Show how you would use a Suzuki reaction to synthesize the follow...

Question

Show how you would use a Suzuki reaction to synthesize the following biaryl compound. As starting materials, you may use the two indicated compounds, plus any additional reagents you need.

Chemical structure of a biaryl compound for Suzuki reaction synthesis.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to describe how a Suzuki coupling reaction can be used to synthesize the compound shown below as starting material. You can use one bromo, two isopropyl benzene and any other required regions. Now let's first of all understand how a Suzuki coupling reaction works. And we would have the following reaction. Where are one responded to Boron which contains two substations OR two, OR two. And we essentially have to react it with our three bonded to ax. Don't worry. We will define the identities of R one, R two, R three and X just in a bit. And the conditions that we're using for this reaction, they will be the following. We want to make sure that we're using palladium Where they zero oxidation state base and our servants. Typically, you may use sodium with oxide for this stuff in the presence of heat. Now, the product that we are producing with essentially couple R1 and R3, right, we're essentially taking our one from here And bonding it to our three. Now, it's time to define the nature of these groups. First of all. First of all are one would be our aerial group. Even though you can use an alkaline or an al Qaeda. In this case, you may notice how we're forming that bond over here between the two benzene rings. So we can think about the aerial groups. So we will have our aerial group over here Thinking about our two that may be hydrogen or an L kill group. R three. Same thing. We essentially want to use an aerial in this case. Right? Because we're bonding two benzene rings and now whenever we come to acts, that would be our hey lied. But we have to be extra careful. We want to make sure that the hale I've used is roman or iodine because chlorine requires special conditions for activation. We want to just stick with these two. Okay, so we understand how the reaction works and let's apply that to our synthesis. First of all, you may understand that if you look at the structure it's basically symmetrical, right? Notice if we break this bond we have to exactly same fragments, meaning we're forming a coupling reaction between R one and R. Three which are exactly the same. Right? Those will be I were Isopropyl benzene fragments. Okay, so we definitely understand what R one and R three are. So we can say that if we label one of them. Let's suppose this one actually let's change the positions. We clearly understand which fragment we're looking at. So that would be our R one and R three. We are not going to worry about our two. We can just leave it as our two saying that okay, will be either hydrogen or I'll kill group and for our acts based on the nature of the material we have our bromo substitutions. We can say that X is our bromo substitution. So we can propose the synthesis plan now based on all this information, let's draw the starting material. So that would be our benzene ring. It says one bromo two is a pro po so let's add our bromo substitution. And at position number two we are going to draw an isopropyl group. Okay, so we definitely understand that we have our our 3X. We but because we want to couple it with itself, we first of all have to form our organa boring and to do that the world first thing we want to perform is lengthy ation. In other words, recall that whenever we use two equivalents of lithium we can replace this rooming with lithium. Okay, so we are going to form our organa lithium regent now because this whole fragment is our our one. That means we essentially want to replace lithium with boring with 20. R. Two groups. Right? So how do we do that? We call that for this stuff. We can essentially use category boring. So we can say that for this reaction. We're using categorical boring. Okay. And this allows us to achieve our organic boring. Now let's draw it out that would be our boron with O. R. Two times. Right? Because we have two equivalent O. R. Groups. No need to worry about this too. You just part of the general scheme here because we're only using one are in our synthesis we are just going to use our but keep in mind that R. Is hydrogen or an L kill group. Right? So our is hae jin war and L kill group. And that said we are ready to couple it right because we have our organic boring. We understand that errol hey light is already there. Right. We're coupling it with itself. So we essentially need to add our regions. Let's see that. As we said, we need to use palladium and generally what you will see in this reaction will be this region we essentially call it Quattro, try fennel, fosse fine palladium and it will go in combination with our solvent sodium with oxide. As we said heat and heat. And as we said, we are essentially coupling it with a soft so the arrow bromide that we're using is the structure. So let's draw it out as our region and that would be at essential. We understand that we are coupling these two fragments right, which are equivalent to each other and we are going to form a bond between this carbon atom and this carbon atom. For simplicity, if we label them, let's suppose that this is our 123456. And let's suppose that these are A B C D E and F. We're forming a bond between carbon number one and carbon A. So let's through our product which would correspond to the given structure. So if we start by drawing the left hand side ring. My policies, that would be the upper ring, right? So we can show our isopropyl group. We understand that this is our carbon number. Once we are now going to form a bond between carbon number one and carbon A. So first of all, we can just draw our ring because we're attaching the whole ring. Okay, this is our carbon A. And you may essentially notice that the adjacent position F contains an isopropyl group, so we are going to add an isopropyl group. This indeed corresponds to our final product. So this would be the outline of our synthesis. Let's label it. And based on the synthesis scheme, we may conclude that the correct answer to this multiple choice question is B. However, if you have any questions, don't hesitate to leave your comment down below and thank you for watching.

Show how you would use a Suzuki reaction to synthesize the following biaryl compound. As starting materials, you may use the two indicated compounds, plus any additional reagents you need.

Key Concepts

Suzuki Reaction

Aryl Halides

Organoboron Compounds

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