysts, we have our coupling product being formed. And then, this is just our byproduct. So, if we're looking at it through a simple lens, what does it look like that's happening with the Kumada coupling reaction? Well, it looks like the X of the carbon halide and the MgX of our Grignard reagent are both being lost. They're both lost. And then, R₁ and R₂ that's left behind, they combine together to give us our final coupling product. We're going to take this simple approach to help us figure out what our products will be for these reactions. After that, we'll go into greater detail on how we can affect the type of stereochemistry our products will have at the end. So, click on the next video and take a look at the example of how we approach it to get to our final answer.

Kumada Coupling Reaction Example 1

So here we need to determine the product from the following Kumada coupling reaction. Now, basically what's happening is we have the halogen that is part of my carbon halide, therefore this is R1. And then we have the MGX part of our Grignard, so this is R2, they're lost and then R1 and R2 are combined together. We do this through the use in this particular case through a nickel catalyst. So, what's going to happen for our final product is we're going to have our R1 group and we're attaching to it our methyl group which is just ME. So, we just made an alkene as our final product here. In this case, we didn't make something that is more substituted or more conjugated rather, but this is the product that we would get in terms of this reaction. Remember, it's pretty common for Grignard reagents to use an alkyl group for its R2 portion. So in this case, we're making an alkene at the end. Now that we've seen this example, click on to the next video and let's take a look at how stereoselectivity plays a role in the Kumada coupling reactions.

Kumada Coupling Reaction

So when it comes to the Kumada coupling reaction, we have to take into account stereoselectivity as well as chemoselectivity. Now, for stereoselectivity, we're going to say the stereochemistry of a vinyl halide is maintained when an alkyl Grignard reagent is used in the reaction. So, in these two reactions, we have ethyl as our alkyl group that is part of our Grignard reagent. That ethyl represents R₂. And then here we have our vinyl halide, so that represents R₁. We're going to maintain our E and Z configuration. So here, we'd have our R₁, that halogen, that Br gets replaced by the ethyl, and we're going to maintain the E the Z configuration. So here goes ethyl here. Then for this one, it's E, so we have to maintain that. So the ethyl's here. So when we're dealing with an alkyl Grignard reagent and a vinyl halide, you're going to maintain whatever the E or Z configuration is of the vinyl halide. Now when you're using vinyl or Aryl Grignard reagent, then you're going to get a mixture of products. Here we have pH which is an abbreviation for benzene. So we're dealing with in an aryl Grignard reagent. And here we have a Z alkene. The pH which represents my R₂ group will replace the Cl. This alkene portion is my R₁ group. So here we're going to get a mixture of products. So we'll get one product where the Z configuration is maintained. But then we're going to get a second product where we get the opposite, in this case, E.

Now, besides stereoselectivity, we have to pay attention to chemoselectivity. We're going to say the Grignard reagent does not readily couple with aryl or arylyl chlorides. So on this benzene ring and the benzene ring represents our R₁ group. We have chlorine and we have bromine. Again, the benzene ring, the R₂ group of the Grignard which is this benzene ring, will not be able to displace this chlorine. So instead, it's going after the bromine. It's going to kick out the bromine, take its place. So what we're going to get here is our benzene ring with the Cl still attached. And that Br has been displaced by this benzene ring. So this would be our answer here.

So remember, for the Kumada coupling reaction, it's great because we're utilizing a reagent that we're accustomed to seeing in the form of a Grignard reagent. But we have to pay close attention to ideas of stereoselectivity and chemoselectivity to get our final coupling product.

Determine the product from the following Kumada Reaction.

So here we need to determine our product from the following Kumada reaction. We have MgBr, which is part of our Grignard, meaning that this whole structure here is R2. Then here we have a halogen, so therefore this must be R1. Remember, fundamentally what's happening is that our halogen from the carbon halide is lost and our MgBr from the Grignard is lost, so that R1 and R2 can combine together. When it comes to alkenes, you have to make sure to maintain E or Z configurations around the double bond. Here, it is an E configuration, so make sure your product is also E at the end. Alright. So, we're going to have our R2 group, and now we are going to attach this here. Make sure it keeps its E configuration. It would look like that. So, that would be our final product where we made something more conjugated, therefore, more stable. And the alkene portion of our R1 structure maintains its E configuration. Now that we've done that, attempt to do this question here. We have to determine the coupling product for the following Kumada reaction. I'll give you guys a hint: all of them are iodine, all of them are equally reactive. What do you think is going to occur now? So once you figure that out, come back and see if your answer matches up with mine.

Determine the coupling product for the following Kumada reaction.

Alright. So we know that when it comes to the Kumada coupling reaction, we have these halogens which are going to be replaced by the R2 group of my Grignard reagent. My R2 group is this ring here. Since they're all iodized, they're all equally reactive. Therefore, all of them are going to get replaced by one of these. So we're assuming that we have more than just one molecule of this Grignard reagent. We'd have multiple. So that they can all replace all 3 iodines. Now, regarding the nickel catalyst, we have NiCl2 and then we have DPE. DPE is just a very bulky, bidentate ligand. So the way it would look, just for reference, it's not a typical type of bidentate ligand, but I wanted to show you guys that when it comes to the Kumada coupling reaction, sometimes they'll use this particular ligand. So it looks like this. We have a phosphorus connected to a benzene ring which we'll abbreviate as Ph, connected to another benzene ring, connected to 2 CH2s, connected to another phosphorus which also has 2 benzene rings. And then that's connected to the nickel which has the 2 Cl attached. So, this whole thing here, that's DPE. Again, it's a bidentate ligand because it's making 2 attachments to the nickel transition metal. It's a lesser-known bidentate ligand. But I just wanted to show you guys the structure of it. Alright. So going back to the question itself, we know that this here is R1 and all that's going to happen here is my R2 is going to replace all the iodines. If they had been different halogens, they would have different reactivity. Therefore, we wouldn't have to replace all of them. Remember, we're replacing where the MgBr was. So this would be our final product in terms of this Kumada coupling reaction. So remember, we're utilizing a Grignard reagent and using either a palladium or nickel catalyst in order to combine R1 with R2 to create a coupling product at the end. This is better than your typical type of Grignard reagent, because with the catalyst, we're able to control the type of stereochemistry that we'd get for our product. And also, we have to pay attention to things such as chemoselectivity as well.