Monday, May 22, 2006

Lecture 021 Alkyl halides part 1

Lecture 021 Alkyl halides part 1

View Lecture

Jean-Claude Bradley: We're going to start off on chapter six today, which involves alkyl halides. You're already familiar with alkyl halides; we've used them in the free radical halogination, that's how we make them, that's one way we saw how to make them. In this chapter, we're going to take a closer look at alkyl halides in detail, how to make them different ways than what we saw so far, and some of the reactions that they do, and then also we're going to look at some additional mechanisms that are completely different from the free radical halogination.

We're going to start off with nomenclature, just to make sure that we all know how to name these things. Again here, no big surprise, you've already been naming alkyl halides, because when you look at the products of free radical halogination you have to know how to name them, but let's just make sure we know two different ways to name alkyl halides. If we have a compound like this, you can use upac format, and you can say I have three carbons as the longest chain, so I have propane, the chlorine is on the second position, so this is 2-chlorl propane. What's an alternative? It'll only work for groups that have four or fewer carbons, though. If we consider the three carbons to be a fragment, then we can consider this to be an isyl propyl chloride. That's the other way that you came name these. It's a little bit less general because it'll only work with a fewer number of carbons, but you should be familiar with that also. Don't get confused by the fact that it almost looks like it should be a salt. Sodium Chloride is a propyl chloride, this is certainly not an ionic compound, it's just nomenclature. Really, it's all covalent. There's really not that much more to do about nomenclature because we've done so much already with alkyl halides.

I'm going to do the preparation, and then I'm going to look at the reactions of alkyl halides, but first the preparation. There are four ways we're going to look at making them. I have to list what we already saw, free radical halogination, we spent a lot of time on that, so I'm not going to give you all the details, I'm just going to summarize it with ethane. As an example, we use the bromination. You now know all the details about that reaction, you know what all the major products are going to be, you know what happens with chlorine, you know what happens with fluorine from the salt activity, so I'm just going to list it to make sure we're complete here.

Second, we have hydrohalogination of alkenes. In organic chemistry, there's a lot of really big words, don't get intimidated by that, just break it down, so if you see a new reaction, basically just take a look at what composes it, so in this word here, hydrohalogination, we have halogen, which we know what it is, it'll be an F, Br, Cl, or an I, and we have hydro. Hydro can sometimes mean water, and sometimes it can mean hydrogen. In this case it happens to mean hydrogen. We're doing a hydrohalogenation, which means that we're going to be adding a hydro halogen, so a hydro halogen would be something like HF, HBr, HCl, HI, of alkenes, so an example of that would be this. Reacting this with HBr, and I'm going to put the H on one side, and I'm going to put the Br on the other. What I've done is I've added HBr, and so this is a special case of a larger group of reactions that are called addition reactions. We add two things together, one plus one, we only end up with one molecule, and there's special rules for this reaction that we'll take a look at a little later, probably in the alkene chapter, but for now it's basically the important thing is that you know you have an alkene, you know you can add HBr, and you know you're going to get an alkyl halide, but again if you have different kinds of alkenes, you'll notice there's two ways you can add the HBr, so there's actually rules you'll need to follow for that, but for now in this chapter, we'll just look over this very generally, and we'll look at the details later.

The third, we have from alcohols. There's a couple ways to do this, but the way that I want to mention here is this. We start with an alcohol, and we react it with HBr, so we're going to substitute the OH for a Br, so we're basically going to swap the OH for Br, and the H will go on the OH, so we'll end up with water and an alkyl halide. Again, there's some rules for this, but we're going to leave that a little later.

The last one is from another alkyl halide. If we start with methyl bromide, and we treat it with Cl-, we can actually swap the Br for Cl, so that means the other product in this reaction will be Br-. A couple of things about this, on the arrow I'm putting Cl-, obviously I can't just add Cl-, I have to add a counter ion, but the counter ion is not important, because it's just going to be a spectator ion, so I may or may not name it. I can say that methyl bromide reacts with Cl-, I can say that methyl bromide reacts with NaCl, it's all the same, and depending on the problem you may see it differently, so just keep in mind that just because we don't mention a cation, there's still a cation there, just like there's a salt there that we don't list here. This here is a substitution reaction, because we're exchanging one atom for another, or one group for another, it's not like an addition where we take two molecules, put them together, and end up with one larger molecule, here we're just swapping a Br for a Cl, so it's a substitution reaction, and we're going to look at this one in detail in this chapter.

Of those four, what do I expect you to know? Again, we didn't look at the mechanisms for number two and number three, but I do expect you to know that those are two additional ways to make an alkyl halide, so you could have things like Ethyl bromide could be prepared from which of the following? Well, we saw that we could make it from an alcohol, so ethyl alcohol would be okay, we saw that we could make it from an alkene, so ethylene would be one possible answer, so that's at the level that I expect you to know it until we look at the mechanisms of these things in more details. Same thing for the reactions, that's how we make it, now what do you do with alkyl halides? They're actually really useful compounds, as we'll see why we're spending so much time on these mechanisms.

There are two reactions we can do. First thing is an elimination reaction, and that's actually just the reverse of what we did when we made an alkyl halide with an alkene and HBr, so I can reverse that process, let's say I start with this ethyl bromide, and all I really need to do is add a reasonably strong base. As an example, hydroxide, OH-, but many other bases, including amines will do the same thing, so basically what we're going to do is reverse the reaction, we're going to lose Br in this case, so we'll end up with the alkene, so there's the Br going backwards. This is another reaction that we'll look at in detail in this chapter.

Let's focus on what we're doing here, we're looking at reactions of alkyl halides, one of the things that we can do is we can create an alkene, another thing we can do with an alkyl halide is I can repeat the substitution reaction, so again if I have any alkyl halide, I can treat it with Br-, and I can do that swap again. Alkyl halides are actually really useful intermediates, these are just two of many many reactions we can do, they're really a good start, because these mechanisms that we're going to look at will be very general for lots of stuff that we're not going to explicitly do here. Let's say that we want to consider this reaction in detail the same way that we looked at the free radical halogination. What we want to do is look at the mechanism, how it actually goes from the left to the right, because if we don't look at the mechanism, if I draw another alkyl halide and I ask you for the products, you can take a guess, but you won't be able to predict with certainty. Once we knew the free radical halogination mechanism, you know what would happen just based on primary, secondary terstry[sp], and based on the stability of the radicals, so that's the level that we want to get at. One of the ways to understand the reactions is to do kinetics, to see how long it take for the two reagents to react with each other, so let's do that with this reaction. What we're going to do is I'm going to draw this a slightly different way, and I'm going to draw, instead of putting it on the arrow, I'm going to make it a little more clear. That's the reaction we're looking at, so if we look at the right of the reaction, one thing we can do is we can change the concentration of the reagents, and see what happens to the rate. So if we know what happens to the concentrations of these two, we can predict how long the whole reaction is going to take. The rate will be equal to a constant K, which you calculate experimentally by doing a couple runs, and it's found experimentally that it's related to the concentration of the methyl bromide, but it's also related to the concentration of Cl-. You guys familiar with these brackets? That should be straight out of first-year chemistry, right? So what's the order of this reaction? Second order. So the order of the reaction is, if it's dependant upon the concentration of two reagents, it'll be two, if it's dependant on one, it'll be one, if it's dependant on the square of one of the reagents, it'll be two, right? Basically this is a second order reaction and it tells us something very important, that basically if I double the concentration of the Cl-, the reaction will double, if I double the concentration of methyl bromide, the speed will double, if I double both of them, the reaction will quadruple. So if we're looking at the rate, if you remember the free radical halogination, it wasn't every single step that actually determined the rate of the reaction, right? There was actually one step in the mechanism that determined the rate, and that's all that you can really measure experimentally, is the rate of the rate-limiting step, so what this is telling us macroscopically, is that in the rate-limiting step, both methyl bromide and chloride are involved, right? Because if methyl bromide was part of the rate-limiting step and chloride wasn't, then chloride would have no effect on the speed of the reaction. So I'm going to rewrite this to say specifically what the conclusion is. So from these experimental results, we conclude that both methyl bromide and Cl- are in the rate-limiting step.

It turns out that this one is a lot simpler than free radical halogination, there's only one step, and it turns out to be the rate-limiting step, right? It works like this. The methyl bromide, we learned, was polar, it's got a polar CBr bond, so we know from that that the bromine will have slightly higher negative charge, and the carbon will have a slight positive charge, which is important. The Cl- has a bunch of electrons; specifically it has four on lone pairs. One of those lone pairs is going to be attracted to the slightly positive center on the carbon, so when we draw an arrow, I'm starting the arrow on the lone pair, remember when I start an arrow it always has to start at the source of electrons, and it has to end up where the electrons are going, so the electrons end up directly on the carbon atom, that's what they're attracted to. Now if I didn't draw any more arrows, then my carbon would end up having five bond, so-called texas carbon, you never want to have that, so what has to happen at the same time as I'm forming that bond, is I want to break a bond, I want to break the CBr bond, and the way I show that is by saying my electrons will start out at the bond, and will end up on the bromine, like that, so simultaneously, as the Cl- is coming in, the Br- is leaving; at no point does my carbon have five bonds, it's going to have three bonds plus two partial bonds, which is okay, we just can't have five full bonds on that carbon. The result of this is that I end up with a Cl, my hydrogen's, and Br-. That's the whole mechanism. This is a substitution reaction, and it's called a nucleophilic substitution reaction. Substitution we already saw, because I'm exchanging one thing from the other, and it's nucleophilic because, very differently from the free radical halogination where it was all free radicals, here we actually have charges, we have Cl- coming in, and there's actually two reactions we have coming in, it's not one electron reactions. So in this case we have one species that is seeking a nucleus, seeking a positive center, that would be a nucleophile, so in this case a nucleophile, which is generally electron-rich, is at least going to have at least one pair to be able to do the attacking, that will be the nucleophile. Phile means to like, nucleophile means to like a nucleus, the reason for that naming is that nuclei are positively charges, so you want to translate nucleophile to seeking positive charge, or slightly positive charge. On the other hand, I have the electrophile, that loves electrons, is looking for electrons, more specifically a lone pair, so that's why this reaction is called nucleophilic substitution, but we have to make one more refinement here, it's a nucleophilic substitution reaction that is second order, so if we want to completely name this reaction so every chemist would know what we're talking about, we would call this S for substitution, N for nucleophilic, and two for second order. That basically for now is what we need to know about the SN2 reaction. The electrophile is actually the whole molecule. The carbon would be the electrophilic center, it would be the point of attack, but if I asked you to point out the electrophile, it would be the methyl bromide. The full arrow means the movement of two electrons.

This is a little more useful, you know a little more about the reaction, and you can start to make predictions, but you will see that your prediction will fail very quickly. Let's take a look at another molecule and try to use the same mechanism. If I try to do the same thing with t-butyl[?] bromide, react it with Cl-, I can in fact substitute the bromine for the chlorine and end up with Br-. Now if I do the kinetics on this, though, something strange happens, I get a very different result than I did the previous time. The rate of the whole reaction, in other ways the rate at which I observe the t-butyl[?] chloride forming is equal to a constant times the concentration of t-butyl[?] bromide, and that's it. In other words, it doesn't matter what the concentration of the chloride is as long as I have some minimum amount, because if the concentration is way too low, obviously I can't do the reaction, but if I have a minimum, it doesn't matter what the concentration of the chloride is, which is very different from the previous reaction, so this reaction is not second order, this is first order, and it tells us that in the rate-limiting step, only the t-butyl[?] bromide is present, so that tells us we absolutely have to have more than one step in this reaction. Let's take a look at what could actually produce this, like I said we have to have at least one step, luckily we have two steps, so there's a lot less steps than free radical halogination. In the first step, what happens is the t-butyl[?] bromide, all by itself, will decompose heterolyticly[sp], so this arrow is saying take those two electrons from that carbon-bromine bond, and but both of them on the bromine, and leave none on the carbon. So if there're no electrons left on that carbon, I have to have a carbyl cation, I have to have the C+, and I have the Br-. In the second step, this carbyl cation will then react with this Cl-. Again I start at the lone pair on the chloride, and I end up on the carbon that's positively charged. So overall, it's basically the same thing, it's just that the SN2 reaction happens in one step, and this, maybe you can guess, would not be an SN2 reaction, it'd be an SN1, because it's first order. Everything else is pretty much the same, we have our nucleophile the same, Cl-, we still have the electrophile, which is the alklohalibe, the nomenclatures very similar, everything's very similar, but as we can see, this kind of thinking will cause a problem. In general, as a chemist, we want to do SN2 reactions, because we can see SN1 reactions have some pretty nasty side products that'll come from it, because we end up with this carbyl cation. That's the overall picture; we'll look at that in a little bit.

What's different between those two compounds that one is going to go by SN1 and the other is likely to go by SN2? It comes back to our primary/secondary tersery[sp]. One is a tersry[sp] alkyl halide, this one right here; this is the rate-limiting step. Again, to compare these two, we have a tesry[sp] alklhalide that's doing an SN1 reaction, and with alkyl bromide that was actually beyond primary, so I use those as two extreme examples, so what we can do is actually write out a little page. So we have a spectrum, and what we can say is that on the left hand side, we're going to have a tendency towards SN1. All the way over on the right side, we would have SN2. Now, in the case of secondary, you would expect that you would have competing SN1 and SN2 in a lot of cases. Now I have to stress that these are tendencies, so what that means is that if I have a compound, I can make you do an SN1 or an SN2 on any alkyl halide, it doesn't mean that it will be the dominant reaction, but still if I ask you what the tendency is, for example if I draw you t-butyl[?] bromide, you would look at it and expect it's probably going to undergo an SN1 reaction, but any question I can definitely ask you to do either one. You'll see that in the quizzes as we go through them. So why is it that we have this tendency? We can actually go back and understand that if we look at what the intermediates are. In case of the tersery[?] alkyl halides, you'll notice that your intermediate is a tersery[?] carbyl cation. Tersery[?] carbyl cations and radicals have a very similar stability. So that's one reason that basically your tersery[?] will favor SN1. In other words, if I had methyl bromide, and I tried to do an SN1 reaction, it would be a lot harder because that methyl cation is relatively unstable when compared to the t-butyl. The other reason has to do with sterikindrence[sp], which we looked at a little bit with Newman projections, with the cyclohexane folding, and it just means that two groups don't want to be in the same place at the same time. If I'm trying to do an SN2 reaction, you'll see that the nucleophile has to come in intimate contact with the electrophile. It actually has to get to that carbon before the reaction can happen, so if I don't have big groups that are around the carbon, it's going to be much easier to do that SN2 reaction, so if I try to picture doing an SN2 on t-butyl bromide, look how different it is... So relative between these two, there's little setrokindrence[sp] in the case of a primary or a methyl alkyl halide, so this would be a primary or methyl, and in the case of tersery[sp] the Cl- is going to have to negotiate those methyl groups in order to make it to the center to that molecule, and that can be difficult, so Cl- is bouncing around. The probability that they get right between those three groups and manage to meet the center of the carbon is far less than in the case of the primary or methyl.

Those are the two basic reasons, and they both go in the same direction, so they're additive, so there's basically two reasons why your primary or methyl alkyl halides will do SN2, but again you should be ready to do either one if you're asked to. When we do the problems, also we'll take a look at these. There will be two molecules, and you'll be asked which one will do an SN2 reaction faster or slower. So SN1 and SN2 have a lot of different consequences and one of them will also be with chirality, so what happens when we do SN1 and SN2 with chiral compounds? In order to understand what happens, we have to look at the details of the SN2 reaction, so I'm going to go back and draw a transition step for that, and then we'll see if we can see what actually happens. I want to draw everything in one line, I want to draw the nucleophile on the left and the electrophile on the right, and this is an SN2, and I'm going to end up with Cl on the left. Drawing a transition state, all we do is we look at the average of the left and the right, the energies here are pretty similar, so basically we can assume that the transition state is going to be in the middle, so your transition state is going to have an equal length of bonds, so Cl- comes in, attack, Br- goes out. I'm going to have a partial bond between the chlorine and the carbon in my transition state, and I'm going to have a partial bond between the carbon and the bromine as well, because I'm going from a full bond to zero bonds. What happens to the other three groups is kind of interesting because if you notice on the right, it's almost like they went over to the other side, and that's exactly what happens actually. Chlorine is coming in on the left, and this is an sp3 hybridized center, so what happens at that transition state, those three groups actually end up all lined up at the middle, so they could go from the right side, and flip flop over to the right side. One more thing about this transition stat is that we have to worry about charges, so the charge on the chlorine would be what in the transition state? We don't need to know exactly, but it's going to be somewhere between minus one and zero, so any transitions that you ever do, you do it the same way. You basically look at where it starts off and where it ends up, and in the middle we assume it's going to be something between the two, but we don't have to say necessarily that it's negative one-half, we just say that it's slightly negative, for the same reason that we say the bromine is slightly negative, but the carbon doesn't at all, because it doesn't go from plus to minus, it goes from neutral to neutral. So this is a transition state, and what happens is that those three groups in the middle flip from one side to the other. It's very similar to what happens to an umbrella when it gets inverted in strong wind, you've got this pyramidal kind of structure, and then it flip flops over to the other side, so this is called an inversion, named after the guy who discovered it, a Walden inversion. The fact that the three groups flip over for methyl bromide really doesn't matter because after you make the molecule it doesn't matter what the hydrogen's went through, you're still going to get the same compound. The time when a Walden inversion is important is when you have a chiral structure, because a Walden inversion will invert the chirality, so let's do a chiral molecule and see what happens.

So if we start out with enantiomerically pure 2-bromyl butane and we treat it with Cl-, and I specify that you do an SN2 reaction, then the Cl- will come from the back. An SN2 reaction is always a backside attack where your leaving group is coming off and the nucleophile is always coming directly behind, so what happens is we basically invert the chirality of that center. So we get a hundred percent conversion of that center. This is the first time something like this has happened, so far whenever we go through the mechanism, we end up with a mixture, this is different. This is also why SN2 reactions are so useful, because you don't get random products, you get 100% inversion, so biology uses this to its advantage. DNA is made like this, proteins are made using SN2 reactions because of the very tight control you have, and there really are very very few side products if any from SN2 reactions, so it's a really good reaction to understand carefully as an organic chemist.

Transcription by CastingWords


Post a Comment

<< Home