Suzuki Reaction - Online Tutor, Practice Problems & Exam Prep

The Suzuki reaction is a cross-coupling process that combines a carbon halide with an organoboron compound to form conjugated products like alkenes and styrenes. The reaction involves three key steps: oxidative addition, where the carbon halide binds to a transition metal catalyst; transmetalation, where the organoboron compound transfers its group to the metal; and reductive elimination, which forms the final product while regenerating the catalyst. This mechanism emphasizes the importance of creating more conjugated products and adhering to the 16 or 18 electron rule for transition metals.

Suzuki Reaction

In this video, we're going to take a look at the Suzuki reaction. Now, the Suzuki coupling reaction involves the coupling between a carbon halide and an organoboron compound. We're going to say the reaction creates conjugated compounds composed of alkenes, styrenes, or bipyridyls. Now remember, a styrene, the generic form of it, is just a benzene ring connected to 2 alkene carbons.

Now, if we look at the generic setup for a cross coupling reaction, we'll see that it's composed of a carbon halide that is represented by R₁X, a coupling agent which is R₂C₂. We have ML_n here. M is a transition metal. L is the ligand attached to it and usually, in terms of the number of ligands, it's usually 2 or 4. Through the use of this transition metal catalyst, we can have the combining of R₁ to R₂ to give us our coupling product. Through this process, we also form some byproducts.

Now taking this generic cross coupling reaction model, we can apply it to the Suzuki coupling reaction. With the Suzuki coupling reaction, we still have a carbon halide. Here, the R₁ group of the carbon halide is represented by a vinyl or aryl group. Next, we have our coupling agent in the form of an organoboron compound. Here, R₂ of the organoboron compound is represented by, vinyl or aryl or an alkyl group. Then, we're going to say here that C, in terms of my coupling agent, would be represented by BR₂ inside the Suzuki coupling reaction. Here, the Y can represent different things: Y could be an OH (boronic acid). Y could be an OR group where R is a carbon. So, it'd be like boronic ester R₂. Y could be an alkyl group, like BR₂, where R could be a methyl or it could be an ethyl.

Then we're going to say that the X group of the carbon halide is represented by our typical really good leaving groups of chlorine, bromine, iodine, or a triflate. Now, what's happening in this reaction is just that the carbon halide loses its X group. The organoboron group loses its BY₂ group. And they are lost as byproducts. And then the R₁ and the R₂ left behind, combine to form our coupling product. That's fundamentally what's going on in terms of the Suzuki coupling reaction. And as long as you can approach it with this simple mindset, you'll be able to identify the types of products you're going to obtain within the Suzuki coupling reaction before we even cover the reaction mechanism.

So for now, guys, click on to the next video and see how I approach this example question. If you think you can, answer the question on your own, attempt it. If you get stuck, again, just click on the next video and see how I answer this Suzuki coupling reaction.

Suzuki Reaction Example 1

So before we do the actual example question, here it states, when creating conjugated products the reaction is observed to be stereo specific with retention of configurations. So what exactly is this saying? Well, what this is saying is that if our R₁ or our R₂ groups, groups are vinyl groups with an E or Z configuration. Then the product has to maintain that E or Z configuration. So, for example, if we're looking at this Suzuki coupling reaction, here is our halogen. So this is our carbon halide. Which would mean this alkene portion is R₁. This alkene is, in terms of configuration an E configuration. That's because the 2 high groups are opposite of one another, so we know it's E. So that means when I give my product, my product has to maintain this E configuration. Now, over here we have boron. And remember this here would have to be my R₂ group. Remember R₂ is connected to BY₂. Y could be represented as an OH, OR, or in this case, 2 R groups. These groups here. And we know that this portion here has to be R₂ because the 2 Y groups have to be identical. Alright. So what we have to do here now is, what is the fundamental thing that's happening within the Suzuki coupling reaction? As we said above, we have the X from the carbon halide in a way merging with the BY₂ of the organoborane to create a byproduct. The R₁ and the R₂ left behind combined together. Now, when we go over the mechanism, we'll see how it really works. We're just thinking of it in terms of this simplicity, so that we can get to our final answer quickly. Alright. So we're gonna say here, that the X group of my out of my carbon halide and the BY₂ group of my organoborane are lost. So now my R₁ and my R₂ have to combine together. So here is my R₁. Remember it is an E alkene so it has to maintain that E configuration. So that means that my R₂ which is this benzene ring, has to be placed in the same position that the halogen was in so that we maintain our E configuration. So this here would be my final answer. I created a product that is more conjugated and therefore more stable. A product that's also having the retention of its E configuration. The alkene as a reactant was E, as a product it stays E. So just remember the fundamental things that we did here in order to find our final product. We just circled the X group of the carbon halide, the BY₂ group of our organoborane or boron compound, and we were able to isolate our coupling product. Now that we've seen this, click on to the next video and let's go through the coupling mechanism together to see how we create these different types of coupling products for the Suzuki reaction.

Suzuki Reaction

So, for the Suzuki reaction, we have our 3 noticeable steps involved in the catalytic cycle: oxidative addition, transmetalation, and reductive elimination. Now with oxidative addition, it involves the addition of the carbon halide to the transition metal complex. So our transition metal, palladium, has a lone pair from its d orbital well. So we're going to have here palladium, still connected to its 2 original ligands, connected to X, and also connected to R₁.

When we go into step 2, where we are dealing with transmetalation, the R₂ group transfers from our organoborane or boron compound to the palladium metal complex. So what happens here is that this bond breaks, attaches to the palladium, and at the same time, the halogen leaves and attaches to the boron portion. So what we get here at this point is we're going to have palladium, still connected to its 2 original ligands, connected to R₁ and now connected to R₂. And then as our waste material we have B(Y₂), and then the X that has just joined it.

Now with reductive elimination, we're going to create the connection between R₁ and R₂ and regenerate the palladium catalyst. So we're going to take this compound here and bring it down to show this process. Okay. So here it goes. What happens here is the R₁ group's going to attach to R₂, and then the bond is going to break and go to palladium. So what we're going to get now is we're going to get R₁ being connected to R₂, plus the regeneration of my palladium catalyst.

Remember, the two driving forces for a lot of these coupling reactions are to create a more conjugated product—we do that in the form of R₁ connecting to R₂—and also for us to follow the 18 or 16 electron rule for the transition metal. At this point, by giving up R₁ and R₂, palladium is no longer following the 18 or 16 electron rule. So it wants the reaction to continue again. The regeneration of the catalyst is basically a key factor that wants us to continue doing this reaction over and over again, which is why it's involved in a catalytic cycle. But just remember, when it comes to this basic Suzuki coupling reaction, remember we have a carbon halide and from the carbon halide we lose our X group, and then we have an organoborane or boron compound where we lose the BY₂ group. They are lost as a byproduct, and then R₁ and R₂ combine together to give us our coupling product at the end.

Determine the product from the following Suzuki Reaction.

So here it says determine the product from the following Suzuki reaction. Alright. So here we have our halogen, so this represents our carbon halide, which would mean that this portion here is R1. And it's a vanillic halide, so that means we have to look at the E or Z configurations if they're present. Here this is an E alkene, which means in our product that configuration must be maintained. Then here we have our BY2 group in the form of BOH2, so this must be my organoborane or boron compound. So that would mean that this portion here is R2. It is also vanillic and it is an E alkene. So that has to be maintained when we draw our product. So we're gonna draw out our R1 group. And then we have to draw our R2 group. So we're connecting to this carbon here and it has to maintain its E configuration. And then we have these remaining carbons. So if we check here, this alkene is still E, and this alkene is still E. So both configurations have to be retained. And we just created a product that is more conjugated and more stable as a result of this. So just remember, we know that this is a Suzuki reaction because we have the presence of an organoborane compound, which is a pretty specific thing attached only to the Suzuki coupling reaction. And we know that fundamentally what's happening here is that we're losing the X group of our carbon halide and we're losing the BY2 group of our organoborane or boron compound. Then just connect R1 and R2 together. If they are vanillic, make sure that they maintain their E or Z configuration and you'll have your final answer. Now here, I'll give you a bit of a hint in terms of this practice question. So, here this compound is catecholborane. You've seen this kind of compound before, at least you've heard of it before. Remember, borane, that's a name that you should remember and it's associated oxidation, we're going to have the syn addition of hydrogen and BY2 to a pi bond, and by anti-Markovnikov's rule. So, some of you might have thought, "Oh no, I forgot all about this. I haven't seen these types of alkene and alkyne reagents in forever." They are still present even now. So, remember, borane, that's part of hydroboration oxidation. So we're dealing with the syn addition of H and BY2 by anti-Markovnikov's rule. That's supposed to give you insight into what A should be. Now, once you have A, you're going to undergo the Suzuki reaction in order to create compound B. So attempt it on your own, but if you get stuck, don't worry. Just come back and see if you get the same answer as I do. So good luck guys.

Predict the structures of organoborane compound A and coupling product B in the following reaction sequence.

So here it says predict the structures of organoborane compound A and couple product B in the Phong reaction sequence. I gave us a little bit of a hint. We have catecholborane here, and because it's catecholborane, that's indicative of hydroboration-oxidation. And remember, within hydroboration-oxidation, we have syn addition of hydrogen and our BY2 group by anti Markovnikov's rule. So I know I'm saying a mouthful here in terms of what we need to do, but let's break this down. We know that this compound here, we're going to add hydrogen. So this hydrogen and BY2, which is this part here, to one of the pi bonds of this triple bond. So that's what we're going to do. And remember, if we're adding to a pi bond we're going to remove that pi bond. So now we'll have just a double bond remaining. Alright, so this carbon and this carbon. So we have a double bond remaining. Syn addition means that this H and then this part here, which stands in for BY2, add on the same side. And we have to do it by anti Markovnikov's rule. So remember, anti Markovnikov says that hydrogen will go to the multiple bonded carbon that is more substituted. So this one here. And BY2 will go to the one that is less substituted. So this one right here. Alright. So, this carbon here is this carbon here which has its H. And we're adding the boron to the less substituted multiple-bonded carbon. And then here goes the other H. So that's what I get initially. So this is my compound organoborane or boron compound. So that would mean that this portion represents R2. And now we have a vinylic halide here. So this is my carbon haLide group, that would mean that this alkene portion here represents R1. And since we're dealing with the Suzuki reaction, what is happening here? Well, in the simplest of terms remember for the Suzuki reaction the carbon halide portion loses its X group which would be this Br. And then the organoborane or boron compound would lose its BY2 group, this compound. So those are gone. Now we have to realize that both of my R1 and R2 groups are vinylic, so we have to worry about E and Z configurations. Right now for R2, this alkene portion is E and for R1 this one is Z. So we have to keep that in mind when we're drawing our products. So here I'm drawing my R2 product. Got to make sure it stays E. So this carbon here, is this carbon here. And it's going to connect to this carbon here. This needs to stay Z. So the only way it can stay Z is if I draw it this way. So by drawing it this way, this alkene stays E and this alkene stays Z. So we have our E and Z configurations being maintained. So this here represents my compound B. So it can get kind of tricky. Sometimes it's not just simply removing the X group of the carbon halide and the BY2 group of the organoborane compound. We have to take into account things such as configuration. E and Z configuration. If the R1 and the R2 groups are vinylic. So here, one was E, that's maintained. The other one was Z, that's maintained. Now that we've seen this practice question, attempt to do this additional practice question. Here it says beginning from 1-pentyne, synthesize the following compound via a Suzuki coupling reaction. So what we've just seen, is supposed to give you inspiration in terms of creating this product here. So attempt it on your own. If you get stuck, don't worry. Come back and see how I approach the same exact question.

Beginning from 1-pentyne, synthesize the following compound via a Suzuki Coupling Reaction. 

So here it says, beginning from 1-pentyne, synthesize the following compound via a Suzuki coupling reaction. Alright. So, 1-pentyne. If we're looking at this structure, we should know that 1-pentyne looks like this. And, if we look carefully. Here are 5 carbons. So this was 1-pentyne and somehow it got connected to this benzene ring. Now realize in the practice above us, we had started out with a triple bond. And what did we do? We used catecholborane or some type of borane derivative to do a hydroboration oxidation, at least the hydroboration portion. We didn't really do the oxidation portion. We added H and By2 by anti-Markovnikov and syn addition. So, we're going to take the same kind of approach here. We're going to start out with our 1-pentyne. We're going to react it with our catecholborane. And we're going to add it by anti-Markovnikov and syn addition. So when we do that, we should get a double bond here now. And then we're going to add H to the more substituted alkyne carbon. And then the By2 of the chemical borane is going to go to the less substituted part. So, the hard part is also drawing the structure. Okay. Something like that. And then, we still have the other hydrogen, this hydrogen still there. So we got this so far, and now all we have to do is find a way of connecting this new alkene to this benzene part here. So, just make it simple, we're going to bring in our benzene ring with this ethyl group already attached. And then remember, what's the fundamental thing that's happening within a Suzuki reaction? Well, the X group of our carbon halide is lost and the By2 of our borane compound is lost. We have our organoborane here. All we need now is our carbon halide. So just place a halogen here, a Br right there. And then what's going to happen is we're going to lose this Br and we're going to lose this By2 group. This right here is R2 and this portion here is R1 and they're going to connect together. So that’s how we’re going to get our answer at the end. And realize here that this is an E-alkene. And here when we drew it, it was still an E-alkene. So everything was done correctly. So this question looked a little bit challenging but it really wasn't that bad. We just had to realize that we're starting out with an alkyne, so we take the same approach that we did in practice question 2. We would use the catecholborane in order to convert our alkyne into an organoborane or boron compound. It would remain an E-alkene, and then all you have to do is introduce a carbon halide that looks very similar to the final product that we want. So we needed that benzene ring with that ethyl, so just put a Br on it as well. We'd lose the Br, we'd lose the boron portion of our organoborane compound. R1 and R2 would combine, and we get our final answer at the end. So, guys, just remember these fundamental steps that we went over, and then you can do other types of Suzuki coupling reactions on your own.

Here’s what students ask on this topic:

What is the Suzuki reaction and what are its key steps?

The Suzuki reaction is a cross-coupling process that combines a carbon halide with an organoboron compound to form conjugated products like alkenes and styrenes. The reaction involves three key steps: oxidative addition, transmetalation, and reductive elimination. In oxidative addition, the carbon halide binds to a transition metal catalyst. During transmetalation, the organoboron compound transfers its group to the metal. Finally, in reductive elimination, the final product is formed, and the catalyst is regenerated. This mechanism emphasizes the importance of creating more conjugated products and adhering to the 16 or 18 electron rule for transition metals.

What types of compounds are typically formed in a Suzuki reaction?

The Suzuki reaction typically forms conjugated compounds such as alkenes, styrenes, and biaryls. Alkenes are hydrocarbons containing carbon-carbon double bonds. Styrenes are characterized by a benzene ring connected to two alkene carbons. Biaryls consist of two aromatic rings connected by a single bond. These products are valuable in various fields, including pharmaceuticals, materials science, and organic synthesis, due to their stability and electronic properties.

What role does the transition metal catalyst play in the Suzuki reaction?

The transition metal catalyst, often palladium, plays a crucial role in the Suzuki reaction. It facilitates the three key steps: oxidative addition, transmetalation, and reductive elimination. In oxidative addition, the carbon halide binds to the metal. During transmetalation, the organoboron compound transfers its group to the metal. Finally, in reductive elimination, the final product is formed, and the catalyst is regenerated. The catalyst's ability to undergo these transformations repeatedly makes the reaction efficient and sustainable.

What are the common leaving groups in the Suzuki reaction?

In the Suzuki reaction, common leaving groups for the carbon halide include chlorine (Cl), bromine (Br), iodine (I), and triflate (OTf). These groups are good leaving groups because they can stabilize the negative charge after dissociation from the carbon halide. Their presence facilitates the oxidative addition step, where the carbon halide binds to the transition metal catalyst, making the reaction more efficient.

What is the significance of the 16 or 18 electron rule in the Suzuki reaction?

The 16 or 18 electron rule is significant in the Suzuki reaction as it helps ensure the stability of the transition metal catalyst. Transition metals tend to be most stable when they have 16 or 18 valence electrons. During the reaction, the catalyst undergoes changes in its electron count through oxidative addition, transmetalation, and reductive elimination. Adhering to this rule helps maintain the catalyst's stability and efficiency, allowing the reaction to proceed smoothly and regenerate the catalyst for repeated cycles.