Buchwald-Hartwig Amination Reaction: Study with Video Lessons, Practice Problems & Examples

The Buchwald-Hartwig amination reaction is a cross-coupling process that forms aryl amines from aryl halides and amines using a palladium catalyst. The mechanism involves four key steps: partial dissociation of the palladium catalyst, oxidative addition of the aryl halide, ligand substitution with the amine, and reductive elimination to yield the final product. The reaction typically employs a base like terbutoxide, facilitating the formation of carbon-nitrogen bonds while releasing byproducts. Understanding this reaction is crucial for applications in organic synthesis and medicinal chemistry.

Buchwald-Hartwig Amination Reaction

In this video, we're going to take a look at the Buchwald-Hartwig amination reaction. Now, this reaction involves the coupling between an aryl halide, an amine, and a palladium catalyst. Now, we're going to say here that the reaction creates a bond between carbon and nitrogen in the formation of an aryl amine. Now we're going to say that the Buchwald-Hartwig amination reaction mirrors the generic formula for a cross-coupling reaction. Now remember, in a cross-coupling reaction, we have our carbon halide. 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 ligands attached to it, usually 2 or 4. And through the use of this catalyst, we have the combining of R₁ and R₂ together. This will form our coupling product, and then the rest is just byproduct that we're not concerned with. Now if we take a look at the Buchwald-Hartwig amination reaction, we're going to say here that our carbon halide is represented by an aryl halide. So that means that the R₁ of my carbon halide group has to be an aryl group. Next, we have our coupling agent here in the generic form. Well, in this coupling reaction, it's represented by an amine. That nitrogen must contain at least 1 hydrogen on it. So the amine is usually a primary or secondary amine. That means that R₂ and R₃ could be hydrogens. They could be alkyl groups. They could be aryl 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 going to say that the X group of the carbon halide is chlorine, bromine, iodine, or a triflate like we're used to seeing. Now, that's a common theme for these coupling reactions because those 4 groups are incredibly great at leaving. They're great leaving groups. Then here, we're going to say that the base being used within this particular coupling reaction is specifically terbutoxide. So OTBu- or you could have it as CH₃₃CO- or you might see it drawn out in its skeletal form. So all those are different ways of writing terbutoxide. Now that we see that we have an aryl halide 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 it in a simple way is that we have the loss of our X group and the hydrogen on a nitrogen. They're lost, so the groups that remain combine together to give us our aryl 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 aryl amine. 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 amination reaction, click on to the next video and see how I approach the example question that's below.

Buchwald-Hartwig Amination Reaction Example 1

So we haven't seen the mechanism for the Buchwald-Hartwig amination reaction yet. So we're going to approach this example question in a very simple way. Remember, what's fundamentally happening? 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 R₁ and R₂ 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 R₁. And then here is the H, and then here is the H, which represents my C group that is part of R₂. We know that those are going to be lost and all that happens is R₁ and R₂ combine together. So what we're going to get here at the end is here's R₁, it combined with R₂. Now here, you can draw the 2 alkyl groups any way you want. They just have to be connected to the nitrogen. So here, this would be our aryl amine 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.

Buchwald-Hartwig Amination Reaction

So the Buchwald-Hartwig amination reaction doesn't begin with oxidative addition as its first step. We actually have a step zero, which is partial dissociation. Here, the coupling mechanism begins with the partial dissociation of the palladium catalyst. One of the ligands attached to palladium comes off, leaving palladium still connected to one ligand, plus the ligand that has left. Based on the equilibrium arrows, these products are not highly favored. Equilibrium favors the reactant side, preferring to keep our ligands attached to the palladium. However, we don't need a lot of the product to be made; just a small catalytic amount to get the process rolling. Once we've done that, we go into step one, oxidative addition.

This involves the addition of the aryl halide to the transition metal complex. Here, "Ar" is basically standing in for the aromatic benzene ring that we have. "Ph" could also be used as the abbreviation for the benzene ring involved. For this coupling reaction, our carbon halide is just an aryl halide; there is no vinyl halide being used. In oxidative addition, the transition metal uses a lone pair from its d orbital electrons to attach to the "X" group. As it does, this bond breaks and attaches to the palladium. We will now have palladium connected to its one remaining ligand, "X," and the aromatic ring.

Next, we have ligand substitution, which can be seen as a transmetalation step, except it is being replaced by using an amine. Normally, transmetalation steps involve the transferring of groups between metals or metalloids. Here it is analogous to a transmetalation step. The amine, the nitrogen with a lone pair, attaches itself to the palladium, kicking out the "X" group. We initially get the aromatic ring connected to palladium, connected to our ligand, with nitrogen connected, still bearing its hydrogen and its two "R" groups. When nitrogen makes four bonds, it is positively charged.

During this process, we use terbutoxide, which uses a lone pair on oxygen to grab the hydrogen from nitrogen, making it neutral. In this ligand substitution step, we create our benzene ring connected to palladium, connected to our ligand, connected to nitrogen which in turn is connected to our two "R" groups. This structure moves us into step three, which deals with reductive elimination. This helps to form our coupling product and also helps regenerate our partial palladium catalyst. Here, the nitrogen group attaches itself to the aromatic ring, and the paired electrons are returned to palladium. We end up with the structure of the benzene ring connected to nitrogen, which has its two "R" groups, plus the regeneration of our partial palladium catalyst.

These are the four steps involved in the Buchwald-Hartwig amination reaction. It starts off a little differently than what we may be accustomed to, not starting with oxidative addition but with the partial dissociation of our palladium catalyst. This gets the ball rolling so that we can then continue through our traditional steps of oxidative addition, ligand substitution (seen as just another form of transmetalation), and reductive elimination. The entire process aims to synthesize an arylamine as the final product.

Determine the product from the following Buchwald-Hartwig Amination Reaction.

So here it says, determine the product from the following Buchwald-Hartwig amination reaction. Alright. So we have our two reactants and then we have our palladium catalyst with our sodium tert-butoxide base. Well, the left compound we have a halogen on it, so because that's our halogen, our X group, this whole structure must be our carbon halide. All of this here must be R1. Then we are next to a secondary amine. So here this is my secondary amine, which would mean the H that we see is our H group. Remember for this coupling reaction, we have the loss of the X group from the carbon halide and the H group from the nitrogen. Then what's left behind connects together. All right. So our R1 portion is written out. It's going to connect to my R2 portion, which is basically my secondary amine portion minus the hydrogen that was lost. This would be our final product for this particular practice question. Now that we've seen this, click on to the next video after you attempt to do practice question 2. I'll give you guys a hint. The reagents being used initially are to help make a good leaving group which will represent A. Once you make that good leaving group, it then reacts with these next set of reagents to help make compound B. So attempt this question on your own, come back and see does your answer match up with mine.

Determine compounds A and B from the following reaction sequence.

So we said that the initial reagents will help to make a good leaving group that will represent compound A. Now, you should see that we have TF, and where have we seen TF? Well, we've seen TF when it comes to triflates. Right? A triflate is a great leaving group just like chlorine, bromine, and iodine. So here we have this structure here which will help convert the OH group into a triflate. Here we just have methylene chloride which is just a solvent involved in the formation of that triflate group. So what we're going to get for compound A is our triflate. So this would be A. Now that we have our triflate, we're going to react it with this aryl amine with our palladium catalyst and of course our sodium terbutoxide base. Now, here our triflate represents the X group, so this is just our carbon halide which would mean that this here represents R one. Then over here, our aryl amine, since it's an amine, that would mean that this ring here represents R two. And one of the hydrogens on the nitrogen represents our C group. So what's going to happen here is remember, we just lose our X group of the carbon halide, we lose the hydrogen of the amine, and then what's left combines together to give us R one connected to R two. So here goes my R1 group, it's going to connect to my R2 group. We just so we're going to have it connected to the NH. Remember, we lost one of the hydrogens. And then we have the benzene ring portion. So this here would represent our final product, our diaryl amine compound. Because remember for R2, it could also be an aryl group that's used so we could be using an arylamine to make a more substituted arylamine. Alright. Now that we've done practice questions 12, attempt to do practice question 3. This one requires you to remember some of the reactions you've seen when it comes to benzene reactions. In particular, EAS reactions. So what kind of EAS reactions will we have to utilize in order to get to our final answer here? Give yourself some time to figure it out. If you get stuck, again, don't worry. Just come back. See, does your setup match up with mine?

Here’s what students ask on this topic:

What is the Buchwald-Hartwig amination reaction?

The Buchwald-Hartwig amination reaction is a cross-coupling process that forms aryl amines from aryl halides and amines using a palladium catalyst. The reaction involves four key steps: partial dissociation of the palladium catalyst, oxidative addition of the aryl halide, ligand substitution with the amine, and reductive elimination to yield the final product. A base like terbutoxide is typically used to facilitate the formation of carbon-nitrogen bonds while releasing byproducts. This reaction is crucial for applications in organic synthesis and medicinal chemistry.

What are the key steps in the Buchwald-Hartwig amination mechanism?

The Buchwald-Hartwig amination mechanism involves four key steps: (1) Partial dissociation of the palladium catalyst, where one ligand detaches from the palladium complex. (2) Oxidative addition, where the aryl halide adds to the palladium complex. (3) Ligand substitution, where the amine replaces the halide on the palladium complex. (4) Reductive elimination, where the aryl amine product is formed, and the palladium catalyst is regenerated. These steps collectively facilitate the formation of a carbon-nitrogen bond.

What types of amines can be used in the Buchwald-Hartwig amination reaction?

In the Buchwald-Hartwig amination reaction, primary and secondary amines are typically used. The nitrogen in the amine must contain at least one hydrogen atom. This means that the amine can have hydrogen atoms, alkyl groups, or aryl groups attached to it. The versatility in the types of amines that can be used makes this reaction highly valuable for synthesizing a wide range of aryl amines.

Why is terbutoxide commonly used as a base in the Buchwald-Hartwig amination reaction?

Terbutoxide (t-BuO⁻) is commonly used as a base in the Buchwald-Hartwig amination reaction because it is a strong, non-nucleophilic base. It effectively deprotonates the amine, facilitating the formation of the carbon-nitrogen bond. Additionally, terbutoxide helps to stabilize the palladium catalyst and improve the overall efficiency of the reaction. Its steric bulk also minimizes side reactions, making it an ideal choice for this coupling process.

What are the common leaving groups in the Buchwald-Hartwig amination reaction?

In the Buchwald-Hartwig amination reaction, common leaving groups include chlorine (Cl), bromine (Br), iodine (I), and triflate (OTf). These groups are excellent leaving groups due to their ability to stabilize the negative charge after dissociation. Their presence facilitates the oxidative addition step, where the aryl halide adds to the palladium catalyst, making the reaction more efficient.