Cross-Coupling General Reactions - Online Tutor, Practice Problems & Exam Prep

Overview of Cross-Coupling General Reactions

In this video, we take an overview of cross-coupling reactions. Now, these reactions involve synthetic transformations that combine a coupling agent with a carbon halide. Now, there are two main driving forces at work that help these reactions occur. The first major driving force is the formation of highly conjugated products. Remember, conjugation here has to do with alternating double and single bonds. The more conjugated a compound is, the more stable that compound will be. The second driving force is that the transition metal catalyst that's involved in these coupling reactions, that transition metal is aiming to follow the 18 or 16 electron rule. Remember, for main group elements, they're trying to follow the octet rule to become more like a noble gas. But for transition metals themselves, they're aiming for 18 or 16 electron rules. Eighteen is the ideal number because that makes them more like a noble gas. But remember, valence groups from 8 to 11 are okay with having just 16 electrons, in particular, palladium and nickel.

Now, here we have the basic setup for a cross-coupling reaction. So, if we take a look here, we have R₁X. X can stand in for halogen. So this is our carbon halide. Then we have R₂−C. R₂−C here, this represents our coupling agent. For this reaction to occur, we need to use a metal catalyst. M here represents a transition metal. L is the ligand attached to it. N just means that the number of ligands attached to the transition metal. Now, the general premise of this reaction and these coupling reactions as a whole is for us to combine R₁ with R₂. So, these two combine together. And that gives us R₁−R₂, and this represents our coupling product. Now, since those two are combining, what happens to our X and to our NR? Well, those are just byproducts, which we're not concerned with. So as we investigate different types of coupling reactions, they'll try to stay as close to this general setup where we have R₁ and R₂ combining and X being byproducts. R₁ and R₂ represent carbon groups, so these coupling reactions are just a new way for us to form carbon-carbon bonds. So, we're just expanding the ways in which we can connect different carbon groups together. Now that you've seen the basic setup for a cross-coupling reaction, click on to the next video and see if we can delve deeper into understanding what is R₁ and what is R₂, what is C, and what is X in terms of all of these reactions.

Overview of Cross-Coupling General Reactions

So we've seen the general layout of a cross coupling reaction, where we have R1 combining with R2 through the use of a transition metal catalyst, and then our byproduct being C and X. But what exactly can R1 be? What exactly can R2 be? Well, we're going to say that the R1 group in the reaction, we know that it's a carbon halide. Well, that R portion could be an alkyl halide, it could be a vinylic halide, so a vinyl halide. And remember, vinyl just means a double-bonded carbon, so an alkene carbon. It could be an aryl halide, so a benzene connected to a halogen. Or it can be a benzylic halide. Now, when it comes to R1, so remember the general setup is R1X, R1 in all these cross-coupling reactions, R1 will be either a vinyl halide or vinylic halide or an aryl halide. And depending on the type of cross-coupling reaction we have, the carbon halide has the potential to be others as well. It has the potential to be an alkyl halide, an alkyl halide, and a benzylic halide. Now, remember one of the driving forces of these reactions is to create a highly conjugated product. When it comes to alkyl halides, they don't have pi bonds involved. So this connecting to the R2 carbon group, it's not going to make a more conjugated product. Therefore, alkyl halides can be present in some of these coupling reactions, but to get it to happen with the alkyl halide, it's very hard. So additional work needs to be done in order for it to happen. So for the most part, when it comes to these cross-coupling reactions, they're going to have these 2 components as possible reactants, and then they may have these 2. One of these 2 or both of them. This is very rarely used. Because, again, we're trying to make more conjugated products. If using an alkyl halide, that's not possible. Now, for R2, R2 in these reactions, so remember R2 is part of our coupling agent, R2 has the potential to be a vinylic or vinyl coupling agent or an aryl coupling agent. Okay. It's usually one of these 2 that are possible, and also, it could represent one of these potential 3 here. But again, when we use an alkyl group, it's not going to make a more conjugated product, so alkyl groups are less likely to be used. Now the cross-coupling agent, the C group in the reaction like we said, it all depends on the type of coupling reaction that is occurring. Okay. So our carbon halide, the R1 and R2 groups, they change with the different types of cross-coupling reactions that we're going to see. But don't worry, we're going to organize a list so that we can see the similarities in cross-coupling reactions, similarities and differences between them. Now that we've laid out the basic groundwork for a cross-coupling reaction, we're going to be ready to do examples where we're going to see can we follow the pattern that we see up here for general cross-coupling reaction. Can we spot the R1 group? Can we spot the R2 group? And then just combine them together to give us our final product. That's going to be the key to all these cross-coupling reactions. Spot those 2 groups, connect them together. C and X will be your byproducts and waste. So we don't really worry about them.

Overview of Cross-Coupling General Reactions Example 1

Before we take a look at the example, let's quickly go over again the general setup of a cross-coupling reaction. Remember, we have our R₁ connected to X, which represents our carbon halide, and then we have R₂ connected to C, which is our coupling agent. The whole idea is to use our transition metal catalyst to get R₁ to connect to R₂, which is our coupling product. Remember, C and X will be our byproduct within these coupling reactions. Now, let's take a look at the example. Here it says, the Heck reaction is a well-known coupling reaction that involves the combining of a carbon halide with an alkene. Based on the example provided, determine a possible coupling product. Alright. So if we take a look here, we can see that here we have bromobenzene, which represents our carbon halide, and here's our alkene. We'd say here that our R₁ would just be our benzene ring. We'd see that our R₂ is just the portion of the alkene that we want to attach to it. Our C here would be the hydrogen. So we've color-coded things so that it's easier for us to spot what's supposed to connect together. And then our X is just RBr here. Remember, the whole idea is to connect R₁ to R₂. So we're going to take this benzene ring here, and it's going to replace this C group so that it can connect to this alkene structure. Okay. So that's what we're going to do. So I'm going to draw my R₂ group, which is my alkene structure. And then let's add the benzene ring here. Now remember, you could write Ph instead of drawing the entire benzene because Ph is an abbreviation for benzene, or you can just draw the benzene. It's not that bad. So this would be our final product. And, remember, the whole premise here is to create an R₁, R₂ product that is more conjugated. So that provides greater stability. Within this Heck reaction, we have all these reagents being used, but our transition metal here is the palladium. So this would be our transition metal catalyst. Later on, when we go over the Heck reaction, we'll talk about how the other components, the other reagents, are being used in tandem in order to produce our final product here. So now that you've seen this example, look to see can you solve the practice question below it. Attempt it on your own, come back and see does your answer match up with mine.

In the Stille reaction, an organostannane compound reacts with a carbon halide in order to form a new carbon-carbon bond. What would be the final product from the following Stille coupling reaction?

The following practice question says, in the Stille reaction, an organostanane compound reacts with a carbon halide to form a new carbon-carbon bond. What would be the final product from the following Stille coupling reaction? Now, these reactions really just use organotin. So organostanane just means that we have organotin compounds. Now if we take a look here, we can see that our carbon halide in this case we're using is a vanillic halide. Specifically, a vanillic iodide compound. We see here that this is our coupling agent that's being used. And then here we have our transition metal catalyst. So, we're going to say here that our R1 equals the vanillic group. Our R2 is also a vinyl group. Remember, that's just an alkene group. We see that in this case, our C is Sn connected to 3 butyls. BU is just shorthand for butyl. So, we have a tin connected to 3 butyl groups. And our X here is the halogen of iodine. And then, again, remember, this is our transition metal complex or transition metal catalyst that's being used for R1 and R2 to combine. Remember, the whole premise is for R1 to combine with R2. So, what we’re going to have here is we're going to say that the C, which is SnBu3, will be replaced with the R1 group here of the carbon halide. So, our product at the end will look like this. Alright. So, we're connecting this carbon here which is this carbon here. It's double bonded and then it's connected to these 2 remaining carbons. So that would be our final product for this reaction. Now remember the whole premise is for R1 to combine with R2. Really, R1 is replacing, whatever the C group is and we've seen that in these two reactions, C has changed. In the first one, C was a hydrogen. But in this one, it represents tin bonded to 3 butyl groups. Just remember, the coupling reactions are different from one another, but they share some similarities where we use a transition metal catalyst and we're connecting R1 with R2. So just remember those fundamental principles and you'll be able to predict your final product.

Cross-Coupling Reaction Mechanisms

In this video, we're going to take a look at cross-coupling reaction mechanisms. Now, the detailed mechanisms for many of these reactions are still debated, but it is accepted that all of them follow 3 stages. These stages include oxidative addition, transmetalation, and reductive elimination. Now all the coupling reactions will have some form of these three stages. Oxidative addition isn't always going to be the first step but it will be found somewhere in the beginning of these reaction mechanisms. Transmetalation will be found somewhere in the middle of all these coupling reactions and then reductive elimination will be near the end of these coupling reactions. Now starting out with oxidative addition, we're going to say a transition metal complex where 𝐱 represents a transition metal and ℓ is just a set number of ligands attached to it right from the beginning, usually 2 or 4, reacts with a carbon halide by inserting itself into the 𝐼₁𝑱 bond. So this represents our carbon halide.

Now, this step can happen by a variety of mechanisms. A concerted effort or concerted process means it happens all at once, so it's a one-step process. So here, if we're taking a look at a generic oxidative addition step, we have our carbon halide and we have our transition metal complex. Now, the transition metal has a lone pair from its 𝑨 orbitals, and it uses them to connect to the halogen or our 𝑱 group. While it's connecting to that 𝑱 group, at the same time, the bond between 𝐼₁ and 𝑱 will break and it will connect to the transition metal. So what winds up happening is we're going to have our transition metal connected to a set number of ligands and then we're going to have 𝐼₁ attached to the transition metal and 𝑱 as well. So this represents the oxidative addition step.

Now, in this process, we're going to say both of the new bonds formed, so the bond here and this bond here, they behave like 𝑱 type ligands. And as a result of behaving like 𝑱 type ligands, they cause the electron count to increase by 2. So as we do examples of oxidative addition, we'll see that the electron count of my transition metal complex is going to increase by 2. We're going to say here also recall this part of the cycle is driven by the 18 or 16 electron rule. Remember, transition metals want to try to reach 18 electrons ideally and 16 in other cases. So they'll welcome the addition of your 𝐼₁ and 𝑱 group to the transition metal complex because it gets them closer to these ideal values of 18 and 16 for the transition metal. Now that we've seen this generic layout of oxidative addition, click on to the next video and see how we approach the example where we're asked to create the new complex for palladium after oxidative addition.

Cross-Coupling Reaction Mechanisms Example 1

So the example states, determine the new palladium complex that forms during this oxidative addition step. Alright. So we have our palladium, it uses its d orbital electrons, and it uses them to attach to the bromine. At the same time, this bond breaks and it attaches right back to the palladium. So our structure after oxidative addition would look like this: We'd have our palladium connected to those 2 carbonyls, and remember, it's the oxygen of the carbonyl that's attaching, and then an attachment to this alkene carbon part. Initially, we started with a 14 electron complex. Now we are going to use the electron count formula to see what our new electron count total would be here. So electron count, remember that is valence of the transition metal minus the charge of the complex plus the number of x ligands plus 2 times the number of l ligands. Alright.

The electron configuration of palladium is Krypton 4d¹⁰. So it has 10 valence electrons. Remember, for a transition metal, the number of valence electrons is your s orbital electrons plus your d orbital electrons together. Palladium only has d orbital electrons which total up to 10. Minus the charge of these complexes that are made in the oxidative addition step, they're neutral. So our charges are 0. Plus, now we need the number of x ligands. Remember, we said up above that the new bonds formed behave like x type ligand bonds. So we'd say that this would be an x type ligand and bromine, of course, is an x type ligand. So that'd be 2 X type ligands plus our carbonyl groups, our L type ligands, and there are 2 of them. So 2 times 2. So that'd be 10 plus 2 plus 4 which gives us 16 electrons. So palladium here is following the 16 electron rule. Remember, valence groups from 8 to 11 tend to be okay with having just 16 electrons instead of the optimal 18 electrons, especially palladium and nickel. So here, this would be our structure after the oxidative addition step.

Now that you've seen this, look to see if you can figure out the practice question below. If you get stuck, don't worry. Just come back and see how I approach the same practice problem.

Determine the new cadmium complex that forms during this oxidative addition step. 

Determine the new cadmium complex that forms during this oxidative addition step. Now, cadmium, we're told that this is a 14-electron complex. But why is that? If we look up the electron configuration of cadmium, we'd see that it is argon, 5s24d10. So, that would mean that our valence electron count for cadmium is 10+2, which is 12. Now, minus the charge of the complex, which has no charge so it's 0, plus the number of x-type ligands. Chlorines are x-type ligands, so there are 2 of them. Plus 2 times the number of l-type ligands, which there are none. So this comes out to 14 electrons. So that's how we have a 14-electron complex. Now one of the main driving forces of coupling reactions is the transition metal trying to obtain an 18 or 16 electron count. Now what's going to happen is our cadmium, which has a lone pair from its d orbital electrons, will use them to connect to our triflate group here which is an excellent leaving group. At the same time, it's going to cause this bond to become destabilized. It's gonna break and attach to the cadmium. At the end of this, we're gonna have our cadmium still connected to the 2 Cl groups. Now connected to our triflate and also connected to our benzene ring. For simplicity, you could have also just written Ph to represent the benzene ring. Now, what effect does this have on our electron count now for this new complex? Well, the valence electron count for cadmium is still 12. The structure is still neutral, so 0. Remember that the 2 groups being added, they act as 4 x-type ligands attached to cadmium. Plus 2 times the number of l-type ligands which is 0. So we have 16 electrons now. So our new electron count is 16. So this is again, one of the driving forces that's pushing these oxidative addition steps. Our transition metal wants to get to 18 or 16 electrons. Now, going back to triflate. So triflate, remember, is just trifluoromethane sulfonate. So it'd be represented as CF3SO3-. We're going to say here that triflate, which is a common term. So trifluoromethanesulfonate here, it comes from the strong acid, triflic acid. So remember, triflic acid is known as a super acid. It's even stronger than sulfuric acid. So this is a very, very strong acid. If it's a very strong acid, that means that this is a very weak conjugate base and remember weak bases, weak conjugate bases make excellent leaving groups. So, we're going to see in a lot of these coupling reactions, Cl, Br, and triflate being used as our leaving groups. They all come from strong acids. Therefore, they're gonna be very weak conjugate bases and therefore very good leaving groups. But going back to oxidative addition, remember the goal is for the transition metal to obtain either an 18 or 16 electron count in order to become closer to like a noble gas in terms of the number of electrons it possesses. Alright. So just remember that when it comes to oxidative addition, we have 2 arrows. 1 extending from the transition metal to the x group or leaving group. And then we have the breaking of the bond and then reconnection with the transition metal. And this helps to increase the electron count of the transition metal.

Cross-Coupling Reaction Mechanisms

The transmetalation step normally occurs after our oxidative addition step. We're going to say within this step we basically have the R₂ group of our coupling agent connecting to the transition metal complex while at the same time our X group leaves. Now our X group represents a good leaving group, so normally X is represented by chlorine, bromine, iodine or a triflate. If we take a look here, we have our R₂ group leaving, breaking this bond and connecting to our transition metal M. At the same time, this bond here breaks and X connects to our C group. So, as products, we're going to have still our R₁ group connected to our transition metal, which is connected to a set number of ligands, which, remember, is normally 2 or 4 at the beginning of the reaction where the X group left are R₂ attaches. Then as a byproduct, which we're really not concerned with but we're going to draw here just so that you know it's still around. We have our C group connected to our X group. So if you really want to think about transmetalation, it really just is one ligand coming in so that another ligand can leave. And those ligands are sometimes attached to 2 metals or to metalloids.

Now that we've seen this basic generic setup for the step, let's apply it to this image here. So, we created this structure here in the oxidative addition step. Now what we're going to do is we are going to expose it to a coupling agent. So here's my coupling agent here. So, the blue portion is my R₂ group. So, what's going to happen here is the bond between my R₂ group and the C group is broken and we're going to attach it to my transition metal, in this case palladium. While it does that, we're going to have this bond here break so that bromine can attach to the carbon. Okay. So, we're going to just have it attached to the carbon. And actually, let's redraw those arrows so it's not as messy. So, we have this bond here break, connect to the palladium, and this bond here breaks and attaches to the C. So what we wind up getting is we're going to have our 2 carbonyls still connected to palladium. And then we're going to still have this alkene group there. And there, where the halogen was, is now where we're going to place the R₂ group. And then we'd have our byproduct, which would just be our C group connected to our Br. So, that's basically how we would set up this transmetalation step. So, just remember, it all boils down to one ligand attaching to the transition metal of our complex so that the X ligand can leave. Okay. So our X group here could represent chlorine, bromine, iodine or a triflate group.

Cross-Coupling Reaction Mechanisms Example 2

So here it says, determine the products formed from the following reaction sequence. Alright. So the first step deals with oxidative addition. We know that in oxidative addition, the transition metal uses its lone pair, which it gets from its d orbital electrons, and it uses them to attach to our X group, which in this case is the bromine. At the same time, it causes this bond here to break and attach right back to the palladium. So for our compound A at this point, we're going to have our palladium, which is connected to the 2 acetate groups. Then we're also going to have the bromine being attached. You could place the bromine to the top or bottom; it doesn't matter. And then palladium is also going to be attached to this alkene group right here. So we're going to attach that like so. So that would be our compound A after the oxidative addition step.

Now, after the oxidative addition step, we move into transmetallation which remember is just really one ligand coming in and attaching to the transition metal so that one of our X groups, which in this case is bromine, can leave. So we're going to continue this. So we're going to write this here, well actually we're going to do it in blue. Alright. So what's going to happen here is remember, we're going to have this bond breaking so that we can attach to the transition metal, the palladium. At the same time, we have this X group leaving and attaching to my C group which in this case is the boron. So what we're going to get here for our compound B is we still have the palladium, still connected to both of its acetates. We still have this portion down here. And now we have this new group that's being added. Okay. So it's this carbon that's attaching now to the palladium. And then we draw the remaining carbon chain. So this would represent my transmetallation product B. We also have a byproduct. We could write down the byproduct as well. So we'd have the boron connected to 3 of these OC(CH₃)₃ groups and then the Br attached to the boron. Okay. So that would just be our simple byproduct.

Just remember, we have our oxidative additional step, which normally is one of our first steps within a coupling reaction, and then normally that's followed up by a transmetallation step which just has the incoming R₂ group attaching to our transition metal M so that we can basically have the X group leave in the process.

Cross-Coupling Reaction Mechanisms

Reductive elimination can be seen as one of the last steps within a typical cross-coupling reaction. Within this step, we have the R₁ group, which was originally part of our carbon halide, and the R₂ group, which is originally part of our coupling agent, finally leaving our transition metal complex and forming a sigma or single bond with one another. Now, this step can be seen as the opposite of oxidative addition. Remember, in oxidative addition, we have ligands bonding to our transition metal complex in order to get that transition metal closer to the 18 or 16 electron rule. Within this step, we have the exact opposite. We have those same ligands finally leaving our transition metal complex, and we're turning it back to a count that's not 18 or 16 under normal cases.

Now we're going to say that elimination usually signifies the formation of a pi bond, but reductive elimination doesn't always do that with every cross-coupling reaction. Now, the way it works is we're going to have our R₁ group leaving our transition metal M and bonding to our R₂ group. At the same time, we have the bond between R₂ and M breaking, and the electrons going to M. This creates R₁ single bonding to R₂. Then we're going to have the regeneration of our transition metal complex or transition metal catalyst. Now, this is a good thing because remember, the catalyst is not consumed within the reaction so it had to return to its original form. Also, this gives extra motivation for the catalyst to undergo another cycle of cross-coupling reactions because, remember, it's no longer following the 18 or 16 electron rule, so it desperately wants to undergo oxidative addition again so that its transition metal can get closer to the 18 or 16 electron rules. So that's how we can keep this reaction going and going. Also, by going through the reaction again, we have a chance to make more conjugated products later on. So those two driving forces can help this reaction go through another cycle.

Now, this step also follows stereochemical rules. So, with stereochemistry, we're going to say the reductive elimination generally results in the retention of stereochemistry when it comes to double bonds. So if your alkene has E or Z configurations, it tries to retain that as much as possible in terms of this last step. So remember, with a typical cross-coupling reaction, we have 3 steps. We have oxidative addition, usually somewhere in the beginning, transmetalation near the middle, and at the end we have our reductive elimination step. Now these three steps will be found in some way within these cross-coupling reactions. Some may contain additional steps, but these three steps here of oxidative addition, transmetalation, and reductive elimination form the foundation for a lot of the reactions we'll see later on. So just keep them in mind and keep in mind some of the rules that we covered in terms of how they operate.

Cross-Coupling Reaction Mechanisms Example 3

So here it says to determine the final product in the following reaction. Alright. So, we have to undergo reductive elimination. Up to this point, we've added our alkene group, which was our R1 group. We've added this alkene group which is our R2 groups. Now, we have to get them to combine together. Alright. So, we're going to say that our R1 group, which is this group up here, breaks off from my transition metal complex and combines with this carbon here. At the same time, this bond here breaks and goes back to the palladium. So, we wind up getting this structure with that alkene group attached to where the palladium was once attached. And we do it this way in order to maintain or retain our stereochemistry. This alkene here had an E configuration. And by adding the R1 group up here, it maintains that E configuration. We also have the regeneration of our palladium catalyst. Those will be our 2 final products.

Now, what we need to realize here is that reduction is seen as the gaining of electrons and a decrease in an element's oxidation number. Now, when the 2 X-type ligands, because remember, the R1 and R2 are acting like ligands but more specifically X-type ligands, when they're lost, the oxidation state of the transition metal decreases by 2. Remember, this bond here breaks and those 2 electrons go back to the palladium. Therefore, the oxidation state decreases by 2 because you gain those 2 electrons back. Now, at the same time, the formation of a conjugated product, our conjugated product here, allows for the unstable catalysts to be regenerated. So here we just regenerated it. It's great that we made something that's more conjugated, therefore, we made a very stable product. But we wound up regenerating our transition metal catalyst which is going to want to go through another wave or another cycle of a cross-coupling reaction in order to get this transition metal back within the 18 or 16 electron count. Okay. So these are our driving forces. We want to make something more stable as a product, but then the transition metal wants to get to 18 or 16 electrons. Those two forces are what pushes and propels these typical cross-coupling reactions.

Cross-Coupling Reaction Mechanisms

So as I mentioned earlier, by the time we get through the reductive elimination step and regenerate the transition metal catalyst, there's a drive or desire by that transition metal to go through another round or cycle. So, it wants to undergo oxidative addition yet again. Now, if we take the 3 stages that we talked about, oxidative addition, transmetalation, reductive elimination, we'll see that they help to form a catalytic cycle where we basically have product formation and catalyst regeneration. So, if we were to put it within a catalytic cycle, we'd start out with our transition metal catalyst here where \(M\) is our transition metal and \(L\) is just the number of ligands attached to it. In this case, 2. Remember, in the first step which is oxidative addition, we have our transition metal with its lone pair that comes from its d orbital electrons. It's going to bond to our \(X\) group causing this bond here to break and it also connects to our transition metal \(M\). So, at this point here we're going to have our transition metal \(M\), connected to the 2 ligands. Now connected to \(R_1\) and connected to our \(X\) group.

Next, we're going to go into the transmetalation step which introduces our coupling agent. Remember our coupling agent is \(R_2\) connected to our \(C\) group. In this reaction, what's the basis? What we're doing here is we're trying to connect \(R_2\) to our transition metal \(M\) and when it does so, we have the \(X\) group leaving and bonding to our \(C\) group. So \(X\) will be bonded to our \(C\) group. And that'll be just byproduct. So where the \(X\) was, \(R_2\) now, will be located. So we still have our transition metal and with its 2 ligands. We still have our \(R_1\) connected, and now we have our \(R_2\) connected. Then finally, we have reductive elimination, where \(R_1\) and \(R_2\) bond together. So \(R_1\) will connect to \(R_2\). And we have the regeneration of our catalyst.

So the 2 driving forces, we have our transition metal obtaining 18 or 16 electron rule at the oxidative addition step to get the whole process rolling and then the reaction is pushed forward till the end so that we can make a more conjugated, more stable product. So remember, these three steps form the foundation for the coupling reaction cross coupling reactions that we're going to encounter. Some may have additional steps in beginning or at the end. But these three steps themselves are the main things that you need to take to heart when it comes to understanding the conjugated, more stable product and for the transition metal of the transition metal complex to obtain the 18 or 16 electron rule. Remembering these steps is key to understanding the cross coupling reactions we'll see later on.