Fukuyama Coupling Reaction - Video Tutorials & Practice Problems
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Fukuyama Coupling Reaction
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in this video, we're gonna take a look at the fukuyama coupling reaction. In this reaction, we have the coupling between a thio ester and an organic zinc, a lead with a palladium catalyst. The reaction itself creates a key tone product. Now, if you look at the generic setup for a cross coupling reaction, we see that we have our carbon khallad which is represented by our one X. We have our coupling agent which is represented by our to see M. L. N. Represents our transition metal complex where M. N. Is a transition metal and then L. Is our Ligon, usually two or four of them attached. Um We then have R. One and R. Two being connected together to form our coupling product. And then again, see exes are byproduct within the Fukuyama coupling reaction. Instead of using a traditional carbon khallad. Instead we have a thio ester. And then for our coupling agent, we have our organic no zinc khallad in this reaction. Um we utilize palladium as a catalyst and then we have to lean as our um solvent. Through this reaction, we have the exile group which are one is part of connecting to the R to group of our organic zinc salad. 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 our one um group of the west is represented by a vinyl or in a real group. The are two of the organic zinc Halide is represented by an owl kill group and then when it comes to see it is represented by zinc, I dyed 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 ethel portion of the thigh wester and the zinc iodide portion of the organic zinc Halide. And then we have the a cell group which contains the R. One group combining with the R. Two group of the organic zinc Halide. To give us our key tone 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.
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example
Fukuyama Coupling Reaction Example 1
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in this example, we need to determine the product from the following fukuyama coupling reaction. So here we have our thio ester. So the group with the benzene ring represents our our one group. Then we have the ethel that's attached to my zinc iodide amalgam portion. So this here represents my R. Two. So remember all that's really happening is we're losing the sulfur with the ethel and the zinc with the iodide and R. One and R two combined together. Through this process, we're using our palladium catalyst and we're using to lean as our solvent in the reaction. Now combining R one and R two gives us as our answer. Here goes our one and then we're connecting our two to that carbon carbon to get our key tone as our final product. Now we also have as a byproduct, the zinc iodide combining with the sulfur and ethel. But again, we don't really concern ourselves with the byproduct. We're more concerned on what exactly is the coupling reaction we're aiming to create now that we've seen the basic, simple view of this coupling reaction, click onto the next video and see how exactly does its mechanism work to give us a ketone as the final product
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concept
Fukuyama Coupling Reaction
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So one big advantage of the Fukuyama coupling reaction is that unlike the graveyard re agent, the organic zinc Halide stops at the ketone phase instead of progressing forward to a tertiary alcohol. Now, a thio ester is just kind of like an analog of of an Esther. Remember Esther is a carbon cilic acid derivative and if you were to use a green yard re agent with it then it would add more than once to help create a tertiary alcohol. Right? So we get our tertiary alcohol. And remember that's because carbon selic acid derivatives undergo nuclear filic castle substitution. So something you've learned earlier this semester with the Fukuyama coupling reaction because we're using um an organic zinc Halide and we're using a palladium catalyst. We can arrest or stop the reaction for progressing beyond the ketone phase. So we never get a chance to get a tertiary alcohol. So 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, our group to create a tertiary alcohol. Now the steps involved with the fukuyama coupling reaction are pretty familiar. They are oxidative addition, We have transmit elation and then we have reductive elimination. So for the first step we have, it involves the addition of the thio ester to the palladium complex. So here we're just using regular palladium neutral palladium as a catalyst. I'm not attaching any Liggins on it, it could have Liggins. Um but you can also use this um naked form where it's just palladium and its neutral form. Now, what happens here is that we have this portion, our thio ester attaching to the palladium. And this portion of our tile Leicester connecting to the palladium. So what we'll have here is palladium connected to S. E. T. And then connected to R. R. one Group. And it's a sound portion. So in oxidative edition, this is what we get. We're not utilizing the traditional type of arrow pushing here because we don't have the similar carbon halid that we're used to seeing with coupling reactions. Now the transmit elation step, the R two group of the organic zinc compound transfers from zinc to the palladium complex. All right, so here we're going to have uh this R. Two coming in and it's going to attach to this palladium and that's gonna cause this S. E. T. To leave. And that S. E. T. When it leaves actually attaches to my zinc here. So, what I'm gonna get is I'm gonna get My are one group still attached to the palladium and we're gonna have The R two group also attached to the palladium plus my byproduct, which is just zinc. I died connected to S. E. T. Then finally come down here where we deal with reductive elimination. So this is where we're gonna make our coupling product and we're gonna regenerate our palladium catalyst. So bring that down here. So what happens here is this a style portion that is are one part of our one is going to connect to my R2 group at the same time. Are two is gonna relinquish or give up its electrons and and and pass them on to palladium. So what we're gonna wind up getting it was going to get our our one a style portion connecting To R two. And that's how we create our ketone plus, we have the regeneration of our catalyst in this reaction. Um we don't necessarily make a more conjugated product because the R two group that we're adding is an hour kill group, not a vinyl or in a real group. Um and also we're going to 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 castle group and the R two group closer to 18 or 16 electron rules. So, remember these are the driving forces that are pushing and propelling these coupling reactions so that we can make our products in the fukuyama coupling reaction. In this case we're just trying to make a key tone through the utilization of a thio ester and an organic zinc halliday
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Problem
Problem
Determine the product from the following Fukuyama Reaction.
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Problem
Determine compounds A, B, and C from the following reaction sequence.
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