Eglinton Reaction: Study with Video Lessons, Practice Problems & Examples

The Eglinton reaction is a coupling process involving two identical terminal alkynes, facilitated by a copper catalyst and pyridine base. The mechanism consists of four steps: deprotonation, where the terminal alkyne hydrogen is removed to form an alkynide ion; substitution, where the alkynide ion reacts with copper acetate; radicalization, leading to the formation of alkynide radicals; and dimerization, where two radicals combine to yield a stable bialkene product. This reaction emphasizes the stability gained through conjugation and the unique role of radicals instead of a catalytic cycle.

Eglinton Reaction

In this video, we're going to take a look at the Eglinton reaction. Now, we're going to say that the Eglinton coupling reaction involves the coupling between two identical terminal alkynes with a copper catalyst and base. We want to make sure that the two alkynes we're using are identical so as to avoid a mixture of possible products. We're going to say here that the reaction uses a copper catalyst in the formation of a bialkynal product. And like coupling reactions, it has the same two driving forces. One driving force is the formation of conjugated products. Remember, the more conjugation our compound possesses, the more stable it is. And, we're going to say that another driving force is that unlike other coupling reactions, it doesn't use a catalytic cycle but instead radicals. The general setup for an Eglinton coupling reaction is that we have alkyne 1, which is identical to alkyne 2. We utilize copper 1 and copper 2 catalysts. And we have pyridine as our base. Remember, pyridine is like benzene except there's a nitrogen within it. And basically, what happens here is that we have the loss of hydrogen from both of these so that the alkyne portions that are left at the end combine together to give us our bialkynal product. We're going to look at the mechanism later on to see how it really occurs. But for simplicity's sake, you could just think of it as the two hydrogens being lost, and what is left combines together. Since our two alkynes are identical, that means that their R groups would be the same. So here we're going to say the R1 and R2 groups of the terminal alkynes can be represented by vinyl or aryl groups. They can also be represented by an alkyl group or even another alkyl group. So that's just the basic setup of the Eglinton reaction. Now that we've talked about the basic steps involved, click on to the next video and see how I approach the example question where we are asked to find the final product.

Eglinton Reaction Example 1

Although we haven't gone over the coupling mechanism for the Eglinton coupling reaction, we can take a simple approach to find our answer. So if we take a look at this example here, it says determine the product from the following Eglinton coupling reaction. Now, here we have 2 identical terminal alkynes. If this were an actual question, you wouldn't have the number 2 here given to you. You'd have to realize that, oh, I'm dealing with the Eglinton coupling reaction. Therefore, I'm dealing with 2 identical terminal alkynes. So it's kind of implied that there isn't just one of these alkynes, there's actually 2 of them. Now all we have to do is, we're going to draw 1 terminal alkyne, then realize that there's going to be an identical one also involved in the reaction. And in terms of simplicity, we know that through a process that we haven't seen yet, that the hydrogens that are on the terminal alkyne carbons will be lost somehow and that the alkyne carbons then would connect with one another. This would give me my final product as, okay. We'd have this terminal alkyne connecting with this terminal alkyne. So this would be our final product. Now, this reaction also involves copper 2 acetate being used here and we have pyridine base also involved in this reaction. But again, even though we haven't done the mechanism yet for this coupling reaction, we know that from what we've seen up above, the hydrogens on both ends of the terminal alkynes would have been lost and so those 2 carbons left behind would connect together to give us our bialkynyl product that we have here. Now that we've seen this simple example question, click on to the next video and see how exactly does the coupling mechanism work for the Eglinton coupling reaction.

Eglinton Reaction

So, the Eglinton coupling reaction can be seen as being comprised of 4 basic steps when it comes to its coupling mechanism. Step 1 involves deprotonation. Step 2 substitution. Step 3 is radicalization and finally step 4 is dimerization. Now, if we start out with deprotonation, we're going to say the slight acidity of the terminal alkyne hydrogen allows it to be deprotonated by the pyridine base. So here we have this slightly acidic hydrogen. Remember that the nitrogen within our pyridine base has a lone pair. It utilizes that lone pair to remove this hydrogen here. When it's removed, carbon holds on to the electrons within the bond. So as a result, we create an alkynide ion. Plus, as a byproduct, we make Pyridium ion as our second product. Now byproduct, we're not really going to be concerned with. Now we head into step 2, substitution. So the alkynide ion that we made in step 1, it's formed during deprotonation undergoes a substitution with copper acetate. So here, this carbon uses its lone pair to attach to carbon which causes the breaking of this bond towards the acetate ion. So, we wind up making is we have our alkyne here that is now connected to this copper. And we have as another byproduct our acetate ion. From here we go into the 3rd step which is radicalization. So the newly formed carbon-copper bond undergoes homolytic cleavage in order to form an alkynide radical. So remember, in homolytic cleavage we have equal splitting of our single bond. So this bond will split. The terminal alkyne carbon will hold on to an electron. This carbon this copper here uses this electron to actually attach itself to this acetate. Now remember, with homolytic cleavage we don't have full arrows. Instead we have half or hook arrows being used because we only have the movement of individual electrons. At the same time that this acetate uses this electron to connect to this copper, this copper in the bond takes its electron and holds on to it. So as a result, what we'll have at this point is our alkynide radical, plus 2, copper's connected to acetates. Now, finally we have dimerization. So the final step involves the dimerization of the 2 alkynide radicals that have been formed. Notice I say 2 alkynide radicals. That's because during this entire process, remember, for this coupling reaction, we're using 2 identical terminal alkynes. So all that's happening here to alkyne number 1, the same exact thing has been happening to alkyne number 2, which is why by the time we get to our 4th step, we have 2 alkynide radicals. Dimerization can be seen as a form of termination where 2 radicals join together to form a new stable compound. So here, half arrow because we're moving 1 electron, half arrow because we're moving 1 electron, and now these 2 are joined together. So this would be our final product. Coupling reactions, that you might be used to. It doesn't use a catalytic cycle in order to form the product, instead we have the utilization of radicals. As long as you can remember the basic setup for this type of coupling reaction, you can get to the product very easily. But if you need to work out the mechanism, just remember that it's comprised of 4 steps within its mechanism. We have deprotonation, substitution, radicalization, and then finally dimerization to get our bialkenal product.

Determine compounds A and B from the following reaction sequence.

So the practice problem states determine compounds A and B from the following reaction sequence. We're starting out first with what's called a Vicinal Dihalide. Remember, in a vicinal dihalide we have neighboring carbons each possessing a halogen. They are in the same vicinity of one another. What we're doing here is reacting our vicinal dihalide with excess strong base, so sodium amide, followed by water. Remember this is an early organic chemistry reaction. Anytime we're reacting a vicinal dihalide with excess base, in particular the amide ion, that means we're trying to form a triple bond. So we're trying to make an alkyne in the first step. So what happens here, if we want to see what happens we have our vicinal dihalide. Now, chlorine is a leaving group. So either carbon could represent our alpha carbon. Remember, the alpha carbon is the carbon possessing the leaving group. If one carbon is the alpha, the other one would be the beta by default. I just chose this to be alpha, but the other carbon could have been alpha as well. Since that's alpha, I'll make this one beta. Our beta carbon has hydrogens. So what happens here is we have 1 mole of sodium amide. This amide comes in, removes a beta hydrogen, the bond breaks, falls here to make a double bond, and kicks out the Cl. At this point, we're going to have this structure, but it's not a, or it's not done yet. Alright, and we still have a hydrogen here and we technically had a hydrogen here the whole time as well. Now a second mole of sodium amide comes in. Now, only this carbon here has a halogen so now it has to be the alpha carbon, and then this has to be beta. So now, this comes in, removes this H here, falls here to make a triple bond and kicks this out. So compound A is this terminal alkyne. Now we use water as a follow-up because we're using excess base here. That excess base is supposed to, in a sense, wash off any excess base that's still lingering around because remember terminal alkynes are acidic so if we have any base remaining it would deprotonate that terminal alkyne and remove this hydrogen. So you can think of water as just helping to make sure that we have a terminal alkyne at the end. So that's our compound A. And now we know from what we've seen previously that this here is important for the Eglinton coupling reaction where we have 2 identical terminal alkynes reacting together. Remember, 2 identical terminal alkynes. That would mean that this reaction that we've seen up to this point happened to two of these. Okay. So there are 2 of these that will now react together in the Eglinton coupling reaction to give us compound B. So here is what we saw we created. And then I'll just do an exact copy of it. Okay. So remember, this happened twice because we need 2 identical terminal alkynes. And then, for simplicity's sake, we know that these hydrogens will be lost. And then the terminal alkyne ends will connect together to give us so they're connected together. Oh, that looks horrible. Let's just clean that up, guys. So we have that. So that would be our final answer as compound B. Now, we know that there's much more involved than just these hydrogens being lost because we've gone over the mechanism together. But just for saving time and not having to go through the whole mechanism, we know that what's going to happen is those 2 hydrogens are lost and then the 2 end carbons are going to connect together and we get our final compound B. So hopefully you guys are able to follow along. This question was great because it brings back things that you might have learned a long time ago in terms of how do you change a vicinal dihalide into a terminal alkyne and that's by using excess base. So from there, the rest of it is just a coupling reaction where we're just combining the 2 terminal alkynes together to give us compound B.

Predict the product formed from the following intramolecular Eglinton reaction.

Here it says predict the product formed from the following intramolecular Eglinton reaction. Now, it's given that there are 3 moles, or 3 molecules, of these di terminal alkynes. The reason three are provided is due to the type of structure we're going to make; we have to use 3 of them to achieve this. Now, they mention that we are conducting intramolecular reactions. Remember, whenever we attempt to perform something by an intramolecular process, it usually means that we're going to obtain some type of cyclic structure. The easiest way to approach this is to draw these three molecules or moles of our di terminal alkyne. We're going to have one that looks like that. Remember there are two carbons in here. That's one of them. Then we're going to have our second mole, and I'll use different colors to show how things are reacting with one another. We have two carbons here, and then we have our third one. We know that the three of them have to react with one another, so just position them all to face one another. Alright. So we have these three here, and you can somewhat see where this is leading us because, at this point, we know what happens: these hydrogens will be lost through the mechanism. We don't need to illustrate the mechanism to get the final answer, but we know that these will all be lost. What's left behind is that they'll start to connect together. So that means our product will look something like this. Okay. We then have these two carbons. Okay. And we have this. So, it looks something like this. The last single bond is pretty long compared to the others, but we would achieve something cyclic like this. Let me show it to you again. So this illustrates that this type of reaction is pretty unique because it allows for sp alkyne carbons to connect together. And if we're trying to do something intramolecularly, it can create some really interesting shapes. Here we get this cyclic structure. When it comes down to it, these reactions are pretty straightforward if you don't have to explore the mechanism. We just draw the terminal ends close to one another. We have the loss of the two hydrogens and then just connect the carbons together, and you're done. Because it's stated we wished to conduct this in an intramolecular fashion, we know that we need all three of them to face one another in order to close up and form a ring at the end.

Here’s what students ask on this topic:

What is the Eglinton reaction in organic chemistry?

The Eglinton reaction is a coupling process involving two identical terminal alkynes, facilitated by a copper catalyst and pyridine base. The mechanism consists of four steps: deprotonation, where the terminal alkyne hydrogen is removed to form an alkynide ion; substitution, where the alkynide ion reacts with copper acetate; radicalization, leading to the formation of alkynide radicals; and dimerization, where two radicals combine to yield a stable bialkene product. This reaction emphasizes the stability gained through conjugation and the unique role of radicals instead of a catalytic cycle.

What are the steps involved in the Eglinton reaction mechanism?

The Eglinton reaction mechanism consists of four steps:

1. Deprotonation: The terminal alkyne hydrogen is removed by the pyridine base, forming an alkynide ion.
2. Substitution: The alkynide ion reacts with copper acetate, forming a copper-alkyne complex.
3. Radicalization: The carbon-copper bond undergoes homolytic cleavage, creating alkynide radicals.
4. Dimerization: Two alkynide radicals combine to form a stable bialkene product.

What role does the copper catalyst play in the Eglinton reaction?

In the Eglinton reaction, the copper catalyst facilitates the formation of the bialkene product. Copper (I) and copper (II) catalysts are used to assist in the substitution and radicalization steps. The copper catalyst helps in the formation of the copper-alkyne complex and the subsequent homolytic cleavage to generate alkynide radicals, which then dimerize to form the final product. Unlike other coupling reactions, the Eglinton reaction does not use a catalytic cycle but relies on radical intermediates.

Why is pyridine used as a base in the Eglinton reaction?

Pyridine is used as a base in the Eglinton reaction because it effectively deprotonates the terminal alkyne hydrogen due to its basicity and the presence of a lone pair on the nitrogen atom. This deprotonation step is crucial for forming the alkynide ion, which then participates in the subsequent steps of the reaction. Pyridine's structure, similar to benzene but with a nitrogen atom, makes it a suitable base for this reaction.

What are the driving forces behind the Eglinton reaction?

The Eglinton reaction is driven by two main forces:

1. Formation of Conjugated Products: Conjugated systems are more stable due to delocalization of electrons, which increases the stability of the final bialkene product.
2. Radical Mechanism: Unlike other coupling reactions that use a catalytic cycle, the Eglinton reaction relies on the formation and dimerization of radicals, which drives the reaction forward.