Conjugated Hydrohalogenation (1,2 vs 1,4 addition) - Online Tutor, Practice Problems & Exam Prep

Conjugated hydrohalogenation involves the addition of HX to a diene, resulting in a carbocation and the formation of two possible products: the 1,2 product (kinetic) and the 1,4 product (thermodynamic). The reaction's temperature influences product distribution; higher temperatures favor the 1,4 product, while lower temperatures favor the 1,2 product. The 1,2 product has a more stable intermediate (secondary carbocation) but yields a less stable product (monosubstituted), whereas the 1,4 product has a less stable intermediate (primary carbocation) but results in a more stable product (disubstituted).

Conjugated Hydrohalogenation - General Mechanism

So now we're going to discuss an allylic site reaction called conjugated hydrohalogenation. For conjugated hydrohalogenation to take place, we're going to need a double bond and a strong halohydric acid like HCl or HBr. Now, we're going to run into a problem though because if you guys recall, there's already a reaction that happens between a double bond and HX. Do you guys remember what that is? It's an addition type reaction where, if you recall, a double bond is nucleophilic and HX is always highly electron deprived, so you're always going to get or very easily going to get the formation of basically a carbocation and a Markovnikov addition. Let's just go through this mechanism really quick, so you guys can remember what this mechanism is. Remember that your double bond would hit the H. Now you have 2 choices. We could either place the carbocation on the primary or on the secondary carbon and we would definitely choose the secondary because that's the Markovnikov addition. That's the Markovnikov carbocation. From there, that carbocation could rearrange if it was unstable. In this case, we're not going to have a rearrange possible, but it's something you have to think about anytime you make a carbocation. Then you would get your X^- attacking to form an alkyl halide. Alright. A Markovnikov alkyl halide. So I'm just going to put here Mark alkyl halide is our product. Okay. Now remember that the name of this reaction is simply hydrohalogenation. Okay. So if I say the word hydrohalogenation, I'm talking about a double bond attacking HX.

So what's different about hydrohalogenation and conjugated hydrohalogenation? They sound so similar and in fact, the reagents look very similar. But there's a huge difference. Let me show you. Well, remember that in the mechanism for hydrohalogenation, you always get rid of the double bond. You start off with a double bond. You make a carbocation and now that double bond is gone forever until you do an elimination reaction later. But notice that for conjugated hydrohalogenation, we keep one double bond around because you're always going to start off with a diene. Instead of starting off just with one double bond, you're always going to start off with two. What does that mean? That means one of the double bonds attacks the HX and one of them is left over to participate in conjugation. Does that make sense? That's the big difference here that we still have an allylic position after the reaction has taken place.

Now really quick, I just want to interject and say that your textbook might not call this conjugated hydrohalogenation. There are a few different names for it. It could also be called hydrohalogenation of dienes. You might also see it called 1,2 versus 1,4 addition to dienes. These are all the same exact concepts. In fact, if you ever see hydrohalogenation and allylic together, again another way to say it. Notice that the biggest difference being that one of the double bonds is going to react, but one of them is going to be left over. We're always going to be left with one left over and that means that we're going to have a conjugated intermediate.

Now, one kind of misconception that I want to take out of here is that you might be thinking about another reaction. But keep in mind that this is not a radical reaction. This is a carbocation reaction, so we're not going to have a radical initiator in this reaction. We're simply going to rely on the fact that you have one double bond and a carbocation present.

Let's draw the mechanism really quick. In this mechanism, we would expect that my HX is going to attack one of the double bonds. I'm sorry, one of the double bonds will attack the HX and it will form a carbocation. Now we have the choice of putting the carbocation on the primary or the secondary position. Obviously, we're going to choose secondary due to Markovnikov's rule, but also because it's allylic. And we know that allylic sites are always more stable than any others, than their respective types that aren't allylic. Now we have that carbocation and we have an HX present and you might already be thinking what's going to happen next. You're thinking this X^- is going to hit the positive charge just like always, just like a normal hydrohalogenation. You're exactly right, but there's one extra complication which is that now this positive charge is actually conjugated, so we have to draw a resonance structure in this mechanism. That resonance structure is going to be of one arrow swinging open like a door hinge. So we're going to get now a double bond here and a positive charge here. Now we have two different reactive intermediates that we could possibly react with. In general, we're going to react with both. What you're going to notice is that the X^- has the option to attack that carbon, but it also has the option to attack this carbon. You should be aware of the fact that two different products are possible. Let's go ahead and draw these. One of the products would be an allylic halogen, an allylic alkyl halide in that position.

And another one would be an allylic halogen in that position. Now, you'll notice that this looks very similar. If you're aware of the allylic halogenation reaction that uses radicals, this product looks very similar to the product that you would get before in using an allylic halogenation. However, the reaction is completely different and the mechanism is completely different because one of the reactions uses radicals and this one uses carbocations, so it's a very different situation.

Now let me go ahead and give you an extra set of tools to be able to understand these. We actually give these two different products different names based on where the halogen attacks. In both of these cases, my hydrogen always attacked one of the carbons. We're going to say that the hydrogen attacked this carbon here and this carbon here. If you're wondering why I picked that carbon, look at the original mechanism. The original mechanism has this double bond attacking the H and the carbocation forming here, meaning that an H must have been attached to that carbon. The site where the H attacks is called the one carbon. That's your number one carbon. From there, we can continue to count carbons to determine where the halogen adds. As you could see, this halogen being right next to the hydrogen would be considered 2. So this would be what we call a 1,2 product or a 1,2 alkyl halide product because of the fact that your hydrogen and your halogen attacked right next to each other on the same double bond.

Whereas we see that it's a little bit different. After you resonate, you get a different distance, right? So this still is my one because no matter what, the HX still attacked the same position. But now notice that 2, 3, this is now attached to the 4th position. This product is called the 1,4 product. It turns out that we don't always have to get an even mixture of both the 1,2 and the 1,4 product. You should be aware that they can form, but it turns out that there is a way to selectively favor one over the other. The way we can do that is by using temperature control. This is a type of reaction that can use temperature control to prefer one product over another. And in fact, you can get a very high yield of just one of the products if you use the correct temperature. What are the temperatures we need to know? The temperatures that we need to know are that temperatures above 40 degrees Celsius, these are going to be hot temperatures, are going to favor the 1,4 product. Temperatures below 0 degrees Celsius, that's going to be really cold.

Think about it. That's below freezing, right? Are going to favor the 1,2 product. Before I read off any of these other lines, I want to help you memorize this. How can you remember that 1,4 correlates to 40 or higher, 1,2 correlates to lower? I just think of the bigger number. The hotter the temperature, I'm going to get the 1,4 product mostly. The lower the temperature, I'm going to get more of the 1,2. So even if you have no clue what's going on, you can at least remember that and use that as a memory trick on your exam. Now, there are a few other words here that I haven't defined yet. Notice that I'm calling the 1,4 product the thermodynamic product. And I'm calling the 1,2 product the kinetic product. This brings up a type of reaction called thermodynamic versus kinetic control. This is actually a really important concept for organic chemistry. This is not the only reaction in organic chemistry that we've learned that uses thermodynamic versus kinetic control. I'm going to assign an entire set of videos. I'm going to do an entire other set of videos just to explain that process. Right now, I'm not actually going to define thermodynamic and kinetic because that would be for another video. What I'm really trying to do here is just get you to memorize it and get you to basically recognize when we're going to have different types of products. Once again, since this is my 1,2 product, this would be my kinetic product.

And since this is my 1,4 product, this would be my thermodynamic product. Now notice that in my original reaction, I had no temperature present.

What happens if your professor doesn't give you a temperature? What do you think you do?

Do you assume that it's 1,4? Do you assume that it's 1,2? No.

If there's no temperature, let's just add a bullet point. This is good learning.

If there's no temperature, then assume both products.

If your professor does not give you temperature information, it's probably somewhere between 40 and 0 degrees and you have to assume that you're just going to get a mixture of products. No thermodynamic control. No kinetic control. Guys, I'm going to go ahead and let you guys try to solve these. We'll start off with the first one. Go ahead and take some time to try to draw the product of that first one and then I'll explain it. Go for it.

1,2 vs. 1,4 products

First, let's just try to recap what type of reaction this is. See how I'm starting off with a diene and a strong halohydric acid HCl and I have a low temperature. I have a temperature of 0 degrees Celsius. So immediately, I should be thinking, well, I've got double bonds. I've got HCl. This should be hydrohalogenation. But wait, since I have 2 double bonds, it needs to be conjugated hydrohalogenation. Okay? Awesome. The 0 part means that it's not just going to be a mixture of products. Right? I'm actually going to prefer the 1, 2 product as my major. I'm going to prefer this is going to be a 12 addition of HX. So now here's another good question for you. How do I know which double bond is going to attack the HCl first? On my prior mechanisms, I was always starting with the one on the right-hand side. Do you think it always has to be that one? Nope. In this case, guys, it's not really going to matter because it's going to make the same product no matter what. But in the case of having an asymmetrical diene, you have to worry about lots of different products. You could get this HCl I mean this double bond attacking the HCl. But the other one could also attack, meaning that you have 2 different cations that are possible. We have a cation possible that would look like this in red, right, with a line there. But we also have the cation possible that looks like oh my gosh. I'm sorry, guys. That was a mistake. It has to be a carbocation. Wow. I just looked like a fool. You've got your 2 different carbocations that are possible. And notice that in both of these cases, they're the same exact thing. That's the same molecule just flipped. But if this were an asymmetrical diene, then you would have to draw both. We said that we're going to favor a 1, 2 product. That means do I resonate this carbocation or just do I just keep it where it is? I keep it exactly where it is. So I'm going to take my Cl and I'm going to attach it to both. And look what I'm going to get from my product. I'm going to get a product that looks like this, Cl, or like this, Cl. And notice that those are the same exact molecule just flipped around, so I wouldn't actually draw both. I would just cross one of them out and I would just keep one of these and this would be my answer to have just the 1, 2 product. I know it's a little tight for space but that's what we would do. Again, guys, keep in mind that it's not really possible for you to predict which double bond in this case would attack. So you have to draw the products of both. In this case, they happen to be the same. All right? So let's move on to the next one. Try to draw the whole mechanism for that one and predict how many products you would find.

1,2 vs. 1,4 products

All right. Due to limited space, I'm just going to go ahead and take myself out of the camera so we can use this whole area to write. All right. So guys, once again, we have HBr and a diene. But notice that my temperature is different. It's 50 degrees, meaning that I'm going to favor my 1,4 product. Now once again, we can't really determine which double bond is going to attack in this case, so I really have to draw both. I have to draw the fact that I would have a carbocation that looks like this. And I would also have a carbocation that looks like this. Fair? It's a little bit low. I'm sorry it's off the screen a bit, but I think you could tell what I was trying to write. I'll try to make it a little higher next time. So now guys, because I'm favoring the 1,4 product, should I resonate this cation, or should I just keep it where it is? I actually have to resonate it. We're going to have to resonate both of these to go to the other position because remember, we're favoring the 1,4 product. I'm going to resonate this guy over here. This one's even easier to tell than the last problem. You can see how that carbocation is the exact same thing just flipped to the other side. So, I'm just going to go ahead and use one of these for the product. I'll just use the red one, and I'll say that the final product is going to be this, Br. Okay? That's my final product because I used 50 degrees Celsius, so this is a thermodynamically controlled product. Okay, guys. So that's it for this part of the lesson. Now I'm actually going to go into more detail on what it means for it to be kinetically controlled or thermodynamically controlled. But thankfully, now you have the basics. You can predict the product no matter what. Now I'm just going to explain a little bit more about theory.

Kinetic vs. Thermodynamic Control

Now I want to go into a little bit more detail on what it means for a reaction to be kinetically controlled or thermodynamically controlled. The reaction called conjugated hydrohalogenation is a really good example of a reaction that has these different types of control. It's not the only one, but it's a great reaction to use as an example to explain this concept. If you guys recall, with this reaction, hot reaction conditions favor the formation of the thermodynamic product which we called the 1,4 product due to the orientation of the hydrogen and the halogen that goes with it. And we learned that cold reaction conditions favored the kinetic product which we called the 1,2 product for the same reasons. Now what I'm here to do is show you guys on an energy diagram what these words kinetic and thermodynamic actually relate to.

First of all, I just want to say a disclaimer here that notice that I put the words "simplified energy diagram" here and that's because this energy diagram isn't perfectly drawn. It's simplified to make it as easy for me to teach you as possible, but this is not the official diagram. In fact, any diagram of a reaction that has a carbocation should actually have 2 humps. Notice that here I only have 1 hump drawn and that's because in order for you to understand what this is about, you don't need the full reaction diagram. You just need to see kind of the intermediate and then the product. That's going to help us to really determine why it's called kinetic or thermodynamic.

First of all, let's just start at the beginning. Notice that I'm starting off this is my energy reaction coordinate at the bottom. This is my x-axis and my y-axis is in free energy or delta g. What we notice is that in the reaction coordinate, we're starting off with a diene. And the diene energy level is starting off right here. Regardless of the type of reaction conditions, we're always starting at the same energy. And notice that there's 2 different pathways that we can take from the get-go. We notice that I have these 2 activation energies. I have an activation energy that's called activation energy K or kinetic and an activation energy that is thermodynamic. And what we notice is that the intermediates that are created in these two different reactions actually have different energies because if you'll notice, remember kinetic means 1,2, right? And remember that thermodynamic means 1,4. Well, what we see when we're looking at the different types of intermediates is that the 1,2 intermediate is a secondary carbocation, whereas the 1,4 intermediate is a primary carbocation. Do you guys remember which one is more stable? Would a primary or a secondary carbocation be more stable?

Secondary. What that means is that the preferred intermediate is actually going to be the secondary or the 1,2 product. Okay. So the 1,2 product is going to flow through an intermediate that is more stable. This intermediate is more stable and this intermediate is less stable. That's the first part. But notice that as we move to the products, things kind of change because what we see is that even though the secondary intermediate is more stable, look at the reaction product that it produces. We wind up getting an alkyl halide that looks like this:

1,2 product: C-CH₃-CH₂-Halogen

Whereas for my thermodynamic product or my 1,4, I get an alkyl halide that looks like this:

1,4 product: C(CH₃)₂Halogen

Now the alkyl halides have equal energies, very similar energy. Halogens do not benefit from being primary or secondary, so that's not the important part. What we do see though is that double bonds have different levels of stability depending on how many R groups are surrounding them. I'm not sure if you guys recall the concept of hyperconjugation which said that carbocations and double bonds are stabilized through R groups. That's why a tetrasubstituted double bond is much more stable than a monosubstituted double bond. This is a concept that we borrow from elimination reactions way back in the day when we were just talking about elimination reactions. We learned that the most stable double bonds are the ones with the most R groups around them.

Well, check out these two products. Which of these two products is the more stable one? For my kinetic product, I have a monosubstituted product, whereas for my 1,4 product, disubstituted. Notice that these things are in conflict with each other because my thermodynamic intermediate was the least stable, but the thermodynamic product is the most stable. It's exactly the opposite for my kinetic product. My kinetic intermediate was most stable but my kinetic product was least stable.

Okay. So then what does this mean in terms of the temperature control? Where does that come into play? Well basically, temperature control allows us to overcome the very high activation energy of the thermodynamic product because guess what? Eventually, all the molecules, all the products want to look like this. They want to be the thermodynamic product. That's the best one. However, if they don't have enough energy, they're not going to be able to overcome the very steep activation energy to become a thermodynamic 1,4 product. If we want to favor the most stable product, we use lots of heat to jack up the energy of the reagents so that we can overcome this very high activation energy and make the most stable product. Whereas if we want to favor the less stable product, then we make the reaction conditions very cold so that the ambient energy is very low so that it's only going to be able to cross the threshold of less activation energy and it's never going to be able to form the less stable intermediate because it's so cold that it can't overcome the very steep activation energy of the thermodynamic product. That has to do with the names, kinetic and thermodynamic. Kinetic means that it's the easiest one to make and it's the one that is formed the fastest. Whereas the thermodynamic product is going to be the one that is overall the most stable at the end. And the thermodynamic product is usually going to require heat to make it possible because of the extra energy that you need to put in for the activation energy. If we were to summarize this, what we could say is that the kinetic pathway has the more stable intermediate, the secondary, but it has the less stable product, the monosubstituted. Correct? Whereas the thermodynamic pathway has the less stable intermediate, primary, but the most stable product which would be disubstituted.

Okay? Monosubstituted and disubstituted. Secondary and primary. These things are in conflict with each other and it allows us, it gives us, the opportunity to use temperature to control the formation of whichever product we want, which is really cool. Again, this isn't the only reaction we learned in Orgo 12 that has this idea of thermodynamic versus kinetic control but it's one of the best examples. Anyway, guys, remember that if anything, you can always just remember that your thermodynamic product, the hot one, is the higher number, 1,4. But now at least hopefully you understand more why this even exists. Okay? Awesome. Let's go ahead and move on to the next topic.

Here’s what students ask on this topic:

What is the difference between 1,2 and 1,4 addition in conjugated hydrohalogenation?

In conjugated hydrohalogenation, 1,2 addition (kinetic product) occurs when the halogen (X) attaches to the carbon adjacent to the carbon where the hydrogen (H) attaches. This results in a monosubstituted product. In contrast, 1,4 addition (thermodynamic product) happens when the halogen attaches to the fourth carbon from the carbon where the hydrogen attaches, resulting in a disubstituted product. The 1,2 product forms faster due to a more stable intermediate (secondary carbocation), while the 1,4 product is more stable overall but forms slower due to a less stable intermediate (primary carbocation).

How does temperature affect the product distribution in conjugated hydrohalogenation?

Temperature plays a crucial role in determining the product distribution in conjugated hydrohalogenation. Higher temperatures (above 40°C) favor the formation of the 1,4 product (thermodynamic product) because the additional energy helps overcome the higher activation energy required for its formation. Lower temperatures (below 0°C) favor the 1,2 product (kinetic product) because the reaction proceeds through a pathway with a lower activation energy, making it easier to form at colder conditions. If no temperature is specified, a mixture of both products is typically assumed.

What is the mechanism of conjugated hydrohalogenation?

The mechanism of conjugated hydrohalogenation involves the following steps: First, one of the double bonds in the diene attacks the hydrogen (H) from HX, forming a carbocation. This carbocation can be stabilized either as a secondary (more stable) or primary (less stable) carbocation. The halogen (X) then attacks the carbocation, leading to two possible products: the 1,2 product (kinetic) and the 1,4 product (thermodynamic). The 1,2 product forms faster due to the more stable intermediate, while the 1,4 product is more stable overall but forms slower.

Why is the 1,4 product considered the thermodynamic product in conjugated hydrohalogenation?

The 1,4 product is considered the thermodynamic product because it is more stable overall compared to the 1,2 product. This stability arises from the disubstituted nature of the double bond in the 1,4 product, which benefits from hyperconjugation and the stabilization provided by additional R groups. Although the intermediate leading to the 1,4 product is less stable (primary carbocation), the final product is more stable, making it the preferred product under thermodynamic control, typically achieved at higher temperatures.

What is the role of carbocation stability in conjugated hydrohalogenation?

Carbocation stability plays a significant role in determining the pathway and product distribution in conjugated hydrohalogenation. The 1,2 product forms through a more stable secondary carbocation intermediate, making it the kinetic product that forms faster. In contrast, the 1,4 product forms through a less stable primary carbocation intermediate but results in a more stable final product. The stability of the carbocation intermediate influences the activation energy required for the reaction, with more stable intermediates having lower activation energies.