Heck Reaction - Video Tutorials & Practice Problems
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1
concept
Heck Reaction
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in this video, we're gonna take a look at the heck reaction. Now in the heck reaction we basically have the coupling between a carbon khallad and are now keen to the use of a palladium catalyst. Now we're going to say here that the are group of the carbon Halide is substituted for raven, Ilic hydrogen of an al keen. 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 our one X where X just represents a good leaving group. We have our coupling agent illustrated by R two C. Along with this we have M L. N where M represents a transition metal. L. Is just the number of Liggins attached to it. Typically for these coupling reactions, the number of Liggins is either two or four. Remember this reinforces the idea that our transition metal wants to get either the 18 or 16 electron role 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 RR one and R two groups to get together. Then we're gonna have here R C. X, 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 Hallett. Next we're gonna say that are all keen 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 all keen 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 say that within the heck reaction, we basically have the loss of our leaving group X. And this vanilla hydrogen they're lost so that my R one and R two groups can combine together to create a more substituted, more conjugated. All keen. Now, if we take a look here, we're gonna say that are are one group of the carbon Hallett here, if we apply to our heck reaction, we're gonna say that our group is represented by a vinyl, a real or Ben's ill group. So when it comes to these text reactions are will be represented by one of those three types of groups. Next we're gonna say here, that are two group of our of our heck reaction is just represented by the ALC in 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 reaction is X. Are leaving group here. X leaving group of the carbon. Hallett is represented by chlorine, bromine, iodine or trife late. So for good leaving groups and in terms of the heck reaction, the bases that are typically used are are acetate ion which is O A C. For short you could use hydrogen uh carbonate or we can say bicarbonate or we can even use triathlon. Mean as a base. Now we have a basic simple layout for heck reaction but we need to remember that heck reaction has some rules that it follows. Its regional selectivity and stereo selectivity are the two things that we must also take into account when dealing with heck reactions with regional selectivity. We're gonna say the reaction is highly regional selective. So the R1 group going to the less substituted position of the Akim. So as we do examples and practice questions we'll see how this is being applied. Then we have stereo selectivity. So if an E. Or Z. Configuration is possible for the al keen then the reaction is highly selective, often giving the E configuration as the major product. All right. So when we're dealing with these heck reactions, just remember in the simplest of ways we just have the X. Group of the carbon khallad and the hydrogen or vanilla hydrogen of the walking being lost in order to combine R. One and R. Two together. Then we also have to take into account regional selectivity and stereo selectivity when approaching any type of heck reactions. Now that we've laid down the basic groundwork for heck reactions will attempt to do some example questions dealing with how to find our final product.
2
example
Heck Reaction Example 1
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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 are all keen. So this is our our two group. Then we're gonna have this second structure here which has the halogen on it. Remember that halogen is a leaving group So it represents our X group. Therefore it's connected to all of this which is our our one group. Remember the basic setup is that we're gonna lose the X group and the hydrogen of the alkaline. So that my R. One and R two groups can combine in this reaction. We also have our palladium catalyst in the form of P. D. O. A. C. Two. We have PPH three couple um in conjunction with it. So you can consider this as your catalyst on top. Then we have on the bottom our base which is triathlon. Mean now remember regional selectivity. We have the reaction. It's highly regional selective with the R. One group going to the less substituted alc in position the substitute position of the talking. So we have hydrogen attached to this walkin. And remember the less substituted position. Just means the alc in carbon with more hydrogen. This alkaline carbon here has one hydrogen and this one here has to so we're gonna lose one of these hydrogen on this carbon, it's gonna combine with the iodine and it's gonna be lost so that we can combine R. One and R. Two together. Now here I'll make it doesn't really matter which one is lost because there is no stereo chemistry associated with this double bond. It's not E or Z. So H. And our ex Are lost to the process. So that R. one and R. two can combine. Now the alcan of our one though has um 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 gonna do here is I'm gonna draw our one first and to maintain its E configuration. All you have to do is just draw the R. Two group in the same position that the iodine was in and in that way maintained its configuration. So pretty simple if you approach it that way. So now we're gonna connect this carbon here. Two. 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 cyclo plantain ring. So, this year would represent my final product. So, again, the al keen of my R. Two group isn't here. Z. So, it doesn't really matter. But the structure of my R. One group has an E. Configuration and that has to be maintained because we're also dealing with stereo selectivity where it maintains um it tries to maintain its stereo chemistry but it greatly favors the E. Configuration as the major. All right, 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 R one and R two are combined together. If you can do that, you'll get your final answer for the heck reaction.
3
concept
Heck Reaction
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So remember that the coupling agent of our heck reaction is basically the alkaline. Now the reactivity of the ALC in can be further increased or decreased depending on what's connected to it. Now we're gonna say the reactivity of the ALC in which remember is our our to group. It decreases with increasing substitution. We're gonna say here the highest yields for the heck reaction happen when we have just a non substituted simple ethylene compound as are all keen. So this year and when we have a mono substituted ethylene with electron withdrawing groups. So here if we take a look we have electron withdrawing groups attached to our AL keen remember the balconies are are two group and if we take a look here we have a carbon Neil within our group our could represent hydrogen, it could represent carbon groups. We have C. N. We have our cell phonic acid group, C X three where X is a halogen, we have an R. Three plus where are could be a hydrogen or carbon. And then we have our nitro group. For those of you 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 going to say here that the first three here on the right there moderately deactivating or are moderately electronic drawing groups and then these three here are very strong and the strength is increasing as we go this way from right to left, the strength is increasing and as the strength increases the reactivity of my al keen increases in terms of the heck reaction we're gonna say here reactivity increases. Okay so take that into consideration when dealing with the heck reaction we wanted. Now keen in terms of our coupling agent we either want it to be not substituted at all. So when it's typical ethylene structure or if we're gonna have it substituted just having mono substituted with an electron withdrawing or meta directing group attached, the more deactivating or withdrawing the meta director, the more reactive that all keen is towards the heck reaction. So keep this in mind when taking a look and comparing the reactivity of different types of heck reactions.
4
example
Heck Reaction Example 2
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So here based on what we've learned about the reactivity of the Al keen. Let's answer the following question here. It says rank the following Elkins in order of increasing reactivity towards the heck reaction. So what we have to do here is we have to identify what kind of groups are attached to our al keen carbons. So we have here a CN group, an NH two group here we just have mental groups. And here we have a C F three group here. We also have methyl groups as carbon groups as well. Okay, attached to our Alpine carbons. Okay now remember we say that the Al king is greater enhanced towards the heck reaction if it has electron withdrawing groups attached to it. So we need to identify if we have any electron withdrawing groups attached to my elkin carbons. If we look we see that CN is an electronic drawing group. We also see that CF three is an electron withdrawing group. For see all we have are methyl groups which are weekly um electron donating groups. And then in B we have an NH two group. Remember nitrogen here as a lone pair. Because of that lone pair, NH two 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 at least reactive to most reactive the least reactive would have to be be because that one is a strong electron donating group. So if strong electron withdrawing groups increase the reactivity of the all keen then putting a strong electron donating group would have the opposite effect. It would decrease reactivity of that all keen towards the heck reaction. So be would be last then we have C. Which is electron donating but weaker. Then we have C N versus C X three. In terms of our scale we can see that CN is a weaker or more moderate electron withdrawing group compared to CF three which is represented here by C X three. So then we'd say a so D would be the Most reactive out keen towards the heck reaction because it has the strongest electron withdrawing group attached to it. He has CF three. So just remember when it comes to the heck reaction we can more greatly increase the reactivity of the walking towards the heck reaction. If we attach electron withdrawing groups onto it or if we keep it as just a simple ethylene molecule.
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Problem
Problem
Determine the product from the following Heck Reaction.
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Problem
Problem
Predict the product formed from the following intramolecular Heck reaction.
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7
Problem
Problem
Using ethyl-2-butenoate as your starting material, prepare the following compound by a Heck reaction.
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8
concept
Heck Reaction Mechanism
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in this video, we're going to take a look at the heck reaction mechanism. Now we're gonna say that this mechanism occurs via a catalytic cycle that is comprised of organic 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 set up for a cross coupling reaction is we have our carbon Hallett represented by our one X. It reacts with a coupling agent represented by R two c. And a transition metal complex which is M. L. N. Through the use of this transition metal catalysts were able to create our coupling product where R one and R two connect together. C. X. 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 halat as well. And in terms of our coupling agent, we're using an owl keen and then we have our palladium catalyst over our base and as a result of this coupling reaction, what we're making is we're making a new usually more conjugated alkaline. So we're just gonna say now keen. And then here this will just be our byproduct. So fundamentally what's happening here is we just have the X group of our carbon khallad and the H group of are all keen being lost as byproduct and the r group of the carbon khallad connecting with this al keen portion to give us our new usually more conjugated talking product. Now that we've seen just the basic setup, click onto the next video and let's start talking about the mechanism that is indicative of a heck reaction.
9
concept
Heck Reaction Mechanism
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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 Hallett 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 we wind 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 three major steps. We have oxidative addition. We have transmit elation and we have reductive elimination when it comes to the heck reaction, there's a small wrinkle instead of doing a traditional transmit elation step instead we do a syn addition step. Now, what's in addition because of the presence of the ALc in, we're going to say the resulting complex reacts with the all keen and ads are one 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. Alright, so we have our X group that got added to the palladium and we have our our group, we're gonna say here that my our group and then my palladium, I'm gonna 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 is gonna have our hydrogen here on this wedge bond. And then we're gonna have our our group and our palladium complex being added onto the same side. Next we have our reductive eliminations, which can be broken down into three A. And three B. The first one, Step three A. Has a new complex undergoes a carbon carbon bond rotation, followed by a syn elimination to give an al keen. So, from the complex we created in step two under Sin edition, we're going to bring that down. Then we have our our here and the palladium complex. We have to undergo a 60° 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 sin elimination. So we're gonna rotate around the single bond. We're gonna 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 ages here and then the r gets moved over here. Now we're gonna have our own reductive elimination that happens here. So, under this reductive elimination, this bond breaks because hydrogen escaping us high dry connects to the palladium and then this bond here breaks carbon being more electro negative than palladium takes the bond to help make a pi bond. So we're gonna make a double bond here and then the our group is right here. So under the first step within step 3 3 a. We have the formation of our talking. What we wanted to do. Step three be is all about regenerating the palladium catalyst. So here the basis that we usually use our 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 onto 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's two major driving forces that come with these coupling reactions were 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 role. So this is the driving force that's gonna want us to redo the whole reaction, get palladium back to its optimal 18 or 16 electron count and also to generate more alkaline product. Now that we've seen this mechanism, we're going to attempt to answer the example question at the bottom of the page in the next video and find a justification of why not just one but two products are formed by following this whole mechanism of the heck reaction.
10
example
Heck Reaction Mechanism Example 1
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So here it says a heck reaction between Z three X. Scene and bromine. Methane creates both Z three ethyl 13 hexi dying and three and E three ethel 14 hexi dying 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 are palladium catalyst would interact with my alcohol Halide. So it would create something like this. So we have both of those Liggins still attached but then we also have our B are 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 Carole king are all keen. Has hydrogen, Is that we don't see that are here. So what would happen? It would add, they would add by sin addition from step two. So what would that would create was we'd have we're gonna draw this is dashed, this is dash. So they're both on the same side because we're starting from a Z. Aachen, then we have these hydrogen. And remember we're doing, we're adding by syn addition. So we'd have this portion here getting attached. And then over here we 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 two hydrogen is here, which I'll draw as dashed. So this is what we've made after sin 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 gonna happen here is this palladium complex is gonna move to where this is, and then this H is gonna move down here, and then this is gonna 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° 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 gonna 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 lose this to create a double bond. So we have our reductive elimination. So that's gonna give me this part here, now connected to double bonded carbon. So it loses its stereo chemistry and this year, and this year 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 year. I don't drop, we know it's there. We don't have to show hydrogen is 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 two. So, option one was this one. Now, let's do option two. So, for option two, 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. Mhm. But here, we're not gonna do the 60° 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 two hydrogen here, there's one hydrogen that's dashed, and then there's one hydrogen that's wedged. And it's this wedge 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 we're gonna draw that double bond. Okay, so when we do that, There goes the double one 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, not just rotated, 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 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 two products here is because we have different hydrogen that can be made to be sin or cysts to the palladium complex. We lose that hydrogen with the palladium complex and we create a double bond here to locations. Therefore we have two different possible products. So just remember the heck reaction mechanism can be a bit tricky. Instead of a traditional transmit elation 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 sin elimination that's being done