Hydrohalogenation of alkynes
Reaction of an alkyne with a hydrogen halide, leading to formation of a halogenated alkene or alkane. Created by Jay.
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- At7:15, what happened to the H that was connected to the halogen? Does that proton go off on its own or what?(5 votes)
- It combines with the X- that leaves to regenerate HX. This HX formed will be one of the HX molecules to be used for the hydrohalogenation of another alkyne molecule present in the reaction mixture.(1 vote)
- At10:00, in the final product, are those halogens added to the same side? Is that always what happens because the example at the top indicates that adding the halogens results in one added to each side or am I confusing the different structures?(3 votes)
- At that point, he is not indicating stereochemistry. He is just showing which atoms add to which.(2 votes)
- so in the single molar equivalent reaction the halogen attaches to the carbon underlined in3:35, isn't acetylene symmetrical previous to the reaction thus making markovnikovs rule for carbon selection arbitrary in that reaction? also does steric hinderence or molecule polarity ever take precedence over markovnikovs rule?(3 votes)
- Quick question, at about5:50after discussing the transition state, I understand how we get the double bond and then end up with the X-, but what happens to the H+? It cannot actually exist as a lone H+, so is it immediately taken up by another X- or are there other bases than can take it?(2 votes)
- The mechanism may not be exactly as shown in the video.
It may involve a cyclic transition state, analogous to the three-membered transition state you get in the addition of Br₂ to an alkene.
In this case, there would be an H⁺ instead of a Br⁺ in the triangle.(3 votes)
- How does the other potential product (the one with two halogens on) get formed? Does the same reaction just happen twice?(2 votes)
- What do you mean more than one or one molar equivalence of my hydrogen halide? Do you dope the solution with equivalent measurements of different elements in addition to the hydrogen halide solution? Or different proportions of the hydrogen halide.(1 vote)
- at9:48, why is the carbocation at the "left side" more stable? the carbon there is bonded to halogen, so its carbocation will be less stable because of inductive effect from halogen atom. So why both halogens at the same carbon?(1 vote)
- It’s still more stable than having a primary carbocation on the right.
Both adding to the same carbon is what we observe the major product to be.(2 votes)
- Just after9:00, "Jay" says that if you don't understand or forgot what he talks about, i.e. Markovnikov's Rule, you should check out his earlier videos. However, I have watched those before watching this video and I still don't understand Markovnikov's Rule. Can someone please explain me this as well as what is defined as "stable"?(1 vote)
- at around1:59, he adds a bond to the right carbon, why is this? does this always happen? Is it to give carbon a full octet? And just to clarify, a methyl group is added, correct?(1 vote)
- The starting alkyne (triple bond) had a methyl group off of the right carbon. The methyl group Sal draws in at1:59is just this same methyl group, still attached in the carbocation alkene.(1 vote)
- Wait just as a clarification, the Cl and H add on the same side at first to make a trans molecule? (3:07molecule)(1 vote)
- The H and the Cl add on opposite sides in a trans addition.(1 vote)
Let's look at the hydrohalogenation reaction of alkynes. We've seen this reaction before with alkenes, and there are some similarities and some differences. So let's look and see what happens here. We start with our alkyne, our triple bond. And we add our hydrogen halide. And we add either one or two more equivalents of our hydrogen halide. If we add one equivalent of our hydrogen halide, the hydrogen and the halogen are going to add from opposite sides. And the halogen is going to add to the most substituted carbon. So this halogen here is going to add to the most substituted carbon, which, in terms of regiochemistry-- so go ahead and write here regiochemmistry-- that was called Markovnikov's rule. We're going to add the halogen to the most substituted carbon in the reaction. So with two molar equivalents, you're going to end up with two halogens on the same carbon, like that. So let's take a look at one of the proposed mechanisms for the hydrohalogenation of alkynes. And it closely parallels the hydrohalogenation of alkenes. And this isn't considered to be the perfect mechanism for alkynes, but we're going to start with that, just to show why this is Markovnikov's, in terms of regiochemistry. So let's say this was the mechanism. We start with our alkyne. And we have our hydrogen halide, so our hydrogen halide like that. And when we did this mechanism for alkenes, we said the first thing that happens was these electrons in this bond are going to form a bond with this proton here. And then these electrons are going to kick off onto your halogen, like that. So let's see we can do. So let's say the hydrogen adds onto the right side. So now there's only a double bond between my two carbons. And the hydrogen added onto the right carbon, like that. And so now, this right carbon has a bond going down like this. And now this carbon on the left has only three bonds to it. So that gives it a plus 1 formal charge, like that. And then our halogen is going to have a negative 1 formal charge. And so our halogen would act as our nucleophile. And our positively charged carbon right here would act as our electrophile. And so you would get the halogen adding onto that carbon. And since a carbocation was involved in the mechanism, that's the reason why Markovnikov's rule is seen, because you want to form the most stable carbocation possible. So the more substituted carbocation is the more stable one. So this really isn't quite the correct mechanism. And the reason why it's not quite the correct mechanism is because if this was the right mechanism, the rate of this reaction would depend on the concentration of two molecules, bimolecular. So if we were to write the rate law for this mechanism, we would say, Oh, OK, I expect the rate of reaction for hydrohalogenation of alkynes-- we have a rate constant in there-- to be proportional to the concentration of your alkyne, and then the concentration of your hydrogen halide. Two things, because that's what we started in this mechanism. We had our hydrogen halide right here, and we had our alkyne. And that is not observed experimentally. So this mechanism is not quite right. So this mechanism is not quite what happens, but it is kind of important to think about this carbon as being the positively charged one and Markovnikov's rule applying. So it helps to think about this mechanism, but this mechanism isn't quite right, because this is not the experimental rate law. The experimental rate law turns out to be second order with respect to your hydrogen halide and third order overall. So if you put a superscript 2 there, it's actually dependent upon two molecules of your hydrogen halide and one molecule of your alkyne. So since it's actually third order overall, we need to come up with a different mechanism. So let's look at the mechanism that is currently proposed as the mechanism for the hydrohalogenation of alkynes. We start with our alkyne here. And we need two molecules of our hydrogen halide. So here's one molecule of our hydrogen halide, like that. And let's go ahead and draw the other one down here. OK, so let's put in those lone pairs of electrons, like that. So in this mechanism, all three of these molecules are reacting at the same time. So as these electrons in here, from the bond between the hydrogen and the halogen, the halogen is going to take those electrons and they're going to move and form a new bond with this carbon. At the same time that's happening, these two electrons are going to form a bond with this proton. And then these electrons are going to kick off onto your halogen, like that. So if we were to draw the transition state of all this stuff happening at the same time-- let's go ahead and put in our brackets here, like that-- so what's going to happen? Well, we're going to have carbon double bonded to another carbon, like that. And then we're going to have a lot of partial bonds in here. So let's use a different color for partial bonds. We have our hydrogen and our halogen. And then we have our hydrogen and our halogen up here. And then let's use blue for partial bonds. So this halogen is forming a bond with this carbon here, at the same time this bond is breaking in here, like that. And we started with a triple bond. But that triple bond is leaving to form a bond over here with this proton, like that. At the same time, this proton's bond is breaking with this halogen, like that. So this is all one giant transition state. And if you think about what's happening, the reason why I wanted to show this mechanism up here is to show you, if this happened stepwise, then this carbon on the left would get your full positive charge. Let me just highlight that. This carbon on the left gets your full positive charge, this one right here. And so if we think about what's happening in this mechanism down here, it's not quite the same thing. But since the bond is leaving this carbon on the left-- the triple bond is leaving this carbon on the left-- this is the one that's going to get some partial carbocationic character. So it's going to be partially positive. And it's hard to see in this mechanism how this carbon could be partially positive. But up here, it's easy to see, because the triple bond is breaking from the carbon on the left. And it's going over here to the carbon on the right. So that's the reason why it's important to think about this mechanism. This gives you your partial carbocationic character, which explains the Markovnikov's regiochemistry. All right, so now we're pretty much done. Once those electrons finish moving, we're going to get-- let's go back to our yellow color here-- we're going to get our alkene. So we're going to form our alkene, like that. And we formed a bond with our halogen down here. And then we formed a bond up here with that proton. And so that is going to be the product of our mechanism here, like that. So let's look at a couple of reactions. So let's look at the hydrohalogenation of alkynes here. So we'll start with an alkyne. So go ahead and draw it like that. And remember, you want to make it linear around your alkyne, since they are linear. So in the first reaction, we'll just add one molar equivalent of our hydrogen halide, like that. So I know I'm going to add on hydrogen and halogen across my triple bond. I know that's going to form a double bond. So the first thing I'm going to do is count how many carbons I have. Let's see, 1, 2, 3, 4, and 5. So I know that I'm going to turn a five-carbon alkyne into a five-carbon alkene. So let's go ahead and draw our five-carbon alkene, so 1, 2, 3, 4, and 5. And the alkene is going to be between these first two carbons over here on the right, like that. So that's part of my product. But now I have to figure out, OK, which one of these two carbons do I add my halogen to? Do I add it to this carbon over here on the left side of my triple bond? Or do I add that halogen over here to the carbon on the right side of my triple bond? And to do that, you need to think about Markovnikov's rule. So if I add it to the right one, that would give me a primary carbocation-- I mean, you have to think about which one will give you the most stable carbocation. And the most stable carbocation would be your secondary one over here on the left. And so that is the one that your halogen is going to add over here on the left side. So again, watch those earlier videos for much more, in terms of detail, about carbocations and stability and Markovnikov's rule. So let's do the addition of two equivalents of hydrogen halide, or just make it in excess here. So you have to think about the fact that, OK, so once again, I'm starting with five carbons, so I'm going to get five carbons for my product here. And go back up here to our original reaction. You can see, if you have two molar equivalents, you're going to end up adding your two halogens to the same carbon. And I probably should have drawn this a little bit differently on the general reaction. It also is going to exhibit Markovnikov's for regiochemistry. So when you're trying to figure out which carbon that is, think about which one would be the most stable carbocation again. So once again, we have two choices, the carbon on the left side, or the carbon on the right side. The carbon on the left side gives us more stable. So we're going to add on two equivalents of our halogen here. So we actually form a dihalide as our product. So that's hydrohalogenation of alkynes with one or two molar equivalents of your hydrogen halide.