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.