Lecture 25. Elimination Reactions. Introduction to E2 Reactions
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Nowick, James S.
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. 
 Good morning. Good morning. I am happy. Welcome to a long and very fast journey through eight chapters of a course that focuses on 24 chapters. We come, we review the structure and bonding. We've learned about some organic compounds, functional roots. We've learned about curve arrows and begun to discuss various types of organic reaction mechanisms. And now we're starting to see that organic reaction mechanisms can become complex in their many different pathways that can proceed upon and also bury many different stereochemical issues that come up. We saw this in chapter seven of SN2 displacements. We saw even having to really be cognizant of what happens to a stereochemical center and looking at the quiz problem that we just had with the first problems on SN2 reactions. And now we're going to continue some of these ideas of many fold courses of reactivity in discussing chapter eight, discussing elimination reactions. We're going to talk about two different mechanisms of elimination reactions during the course of this week. We're going to discuss E2 eliminations and then E1 eliminations. And we're going to be learning more about the stereochemical course of these reactions and particularly the requirement that E2 eliminations occur with what's called an anti-curry-cliner relationship. In the second lecture, Wednesday's lecture, we'll be seeing the stereochemical implications of this. The homework in the discussion problems will actually be getting to explore the antithesis of an anti-curry-cliner relationship and a sin-curry-cliner relationship that comes in a very special concerted reaction that you'll be seeing. So there's a lot of me here. We'll be learning about which types of electrophiles, which types of alkyl halides participate in E2 reactions and under one condition and which types participate in E1 reactions. So let's start by looking at a reaction that I talked about or at least alluded to in the discussions of chapter seven, but I didn't really tell you where this reaction went. So let's take tert-butyl bromide and let's envision treating this with sodium anthoxide. The first thing you might think when you see this reaction, you might envision some type of displacement reaction. Perhaps you'd even envision an SN1 displacement reaction. And this is not what occurs. Big X here means that this doesn't occur. So the key is sodium anthoxide. We've talked about bases. We've talked about nucleophiles. Sodium anthoxide is the strong base. PKA is the basicity of the anhydride. It's very similar to the basicity of the hydroxide anhydride. Remember, in all quarter making and analogy between alcohol groups and hydrogen, between the ROH and HOH water functionality, the PK of water associated with water being an acid loss of protons 15.7, and PK of ethanol is very similar. It's about 17. And instead, what happens is an elimination reaction. And the product that we observe is isobutylene. If we want to write a complete set of products from the reaction, then our sodium anthoxide is going to give us ethanol. And we'll also get sodium bromide as an organic byproduct of the reaction. I'll put these in parenthesis because organic chemists are often so focused on one reactant that they'll often ignore other products. But if we're trying to be complete, here are other products. This is called an E2 elimination reaction, or E2 reaction. E2 stands for elimination bimolecular. Just like SN2 stood for substitution nucleophilic bimolecular. SN1 stood for substitution nucleophilic unimolecular. And remember when we looked at substitution reactions, of course, in both an SN1 and an SN2 substitution reaction, ultimately two molecules were involved, the electrophile and the nucleophile. But the big difference between SN2 and SN1 was that in the transition state, in the rate determining step, we had one molecule involved in SN1, two molecules involved in SN2. And this is exactly the case with both the E2 elimination that I'm going to talk about now and the E1 elimination that we'll discuss on Friday. So what's occurring in this reaction is our ethoxide, and I'll write it as ETO minus, and I'll draw in some long pairs so I can kind of push some arrows here, is coming in in the rate determining step and is pulling off a proton at the same time we are spitting out bromide anion. So here's our ethoxide and here's our tert-butyl bromide. I have one methyl group, so I need to draw hydrogens in mice first, and here's a bromide, here are our methyl groups. So in the transition state, we are having the ethoxide come in with a pair of electrons and start to pull off this proton. We're going to push electron density in between the two carbons going from a single bond to a double bond, and we're going to spin out bromide. So as we go through the reaction, and I'm just going to start and arrow here and make the drawing of the transition state. Now I didn't do anything as fancy as old overhead animation or SN2, but this is very much the same process. Ethoxide starts to come in, it starts to form a bond to hydrogen. At the same time we're starting to push electron density between the two carbons, which is starting to push out bromide. Partway through the reaction, we're halfway there, we have our ethoxide, which is now partially bonded to hydrogen. That hydrogen is partially bonded to carbon. We of course have our single bond between carbon, but we're starting to form a double bond. You want to be technical just as we started to change geometry about the carbons in an SN2 reaction, we're starting to change geometry in our E2 reaction. We're going from being tetrahedral to being planar as we progress into the transition state. We're starting to break our bond to bromine. This is my drawing of the transition state. Remember a little double dagger there is what organic chemists often use to symbolize the transition state. I'll draw it out. Two electrons here on the ethoxide. We have part of our negative charge on the ethoxide. We haven't yet formed a full bond to hydrogen. Remember a transition state is that species that exists for a fleeting moment for the length of a vibration per second as the reaction is occurring. We can't go ahead and have this as a stable configuration. It's not an intermediate, it's just that high point as we go over the reaction coordinate that we proceed from reactants to intermediates or intermediate products or in this case directly from reactants to products. So we have a partial negative charge on ethoxide. We have a partial negative charge on bromide. That fleeting moment of the transition state passes as we go through the reaction and now we have our ethanol. We've planarized, is that a word? We've made planar. Our carbon atoms as we've changed hybridization from sp3 to sp2 associated with a double bond. We now have a fully formed kaibond. This is my very head drawing of isobutylene trying to represent it in the plane that's orthogonal to the blackboard with a methyl group coming out and a methyl group coming back and a hydrogen coming out and a hydrogen going back. And finally we have our bromide over here. So those are our products of our reaction. Thoughts or questions about this process? So here's the delta negative charge. So is there to be some positive charge as well? Not necessarily. We started the overall. So the question was there's delta negative charges. Do there need to be delta positive charges? All the charges need to add up to the net charge that we have. We start with a negative charge in the ethoxide. So at all points we've got to have a net negative charge. And so there's delta negative charge on the ethoxide group and the bromobrub in the transition state. Now your point is well taken. We're only catching a little snatch. We're only catching part of the picture in any sort of drawing that we can take. And you're right in that the carbon bromine bond is already polarized. So you're right. So there's a little bit of a delta negative here and a little bit of a delta positive. A little bit of a delta negative on the bromine. A little bit of a delta positive on the carbon which we're overlooking in the process. That's one of the reasons why bromine is so happy to go away. So I'm just showing part of the picture here. But coming back to the question, do we need to have a delta positive when you have a delta negative? Absolutely not. So we're just using that to represent the main parts of the charges. Other questions. That's a good question. Back row. What is the delta charge on the H with two bonds? The negative is a bit of a negative. Would that H with two bonds also have a negative? Probably not here. Probably not any significant negative charge. You can think of it in the transition state really as half a bond to one half a bond to the other. And remember, that's not a stable configuration. That's only occurring transitively. Good question. Other questions. All right, just as we thought about reaction energy diagrams for reactions of the SN1 and SN2 variety, we can think about energy diagrams to help us think our way through an E2 reaction. And later on we'll do this for E1 reaction. So the reaction is downhill in energy. So here's our energy diagram. Here's our reaction coordinate. Remember, the reaction coordinate you can think of as a sort of a time axis as the reaction progresses. Or you can think of it as representing, say, the distance between the adoxy group and the hydrogen. It's just showing the progression of the reaction. The reaction is downhill in energy. In other words, the equilibrium, if you want to even talk about equilibrium for this reaction, lies very far to the right. So I can start our reactions up high in energy, our products down low in energy. I like to go ahead and try to write out all of my reactants and products explicitly in my intermediate. So I know what I'm talking about. So I'll draw a tert butyl bromide on the left. I'll draw an ethoxide on the left to remind us that these are our reactants. For our products I'll draw isobubling and ethanol and bromide. And so in a single step, without the formation of any intermediates, without the formation of any carbocation, we go from reactants, we cross over that energy barrier through our transition state, the high point that we've represented on the blackboard over here, to our products. And since we have two reactants intimately involved in the rate-determining step, our rate law is going to depend on the concentration of both of those reactants. In other words, we'll have bimolecular kinetics. E2 told us we have bimolecular kinetics, just like SN2 tells us we have bimolecular kinetics. In other words, our rate is going to equal rate constant K times the concentration of tert butyl bromide times the concentration of the bimolecular kinetics. And just like in SN2 reaction, if you double the concentration of adoxide, the reaction proceeds twice as fast. And if you double the concentration of E2 reaction, of course, the reaction goes four times as fast. So the whole part of an E2 elimination reaction is you need a strong base to get in the adobt and proton. If you had a very weak base, like iodide, I-, we said it's a pKa of HI, is negative 10, I- is a very weak base. We wouldn't have a species that's able to get in and pull off the proton. But adoxide, pKa of ethanol is 17. Adoxide is 10 to the 27 times stronger base. It's able to get in and pull off that proton. Now we've still seen that there are plenty of issues that come up. We've seen in SN2 reactions and SN4 reactions that we get certain conditions that tend to lead to SN2 and certain conditions that tend to lead to SN1. One of the big issues here for whether we're getting SN2 or E2 is going to be the substitution pattern of the alkyl halide. We might have E2 versus SN2 or E2 elimination versus SN2 substitution or SN2 displacement. So generally, if you take a tertiary alkyl halide, and of course that tertiary alkyl halide has to have some beta protons like we've seen here, a proton that can be pulled off, plus an alkyl oxide. And I'll write this generically as tertiary alkyl halide plus an alkyl oxide. Now we're going to get mainly E2. Tertiary alkyl halides also with alkyl oxide bases and other very strong bases like hydroxide tend to give mainly E2. So in other words, let's say I took, oh, bromo cyclohexane. And let me take a different alkyl oxide base. Let me take potassium tert butoxide. And I'll indicate our solvent. Usually when one works with potassium tert butoxide, one generates it in tert butanol. And so often one uses tert butanol as a solvent. Sometimes as you saw in the last chapter, one can use a dipolar or aaprodic solvent like dimethyl sulfoxide. But if one treats our bromo cyclohexane with potassium tert butoxide, the main product is going to be cyclohexane. Perhaps you will get a very, a very minor amount of the tert butoxide of the SN2 displacement product, cyclohexyl O tert butyl. I'm going to put minor, but I'll put parenthetically very minor. By that I mean that you would get, oh, a few percent of the ether product, a few percent of the SN2 displacement product at best. In general, tertiary, in general, secondary alkyl halides tend to eliminate predominantly mistorm bases. Just to remind us what this means here, I'll draw out the structure. So this is potassium tert butoxide drawn out as a full structure. We call it tert butoxide. I'll write out the name. We call it tert butoxide or tertiary butoxide because it is the species that results from deprotonation of tertiary butanol or t-butanol. Don't answer questions. All right. So we've talked about tertiary halides, talked about secondary halides. Let's take a moment to talk about primary halides. The first question. Ah, we're comparing the E2 to SN2. But what about SN1? Okay, this is a great question because this is where people start to say, wait a second, there are all these matter folds in the bottom. All right. So SN1 reaction, great question. SN1 reaction, tert halides do indeed, tertiary halides like tert-butyl bromide, do indeed ionized. So there is a rate of reaction for ionization before mature tertiary carbocation followed by what's possibly reaction with tert-butanol or tert-butoxide. But the rate of reaction with a strong base of pulling off that beta proton is so much greater that for all intents and purposes we never get a chance to have reaction by way of an SN1 pathway when you have a strong base present. So indeed, that pathway is possible if we didn't have tert-butoxide just like a tert-butanol, you would indeed see an SN1 pathway which would give rise to a substitution product. But once you have that base, it's so much lower in energy to just yank off that proton. Good question. Other questions? All right. Let's look at primary halides. And again, we're really looking here at reaction with alkoxide or other bases of comparable or greater strength. So hydroxide would be comparable in strength. NH2 minus MIN would be stronger in strength. So mainly SN2, but I'm going to put a caveat on that. And that star is just a reminder that we're going to see that even here there's some complexity. So let's start with a simple case. We'll take bromopropane. And again, we'll use sodium ethoxide, let's say, inethanol. So the main product of this reaction is going to be our SN displacement product. You'll also get a little bit, so I'll write SN2 here, and I'll write major. We'll also get a little bit of the E2 elimination product, so I'll write E2 and minor here. Now chemists know this from actual experiment. I'm not just pulling these numbers out of thin air or making this up. This is what people in the 1930s and 40s investigated experimentally to learn about. And so when this was investigated it was found that there was a 90% conversion to the SN2 displacement product and a 10% conversion to the E2 product. You wouldn't know that. You wouldn't have to know that. You don't have to know those exact numbers, but this helps us see some trends. So we sort of tear on the edge of an equilibrium. And I've tried to give us a feeling for energetics in equilibria. A 10 to 1 ratio corresponds at room temperature to a 1.36 kilocalorie per mole energy difference. That's not a big difference in energy. And some factors can tip the balance in going one way versus going the other way. And so even something as simple as steric hindrance can make a difference and push things into the E2 pathway. Specifically, if we go and we have beta substituents, this makes it very hard, makes it, let us say, harder for the nucleophile to get in close for that trigonal bond. For that trigonal bifuraminal transition state associated with an SN2 displacement reaction. Remember when I projected the reaction up there, projected the transition state, how crowded it was, how everything had to get in close. Even a nearby substituent that's in the way can help block things and make it more difficult. So if I add a beta methyl group, this isn't a methyl group on the carbon with the halide. I've been alphabet. That would take us from primary to secondary. But if I add a beta methyl group, just the spectator here, that extra steric bulk in going from a simple chain to a branch chain makes it more difficult for the nucleophile to come in. And the result is that it favors E2 elimination. Just a little bit. So if I take, instead of our bromocropate, our isopropyl, our isobutyl bromide, and again I treat it with sodium, methoxide, and ethanol. And just for the sake of variety, I'll remind us that there are many ways to write this reaction. This time I'll write the reaction over the arrow. Now we also get a mixture of SN2 displacement and E2 elimination. But now the E2 elimination predominates. So we get a mixture of the ether and the alkene, and that mixture is a 38 to 62% mixture. In other words, we've switched from the elimination pathway being the minor pathway to the elimination pathway being the major pathway. Not only does steric hindrance of the electrophile affect the pathways, but also steric hindrance of the base affects the pathways. So a bulkier base leads to more elimination as well. And so I'll keep our same electrophile in this comparison. I'll keep our same alkyl halide. But now instead of having sodium ethoxide and ethanol, I'll switch to potassium tertbutoxide. In tertbutanol, tertbutanol makes the alkoxybrom a little bit more sterically hindered. It also makes it a little harder to salvate, a little bit more basic. PK, or tertbutanol, usually people use a number about 18 instead of say 17 for ethanol. And as a result of the greater steric hindrance and the greater base isamine, it's harder, even harder, for the nucleophile to get in and do an SN2 displacement. And it's more favored to have the beta proton hold off. So now our ratio of SN2 displacement to E2 elimination goes all the way up to 8 to 92. In other words, now we've switched from 38 to 62 to 8 to 92. We've really pushed the reaction. And this is good, because organic chemists like to be able to control which way a reaction goes. And this begins to show us how an organic chemist can choose conditions that favor either an alkane product or an alkene product or an ether product. Last point I want to make, just on the generation of alkoxide bases, just because it's useful to know where they come from and why they can be used in the same alcohol. In general, if you take an alcohol plus sodium or potassium, the metal, the alkali metal will react to give the corresponding alkoxide plus hydrogen gas. And you know if you throw sodium into water, it catches fire. You know if you throw potassium into water, it explodes. So organic chemists often choose their metal carefully. They'll use sodium for more sterically unhingered alcohol, like methanol or ethanol. Because nobody wants to have their laboratory catch fire. And they'll use potassium for more sterically hindered alcohols that react slowly like tributanol. And that's one of the reasons why I've been switching around from sodium to potassium.
 Good morning.  Good morning.  I am happy.  Welcome to a long and very fast journey through eight chapters of a course that focuses on 24 chapters.  We come, we review the structure and bonding.  We've learned about some organic compounds, functional roots.  We've learned about curve arrows and begun to discuss various types of organic reaction mechanisms.  And now we're starting to see that organic reaction mechanisms can become complex in their many different pathways  that can proceed upon and also bury many different stereochemical issues that come up.  We saw this in chapter seven of SN2 displacements.  We saw even having to really be cognizant of what happens to a stereochemical center and looking at the quiz problem  that we just had with the first problems on SN2 reactions.  And now we're going to continue some of these ideas of many fold courses of reactivity in discussing chapter eight, discussing elimination reactions.  We're going to talk about two different mechanisms of elimination reactions during the course of this week.  We're going to discuss E2 eliminations and then E1 eliminations.  And we're going to be learning more about the stereochemical course of these reactions  and particularly the requirement that E2 eliminations occur with what's called an anti-curry-cliner relationship.  In the second lecture, Wednesday's lecture, we'll be seeing the stereochemical implications of this.  The homework in the discussion problems will actually be getting to explore the antithesis of an anti-curry-cliner relationship  and a sin-curry-cliner relationship that comes in a very special concerted reaction that you'll be seeing.  So there's a lot of me here.  We'll be learning about which types of electrophiles, which types of alkyl halides participate in E2 reactions  and under one condition and which types participate in E1 reactions.  So let's start by looking at a reaction that I talked about or at least alluded to  in the discussions of chapter seven, but I didn't really tell you where this reaction went.  So let's take tert-butyl bromide and let's envision treating this with sodium anthoxide.  The first thing you might think when you see this reaction, you might envision some type of displacement reaction.  Perhaps you'd even envision an SN1 displacement reaction.  And this is not what occurs.  Big X here means that this doesn't occur.  So the key is sodium anthoxide.  We've talked about bases.  We've talked about nucleophiles.  Sodium anthoxide is the strong base.  PKA is the basicity of the anhydride.  It's very similar to the basicity of the hydroxide anhydride.  Remember, in all quarter making and analogy between alcohol groups and hydrogen,  between the ROH and HOH water functionality,  the PK of water associated with water being an acid loss of protons 15.7,  and PK of ethanol is very similar.  It's about 17.  And instead, what happens is an elimination reaction.  And the product that we observe is isobutylene.  If we want to write a complete set of products from the reaction, then our sodium anthoxide  is going to give us ethanol.  And we'll also get sodium bromide as an organic byproduct of the reaction.  I'll put these in parenthesis because organic chemists are often so focused on one reactant  that they'll often ignore other products.  But if we're trying to be complete, here are other products.  This is called an E2 elimination reaction, or E2 reaction.  E2 stands for elimination bimolecular.  Just like SN2 stood for substitution nucleophilic bimolecular.  SN1 stood for substitution nucleophilic unimolecular.  And remember when we looked at substitution reactions, of course, in both an SN1 and an SN2  substitution reaction, ultimately two molecules were involved, the electrophile and the nucleophile.  But the big difference between SN2 and SN1 was that in the transition state,  in the rate determining step, we had one molecule involved in SN1,  two molecules involved in SN2.  And this is exactly the case with both the E2 elimination that I'm going to talk about now  and the E1 elimination that we'll discuss on Friday.  So what's occurring in this reaction is our ethoxide, and I'll write it as ETO minus,  and I'll draw in some long pairs so I can kind of push some arrows here,  is coming in in the rate determining step and is pulling off a proton  at the same time we are spitting out bromide anion.  So here's our ethoxide and here's our tert-butyl bromide.  I have one methyl group, so I need to draw hydrogens in mice first,  and here's a bromide, here are our methyl groups.  So in the transition state, we are having the ethoxide come in with a pair of electrons  and start to pull off this proton.  We're going to push electron density in between the two carbons going from a single bond to a double bond,  and we're going to spin out bromide.  So as we go through the reaction, and I'm just going to start and arrow here  and make the drawing of the transition state.  Now I didn't do anything as fancy as old overhead animation or SN2,  but this is very much the same process.  Ethoxide starts to come in, it starts to form a bond to hydrogen.  At the same time we're starting to push electron density between the two carbons,  which is starting to push out bromide.  Partway through the reaction, we're halfway there,  we have our ethoxide, which is now partially bonded to hydrogen.  That hydrogen is partially bonded to carbon.  We of course have our single bond between carbon, but we're starting to form a double bond.  You want to be technical just as we started to change geometry about the carbons in an SN2 reaction,  we're starting to change geometry in our E2 reaction.  We're going from being tetrahedral to being planar as we progress into the transition state.  We're starting to break our bond to bromine.  This is my drawing of the transition state.  Remember a little double dagger there is what organic chemists often use to symbolize the transition state.  I'll draw it out. Two electrons here on the ethoxide.  We have part of our negative charge on the ethoxide.  We haven't yet formed a full bond to hydrogen.  Remember a transition state is that species that exists for a fleeting moment  for the length of a vibration per second as the reaction is occurring.  We can't go ahead and have this as a stable configuration.  It's not an intermediate, it's just that high point as we go over the reaction coordinate  that we proceed from reactants to intermediates or intermediate products  or in this case directly from reactants to products.  So we have a partial negative charge on ethoxide.  We have a partial negative charge on bromide.  That fleeting moment of the transition state passes as we go through the reaction  and now we have our ethanol.  We've planarized, is that a word?  We've made planar.  Our carbon atoms as we've changed hybridization from sp3 to sp2 associated with a double bond.  We now have a fully formed kaibond.  This is my very head drawing of isobutylene trying to represent it in the plane  that's orthogonal to the blackboard with a methyl group coming out  and a methyl group coming back and a hydrogen coming out and a hydrogen going back.  And finally we have our bromide over here.  So those are our products of our reaction.  Thoughts or questions about this process?  So here's the delta negative charge.  So is there to be some positive charge as well?  Not necessarily. We started the overall.  So the question was there's delta negative charges.  Do there need to be delta positive charges?  All the charges need to add up to the net charge that we have.  We start with a negative charge in the ethoxide.  So at all points we've got to have a net negative charge.  And so there's delta negative charge on the ethoxide group  and the bromobrub in the transition state.  Now your point is well taken.  We're only catching a little snatch.  We're only catching part of the picture in any sort of drawing that we can take.  And you're right in that the carbon bromine bond is already polarized.  So you're right.  So there's a little bit of a delta negative here and a little bit of a delta positive.  A little bit of a delta negative on the bromine.  A little bit of a delta positive on the carbon which we're overlooking in the process.  That's one of the reasons why bromine is so happy to go away.  So I'm just showing part of the picture here.  But coming back to the question, do we need to have a delta positive when you have a delta negative?  Absolutely not.  So we're just using that to represent the main parts of the charges.  Other questions.  That's a good question.  Back row.  What is the delta charge on the H with two bonds?  The negative is a bit of a negative.  Would that H with two bonds also have a negative?  Probably not here.  Probably not any significant negative charge.  You can think of it in the transition state really as half a bond to one half a bond to the other.  And remember, that's not a stable configuration.  That's only occurring transitively.  Good question.  Other questions.  All right, just as we thought about reaction energy diagrams for reactions of the SN1 and SN2 variety,  we can think about energy diagrams to help us think our way through an E2 reaction.  And later on we'll do this for E1 reaction.  So the reaction is downhill in energy.  So here's our energy diagram.  Here's our reaction coordinate.  Remember, the reaction coordinate you can think of as a sort of a time axis as the reaction progresses.  Or you can think of it as representing, say, the distance between the adoxy group and the hydrogen.  It's just showing the progression of the reaction.  The reaction is downhill in energy.  In other words, the equilibrium, if you want to even talk about equilibrium for this reaction,  lies very far to the right.  So I can start our reactions up high in energy, our products down low in energy.  I like to go ahead and try to write out all of my reactants and products explicitly in my intermediate.  So I know what I'm talking about.  So I'll draw a tert butyl bromide on the left.  I'll draw an ethoxide on the left to remind us that these are our reactants.  For our products I'll draw isobubling and ethanol and bromide.  And so in a single step, without the formation of any intermediates,  without the formation of any carbocation,  we go from reactants, we cross over that energy barrier  through our transition state, the high point that we've represented on the blackboard over here,  to our products.  And since we have two reactants intimately involved in the rate-determining step,  our rate law is going to depend on the concentration of both of those reactants.  In other words, we'll have bimolecular kinetics.  E2 told us we have bimolecular kinetics, just like SN2 tells us we have bimolecular kinetics.  In other words, our rate is going to equal rate constant K times the concentration of tert butyl bromide  times the concentration of the bimolecular kinetics.  And just like in SN2 reaction, if you double the concentration of adoxide,  the reaction proceeds twice as fast.  And if you double the concentration of E2 reaction, of course, the reaction goes four times as fast.  So the whole part of an E2 elimination reaction is you need a strong  base to get in the adobt and proton.  If you had a very weak base, like iodide, I-, we said it's a pKa of HI, is negative 10, I- is a very weak base.  We wouldn't have a species that's able to get in and pull off the proton.  But adoxide, pKa of ethanol is 17.  Adoxide is 10 to the 27 times stronger base.  It's able to get in and pull off that proton.  Now we've still seen that there are plenty of issues that come up.  We've seen in SN2 reactions and SN4 reactions that we get certain conditions that tend to lead to SN2  and certain conditions that tend to lead to SN1.  One of the big issues here for whether we're getting SN2 or E2 is going to be the substitution pattern of the alkyl halide.  We might have E2 versus SN2 or E2 elimination versus SN2 substitution or SN2 displacement.  So generally, if you take a tertiary alkyl halide, and of course that tertiary alkyl halide has to have some beta protons  like we've seen here, a proton that can be pulled off, plus an alkyl oxide.  And I'll write this generically as tertiary alkyl halide plus an alkyl oxide.  Now we're going to get mainly E2.  Tertiary alkyl halides also with alkyl oxide bases and other very strong bases like hydroxide tend to give mainly E2.  So in other words, let's say I took, oh, bromo cyclohexane.  And let me take a different alkyl oxide base.  Let me take potassium tert butoxide.  And I'll indicate our solvent. Usually when one works with potassium tert butoxide, one generates it in tert butanol.  And so often one uses tert butanol as a solvent.  Sometimes as you saw in the last chapter, one can use a dipolar or aaprodic solvent like dimethyl sulfoxide.  But if one treats our bromo cyclohexane with potassium tert butoxide, the main product is going to be cyclohexane.  Perhaps you will get a very, a very minor amount of the tert butoxide of the SN2 displacement product, cyclohexyl O tert butyl.  I'm going to put minor, but I'll put parenthetically very minor.  By that I mean that you would get, oh, a few percent of the ether product, a few percent of the SN2 displacement product at best.  In general, tertiary, in general, secondary alkyl halides tend to eliminate predominantly mistorm bases.  Just to remind us what this means here, I'll draw out the structure.  So this is potassium tert butoxide drawn out as a full structure.  We call it tert butoxide. I'll write out the name.  We call it tert butoxide or tertiary butoxide because it is the species that results from deprotonation of tertiary butanol or t-butanol.  Don't answer questions.  All right. So we've talked about tertiary halides, talked about secondary halides.  Let's take a moment to talk about primary halides.  The first question.  Ah, we're comparing the E2 to SN2.  But what about SN1? Okay, this is a great question because this is where people start to say, wait a second, there are all these matter folds in the bottom.  All right. So SN1 reaction, great question.  SN1 reaction, tert halides do indeed, tertiary halides like tert-butyl bromide, do indeed ionized.  So there is a rate of reaction for ionization before mature tertiary carbocation followed by what's possibly reaction with tert-butanol or tert-butoxide.  But the rate of reaction with a strong base of pulling off that beta proton is so much greater that for all intents and purposes we never get a chance to have reaction by way of an SN1 pathway when you have a strong base present.  So indeed, that pathway is possible if we didn't have tert-butoxide just like a tert-butanol, you would indeed see an SN1 pathway which would give rise to a substitution product.  But once you have that base, it's so much lower in energy to just yank off that proton. Good question. Other questions?  All right. Let's look at primary halides. And again, we're really looking here at reaction with alkoxide or other bases of comparable or greater strength.  So hydroxide would be comparable in strength. NH2 minus MIN would be stronger in strength. So mainly SN2, but I'm going to put a caveat on that.  And that star is just a reminder that we're going to see that even here there's some complexity. So let's start with a simple case. We'll take bromopropane.  And again, we'll use sodium ethoxide, let's say, inethanol. So the main product of this reaction is going to be our SN displacement product.  You'll also get a little bit, so I'll write SN2 here, and I'll write major. We'll also get a little bit of the E2 elimination product, so I'll write E2 and minor here.  Now chemists know this from actual experiment. I'm not just pulling these numbers out of thin air or making this up. This is what people in the 1930s and 40s investigated experimentally to learn about.  And so when this was investigated it was found that there was a 90% conversion to the SN2 displacement product and a 10% conversion to the E2 product. You wouldn't know that. You wouldn't have to know that.  You don't have to know those exact numbers, but this helps us see some trends. So we sort of tear on the edge of an equilibrium.  And I've tried to give us a feeling for energetics in equilibria. A 10 to 1 ratio corresponds at room temperature to a 1.36 kilocalorie per mole energy difference. That's not a big difference in energy.  And some factors can tip the balance in going one way versus going the other way. And so even something as simple as steric hindrance can make a difference and push things into the E2 pathway.  Specifically, if we go and we have beta substituents, this makes it very hard, makes it, let us say, harder for the nucleophile to get in close for that trigonal bond.  For that trigonal bifuraminal transition state associated with an SN2 displacement reaction.  Remember when I projected the reaction up there, projected the transition state, how crowded it was, how everything had to get in close.  Even a nearby substituent that's in the way can help block things and make it more difficult.  So if I add a beta methyl group, this isn't a methyl group on the carbon with the halide.  I've been alphabet. That would take us from primary to secondary.  But if I add a beta methyl group, just the spectator here, that extra steric bulk in going from a simple chain to a branch chain makes it more difficult for the nucleophile to come in.  And the result is that it favors E2 elimination.  Just a little bit. So if I take, instead of our bromocropate, our isopropyl, our isobutyl bromide, and again I treat it with sodium, methoxide, and ethanol.  And just for the sake of variety, I'll remind us that there are many ways to write this reaction. This time I'll write the reaction over the arrow.  Now we also get a mixture of SN2 displacement and E2 elimination.  But now the E2 elimination predominates. So we get a mixture of the ether and the alkene, and that mixture is a 38 to 62% mixture.  In other words, we've switched from the elimination pathway being the minor pathway to the elimination pathway being the major pathway.  Not only does steric hindrance of the electrophile affect the pathways, but also steric hindrance of the base affects the pathways.  So a bulkier base leads to more elimination as well.  And so I'll keep our same electrophile in this comparison. I'll keep our same alkyl halide. But now instead of having sodium ethoxide and ethanol, I'll switch to potassium tertbutoxide.  In tertbutanol, tertbutanol makes the alkoxybrom a little bit more sterically hindered. It also makes it a little harder to salvate, a little bit more basic.  PK, or tertbutanol, usually people use a number about 18 instead of say 17 for ethanol. And as a result of the greater steric hindrance and the greater base isamine, it's harder, even harder, for the nucleophile to get in and do an SN2 displacement.  And it's more favored to have the beta proton hold off. So now our ratio of SN2 displacement to E2 elimination goes all the way up to 8 to 92.  In other words, now we've switched from 38 to 62 to 8 to 92. We've really pushed the reaction. And this is good, because organic chemists like to be able to control which way a reaction goes.  And this begins to show us how an organic chemist can choose conditions that favor either an alkane product or an alkene product or an ether product.  Last point I want to make, just on the generation of alkoxide bases, just because it's useful to know where they come from and why they can be used in the same alcohol.  In general, if you take an alcohol plus sodium or potassium, the metal, the alkali metal will react to give the corresponding alkoxide plus hydrogen gas.  And you know if you throw sodium into water, it catches fire. You know if you throw potassium into water, it explodes.  So organic chemists often choose their metal carefully. They'll use sodium for more sterically unhingered alcohol, like methanol or ethanol.  Because nobody wants to have their laboratory catch fire. And they'll use potassium for more sterically hindered alcohols that react slowly like tributanol.  And that's one of the reasons why I've been switching around from sodium to potassium.
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