Lecture 4: Unimolecular Elimination
King, Susan
This is the second quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 Okay, alright, so here's where we left off last time. When we were talking about the mixture of products from this particular alkyl halide, and we saw that we got 41% of the trans, so let's label that. 41%, that's not a good sign what it just did over there, huh? Trans. You know, you can talk about overhead projector being old technology, but it's a lot more dependable than this, right? Okay, so that's trans. This one is cis, and this one here is least substituted. Alright, so let's show how we get the, we can easily see how we can get the least substituted by eliminating right here this carbon-carbon bond that I'm pointing to. Alright, let's see how we get the cis and the trans. So here's, I just put this in this orientation, I put it into a new one projection so you can see where that comes from, and so let's scroll down a little bit. So here's, let's, here's our leaving group. Here's our beta hydrogen. So leaving group, and here's our beta hydrogen. So that's on the back carbon, the way I have it drawn. But if we eliminate the leaving group and the beta hydrogen, we get this alkene. This one is going to be the trans. So it's going to look like that. And now, if we rotate, so notice we have, we have two beta hydrogens here. We have this one here, and we have this one here. And so if we rotate this down, let me change the color for that. I want to just rotate, I'm going to keep the bromine on the top so I'm just going to rotate this hydrogen right here, that's going to rotate down. So it's anti to the leaving group, and that's what we want. So now that hydrogen's right here, and here's our leaving group staying the same. And it's just doing that again. Okay. Okay, so there's our leaving group. We have our beta hydrogen, and if we eliminate there, you can see that now you can see that the ethyl and the methyl are now going to be on the same side. So we have our carbon-carbon bond here. The two hydrogens are on the same side, and over on this side we have the ethyl, and this side we have the methyl. So that's how we got the cis and the trans. And we want to have the bulkiest groups on opposite sides, so this one's going to be more stable. Most groups on opposite side. And therefore it's going to be the major product. And so we definitely see that that one is the major product up here. Questions about that problem? Anybody? How we got that? Yes? Show the arrows for elimination, it's kind of hard to do on a Newman projection. The least substituted one? Yes, let's do that right here. Since there's no stereochemical issues or no different ways to draw the, there's no cis and trans when it's on the end, we'll just, I'm not going to put it into a Newman projection. So we'll just draw it right here. And the way I usually draw this when I'm not in a Newman projection is I definitely going to put the hydrogen on the opposite side of the leaving group, anti-periplanar. So our base is a foc side, so I'm just going to abbreviate that ETO minus. And that should do the command and grab the beta hydrogen. We're going to move electrons here to between those two carbons. Leaving group's going to leave, so I'm giving it its lone pairs. And that's what that looks like. So that's all three products. So if the reactant only has one hydrogen bonded to the beta carbon, there's only one conformer in which the groups to be eliminated are anti. The particular isomer that's formed depends on the configuration of the reactant. So we don't automatically draw the trans because you might not be able to get the trans. If you're showing stereochemistry in the reactant, you might not be able to get into an orientation to give the trans product. So this is a good example right here. This one only has one beta hydrogen here on the least substitute. That's going to give the most substituted alkene. Okay, so this is our beta hydrogen here. So you eliminate a hydrogen on the least substituted, on the most substituted carbon that will give you the most substituted alkene. So we certainly can eliminate one on this, this, one of these beta hydrogens, but that's not going to give us the most substituted alkene. And I'm just trying to draw the ZYTESF product. All right, so we only have one. And so when the base comes in, a foxide ion to remove that beta hydrogen. So here's some more arrow pushing. We need a lot of practice with this particular mechanism because it's just a strange reaction looking reaction. All right, so you can see there is only one product that we can get. So we can't even, we can't really even get any of the product where the phenyl and the hydrogen are trans to each other. That's the only product that we can get. All right, and so that moves us to competition between SN2 versus E2. All right, so we're done talking about all the aspects of the E2, and now we're done to talk about the competition between the SN2 and the E2 because SN2 and E2 reactions are in competition with each other. Is there a way to favor one or the other? Let's show both of the reactions here. So this is our nucleophile slash base. Just put base here. It's the same species. Most of the examples we've done have used a foxide. It's the same species. We're just calling it something different. But both terms can be used interchangeably. So we have our base slash nucleophile. We're going to have a red pathway here. We'll call that A. Where the nucleophile comes in and removes the beta hydrogen, we break the carbon-hydrogen bond to make a new bond between carbon and carbon, and then the leaving group leaves. So A is the E2 pathway right here, and our product is an alkene. So this right here is the conjugate base. And then we have our leaving group X minus. All right, alternatively, we'll call this one pathway B down here. The nucleophile can come in and do backside attack. So that would be coming in from backside here, attack the carbon, break the carbon X bond here. So we look like that. Those are really the two things that we're comparing. Yes? I think it's conjugate acid. Acid? Yes, it is. Thank you. Thank you for catching up. All right. And as you can imagine, it's much easier for the nucleophile to reach this hydrogen here that's on the periphery of the molecule, much, much easier than it is for it to do backside attack, especially when we have a very hindered carbon bonded to the leaving group. Okay, so that's really going to be the whole entire thing. It's all about steric hindrits in the alkyl halide. It's also about strength of the base slash nucleophile. The more basics the nucleophile get, the more favorite E2 will be over SN2. All right, so what determines which is going to be the major pathway? It is the structure of the alkyl halide, the structure of the base, and temperature. So as we're going to learn, temperature favors elimination. Definitely temperature favors elimination. So this is the structure of the base slash nucleophile here. So let's first look at the structure of the alkyl halide. All right, so you can see in this first substrate, we have a really nice primary alkyl halide. And so it's going to be really easy for the nucleophile to come in and attack. Not very much steric hindrance. The carbon that's bonded to the leaving group has two hydrogens and one R group attached to its primary. So it's going to be very easy for the nucleophile to come in and do backside attack and kick off the leaving group, which for now our leaving groups are all halogens. When we get to chapter nine, you'll see that they don't always have to be halogens. Or for right now, they are. All right, so what we see here is that attack at the carbon is unhindered. So therefore SN2 is going to occur. In the second example, we have a little bit different issue here. We have a very hindered alkyl halide. It's tertiary. And we talked about the fact last quarter that we don't even get SN2 on a tertiary alkyl halide, but we do get E2 on a tertiary alkyl halide. So this is way too hindered for a substitution, but it is actually, that whole idea is going to favor, and this is what we've got to stop doing here. This is going to favor the base in acting as a base. So the base slash nucleophile is going to act as a base instead. And remove a beta hydrogen. So we've got nine different beta hydrogens to remove. We can remove any one of those nine. I'm going to pick the one that is antiperiplanar. All of these bonds can rotate, so really any of these hydrogens can be removed. I'm going to pick the one that's antiperiplanar. We're going to make a new double bond here, and then the leaving group's going to leave. So what you see is that a tack at carbon is blocked so that E2 occurs. So it's really easy to make a decision when we have a primary or a tertiary substrate. It is not easy to make a decision when we have a secondary, because now the rates of those two reactions are going to be much closer together. So let's look at some examples. That tends to be a good way to see what's going to happen. Let's look at some real examples that have been run with primary, secondary, and tertiary alkylheylides. And what we're doing is we're choosing a thawkside ion, so that's a good nucleophile. It's also a good base. And the temperature is 55 for all of these. So it's a little bit elevated temperature. Elevated temperature favors E2, but let's just see what our actual outcome. So for a primary alkylheylide, here we have ethyl bromide. You're going to get mostly SN2 with a thawkside. So let's see what numbers we get here. So our product here for the SN2 is an ether. Again that's going to take a little practice to be able to draw these products quickly and easily. It's an SN2, and we get 90%. The E2 reaction gives ethylene as a product, and that's 10%. So when I'm asking for the major product, which is what I do most often, I want to see only the SN2 product for a primary alkylheylide. It happens with the secondary. Draw the products, and then we'll just kind of summarize here. This is our SN2, 21%. Not very good, huh? 21% SN2. It should, yes. Let's fix that. Here we go. Okay, so we get that, plus. This one only has one E2. One possible E2 product, and that's 79%. All right, so for secondary alkylheylides, E2 is favored with ethoxide. So ethoxide is a good nucleophile, but it's also a fairly strong base. How do we know it's a fairly strong base? Well, what we do is we look at the pKa of the conjugate acid. Okay, so let's look at the pKa of the conjugate acid. I'll just take these pleasant little pauses here. We're going to be calling about that pKa of the conjugate acid. That's one of the ones I want you to know rounded to the nearest 15. It's an alcohol, so it's about 15. Okay, so we want things that are fairly strong bases. It's about above pKa of 8-ish for the conjugate acid, 8-9-ish, and that's well above that, so that's a fairly strong base. With a secondary alkylheylide and ethoxide ion, E2 is favored. So if I'm asking for the major product, you would only draw the elimination product. Questions so far? Anybody? Yes, up at the top. I will provide any pKa's that are not the 8 that I want you to memorize. Okay? Yes? It still would be, and the major would be E2. Yeah, absolutely. That's a good question. All right, let's look at tertiary alkylheylides. Well tertiary alkylheylides don't do SN2, right? So no SN2 for tertiary halides. Therefore, E2 only. All right, so we only have one possible product with this particular alkylheylide. 100% of the alkylheylide. All right, so what happens if we change to a bulky base? So what if we take a primary alkylheylide and we use a bulky strong base? So how about terputoxide better rather than ethoxide? And I would recommend that you keep going back to this page because these things come back. And this is a page I often refer to when students ask me questions. So we have a primary alkylheylide. We have methoxide ions, so a little tiny bit less hindered than ethoxide. And we have terputoxide. What do we actually get? So this one we get 96% SN2 and 1% E2. Don't know the temperature on this one. It's probably a lower temperature. It would make sense, huh? Probably a lower temperature. This particular experiment I don't have the temperature for. You'll also notice that the numbers are not adding up to 100, so we've got 96% and 1%. So when that happens, it means these are isolated yields rather than, you know, for what I want to say, ideal yields or, you know, these are isolated yields when they don't add up to 100%. And then in the second example, when we do the same alkylheylide and terputoxide, we get only 12% SN2 and 85% E2. So the message here is that you can actually determine which product that you want by choosing the correct base. If you have a primary alkylheylide and you want elimination as your major product, you better choose a bulky base, okay? If you don't want elimination on a primary, then you want to use an unhindered. If you want elimination on a primary, then you're just going to use a normal base. I mean, I said that wrong. If you want substitution, you would just use it. You know, you're not going to have to worry too much about elimination with a primary alkylheylide. Yes? Yes? What happens on the top part of this point? Why do you choose to do with, like, the CS3, CS2, and A and CS3, CS2, and OH? Like, why do you choose to do with your differences? The alcohol is the solvent. And it's very common with a thoxide or alkoxide bases to have the alcohol as a solvent. In fact, you can make those by taking an alcohol and adding a chunk of sodium to them, okay? So that's the most common solvent used for that. All right? More questions? Anybody? Yes? I have a question about the primary alkylheylide. Okay. Why do you choose to do with a polyaprodic solvent? I thought you'd choose to do with polyaprodic solvent. Okay, really good question. So E2 is faster in a polyaprodic solvent, but polyaprodic solvent is often not the most convenient. So some of the polyaprodic solvents, like, let me give you some examples. The two most common that we use in this class are DMF and DMSO. Those solvents are a little bit problematic because they have higher boiling points. So once you isolate your product, you can't easily remove the solvent. That's a problem. The second problem is have you guys done an extraction yet in the lab? You did it in T-chem though, right? In a sept funnel? You know how you had the two layers, and that's how you often isolate products in organic chemistry. If you have DMSO and DMF, they don't form two layers. It just will make, it'll make, the water will not form a separate layer. So it makes it a lot more difficult to isolate your product. So that's why we don't use acetone. And like, let's say for this one, ethoxide ion in acetone, why not just use acetone instead? You know, why not? And the reason is that the sodium ethoxide doesn't dissolve in acetone. You see what I'm saying? So there's all sorts of problems. So for the SN1 and the E1, we have to use protic solvent for the SN2 we don't. Okay? All right. We've got a little bit left here. So to favor E2 over SN2, use secondary or bulky halides and strongly basic and or bulky nucleophiles. Increasing temperature also favors elimination over substitution. So there's where we talk about the increasing temperature. To favor SN2 over E2, and we're especially going to have to worry about this with secondary alkyl halides. We want low temperature, so room temperature for example, and weakly basic unhindered nucleophile. That's a tall order there, really. Because a lot of nucleophiles are also strong bases. So what kind of nucleophiles are we looking for? Here's some examples of things that are good. Nucleophiles that are weak bases. All of the halogens except fluoride's not so good, right? I died. Bromide. Another one is N3 minus. Chloride can also be okay. These are sort of like go-to, good nucleophiles, weak bases. So these guys are good nucleophiles and weak bases. So when we're doing secondary alkyl halides, we're really going to have to choose from these guys, because as you saw with the thoxide ion, your major product is elimination. That leads us directly into the E2, which we'll just start today. We're not going to spend very much time on the E1, but a couple of things I want to point out is that number one, the rate law is exactly the same as the SN1. And step one, the first step of the SN1 is actually exactly the same as the E1. So in the first step of an SN1, we have leaving group leave. So draw lone pairs on your leaving group. We're waiting for this to do whatever it does. Okay. So we have our leaving group leave. First step, we make a carbocation, right? So that should look very familiar to you. So the leaving group leaves, and that's why we only have the alkyl halide appearing in the rate determining step. That is the rate determining step. In the second step, a basic abstracts of beta hydrogen. So I'm going to redraw this species here. It's carbocation. Just looks like that. So we're eliminating a beta hydrogen. Let's label that for you. Here's alpha. There's beta. So we get BH plus, and we get our alkene. All right, so that's the same carbocation seen in the SN1 reaction. The E1 and the SN1 reactions are in competition with each other, and generally you will get mixtures of products. It's that we have very little control because the rate determining step is the same for both. The other two steps for the SN1 versus the E1, low energy of activation, very competitive. We'll talk more about that next time. Thank you.
 Okay, alright, so here's where we left off last time.  When we were talking about the mixture of products from this particular alkyl halide,  and we saw that we got 41% of the trans, so let's label that.  41%, that's not a good sign what it just did over there, huh?  Trans.  You know, you can talk about overhead projector being old technology, but it's a lot more  dependable than this, right?  Okay, so that's trans.  This one is cis, and this one here is least substituted.  Alright, so let's show how we get the, we can easily see how we can get the least substituted  by eliminating right here this carbon-carbon bond that I'm pointing to.  Alright, let's see how we get the cis and the trans.  So here's, I just put this in this orientation, I put it into a new one projection so you  can see where that comes from, and so let's scroll down a little bit.  So here's, let's, here's our leaving group.  Here's our beta hydrogen.  So leaving group, and here's our beta hydrogen.  So that's on the back carbon, the way I have it drawn.  But if we eliminate the leaving group and the beta hydrogen, we get this alkene.  This one is going to be the trans.  So it's going to look like that.  And now, if we rotate, so notice we have, we have two beta hydrogens here.  We have this one here, and we have this one here.  And so if we rotate this down, let me change the color for that.  I want to just rotate, I'm going to keep the bromine on the top so I'm just going to rotate  this hydrogen right here, that's going to rotate down.  So it's anti to the leaving group, and that's what we want.  So now that hydrogen's right here, and here's our leaving group staying the same.  And it's just doing that again.  Okay.  Okay, so there's our leaving group.  We have our beta hydrogen, and if we eliminate there, you can see that now you can see that  the ethyl and the methyl are now going to be on the same side.  So we have our carbon-carbon bond here.  The two hydrogens are on the same side, and over on this side we have the ethyl, and this  side we have the methyl.  So that's how we got the cis and the trans.  And we want to have the bulkiest groups on opposite sides, so this one's going to be  more stable.  Most groups on opposite side.  And therefore it's going to be the major product.  And so we definitely see that that one is the major product up here.  Questions about that problem?  Anybody?  How we got that?  Yes?  Show the arrows for elimination, it's kind of hard to do on a Newman projection.  The least substituted one?  Yes, let's do that right here.  Since there's no stereochemical issues or no different ways to draw the, there's no cis  and trans when it's on the end, we'll just, I'm not going to put it into a Newman projection.  So we'll just draw it right here.  And the way I usually draw this when I'm not in a Newman projection is I definitely  going to put the hydrogen on the opposite side of the leaving group, anti-periplanar.  So our base is a foc side, so I'm just going to abbreviate that ETO minus.  And that should do the command and grab the beta hydrogen.  We're going to move electrons here to between those two carbons.  Leaving group's going to leave, so I'm giving it its lone pairs.  And that's what that looks like.  So that's all three products.  So if the reactant only has one hydrogen bonded to the beta carbon, there's only one conformer  in which the groups to be eliminated are anti.  The particular isomer that's formed depends on the configuration of the reactant.  So we don't automatically draw the trans because you might not be able to get the trans.  If you're showing stereochemistry in the reactant, you might not be able to get into an orientation  to give the trans product.  So this is a good example right here.  This one only has one beta hydrogen here on the least substitute.  That's going to give the most substituted alkene.  Okay, so this is our beta hydrogen here.  So you eliminate a hydrogen on the least substituted, on the most substituted carbon that will  give you the most substituted alkene.  So we certainly can eliminate one on this, this, one of these beta hydrogens, but that's  not going to give us the most substituted alkene.  And I'm just trying to draw the ZYTESF product.  All right, so we only have one.  And so when the base comes in, a foxide ion to remove that beta hydrogen.  So here's some more arrow pushing.  We need a lot of practice with this particular mechanism because it's just a strange reaction  looking reaction.  All right, so you can see there is only one product that we can get.  So we can't even, we can't really even get any of the product where the phenyl and the  hydrogen are trans to each other.  That's the only product that we can get.  All right, and so that moves us to competition between SN2 versus E2.  All right, so we're done talking about all the aspects of the E2, and now we're done  to talk about the competition between the SN2 and the E2 because SN2 and E2 reactions  are in competition with each other.  Is there a way to favor one or the other?  Let's show both of the reactions here.  So this is our nucleophile slash base.  Just put base here.  It's the same species.  Most of the examples we've done have used a foxide.  It's the same species.  We're just calling it something different.  But both terms can be used interchangeably.  So we have our base slash nucleophile.  We're going to have a red pathway here.  We'll call that A. Where the nucleophile comes in and removes the beta hydrogen, we break  the carbon-hydrogen bond to make a new bond between carbon and carbon, and then the leaving  group leaves.  So A is the E2 pathway right here, and our product is an alkene.  So this right here is the conjugate base.  And then we have our leaving group X minus.  All right, alternatively, we'll call this one pathway B down here.  The nucleophile can come in and do backside attack.  So that would be coming in from backside here, attack the carbon, break the carbon X bond  here.  So we look like that.  Those are really the two things that we're comparing.  Yes?  I think it's conjugate acid.  Acid?  Yes, it is.  Thank you.  Thank you for catching up.  All right.  And as you can imagine, it's much easier for the nucleophile to reach this hydrogen here  that's on the periphery of the molecule, much, much easier than it is for it to do backside  attack, especially when we have a very hindered carbon bonded to the leaving group.  Okay, so that's really going to be the whole entire thing.  It's all about steric hindrits in the alkyl halide.  It's also about strength of the base slash nucleophile.  The more basics the nucleophile get, the more favorite E2 will be over SN2.  All right, so what determines which is going to be the major pathway?  It is the structure of the alkyl halide, the structure of the base, and temperature.  So as we're going to learn, temperature favors elimination.  Definitely temperature favors elimination.  So this is the structure of the base slash nucleophile here.  So let's first look at the structure of the alkyl halide.  All right, so you can see in this first substrate, we have a really nice primary alkyl halide.  And so it's going to be really easy for the nucleophile to come in and attack.  Not very much steric hindrance.  The carbon that's bonded to the leaving group has two hydrogens and one R group attached  to its primary.  So it's going to be very easy for the nucleophile to come in and do backside attack and kick  off the leaving group, which for now our leaving groups are all halogens.  When we get to chapter nine, you'll see that they don't always have to be halogens.  Or for right now, they are.  All right, so what we see here is that attack at the carbon is unhindered.  So therefore SN2 is going to occur.  In the second example, we have a little bit different issue here.  We have a very hindered alkyl halide.  It's tertiary.  And we talked about the fact last quarter that we don't even get SN2 on a tertiary alkyl  halide, but we do get E2 on a tertiary alkyl halide.  So this is way too hindered for a substitution, but it is actually, that whole idea is going  to favor, and this is what we've got to stop doing here.  This is going to favor the base in acting as a base.  So the base slash nucleophile is going to act as a base instead.  And remove a beta hydrogen.  So we've got nine different beta hydrogens to remove.  We can remove any one of those nine.  I'm going to pick the one that is antiperiplanar.  All of these bonds can rotate, so really any of these hydrogens can be removed.  I'm going to pick the one that's antiperiplanar.  We're going to make a new double bond here, and then the leaving group's going to leave.  So what you see is that a tack at carbon is blocked  so that E2 occurs.  So it's really easy to make a decision when we have a primary or a tertiary substrate.  It is not easy to make a decision when we have a secondary, because now the rates of  those two reactions are going to be much closer together.  So let's look at some examples.  That tends to be a good way to see what's going to happen.  Let's look at some real examples that have been run with primary, secondary, and tertiary  alkylheylides.  And what we're doing is we're choosing a thawkside ion, so that's a good nucleophile.  It's also a good base.  And the temperature is 55 for all of these.  So it's a little bit elevated temperature.  Elevated temperature favors E2, but let's just see what our actual outcome.  So for a primary alkylheylide, here we have ethyl bromide.  You're going to get mostly SN2 with a thawkside.  So let's see what numbers we get here.  So our product here for the SN2 is an ether.  Again that's going to take a little practice to be able to draw these products quickly  and easily.  It's an SN2, and we get 90%.  The E2 reaction gives ethylene as a product, and that's 10%.  So when I'm asking for the major product, which is what I do most often, I want to see  only the SN2 product for a primary alkylheylide.  It happens with the secondary.  Draw the products, and then we'll just kind of summarize here.  This is our SN2, 21%.  Not very good, huh?  21% SN2.  It should, yes.  Let's fix that.  Here we go.  Okay, so we get that, plus.  This one only has one E2.  One possible E2 product, and that's 79%.  All right, so for secondary alkylheylides, E2 is favored with ethoxide.  So ethoxide is a good nucleophile, but it's also a fairly strong base.  How do we know it's a fairly strong base?  Well, what we do is we look at the pKa of the conjugate acid.  Okay, so let's look at the pKa of the conjugate acid.  I'll just take these pleasant little pauses here.  We're going to be calling about that pKa of the conjugate acid.  That's one of the ones I want you to know rounded to the nearest 15.  It's an alcohol, so it's about 15.  Okay, so we want things that are fairly strong bases.  It's about above pKa of 8-ish for the conjugate acid, 8-9-ish, and that's well above that,  so that's a fairly strong base.  With a secondary alkylheylide and ethoxide ion, E2 is favored.  So if I'm asking for the major product, you would only draw the elimination product.  Questions so far?  Anybody?  Yes, up at the top.  I will provide any pKa's that are not the 8 that I want you to memorize.  Okay?  Yes?  It still would be, and the major would be E2.  Yeah, absolutely.  That's a good question.  All right, let's look at tertiary alkylheylides.  Well tertiary alkylheylides don't do SN2, right?  So no SN2 for tertiary halides.  Therefore, E2 only.  All right, so we only have one possible product with this particular alkylheylide.  100% of the alkylheylide.  All right, so what happens if we change to a bulky base?  So what if we take a primary alkylheylide and we use a bulky strong base?  So how about terputoxide better rather than ethoxide?  And I would recommend that you keep going back to this page because these things come  back.  And this is a page I often refer to when students ask me questions.  So we have a primary alkylheylide.  We have methoxide ions, so a little tiny bit less hindered than ethoxide.  And we have terputoxide.  What do we actually get?  So this one we get 96% SN2 and 1% E2.  Don't know the temperature on this one.  It's probably a lower temperature.  It would make sense, huh?  Probably a lower temperature.  This particular experiment I don't have the temperature for.  You'll also notice that the numbers are not adding up to 100, so we've got 96% and 1%.  So when that happens, it means these are isolated yields rather than, you know, for what I want  to say, ideal yields or, you know, these are isolated yields when they don't add up to  100%.  And then in the second example, when we do the same alkylheylide and terputoxide, we  get only 12% SN2 and 85% E2.  So the message here is that you can actually determine which product that you want by choosing  the correct base.  If you have a primary alkylheylide and you want elimination as your major product, you  better choose a bulky base, okay?  If you don't want elimination on a primary, then you want to use an unhindered.  If you want elimination on a primary, then you're just going to use a normal base.  I mean, I said that wrong.  If you want substitution, you would just use it.  You know, you're not going to have to worry too much about elimination with a primary  alkylheylide.  Yes?  Yes?  What happens on the top part of this point?  Why do you choose to do with, like, the CS3, CS2, and A and CS3, CS2, and OH?  Like, why do you choose to do with your differences?  The alcohol is the solvent.  And it's very common with a thoxide or alkoxide bases to have the alcohol as a solvent.  In fact, you can make those by taking an alcohol and adding a chunk of sodium to them, okay?  So that's the most common solvent used for that.  All right?  More questions?  Anybody?  Yes?  I have a question about the primary alkylheylide.  Okay.  Why do you choose to do with a polyaprodic solvent?  I thought you'd choose to do with polyaprodic solvent.  Okay, really good question.  So E2 is faster in a polyaprodic solvent, but polyaprodic solvent is often not the most  convenient.  So some of the polyaprodic solvents, like, let me give you some examples.  The two most common that we use in this class are DMF and DMSO.  Those solvents are a little bit problematic because they have higher boiling points.  So once you isolate your product, you can't easily remove the solvent.  That's a problem.  The second problem is have you guys done an extraction yet in the lab?  You did it in T-chem though, right?  In a sept funnel?  You know how you had the two layers, and that's how you often isolate products in organic  chemistry.  If you have DMSO and DMF, they don't form two layers.  It just will make, it'll make, the water will not form a separate layer.  So it makes it a lot more difficult to isolate your product.  So that's why we don't use acetone.  And like, let's say for this one, ethoxide ion in acetone, why not just use acetone  instead?  You know, why not?  And the reason is that the sodium ethoxide doesn't dissolve in acetone.  You see what I'm saying?  So there's all sorts of problems.  So for the SN1 and the E1, we have to use protic solvent for the SN2 we don't.  Okay?  All right.  We've got a little bit left here.  So to favor E2 over SN2, use secondary or bulky halides and strongly basic and or bulky  nucleophiles.  Increasing temperature also favors elimination over substitution.  So there's where we talk about the increasing temperature.  To favor SN2 over E2, and we're especially going to have to worry about this with secondary  alkyl halides.  We want low temperature, so room temperature for example, and weakly basic unhindered nucleophile.  That's a tall order there, really.  Because a lot of nucleophiles are also strong bases.  So what kind of nucleophiles are we looking for?  Here's some examples of things that are good.  Nucleophiles that are weak bases.  All of the halogens except fluoride's not so good, right?  I died.  Bromide.  Another one is N3 minus.  Chloride can also be okay.  These are sort of like go-to, good nucleophiles, weak bases.  So these guys are good nucleophiles and weak bases.  So when we're doing secondary alkyl halides, we're really going to have to choose from  these guys, because as you saw with the thoxide ion, your major product is elimination.  That leads us directly into the E2, which we'll just start today.  We're not going to spend very much time on the E1, but a couple of things I want to point  out is that number one, the rate law is exactly the same as the SN1.  And step one, the first step of the SN1 is actually exactly the same as the E1.  So in the first step of an SN1, we have leaving group leave.  So draw lone pairs on your leaving group.  We're waiting for this to do whatever it does.  Okay.  So we have our leaving group leave.  First step, we make a carbocation, right?  So that should look very familiar to you.  So the leaving group leaves, and that's why we only have the alkyl halide appearing in  the rate determining step.  That is the rate determining step.  In the second step, a basic abstracts of beta hydrogen.  So I'm going to redraw this species here.  It's carbocation.  Just looks like that.  So we're eliminating a beta hydrogen.  Let's label that for you.  Here's alpha.  There's beta.  So we get BH plus, and we get our alkene.  All right, so that's the same carbocation seen in the SN1 reaction.  The E1 and the SN1 reactions are in competition with each other, and generally you will get  mixtures of products.  It's that we have very little control because the rate determining step is the same for  both.  The other two steps for the SN1 versus the E1, low energy of activation, very competitive.  We'll talk more about that next time.  Thank you.
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Okay, alright, so here's where we left off last time. When we were talking about the mixture of products from this particular alkyl halide, and we saw that we got 41% of the trans, so let's label that. 41%, that's not a good sign what it just did over there, huh? Trans. You know, you can talk about overhead projector being old technology, but it's a lot more dependable than this, right? Okay, so that's trans. This one is cis, and this one here is least substituted. Alright, so let's show how we get the, we can easily see how we can get the least substituted by eliminating right here this carbon-carbon bond that I'm pointing to. Alright, let's see how we get the cis and the trans. So here's, I just put this in this orientation, I put it into a new one projection so you can see where that comes from, and so let's scroll down a little bit. So here's, let's, here's our leaving group. Here's our beta hydrogen. So leaving group, and here's our beta hydrogen. So that's on the back carbon, the way I have it drawn. But if we eliminate the leaving group and the beta hydrogen, we get this alkene. This one is going to be the trans. So it's going to look like that. And now, if we rotate, so notice we have, we have two beta hydrogens here. We have this one here, and we have this one here. And so if we rotate this down, let me change the color for that. I want to just rotate, I'm going to keep the bromine on the top so I'm just going to rotate this hydrogen right here, that's going to rotate down. So it's anti to the leaving group, and that's what we want. So now that hydrogen's right here, and here's our leaving group staying the same. And it's just doing that again. Okay. Okay, so there's our leaving group. We have our beta hydrogen, and if we eliminate there, you can see that now you can see that the ethyl and the methyl are now going to be on the same side. So we have our carbon-carbon bond here. The two hydrogens are on the same side, and over on this side we have the ethyl, and this side we have the methyl. So that's how we got the cis and the trans. And we want to have the bulkiest groups on opposite sides, so this one's going to be more stable. Most groups on opposite side. And therefore it's going to be the major product. And so we definitely see that that one is the major product up here. Questions about that problem? Anybody? How we got that? Yes? Show the arrows for elimination, it's kind of hard to do on a Newman projection. The least substituted one? Yes, let's do that right here. Since there's no stereochemical issues or no different ways to draw the, there's no cis and trans when it's on the end, we'll just, I'm not going to put it into a Newman projection. So we'll just draw it right here. And the way I usually draw this when I'm not in a Newman projection is I definitely going to put the hydrogen on the opposite side of the leaving group, anti-periplanar. So our base is a foc side, so I'm just going to abbreviate that ETO minus. And that should do the command and grab the beta hydrogen. We're going to move electrons here to between those two carbons. Leaving group's going to leave, so I'm giving it its lone pairs. And that's what that looks like. So that's all three products. So if the reactant only has one hydrogen bonded to the beta carbon, there's only one conformer in which the groups to be eliminated are anti. The particular isomer that's formed depends on the configuration of the reactant. So we don't automatically draw the trans because you might not be able to get the trans. If you're showing stereochemistry in the reactant, you might not be able to get into an orientation to give the trans product. So this is a good example right here. This one only has one beta hydrogen here on the least substitute. That's going to give the most substituted alkene. Okay, so this is our beta hydrogen here. So you eliminate a hydrogen on the least substituted, on the most substituted carbon that will give you the most substituted alkene. So we certainly can eliminate one on this, this, one of these beta hydrogens, but that's not going to give us the most substituted alkene. And I'm just trying to draw the ZYTESF product. All right, so we only have one. And so when the base comes in, a foxide ion to remove that beta hydrogen. So here's some more arrow pushing. We need a lot of practice with this particular mechanism because it's just a strange reaction looking reaction. All right, so you can see there is only one product that we can get. So we can't even, we can't really even get any of the product where the phenyl and the hydrogen are trans to each other. That's the only product that we can get. All right, and so that moves us to competition between SN2 versus E2. All right, so we're done talking about all the aspects of the E2, and now we're done to talk about the competition between the SN2 and the E2 because SN2 and E2 reactions are in competition with each other. Is there a way to favor one or the other? Let's show both of the reactions here. So this is our nucleophile slash base. Just put base here. It's the same species. Most of the examples we've done have used a foxide. It's the same species. We're just calling it something different. But both terms can be used interchangeably. So we have our base slash nucleophile. We're going to have a red pathway here. We'll call that A. Where the nucleophile comes in and removes the beta hydrogen, we break the carbon-hydrogen bond to make a new bond between carbon and carbon, and then the leaving group leaves. So A is the E2 pathway right here, and our product is an alkene. So this right here is the conjugate base. And then we have our leaving group X minus. All right, alternatively, we'll call this one pathway B down here. The nucleophile can come in and do backside attack. So that would be coming in from backside here, attack the carbon, break the carbon X bond here. So we look like that. Those are really the two things that we're comparing. Yes? I think it's conjugate acid. Acid? Yes, it is. Thank you. Thank you for catching up. All right. And as you can imagine, it's much easier for the nucleophile to reach this hydrogen here that's on the periphery of the molecule, much, much easier than it is for it to do backside attack, especially when we have a very hindered carbon bonded to the leaving group. Okay, so that's really going to be the whole entire thing. It's all about steric hindrits in the alkyl halide. It's also about strength of the base slash nucleophile. The more basics the nucleophile get, the more favorite E2 will be over SN2. All right, so what determines which is going to be the major pathway? It is the structure of the alkyl halide, the structure of the base, and temperature. So as we're going to learn, temperature favors elimination. Definitely temperature favors elimination. So this is the structure of the base slash nucleophile here. So let's first look at the structure of the alkyl halide. All right, so you can see in this first substrate, we have a really nice primary alkyl halide. And so it's going to be really easy for the nucleophile to come in and attack. Not very much steric hindrance. The carbon that's bonded to the leaving group has two hydrogens and one R group attached to its primary. So it's going to be very easy for the nucleophile to come in and do backside attack and kick off the leaving group, which for now our leaving groups are all halogens. When we get to chapter nine, you'll see that they don't always have to be halogens. Or for right now, they are. All right, so what we see here is that attack at the carbon is unhindered. So therefore SN2 is going to occur. In the second example, we have a little bit different issue here. We have a very hindered alkyl halide. It's tertiary. And we talked about the fact last quarter that we don't even get SN2 on a tertiary alkyl halide, but we do get E2 on a tertiary alkyl halide. So this is way too hindered for a substitution, but it is actually, that whole idea is going to favor, and this is what we've got to stop doing here. This is going to favor the base in acting as a base. So the base slash nucleophile is going to act as a base instead. And remove a beta hydrogen. So we've got nine different beta hydrogens to remove. We can remove any one of those nine. I'm going to pick the one that is antiperiplanar. All of these bonds can rotate, so really any of these hydrogens can be removed. I'm going to pick the one that's antiperiplanar. We're going to make a new double bond here, and then the leaving group's going to leave. So what you see is that a tack at carbon is blocked so that E2 occurs. So it's really easy to make a decision when we have a primary or a tertiary substrate. It is not easy to make a decision when we have a secondary, because now the rates of those two reactions are going to be much closer together. So let's look at some examples. That tends to be a good way to see what's going to happen. Let's look at some real examples that have been run with primary, secondary, and tertiary alkylheylides. And what we're doing is we're choosing a thawkside ion, so that's a good nucleophile. It's also a good base. And the temperature is 55 for all of these. So it's a little bit elevated temperature. Elevated temperature favors E2, but let's just see what our actual outcome. So for a primary alkylheylide, here we have ethyl bromide. You're going to get mostly SN2 with a thawkside. So let's see what numbers we get here. So our product here for the SN2 is an ether. Again that's going to take a little practice to be able to draw these products quickly and easily. It's an SN2, and we get 90%. The E2 reaction gives ethylene as a product, and that's 10%. So when I'm asking for the major product, which is what I do most often, I want to see only the SN2 product for a primary alkylheylide. It happens with the secondary. Draw the products, and then we'll just kind of summarize here. This is our SN2, 21%. Not very good, huh? 21% SN2. It should, yes. Let's fix that. Here we go. Okay, so we get that, plus. This one only has one E2. One possible E2 product, and that's 79%. All right, so for secondary alkylheylides, E2 is favored with ethoxide. So ethoxide is a good nucleophile, but it's also a fairly strong base. How do we know it's a fairly strong base? Well, what we do is we look at the pKa of the conjugate acid. Okay, so let's look at the pKa of the conjugate acid. I'll just take these pleasant little pauses here. We're going to be calling about that pKa of the conjugate acid. That's one of the ones I want you to know rounded to the nearest 15. It's an alcohol, so it's about 15. Okay, so we want things that are fairly strong bases. It's about above pKa of 8-ish for the conjugate acid, 8-9-ish, and that's well above that, so that's a fairly strong base. With a secondary alkylheylide and ethoxide ion, E2 is favored. So if I'm asking for the major product, you would only draw the elimination product. Questions so far? Anybody? Yes, up at the top. I will provide any pKa's that are not the 8 that I want you to memorize. Okay? Yes? It still would be, and the major would be E2. Yeah, absolutely. That's a good question. All right, let's look at tertiary alkylheylides. Well tertiary alkylheylides don't do SN2, right? So no SN2 for tertiary halides. Therefore, E2 only. All right, so we only have one possible product with this particular alkylheylide. 100% of the alkylheylide. All right, so what happens if we change to a bulky base? So what if we take a primary alkylheylide and we use a bulky strong base? So how about terputoxide better rather than ethoxide? And I would recommend that you keep going back to this page because these things come back. And this is a page I often refer to when students ask me questions. So we have a primary alkylheylide. We have methoxide ions, so a little tiny bit less hindered than ethoxide. And we have terputoxide. What do we actually get? So this one we get 96% SN2 and 1% E2. Don't know the temperature on this one. It's probably a lower temperature. It would make sense, huh? Probably a lower temperature. This particular experiment I don't have the temperature for. You'll also notice that the numbers are not adding up to 100, so we've got 96% and 1%. So when that happens, it means these are isolated yields rather than, you know, for what I want to say, ideal yields or, you know, these are isolated yields when they don't add up to 100%. And then in the second example, when we do the same alkylheylide and terputoxide, we get only 12% SN2 and 85% E2. So the message here is that you can actually determine which product that you want by choosing the correct base. If you have a primary alkylheylide and you want elimination as your major product, you better choose a bulky base, okay? If you don't want elimination on a primary, then you want to use an unhindered. If you want elimination on a primary, then you're just going to use a normal base. I mean, I said that wrong. If you want substitution, you would just use it. You know, you're not going to have to worry too much about elimination with a primary alkylheylide. Yes? Yes? What happens on the top part of this point? Why do you choose to do with, like, the CS3, CS2, and A and CS3, CS2, and OH? Like, why do you choose to do with your differences? The alcohol is the solvent. And it's very common with a thoxide or alkoxide bases to have the alcohol as a solvent. In fact, you can make those by taking an alcohol and adding a chunk of sodium to them, okay? So that's the most common solvent used for that. All right? More questions? Anybody? Yes? I have a question about the primary alkylheylide. Okay. Why do you choose to do with a polyaprodic solvent? I thought you'd choose to do with polyaprodic solvent. Okay, really good question. So E2 is faster in a polyaprodic solvent, but polyaprodic solvent is often not the most convenient. So some of the polyaprodic solvents, like, let me give you some examples. The two most common that we use in this class are DMF and DMSO. Those solvents are a little bit problematic because they have higher boiling points. So once you isolate your product, you can't easily remove the solvent. That's a problem. The second problem is have you guys done an extraction yet in the lab? You did it in T-chem though, right? In a sept funnel? You know how you had the two layers, and that's how you often isolate products in organic chemistry. If you have DMSO and DMF, they don't form two layers. It just will make, it'll make, the water will not form a separate layer. So it makes it a lot more difficult to isolate your product. So that's why we don't use acetone. And like, let's say for this one, ethoxide ion in acetone, why not just use acetone instead? You know, why not? And the reason is that the sodium ethoxide doesn't dissolve in acetone. You see what I'm saying? So there's all sorts of problems. So for the SN1 and the E1, we have to use protic solvent for the SN2 we don't. Okay? All right. We've got a little bit left here. So to favor E2 over SN2, use secondary or bulky halides and strongly basic and or bulky nucleophiles. Increasing temperature also favors elimination over substitution. So there's where we talk about the increasing temperature. To favor SN2 over E2, and we're especially going to have to worry about this with secondary alkyl halides. We want low temperature, so room temperature for example, and weakly basic unhindered nucleophile. That's a tall order there, really. Because a lot of nucleophiles are also strong bases. So what kind of nucleophiles are we looking for? Here's some examples of things that are good. Nucleophiles that are weak bases. All of the halogens except fluoride's not so good, right? I died. Bromide. Another one is N3 minus. Chloride can also be okay. These are sort of like go-to, good nucleophiles, weak bases. So these guys are good nucleophiles and weak bases. So when we're doing secondary alkyl halides, we're really going to have to choose from these guys, because as you saw with the thoxide ion, your major product is elimination. That leads us directly into the E2, which we'll just start today. We're not going to spend very much time on the E1, but a couple of things I want to point out is that number one, the rate law is exactly the same as the SN1. And step one, the first step of the SN1 is actually exactly the same as the E1. So in the first step of an SN1, we have leaving group leave. So draw lone pairs on your leaving group. We're waiting for this to do whatever it does. Okay. So we have our leaving group leave. First step, we make a carbocation, right? So that should look very familiar to you. So the leaving group leaves, and that's why we only have the alkyl halide appearing in the rate determining step. That is the rate determining step. In the second step, a basic abstracts of beta hydrogen. So I'm going to redraw this species here. It's carbocation. Just looks like that. So we're eliminating a beta hydrogen. Let's label that for you. Here's alpha. There's beta. So we get BH plus, and we get our alkene. All right, so that's the same carbocation seen in the SN1 reaction. The E1 and the SN1 reactions are in competition with each other, and generally you will get mixtures of products. It's that we have very little control because the rate determining step is the same for both. The other two steps for the SN1 versus the E1, low energy of activation, very competitive. We'll talk more about that next time. Thank you.
that's formed depends on the configuration of the trans. If you're showing stereochemistry in the between carbon and carbon, and then the leaving what we've got to stop doing here. This is going often isolate products in organic chemistry. If determining step. That is the rate determining step. In the second step, a basic abstracts of SN1 reaction. The E1 and the SN1 reactions are in