Sonogashira Coupling Reaction - Online Tutor, Practice Problems & Exam Prep

The Sonogashira coupling reaction involves the coupling of an aryl or vinyl halide with a terminal alkyne, utilizing palladium as the primary catalyst and copper(I) as a co-catalyst. The reaction proceeds through four key steps: preparation of the coupling agent, oxidative addition, transmetalation, and reductive elimination. The terminal alkyne, which has a hydrogen atom, is activated by triethylamine, allowing for the formation of a coupling product. This unique mechanism highlights the cooperative nature of the catalysts in enhancing reaction efficiency.

Sonogashira Coupling Reaction

In this video, we're going to take a look at the Sonogashira coupling reaction. Now, in the Sonogashira coupling reaction, we have the coupling between an aryl halide or vinyl halide with a terminal alkyne. Unlike other coupling reactions, cooperative catalysts help to improve one another's performance within the coupling reaction. We will say within this reaction we use palladium as our primary or main catalyst and then we use copper as the cocatalyst, to form a biaryl or bivinyl alkenyl product.

If we take a look on the left at the generic setup for a cross-coupling reaction, we have our carbon halide which is represented by R1X reacting with our coupling agent R2C with the help of our transition metal catalyst so that we can form our coupling product of R1R2 with CX as a byproduct.

Applying this generic setup for a cross-coupling reaction, we're going to use the Sonogashira coupling reaction. Here, we still have our carbon halide. Our coupling agent in this case, though, is a terminal alkyne. Remember, terminal alkyne just means that at least one of the triple-bonded carbons possesses a hydrogen. We're going to have our palladium catalyst working cooperatively with our copper(I) catalyst and some base so that we can form our coupling product here, which just has our R1 and R2 groups connected by the use of a triple bond. And then, of course, we're going to have our byproducts here.

When it comes to describing what each of these groups are, we say that our R1 group is just represented by a vinyl or an aryl group. R2 could also be a vinyl and an aryl group here. Our C group in this case has to be the hydrogen that is part of the terminal alkyne. X, as usual, when it comes to our carbon halide, is one of the four good leaving groups, so chlorine, bromine, iodine, or triflate. The base that we typically use in this reaction is triethylamine. So, we have the general layout of a Sonogashira coupling reaction and we're going to use this basic setup to understand how to get to our final product quickly before we get to the actual mechanism for this particular cross coupling reaction.

Sonogashira Coupling Reaction Example 1

Looking at the general setup for Sonogashiri coupling reaction, we can see that fundamentally what's happening. We have our x group and our h group being lost so that this terminal alkyne part which is attached to R₂ can combine with R₁. Use that in order to determine what our product would be in terms of this particular Sonogashira coupling reaction question. Alright. So here we have our terminal alkyne which has a hydrogen we don't see. In this case, the x group is represented by our triflate. Alright.

So this would represent so this would be h green. Since this is our X group, that would mean that what's attached to it would represent our R₁. And then here, this would be our R₂, this portion here. So we know that all that's going to really happen, if we think about it in the simplest of terms, is we know that we're going to have them being lost, and then our R₁ and R₂ groups bonding together. So our product would look like this: Stronger R₁ group first. And then we're going to connect the terminal alkyne which is still connected to this benzene ring and the methyl group attached. Okay.

And realize here that the alkene had an E configuration and here it's still maintaining that same E configuration. So, this is pretty common when it comes to the reductive elimination step, where we have the retention of our stereochemistry when we first talked about the basic layout for a cross-coupling reaction mechanism. Now that we know just the basic way of looking at a Sonogashira coupling reaction to produce our product, we can now look at the mechanism involved to create these types of more conjugated products.

Sonogashira Coupling Reaction

The Sonogashira coupling reaction is pretty unique because it utilizes cocatalysts. Now, we're going to say that this coupling reaction is a palladium catalyzed reaction of a copper(₁) alkyne complex with a vinyl or aryl halide. Now, we need to realize what's the function of these 2 catalysts. Well, we're going to say here that the palladium catalyst, all it's going to do is it's going to partake in the oxidative addition step. So it's prepping our R₁X group for the coupling reaction. So it's working with the carbon halide. The copper(one) catalyst, what it does is it's prepping our coupling agent. So, our R₂C group. And by prepping the R₁ and R₂ group, that's going to lead them to combining later on to create our coupling product. So we're going to say here that the Sonogashira coupling reaction starts out with a step 0, where we have the preparation of our coupling agent through the use of our copper(one) catalyst. So what's going to happen here is that our alkyne carbon is weakly acidic. Therefore, the triethylamine that we use will deprotonate it. We wind up getting here R₂ connected to our triple bonded carbon which now possesses a negative charge.

Here we have copper(₁) iodide. Remember, this is an ionic compound. So it's not really together. It's just 2 free-floating ions that are close together because of opposite charges. So as a result of this, this negative carbon will grab onto this positive copper. So what we're going to get at this point is we have our R₂ group connected to our alkyne which is now connected to our copper. We're going to say here that our R₂ group has been primed or it's been activated so that it can later on react within this coupling reaction. Now, we head into step 1 here which is our oxidative addition step where things are what we are accustomed to seeing in terms of a cross-coupling reaction. We have our palladium with its d orbital electrons with that lone pair. They're going to attach to my X group here and this is going to break and attach right back to the palladium. So we're going to get our palladium still connected to its 2 ligands, connected to the X group and now connected to R₁.

This leads us into step 2 transmetalation where our primed R₂ group or our copper(₁) alkyneal compound is going to react now with the transition metal complex from the oxidative addition step. So, here we're going to have our R₂ group being transferred from the copper to the palladium. Remember that's going to cause this bond here to break. It's going to attach to the palladium at the same time this X leaves and attaches to the copper. So what we wind up getting is we're going to get R₁ still connected to palladium with its 2 ligands attached. And now my R₂ group is attached.

And then finally, we have our final step which is reductive elimination, We're going to have the R₁ group attaching to this carbon and then having this bond breaking going to the palladium. So what we're winding up getting is our R₁ attaching to the alkyne which is part of R₂, and then the regeneration of our transition metal catalyst which is a driving force for this to go through another cycle in terms of the Sonogashira coupling reaction. So remember, this reaction is pretty unique through the use of the 2 catalysts. So they work in conjunction with one another to help prime my R₁ group and R₂ group so at the end they combine together to form my coupling product. So we have the addition of that, step 0 which we usually don't see in terms of coupling reactions but in this case, we do have it. And then followed by our 3 normal steps of oxidative addition, transmetalation, and reductive elimination to round out the 4 steps involved in the Sonogashira coupling reaction.

Determine the product from the following Sonogashira Reaction.

Determine the product from the following Sonogashira reaction. Alright, so here we have our halogen I, so that represents our X group. Which means what it's connected to would have to be my R1 group. And then here we have our terminal alkyne, so this has to be R2. We know that we're using our palladium catalyst as the main catalyst followed by our copper(I) cocatalyst over our base. In this case, we're using diethylamine. It's still an amine, so it still acts effectively like triethylamine. Now, we know that, in the simplest of ways, what's going on is we're going to have the loss of our I and the loss of our H groups, and then we're going to have the combining of our R1 and R2 through a sigma bond. Alright. So we're going to draw our R1 group, and then we know that it's going to connect to this R2 group here when the hydrogen is lost. So this is our product. We don't really care about the waste material, but if you want to see it, it would be the hydrogen, which represents our C group, connected to iodine, which is our X group. So here we wind up making a more conjugated, more stable product. We'd have the regeneration of our catalyst in order to have this reaction go through another cycle to create even more conjugated product, and then it goes on and on and on in terms of this catalytic cycle. So remember, the Sonogashira reaction, as long as we apply what we learned in terms of the generic layout of a cross-coupling reaction, we can always figure out what our final product will be. This one is unique because it utilizes a cocatalyst in the form of copper(I). But again, look for the R1 group, look for the R2 group, lose the X and the C group, and you'll be able to isolate your final product at the end.

Determine compounds A, B, and C from the following reaction sequence.

Determine compounds a, b, and c from the following reaction sequence. Alright. So we're starting out with a basic alkane. And in step 1, we're doing Br2 over ultraviolet light. Now remember, alkanes are paraffins. They're called this because of their low reactivity. In fact, they only do 2 real reactions. One is combustion, which doesn't help us, and the other is radical-catalyzed halogenation, which is what this first step here is. Remember, what that does is it replaces one of the hydrogens with a halogen. When it comes to bromine, it's very selective. If it finds a tertiary hydrogen, it'll always replace that with a bromine. If a tertiary isn't present, then it goes after a secondary hydrogen. So what we're gonna get initially is one of the secondary hydrogens being replaced with a Br. Since it's a symmetrical terp butoxide, which is a strong bulky base, is Sodium Terp butoxide, which is a strong bulky base. Remember, a bulky base wants to follow Hoffman's rule over Zaitsev's rule. Right. So here, this would be our alpha carbon, this is beta, and this is beta. Under Hoffman's rule, we go after the less substituted beta carbon. So this one here on the end, which has more hydrogens. So here, this bulky base removes this hydrogen, which falls here to make a double bond and kicks out the bromine. So compound a is an alkene. So this would be our compound a. Next, we're gonna go through these next set of reagents. So Br2 doesn't have ultraviolet light involved with it. Here we're dealing with an alkene. So this would just be a simple bromination reaction where we add 2 halogens by anti addition. So we wind up making a vicinal dihalide. Next, we have excess strong base followed by water. So remember, when you have a dihalide and you're using excess strong base, what are you trying to create? You're trying to create an alkyne. Okay. So steps 2 and 3 will change this vicinal dihalide into an alkyne. So we wind up getting a triple bond between these two carbons here. So that would look like this. K. Now that we have a terminal alkyne, that's what connects it to the Sonogashira coupling reaction. Because now we have our terminal alkyne here, we have here a vinyl group or actually a real group connected to our x group which is our leaving group. Okay. So this is x. That means that the benzene ring itself acts as r one. This terminal alkyne here is r two. And through the use of our palladium catalyst with our copper one co-catalyst and our triethylamine base, we'll be able to create our product. Alright, so remember what happens is we have the H group from the terminal alkyne and then the X group being lost so that r one and r two can combine. So this would be oh, so actually we're gonna say here that this is compound b. And now we're writing down compound c here where r one and r two combine. Okay. So now it's gonna connect to r two, so this would be our final product. Our final product. So this would be compound c within this reaction. So, just remember, when it comes to the Sonogashira coupling reaction, we have a carbon halide reacting with a terminal alkyne. We do this through the utilization of a palladium catalyst and a copper one co-catalyst followed by some type of amine acting as a base. Through this, we are able to create our more conjugated, more stable product in which our r one and r two are connected together by a triple bond. K. So we wind up making a more conjugated, more stable product at the very end.

Here’s what students ask on this topic:

What is the Sonogashira coupling reaction?

The Sonogashira coupling reaction is a chemical reaction that involves the coupling of an aryl or vinyl halide with a terminal alkyne. This reaction uses palladium as the primary catalyst and copper(I) as a co-catalyst. The process proceeds through four key steps: preparation of the coupling agent, oxidative addition, transmetalation, and reductive elimination. The terminal alkyne, which has a hydrogen atom, is activated by triethylamine, allowing for the formation of a coupling product. This reaction is unique due to the cooperative nature of the catalysts, which enhances the reaction's efficiency.

What are the key steps in the Sonogashira coupling reaction?

The Sonogashira coupling reaction involves four key steps: (1) Preparation of the coupling agent, where the terminal alkyne is activated by triethylamine and copper(I) iodide; (2) Oxidative addition, where the palladium catalyst interacts with the aryl or vinyl halide; (3) Transmetalation, where the activated alkyne is transferred from copper to palladium; and (4) Reductive elimination, where the final coupling product is formed, and the palladium catalyst is regenerated for another cycle.

What role do the palladium and copper catalysts play in the Sonogashira coupling reaction?

In the Sonogashira coupling reaction, the palladium catalyst is responsible for the oxidative addition step, where it interacts with the aryl or vinyl halide to prepare it for coupling. The copper(I) catalyst, on the other hand, activates the terminal alkyne by forming a copper-alkyne complex. This cooperative action between the two catalysts enhances the efficiency of the reaction, allowing for the successful formation of the coupling product.

Why is triethylamine used in the Sonogashira coupling reaction?

Triethylamine is used in the Sonogashira coupling reaction as a base to deprotonate the terminal alkyne. This deprotonation step is crucial because it generates a negatively charged alkyne, which can then form a complex with the copper(I) catalyst. This activation of the alkyne is essential for the subsequent transmetalation step, where the alkyne is transferred to the palladium catalyst, leading to the formation of the final coupling product.

What are the typical leaving groups in the Sonogashira coupling reaction?

In the Sonogashira coupling reaction, the typical leaving groups are halides such as chlorine (Cl), bromine (Br), and iodine (I), as well as triflate (OTf). These leaving groups are attached to the aryl or vinyl halide and are displaced during the oxidative addition step, allowing the palladium catalyst to form a complex with the aryl or vinyl group, which is essential for the subsequent steps of the reaction.