Heck Reaction - Online Tutor, Practice Problems & Exam Prep

The Heck reaction involves the coupling of a carbon halide with an alkene using a palladium catalyst, resulting in a more substituted alkene. Key steps include oxidative addition, syn addition, and reductive elimination. The reaction is regioselective, favoring the less substituted position, and stereoselective, often yielding the E configuration. Reactivity of the alkene increases with electron-withdrawing groups, while the best yields occur with non-substituted alkenes. Understanding these principles is crucial for predicting products in organic synthesis.

Heck Reaction

In this video, we're going to take a look at the Heck reaction. Now, in the Heck reaction, we basically have the coupling between a carbon halide and an alkene through the use of a palladium catalyst. Now, we're going to say here that the R group of the carbon halide is substituted for a vinylic hydrogen of an alkene.

If we take a look on the left, we have the generic layout for a cross-coupling reaction. So in a typical cross-coupling reaction, we have our carbon halide, which is represented by R₁X, where X just represents a good leaving group. We have our coupling agent illustrated by R₂C. Along with this, we have ML_n where M represents a transition metal, L is just the number of ligands attached to it. Typically for these coupling reactions, the number of ligands is either 2 or 4. Remember, this reinforces the idea that our transition metal wants to get either the 18 or 16 electron rule being followed. So that's one of the main driving forces for these types of coupling reactions. Through the use of these three pieces, we have the creation of our coupling product, which remember is just the connecting of our R₁ and R₂ groups to get together. Then we're going to have here our CX, which is just our by-products that will be forming in these different types of coupling reactions.

Now by applying this generic layout for a cross-coupling reaction, we can take a look at the Heck reaction itself. In the Heck reaction, we still have a carbon halide. Next, we're going to say that our alkene serves as our coupling agent. Then we have the use of a palladium catalyst with some type of base. This leads to the creation of our more substituted alkene. Remember, creating more substituted, more conjugated products is our second type of driving force that pushes these coupling reactions forward. So here, this would be our coupling product. And then what we have here at the end is just our byproduct. If we're looking at this in the simplest type of way, we'd see that within the Heck reaction, we basically have the loss of our leaving group X and this vinylic hydrogen. They're lost so that my R₁ and R₂ groups can combine together to create a more substituted, more conjugated alkene.

Now if we take a look here, we're going to say that our R₁ group of the carbon halide here, if we apply it to our Heck reaction, we're going to say that R group is represented by a vinyl, aryl, or benzyl group. So when it comes to these Heck reactions, R will be represented by one of those 3 types of groups. Next, we're going to say here that R₂ group of our Heck reaction is just represented by the alkene group. When it comes to our C group with the Heck reaction, that's just represented by H. Now, what's going to be a common theme within all these coupling reactions is X, our leaving group. Here, X leaving group of the carbon halide is represented by chlorine, bromine, iodine, or triflate. So four good leaving groups. And in terms of the Heck reaction, the bases that are typically used are our acetate ion which is OAc for short. You could use hydrogen carbonate or we can say bicarbonate or we can even use triethylamine as a base.

Now we have a basic simple layout for a Heck reaction, but we need to remember that Heck reaction has some rules that it follows. Its regioselectivity and stereoselectivity are the 2 things that we must also take into account when dealing with Heck reactions. With regioselectivity, we're going to say the reaction is highly regioselective, so the R₁ group, going to the less substituted position of the alkene. So as we do examples and practice questions, we'll see how this is being applied. Then we have stereoselectivity. So if an E/Z configuration is possible for the alkene, then the reaction is highly selective often giving the E configuration Heck reactions, just remember, in the simplest of ways, we just have the X group of the carbon halide and the hydrogen or vinylic hydrogen of the alkene being lost in order to combine R₁ and R₂ together. Then we also have to take into account regioselectivity and stereoselectivity when approaching any type of Heck reactions. Now that we've laid down the basic groundwork for Heck reactions, we'll attempt to do some example questions dealing with how to find our final product.

Heck Reaction Example 1

So here it says, determine the product from the following Heck reaction. Alright, so if we take a look here, we have this as our first compound, which is our alkene. So this is our R₂ group. Then we're going to have this second structure here, which has the halogen on it. Remember that the halogen is a leaving group, so it represents our X group. Therefore, it's connected to all of this, which is our R₁ group. Remember, the basic setup is that we are going to lose the X group and the hydrogen of the alkene so that my R₁ and R₂ groups can combine. In this reaction, we also have our palladium catalyst in the form of Pd(OAc)₂. We have PPh₃, employed in conjunction with it. So, you can consider this as your catalyst on top. Then we have on the bottom our base which is triethylamine. Now remember regioselectivity. We have the reaction that's highly regioselective with the R₁ group going to the less substituted alkene position - the substituted position of the alkene. So we have hydrogens attached to this alkene. And remember, the less substituted position just means the alkene carbon with more hydrogens. This alkene carbon here has 1 hydrogen, and this one here has 2. So we're going to lose one of these hydrogens on this carbon. It's going to combine with the iodine and it's going to be lost so that we can combine R₁ and R₂ together. Now, here, it doesn't really matter which one is lost because there is no stereochemistry associated with this double bond. It's not E or Z. So H and our X are lost through the process so that R₁ and R₂ can combine. Now, the alkene of R₁ though has E or Z configurations associated with it. It is an E configuration which would mean that it needs to stay E at the end of this reaction.

So what I'm going to do here is I'm going to draw R₁ first. And to maintain configuration, all you have to do is just draw the R₂ group in the same position that the iodine was in, and in that way, it maintains its E configuration. So pretty simple if you approach it that way. So now we're going to connect this carbon here to this carbon here. Okay. So there goes my carbon, which is still connected to the other double-bonded carbon, which is still connected to this cyclopentane ring. So this here would represent my final product. So, again, the alkene of my R₂ group isn't E or Z, so it doesn't really matter. But the alkene structure of my R₁ group has an E configuration, and that has to be maintained because we're also dealing with stereoselectivity where it maintains, it tries to maintain its stereochemistry, but it greatly favors the E configuration as the major. Alright. So this would be our final product for this particular question. And that's the only approach you need to take if they're not asking for the mechanism. Later on, we'll go into greater detail on how the Heck, reaction mechanism exactly works to get our final product. But for now, just remember, look at it in a simple way. X and H are lost so that R₁ and R₂ are combined together. If you can do that, you'll get your final answer for the Heck reaction.

Heck Reaction

So remember that the coupling agent of our Heck reaction is basically the alkene. Now the reactivity of the alkene can be further increased or decreased depending on what's connected to it. Now we're going to say the reactivity of the alkene, which remember is our R2 group, it decreases with increasing substitution. And we're going to say here the highest yields for the Heck reaction happen when we have just a non-substituted simple ethylene compound as our alkene. So, this here. And when we have a mono-substituted with withdrawing groups. So here if we take a look, we have electron withdrawing groups attached to our alkene. Remember the alkene is our R2 group. And if we take a look here, we have a carbonyl with an R group. R could represent hydrogen, it could represent carbon groups. We have CN. We have our sulfonic acid group. ₂ₓ₃ where X is a halogen. We have NR³⁺ where R could be a hydrogen or a carbon. And then we have our nitrile group. For those of you who remember, these groups have something in common. Remember, these are meta groups, meta-directing groups that you've learned before when dealing with benzene reactions. And we're gonna say here that the first three here on the right, they're moderately deactivating or or moderately electronically drawing groups. And then these three here are very strong. And as the strength increases the reactivity of my alkene increases in terms of the Heck reaction. So we're going to say here that reactivity increases. Okay. So take that into consideration when dealing with a Heck reaction. We want an alkene in terms of our coupling agent. We either want it to be not substituted at all, so in its typical ethylene structure, or if we're going to have it substituted, just have it mono-substituted with an electron withdrawing or meta-directing group attached. The more deactivating or withdrawing the meta-director, the more reactive that alkene is towards the Heck reaction. So keep this in mind when taking a look and comparing the reactivity of different types of Heck reactions.

Heck Reaction Example 2

So here, based on what we've learned about the reactivity of the alkene, let's answer the following question. Here it says to rank the following alkenes in order of increasing reactivity towards the Heck reaction. So, what we have to do here is to identify what kind of groups are attached to our alkene carbons. So we have here a CN group, an NH₂ group. Here we just have methyl groups, and here, we have a CF₃ group. Here, we also have methyl groups as carbon groups as well.

Now remember, we say that the alkene's reactivity towards the Heck reaction is enhanced if it has electron-withdrawing groups attached to it. So, we need to identify if we have any electron-withdrawing groups attached to my alkene carbons. If we look, we see that CN is an electron-withdrawing group. We also see that CF₃ is an electron-withdrawing group. For C, all we have are methyl groups, which are weak electron-donating groups. And then in B, we have an NH₂ group. Remember, nitrogen here has a lone pair. Because of that lone pair, NH₂ is considered to be a strong electron-donating group. So here we have a strong electron-donating group, and here we have weak electron-donating groups.

So we want to go in order of increasing reactivity, so least reactive to most reactive. The least reactive would have to be B because that one is a strong electron-donating group. So if strong electron-withdrawing groups increase the reactivity of the alkene, then putting a strong electron-donating group would have the opposite effect; it would decrease the reactivity of that alkene towards the Heck reaction. So B would be last. Then we have C, which is electron-donating but weaker.

Then we have CN versus CF₃. In terms of our scale, we can see that CN is a weaker or more moderate electron-withdrawing group compared to CF₃, which is represented here by CF₃. So then we'd say A. So D would be the most reactive alkene towards the Heck reaction because it has the strongest electron-withdrawing group attached to it, CF ₃. So just remember, when it comes to the Heck reaction, we can more greatly increase the reactivity of the alkene towards the Heck reaction if we attach electron-withdrawing groups to it or if we keep it as just a simple ethylene molecule.

Determine the product from the following Heck Reaction.

So here it says determine the product from the following Heck reaction. Alright. We know that this is a Heck reaction because we have here our halogen. So the halogen or our triflates are always the easiest to spot. That would represent our X group. And if that's our X group, what it's attached to would have to be our R1 group. So all of this here represents R1. Next, we're going to say if that's R1, then over here this has to be R2. Remember, we're trying to replace one of the vinylic hydrogens on R2 with the R1 group. What you need to do next is you need to identify the vinylic carbons. So here's our double-bonded carbon, here's our vinylic carbon here, double bonded carbon. Look at the hydrogens that they have. This one has 1. This one here has 2. Now, remember, we're going to deal with regioselectivity so we're targeting the less substituted alkene carbon, so the alkene carbon with more hydrogens. So we're looking at this alkene carbon. Now, that double bond doesn't have E or Z configurations tied to it so we don't have to worry about that. But for the carbon halide group over here, this has an E configuration attached to it so we have to keep that in mind. Now, we're going to say here, the simplest way of looking at it is we know that one of the hydrogens on the vinylic, on the alkene carbon or vinylic carbon is going to switch out with the halogen, so they're going to be lost. And R1 and R2 are going to combine. Alright. So let's draw R1 first. So there goes R1. Then we know that we have to maintain this E configuration. And to do that, where the halogen is, R2 is going to replace it. So we're going to attach this carbon here to where the halogen was. So there it goes right there. That alkene carbon is still connected to a double-bonded carbon. And that double-bonded carbon here is still connected to this ester component. So draw that ester component. Here, it doesn't matter that I'm configuration is maintained in terms of my alkene. So this would be my final Heck reaction product. Alright. Now that we've seen this practice question, we'll move on to the next practice question where we're applying what we've learned to an intramolecular Heck reaction. Remember, intramolecular means that the reaction is happening on the same molecule and usually with intramolecular reactions, we make cyclic products. So, attempt that practice question on your own. If you get stuck, don't worry. Come back and see if your answer matches up with mine.

Predict the product formed from the following intramolecular Heck reaction.

So this question is a bit trickier because we're dealing with intramolecular Heck reactions. So we're going to say here that here goes our halogen which is our X component. And if that's the X, that means that this portion here has to be R1. And if that portion is R1, then over here is R2. Remember, we know that what happens is that one of the vinylic hydrogens on the less substituted alkene carbon for the R2 group is lost with the halogen. Now, when we learn the mechanism later, we'll learn in greater detail how that works, but remember, we're just looking for the H, the right H, and the halogen or triflate to combine together and be lost. And then what we have left is our Heck reaction product. Now to answer this question correctly, we're going to basically redraw this structure to see how easily the H and the Br are lost within the reaction. Alright. So we're going to draw this out. We're going to have this portion. So we're going to have this portion still be the same. Alright. So all I've done is drawn this part here. Now instead of drawing this carbon in the same method we have here, we're going to draw it down because remember, we're making a cyclic product. So this carbon here is this carbon here which is connected to this carbon, which is connected to this carbonyl group, which is connected to these two double-bonded carbons here. Now, remember, we're targeting the less substituted alkene carbons, so this end carbon here, which has 2 hydrogens, as opposed to this one that only has 1. And the simplest way to look at it is we know that in the process these two are lost. So then because those two are lost, now we're going to draw our new product. Okay. Try your best to draw this; you know, it's going to be a cyclic product. Okay. Okay. Then we have that. So this double bond stays where it is. And then realize here that, this carbon here is this carbon here. And this carbon here is this carbon here. So they're connected together now and then we still have this double bond here which we draw in. So there goes my final answer. Okay? So with a molecule that possesses both the R1 and R2 group, we enter the territory of an intramolecular reaction which typically creates a cyclic product. So here we have our final cyclic Heck reaction product as a result. So just keep in mind, find where the X group is, which is our halogen or triflate group. If that's X, what's connected to it has to be R1. Then if that's if you have R1, then the other alkene that you see somewhere else would be R2. It's your goal to connect R1 and R2 and in the process lose your vinylic hydrogen from R2 and the halogen or triflate from R1. The more of these that you do, the easier it'll get and the more second nature will kick in and you'll be able to get your final product.

Using ethyl-2-butenoate as your starting material, prepare the following compound by a Heck reaction. 

Using ethyl 2-butanoate as your starting material, prepare the following compound by a Heck reaction. Alright. So we're told what one of our two components is for this Heck reaction. All we have to do is figure out what the other compound is. Now 𝛺8 means that we're dealing with an ester. It doesn't have a halogen in its name. It doesn't make reference to a triflate in its name, so it doesn't possess an X group. Therefore, this is not my carbon halide as my starting material. This is my coupling agent. This represents R2. All we have to do now is draw what the carbon halide looks like. Remember, the carbon halide itself, remember your carbon halide is R1X. And remember, R1 is aryl, or benzyl group. So that means that the halogen is going to be attached to one of these three types of carbons. What we have to do here is we have to make a cut here. And we're making a cut there because making a cut there means that here is the FO₂B10O8 that we had as our starting material. And also, when I make the cut there this stuff here becomes my carbon halide. And where I made the cut, that's where I place my X group. In this case, I'll make it an iodine. But, again, remember, it could also be chlorine, bromine, or triflate. So this would be my carbon halide. Then my R2 group, again, is the ethyl 2-butanoate. And then all we have to do is supply the reagents necessary for this. So we can use palladium(II) acetate, and we can do it over our base. So let's do NaOAc, and then we know that for these two reactions to have combined, the vinylic hydrogen here and the halogen itself were both lost and that's how they combine together to give me this product here. So, just remember, when it comes to the Heck reaction, remember we have our R1 combining with R2. R1 is part of our carbon halide. In this case, R2 is just part of an alkene. We lose the X group of the carbon halide and the vinylic hydrogen from our R2 group so that they can combine R1 and R2 together to give us our final Heck reaction product. Keep these basic steps in mind as we go deeper and deeper into the Heck reaction in order to understand how we make our more substituted, more conjugated product.

Heck Reaction Mechanism

In this video, we're going to take a look at the Heck reaction mechanism. Now, we're going to say that this mechanism occurs via a catalytic cycle that is comprised of organo, palladium, intermediates. Now if we're comparing the generic setup of a cross coupling reaction, we can kind of compare it to a Heck reaction. Now the general format or setup for a cross coupling reaction is we have our carbon halide represented by R₁X. It reacts with a coupling agent represented by R₂C and a transition metal complex which is ML_n. Through the use of this transition metal catalyst, we're able to create our coupling product where R₁ and R₂ connect together. CX is just a byproduct that is formed during these coupling reactions. Now, if we take a look here at the Heck reaction, in the Heck reaction we have a carbon halide as well. And in terms of our coupling agent, we're using an alkene. And then we have our palladium catalyst over our base and as a result of this coupling reaction what we are making is we're making a new, usually more conjugated alkene. So we're just going to say an alkene. And then here, this would just be our byproduct. So fundamentally, what's happening here is we just have the X group of our carbon halide and the H group of our alkene being lost as a byproduct and the R group of the carbon halide connecting with this alkene portion to give us our new usually more conjugated alkene product. Now that we've seen the basic setup, click on to the next video and let's start talking about the mechanism that is indicative of a Heck reaction.

Heck Reaction Mechanism

So, the first step of our Heck reaction mechanism begins with oxidative addition. This involves the addition of the carbon halide to the transition metal complex. Now, here we have our palladium catalyst. It possesses a lone pair which comes from the d orbital electrons of palladium. By utilizing this lone pair, we're going to connect to our X group of the carbon halide. When we connect to that X group at the same time, we have the breaking of this bond here, where it connects to palladium. So what we end up making is our transition metal complex where R and X have attached to the palladium.

Now, normally with a lot of these coupling reactions, we have 3 major steps: oxidative addition, transmetalation, and reductive elimination. When it comes to the Heck reaction, there's a small wrinkle. Instead of doing a traditional transmetalation step, instead we do a syn addition step. Now with syn addition, because of the presence of the alkene, we're going to say the resulting complex reacts with the alkene and adds R₁ and palladium across the pi bond on the same side. Because they're adding on the same side that's why it's called syn addition.

So, we have our X group that got added to the Palladium, and we have our R group. We're going to say here that my R group and then my Palladium. I'm going to say our Palladium complex. Two separate things are getting attached on the same side onto the pi bond. So that means that our new structure R group and our palladium complex are being added to the same side.

Next, we have our reductive eliminations which can be broken down into 3a and 3b. The first one, step 3a, has a new complex undergo a carbon-carbon bond rotation followed by a syn elimination to give an alkene. So from the complex we created in step 2 under syn addition, we're going to bring that down. Then we have our R here and the palladium complex. We have to undergo a 60 degree bond rotation because we need both the hydrogen and this palladium complex to be on the same side so that they can be eliminated by syn elimination. So we're going to rotate around this single bond, we're going to keep the palladium complex in the same position. All that's rotating is the Hydrogen and the R group. So Hydrogen will rotate when we rotate this way, H moves over here. So now, the H is here and then the R gets moved over here. Now, we're going to have our reductive elimination that happens here.

So, under this reductive elimination, these bonds break because hydrogen escaping as hydride connects to the palladium. And then this bond here breaks, carbon being more electronegative than palladium, takes the bond to help make a pi bond. So, we're going to make a double bond here, and then the R group is right here. So under the first step within step 3a, we have the formation of our alkene. What we wanted to do.

Step 3b is all about regenerating the palladium catalyst. So here the bases that we usually use are our acetate ion or hydrogen carbonate or bicarbonate. So what happens here is that they represent a base, they're going to remove the hydrogen which causes this bond to break in palladium to hold on to the electrons and at the same time for the X group to leave because it represents a good leaving group. So we wind up getting the regeneration of our palladium catalyst.

Remember there are 2 major driving forces that come with these coupling reactions. We're trying to make a more conjugated product, and also we're trying to get the transition metal to follow either the 18 or 16 electron rule. At this point, by regenerating the catalyst in its original form, the palladium no longer follows the 18 or 16 electron rule. So this puts pressure on it to go through the cycle again to get closer to the 18 or 16 electron rule. So this is the driving force that's going to want us to redo the whole reaction, get palladium back to its optimal 18 or 16 electron count, and also to generate more alkene product. Now that we've seen this mechanism, we're going to attempt to answer the example question by following this whole mechanism of the Heck reaction.

Heck Reaction Mechanism Example 1

So here it says, a Heck reaction between Z-3 Hexene and Bromine Ethene creates both Z-3-Ethyl-1,3-hexadiene and E-3-Ethyl-1,4-hexadiene. Here we have to illustrate how both products can be formed. Now, remember when it comes to the Heck mechanism, we know that the first step is oxidative addition. So here our palladium catalyst would interact with my alkyl halide. So it would create something like this. So we'd have both of those ligands still attached. But then we'd also have our Br attached, and this attached. So we've seen this before. Then what would happen is that this complex that we just created would react with our alkene. Our alkene has hydrogens that we don't see that are here. So what would happen is it would add, they would add by syn addition from step 2. So what we would create is this: we're gonna draw this as dashed, this is dashed, so they're both on the same side because we're starting from a Z alkene. Then we have these hydrogens. And remember, we're adding by syn addition. So we'd have, this portion here getting attached. And then over here we'd have the Palladium complex with the halogen still attached to it.

Okay. So just to make this easier and keep it like that. And then we still have these 2 hydrogens here which I'll draw as dashed. So this is what we've made after syn addition. Now, what we can do here is remember, we want the Palladium complex to be on the same side with this hydrogen, so that we can make a double bond right here and give us this first product. So to do that, we need to have a 60-degree carbon-carbon bond rotation. So, we're gonna rotate around this. So when I do that, I'm actually gonna move the Palladium into position. So this doesn't move so it's staying where it is. And, what's going to happen here is this Palladium complex is going to move to where this is and then this H is going to move down here and then this is going to move to this position where the Palladium complex was. So this left side is untouched and we're gonna implement all the changes we did because of the 60-degree bond rotation. So now my Palladium complex is here on the same side with the hydrogen And then the H that I rotated is over here now. And then this here is going to rotate to the same position that the Palladium complex was originally, so now it's going to be wedged.

Alright. So now, we're gonna have reductive elimination. So that's gonna give me this part here. Now, connected to double-bonded carbon so it loses its stereochemistry. And this here and this here are both wedged. They're both on the same side so that's why in the product you see them both on the same side. The H here, I don't draw, we know it's there, we don't have to show hydrogens connected to carbon. So this is the justification for the first product that we made. But how do we make the second product? Well, option 2. So option 1 was this one. Now let's do option 2. So for option 2, we have the same structure here. So what I'm gonna do is I'm going to copy it. And move it down here. Hopefully, you guys can see that. But here we're not gonna do the 60-degree bond rotation. What we're gonna do instead is that this palladium complex will not be on the same side as this hydrogen. But that's okay because there's another hydrogen here. There's a hydrogen, there's 2 hydrogens here. There's 1 hydrogen that's dashed, and then there's 1 hydrogen that's wedged. And it's this wedged hydrogen here that's on the same side as this Palladium complex. So we can lose both of these guys here, and create a double bond right here. Which explains how we get this product over here.

Okay. So when we do that, there goes the double bond that we created. And then, we're gonna have we still have this portion over here. But this product doesn't quite look like what we have up here. But remember there's free rotation around a single bond. Now just rotate it so that this group moves up here and it will look exactly like this product here. And then we have this portion here. So, hopefully, you guys are able to see through my chicken scratch what's going on in terms of the justification for both products. So, the reason that we get 2 products here is because we have different hydrogens that can be made to be syn or cis to the Palladium complex. We lose that hydrogen with the Palladium complex and we create a double bond, here, in 2 locations. Therefore, we have 2 different possible products. So just remember, the Heck reaction mechanism can be a bit tricky. Instead of a traditional transmetalation step, we have a syn addition step and then we also have a syn elimination step, where the Palladium is lost and then the carbon next to it, the hydrogen on that carbon that's on the same side as the Palladium complex is lost with it. So we have syn elimination that's being done.

Here’s what students ask on this topic:

What is the Heck reaction in organic chemistry?

The Heck reaction is a palladium-catalyzed coupling reaction between a carbon halide (such as an aryl or vinyl halide) and an alkene. This reaction results in the formation of a more substituted alkene. The key steps in the mechanism include oxidative addition, syn addition, and reductive elimination. The reaction is both regioselective, favoring the less substituted position of the alkene, and stereoselective, often yielding the E configuration. The reactivity of the alkene can be influenced by electron-withdrawing groups, which increase the reaction's efficiency.

What are the key steps in the Heck reaction mechanism?

The Heck reaction mechanism involves three key steps: oxidative addition, syn addition, and reductive elimination. In oxidative addition, the carbon halide adds to the palladium catalyst. During syn addition, the alkene reacts with the palladium complex, adding the R group and palladium across the π bond on the same side. Finally, in reductive elimination, the complex undergoes a bond rotation and syn elimination to form the alkene product and regenerate the palladium catalyst. This cycle ensures the formation of a more substituted, often more conjugated, alkene.

What factors influence the reactivity of alkenes in the Heck reaction?

The reactivity of alkenes in the Heck reaction is influenced by their substitution and the presence of electron-withdrawing groups. Reactivity decreases with increasing substitution of the alkene. The highest yields are observed with non-substituted ethylene compounds. Electron-withdrawing groups, especially those that are meta-directing, increase the reactivity of the alkene. Examples include carbonyl groups, nitriles, and sulfonic acids. The more deactivating or withdrawing the group, the more reactive the alkene is towards the Heck reaction.

What are the common bases used in the Heck reaction?

Common bases used in the Heck reaction include acetate ion (OAc⁻), hydrogen carbonate (HCO₃⁻), bicarbonate (HCO₃⁻), and triethylamine. These bases facilitate the reaction by deprotonating intermediates and helping to regenerate the palladium catalyst. The choice of base can influence the reaction's efficiency and the yield of the desired alkene product.

How does regioselectivity and stereoselectivity affect the Heck reaction?

Regioselectivity in the Heck reaction ensures that the R group from the carbon halide attaches to the less substituted position of the alkene. This selectivity is crucial for forming the desired product. Stereoselectivity often results in the E configuration of the alkene, which is more stable due to less steric hindrance. These selectivities are important for predicting and controlling the outcome of the reaction, making the Heck reaction a valuable tool in organic synthesis.