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.

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.

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