Fukuyama Coupling Reaction - Online Tutor, Practice Problems & Exam Prep

Fukuyama Coupling Reaction

In this video, we're going to take a look at the Fukuyama coupling reaction. In this reaction, we have the coupling between a thioester and an organozinc halide with a palladium catalyst. The reaction itself creates a ketone product. Now, if we look at the generic setup for a cross-coupling reaction, we see that we have our carbon halide, which is represented by R₁X. We have our coupling agent which is represented by R₂C. M_nL_n represents our transition metal complex where Mn is a transition metal. And then L is our ligand, usually 2 or 4 of them attached. We then have R₁ and R₂ being connected together to form our coupling product, and then again CX is our byproduct.

Within the Fukuyama coupling reaction, instead of using a traditional carbon halide, we have a thioester. And then for our coupling agent, we have our organozinc halide. In this reaction, we utilize palladium as a catalyst and then we have toluene as our solvent. Through this reaction, we have the acyl group which R₁ is part of, connecting to the R₂ group of organozinc halide. That in turn creates our ketone as a product and then we have this byproduct.

Now, in terms of the R groups involved in the Fukuyama coupling reaction, we have the R₁ group of the thioesters represented by a vinyl or aryl group. The R₂ of the organozinc halide is represented by an alkyl group. And then when it comes to C, it is represented by a zinc iodide amalgam. So, basically, they're connected together.

Now, this is just the basic breakdown of what's going on with the Fukuyama coupling reaction. So just see it as us losing the sulfur ethyl portion of the thioester and the zinc iodide portion of the organozinc halide, and then we have the acyl group which contains the R₁ group, combining with the R₂ group of the organozinc halide to give us our ketone as our product. From this basic pattern, we're going to attempt to answer the example question in the next video. So take a look at what the question states and see if you can figure out what the answer is before you click on the next video.

Fukuyama Coupling Reaction Example 1

In this example, we need to determine the product from the following Fukuyama coupling reaction. So, here we have our thioester, so the acyl group with the benzene ring represents our R₁ group. Then we have the ethyl that's attached to my zinc iodide amalgam portion. So, this here represents my R₂. Remember, all that's really happening is we're losing the sulfur with the ethyl and the zinc with the iodide, and R₁ and R₂ combined together. Through this process, we're using our palladium catalyst and we're using toluene as our solvent in the reaction. Now, combining R₁ and R₂ gives us our answer; here goes R₁ and then we're connecting R₂ to that carbonyl carbon to get our ketone as our final product. We'd also have as a byproduct, the zinc iodide combining with the sulfur and ethyl. But, again, we don't really concern ourselves with the byproduct. We're more concerned with exactly what the coupling reaction we're aiming to create is. Now that we've seen the basic simple view of this coupling reaction, click on to the next video and see how exactly its mechanism works to give us a ketone as the final product.

Fukuyama Coupling Reaction

So one big advantage of the Fukuyama coupling reaction is that, unlike the Grignard reagent, the organozinc halide stops at the ketone phase instead of progressing forward to a tertiary alcohol. Now, a thioester is just kind of like an analog of an ester. Remember, an ester is a carboxylic acid derivative. And if you were to use a Grignard reagent with it, it would add more than once to help create a tertiary alcohol. Right? So we get our tertiary alcohol. And remember, that's because carboxylic acid derivatives undergo nucleophilic acyl substitution, something you've learned earlier this semester. With the Fukuyama coupling reaction, because we're using an organozinc halide and a palladium catalyst, we can arrest or stop the reaction from progressing beyond the ketone phase. So we never get a chance to get a tertiary alcohol. That could come in handy if we want to have a reaction where we just make a ketone at the end and not add another mole of this R group to create a tertiary alcohol.

Now, the steps involved with the Fukuyama coupling reaction are pretty familiar. They are oxidative addition, transmetalation, and reductive elimination. In the first step, it involves the addition of the thioester to the palladium complex. Here, we're just using regular palladium, neutral palladium as a catalyst. I'm not attaching any ligands on it. It could have ligands, but you can also use this naked form where it's just palladium in its neutral form. What happens here is that the thioester attaching to the palladium and this portion of our thioester connecting to the palladium. So what we'll have here is palladium connected to SET, and then connected to our R1 group and its acyl portion. In oxidative addition, this is what we get. We're not utilizing the traditional type of arrow pushing here because we don't have the similar carbon halide that we're used to seeing with coupling reactions.

In the transmetalation step, the R2 group of the organozinc compound transfers from zinc to the palladium complex. Here, we're going to have R2 coming in and it's going to attach to this palladium, causing the SET to leave. The SET, when it leaves, actually attaches to my Zinc here. So what I'm going to get is my R1 group still attached to the palladium and we're going to have the R2 group also attached to the Palladium, plus my byproduct which is just Zinc Iodide connected to SET.

Finally, we come to reductive elimination. This is where we’re going to make our coupling product and we’re going to regenerate our palladium catalyst. What happens here is the acyl portion that is part of R1, is going to connect to my R2 group. At the same time, R2 is going to relinquish its electrons and pass them on to palladium. So what we're going to end up getting is our R1 acyl portion connecting to R2. That's how we create our ketone, plus we have the regeneration of our catalyst. In this reaction, we don't necessarily make a more conjugated product, because the R2 group that we're adding is an alkyl group, not a vinyl or aryl group. We also say that we regenerate our palladium catalyst, which means that it's no longer following the 18 or 16 electron rule, which means that there's a drive for it to go back into the catalytic cycle to reattach itself with the acyl group and the R2 group to get closer to 18 or 16 electron rules. So remember, these forces are pushing and propelling these coupling reactions so that we can make our products. In the Fukuyama reaction, in this case, we're just trying to make a ketone through the utilization of a thioester and an organozinc halide.

Determine the product from the following Fukuyama Reaction.

So here it says determine the product from the following Fukuyama reaction. Alright. So here we have our thioester. So this portion here represents our R1 group. And then here we have our organozinc, halide or organozinc iodide group with the methyl group here representing R2. Remember, all that's really happening is we have the elimination of the SET group and the zinc iodide group. And then we have the connections between R1 and R2 to help make our ketone in this case. So that means our product at the end would be here. We'd have our R1 group, and it would attach to my R2 group of me, which is just a way of writing methyl. And in this case, we don't just use a naked, palladium catalyst. We actually have ligands attached to it. We have 2 chloroanions and then we have 2 of those L ligands in the form of PPH3, where PH is an abbreviation for benzene. But remember, just approach it with a simple understanding of what's going on. We have R1 and R2 combining, we have SET and Zinc Iodide being eliminated. Alright. So now that we've done that practice, I want you guys to attempt to do this practice left on the bottom of the page. I want us to determine what a, b, and c compounds will be. This question is a combination of reactions you've seen this semester and it's all coming to this point where we're now going to utilize the Fukuyama coupling reaction to get our ketone product. So attempt it on your own and once you do that, come back and see does your answer match up with mine.

Determine compounds A, B, and C from the following reaction sequence.

Alright. So here we have to determine compounds a, b, and c from the following reaction sequence. So we're beginning with a carboxylic acid. And to this carboxylic acid, we're using thionyl chloride and we have pyridine base. Now, remember, what's one of the functions of Thionyl Chloride, SOCl2? Remember, one of its main functions is to change an OH group into a Cl. This is going to help change our carboxylic acid into an acyl chloride. So this would be our compound a, we just created an acyl chloride. Now that we have an acyl chloride, we now have this thiol. And then we just have, TS-OH on the bottom as just a reservoir of H+ ions if needed. We're not concerning ourselves with the mechanism in terms of how to determine what b is, but remember, we have an acyl chloride. When an acyl chloride reacts with an alcohol, what do we make? Well, when we have an acyl chloride react with an alcohol, so you'd lose this portion, you know that we would make an ester. In this case, we're not using an alcohol, we're using a thiol. So what are we going to make instead? Well, same premise. In this process, we're losing these 2. They're not combining. We're just losing them. Remember, we're not looking at doing the mechanism. We need to get to what b will be. So when that's lost, what do I have remaining? I have, SEt now combined with the carbonyl. So this is our compound B and we just made a thioester. So now, we're where we're familiar with. We know that with the Fukuyama coupling reaction we have a thioester and we react it with an organozinc halide which is this structure here. So this thioester here represents my R1 compound. And then here, this organozinc halide represents my R2. And we're using a palladium catalyst and toluene as our solvent. So then we just have to remember what is happening basically in a Fukuyama coupling reaction. Well, in terms of simplicity all that's happening is we're losing the SEt portion and the zinc iodide portion and then R1 and R2 combined together to give us our ketone. So my R1 portion is this and it's going to connect with my R2 portion which is this benzene ring. Now in this case, we do make a ketone and we do make something that is more conjugated. So, our product's more stable. Usually the R2 group will be an alkyl group but it does have the possibility of being an aryl group as well because that's a carbon group. So, just remember this question is interesting because it's a combination of reactions that you've seen, earlier in the semester or even earlier in Organic 1. We're bringing everything back here so that we can know how to go from a carboxylic acid to our final answer of a ketone. Remember, one great advantage of the Fukuyama coupling reaction is that we can stop at the ketone phase instead of going all the way into a tertiary alcohol phase. If we hadn't used an oroganozinc halide and used a regular Grignard, then we would've added 2 moles of the Grignard to create a tertiary alcohol at the end. With this type of coupling reaction, we don't have to worry about that. We stop at the ketone phase.