in this video, we're gonna take a look at the Suzuki reaction. Now, the Suzuki coupling reaction involves the coupling between a carbon halid and an organic oh, boron compound, we're gonna say the reaction creates conjugated compounds composed of al Keen's style Irene's or buy a reels. Now remember a styrene, The generic form of it is just a benzene ring connected to to al king carbons. Now, if we look at the generic setup for a cross coupling reaction, we'll see that it's composed of a carbon Halide that is represented by our one X uh coupling agent, which is our to see we have M. L. N. Here, M is a transition metal. L. Is the ligand attached to it. And usually in terms of the number of Liggins, it's usually two or four. Through the use of this transition metal catalyst, we can have the combining of R one to R two to give us our coupling product through this process, we also form some byproduct. Now, taking this generic cross coupling reaction model, we can apply to the Suzuki coupling reaction with the Suzuki coupling reaction. We still have a carbon Hallett Here. The R one group of the Carbon Hallett is represented by a vinyl owner real group. Next we have our coupling agent in the form of a organic no boron compound. Here are two of the organo boron or boraine compound is represented by uh vinyl in a real or an al kiel group. Then we're going to say here that C. In terms of my coupling agent would be represented by B. Y two within the Suzuki coupling reaction here, the Y can represent different things, why could be an O. H. So that would mean that we would have B O. H. Two, which would be Baranek acid. Then why could it be an O. R group where R. Is a carbon? So in that case it would be be parentheses are too, So we'd have something like a Baranek Esther or why could be an hour keel group. So it'd be like B. R two where R could be a metal or it could be an ethel. Then we're going to say that the X group of the carbon khallad is represented by our typical, really good leaving groups of chlorine, bromine, iodine or a trifle late now, what's happening in this reaction is just that the carbon khallad loses its X group. The organic boraine or boron group uses loses its B. Y two group and that they're lost as by product And then the R. one and the R two left behind combined to form our coupling product. That's fundamentally what's going on in terms of the Suzuki coupling reaction. And as long as you can approach it with this simple mindset, you'll be able to identify the types of products you're going to obtain within the Suzuki coupling reaction before we even cover the reaction mechanism. So for now guys, click onto the next video and see how I approach this example question, if you think you can um answer the question on your own, attempt it. If you get stuck again, just click on the next video and see how I answer this Suzuki coupling reaction.

So for the Suzuki reaction, we have our three uh noticeable steps involved in the catalytic cycle, we have oxidative addition. We have transmit elation and we have reductive elimination. Now with oxidative addition? It involves the addition of the carbon halid to the transition metal complex. So our transition metal palladium has a lone pair from its D orbital electrons. It uses them to connect to the X. Group, which then causes this bond here to break. And from our our one group to connect to the palladium as well. So we're gonna have here palladium Still connected to its two original Liggins connected to X. And also connected to our one. When we go into step two, what we're dealing with transmit elation. The R two group transfers from our organic oh boraine or boron compound to the palladium metal complex. So what happens here is that this bond breaks, attaches to the palladium at the same time the halogen leaves and attaches to the boron portion. So what we get here at this point is we're gonna have palladium Still connected to its two original Liggins Connected to our one And now connected to R. two. And then as our waste material we have B. Y. two. And then the X. That has just joined it. Now with reductive elimination, we're going to create the connection between R one and R two and regenerate the palladium catalyst. So we're gonna take this compound here and bring it down to show this process. Okay, so here it goes. So what happens here is the R one group is gonna attach two R. Two. and then the bond is gonna break and go to palladium. So what we're gonna get now is we're gonna get our one Being connected to our two Plus the regeneration of my palladium catalyst. So remember the two driving forces for a lot of these coupling reactions is to create a more conjugated product. We would do that in the form of our one connecting to R2. And also for us to follow the 18 or 16 electron role for the transition metal at this point by giving up R one and R two, palladium is no longer following the 18 or 16 electron rule. So it wants the reaction to continue again. The regeneration of the catalyst is basically a key factor that wants us to continue doing this reaction over and over again. Which is why it's involved in a catalytic cycle. But just remember when it comes to this basic Suzuki coupling reaction. Remember we have a carbon Halide and from the carbon halid we lose our X group and then we have an organic boring or boron compound where we lose the B. Y two group. They are lost as byproduct and then R one and R two combined together to give us our coupling product at the end