Alkynes contain two π-bonds, so when they are exposed to electrophiles, they do exactly what you would expect them to do: they react TWICE with them.
Concept: General properties of double addition reactions to alkynes.2m
It turns out that alkynes can react with halogens a lot the same way that double bonds can. Except for the fact that double bonds, remember, have one pi bond and triple bonds have two. What that means is that some reagents that we expose to alkyne are actually going to react twice. And that's what we're going to study right here.
Basically, there's two different reactions that alkynes will react twice with and these are going to produce double addition products. That just means that anything I was expecting to get from my double bond, just double that and that's going to be my expected product for a triple bond.
Now keep in mind, if we have carbocations in any of these mechanisms, which we will, that vinyl carbocations – now remember what the work vinyl means. It means something directly on a double bond. That means a carbocation that looks like this. Vinyl carbocations cannot easily rearrange. What that means is that don't be thinking about shifts in these mechanisms because they're really just not going to happen.
Keep in mind that for some of these reactions, vinyl carbocations will be created. While unstable, these cannot rearrage, so don’t worry about carbocation rearrangements when reacting with alkynes!
Concept: Double hydrohalogenation of alkynes.3m
Let's go ahead and look into the first one. The first one is a hydrohalogenation of an alkyne. Now a hydrohalogenation, remember, is just an HX with a triple bond. If I were to react that one time, I would expect to get a Markovnikov halogen added. But it turns out if you react this twice with alkynes, what you're going to wind up getting is actually gem-dihalides. Now remember that the word gem stands for geminal and that means that they're both on the same carbon.
Let's go ahead and see how this works. If I were to react this with one equivalent, I would expect to get that the triple bond attacks the H, kicks out the X, so I wind up getting a carbocation that looks like this with a positive charge here and an H here. Then that positive charge would grab – or I'm sorry, the X would grab the positive charge and I would get my intermediate, my first product.
But if I expose it to more than one equivalent, for example, let's say that I exposed it to two equivalents total, then it could react again. So then I would get this double bond attacking the H, kicking out the X. What I would wind up getting is a carbocation that now looks like this, where my X is there, my H is there. And once again, I would get my X – and now, by the way, there's two H's there. And now my X would attack there again.
So what I would wind up getting is a product that looks like this, where I have a five-membered ring with two X's in the Markovnikov position. That's the geminal dihalide part and that's a Markovnikov part. Then I would have two H's that came from my addition reaction.
And you might be wondering well, what about that carbon? Shouldn't it have 3 H's? Shouldn't it be CH3? Yeah. Well, that H was always there, though. That kind of looks messy. I'm just going to redraw that. That H was always there because that's the H that was originally on that triple bond anyway. I'm just saying that through this reaction, you wind up adding H twice and X twice. Does that make sense?
I hope so. I hope that that flows. It's really not a complicated reaction as long as you know what hydrohalogenation is.
Concept: Double halogenation of alkynes.3m
Now let's go to halogenation. Halogenation of alkynes is going to be a very similar ordeal, where usually what would we expect to get for a halogenation of an alkyne? You would expect to get vicinal dihalides. But instead of getting just two halogens, I'm going to react this twice, so I would actually expect to get a tetrahalide product. Obviously, that's going to be a whole lot of halogens. Let's see how that works.
Basically, remember that the way this mechanism looks is that the triple bond or the double bond grabs the X, the X grabs the double bond or the triple bond, the pi bond, kicks out the X. So what we wind up getting is we wind up getting a basically, a bridge. What that bridge looks like is like this. So I'd have X, like that and like that. Then another X comes around and goes to the Markovnikov position and kicks out the X that's on the bridge.
So what I'd wind up getting is this original product where I have vicinal dihalides. These are the vicinal dihalides I was talking about. And that would be the product if I only reacted it with one equivalent. If I specifically said one equivalent, this is your product. You're done.
But what if I said two equivalents, one plus one. Or how abut if I just said excess, meaning more than one. Then I would react it again because I still have a double bond, so what I would wind up getting is the same thing all over again, where I'd react this again with X, X. I would form my bridge. And eventually what you wind up getting is a tetrahalide where it looks like this, where I have an X and an X from the first addition and then I would get another X and another X from the second addition. I'd wind up getting four X's next to each other.
Then finally, what's at the end here? This would still be an H, but that H came from the H that's here. The important part is that now I'm getting four halogens just from the addition of one reagent because it's reacting twice.
None of this tough. None of this is really hard, it's just something that you should know how to do. Let's go ahead and move on.
Vinyl alcohols (alcohols directly on a double bond) undergo a process called tautomerization. We will devote an entire chapter to this process next semester, so you aren’t expected to fully understand it yet.
For now, just memorize what the enol and keto forms look like, so you can predict the products that form when you add alcohol to an alkyne.
Concept: Vinyl alcohols yield tautomers.4m
Back when we talk about how we could add groups to double bonds, we discussed how there were three different ways to add water to a double bond to make an alcohol. Well, it turns out that if we add water to a triple bond, we still are going to get that alcohol. The thing is that we're going to get a slightly different product. Instead of just having a single bond with an alcohol, we're now going to have a double bond with an alcohol attached to it. Even though that sounds like a very minor difference, that's actually going to translate into a huge difference in the functional group that we get afterwards. Let's go into this right now. This is going to be the hydration of triple bonds.
It turns out that anytime that you make a vinyl alcohol, that's the name of basically having an alcohol directly on a double bond, that is going to react very uniquely. It's not going to react like the addition reactions that we saw with double bonds. In fact, this is going to do a phenomenon called tautomerization. Now this is a phenomenon that we're not going to fully understand the mechanism for until orgo two, so it's kind of unfortunate that we have to talk about it now, but I'm just going to give you guys like a really quick refresher on what this is, so you guys know what tautomerization is.
Basically, if I were to summarize it, I'm not going to teach you the full mechanism because that would be a whole separate lesson, but all you really need to know is that there you're going to reversibly swap the position of a hydrogen and a pi bond. This is what I'm saying, anytime that you make a vinyl alcohol, this is something special. This is not a regular alcohol. This is an alcohol that is now subject to a phenomenon called tautomerization.
Here, I'll show you. Here would be an alcohol that's directly attached to a double bond. This is vinyl alcohol. And through the tautomerization process, that I'm no going to show you the mechanism for, this is going to turn into a completely different functional group, where basically, my double bond is going to move over here, my H is going to move down here, so these are going to switch places. What's going to wind up happening is that you get a carbonyl formed and instead of this being a CH2, now this is going to turn into a CH3.
What winds up happening is that this turns from a vinyl alcohol to a ketone. How did that happen? Like I said, unfortunately, it would take me 20 minutes to explain this whole thing to you, so instead I'm just going to tell you guys to memorize that a double bond and a hydrogen switch places anytime that you have a vinyl alcohol.
Now we do have some fancy words for this because this is its own thing. Basically, when it's in the vinyl alcohol phase, that's called the enol. That makes sense because ene stands for alkene, -ol stands for alcohol. So it's basically whenever you have an alcohol on that alkene, that would be called an enol. Well, the enol rapidly tautomerizes to a keto form. The keto form is just the ketone or the aldehyde that's produced after tautomerization takes place.
Now, what you notice is that I didn't draw these equilibrium arrows evenly. This is a phenomenon that's constantly in equilibrium, but one of the arrows is much bigger than the other. That's because it turns out that the keto form is going to be favored in almost all cases, highly favored over the enol form. What that means is that immediately upon making any vinyl alcohol or most vinyl alcohols, I can expect it to rapidly transform into the keto phase and the keto side of the equilibrium looks like a ketone or an aldehyde.
Basically, the whole gist of what I'm trying to say is that any time that you hydrate a triple bond, you're actually going to get a ketone or an aldehyde as the product. It's through this process of tautomerization. Now exactly which ones do we get? Let's go ahead and look at each specific reagent.
Concept: Markovnikov addition of alcohols yields ketones.3m
There is oxymercuration of alkynes and there's hydroboration of alkynes. When we do an oxymercuration of an alkyne, what we're really doing is we're doing it a Markovnikov addition of alcohol. Remember that oxymercuration is the most popular way, one of the most popular ways to add a Markovnikov alcohol to a double bond. Well, the same thing applies for a triple bond as well.
What that means is that if I have two sites, I have let's say the blue site and the red site and I'm trying to figure out where the alcohol is going to go, it's going to go in the more substituted position, so I would expect that after an oxymercuration, I'm going to get an alcohol right here in the more substituted position.
Now notice that I put the oxymerc reagents down here, but I also included HA over H2O, do you guys remember what that was? That's hydration. This would be an acid catalyzed hydration. Just so you know, both of these create Markovnikov additions. Both of them favor the Markovnikov alcohol, so actually I can use both of them. Even though oxymerc is maybe more commonly used, hydration is still a great choice. Both of these reagents really lead to the same intermediate structure which is going to be this enol.
All right? Are you getting that so far? The reason I'm calling it an enol is because now I have a Markovnikov alcohol on a double bond. But what we know that it's not going to stay like that because enols are not stable. They like to tautomerize. After the tautomerization process, what's the product going to look like?
Well, the product is going to be the same ring, but now instead of having a single bond to O, I'm going to get a double bond to O. Instead of having a double bond to the carbon, I'm now going to have a single bond to the carbon. It turns out that the product of oxymercuration or even hydration is going to be ketones. Anytime that I'm Markovnikov hydrating a triple bond, I'm going to get a ketone as the product.
Now what part of this mechanism should you be able to draw? The first part. The second part, you are fine just to say tautomerization, just label it and then draw the product. Like I said, I'm not going to teach you that full mechanism until we get to orgo two. But for right now, you know at least the general idea of what's going on.
Both acid-catalyzed hydration and oxymercuration-reduction of any alkyne leads to formation of a ketone. These reactions both yield a Markovnikov vinyl alcohol, which then tautomerizes.
Concept: Anti-Markovnikov addition of alcohols to terminal alkynes yields aldehydes.3m
Alright so remember that we have 3 different ways to add alcohol we had hydration we had Oxymerc we had one more and that was hydroboration, remember what was kind of interesting about hydroboration is that it did everything opposite, hydroboration is actually going to be anti-markovnikov addition of alcohol so what that means is that if once again I have the blue site and I have the red site which one is it going to add to? Well it would actually add to the less substituted position so would it would add right here, OK? So once again notice that I'm getting an enol I'm still getting a vinyl alcohol the difference is that it's just in a different position, OK? Now notice my reagents really quick just notice that I said BH3 or another Boron source that's because hydroboration there's a lot of different borons such as that your professors could use so you just have to be aware of the one that your professor likes to use, OK? And then obviously the bottom part was the oxidation step so now I've got the enol how do I figure out what the product looks like? Same process, I'm going to switch the position of the double bond and the H so what that means is that in my final product what I'm going to get is now instead a double there I'm just going to get a single bond and instead of a single bond to the O, I'm going to get a double bond to the O, OK? So what this means is that where did the extra H go? The extra H went here because for right now there was only one H here and now there's going to be two Hs so there's that original H and then there's that extra H, that extra blue H or whatever was this H here and it transferred, OK? So now notice what kind of molecule this is I just drew a terminal Carbonyl this actually has a hydrogen at the end this is actually going to lead to the formation of Aldyhides, OK? So when you do a markovnikov addition of water that's going to be a ketone product, OK? When you do an anti-markovnikov addition of water what you're going to lead to is an anti-markovnikov alcohol or enol which then turns into an aldihyde, now notice that in general the still called the keto form, OK? If you were to say just in terms of totamers this is the enol totomer this is the keto totomer but this specific molecule happens to be an aldehyde because the carbon is right at the edge and it has one H on it, OK? So just something that you guys need to be aware of it's something that could definitely come up so you guys just need to know what's going on, alright? So I hope that made sense let me know if you have any questions.
Hydroboration-oxidation of terminal alkynes leads to formation of aldehydes. (If the alkyne is not terminal, it will just yield a ketone). This reaction yields an anti-Markovnikov vinyl alcohol, which will tautomerize into a carbonyl on the terminal position, which is the definition of an aldehyde.
For the following reaction, show the product of one mole of reagent adding across the triple bond, and then show the product of a second mole of reagent.
Provide the reagents required for the following transformations. Note, that some of these methods will require more than one chemical step. All the steps and reagents must be included.
Propose a detailed mechanism that explains the formation of compound 2 from alkyne 1. (No additional reagents or reactions are needed).
Reaction of an asymmetrical terminal alkyne with mercury sulfate and sulfuric acid in water would produce which of the following?
Synthesize the following compound starting with ethyne and 1-bromopentane as your only organic reagents (except for solvents) and using any needed inorganic compounds.