in this video, we're going to take a look at the steely reaction. Now in the stealing reaction and organize standing compound reacts with the carbon Hallett in order to form a new carbon carbon bond, we're gonna say the reaction creates conjugated compounds composed of Al Keen's style Irene's or by a real compounds. If we look at the general format for cross coupling reaction, we know that it's composed of carbon khallad in the form of usually R one X. Reacting with a coupling agent in the form of our to see. And here we use a transition metal complex in order to create a coupling product and then we'd have byproduct waste. Now, if we compare this to the steely reaction, we still have a carbon khallad being used. Then our coupling agent is our organic standing in react compound. I remember standing is just another term we can use for tin. So it's really just an organic tin compound here. We use a palladium catalyst and that's gonna help us to create our coupling product. And then we have a by product here. So in terms of R one R, one of the carbon halid is represented by a vinyl, a real or an elite group. R two can be represented by a vinyl Oreo and also an elite group. When it comes to see for stealing reaction, it's in the form of S and connected to three R groups. These three R groups are in the form of al kiel groups. So like Meth Als or ethel's or bugles an X. Like we're used to seeing when it comes to the carbon khallad? It just represents a good leaving group in the form of chlorine, bromine, iodine or a trifle. Now here we have the general layout for the steely reaction. So what's really going on if we look at it in terms of um simplicity, we're saying here that we have the X group of our carbon Hallett and then the tin portion of our organic tin compound there being lost together. And what's happening is that R one and R two are bonding together. So that's the basic premise of the steely reaction and that's what we're going to take and use to get answers for steely reaction questions. Now, when we get to the mechanism itself, we'll see how exactly do we create the product. But for now guys, click onto the next video and just take a look at the practice question, see if you can identify what the answer will be at the end and one huge thing in terms of these types of reactions, we say that when creating conjugated dying's the reaction is observed to be stereo Specific with retention of configurations. So what does that mean? That means that if your R one or R two is a vinyl group then it maintains it's E or Z configuration If it has one. So keep that in mind if if R one or R two R of a vinyl group, then you have to pay attention to the stereo chemistry at the out keen sight. So is it an E configuration? Or is it a Z configuration? This will dictate what kind of product you're going to make now that we've gone over the basic layout of a stealing reaction, attention to the practice question come back and see, does your answer match up with mine?

So the stealing reaction contains the three major steps that will grow accustomed to seeing with a typical coupling reaction. So we have step one with oxidative addition step to transmit elation followed by step three with reductive elimination. So with oxidative addition it involves the addition of the carbon Hallett to the transition metal complex. So here we have the lone pair on the palladium which comes from its d orbital electrons. We know that's going to attach to our X group of the carbon Halide at the same time. This bond here breaks and connects to the palladium. So we know that our new complex has the palladium still with its original Liggins, but now having attached to it RR one Group and our X group from there we move on to step two which is transmit elation. So here the R two group transfers from the organic standing um compound to the palladium metal complex. So what's going to happen here now is this bonds going to break attached here to the palladium? Cause this halogen to leave and attached to the tin structure. So what we're gonna get now at this point is our palladium Having still its two original Liggins, but now being connected to R one and R two. And then as a byproduct we have our tin structure with the halogen attached. Then finally for step three we have our reductive elimination step where our coupling product is formed and we have the regeneration of our palladium catalyst. So we're gonna bring this down. So my are one group is going to attach to my R2 group, this bond will break and the electrons go back to palladium. That's what we wind up making is our coupling product, Which is our one attached to our two plus the regeneration Of our palladium catalyst. So this is gonna be a common theme that we're seeing with coupling reactions. We are trying to make a product that is more conjugated, therefore more stable and that's one driving force. At the same time, the reaction is occurring because our transition metal catalyst is trying to reach an 18 or electron uh count to be more stable At the end. We have the regeneration of our palladium catalyst, so it no longer has an 18 or 16 electron count. Therefore it wants to undergo this process yet again, in order to get closer to the 18 or 16 electron count by going through the process again, we can therefore make even more of these stable conjugated products. So, just remember when it comes to this reaction, the general idea is we're trying to lose the X group from our organic Halide or carbon khallad and we're trying to lose the S N R three group for our organic standing compound. As a result of this, we're going to connect r one to R two to make our more conjugated product