Suggest a synthesis of the following molecule starting with th...

Question

Suggest a synthesis of the following molecule starting with the reagents shown, using cuprate cross-coupling as the key step.

Chemical reaction diagram showing the synthesis of a molecule using cuprate cross-coupling with specified reagents.

Accepted Answer

Hi, everyone and welcome. Our next problem says to propose a synthesis of the following molecule starting with the initial molecule shown using a crate cross coupling reaction. And our starting molecules we have this aerial Halide. So we have a benzene ring and it's attached to a cyclops ring. And then coming from the benzene ring, we have an iodide. It's reacting with an alcohol, which is our alcohol group attached to a 12345 carbon chain where there's two methyl groups on the fourth carbon in the chain. And then the reaction has the two of these linked together with the alcohol and the iodide being eliminated and the aerial chain and the alkyl chain being linked together at that point. So this is this coupling reaction of these two molecules forming a new CC bond. And it says we need to use a coup ate cross coupling reaction. Well, for that to happen, we need a Gilman reagent and we can recall that a Gilman reagent is one in the format of R two cul I. So our, our group, two of our groups attached to a copper and then lithium and we react that with an, well, al el or in this case, aerial hide and that will couple those two, our groups together. So how do we get this Gilman reagent? So if we look at our, we've already got our aerial Halide already is in the right form. So we don't need to do anything to that a little check mark there, but we do need to generate our Gilman reagent from this alcohol group. We're gonna put brackets around this chain here, that five carbon chain. And we'll just call that R to take up less space here. But just for the purposes of our first step here, we'll draw up that whole group to see what our Gilman reagent will look like. You can see we actually have two of that R group, two of that five carbon chain with two methyl groups. And in the center joining those two groups, we have a negatively charged copper. So cu minus with a positively charged lithium ion L I plus. So that's our Gilman reagent. There's the copper so that we can have this cup rate cross coupling reaction. But we need to generate that from the alcohol. Well, you can generate the Gilman reagent from an organo lithium using Cubr. So I've drawn an arrow pointing backwards to our Gilman reagent with Cubr over the arrow. And that would come from our, our group attached to a lithium. And of course, we would need to start with two equivalents of that since we have the two of the R group attached but two in parentheses there. And then we'd have to, how do we generate that lithium again? We're starting from alcohol. So how do we generate the organic lithium from the alcohol? Well, we can't do that directly, but we can change the Halide into that organo lithium by reacting with lithium. So we could have are our group attached to a bromine and react that with lithium again, two and then two equivalents of lithium. So now we're done to hide and then we can get to hide from our alcohol using the reagent PB R three. Again, I'll say two equivalents for everything. So we start with our alcohol, two molecules of that react with two equivalent equivalents of PB R 31 up with two molecules of our our group. Now attached to bromine, react it with two equivalents of lithium. We'll get this organo lithium R attached to the lithium. Two of those react with cur BC UBR and you'll end up with this Gilman reagent. So then we'll have two of our R chains attached to the CU minus with the L I plus ion there as well. And we can react that with our aerial Halide in the original reaction to get the final product. So if we want to draw this to this out in the correct order sort of nicely in its final form, scroll up a little bit to make more room. I've drawn out the whole structure of my alcohol chain. So I have two molecules of his alcohol, I'm reacting with two equivalents of PB R three. And that's going to give me two equivalents of this molecule. Now, as a hide, so bromine has replaced the O group and now we'll react it with two equivalents of lithium. Actually went to run out of space and went to scroll down kind of make this going oops diagonally down. So with lithium two equivalents, so now I have two molecules of my organo lithium. And if I react that with Cubr, so copper bromate, I will generate my Gilman reagent. And now we have the cross coupling reaction itself. Our Gilman reagent has two of those chains on there, three of those alkyl chains. So now we just react it. So I've added plus and then our aerial iodide there and we get our final desired product with our aerial group and our alkyl group linked together at the point where the iodide and the alcohol groups were. So those have been eliminated and we've had this cross coupling reaction. So there is our mechanism highlight this starting with the alcohol reacting to generate a bromide reacting with lithium to generate organo lithium. And then with Cubr to make this Gilman reagent that Gilman reagent can react with the iodide and generate our final desired product. See you in the next video

Suggest a synthesis of the following molecule starting with the reagents shown, using cuprate cross-coupling as the key step.

Key Concepts

Cuprate Cross-Coupling

Reagents and Reaction Conditions

Synthetic Strategy and Retrosynthesis

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