Understanding Elimination Reactions

In chapter 8, we are going to look at alkyl halide and elimination reactions. So these are very similar related to the nucleophilic substitution reactions because, again, there are going to be two different types, the E1 and E2, just like the SN1 and SN2, and they go through similar mechanisms. This compound here is quainine. It was kind of one of those things where the synthesis of this... began the modern day organic synthesis of different compounds. So this is a natural product from the bark of the synchoid tree in the Andes Mountains, and it's a fever reducer, and it also was a treatment for malaria. It began to be put into different types of waters. Tonic water was included in that type of water. You can still buy this today with quinine in it, and it gives it a kind of a bitter flavor. Also it fluoresces under UV light. So if you have tonic water and you put a black light up to it, it kind of glows. It's pretty neat. So one of the steps in the synthesis of quinine is an elimination reaction. So that's why it's included in the chapter header of this chapter. On the top here, we have a generic version of an elimination reaction. And some of the big things that happen here is that we are going to, now we have an elimination. So we are going to remove some compounds from here and we're going to make a double bond. So what we're going to be doing is we're actually going to use this as instead of a nucleophile. So now instead of a nucleophile, we are going to have a base. So this is the big difference here. And now base is going to be attacking the hydrogen, the hydrogen on here. And that's going to, then the hydrogen is going to leave. the electrons behind and we're going to also have the leaving group still leave with the electrons in the bond. So now we're going to form a new bond here you can see in the middle we have our leaving group left with the electrons and now we've formed a new compound formerly the space now it's the conjugate acid is formed. And that's kind of the difference here. So now the elimination, again, we're going to remove things. We're going to form a double bond. And this is the difference. Here's some more specific examples on the bottom where we have potassium. Remember, we've talked about this a few times where we have this metal and a nonmetal, typically an organic compound. So those will ionize. This becomes a strong base now because it's negatively charged. It is going to be able to attack the, we have some electrons on here, it's going to be able to attack that proton. The proton creates that double bond and the leaving group leaves with our ions from this bond here. So, or electrons from that bond. So now we have the conjugate acid, we have our leaving group with our metal ion, and that's kind of the double bond product as well. Here's another example. The sodium is the metal. That means we have a strong base. It's going to attack our proton, and that proton is going to leave the electrons to form that double bond. Our leaving group is leaving with the electrons. So we make a salt here, a conjugate acid, and our new double bond. So that's kind of the overall gist of an elimination reaction. We're going to have a base pull off a proton, the proton leaves electrons, and then a leaving group takes the electrons. the carbon that forms the new double bond. In the past I used to run an elimination reaction as part of an aldol condensation in the organic chemistry lab. We didn't run that this year because we had a little less time for this. But the first step is this addition reaction where basically gluing this aldehyde and ketone together to form this aldol group. And then this next step what we do is we heat it and we have We get this water to be a leaving group, and then we form a double bond here. So this is really an elimination reaction that we're doing in the second step. Again, we didn't have time to run this this year, but this is a pretty common type of chemical reaction that's done in organic chemistry labs as well. Some terminology that's important to recognize here is that we have a... our base rather than again our nucleophile and we are going to have our proton that's going to attack. So a lot of times we're going to be able to and want to define the different carbons that our groups are attached to. So typically the leaving group carbon, the carbon that's attached to the leaving group and here in this chapter it's typically going to be a halogen or an X here in this instance. So this carbon is going to be our alpha carbon so that first carbon that we're going to talk about. So a lot of times you kind of want to identify the leaving group and that's going to be alpha. Then next to it is an adjacent carbon. So maybe any of these are adjacent carbons. Those are your beta carbons. That's your B carbon. The beta carbon in this reaction is where the base is going to grab a proton from a hydrogen. So it could grab it from this beta carbon. There's actually a beta carbon here and a big beta carbon here as well. It's just any carbon that's attached to that alpha carbon. So this base has options for which hydrogen to grab, but we'll talk about how to identify that one. What's going to happen is that we're going to be able to identify the alpha and beta. Between alpha and beta, that's where that double bond is going to form. And then again, we have that conjugate acid and then the leaving group compound as well. Here are some examples with 2-bromone, 2-methylpropane. So just specific examples. And here's the alpha. and it's really the same compound that's up here right but i was showing here we have a beta a beta beta there are three different options for where that double bond can form here so it's helpful to use that terminology and understanding that oh alpha and beta this is this is where the bond goes between alpha and beta and this type of chemical reaction here's another example with a strong base and then we're going to have multiple options we have a beta 1 or a beta 2. so sometimes These beta positions are all equivalent, so it doesn't really matter because it's just a methyl group. Either way you go, anywhere you go. These are maybe a little different, right? This beta is a primary carbon, and this is going to be a secondary carbon. So these are different positions because they're different types of carbon systems, and that's going to dictate if you have your double bond on the end or double bond in the middle. And we're going to talk about that, but one is better than the other. One is going to be more stable. you'll have one of these that'll form in higher concentrations than the other, but we'll get there when we get there. In this chapter, these are the types of bases that you'll commonly see, and you'll notice that sodium, potassium, right, those are going to be the metals typically that are used in these types of bases, and then we either have hydroxide or we have different organic compounds that might be involved, and these are all pretty strong bases, and these are smaller bases. but then we can start to make bigger and bigger bases potassium tert butoxide for example is going to be quite different than sodium hydroxide because this group is going to have a lot of steric hindrance and it's going to influence where it's going to be able to take a proton from right remember these are bases it's going to be looking for acids and typically protons we're talking about more of a bronstad type of base that we're talking about so this is a little bit of a learning check i would like you to make sure that your label the Find the alpha and beta carbons. If you don't remember what those are, go back a few slides, go rewind this, go back to the beginning of this lecture and figure out how to define those. And then draw all of the possible elimination products. So make sure you're drawing all of them that are possible. There might be more than one in some of these. So pause here, get a piece of paper out, try to do these questions. And what I'm going to do is when we meet again, we'll... I'll go through some of these answers with you just to make sure you're on the right track. Now this compound here is DDT and it used to be used as an insecticide but what was found it was getting into the environment and then this got into different animal species into the fats of animals and then those animals were eaten by birds of prey a lot of times so eagles and hawks. And what happened is it actually resulted in the development of really thin shells that are easily crushed. So then when these large birds would sit on the eggs, then the eggs would crush them and they would kill the babies. So it was pretty sad. So this has been since eliminated. And we're also going to look at an elimination reaction with them. So what I want you to do is when we, the first step for the degradation of this. pesticide is actually shown here what happens is we get we actually have a base is going to pull this proton off. So this carbon is the alpha carbon. Here's the beta carbon. So this base pulls this and then we eliminate one of the chlorine and then DDE, that's the other compound that's generated here. We have the double bond and then the elimination reaction. So these are the compounds that actually build up both of these in the fatty tissues. So the next thing we want to do in order to start to understand the stability and the different types of compounds that are generated from these elimination reactions where we want to be able to identify and classify the double bonds. So when we have mono substituted in terms of just defining what that means, here's our double bond and it just means that there is one carbon that's associated off of or substituted off of the carbons that are involved in the double bond. So this is just mono substituted. Here's an example of a ring structure where we have a double bond in this ring. And now we say there are di substituted because here's the double bond is a carbon on both sides of that double bond. So we have a carbon here and a carbon here. And then our other carbons that are attached to that, the double bonded carbons are here. So that's why we have a di substituted. Here we have on the third one here, we have the double bond. We have again a carbon and carbon on both sides. And then associated with all of those carbons, we have a tri-substituted double bond. So we can have mono, di, tri, and tetra-substituted groups that are associated. And the reason we need to identify those and define those and provide vocabulary for that word is because we are going to now, next, the next step is to understand which one's going to be more stable. And it determines where, you know, what... beta carbon, essentially, that we're taking our hydrogen from typically. All right, so here's another learning check. And I want you to take a moment to classify each double bond. So these are referred to as alkenes, E-N-E ending. So you remember we have alkanes, A-N-E, that's single bond, E-N-E, we're talking about double bonds. So all of the double bonds, we've talked about the mono, di, tri, tetra. so i want you to identify there are multiple ones in a and b so pause take a second identify what those all are and then and then move on and then we'll go from there so these are fairly easy to identify we find the double bond first find the carbons associated with them and then we count how many things are attached so this one is four so it's tetra here we have a double bond we have one two so it's die Here we have our double bond here and here. One, two, three. So it's tri. Double bond, double bond. One, two, it's di. And then here we have our double bond. One, two, three. So that's tri. And here we just have three different positions. Carbon here and a carbon here. One, two, three. That's tri. Carbon here, carbon here. One, two, three. That's tri. Carbon here and here. One and two. That's di. So these vitamin A and D3. These are how we would classify each of those alkenes. A few chapters ago, we identified these as conformational isomers. So we have this group coming down on one side and up on the other. But this ring has free rotation around that center carbon-carbon bond. We're able to rotate those freely. And we talked about different... gauch versus the eclipsed versions in which is you know this is the higher energy and this is the lower energy version but they have free rotation and they're able to rotate around and depending on how much energy or heat's in the system they're able to be in different positions but when we start talking about double bonds when we talk about e and e endings so we can have the same side or different side opposite sides we're talking about here are carbons and you notice that this bond up and up, that means the cis bond. And this is a hydrogen down and down. So those, the bulky groups, that's what we're defining as the same side as cis. If the large groups are going to the same side, then we would define it as cis here. And over here, here's our carbon and carbon, a double bond. One is going down, the big one, the one that's going up, and our hydrogens are on either side as well. So that's our trans, trans butytoene. versus cis-2, cis-but-2-ene. So this two number is telling us what carbon has the E and E are the double bond. So it'd be a one. two, three, four, and that's why it's the methyl ethyl propyl butyl bute, and then the cis because we're on the same side, and the two tells us that this is the second carbon is where we have that double bond. These, the big thing about this is that we cannot interconvert these. So this does not easily, the cis bute 2-ene does not easily convert to the trans bute 2-ene. It can happen if there's enough energy in a system where we can temporarily break the double bond and reform. And that's a lot of times where we would generate trans fats, which you may have heard are not really great for your body. But it can happen. And it's usually when we are processing foods. We're doing ultra processed foods and we get trans fats. And they're just not natural for our bodies. And we don't have enzymes to break them down. So they can cause some issues. But these typically don't interconvert under normal conditions. And this is just a different way to show that with these stick structures. And here's our carbon-carbon double bond on the top. And then we have our, sorry, the carbon-carbon. And then on the same side, these are both going up. These are carbon groups up here and here. And this is cis for same. Here is trans for opposites, so our carbon and carbon. We're talking about the opposite side of the double bond, so up versus down. So make sure that is the terminology, that's how they're defined. Just a learning check here, I'd like you to go through and we'll come back to this in class, but for which double bonds are stereoisomers possible? So A, B, and C, identify those. If you're not sure, go back what that means and we'll check this again in class. And then once you're done with that... Then go down to problem 8.4 and walk through that, and then do your best to answer those questions, and then we'll again cover that in class. So pause here, take some time to try these out, and then bring your answers to our next meeting. Okay, so here's another learning check, and here you're going to label each pair of alkenes as constitutional isomers, stereo isomers, or identical compounds. Here is A, B, C, and D. Take your best guess as to what those are. If you're not sure, take some time to go look those up and identify the proper terminology and definition. And then again, bring these answers to our next class period. So this is all I'm going to, this will be the end of the first component. I like to make these about 20 minutes just so you get a little bit of a taste for this. And then there'll be another lecture for the remainder of this other parts of this chapter.

began the modern day organic synthesis of different compounds. So this is a natural product from the bark of the synchoid tree in the Andes Mountains, and it's a fever reducer, and it also was a treatment for malaria. It began to be put into different types of waters.

Tonic water was included in that type of water. You can still buy this today with quinine in it, and it gives it a kind of a bitter flavor. Also it fluoresces under UV light. So if you have tonic water and you put a black light up to it, it kind of glows. It's pretty neat.

So one of the steps in the synthesis of quinine is an elimination reaction. So that's why it's included in the chapter header of this chapter. On the top here, we have a generic version of an elimination reaction. And some of the big things that happen here is that we are going to, now we have an elimination.

So we are going to remove some compounds from here and we're going to make a double bond. So what we're going to be doing is we're actually going to use this as instead of a nucleophile. So now instead of a nucleophile, we are going to have a base. So this is the big difference here.

And now base is going to be attacking the hydrogen, the hydrogen on here. And that's going to, then the hydrogen is going to leave. the electrons behind and we're going to also have the leaving group still leave with the electrons in the bond.

So now we're going to form a new bond here you can see in the middle we have our leaving group left with the electrons and now we've formed a new compound formerly the space now it's the conjugate acid is formed. And that's kind of the difference here. So now the elimination, again, we're going to remove things. We're going to form a double bond.

And this is the difference. Here's some more specific examples on the bottom where we have potassium. Remember, we've talked about this a few times where we have this metal and a nonmetal, typically an organic compound. So those will ionize. This becomes a strong base now because it's negatively charged.

It is going to be able to attack the, we have some electrons on here, it's going to be able to attack that proton. The proton creates that double bond and the leaving group leaves with our ions from this bond here. So, or electrons from that bond. So now we have the conjugate acid, we have our leaving group with our metal ion, and that's kind of the double bond product as well.

Here's another example. The sodium is the metal. That means we have a strong base. It's going to attack our proton, and that proton is going to leave the electrons to form that double bond. Our leaving group is leaving with the electrons.

So we make a salt here, a conjugate acid, and our new double bond. So that's kind of the overall gist of an elimination reaction. We're going to have a base pull off a proton, the proton leaves electrons, and then a leaving group takes the electrons.

the carbon that forms the new double bond. In the past I used to run an elimination reaction as part of an aldol condensation in the organic chemistry lab. We didn't run that this year because we had a little less time for this.

But the first step is this addition reaction where basically gluing this aldehyde and ketone together to form this aldol group. And then this next step what we do is we heat it and we have We get this water to be a leaving group, and then we form a double bond here. So this is really an elimination reaction that we're doing in the second step.

Again, we didn't have time to run this this year, but this is a pretty common type of chemical reaction that's done in organic chemistry labs as well. Some terminology that's important to recognize here is that we have a... our base rather than again our nucleophile and we are going to have our proton that's going to attack. So a lot of times we're going to be able to and want to define the different carbons that our groups are attached to.

So typically the leaving group carbon, the carbon that's attached to the leaving group and here in this chapter it's typically going to be a halogen or an X here in this instance. So this carbon is going to be our alpha carbon so that first carbon that we're going to talk about. So a lot of times you kind of want to identify the leaving group and that's going to be alpha.

Then next to it is an adjacent carbon. So maybe any of these are adjacent carbons. Those are your beta carbons. That's your B carbon. The beta carbon in this reaction is where the base is going to grab a proton from a hydrogen.

So it could grab it from this beta carbon. There's actually a beta carbon here and a big beta carbon here as well. It's just any carbon that's attached to that alpha carbon. So this base has options for which hydrogen to grab, but we'll talk about how to identify that one. What's going to happen is that we're going to be able to identify the alpha and beta.

Between alpha and beta, that's where that double bond is going to form. And then again, we have that conjugate acid and then the leaving group compound as well. Here are some examples with 2-bromone, 2-methylpropane.

So just specific examples. And here's the alpha. and it's really the same compound that's up here right but i was showing here we have a beta a beta beta there are three different options for where that double bond can form here so it's helpful to use that terminology and understanding that oh alpha and beta this is this is where the bond goes between alpha and beta and this type of chemical reaction here's another example with a strong base and then we're going to have multiple options we have a beta 1 or a beta 2. so sometimes These beta positions are all equivalent, so it doesn't really matter because it's just a methyl group. Either way you go, anywhere you go.

These are maybe a little different, right? This beta is a primary carbon, and this is going to be a secondary carbon. So these are different positions because they're different types of carbon systems, and that's going to dictate if you have your double bond on the end or double bond in the middle.

And we're going to talk about that, but one is better than the other. One is going to be more stable. you'll have one of these that'll form in higher concentrations than the other, but we'll get there when we get there.

In this chapter, these are the types of bases that you'll commonly see, and you'll notice that sodium, potassium, right, those are going to be the metals typically that are used in these types of bases, and then we either have hydroxide or we have different organic compounds that might be involved, and these are all pretty strong bases, and these are smaller bases. but then we can start to make bigger and bigger bases potassium tert butoxide for example is going to be quite different than sodium hydroxide because this group is going to have a lot of steric hindrance and it's going to influence where it's going to be able to take a proton from right remember these are bases it's going to be looking for acids and typically protons we're talking about more of a bronstad type of base that we're talking about so this is a little bit of a learning check i would like you to make sure that your label the Find the alpha and beta carbons. If you don't remember what those are, go back a few slides, go rewind this, go back to the beginning of this lecture and figure out how to define those.

And then draw all of the possible elimination products. So make sure you're drawing all of them that are possible. There might be more than one in some of these. So pause here, get a piece of paper out, try to do these questions.

And what I'm going to do is when we meet again, we'll... I'll go through some of these answers with you just to make sure you're on the right track. Now this compound here is DDT and it used to be used as an insecticide but what was found it was getting into the environment and then this got into different animal species into the fats of animals and then those animals were eaten by birds of prey a lot of times so eagles and hawks. And what happened is it actually resulted in the development of really thin shells that are easily crushed.

So then when these large birds would sit on the eggs, then the eggs would crush them and they would kill the babies. So it was pretty sad. So this has been since eliminated.

And we're also going to look at an elimination reaction with them. So what I want you to do is when we, the first step for the degradation of this. pesticide is actually shown here what happens is we get we actually have a base is going to pull this proton off.

So this carbon is the alpha carbon. Here's the beta carbon. So this base pulls this and then we eliminate one of the chlorine and then DDE, that's the other compound that's generated here. We have the double bond and then the elimination reaction.

So these are the compounds that actually build up both of these in the fatty tissues. So the next thing we want to do in order to start to understand the stability and the different types of compounds that are generated from these elimination reactions where we want to be able to identify and classify the double bonds. So when we have mono substituted in terms of just defining what that means, here's our double bond and it just means that there is one carbon that's associated off of or substituted off of the carbons that are involved in the double bond. So this is just mono substituted.

Here's an example of a ring structure where we have a double bond in this ring. And now we say there are di substituted because here's the double bond is a carbon on both sides of that double bond. So we have a carbon here and a carbon here.

And then our other carbons that are attached to that, the double bonded carbons are here. So that's why we have a di substituted. Here we have on the third one here, we have the double bond.

We have again a carbon and carbon on both sides. And then associated with all of those carbons, we have a tri-substituted double bond. So we can have mono, di, tri, and tetra-substituted groups that are associated.

And the reason we need to identify those and define those and provide vocabulary for that word is because we are going to now, next, the next step is to understand which one's going to be more stable. And it determines where, you know, what... beta carbon, essentially, that we're taking our hydrogen from typically. All right, so here's another learning check.

And I want you to take a moment to classify each double bond. So these are referred to as alkenes, E-N-E ending. So you remember we have alkanes, A-N-E, that's single bond, E-N-E, we're talking about double bonds.

So all of the double bonds, we've talked about the mono, di, tri, tetra. so i want you to identify there are multiple ones in a and b so pause take a second identify what those all are and then and then move on and then we'll go from there so these are fairly easy to identify we find the double bond first find the carbons associated with them and then we count how many things are attached so this one is four so it's tetra here we have a double bond we have one two so it's die Here we have our double bond here and here. One, two, three.

So it's tri. Double bond, double bond. One, two, it's di.

And then here we have our double bond. One, two, three. So that's tri. And here we just have three different positions. Carbon here and a carbon here.

One, two, three. That's tri. Carbon here, carbon here. One, two, three. That's tri.

Carbon here and here. One and two. That's di.

So these vitamin A and D3. These are how we would classify each of those alkenes. A few chapters ago, we identified these as conformational isomers.

So we have this group coming down on one side and up on the other. But this ring has free rotation around that center carbon-carbon bond. We're able to rotate those freely. And we talked about different... gauch versus the eclipsed versions in which is you know this is the higher energy and this is the lower energy version but they have free rotation and they're able to rotate around and depending on how much energy or heat's in the system they're able to be in different positions but when we start talking about double bonds when we talk about e and e endings so we can have the same side or different side opposite sides we're talking about here are carbons and you notice that this bond up and up, that means the cis bond.

And this is a hydrogen down and down. So those, the bulky groups, that's what we're defining as the same side as cis. If the large groups are going to the same side, then we would define it as cis here.

And over here, here's our carbon and carbon, a double bond. One is going down, the big one, the one that's going up, and our hydrogens are on either side as well. So that's our trans, trans butytoene. versus cis-2, cis-but-2-ene.

So this two number is telling us what carbon has the E and E are the double bond. So it'd be a one. two, three, four, and that's why it's the methyl ethyl propyl butyl bute, and then the cis because we're on the same side, and the two tells us that this is the second carbon is where we have that double bond. These, the big thing about this is that we cannot interconvert these. So this does not easily, the cis bute 2-ene does not easily convert to the trans bute 2-ene.

It can happen if there's enough energy in a system where we can temporarily break the double bond and reform. And that's a lot of times where we would generate trans fats, which you may have heard are not really great for your body. But it can happen.

And it's usually when we are processing foods. We're doing ultra processed foods and we get trans fats. And they're just not natural for our bodies. And we don't have enzymes to break them down. So they can cause some issues.

But these typically don't interconvert under normal conditions. And this is just a different way to show that with these stick structures. And here's our carbon-carbon double bond on the top.

And then we have our, sorry, the carbon-carbon. And then on the same side, these are both going up. These are carbon groups up here and here. And this is cis for same.

Here is trans for opposites, so our carbon and carbon. We're talking about the opposite side of the double bond, so up versus down. So make sure that is the terminology, that's how they're defined. Just a learning check here, I'd like you to go through and we'll come back to this in class, but for which double bonds are stereoisomers possible? So A, B, and C, identify those.

If you're not sure, go back what that means and we'll check this again in class. And then once you're done with that... Then go down to problem 8.4 and walk through that, and then do your best to answer those questions, and then we'll again cover that in class. So pause here, take some time to try these out, and then bring your answers to our next meeting.

Okay, so here's another learning check, and here you're going to label each pair of alkenes as constitutional isomers, stereo isomers, or identical compounds. Here is A, B, C, and D. Take your best guess as to what those are.

If you're not sure, take some time to go look those up and identify the proper terminology and definition. And then again, bring these answers to our next class period. So this is all I'm going to, this will be the end of the first component.

I like to make these about 20 minutes just so you get a little bit of a taste for this. And then there'll be another lecture for the remainder of this other parts of this chapter.

Transcript for:Understanding Elimination Reactions

Transcript for:
Understanding Elimination Reactions