Monday, May 22, 2006

Lecture 029: Alkenes 2

Lecture 029: Alkenes 2

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Jean-Claude Bradley: We looked at SN1, SN2 reactions. Those were called nucleophilic substitution reactions, because the reagents had a pair of electrons that was attacking the compound. In our case it was the alkyl halide. Those were called nucleophilic reactions. In the case of an alkene, it is the opposite. The double bond is actually electron rich. So instead of having nucleophilic reactions, we'll have electrophilic reactions. [writing]

Jean-Claude Bradley: And one of those would be electrophilic addition. So if I want to add HBr to an alkene, the way that happens is by first taking the electron out of the double bond and putting a proton on there. Let's take a look at an example of that. [drawing] If I react ethylene with HBr, let's think of the HBr as being dissociated, so the first thing, just like the proton goes on an oxygen in an alcohol, the proton will go on the alkene like this. [drawing]

Jean-Claude Bradley: I end up with a carbonation, and, again, if you make a carbonation, you've got to worry about 1, two shifts, so the molecule I chose here won't have that problem. The next step is going to be the bromine coming in and finishing the job. [drawing] So in two steps we have the electrophilic addition of HBr to an alkene.

Jean-Claude Bradley: Now by using ethylene, there was only one product that I could get from that, because it didn't really matter which way the HBr added. Let's take a look at an example where there is a difference. There are two different ways you can add it. If I have something unsymmetrical, like propane, then I can think that the Br could either go on the end of the molecule, or it could go in the middle. Let's look at both scenarios.

Jean-Claude Bradley: Now when I'm drawing the proton coming in on the alkene, you want to think of it almost like opening a door. The door will open either to the left or it will open to the right. If I put the proton on the left-hand side to show that the carbonation is ending up on the right, I'm going to be opening it up like this. [drawing] And I've got my carbonation in the middle. And then as a last step, we add the bromine, so then we get 2-bromo-propane. Ok the alternative is to do it the other way. [drawing] So now the carbonation would go on the end. [drawing]

Jean-Claude Bradley: Okay, so let's stop at this point right here. Which one of these two is more likely, do you think? [inaudible response] The one on the left? [inaudible response] Yes, it just comes down to everything we've seen before: primary, secondary, tertiary. So that's secondary, this is primary. So what ends up happening is that given that choice, you really don't get any of the right-hand side. All of the product will come from that secondary carbonation intermediate. [drawing]

Jean-Claude Bradley: So, you know, we can go through the mechanism every time and show, you know, I get secondary vs. Primary or tertiary vs. Secondary, whatever you want, but the pattern that actually comes out of that, and that has to do with where the hydrogen ends up going. If you notice in this case, as I'm adding HBr, the H goes where there are already more hydrogens. In other words, in this double bond on the left-hand side I have two hydrogens; on the right-hand side I have one hydrogen. So the rule is that in electrophilic addition, the H will go where there are already more H's. That's an empirical rule and that's called Markovnikov's rule. [writing]

Jean-Claude Bradley: So that's a way of explaining things without having to invoke the mechanism. I'll give you another example here. We want to apply Markovnikov's rule. Ok so if I have this compound, and I react it with HBr. So the first thing you need to recognize is that that is an electrophilic addition, that's the first step. Then you determine, how is that HBr going to add? You can go through the mechanism, but if you don't want to go through all that, you simply look at how many hydrogens are on each side of the double bond. On the top portion of the double bond, I have zero hydrogens. On the bottom portion, I have one hydrogen. So to apply Markovnikov's rule, we simply add the hydrogen where there are more hydrogens on the bottom. So I can draw the product then [drawing] by doing that. Okay, that's following Markovnikov's rule. Everybody clear on that? You're going to have a bunch of examples to do anyway, in the problem set for that. So Markovnikov's rule is pretty useful but there is one caveat, and that is if you have 1, two shifts, you're going to have a problem with applying Markovnikov's rule, so you need to be a little bit careful about that. But as long as you don't have any 1, two shifts, the rule will apply exactly like this.

Jean-Claude Bradley: So as I said?? [question from student, inaudible] Right, that's why I didn't finish that one, because I didn't want to get into that whole debate about the shift. I mean, basically what we're deciding at this point is just, where is the proton going to go? Is it going to go on the left or the right? So you're right, after we put it on there, other things can happen. Normally, though, you're not likely to get a shift a lot of the time, simply because you already have a selection of a more stable carbonation. But it is possible to design a problem where you will have going from secondary to tertiary afterwards. But typically because you have that initial choice, you're going to get probably the most stable carbonation, which is why Markovnikov's rule will apply almost all of the time. But that is a good question, and that is why I stopped at that point.

Jean-Claude Bradley: Okay, so we can call these products Markovnikov products, and that's what that means. [writing] If we can have Markovnikov products, then if you want to do the opposite, if I wanted to actually put the bromine on the bottom of the ring and put the H next to the methyl, that would be called an anti-Markovnikov addition. And you just simply can't do it with this particular approach, there's no way that just by adding HBr you can do it. But there is a trick, and that is if you add peroxide, you can actually have anti-Markovnikov addition. So let's take a look at that. [drawing]

Jean-Claude Bradley: Key things about this: the peroxide that I add does not have to be equimolar. I can have catalytic amounts of peroxide. I can have a trace. I think often in the problems, they write "trace" or something like that. It won't hurt it if you have more peroxide, but you don't need a full equivalent. The other thing about the peroxide is that it doesn't have to be hydrogen peroxide, it can be any R-O-O-R, where R can be any alkyl group. So sometimes you'll have dimethyl peroxide, or whatever. It doesn't matter, as long as you have a peroxide. What will happen is you will get the anti-Markovnikov addition. [drawing]

Jean-Claude Bradley: So anti-Markovnikov addition means the H will go where there are fewer hydrogens. There are fewer hydrogens on the top, so that's where the H will go, and the Br will go on the bottom. [drawing] Also this reaction will really only work with HBr, so you can't do it with HCl. If you wanted to put a chlorine in there instead, then you'd have to use some of the other knowledge you have, like you could do an SN2 reaction, on that alkyl halide, and put a chlorine, but you do need to remember that this only applies to HBr. This mechanism is actually pretty extensive, and it's going to be similar to another one we did previously. And I think that is where I will start the next class, at the mechanism for this anti-Markovnikov. That's it.

Transcription by CastingWords


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