Conjugated hydrohalogenation, also known as hydrohalogenation of dienes, or 1,2 vs. 1,4 addition to dienes, is the same reaction as hydrohalogenation, except with a possibility of multiple products due to the presence of a conjugated intermediate.
Concept: Conjugated Hydrohalogenation - General Mechanism13m
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.
We're going to run into a problem though because if you guys recall, there's already a reaction that happens between the double bond and HX. Do you guys remember what that is? It's an addition type reaction where if you guys recall, a double bond is nucleophilic and HX is always highly electron-deprived. 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 two choices. We could either place the carbocation on the primary or on the secondary carbon. 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 rearrangement possible but it's something you have to think about any time you make a carbocation. Then you would get your X negative attacking to form an alkyl halide, a Markovnikov alkyl halide. I'm just going to put here a mark alkyl halide is our product.
Remember the name of this reaction is simply hydrohalogenation. If I say the word hydrohalogenation, I'm talking about a double bond attacking HX. What's the difference 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.
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. 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.
Really quick, I just want to interject and say that your textbook might not call this conjugated hydrohalogenation. There's 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 stay left over. We’re always going to be left with one left over. That means that we're going to have a conjugated intermediate.
One kind of misconception that I want to take out of here is that you might be thinking about another reaction that’s similar to this called allylic halogenation. In allylic halogenation, you always had a radical initiator. 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, what we would expect is that 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 other sites than the respective types that aren’t allylic.
Now we have that carbocation and we have an HX present. Your wheels might be turning of what's going to happen next. You’re thinking this X negative 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. 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. We're going to get now a double bond here and a positive charge here.
Now we have two different reactive intermediates that we have to react 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 negative 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. Another one would be an allylic halogen in that position.
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 carbocation. It’s a very different situation.
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 1-carbon. That’s your number 1 carbon. From there, we can continue to count carbons in order to determine where the halogen adds. As you could see, in the first product, this halogen being right next to the hydrogen would be considered 2. 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? This still is my 1-carbon because no matter what the HX still attack 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 if they conform. 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. 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, so 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. They 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.
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. There's a few other words here that I haven't defined yet. Notice that I'm calling the 1,4 product the thermodynamic product. 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. Since this is my 1,4 product, this would be my thermodynamic. 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 temperatures, 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 0 and 40. You have to assume that you're just going to get a mixture of products. No thermodynamic control, no kinetic control.
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 products of that first one and then I’ll explain it. Go for it.
Example: 1,2 vs. 1,4 products4m
So 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 halo hydric acid HCL and I have a low temperature, I have a temperature of 0 degrees Celsius. So, immediately I should be thinking, I've got double bonds I've got HCL this should be hydrohalogenation but wait since I have two double bonds it needs to be conjugated hydrohalogenation, okay?
Awesome, the 0 degrees part means that it's not just going to be a mixture of products, right? I'm actually going to prefer the one, two product as my major, okay? So, I'm going to prefer, this is going to be a one two addition of HCL, okay? 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, okay? So, you could get this HCl, I mean this two bond attacking HCl but the other one could also attack, meaning that you have two different cations that are possible, we have a cation possible that would look like this in red, right? This is with a line there, but we also have the cation possible that looks like, oh my gosh I'm sorry guys, it was a mistake, it has to be a carbocation, I just looked like a fool, okay? So, you've got your two different carbo cations 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, okay? So, we said that we're going to favor a one, two product so that means that do I resonate this carbocation or 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 one two products, okay? I know it's a little bit cramped for space but that's what we would do. Alright, so again guys keep in mind that you, it's not really able, you're not able to predict which double bond in this case would attack, you have to draw the products of both, in this case they happen to be the same. Alright, 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.
Example: 1,2 vs. 1,4 products3m
Alright, so 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. Alright, 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, okay? 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 carbo-cation that looks like this and I would also have a carbo-cation that looks like this, it's a little bit Lone, sorry, it's off the screen a little bit, I think you can tell what I was trying to write, I'll try to make it a little higher next time, okay? So, now guys because I'm favoring the 1, 4 product I resonate this cation or should I just keep it where it is? I actually have to resonate it. So, we're going to resonate both of these to go to the other position because remember, we're favoring the 1, 4 product. So, I'm going to resonate this guy over here, okay? Now, even, this one even easier to tell them 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? And, that's my final product because I used 50 degrees Celsius, so this is a thermal dynamic controlled product, alright? 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 controls and 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.
Concept: Kinetic vs. Thermodynamic Control9m
Now we want to go into a little bit more detail and 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 call the 1,4-product due to the orientation of the hydrogen and the halogen that goes with it. We learned that cold reaction conditions favor the kinetic product, which we call the 1,2-product for the same reasons. 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. Notice that I put the words simplified energy diagram here. That's because these energies in the 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 two humps. Notice that here you only have one hump drawn. 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. 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. The diene energy level is starting off right here.
Regardless of the type of reaction conditions, we’re always starting at the same energy.
Notice that there's two different pathways that we can take from the get-go. We notice that I have these two activation energies. I have an activation energy that's called activation energy K or kinetic, and an activation energy that is thermodynamic.
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? Remember that thermodynamic means 1,4. 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. The 1,2 product is going to flow-through an intermediate that is more stable. This intermediate is more stable. 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. Whereas for my thermodynamic product or my 1,4, I get an alkyl halide that looks like this.
The alkyl halides have equal energies, very similar energy. Halogens do not benefit from being primary or secondary. 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 tetra-substituted double bond is much more stable than a mono-substituted 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.
Check out these two products. Which of these two products is the more stable one? For my kinetic product, I have a mono-substituted product. Whereas for my 1,4 product or my thermodynamic, I have a di-substituted. 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.
What does this mean in terms of the temperature control? Where does that come into play? Basically, temperature control allows us to overcome the very high activation energy of the thermodynamic product. Because 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. 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. 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. Let’s actually just write that in. It has the more stable intermediate, the secondary, but it has the less stable product, the mono-substituted. Whereas the thermodynamic pathway has the less stable intermediate, primary, but the most stable product, which would be di-substituted. Mono-substituted and di-substituted. 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 that you want which is really cool. Again, this isn't the only reaction we learned in Orga 1 and 2 that has this idea of thermodynamic versus kinetic control but it’s one of the best examples.
Anyway guys, remember that if anything you could 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.
Awesome. Let's go ahead and move on to the next topic.
Give the major organic product for the reaction:
Draw the two major products pbtained in the reaction shown.
Predict the major product of the following reaction
Predict the products of the following reaction.
Predict the products of the following reaction.
Predict the products of the following reactions.
Draw the two major products obtained in the reaction shown.
Draw the two major products obtained in the reaction shown. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms
Draw the two major products obtained in the reaction shown. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Draw one structure per sketcher. Add additional sketchers using
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Authentic skunk spray has become valuable for use in scent-masking products. Show how you would synthesize the two major components of skunk spray (3-methylbutane-1-thiol and but-2-ene-1-thiol) from any of the readily available butenes or from buta-1,3-diene.
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Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw the answer in skeletal form.
Electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under:
C) Kinetic Control
D) Thermodynamic Control
E) None of these
Answer the following questions using the image below.
The nucleophile in this reaction is ____:
The electrophile in this reaction is ____:
The kinetically controlled product in this reaction is _____:
The product that results from 1,4-addition is _____:
Draw the structures of the major products of the following reactions.
Using arrows to show the flow of electrons, write a stepwise mechanism for the reaction shown below. Give all important resonance structures for each reactive intermediate. For your mechanism, concisely explain why X = 81% and Y = 19% when the reaction is performed at -80°C and why X = 44% and Y = 56% when the reaction is performed at room temperature (25°C).
A product of thermodynamic control.
a) is formed fastest.
b) is favored by lower temperature.
c) has the lowest energy transition state.
d) none of the above.
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What compound results from the 1,4-addition of one equivalent of HBr to 1,3-butadiene
d. 1 bromo-2-butene
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Write a mechanism for the reaction of 2,3-dimethyl-1,3-cyclohexadiene with one equivalent of HBr. Provide structural formulas for the products and label each as "1,2-adduct" or "1,4-adduct.”
Fill in the structure of the missing module of the reaction by paying attention to regio - and stereochemistry where necessary.
Consider the reaction in the box, which is carried out under thermodynamic conditions, and circle the most likely product from the choices provided.
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