Monday, May 22, 2006

Lecture 022: Alkyl halides part 2

Lecture 022: Alkyl halides part 2

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Jean-Claude Bradley: So last time we looked at the SN1 and the SN2 reactions. We also looked at what happens with chirality for SN2 reactions; we get inversion. So we're going to kick it off from there and we're going to look at solvent effects for these reactions. We talked about if I have a tertiary Alkyl halide, I'm more likely to have an SN1 reactions. If I have a primary I'm more likely to have SN2. That's just a tendency. For secondary you can have both easily. So it turns out, as a chemist, you have some control over that. By selecting a solvent you can favor more SN1 than SN2. So we're going to look at the solvent effects and we're going to try to understand them and won't have to memorize them. Let's look at doing an SN2 reaction first and two different solvents. SN1.

So if we do an SN1 reaction and we want to talk about solvent effects on speeding up or slowing down, remember that it's only the rate-limiting step we have to worry about. OK? Because anything else we do is not going to have an effect on the rate. So the rate limiting step in an SN1 reaction is the first step. It's that first step where you generate the carbocation. So say I use the same example that we did before, [xx] bromide. The first step is always the same thing, I'm going to lose the R minus and form a carbocation. So we're going to consider two different types of solvents, the very polar, hydrogen bonding solvent like methanol and we'll consider something that's less polar like chloroform. And we'll see if we can predict whether the SN1 reactions go faster in one solvent versus the other. If I were to view this in methanol and we want to find out if it's going to go faster; if there's going to be more stabilization on the right. If it's going to inhibit it we might see more on the left-hand side.

So let's go through the left and the right. On the left do we expect a strong interaction between methanol and the alkyl halide? It's just going to be dipole dipole and there's going to be solvation. We also want something that will dissolve the material we're doing. Ultimately, there's not any strong stabilization that I can draw here on the left but on the right however, I have charged species. When I have charged species, then it becomes very easy to bond with the methanol. So I can actually draw a hydrogen bond between this bromide and methanol like this. So normally, with a neutral bromine, I still have lone pairs but we never really draw a hydrogen bond. I have a bromine on the left but it's neutral so I just don't draw a hydrogen bonding in the bromine and the methanol. However, when I put that extra lone pair in, that now becomes negative so that's going to be strongly attracted to the slightly positive hydrogen on the methanol.

That's why we have hydrogen bonding on the right and not on the left. If I have this kind of bonding I'm going to get a stabilization of the Br minus. I could draw something similar for the carbocation but just for simplicity if I have an ion that's generated I'm going to get stabilization in a very polar or hydrogen bonding environment. That's the first conclusion here that in methanol the right-side is more stabilized. So if it's more stabilized that means it's going to be easier to form a Br minus and a carbocation in methanol than it would be in say chloroform. What we're predicting here is that for highly polar and hydrogen bonding solvents, the SN1 will be faster. So just to put my comparison here, in chloroform, CHCL3 we have less stabilization. The prediction is that in highly polar or H bonding solvents, SN1 is faster. So that's useful to know because if you wanted, for whatever reason, to get SN1 versus SN2 you could control that by just trying to get as polar a solvent as you could but still having the ability to dissolve your compound because that's one of the requirements of having the solvent effect work; your compound still has to be soluble. Any questions on this rationale?

OK, let's do SN2 and see if we can come up with something similar. For SN2, the rate-limiting step; the only step in the reaction has to be the rate-limiting step. So here we're going to look at what's on the left and what's on the right. And again, just to be consistent, I'll use my previous example; I think I used methyl bromide. So we're going to do methyl bromide plus Cl minus. This is SN2, Cl minus, methyl bromide. After the SN2 we lose the Br minus. We have those two arrows. Now I'm going to do the same kind of analysis. On the left hand side we would expect stabilization of anything with methanol. Same thing, we still have an ionic species so it's going to get stabilized. Also on the right we have a Br minus so we can expect stabilization in both places. We see stabilization on both sides so if we try to apply the same argument we did for the SN1, what do we predict would be the effect on the rates? You think it would stay the same? The thing is we have something else going on here.

If we look at the absolute energies that may be true, they both may go down to the same place but notice that if I have four hydrogens bonding to the Cl minus, it's tying up the lone pairs that are required to do the SN2 reaction. It's a different argument but it's kind of related. The effect of the methanol is to block one of the lone pairs that are required for the attack so the effect is the opposite. It's actually going to slow down an SN2 reaction. What happens on the right-hand side is not a concern at all because once the SN2 reaction happens, it's over and we go on to the next molecule. The fact that it's stable on the right really is not that big of a deal in this case. So we can say, it traps the lone pairs, which slows down the SN2 attack. So to write this more generally, we would say that highly polar or hydrogen bonding solvents slow down SN2. So the effect is in the opposite direction. That's good. It means now there's a reason why it's going to go faster and a reason why it's going to go slower in both cases.

So for the secondary alkyl halides now you can go one way or the other. Again, these are just tendencies. It's not so much that you're going to go from SN1 to SN2 but you might change the ratio and get fewer side products. And the problem set has a couple things on solvent effects; that's how we're going to rationalize that. Questions on that?

OK, it looks like we've covered everything we can on SN1 and SN2 so let's try to make sense out of this reaction. Here's a secondary alkyl halide and we're going to react it with Cl minus. I'm going to ask you to do it by SN2. So using the information about the mechanisms, we know that SN2 is a single step reaction and two electrons are going to start from the Cl minus. We'll attack the electrophilic and kick out the Cl minus. I didn't specify the chirality so we don't worry about it, but obviously we're going to get into an inversion here at the chiral center. In that transformation, we are going to expect to get 100% of the product. That is the case. So SN2 reaction goes exactly the way we've seen it before. Let's try the same thing and do an SN1.

I can't do this in one step so I'm going to have to do some work down here to go around. The first step in the SN1 reaction is the loss of the Cl minus and the formation of the carbocation. In terms of arrows, it's always the same, just one double headed arrow. And because I'm going to do a couple of steps here, instead of writing the carbocation plus the Br minus, I'm going to put the Br minus on the arrow but because I'm not adding Br minus, I'm losing it, I'm going to write minus Br minus. So that's a little shorthand when we have multiple steps and don't want to worry about side products. So we'd expect to get that and if we then follow what we're supposed to do at this step (this is the fast step), the step where the Cl minus attacks, we'd expect to get the same product that we did for SN1. However, when we do this reaction, that's not what happens. You don't get any of that product. The problem is that this carbocation that we just made can rearrange which is not the case in SN2. SN2 reaction as I said is the preferred reaction if you want to make carbon-carbon bonds because you're only going to get one product, it's going to be a clean reaction and you're extremely unlikely to get side products but when you have SN1 reactions, you have a lot of opportunity to get reactions that you're not looking for. The reason for that has to do with this carbocation.

What I'm going to say now is going to apply to any organic chemistry or any course you take in the future. Whenever you generate a carbocation, the first thing you have to look for is, can it rearrange to a more stable carbocation? We already looked at the stability of the carbocations. Tertiary is more stable than secondary, is more stable than primary. Same order as the radicals. The thing is the radicals didn't have a mechanism to rearrange. If I got a primary radical, it just stayed a primary radical, that's not the case for the carbocations. We had a secondary carbocation that I just drew here. If there's a way for it to become a tertiary carbocation it will. It can't do whatever it wants. There are rules in organic chemistry. One rule is when you have a carbocation you can do a 1, two shift. You can take a bond that is one carbon away and move it over as long as it gives you a more stable carbocation. So the group that I'm going to move is this hydrogen with its two lone pairs.

So I'm going to move the hydrogen and the bond over by one. The way that I'm going to show that (let me copy this molecule over so the arrows make sense) is by starting the movement of electrons at the CH bond and I'm going to end up on the carbocation. What that tells me to do is take the hydrogen and its two electrons and move it over to the carbon and don't leave any electrons behind. Now the carbocation on the right is no longer a carbocation, it's just a CH2 group. What I have left behind is no electrons so that carbon is going to be C plus. The driving force for this is that I generate a tertiary carbocation, something more stable than what I started off with. So we're going from secondary to tertiary.

Again, I can't move any bond I want, I have to move a bond that is immediately adjacent to that carbocation. This is called a 1, two hydride shift. There are many shifts that are possible. You can have 3, 3. You can have all kinds of things. The 1, two specifies that relative to the carbocation, the bond that's moving is second position. That's why you call it 1, two to specify they are adjacent, they have to be adjacent. It's a hydride shift because what's moving is not a proton. A proton is a hydrogen with no electrons. A hydrogen atom is a neutral hydrogen; a hydrogen with one electron. But this is a hydrogen with two electrons and that's called a hydride, which is an H minus. So you always do that whenever you make a carbocation, in any context. Now the mechanism resumes, so after we do the rearrangement we do the last step of the SN1 which is the attack of the nucleophile. I think we're using chlorine here and that just completes normally. That would be the only product that you would get from that SN1 reaction.

A common mistake here is to think that we are going to get a minor and a major product. Like I said, each chapter is going to have clear definitions about what's major and what's minor. Do not do any (see where the 'X' is here). Do not do any of this. Always do the hydride shift before any other product. The reason is the hydride shift is going to be faster the second step. Anybody have any idea why that should be. Why is a hydride shift faster than with Cl minus? It's closer. All you have to do in a 1, two shift is simply have the molecule rotated at the right angle and bingo, it happens. In order for the Cl minus to react the Cl minus has to find that carbocation so it could be traveling for quite a while relative to how long it takes to rotate. That's why the 1, two shift is always going to be faster than any intermolecular reaction that we can do. So remember that reasoning when you work these problems out.

So basically, that's it for SN1 and SN2 in terms of the complexity of the reaction. We've seen the effect from chirality, we've seen the effect of solvents and we've seen the effect from 1, two shifts. There's one other scenario I want to work out. What if we replaced this hydrogen, this hydride that moved with a methyl group? Is that a way we can avoid the 1, two shift? Let's take a look. So far, it's almost the same molecule. Again, I do Cl minus SN1 conditions. Let's take a look. First step will always be the same; we're going to lose a bromine ion. So now the question is do we do a 1, two shift? There's no hydride that can move here so the question is, can I move the methyl group just as easily as a hydride? That would make it easier, right? Because then you don't have to worry about hydrides but the reality is you can move any alkyl group. So yes, in this case the same thing will happen. It won't be a 1, two hydride shift; it will be called a 1, two alkyl shift but it's going to have the same probability of happening if it can make a more stable carbocation. This is a 1, two alkyl shift. So we go from secondary to tertiary. Ok, so that would be the only product in this reaction.

You notice how long it took to do the problem. That's how long it will take you on any quiz or exam. This is probably the longest problem that you're going to have to do. Just take your time. It's always the same steps you need to apply, just be careful about those issues. We're going to do a bunch of examples in the problem set also. This is basically it. So how do you know what group you could do a 1, two shift? An easy way to do that is to count the two from the carbocation so if I put my pen on the C plus I'll count to two and see what I fall on. So if I count one, two, I'm in the methyl group, so that methyl group could be a 1, two shift. Same thing, I could do one, two and hit all these methyl groups. We already spent a lot of time, so hopefully it's clear those are all the same shift so we don't repeat those. The hydrogen is directly connected to the C plus. I only count the one and I hit it. The only other group that I could move is one of the hydrogens on that methyl group. So if I start at the carbocation and count one, two - I could think about moving that hydride over but that won't happen because that would generate a primary carbocation. You should be prepared to do all the possible shifts and then determine if any is going to give you something more stable. You can generate something of equal stability; secondary to secondary, don't worry about it. I'm interested in seeing the major shifts so I'm interested in primary to secondary, primary to tertiary, secondary to tertiary or we also saw [xx] was stable. So anything that is much more stable.

In reality do you have secondary to secondary shifts? Yeah, you do and that's what complicates SN reactions a lot more; but what I want to see is that you show me the major shifts that are going to happen. If you have two tertiary products then you need to go both ways. You're going to have both products. Questions on this? This is just practice, you've got to get in there and do some problems. So far I've used Cl minus as the nucleophile but if you tried to do the quiz, I often used OH minus and there's a good reason for that. Let's take a look at a simple example of 2-bromopropane reacting with OH minus. And again, when reacting with OH minus I have to tell you what kind of reaction to do. Let's take a look at SN2.

OH minus is very similar to Cl minus; they are both pretty good nucleophiles. I have a negative charge on an electronegative atom; that makes it nucleophilic. Now, I'm going to do the standard SN2 reaction that we've done several times. So my only product would be 2-isopropanol. Similarly I could ask you to do the same reaction; OH minus and SN1. We've done this a number of times. Get a carbocation. Now you know, once you generate a carbocation, you have to look for 1, two shifts. This is a secondary carbocation. If we count to two from the carbocation, one, two, we end up on one of the end hydrogens. However, if I tried to do a 1, two shift I would go from secondary to primary so nothing will happen. So, we checked it and there's no 1, two shifts.

Now, we move on to the last step. In this case we also get isopropanol. So whenever you don't have a 1, two shift odds are you're going to get the same product for SN1 or SN2. The exception of course being with chirality. SN2 does inversion and SN1 will randomize the chiral center but aside from that they are pretty similar. So that would be the end of the chapter except that there are other reactions that will happen between OH minus and electrophiles such as this. There are two other cases. The alternative if we look at the SN2 reaction, OH minus is acting as a nucleophile which means it's coming, attacking with the carbon, and is displacing the leading group. But OH minus is also a pretty good base so if OH minus acts as a nucleophile, you'll get your substitution but if it acts as a base you're going to get an alkene from that. The difference is pretty minor arrow-wise. Instead of attacking the carbon, we're going to attack one of the hydrogens on the methyl group. Let's look at how that works.

It starts off the same way but it's going to attack a hydrogen that is three bonds away from the bromine. So as it's grabbing on to that proton the bond is breaking between the H and the C. So I start from the middle of the CH bond and end up in the middle of the CC bond. I can't stop there because I would have a carbon that has five bonds so in order to complete this, at the same time the bromine has to leave with two electrons. So if I follow everything it is telling me to do it's saying make the carbon carbon double bond between those two carbons and on the left side of the alkene I have methyl and a hydrogen and on the right side I have two hydrogens. This reaction is also second order. It happens all at once, just like the SN2 reaction but it's not a substitution reaction, it's an elimination reaction so instead of SN2 it's called E2. Two meaning second order. The other products we have here are water and Br minus. So if there's an E2, you can imagine there's an E1 and it's going to work in a very similar way...

So now I'm specifying that I want to do this by E1. That tells me, because it's an E mechanism, it's going to start off the same way as an SN1. Once again, any context where you have a carbocation, the first thing you do is look for 1, two shifts. We don't have any in this case so if we were to finish this reaction with an SN1 we would just clip on the OH minus, but we are not, we're doing an E1 so what happens instead is that we are going to lose a proton. The OH minus is going to come in and instead of going to the C plus it's going to go to this end proton and then we are going to get the double bond. So in this case E1 and E2 gave the same product and there is only one product in this case. That's basically the four reactions we cover in this chapter and that is the core mastery material and being able to take into account all of those situations, what happens with chirality, what happens with solvents, and being able to predict the products. On Friday, I will finish up the theory part and we're going to get into some problems.

Transcription by CastingWords


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