Monday, May 22, 2006

Lecture 030: Alkenes 3

Lecture 030: Alkenes 3

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Jean-Claude Bradley: So we left off with the Anti-Markovnikov addition of HBr to an alkene last time, so this time we'll talk about the mechanisms, so let's do that.

The way that works is if we've got an alkene that's unsymmetrical, in this sense, where we've got a methyl group on one side, we have two hydrogens on the other" there are a different number of hydrogens on either side. So anti-Markovnikov addition, the hydrogen would go on the side there are fewer hydrogens; it's the opposite of the Markovnikov rule. And to do that, we mention that we need peroxide" any peroxide will do.

"R" could be an alkyl group, "R" could be a hydrogen, it's all going to work.

If the H goes on the side were there are fewer hydrogens, the H will go in the middle. And the Br will go at the end.

The mechanism for this, as I said, is similar to one we've seen before, which involves free radicals. So let's take a look at that.

As in all free radical mechanisms, we're going to have initiation step, propagation, and termination. This time, there's more than one initiation step. There's no light in this reaction. So it's not like bromine separates into two bromine radicals. So what's happening in this case is that the peroxide is first going to break into radicals, and that doesn't need light. That just happens with heat. If you take a solution of hydrogen peroxide and heat it, you'll see oxygen coming out of it very quickly. So that reaction is pretty easy. And I'm going to use hydrogen peroxide as an example, but it would be the same with dimethyl peroxide or any other peroxide.

This is just a homolytic cleavage of the oxygen-oxygen single bond, this is actually really weak. Okay, so that's the first step, it's going to generate the OH radicals. Now, this OH radical is pretty reactive in itself, and it's going to react with the HBr.

When you're breaking a bond with free radicals, you're always going to have the same three arrow pattern, that we've seen before. So a good way to remember this is that you're making something really stable, in the second step, you're making water. The only way to make water from the reagents is by reacting the peroxy radical with the HBr. So it's the Br radical that is actually going to add onto the alkene. So those two steps are what we consider to be the initiation.

Next, the bromine radical is going to add onto the alkene. Now, that the alkene is unsymmetrical, we're going to end up with a radical, or two different possibilities for a radical. One of these radicals is going to be more stable than the other, and that's what going to decide the product. So this step right here.

Here's my alkene. The bromine could add onto the left or the right. So let's add it to the left, and see what happens.

Again, I'm using the same three arrow pattern, but now I'm actually breaking a double bond. So I haven't completely cleaved the carbon-carbon bond. I still have a single bond left there. But I have created a carbon-bromine bond on the left. So that generates that radical.

Now, let's consider the alternative, which I'll call three-prime (3'). The alternative is that the bromine would add on the right side. Okay, so there are our two alternatives. So you can see from this, the first reaction, we generate a secondary radical. And in the 3" we generate a primary radical.

Going back to the stability order, from the previous chapter, the secondary radical is always going to be more stable than the primary.

Because of this, step 3' simply doesn't happen. So it turns out you're going to get anti-Markovnikov products whenever you're faced with the difference in different kinds of radicals. So when you do anti-Markovnikov addition, you count the hydrogens on each side of the double bond, but this is ultimately the reason why it happens.

Basically, the product is decided. We just have to finish it. So in step four I'm going to take the radical that I generated in step 3, and now we're going to react it with HBr.

Actually we've generated the anti-Markovnikov product in this step. And we've also generated a bromine radical. So steps three and four then are the propagation steps, because the radical generated in step four will then be used in another step 3. So if you didn't run out of material or you didn't have any termination steps, then you wouldn't need any more initiation once you had that stuff gone through, right? That's how propagation works.

Of course, you do have termination steps, and that's going to work the same way we did free radicals. Each free radical coming together would be a termination step. So what do we have available here?

We have a bromine radical, and we have a carbon radical that has a bromine on it. That's pretty much all we have to work with in terms of terminations. We have two possibilities, two possible radicals I should say. The easiest one is the two bromines coming together.

Another example would be the bromine-carbon radical combining with a bromine radical. And finally we can have two carbon radicals coming together. Yes?

Student 1: [inaudible question]

Jean-Claude Bradley: Yeah, basically the termination involves terminating something involved in the propagations steps. So as you get it started, if you capture a hydroxyl radical, yeah if a hydroxyl radical reacted with a bromine, yes, that would be a termination. But that would create something unstable, so. Just focus on the one's that we did, that are analogous. Yes, any two radicals involved in the propagation.

Again with the termination, you get extremely small amounts of these compounds, you will get just trace amounts because so many cycles of propagation relative to termination, but that's what they would be.

Any questions on anti-Markovnikov addition of HBr?

Now, though you can add HCl Markovnikov, you can't add HCl anti-Markovnikov just by dumping in some peroxide. So this particular reaction is for HBr. And the reason for that has to do with step number 2. It's that the energies are right for the hydroxyl radical to make a bromine radical. It won't be right for HCl or HF.

Remember that when you're doing problems. If you add peroxide, you're not going to do much. Just Markovnikov addition for HCl. And we're going to do some of that in the problem set.

Let's talk about Markovnikov addition of water: Instead of adding like HBr, we want to add an H and an OH, we can still do that on an alkene. For adding water, all I have to do is add a reasonably strong acid.

If I have water, and I'll just put H+ here, but that can be sulfuric acid, that can be any strong acid you add to this, you just need to catalyze this process. So acid and water will give you Markovnikov addition of H2O. This means that the H goes where there are already more hydrogens, In this molecule on the left, it means the H would go in the middle.

That's all we mean by Markovnikov.

The mechanism for this is going to be really similar for what we did for HBr: We have the proton that can go either on the left or right, we already did that, so we're just going to show the one that happens. The proton is going to go on the carbon here to generate the secondary carbon cation. You form the secondary carbon cation, you look for the one that shifts, we don't have any in this example, and then we add the water to this.

You can't add OH- at this point because you're under acidic conditions. So, there will be very very little OH- floating around. Okay, so the nucleophile in this case will be water. Generally water is not that great of a nucleophile, but a carbyl cation is very tantalizing for a poor nucleophile like water, so that's what will happen.

So that means we need an additional step to complete the reaction, we now have to lose the proton that's on the alcohol.

Now, we have the neutral alcohol. So that's how Markovnikov addition with water works, similar to what we had with HBr.

One of the problems with this Markovnikov addition, using acid and water, is that if you do have the possibility of a 1, two shift, then you've got a problem, because you're not going to get the addition that you want. So it turns out there's a way around that, that you will get Markovnikov addition no matter, even if there's a possibility of a 1, two shift or not.

And that's using an oxymercuration approach, so let's do that one next.

We're not going to look at the full mechanism of this, we're just going to look at the reagents.

So the first step will be mercuric acetates and water, and the second step will be sodium borohydrate. So the OAC here stands for acetate, which is just that. But that's how we're typically going to abbreviate that when we do reactions. So those two steps, again when you have one and two on the same arrow, that means you have to do this sequentially, those two steps will always give you Markovnikov addition of water, no matter what. So that means I will end up with OH in the middle and not have to worry about 1, two shifts.

Any questions on that?

Student 2: [inaudible question]

Jean-Claude Bradley: Yes, the mechanism's in the chapter, that's not one I'm going to expect you to know. But if you look, the mercury adds, there's not a possibility, you never get a full carbyl cation. Some, but not enough to promote a 1, two shift.

We're going to look at reactions where we go through every step of the mechanism, like the one we just did this morning, and then you're increasingly going to see reactions where we don't look at every step of the mechanism, but you are expected to know what it does. I think a lot of these, if you're really curious, will be in your book. If you want to get down and see what it's like.

The only thing we haven't covered is anti-Markonikov addition of water.

It would be tempting to just think we could add peroxide to the acidic water, but that doesn't work -- again -- because the energies don't make any sense. The peroxide trick will work only for HBr.

So, in this case, we're going to be using again two sets of reactions that we're not going to look at the mechanism of, but that will always give you anti-Markovnikov addition of water.

The first of these is to add borane, BH3, and then we'll add a peroxide. The peroxide here is acting in a completely different way from the peroxide we had with HBr. Actually here it's a base, so, typically, that's written H2O2, and hydroxide, you have to be in a basic environment.

Those two steps will give you anti-Markovnikov addition of water, so that means the H will go where there are fewer Hs. So the H will go in the middle and the OH on the end.

Again, with the anti-Markovnikov additions, there is no need to worry about 1, two shifts. Because you know that radicals don't shift. Even though a secondary radical may be more stable than a primary radical, it's just not possible to do 1, two shifts as with carbyl cations. So again we don't have to worry about any kind of rearrangement.

That covers the reactions with alkenes, but there are additional reactions we are going to look at. The next one is catalytic hydrogenation.

Basically involves adding hydrogens across an alkene. For that you actually use molecular hydrogen, H2, but you need a catalyst. Here, we'll think of a heavy metal like platinum or palladium or iridium or something like that. That's pretty simple to do: It will happen for any normal, unconjugated alkene.

If I just have a carbon-carbon double bond, that will go fine, but if I have a hyper-stabilized double bond, like in an aromatic, that won't be enough. You'll have to use partial reduction.

For example, if I had styrene, and I used a normal catalyst, it's not going to do this, not the catalyst we're going to look at. But if we just took hydrogen and palladium, you would selectively reduce the carbon-carbon double bond that's outside the ring.

I'm writing one atmosphere here to show normal conditions, we're not trying to force it, then you get selectivity.

The Simmons-Smith reaction is the way of making a cyclopropane. So you would use methylene iodide, CH2I2, in the presence of zinc. And that CH2 unit would then become attached to the double bond, and make a three-member ring. So that's one way to make a cyclo propane.

Next is halogenation. Yes?

Student 3: [inaudible question]

Jean-Claude Bradley: You mean in the product? Let's make it clear by putting in the hydrogens. So you've got a double bond and the CH2 getting stuck on the top or bottom of the ring for that matter.

So, halogenation: Hydro-halogenation means to add Hx, halogenation just means to add x2. So, for example, adding Br2. I can add bromine across a double bond, just simply by adding it. You have to be careful with these reactions because this is in the dark: We know that bromine can react with basically alkyl groups, if you have light, so you have a complication there.

If you have a double bond, and you have other positions that would create stable radicals, then if you really want to do this, you have to do this in the dark. Without any light.

In this case, it wouldn't make any difference, because there are no stable radicals. If it just says Br2, and it doesn't say if it's light or dark, you can assume that it's in the dark. When we want to add light, we'll add it on purpose.

We are going to look at the mechanism of this reaction. This is kind of interesting: What happens here instead of opening the double bond, to the left or to the right, the way that we did a proton, we're actually going to go directly from the middle. We'll do something like this.

What happens here, we form this, and end up with bromine in a three-membered ring, with two other carbons: That's what's known as a bromonium ion. This is reactive as an electrophile, and what's going to attack it is going to be the Br- I generated at this step. So I kick out the Br- and the Br- comes back and reacts with the bromonium ion.

What is interesting about this mechanism is that, because it involves an Sn2 reaction, Sn2 reactions are always a backside attack.

That means that the two Brs are always going to end up on opposite sides, they'll always end up trans to each other, if there's a cis and trans possibility.

In ethylene, of course, there is no cis and trans in the final product, so regardless of what the mechanism would be I would end up with 1, two dibromyl ethane. So let's look at the possibility where you do have cis and trans, like in a ring.

Reacting cyclohexene to bromine in the dark, I would end up with two bromines on the opposite sides. So trans addition because of the mechanism.

With this mechanism, also, there are some competitive reactions that will happen. If I mix bromine with another nucleophile, like let's say water, the first step will happen, I'll still get the bromonium ion, but then I'll have a competition between the Br- and the water. So if we do the same reaction in the presence of water, we will end up with competitive reactions at the same time.

Let's take a look at that: We have considerably more water than bromine in this example, because we want competition to be one-sided in favor of water.

Let's take a look at the mechanism we would end up with: First, formation of the bromonium ion. Then, the water would come in and attack the electrophile from the opposite side of the ring. That's the final result: You get Br and OH added. Again, trans.

Lots of other things we could do with alkenes. We can also put an oxygen on it, the same way we made a bromonium ion.

In order to do that, we need what is called a peracid, it has CO3H. It's kind of a hybrid between a carboxylic acid and a peroxide. So whenever you see an oxygen-oxygen single bond, you know that's weak, so it's going to be easy to break that bond. And what the peroxy acid is going to do, it's going to deliver one of the oxygens on top of the carbon-carbon double bond.

That's called an epoxide. Epoxide is just a three-membered ring.

The thing about epoxides is that why don't we just call them ethers? Call a three-membered ring an ether? Because the nomenclature in organic chemistry is based upon behavior.

That means it has a different name because it doesn't act like a normal ether.

The fact that I have a three-membered ring means that it's strange. I have three SP3-hybridized atoms there, the angle wants to be 109.5 degrees, but it's forced into a triangle. So it's forced into being 60 degrees. For that reason, it's unstable. So epoxides can open up with a variety of nucleophiles. I think in the problem set we do a couple examples of that.

The "R" group here can really be anything, so as long as you have the CO3H, you can expect you're going to get epoxidation. In practice, typically you would use a compound called McPBA, which is metachloroperoxybenzoic acid. That would basically be an example of a peracid. It's common enough I expect you to recognize it, McPBA, in a problem. For epoxidation.

The next thing we'll look at a reaction with permanganate. There are two conditions we'll look at. The first one is cold and dilute. So that means you have potassium permanganate, but the conditions are very mild. And that will do something very different from when you heat it up. So in the cold, you're simply going to add OH groups on the double bond.

Now if you do the same thing under hot conditions, you will actually break the carbon-carbon double bond. "Conc" stands for concentrated, so if you have hot, concentrated potassium permanganate, you will break the carbon-carbon double bond, and furthermore you will oxidize either end as far as it can be oxidized.

In this example, when we break the double bond, first we replace C-double bond-C [C=C] with C-double bond-O [C=O]. So the initial products we get would be this. I'm going to put these in square brackets, this means that they're going to be formed transiently. In the reaction, you never actually isolate them, because soon as they're formed, they get oxidized. Aldehydes are easily oxidized through carboxylic acids.

So if you form an aldehyde at this step, you need to carry it through all the way to the carboxylic acid.

Permanganate is a pretty strong oxidizer, and it's going to do that. Okay, there are various permutations here, if you produce ketones, ketones will not be oxidized by permanganate. We'll do some of that when we do the problems.

While we're on this topic here, this top reaction, this hydroxylation reaction, can either be done with potassium permanganate under cold conditions, or it can be done with osmium salt and hydrogen peroxide.

Finally, if we treat the alkene with first ozone O3, and then dimethyl sulfide, you'll do pretty much what you did with the hot potassium permanganate, but you will not oxidize the carboxylic acids. So that means any double bond, you will break it here, and put oxygens on either side.

That's basically it for chapter 8, so we're at the problems which we'll do the next time we have class.

Transcription by CastingWords


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