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 Hallett. 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 catalysts 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, 18 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 um set up for a cross coupling reaction. So if we take a look here, we have our one dash X X. Can stand in for halogen. So this is our carbon Hallett, Then we have two for this reaction to occur. We need to use a transition metal catalyst. M here represents a transition metal. L is the Ligon attached to it and just means that the number of Liggins attached to the transition metal can vary for many of these coupling reactions. we traditionally see two or four Liggins attached to the transition metal. Now the general premise of this reaction and these coupling reactions as a whole is forced to combine our one with our two. So these two combined together And that gives us our 1.2 and this represents our coupling product. Now, since those two are combining, what happens to our X. And two. R N R. C. Well, those are just byproducts which we're not really 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 R. One and R. Two combining and cnx being byproducts R one and R. To 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 cross coupling reaction. Click onto the next video and see if you can um we can delve deeper into understanding what is R. One and what is our to what is C. And what is X. In terms of all of these reactions

So we've seen the general layout of a cross coupling reaction where we have our one combining with our two through the use of a transition metal catalyst and then our byproduct being cnx. But what exactly can R. One B. What exactly can our to be? Well we're going to say that the R one group in the reaction, we know that it's a carbon Halide. Well that our portion could be an al kiel khallad, it could be a vanilla khallad. So a vinyl Halide. And remember vinyl just means a double bonded carbon. So when our king carbon um it could be an alpa Neil hala. So it's a carbon that's triple bonded, it could be in a real khallad. So a Benzene connected to allergen or it can be a Benz Ilic Ben's Illich Khallad. Now when it comes to our one, so remember the general setup is our one X. R. One. And all these cross coupling reactions are one will be either a vinyl howard or vanilla Callard or in a real Hallett. And depending on the type of cross coupling reaction we have the carbon. Hallett has the potential to be others as well. It has the potential to be and I'll kill howard and al Kamil Khallad and Ben's Illich Hallett. Now remember the driving one of the driving forces of these reactions is to create a highly conjugated product when it comes to our kill mallards, they don't have pi bonds involved. So this connecting to the r to carbon group, it's not going to make a more conjugated product. Therefore alcohol holidays can be present in some of these coupling reactions but to get it to happen with the alcohol 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 gonna have these two components as as possible react ints and then they may have these too. One of these two are both of them. This is very rarely used because again, we're trying to make more conjugated products if using an alcoholic, that's not possible Now, for our two are two of these reactions, remember R two is part of our coupling agent. R two has the potential to be a vanilla core vinyl coupling agent or in a real coupling agent. Okay, it's usually one of these two that are possible. And also it could represent one of these potential three here. But again, when we use in our kill group, it's not gonna make a more conjugated product. So how kill groups are are less likely to be used. Now, the cross cup, the coupling agent C group in the reaction. Like we said, it all depends on the type of coupling reaction that is occurring. Okay, so our carbon halid the R one and R two groups, they change with the different types of cross coupling reactions that we're going to see. But don't worry, we're gonna organize a list so that we can see the similarities between cross coupling reactions, similarities and differences between them. Now that we've laid out the basic groundwork for a cross coupling reaction, we're gonna be ready to do examples where we're gonna see. Can we follow the pattern that we see up here for general cross coupling reaction? Can we spot the R. One group? Can we spot the R. Two group and then just combine them together to give us our final product? That's gonna be the key to all these cross coupling reactions. Spot those two groups, connect them together, C. N. X. Will be your byproducts and waste, so we don't really worry about them.

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 our one connected to X. Which represents our carbon khallad and then we have our 22 connected to see which is our coupling agent. The whole idea is to use our transition metal catalyst to get our one to connect to our two, which is our coupling product. Remember, C. N. 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 halid with an all keen based on the example provided determine a possible coupling product. All right, So, if we take a look here, we can see that here, we have bravo benzene which represents our carbon khallad. And here's our alkaline. We'd say here that are are one would just be our benzene ring. We see that our R. two is just a portion of the alc in that we want to attach to it. R. 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 exes, just R. B. R. Here. Remember the whole idea is to connect our one, two R. Two. So we're gonna take this benzene ring here and it's going to replace this C group so that it can connect to this al keen structure. Okay, so that's what we're gonna do. So I'm gonna draw my R. Two group, which is my ALC in structure. And then let's add the benzene ring here. Now remember you could write ph instead of drawing the entire benzine because P. H. 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 R one R two product that is more conjugated. So that provides greater stability within this heck reaction. We have all these re agents 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 re agents 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 attempted on your own. Come back and see does your answer match up with mine

in this video we're gonna 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 three stages. These stages include oxidative addition, transmit elation and reductive elimination. Now all the coupling reactions will have some form of these three stages. Oxidative addition isn't always gonna be the first step but it will be found somewhere in the beginning of these um reaction coupled mechanisms transmit elation 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 gonna say a transition metal complex where M. Represents a transition metal and L. Is just a set number of Liggins attached to it right from the beginning. Usually two or four reacts with a carbon halid by inserting itself into the R. One X bond. So this represents our carbon Hallett. Now this stuff can happen by a variety of mechanisms but a concerted process is most common. Remember 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 khallad and we have our transition metal complex. Now the transition metal has a lone pair from its d orbital's and it uses them to connect to the halogen or X group while it's connecting to that X group at the same time the bond between R. One and X will break and it will connect to the transition metal. So what winds up happening is we're gonna have our transition metal connected to a set number of Liggins and then we're gonna have our one attached to the transition metal and X as well. So this represents the oxidative edition step. Now in this process we're gonna say both of the new bonds formed. So the bond here and this bond here, they behave like X. Type Liggins. And as a result of behaving like X. Type Liggins, uh they cause the electron count to increase by two. So as we're doing examples of oxidative addition, we'll see that the electron count of my transition metal complex is going to increase by two. We're gonna 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 R. One and X 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 onto the next video and see how we approach the example. We were asked to create the new complex for palladium after oxidative addition

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 have our palladium. It's connected to those two carbon deals. Remember? It's the oxygen of the carbon. Well that's attaching to the palladium, then we have an attachment to the bromine and then an attachment to this, our king carbon part. Initially we started with a 14 electron complex. Now we're gonna use electron 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 charge of the complex plus the number of X elegans plus two times the number of elegance. Alright, so the electron configuration palladium is krypton four D. 10. So it has 10 valence electrons. Remember for 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, Their neutral. So our charges are zero plus. Now we need the number of X Liggins. Remember we said he 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 Ligon. So that'd be two X type legans plus our carbon eel groups are L type Liggins and there's two of them. So two times 2. So that be 10 Plus two 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 edition 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.

the trans modulation step normally occurs after our oxidative addition step. We're gonna say within this step, we basically have the R. Two 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 trifle eight. If we take a look here, we have our our two group leaving, breaking this bond and connecting to our transition metal. M. At the same time, this bond here breaks and ex connects to our C. Group. So as products we're gonna have still are are one group connected to our transition metal which is connected to a set number of Liggins which remember is normally uh two or four at the beginning of the reaction Where the X Group left are two attaches then as a byproduct which we're really not concerned with. But we're gonna draw here, draw it 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 transmit elation, it really just is one leg in coming in so that another Ligon can leave. And those Liggins are sometimes attached to two metals or two metal Lloyd's. Now that we've seen this basic generic setup for translation step, let's apply to this image here. So we create, we created this structure here in the oxidative addition step. Now what we're gonna do is we're going to expose it to a coupling agent. So here's my coupling agent here. So the blue portion is my R. two group. So what's going to happen here is the bond between my R. Two group and the C. Group is broken. And we're gonna attach it to my transition metal in this case palladium. Well it does that we're gonna have this bond here break. So that browning can attach to the carbon. Yes we're gonna just have attached to the carbon. Um And actually let's redraw those arrows. So it's not as messy. So we have this bond here break connected the palladium and this bond here breaks and attaches to the sea. So what we wind up getting is we're gonna have our two carbon heels still connected to palladium and then we're gonna still have this our king groups still there and there where the halogen was is now where we're gonna place the R. Two group and then we'd have our by product which would just be our C. Group connected to R. B. R. So that's basically how we would set up this transmit elation step. So just remember it all boils down to one leg in attaching to the transition metal of our complex so that the X. Ligon can leave. Okay so our X. Group here could represent um chlorine bromine iodine or a trifle eight group

So here it says determine the products form from the following reaction sequence. Alright, so the first step deals with oxidative addition. We know that an 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 browning at the same time. It causes this spot here to break and attach right back to the palladium. So for our compound A. At this point we're gonna have our palladium Which is connected to the two acetate groups. Then we're also gonna have the bro men being attached, you could place the broom into the top or bottom, It doesn't matter. And then palladium is also gonna be attached to this working group right here. So we're gonna attach that like so so that would be our compound A after the oxidative edition step. Now after the oxidative addition step, we move into trans modulation which remember is just really one leg in 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 gonna continue this. So we're gonna write this here, we're actually gonna do it in blue. Alright. So what's gonna happen here is remember we're gonna have this bond breaking so that we can attach to the transition metal 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 gonna get here for our compound B. Is we still have the palladium still connected to both of its acetate. It's 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 out to the palladium and then we draw the remaining carbon chain. So this would represent my trans methylation product. B. We also have a by product. We could write down the by product as well. So we have the we have the boron connected to three of these O. Ch three's and then the B. Are attached to the boron. Okay, so that would just be our simple byproduct. So 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 transmit elation step which just has the incoming are two group attaching to our transition metal M. So that we can basically have the X group leave in the process

reductive elimination can be seen as one of the last steps within a typical cross coupling reaction. Within this step, we have the R one group which was originally part of our carbon Hallett and the R two group which was 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 Liggins 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 Liggins finally leaving our transition metal complex and we're turning it back to account 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 gonna have our our one group leaving our transition metal em and bonding to our our two group at the same time, we have the bond between our two and m breaking and the electrons going to em This creates our one single bonding to R two. Then we're gonna have the regeneration of our transition metal complex or transmission transition metal complex or catalyst. Now this is a good thing because remember catalyst is not consumed within a 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 edition again So that its transition metal can get closer to 18 or 16 electron rules. So that's how we can keep this reaction going and going. Also by going to 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 stereo chemical rules. So with stereo chemistry we're gonna say the reductive elimination generally results in the retention of stereo chemistry when it comes to double bonds. So if you're all keen 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 three steps. We have oxidative addition, um usually somewhere in the beginning and we have transmit elation near the middle, and at the end we have a reductive elimination step. Now these three steps will be found in some way. And within these cross coupling reactions. Some may contain additional steps, but these three steps here of oxidative addition, transmit elation 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

So here it says, determine the final product in the following reaction. All right, so we have to undergo reductive elimination. So, up to this point we've added our our king group which was our our one group. We've added this our king group, which is our two groups. Now we have to get them to combine together. All right. So we're going to say that our our one 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 our king group attached to where the palladium was once attached. And we do it this way in order to maintain or retain our stereo chemistry. This our keen here had an E configuration And by adding the R one group up here, it maintains that e configuration. We also have the regeneration of our palladium catalyst. So those will be our two final products. Now, what we need to realize here is that um reduction is seen as the gaining of electrons and a decrease in an element's oxidation number. Now when the two X type Liggins because remember the R one and R two are acting like Liggins but more specifically X type Liggins. When they're lost the oxidation state of the transition metal decreases by two. Remember this bond here breaks And those two electrons go to the palladium therefore decreasing the oxidation state by two because you gain those two electrons back Now at the same time, the formation of a conjugated product are conjugated product here allows for the unstable catalyst to be regenerated. So here we just regenerated this. So 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.

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 because it wants to retain that 18 or electron rule count. So it wants to undergo oxidative edition yet again. Now if we take the three stages that we talked about oxidative edition, trans methylation, reductive elimination, we'll see that they helped 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 al is just a number of Liggins attached to it in this case too. Remember in the first step which is oxidative edition 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 pond here to break. And it also connects to our transition metal M. So at this point here we're gonna have our transition metal M. Connected to the two Liggins Now connected to our one and connected to our X group. Next we're gonna go into the trans methylation step which introduces our coupling agent. Remember our coupling agent is our two connected to our C group in this reaction, What's the basis, what we're doing here is we're trying to connect our two to our transition metal M. And when it does so we have the X group leaving and bonding to R. C group. So X. Will be bonded to R. C. Group. And that'll be just by product. So where the X. was R. two now will be located. So we still have our transition metal and with its two Liggins we still have our our one connected. And now we have our our two connected. Then finally we have reductive elimination where R. one and R. two bond together so our one will connect to our two And we have the regeneration of our catalyst. So the two driving forces. We have our transition metal obtaining 18 or 16 electron rule at the oxidative edition step to get the whole process rolling and then the reactions pushed forward to 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 the 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 overall mechanism of any type of cross coupling reaction. Remember our two driving forces are creation of a more 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