Monday, May 22, 2006

Lecture 035: alkynes

Lecture 035: alkynes

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Jean-Claude Bradley: Let's get started with chapter nine, which are alkynes, carbon-carbon triple bonds. So, this chapter is going to be pretty similar to the alkene chapter. Alkynes and alkenes are pretty similar, but there are some notable differences, but you'll notice that a lot of the reactions are very, very similar. You're likely going to redo chapter nine if you take 242, so this will be kind of an introduction. We're not going to go in as much depth as you will when you redo this chapter later.

Let's start off with the basics, nomenclature. So the simplest alkyne would be just two carbons and hydrogens. So, this molecule is acetylene, but you can also name it by UPAC standards, where you would take the 'eth', for two carbons, and put a 'yne' at the end to show that it's an alkyne, so I can call that ethyne. It also has a common name ethylene, no actually that's fine ethyne or acetylene.

Okay, so you don't have a numbering system here because there's no other place a double bond could be, same thing with three carbons. There's only one place that I can have a triple bond, so there I would have propyne. I wouldn't have to use a numbering system.

Now, when I have four carbons, just like with alkenes, I have to specify where the triple bond is. You do it the same way as you do for alkenes. You start at the position where the triple bond starts. So, this would be position two, so this one would be 2-butyne. It's also simpler than alkenes because there's no scissor trans. We have two sp hybridizeds centers, so the acetylenes are always going to be straight. The nomenclature is simpler for that reason as well.

We spent some time in the first week looking at the orbitals that are involved here, so you have your sp hybridized carbon and you have two leftover p orbitals on each carbon. You have one sigma bond and two pi bonds that constitute that.

It turns out that having that triple bond there, having a high percentage s character on the carbon, actually can stabilize the negative charge on the carbon. It turns out if I treat a terminal acetylene - let me get this nomenclature out of the way here - this will be a terminal alkyne, meaning that it has at least one end with a hydrogen. If I had two alkyl groups, I would have an internal alkyne. That's our definitions for internal and terminal alkyne.

Now, back to the acidity, internal alkynes don't have hydrogen, so they're not going to be acidic, but if I have a terminal alkyne, then it is possible to remove that proton and put a negative charge. When I say that it's acidic, that's relative to an alkene or relative to an alkane, it's certainly not acidic in the sense of water or any carboxylic acid even. These are way beyond all these, so we need to use an extremely strong base. The traditional base that we use here is called sodamide, NaNH2. That's basically ammonia that acted as an acid. We removed the proton from ammonia, so you end up with NH2-. That's extremely basic and it is basic enough to be able to abstract the proton on terminal alkynes. The way this works is like this, you just abstract that.

This is advantageous because by putting a negative charge on the acetylide, we can make it nucleophyllic. Now, this can act just like any other nucleophile, like Cl- or OH-, yet we can do SN2 reactions in the same way. An example of that would be, let's say I react this with methyl bromide. That would give you a classic SN2 reaction. So that's a good way to synthesize alkynes. That would fall under the first preparation of alkynes.

Another way we can make alkynes, if we don't have one to start with, is to dehydrohalogenate, much in the same way that we made alkenes. But, instead of losing one HBr molecule, we're going to have to lose two HBr molecules to make the alkyne. Losing one HBr molecule isn't very difficult to create an alkene, however if we want to put that second double bond in, it turns out to be a little bit more difficult. We need to use a strong base, and in this case, we will use sodamide and furthermore, we're going to do it at high temperature.

So we're going to do it at 150 degrees. Previously, when we used sodamide, we were generally doing that in liquid ammonia, so about -33, this is different, this is much more intense. What that will do is remove two HBr molecules and we can create a triple bond.

There's a lot more to this reaction, and like I said, if you follow along to 242, you're going to start with this chapter. You'll go into a lot more detail about how this whole process happens.

Let's look at a few reactions of alkynes. The first is hydrogenation. If we use the same catalytic system as we did for alkenes, like palladium and hydrogen. Here 'xs' means literally that you have an excess of hydrogen, so you use up as much hydrogen as you need in the reaction. What will happen is that you will use up two equivalents of hydrogen to reduce the triple bond to a single bond, so you'll form a alkane if you use a standard catalyst like palladium.

If you want to do this reaction and you want to stop at the double bond and you don't want to go further, one tactic that you could use, which you'll see used again and again in chemistry, is to take a really good catalytic system and to make it less efficient. What happens is that it becomes less and less able to do all the competing reactions, and if you're lucky, you'll get the energies just right so it will do the reaction that you want, but not the ones you don't want.

Luckily, it's easier to reduce the triple bond to a double bond, than it is to reduce the double bond to a single bond. That means if we poison the catalyst, add a poison like quinoline, you can actually make the catalyst unable to reduce a double bond, but still reduce be able to reduce a triple bond.

Let's take a look at that. Same alkyne, palladium on barium sulfate and quinoline. This mixture is also called Lindlar's catalyst. What will happen is it will reduce to a double bond and it will stay there. That will be an example of partial hydrogenation.

Another thing we can do is add halogens across triple bonds. Adding an excess of bromine, we'll add the first bromine molecule across the triple bond to the double bond, and then all the way to having four bromine atoms added across. The mechanism is a little bit different than the addition onto an alkene. If you remember, when you added on an alkene, you would get trans addition, here you end up getting mixtures of cis and trans initially, but it doesn't matter because if you have excess bromine, you're only going to get one product.

Let me show you what I mean by that. If we get an addition like this - I'm going to put these in square brackets to indicate that those are not isolated; they're just transiently produced in the reaction mixture - both of these alkenes will act like normal alkenes and we will add bromine across the double bond and they will give the same product.

Another reaction we will revisit is Markovnikov addition of HBr. It will also work with alkynes. Let's say that we do Markovnikov addition of HBr, we just add HBr to this. If we have an excess, because it is hard to stop it after the first stage, so if you get just one product, we'll have an excess. We will first give it a Markovnikov addition, which means that the H will go where there are more hydrogens. In this case there is one hydrogen on the left, zero hydrogens on the right, so the H will go on the left and the Br will go in the middle. That is a transient product. That will also undergo an Markovnikov addition. The H, again, will go where there are more hydrogens on the left, so we will get 2Br bromopropane from that.

If we do Markovnikov addition of water, it's a little different. We're going to use an acidic aqueous solution again, the traditional mixture used is mercuric sulfate, HgSO4, and sulfuric acid, H2SO4, with water. Basically, you need to translate this into a Markovnikov addition with water. It's not exactly the same as the previous case because something interesting happens. The first step will be exactly as you would expect, Markovnikov addition of water, where we get H on the left and the OH in the middle. The thing is, is that this is not a normal alcohol; it is an alcohol that is on an alkene. That's a special kind of relationship called an enol. The enol doesn't survive very long, it basically will tautomerize, which is a word that refers to rearrange, in a sense of going from an enol to a ketone. In the mixture, what will happen is the hydrogen that is on the oxygen will leave from there and go on the double bond, on the carbons on the double bond. That's basically the movement of a proton.

All we did here is we moved the double bond to the oxygen and moved the proton from the OH onto the carbon, down here. The end result of the hydration of an alkyne is you end up with a ketone. If you are looking for the final product, what you need to remember is that you are going to get a methyl ketone in this case. If you have a terminal alkyne, you always get a methyl ketone like this.

If you want to do anti Markovnikov addition of water, you do something similar as what we did for alkenes. For alkenes, we use borane, BH3, and then we use peroxide to oxidize away the boron. Here we are going to do something similar, except that if we were to use borane, there would be a chance that we would add twice across the triple bond, because borane is a pretty small molecule, so I would end up with a alkene and then I would end up with two borons on that structure. To avoid that, what we use is a huge boron reagents that has two pentol groups on it, and the specific pentol groups are called siamyl. The traditional reagent will be this, sia 2BH, siamylborane. Again, the siamyl groups are just pentol groups; they are just really, really big, so because of this spheric hindrance, I can only put one.

If you remember, after you add the boron, you got oxidized, so we would do exactly the same thing in this case, H2O2 basic. Anti Markovnikov addition of water, the OH will go on the left, but again, we have an enol, so that is not the product we isolate. We end up with an aldehyde in this case. You can see that the reactions are very analogous, but there are slight twists to each and every one of these reactions. Any questions on this addition of water?

Next we have the permanganate oxidations. Cold end dilute will be similar to alkenes in that there is not any breakage of the carbon-carbon bond, and it is also similar in that we deliver oxygens, but the difference is instead of getting alcohols, we get ketones, or, I should say, carbon enol. We just replace the triple bond with two carbon enol groups. That's called end dilute. Warm and concentrated would be very similar to alkenes, in that we break the carbon-carbon triple bond and then oxidize each carbon as far as it can go. If we break this triple bond, then on the left we have one carbon. Whenever you have one carbon with concentrated permanganate, it oxidizes all the way to CO2. On the right, that carbon will be oxidized to carboxylic acid. That's very similar to the alkenes.

That's basically the theory that I want to cover for chapter nine. Did anyone bring their books?

Transcription by CastingWords


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