Monday, May 22, 2006

Lecture 027: Alkenes 1

Lecture 027: Alkenes 1

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Jean-Claude Bradley: Just before we look at different ways of making alkenes, let's look at a bit of nomenclature. An alkene will be any compound that has a "Carbon=Carbon" double bond, as we've seen many times. And the way we name these is, if we just have a simple alkene, we start with the alkane. The corresponding alkane to this would be ethane, so I change the A to an E and I have ethene. So you can always do that, but a lot of the smaller alkenes have more common names. So this one would be the same as ethylene. But the proper IUPAC name would be ethene for this compound. Same thing for this, IUPAC name would be propene, but it could also be called propylene. When we have more carbons, then we have the possibility of the double bond in different locations. There's no reason to number the ethylene and propene, because there's only one place the double bond can be. Now we have four carbons - four carbons, so it would be butene - so I have to specify where the double bond is. I'm going to start counting from the side that gives me the lower number. Now so far the compounds I've drawn don't have cis/trans isomers, which we spent some time evaluating, so there's no other descriptor we need to add to the name.

Now when we get to compounds which do have cis/trans isomers, then we absolutely have to specify whether the methyl groups are on the same side or opposite sides. The only way we have to do that right now is using cis/trans, so for the top compound, we're looking for trans. How do we know it's trans? We have two groups that are the same on opposite sides. Another way of saying that is I can say that the methyl groups are trans and the hydrogens are trans, and that's unambiguous so we can say it's trans. And specifically it's trans-2-butene. Same thing down here, the two H are on the same side, or the two methyl groups are the same side, so this would be cis-2-butene. We also came across some examples where you did have cis/trans isomers and we couldn't name them. The best we could do was to say this group was cis, but that's not really a way of naming a compound if you're looking for it. So let's see what happens in those kinds of situations.

All right, so if I have a compound like this, it has four different groups on it - I, Cl, F, Br - this is ethene, it's name is actually bromo-1, chloro-2, fluoro-2, iodoethene. So that compound would be two different possibilities, because we do have cis/trans isomers. So we do something similar to what we did to R/S. We divide the alkene into two parts, and we determine the relative priority of each group. So on the left we have I and F, so the I has higher priority than the F. We use the same rules as for R and S. So that would be one and then 2. So now we're done with the left side, we forget about it and go to the right side. Between the Br and a Cl, the Br will be higher. So you're always going to end up with two possibilities, either the two high priority groups are on the same side or different sides. We can't use cis and trans for this because they have the condition that two same groups are on same or opposite sides. We don't have the same groups here. So instead we use an E and Z nomenclature. So where the two highest priority groups are on the same side, that would be Z. If they're on the opposite side, that would be E. So now you can name any alkene unambiguously. Now the E and the Z actually derive from German words so it's hard to remember how to remember. So one of the tricks we can use is if you look at the E, if you picture the top part of the E flipping back, now you can see they have to be on opposite sides. So that's kind of an unusual thing to remember, so you're likely to remember. It would be nice if the letters look the same, but it doesn't. It's the same with Z, it's got two groups pointing in opposite directions, but it's really on the same side. So that's basically all about how you name alkenes.

The other issue about alkenes that you need to worry about that we did discuss - the eight carbon rule, that is- in a ring, we can't have a trans double bond unless we have at least eight carbons. So this would be the smallest cycloalkene that you could make that would be trans, otherwise it's not possible. Okay, so this would be, if you wanted to name this guy, it would be trans-cyclooctene. And questions about that?

So let's actually get into the preparation. There are five methods we have to prepare alkenes. Some of these will be review, but we'll put them in one place: dehydrohalogenation. Let's break down the word. So halogene would be X (F, I, Br, Cl), hydro means H or water, in this case H, and de- means to remove. So dehydrohalogenation is an easy elimination reaction, so E1 or E2. We spent a lot of time on this, so let me just put one example. You need a base, typically to do E1 or E2. And you would get your alkene out. And there's the HBr that you lose. So we spent a lot of time on that, enough to say that you have to keep it in mind, because one of the things you're responsible for here - we're not going to look at the mechanisms of all of these, but you have five ways of making alkenes, so you could have a questions where I draw an alkene and that could be made from a bunch of different things. So even if you didn't see the mechanism, you should know five ways of making alkenes. So I could make an alkene from ethyl bromide.

Let's see how else we could make an alkene: dehalogenation. That's one we haven't looked at yet. So we break down the word: halogene could be X (F, Cl, Br, I), and de- means remove, so here we're removing only a halogen. The way we do that is by actually having two halogens, and there are two reagents we're going to look at, one of them is zinc, or you could use I-, and that's specific to I-, you can't substitute Br- or something, it's something special about the I itself. So with those reagents, you lose the two Br's and form an alkene. So it's not any dihalide that will do this, it's only dihalides that are vicinal. Vicinal means a 1, two relationship, so these two Br's are vicinal because they have a 1, two relationship. If they're on the same carbon, it's not vicinal, if they're another carbon away, it's not vicinal - they have to be next to each other. That's the only condition for this kind of dehalogenation reaction. Next we have the dehydration of alcohols. So dehydration means lose of water. Okay, we have just covered this today at the beginning of class. So if we want to lose an OH, we can't do that directly, we have to protonate the OH, then we can do E1 or E2. Typically you use a strong acid, usually sulphuric acid. You can use other acids, but this one have the advantage of not being nucleophilic very much. So if you're were to use something like HBr, you have the complication of the Br doing SN1 or SN2 on your compound. So if you don't want the complication of substitution reactions, you use something like sulphuric acid, which is an excellent acid but a poor nucleophile. So we actually looked that the mechanism of this earlier. Bottom lines if you end up with your alkene. You do have to be wary though, because you could have 1, two shifts. Because if you do an elimination by E1, you have to make sure there's not any carbocation that can rearrange. In this example there isn't any, but that is one small thing you need to worry about. And we revisit something we saw first week, when we look at reactions of alkanes, when we could break alkanes into smaller pieces, and some of those pieces were alkenes. So we're going to add this as a way of making alkenes. Use the same example as I used earlier. So propane in some catalyst, palladium, platinum, whatever, some heavy metal catalyst, and high temperature and usually high pressure, will actually break this molecule into smaller pieces. One of the pieces would be methane, and the other piece will be ethylene. The reason I get an alkene in this case if because if we break a C-C bond, we can't put a H there because we don't have enough H's, so you notice that breaks in such a way that we have the same number of H's on both sides. Industrially this is a pretty good way to make small alkenes like ethylene, but in the lab it's impractical. For one thing, the conditions are very harsh, so if you have anything else on this molecule on this molecule besides the alkane, this would destroy it. And it's not very specific, so if I have eight carbons, it would be random where it would break, so that's not good if you're making a specific alkene. But industrially it's very good because these are cheap processes, so everything has its place, and this is not going to have a place in the lab where you're going to be working. On the other side of that, we have a reaction that's much more expense, but is very specific, so if a chemist wants to make an alkene, this is one of the best ways to do it. This is called the Wittig synthesis. And what you need is an alkyl halide and a ketone or aldehyde. So let's start with methyl bromide. I'll take you through the whole mechanism; in the future I'll summarise it, but in the beginning let's go through it carefully. This is actually going to be an SN2 reaction, we have triphenylphosphene and methyl bromide. P is right under N, so it has a lone pair, still reasonable nucleophilic and is going to attack an alkyl halide the same way you would use and amine to attack it. So we end up with - these phi symbols are benzene rings, same as Ph or drawing out the whole thing. The reason we write triphenylphosphene like that is because it's really a pain to have to draw three benzene rings, so typically it's abbreviated like this. So just like the reaction of an amine with an alkyl halide, I have a positive charge on the P, and I've also lost Br- here. What we want to do is basically make the carbon that's attached to the P nucleophilic, so we're going to remove a proton from it. If I remove a proton from this, I put a negative charge on a C. Normally it would be difficult to put a negative charge on C, but here we have positive P here that would stabilise it a lot. You still can't use a mild base, but you can use a base that's available, and that base is typically N-butyl lithium. Now the way you want to think about N-butyl lithium is like this, you have a CH2- and a Li+. Now there is a covalent bond between the C and the Li, but it acts as if we have a negative charge on the C. So we draw the arrow, the CH2 is extremely basic, so it's going to extract a proton from the C next to the P. So now I have positive charge on P, negative on the C, and when you have that kind of situation, where there is a negative charge next to a positive charge, that is called an ylide. This is your Wittig reaction, we spent time preparing it. When we talk about the Wittig reaction, it's going to be the phosphorous ylide. So one way to draw the ylide is like this, with the + on the positive and the - on the carbon, and you'll notice that this follows the octet rule nicely. But I'm also going to draw it in a way where you might see it in problems, where the octet rule is not followed. So you can also draw it with five bonds on the P, and you don't have to draw a charge, but it's understood that if you're going draw a correct Lewis structure for this, you're going to draw this octet, but this is more convenient because you don't have to draw the charges. So I've created the Wittig reagent. The Wittig reagent is actually half of your alkene. The other half is going to come from either an aldehyde or a ketone. So let's say we just use an aldehyde as an example. The Wittig reaction will react with the aldehyde in such a way that the C=O will be removed, and exchanged for C=C. And a side product from the Wittig reaction proper will be triphenylphosphene oxide. What makes this a particularly useful synthesis is that it's extremely general to pretty much any alkene you can draw. So you'll have a problem where you have an alkene and try to figure out what original alkyl halide you could use, what original aldehyde or ketone you can use to make the alkene. And when you work backwards with the Wittig, is you just need to remember you stitched the double bond in the end. So you break it off, and you see what groups you need in the alkyl halide and what groups you need in the ketone or aldehyde. So that's something you're going to need to know fairly where; there are a couple of quiz questions on that alone. So that's pretty much it for the theory part of chapter 7, and we're going to practise this next time doing problems.

Transcription by CastingWords


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