Buchwald-Hartwig Amination Reaction - Video Tutorials & Practice Problems
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Buchwald-Hartwig Amination Reaction
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in this video, we're gonna take a look at the Buchwald Hartwig emanation reaction. Now this reaction involves the coupling between and a real Hallett and a mean and a palladium catalyst. Now we're gonna say here that the reaction creates a bond between carbon and nitrogen in the formation of in a real I mean, now we're gonna say that the Buchwald Hartwig ammunition reaction mirrors the generic formula for a cross coupling reaction. Now, remember in a cross coupling reaction we have our carbon halid, we have our coupling agent, we have our transition metal complex here where m is our transition metal. L is just a set number of Liggins attached to it, usually two or four. And through the use of this catalyst, we have the combining of R one and R two together. This will form our coupling product and then the rest is just by product that we're not concerned with. Now, if we take a look at the Buchwald Hartwig emanation reaction, we're gonna say here that are carbon Halide is represented by a real khallad. So that means that the R1 of my carbon Halide group has to be in a real group. Next we have our coupling agent here in the generic form. Well, in this coupling reaction, it's represented by any mean that nitrogen must contain at least one hydrogen on it. So the mean um is usually a primary or secondary amine. That means that R two and R three could be hydrogen, they could be our kill groups, they could be a real groups. Then we're going to say that our C group in terms of this particular coupling reaction is represented by this hydrogen. Then we're gonna say that the X group of the carbon Halide um is chlorine, bromine, iodine or a trifle late. Like we're used to seeing now that's a common theme for these coupling reactions because those four groups are incredibly great at leaving their great leaving groups. Then here we're gonna say that the base being used within this particular coupling reaction, it is specifically turpitude oxide. So O. T. B. U minus. Or you could have it as ch 33 C O minus. Or you might see it drawn out in its skeletal form. So all those are different ways of writing Turpin tac side now that we see that we have in a real halid with a primary or secondary amine. What's the fundamental thing that's happening here? Well, all that's really happening, if you look at look at it in a simple way, is that we have the loss of our X group and the hydrogen on the nitrogen they're lost. So the groups that remain combined together to give us are a real amine here and then of course we have our byproduct. So just remember the fundamental thing that's happening is the X group of my carbon Halide combines with the H. For my nitrogen we have the loss of those two groups and then we have the formation of our a really mean Now this isn't what's really happening. We haven't talked about the mechanism yet, but this is just a quick way of looking at it in order to isolate your final product. Now that we've seen the general layout of the Buchwald Hartwig emanation reaction, click onto the next video and see how I approach the example question that's below.
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example
Buchwald-Hartwig Amination Reaction Example 1
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So we haven't seen the mechanism for the Buchwald Hartwig emanation reaction yet. So we're gonna approach this example question in a very simple way, remember what's fundamentally happening? Um We're seeing that we have the loss of our X group from the carbon Halide and the H group from our amine. And then we have the R. One and R. Two groups combining together to give us our coupling product. So here if we take a look here goes the halogen, that is part of my R1 Group. So this is our one and then here is the H. And then here is the H. Which represents my C. Group, that is part of our two. We know that those are going to be lost. And all that happens is R. One and R. Two combined together. So what we're going to get here at the end is here's our one, It combined with our two. Um now here you can draw the two hour kill groups any way you want, they just have to be connected to the nitrogen. So here this would be our a really mean, that's isolated from this coupling reaction. Now that we've seen a simple approach to solving an example question, click on the next video to take a look at the coupling mechanism for this coupling reaction.
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concept
Buchwald-Hartwig Amination Reaction
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So the Buchwald Hartwig emanation reaction doesn't begin with oxidative addition as its first step. We actually have a step zero, which is partial dissociation. So here the coupling mechanisms begins with the partial dissociation of the palladium catalyst. So what happens here is one of the legends attached to palladium comes off. So we'll have palladium still connected to one Ligon plus the login that has left. Now we can see based on the equilibrium arrows that these products are not highly favored equilibrium favors the react inside. So we prefer to keep our Liggins attached to the palladium. But remember, we don't need a lot of the product to be made. We just need a small catalytic amount to get the process rolling. Once we've done that we go into step one oxidative addition. So here this involves the addition of the real Halide to the transition metal complex here. Um A R is basically standing in for the aromatic benzene ring that we have. We could also use ph as the abbreviation for the benzene ring involved because remember for this coupling reaction, our carbon Hallett is just a real khallad. There is no vinyl Halide being used. All right, So here remember for oxidative addition the transition metal has a lone pair which comes from its d orbital electrons that it uses to attach to the X group when it does that at the same time, this bond breaks and attaches the palladium. So we're gonna have the palladium connected to its one Liggan that's still around, it's connected to X. And it's connected to the benzene ring, which is our aromatic ring. Next we have ligand substitution which can be seen as a trans methylation step. Um Except it's being replaced by using Inamine normally, when we're dealing with trans methylation steps, it's really the transferring of groups between um metals or metal Lloyd's. But here this is a little bit different. It's it's seen as just analogous to a trans methylation step. All right, So what's gonna happen here is we have the amine nitrogen has a lone pair which is going to use to attach itself to the palladium. When it attaches to the palladium, it's gonna kick out the X group. So, what we're gonna get initially is we're gonna get the argon. I keep saying argon aromatic ring connected to palladium connected to our Ligon. Then we have our nitrogen connected. So it has still on it, the hydrogen and it's to our groups. When nitrogen makes four bonds, it's positively charged. So this is what we make initially during this process, we have this base here which remember represents terp it oxide. So what Turpin toxic is going to do, It is going to use the lone pair one of the lone pairs on oxygen and grab this hydrogen here so that nitrogen is neutral. So in this ligand substitution step, we create our benzene ring or aromatic ring connected to palladium connected to our Ligon connected to N. Which is connected to our to our groups. So this group here follows us into step three, which deals with reductive elimination, which helps to form our coupling product and also to help regenerate our partial palladium catalyst. So here we're gonna take this structure, we're gonna bring it down. Alright, So now what happens here is the nitrogen group is going to attach itself to the benzene ring, the aromatic ring. And then here we're gonna relinquish give up these electrons and give them back to palladium. So we're gonna have here our benzene ring. So here I'm drawing it connected to the nitrogen Which has its two r groups, plus the regeneration of our partial palladium catalyst. So these are the four steps involved in the Buchwald Hartwig emanation reaction. It starts off a little bit different than what we may be used to. We don't start out with oxidative addition, but instead the partial dissociation of our palladium catalyst that gets the ball rolling so that we can then go into our traditional state steps of oxidative addition, ligand substitution, which can be seen as just another form of transmit elation. And then reductive elimination. Remember the whole process is trying to get in a real mean as the final product
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Problem
Problem
Determine the product from the following Buchwald-Hartwig Amination Reaction.
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Problem
Determine compounds A and B from the following reaction sequence.
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Problem
Problem
Outline the synthetic pathway for the creation of p-dimethylaminoacetophenone from bromobenzene.
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