Hello organic chemistry students. In this video we're going to talk about substitution chemistry. Now, before I get into the blood and guts of this, I want to make sure to stress that we are not going to cover all of the blood and guts of substitution chemistry. I know you're like, wow, that sounds interesting. And why do I say that?
There are so many other facets about substitution chemistry that we're not going to cover in this video, such as solvent effects, temperature, isoelectric constants, dielectric constants, and other things like that. All we're going to get into is the... bare basics about substitution chemistry that allows us to understand how this happens in biological systems.
So we're going to really focus in on primary, secondary, tertiary carbons, and what type of substitution pathway they go down. And it's going to be based mostly upon the stability of carbocations and the strength of the nucleophile. Now the overall general trend here is what we're going to see is, is that we're going to see an X, and we're going to be looking at halogens mostly, but it can be any.
other good leaving group. So please keep in mind X can be any other good leaving group. And it's going to be attached to a carbon that has two R groups attached next to it.
Where those R groups are either going to be carbon or hydrogen. Now implicitly for this class we're also going to say that this carbon that the leaving group is attached to, even though I'm saying X, I want to stress leaving group. It has to be sp3 hybridized, which we'll talk about here in a little bit. Not sp2 or sp, we're only going to focus in on the sp3 hybridized carbons with leaving groups. So in this, a nucleophile that either could have a full-out negative charge, or it can be 100% neutral or partial negative, it's ultimately going to come around, attack this carbon, and push out this X group.
And in doing this, we're going to substitute the nucleophile for the leaving group. Now in this, the way I've just drawn this was, is that the nucleophile attacks that carbon, pushing out the leaving group. So two things are happening simultaneously right now. And this is what we call an SN2 reaction.
This 2 right here refers to bimolecular. two things are happening. That leaving group does not leave until the nucleophile attacks that carbon and pushes it out of the carbon.
Another way that we can talk about this, and I'm going to use the same exact picture here with a leaving group on a carbon with two R groups on it, is if the leaving group was to leave on its own and we form a carbocation. Once the carbocation forms, the nucleophile being negative, partial negative or neutral, can come in and attack that carbocation and form our substituted product. Here's the nucleophile and X minus right out here. So in this case, did the nucleophile play any part of the leaving group leaving the molecule?
And the answer is no. And this is what we call an SN1 substitution. One thing is happening at a time and it's called unimolecular. So what's the big difference between an SN2 and SN1 reaction?
In an SN2, the nucleophile is directly responsible for attacking the carbon, pushing out the leaving group. Two things happening at once. In an SN1, the leaving group leaves on its own, forms a carbocation, to which the nucleophile attacks.
So the nucleophile only attacks this carbon because the leaving group left on its own. I'm sorry, the leaving group left on its own. And that's why it's a unimolecular reaction.
Now, when we're all said and done, it looks like we're getting the same products here. In a lot of cases, we will, but we'll talk about cases that we don't. The important thing here is we are looking at a substitution done in a bimolecular, the nucleophile causing the displacement, or a unimolecular process where the leaving group leaves on its own and we attack the subsequent group. carbocation.
So if we mix a couple other big, big, big, big observations here is in an SN1 reaction, we form carbocations. SN2, there are no carbocations. In an SN2 reaction, the nucleophile attacks and displaces and we form our products all in one step. Down below with an SN1, the leaving group leaves to form the carbocation.
The nucleophile then attacks to form the product. Two steps. Now, why am I saying this?
This is a common thing that students will get confused on. They'll think, oh, two steps, so SN2. No, no, no. It's SN1 because one thing is happening in each step.
SN2, everything happens at the same time. So we've now just covered the basics of what's happening in an SN1, SN2 carbon. What I want to go over right now is carbon types again.
So we can have a leaving group, and once again, I'm just going to use X, even though X stands for halogen that we're going to see mostly in this class, but it can be any leaving group. This carbon that the leaving group is on is what we call a primary carbon because there is one carbon directly attached to it. So this is the carbon with the leaving group attached to one carbon. We can also have the leaving group on a carbon that's secondary.
We can also have it where it's on a, oh, you got it, a tertiary carbon. Now, the one that we didn't mention up above is, what if it's just a carbon with hydrogens on it and a leaving group? That is what we call a zero-order carbon. Now, these are the basic types of carbons we're going to be dealing with in substitution. Now the first thing I want to go over is, out of all of these, a primary, secondary, tertiary, and even a zero-ordered, carbon down below, which one can form a carbocation more easily?
Meaning, which one could allow the leaving group to leave and form a carbocation? That's the big question. In a tertiary, we know that if that leaving group leaves, we form a tertiary carbocation, which is highly stable. A secondary carbocation, it's possible to form.
It's not that possible, probable. It's going to depend upon some other factors, but it's still probable. In a primary, if we formed a primary carbocation, this is not going to happen.
We are not going to see primary carbocations forming all that readily in this class. And at this level of chemistry, we are not going to see primary carbocations form whatsoever. Now, once again, just because I am a chemistry professor, I have to tell you the truth. We can form them.
They're rare and unique circumstances, but we are not going to cover them in this class, which means the zero-ordered carbocations never form whatsoever either. So we can only form carbocations on tertiary or secondary carbons with a leaving group attached. So can we come up with a trend here? Ah, so on these secondary and tertiaries, Sn1 chemistry is possible.
On the primary and the zero order, since Sn1 is not possible, Sn2 is going to be the major pathway in those two right there. Now what I want to do is go and erase this bracket for one second and Sn1 and go ahead and rewrite Sn1 here and Sn1 here. Now we've covered our basic framework right now. Zero and primaries are Sn2. Tertiary and secondaries are SN1.
Secondaries though, remember we just talked about how the carbocation, while it can form, it's not necessarily a stable. We can also do SN2 chemistry on secondary carbons. This is where it gets a little interesting.
Everything else is pretty much straight cut. Tertiary does SN1. Primary or zero order does SN2 chemistry. Secondary is where several things can happen.
And we're going to cover that one last. Let's go ahead and get into an SN1 reaction and why it occurs on tertiary carbons very readily. So tertiary carbons with a leaving group. Why on earth does a nucleophile wait to attack that carbon until we form a carbocation?
Now, When we're all said and done, just trying to show the full product here, there's the Nucleofile right there and the X- left over. What we have to ask ourselves is, this nucleophile has electrons in it. This carbon has three carbon groups next to it, each with three atoms on it.
So there's a lot of electron bulk on this tertiary carbon. So if the nucleophile that has electrons approaches this carbon that has high electron density around it, is that going to cause a stabilizing or destabilizing effect? And it's going to be destabilizing because of the electron repulsion in the system. And this is why we do not do SN2 chemistry on a tertiary carbon. There are cases that we can.
In this class, we are not going to cover them whatsoever. The main reason why we do not do SN2 chemistry on a tertiary carbon is because of the high electronics or electron density in these groups that repulses the nucleophile trying to approach in. So when this is all said and done, tertiary carbons go through an SN1 process, and that's why we see SN1 on tertiaries. We can see it in secondaries as well, but we're going to cover that in a little bit. But what I want to do now is talk about primary carbons.
So if we look at a carbon that has a leaving group on it, here are our two hydrogens implicitly showing that in this case, and here's the nucleophile. In this case right here, There is no way that this leaving group can leave on its own and form a primary carbocation. One inductive and possibly one hyperconjugation effect will not stabilize this.
Not going to happen. So what's going to happen is, is that the nucleophile will come around and attack from the opposite side of the leaving group and push it out of the molecule. So what we're seeing that happens here is, I'll go ahead and show the interim, not the intermediate, but the transition state of this reaction, noted by that little prime error that I just said up above. Here is the nucleophile right here.
So what's happening is that here's our carbon, here's our halogen trying to leave the molecule or leaving group, and here is the nucleophile approaching from the opposite side. And here are the two hydrogens in this system. Let me go ahead and show this in the three-dimensional plane. One hydrogen going into the plane, one hydrogen coming out.
So as the nucleophile approaches this carbon, the leaving group starts to leave. So it must attack from the back side. An SN2 reaction always involves backside attack.
It's a process called Walden inversion. And I'm going to go ahead and write that term down. Not that I'm ever going to test you on this whatsoever.
I'm not going to ask you what is the name of this process right here. I'm giving you this name in case you want to learn more about this process. It goes into more of the molecular orbitals of this reaction.
that we're not going to cover in this class because you don't need it for any of your majors. The important thing to take home here is that the nucleophile attacks from the opposite side of the leaving group, and that's how we get this reaction to occur. We'll see implications of this later on.
So in a tertiary carbon, we form carbocations to which the nucleophile attacks, and we form our product. In a primary or zero order, the nucleophile attacks the carbon, pushes out the leaving group to form this compound here. Now, out of these two reactions, which one do you think goes faster? Do you think it's faster for a leaving group to leave and form a high-energy carbocation to which the nucleophile attacks and form our product? Or does the nucleophile attack pushing out a leaving group to form our products?
The big answer here is it depends upon the nucleophile. If we have a strong nucleophile, it can attack the backside and push out the leaving group easily. If the nucleophile is relatively weak, it's going to be harder to attack this carbon, right? So that's going to allow the leaving group to leave on its own, form the tertiary carbocation, and then do our substitution process.
Now, I'm just giving you the idea of nucleophilicity right now. This is going to help us in secondary carbons. For right now in this class, it doesn't matter the strength of the nucleophile. Tertiary carbons with leaving groups undergo SN1. Primary carbons with a leaving group on it will undergo an SN2 reaction.
For this class, those are our trends. I just introduced the concept of nucleophilicity and how it can bias a chemical reaction. All right, so now with that in mind, let's go ahead and get right into a secondary carbon where an SN1 or an SN2 process can occur.
So if an SN1 process was to occur, the halogen or the leaving group will leave on its own, and we form our carbocation, in which a nucleophile will attack the carbocation, and we form our substituted product. If we undergo an SN2 reaction, let me go ahead and just move this all the way over here, the nucleophile attacks on the backside of the leaving group, and we form our substituted product. So here's SN1, here is SN2. Now, what's going to decide if we do an SN1 or SN2 on a secondary carbon?
It's going to be the strength of the nucleophile. The strength of the nucleophile is really going to dictate if we do SN1 or SN2. So if it's a strong nucleophile, we're going to be favoring SN2 chemistry because it's quicker.
The nucleophile is highly unstable, therefore it attacks the carbon, pushes out the leaving group. If it is a weak nucleophile, we're going to be doing SN1 chemistry. Now, how do we determine if something is a strong or weak nucleophile? It relates all back down to stability.
And right now you probably hear my cat crying a little bit. She's hopping up on the desk to kind of see what I'm doing right now. So please forgive me if you hear her.
So we're going to talk about the strength of nucleophiles right now. So if I was to give you a Cl- versus OH-, can we clearly say that one of them is a stronger nucleophile than the other? And we can.
We've learned that halogens are stronger than the other. have a large valence shell and therefore can stabilize its negative charge. Oxygen is smaller.
So therefore, this negative charge is in a smaller valence shell and it is more reactive. So if we had to compare which one was more nucleophilic, it's the OH-. So this goes right back down to our basic stabilization or stability inside molecules that we talked about in previous chapters. And this is why we talked about it before.
It wasn't that just talking about stability was important. It's important in terms of chemical reactivity that we see right here. So if you have a strong nucleophile, we favor SN2. If we have a weak nucleophile, we're going to favor an SN1 on secondary carbons.
The nucleophile strength does not matter on tertiaries in this class. Tertiaries undergo SN1. Primary or zero order undergo SN2 only.
Now, what does this mean in terms of product distribution? Let's go ahead and look at this reaction to see what is the implications of the final product. Here, if we were to invoke an SN1 reaction, so if SN1 occurs, we're going to form our secondary carbocation from which the methanol, this oxygen right here, will attack that carbocation. And we've now learned that carbocations have... absolutely no facial selectivity.
So we are going to form that alcohol attacking from the top side of this molecule, or it can attack from the back side of this molecule. And that's when we do an SN1 reaction, forming a carbocation. Now, what if we were to invoke an SN2 mechanism? The methanol has to attack from the back side of that Cl, which means we form one compound.
right here. So if we notice with that Walden inversion, in an SN2 reaction, we attack from the backside and we form one product. In carbocations, we can attack from the top or back face, and that's when we get multiple products out.
Now we can get other bias in the molecule that biases the front or backside attack for SN1, but we're still forming two compounds, and that's the important thing in most cases. So here in the SN2, we just form this one. Now, these are our two probable reactions.
It is. So these are our two pathways that we can do. But which one is it?
Methanol. Is it stable or unstable as shown? So we have lone pairs of electrons. That's what's doing the nucleophilic attack.
And it is 100% neutral. So this molecule is stable. So it is a weak nucleophile, which means we are doing SN1 chemistry and not. SN2 chemistry.
What if we wanted to do SN2 chemistry? What nucleophile would I have to add in to make sure I did this type of SN2 chemistry? And right now you're thinking a strong nucleophile. You're exactly right.
So I have to add in the ME, the methyl group, and the oxygen. So we need that to be negative. Now we can't just add in a negatively charged compound.
We have to stabilize it with something, a counter ion. So I'm going to go ahead and show potassium because all group 1 elements dissociate in solution. So when you have this strong nucleophilic oxygen in this potassium methoxide, it will undergo the SN2 chemistry to form this product, and we will not form SN1 products.
So what we're focusing on right here in this video is primary and zero-ordered carbons do SN2, tertiary does SN1. Secondary carbons depend upon the strength of the nucleophile. Is it stable? SN1. Is it unstable?
SN2. And that is how we predict our basic substitution reactions. Now, once again, there is more details on this reaction type, these substitution reactions. We can look at solvent and temperature and dielectric constants and other things like that.
We are not... focusing on that in this class. So ignore that part of your textbook.
All we care about is zero and primary is SN2 always, tertiary is SN1, and secondary depends upon the strength of the nucleophile that we talked about right here. So with this, I highly recommend go back, watch this video again, take more notes on it, send me emails with questions, come to our discussion section and ask any questions that you might have. So with this... I hope each of you are doing well.
I look forward to seeing you in discussion sections, and I hope you all have a wonderful day. Hello organic chemistry students! In this video we're going to cover elimination chemistry, as well as a combination of elimination and substitution chemistry at the end of this video. But now, what is elimination chemistry? Elimination chemistry is ultimately taking a compound that has a good leaving group on it.
Here we're using X for halogens because we know they're great leaving groups, except fluorine. Fluorine is a horrible, horrible leaving group. And what we're going to see here, instead of a nucleophile substituting with a leaving group, We're going to see a base come in and pull a proton on the adjacent carbon of where the carbon with the leaving group is.
And what we're going to form is a double bond right here. So elimination chemistry will always result in a carbon-carbon double bond being formed. Now I just said something very important that I want to go back to.
The base is going to come in and pull a proton on a carbon that is directly attached to the carbon. with the leaving group. So that is the relationship that we must have.
Now I want to state that over on this other carbon, there are hydrogens over here. Absolutely so. But this hydrogen is on a carbon that is not directly attached to the carbon with the leaving group.
We have to have this direct attachment. This is going to be very key. So now, how do these reactions work?
What type of elimination chemistry is there? And what on earth is a... base and what type of base are we going to use in this class to promote an elimination reaction. That's what we're going to be getting into in this video. So now, first thing I want to do is very similar to what we saw in substitution chemistry.
What type of carbon is this leaving group on? That's a primary. What type of carbon is this leaving group on?
It's a secondary attached to two carbons. And what type of carbon is this leaving group on? And that's a tertiary. So I can go ahead and write primary here, secondary, and tertiary. Now what we are going to see here is different types of elimination depending upon the carbon type and it's going to be a little different than substitution because in substitution we said oh SN2, SN2, SN1, SN1 only.
Where here we might be thinking oh this will be an E1 pathway or E2 pathway but we're actually going to see a combination of the two. Both can happen. Down here E1, E2, E1, E2.
Mostly E1 but E2 is still. 100% possible and actually favored in some cases. That seems a little weird and don't worry, we're going to get into that in this video. So now, before we get too far into this, let's go ahead and classify what base we are going to use in this class.
Now, there are a large variety of bases that can be used here. They can. What we're going to use here is a base that's going to have some type of R group on it and this carbon right here. And we're going to have an OZ group, where this Z can either be a negative charge or a hydrogen.
So basically what I'm saying is that we can have an oxygen with a negative charge or an oxygen that has a hydrogen on it. Those are going to be the bases that we use in this course. I want to stress once again, there are many other types of bases.
There are hundreds of them. But just to confine it for this class, we're going to keep it to just an oxygen with a negative charge called an alkoxy or an alcohol. And that's it. Once again, for this class. Let me go ahead and write this.
For this class only. These will be the bases. for E1 and E2.
What does that mean? E1, E2? We'll get into that in a second. But I want to stress one more time.
We're only going to be looking at oxygen types of bases. If I'm going to give you anything else that can do this type of elimination chemistry that does not have oxygen in it, I will implicitly tell you this is a base. If it's strong or weak, you'll have to figure that out depending upon if it's charged or neutral. But I will implicitly say this is a base. From here on out.
the only types of bases that can do elimination chemistry in this class are ones that are based in oxygen. Wonderful. All right.
So now let's go ahead and look at this primary carbon right here with the chloride. Now, can we have that chlorine leave on its own and form this primary carbocation? How likely is that? And we know that's not very likely.
because primary carbocations do not have enough induction or hyperconjugation effects to stabilize the resulting carbocation. But what if the chlorine was on a secondary carbon? Could that chlorine leave on its own and form a secondary carbocation? Yes, this is possible.
Secondary carbocations are stabilized by two inductive and sometimes two hyperconjugation or one or zero, depending on if there's hydrogens on the neighboring carbons. But the inductive effects allow for stabilization. And then lastly, of course, we can have that chloride leave on a tertiary carbon to form a tertiary carbocation, which we know is very, very stable. So now, if we can't form the primary carbocation, that means we cannot do any type of elimination that results in carbocations.
And if we think back to our substitution chemistry, SN1, what does that mean? SN1 means it is a unimolecular reaction step by step, where something leaves and then something attacks. And that is where carbocations were formed. So here we're asking can we do elimination or E1 chemistry on these types of carbons, thus primary, secondary, or tertiary. Now E1 is just like SN1.
We have to form a carbocation intermediate. The differences between SN and E is E is elimination, SN is substitution. We cannot do an Elim E1 reaction on a primary carbon. Not going to happen.
But now, we can do this right here, or we can also look at an E2 mechanism. Wait, what's an E2? It's very analogous to what we saw in substitution chemistry. Something is going to come in, force a reaction to occur, and we're going to get a leaving group to leave the molecule. So if we look at a primary carbon right here, here is a hydrogen.
base could come in, pull this proton, donate electrons down, and push out the chlorine, giving us the double bond. Here, the chlorine did not leave on its own. The proton being pulled facilitated the chlorine leaving, and this is what's called an E2 mechanism. Now, we can also do this on a secondary carbon, and we form this double bond here. Or we can also do it on a tertiary carbon.
Now this looks a little weird because tertiary carbons for substitution do SN1 only, not SN2. Now this is that very rare, super, super, super strong nucleophile. But why on earth can an E2 reaction happen here so easily? So here we're seeing E2s across the board.
For substitution... The nucleophile had to come in and attack this carbon and push out the chlorine. There's a lot of groups to get by.
In elimination chemistry, the base has to come in and pull the adjacent proton. That's outside the molecule, isn't it? We're not going to the sterically encumbered carbon center.
We're pulling a proton outside the molecule. So when that proton is pulled, the electrons can donate down and we can eliminate the chlorine just like this. And that's why E2 is possible. on a tertiary carbon center.
All right, so now we've talked about E1, E2. We know that E1 cannot happen on a primary carbon, but can happen on a secondary or a tertiary. That's what we're showing right over here.
We know E2 can happen on all of them. So how on earth do we distinguish between E1 and an E2 if we're, pardon me, if we're just looking at the starting material? It's going to be the strength of the base that we use.
Let's go ahead and jump right down here to the tertiary carbons right here. In an E1 mechanism, we have to go through and form a carbocation-based intermediate. This is going to take time to form, isn't it?
And it will. The chlorine has to leave, form the carbocation. So we want a base that's kind of basic to pull a proton, but not so basic that it promotes pulling the proton before the chlorine leaves.
So in order to help promote... An E1 mechanism, we want to use a weak base over here. Now keep in mind that we're only using oxygen-based bases in this class, so that's going to be an oxygen that has a proton on it.
It doesn't need to pull a proton, and it will only do so under distress. So once the carbocation forms, this proton's pKa starts to drop, and then the alcohol can pull the proton and form the double bond. We're going to go through more of the mechanisms here in a little bit. I just want to go through why E1 versus E2 and how to tell the difference.
Now, if we want to force an E2 mechanism, we want to use an oxygen that has strong A high amount of electron density. And let's go ahead and just put that potassium back in. It can be any group one element, the lithium all the way down through the francium. Because in solution, this will become positive, this will become negative.
And this now becomes a strong base. When it's a strong base, it will come in, pull a proton on the adjacent carbon that has the leaving group on it, force electrons to donate down, kick out the chlorine, to form our E2 eliminated product. So we need a strong base to favor E2. We need a weak base to favor E1 chemistry. That's the big thing here.
In order to favor E2, strong base. In order to favor E1, a weak base. Why?
And this is going to lie in the mechanism of the reaction. So let's go ahead and look at that mechanism. So If we look at this secondary carbon right here, and let's go ahead and, ah, we'll leave it right there.
In this, we'll do this in red, the chlorine can leave, and that favors the formation of a carbocation. Now, with this carbon right here being positively charged, the adjacent carbon is going to start donating electrons to that carbocation to help stabilize it. That's the inductive effect. What happens to the acidity of the protons on that carbon?
it becomes more acidic. So if we look at the hydrogen over here versus the one on the right hand side, the hydrogen now on the right side is more acidic than the hydrogen on the left side because of the resulting carbocation. This carbocation pulling electrons from this carbon, therefore pulling electrons from this hydrogen-carbon bond, makes this hydrogen more acidic. And that's what allows a base, a weak base or a strong base, it doesn't matter here, to pull this proton. So we pull that proton and we ultimately end up forming this system right here.
Now notice I just said a weak base or a strong base. If the carbocation is formed, a strong base is going to pull that proton just like a weak base is. But the question is, if a strong base is present, will we pull that proton first or allow carbocation formation?
If there's a strong base, We are going to pull the proton first because it's a strong base and therefore it needs a proton. So now coming over here we'll put in our strong base this time. Here's negative and positive and I'll draw this hydrogen right here.
There's no difference between these two hydrogens just showing them on opposite sides to give a different type of mechanism or different types of colorations to show the mechanisms types. And what we end up forming is this carbon-carbon double bond. Notice in red.
we're going through two steps. You need molecular. That's the E1 mechanism.
Over on the left-hand side, one step giving us our final product. That is the E2 mechanism. Both of these lie in acidity of the protons on the carbon directly attached to the carbon with the leaving group.
In the E2 elimination pathway, this proton can be pulled even though it's less basic than this or less acidic than this proton over here. Because of the strong basicity of this base right here. The base will pull this proton, dump electrons down, kicking out the chlorine. How does that happen?
Chlorine is more electronegative than carbon, so we have a partial, oops, that is not the tool I wanted to use there. We have a partial negative charge here, partial positive charge here. So what happens? Inductively, we send some electrons to this carbon, lowering the pKa of this proton, making it slightly more acidic. And that's why the base can pull it.
Now is this induction going to be weaker or stronger than the induction here? It's going to be far weaker and that's why we need a strong base to pull this proton, not a weak base like we do with the carbocations. It's all about the full positive charge or partial charge differences that we're seeing between E1 and E2.
Now in an E1 mechanism, when we look at this carbon right here, we can also draw the other hydrogens on this carbon. Any one of those protons could be pulled. Any which one. Because carbocations have absolutely no facial selectivity. So any one of those can be pulled, dump the electrons down, form our double bond system.
Wonderful. So let's say that again. In an E1 mechanism, any of the adjacent hydrogens can be pulled. There is no selectivity whatsoever because carbocations have no facial selectivity.
Let's go ahead and come over here. And now going back to the starting material, I'm going to make that chloride coming straight out of the... Now, Looking at this blue proton right here, I'm going to put this dashed line going into the plane. Here's the hydrogen coming out, and there's the hydrogen in the plane. Here, we can only pull the hydrogen that's on the opposite side of the chlorine.
This is important, and this is going to be a little confusing, but we're going to go through it step by step. Let me say this again. We can only pull the hydrogen that's on the opposite side of the leaving group. So I can't pull this proton.
or this one, only this one going into the plane. Why is that? In order to push out this chlorine, we have to have these electrons attack from the back side to push that leaving group out.
If it's on the same side as that chlorine, we can't dump electrons down because this is where the bonding orbital is. So the basic rule of thumb is, is that the hydrogen has to be on the opposite side of where the chlorine or the leaving group is. And this is what we call the antiperiplanar effect. Now the antiperiplanar effect, once again, is we can only pull the hydrogen on the opposite side compared to the leaving group.
So if I was to show you This is kind of jumping ahead a little bit, but this is going to be important. If I was to show you this molecule right here, I want to put two methyl groups here, which means only this hydrogen or this hydrogen can be pulled, H1, H2. Which hydrogen is pulled if we are using a strong base in this reaction? So we're ultimately going to form this carbon-carbon double bond system, but is it by pulling H2 or H1? And it has to be...
H2. The negative charge will pull H2. We donate electrons down on the opposite side of the leaving group and displace the leaving group. So the anti-periplanar effect only applies to E2 mechanisms.
And once again, it states that we must pull the proton on the opposite side of where the leaving group is. So if the leaving group is going into the plane, we pull the protons coming out of the plane. They must be opposite of one another, and that's because of something called orbital symmetry.
We have to access the anti-bonding orbital on the carbon with the leaving group, and the anti-bonding orbital is 180 degrees from the bonding orbital. Now, I'm going to go ahead and talk about this right now. I don't want you to learn this in too much of a great detail. It's not needed for your majors, but this is going to justify why E2, we have to have this anti-periplanar effect. So if we're to look at that carbon that is sp3 hybridized, and like we saw with substitution, all of this happens on sp3 hybridized carbons, and I'm going to write leaving group right here and just make it a big giant s orbital.
There's the electrons. If a base or something needs to come in and attack that carbon to push the leaving group out, we have to do it from the back side of the molecule, and that's where this anti-bonding orbital is present. So the electrons can be donated into this orbital and allow these electrons to go with the leaving groups to leave the molecule. Let's add in, oops, let's add in our hydrogens now.
Here's the hydrogens. Now the only hydrogen that allows access to this anti-bonding orbital is the hydrogen on the opposite side of the leaving group. This one, too far away, this one's on the same side. So in base comes in to pull this proton, these electrons directly access the anti-bonding orbital, allowing these electrons to go to the leaving group to form our carbon-carbon double bond and the leaving group now exiting the compound. So this is why an E2 must have the anti-periplanar effect.
Now, We talked about a whole bunch of different types of chemistry here with E1 versus E2, and let's see if I can shrink this down for a moment. There it is. So now, we saw here and here that we can have both E1 and E2. Luckily, on primaries, we're only going to have E2 chemistry.
How can we favor one versus the other? We talked about the strength of the base again. Now, if you notice, what I'm doing here is trying to recap this and give us this nice big overview. On the E1 side... we want to use weak bases to help promote this.
And for E2, we want to use strong bases to promote this. Now, we are only using oxygen-based bases. in this class.
We're not going to be looking at other types unless, if I do, I will implicitly tell you it's a base. It'll be up to you to determine strong or weak. What makes something a strong base?
Something that is very unstable. High negative charge with no stabilization. What makes it a weak base?
Stabilized. More than likely a neutral molecule or a negative charge that has high stabilization. Let's look at an example of one.
So here, strong bases, I've been saying oxygen with a negative charge. So does that mean any oxygen with a negative charge is a strong base? No.
This here is a weak base. Why is this a weak base? Because of resonance.
And because we can resonate that lone pair of electrons that we've just put on the oxygen, it stabilizes it and makes it a weak base. So please do not forget to consider stabilization of charge. And this is our basic elimination chemistry.
Now, there's other things that help favor this, such as solvents and temperatures. We're not going to cover that whatsoever. The next thing I want to talk about is if we have multiple protons that we can pull. Let me open up another slide to talk about that. So, if we were to look at this molecule right here, and I'm going to go ahead and treat it with methanol, a weak base.
That means we can pull this proton right here, or here. I'll show these two protons. And let me show all the protons on each one of these carbons right here. A, B, D, E, and F.
So, if we have methanol, we can pull one of these protons here, dump the electrons down, kick out the chlorine, and put a double bond here. Or we can pull HE or HF, dump the electrons down after that chlorine leaves. So in an E1 mechanism, the chlorine leaves first.
I kind of misspoke there for a second when I said the hydrogen is pulled, and then we kick out the leaving group. That's an E2. In an E1, we pull A, B, or D after the chlorine has left, and we can form these two molecules right here.
This one or this one. But there's another one that we can form, and it is the cis double bond. Wait a second. How on earth did that happen? Once we lose that...
chlorine, we form this carbocation. We still have free rotation around this bond, and this allows for a mixture to form of these two compounds, to which we can pull the adjacent protons on the carbon, or the adjacent carbon, and form our carbon-carbon double bond. We have a mixture of double bonds here. Which one's major, which one's minor? We'll get to in a second.
This is our E1. What if we were to do an E2 mechanism? In an E2 mechanism, we have to use a strong base, so there's my strong base right there.
And here, we can only pull protons that are on the opposite side of the leaving group. So when we look at A, B, and D, the only one that we can pull here is A. So here we pulled HA.
Now on E and F, which one can we pull? That's right, all we can do is pull F. And we get this one right here, pulled F. Oops, HF right there.
Now, this does not allow for the cis compound to form in this present orientation. In E2, we can only pull the protons on the opposite side of the leaving group. Hopefully you're noticing how important that is.
Because with an E2, we form these two compounds, that's it. E1, we're forming these multitude of compounds. Now out of the E1s, which one is the most stable or the most highly populated? If we have a terminal double bond, this is monosubstituted, which is less stable than disubstituted. And these two disubstituted right here is a trans versus cis.
What's more stable? Cis versus trans. And it's going to be trans to keep them farther apart from one another. And this is called Zatzkiew's rule.
The more substituted the double bond, the more stable it is. So what? tetrasubstituted olefin is more stable than a trisubstituted, a trisubstituted is more stable than a disubstituted, and disubstituted is more stable than monosubstituted.
Now out of the dis and the tris, we always want the biggest groups on the opposite side. So trans is more favored than cis, E is more favored than Z when we look at those types of olefin systems. So this will be our most major right here with these bases that we're using.
Over here, which ones are we going to form the greatest amount of? This top one or this bottom one? The bottom one because it is a disubstituted olefin.
And it's far more stable and therefore it's going to be our highest population. But now, what if I want... So let's do this.
I want this compound right here. That's the one I want. The only one way that we can really help favor this is by using big bulky bases. We want to make something so big and so bulky that it's going to get harder to get into that H, E, and F.
So if I was to use this chicken foot that we've seen previously and attach an oxygen to it, this is a big bulky base. If I put a negative charge on it, this is a strong non-nucleophilic base. If I put a hydrogen on it, this is a big non-nucleophilic weak base. The important thing here is that both of these are non-nucleophilic. Why on earth does that matter?
That's going to matter when we get into the next slide. But here, this is going to help us promote to pull these protons because of steric encumbrances. And that's going to help us favor pulling, making this compound or this one over here, which they are identical in this case.
So we'd want to use big bulky bases that are non-nucleophilic. All right, so now kind of getting towards the end of this video, what we want to talk about now is bringing substitution and elimination chemistry together. If I gave you This material right here, the same material that we just used on the previous slide. And I said, here is our starting material, or our reagent there, and over here, oops, here is the reagent.
For the reaction on the right-hand side, we are looking at a secondary carbon, which means SN1, SN2 is possible, or E1 and E2. Oh great, everything's possible. Look at the reagent over the arrow.
This is a sulfur. It is not an oxygen-based base, which means we can get rid of elimination chemistry in this class. Not in true all chemistry, but in this class, no elimination unless we have an oxygen-based base.
So, SN1 or SN2 here. What's going to be our answer? This is a weak nucleophile, and therefore, we are going to go through an SN1 reaction. and form our mixture of enantiomers because we form a carbocation and the sulfur attacks from the top or bottom.
Wonderful. Now, let's go ahead and look at the other one. We have a strong base. This strong base is still on a secondary carbon, so I'm going to say SN1, SN2, E1, E2.
It's a strong base, which eliminates SE1. Now, it's also a strong nucleophile, which eliminates SN1, and we have SN2 and E2 being possible. Uh-oh, both are possible, which means we're going to get both products. We are going to see the formation of this double bond here, this double bond right here, as well as the substitution to form this single compound right here. All of them form.
When we have something that can act as a nucleophile and a base, we see both substitution and elimination chemistry. We're combining all of this together. Now, which is major, which is minor.
That's a little tricky. Here, just at room temperature, they're all going to roughly be equal to one another. Technically, the elimination is going to be slightly favored because it's acid-based chemistry.
But now, if I want to favor elimination chemistry, we're going to increase the temperature. If you want to favor substitution chemistry, you cool the reaction. So temperatures that are greater than 40 degrees Celsius favor our elimination chemistry, and temperatures below 0 degrees Celsius will favor substitution chemistry. So here, no temperature is listed, so we're saying we form all of these compounds right here. If we wanted to favor substitution, I'd write 0 degrees or cooler.
If I wanted to favor elimination chemistry, I'd show that temperature at 75 degrees or 100 degrees Celsius to make this reaction go. And that is how temperature affects these reactions. There's other conditions that affect these reactions.
We're only looking at temperature. That's it. Now, let's take that same material and practice this one more time. Now, I'm going to go ahead and give you this reaction right here.
What happens? We're on a secondary carbon. SN1, SN2, E1, E2 are all possible. Oops, SN2, E1, E2. We look at the reagent.
It is a weak base. Oh, so E1 over E2. It's also a...
weak nucleophile, SN1 over E1. What happens? We form a carbocation intermediate in this reaction, and then we can pull protons to form this compound here, this compound here, or this terminal olefin, or we can attack that carbocation in an equal amount and form this mixture of products right there.
So in an E1, SN1 type of reaction, we're going to form everything just like we did up above. How can we favor the elimination over the substitution? You got it.
We can go right back to our temperature right here to favor substitution over elimination chemistry or vice versa. So with that, this comes to the end of this video right here. We've covered the basics of elimination chemistry that we're covering in this class.
The key thing I want to stress one more time is we're looking at oxygen-based bases. That's it in this class. Nothing else. If I give you some other type of base, I will implicitly tell you it's a base.
It'll be up to you to determine strong or weak. I am not saying that a sulfur with a negative charge is not basic. It is.
But for this class, you do not need to worry about that. If you go on to other classes where you have to learn more about basicity, please keep that in mind. But for this class and for your majors, we're just going to keep this to oxygen and oxygen alone.
I hope each of you are doing well. If you have any questions over this, please feel free to email me or come to our discussion section. I would recommend go back and watch this video again and make sure you're taking notes over this. Go back and watch your substitution video again as well because now, on this slide, we just saw how substitution and elimination chemistry work hand in hand.
They're always occurring together in a lot of cases. You have to figure out which one is the predominant pathways. I hope each of you are doing well, and I look forward to seeing you all soon.