Lecture 3: Stereochemistry of Elimination Reactions
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00:00
Elimination reactionFunctional groupIce frontBase (chemistry)Optische AktivitätHydrogenDoppelbindungTiermodellAlkeneWursthülleCarbon (fiber)MethylgruppePhenyl groupChemical reactionAlpha particleChemical structureWine tasting descriptorsHardnessHydrogen bondBeta sheetHalideCarcinoma in situChlorineChemical clockComplication (medicine)StereochemistryAlkaneHaloalkaneSubstrat <Chemie>Lone pairÜbergangszustandConformational isomerismChemistryMixtureSubstitutionsreaktionPhase (waves)StuffingAtomic numberCyclohexanAttachment theoryHalogenationAtomTablet (pharmacy)BromideElectronProtonationKupplungsreaktionChlorideColourantSet (abstract data type)Hope, ArkansasEliminierungsreaktion <beta->Stream bedJoint (geology)SeawaterPedosphäreReaction mechanismFermentation starterTrauma (medicine)Human body temperatureBiomolecular structureDyeingOctane ratingCooperativityIslandPeriodateSpring (hydrology)Separation processSandMetalMethylphenidateSunscreenKlinisches ExperimentBronzeMoleculeSong of SongsLightningWeinfehlerOxideProcess (computing)IonenbindungCancerWattNecking (engineering)Alkoholfreies GetränkIsomerMedical historyAreaWasserwelle <Haarbehandlung>AdamantaneClaus processOperonWalkingStiffnessLeakPrimer (film)Grading (tumors)Weizenbier-Brauerei J. BayerSetzen <Verfahrenstechnik>Computer animation
Transcript: English(auto-generated)
00:08
In some elimination reactions we're talking about most of the time the most stable product predominates. In some elimination reactions the less stable alkene is the major product.
00:22
And when that happens is when we have a very hindered base and a very hindered alkyl halide. Now your book will tell you that the alkyl halide has to be tertiary and the alkoxide, the base has to be very hindered also but I think in the lab you're going to
00:41
be doing an elimination experiment. My students did it last week and I think what students found is that even with a secondary alkyl halide and a hindered base the major product is the least substituted to do. You looked at it already? So yeah so it's pretty much secondary and tertiary alkyl halides and a very hindered
01:01
base. So tert butoxide qualifies. You get not the most substituted but the least substituted and it just has to do with the fact that the base has a hard time coming in, let me show you, I'm prepared with everything except a voice but the base, this very hindered base has to come
01:23
in if it's going to make the most substituted alkene it has to remove one of these two beta hydrogens and that's hard for it to do because it's so hindered. So it's much easier for it to grab one of these hydrogens that are on the outside and it turns out that that's the major product you get for this very hindered base
01:41
removing one of these and instead of that one in the middle it's harder to reach. So let's draw the product here.
02:05
Typically do get two products but the major one should be, oh I don't want that, alright
02:33
so that is trisubstituted product and then we also get disubstituted product so that
02:43
would look like this. So some of you are still at the phase in this chapter where you're even having trouble seeing what the product is going to be or being able to predict it and so you want to move past that point rather quickly because this other stuff is much more complicated.
03:01
So CH3 and this would be ethyl. So this is the most substituted, it's trisubstituted and this is disubstituted. So we're looking for R groups here.
03:21
Some of you have a lot of trouble seeing this, don't feel bad if you do because a lot of people feel the same way. I think it's easiest if you draw the hydrogens in since it's easier to see R groups when you have the hydrogens actually in. So this would be a geminal disubstituted product and this would be a trisubstituted product and it turns out that we get 28% and 72% and it's because the base removes
03:53
the most accessible hydrogen when the base and the alkyl halide are very hindered.
04:31
So like we said I think as you'll see when you get into lab that this Smith will tell you that it's only tertiary but even with secondary the major product is the least
04:42
substituted alkane. Alright so compare that with this one here where we take the same substrate so we still
05:07
have our tertiary alkyl halide which is very hindered and but now we're going
05:25
to use an unhindered base so ethoxide is a good unhindered base that does really good in the E2 reaction and now we're going to get the same two products but this time
05:42
the most substituted is going to be the major. So that's the same first product we got above trisubstituted and then the second
06:01
product is the same one we got above and we do get both but the major one is going to be the least substituted so this would be disubstituted.
06:23
So let's see the actual numbers we get here, 69% and 31%. So they don't use the terminology in the book, oh look it was all, I didn't even have
06:44
to draw those it's on the next page, I'm going to go fix, I'm going to fix the next chapter so that it doesn't scroll weird like this. So you guys can just add that in yourself. There is a name, your book doesn't use the name but there is a name for both of these.
07:01
They use Zytcef, so this is Zytcef, the least substituted is called the Hoffman product. So the Zytcef is the most substituted, the Hoffman is the least.
07:22
So most substituted and Hoffman is the least substituted. Alright questions, that's it, all we did was change the base, yes.
07:47
Alright we got one more complication with the E2 and that is the stereochemistry.
08:03
For an E2 elimination to occur the beta hydrogen and the leaving group must be in the same plane, okay. Beta hydrogen leaving group must be in the same plane.
08:21
The best is antiperiplanar and this is the only one you will see in this class. Only one you will see in this class.
08:49
So there's two possibilities, there's antiperiplanar which is the most common. In very rare circumstances you can get synperiplanar and that's with rings and things like that that cannot do antiperiplanar.
09:04
We will never see it in this class, if you are a chemistry major you will see examples of this but if you're not a chemistry major this is all you're going to see so I don't want you to show synperiplanar. Okay so I'm going to put a big X through this, you won't see in this class.
09:30
Alright so antiperiplanar, beta hydrogen and the leaving group, I'll just
09:40
abbreviate LG, are in the same plane and anti to each other and that's
10:10
where we're going to see synperiplanar is higher energy only under special
10:23
circumstances, not in this class.
10:48
Alright so let's do an example here. So what we see is here and what's going to be really helpful in this chapter and hopefully you remember how to do this, is to take three
11:04
dimensional structures and put them into Newman projections. That will help you greatly in this class and that's something that I want you to know how to do. So if you thought you were done with that you're wrong. Do need to know how to do that. So what we have here is this is the beta hydrogen, here's the leaving
11:24
group, so we're going to basically put this into a Newman projection. Here's the leaving group, the beta hydrogen to be removed is on the adjacent carbon, it's not on the same carbon, it's on the adjacent carbon
11:42
so here's the beta hydrogen to be removed.
12:01
When it's drawn nicely like this with the beta hydrogen and the leaving group in the same plane and anti to each other, you don't have to put it into a Newman projection. So you can see here, this is our beta hydrogen and here's our leaving group. So easy to pick out the leaving group, not necessarily easy to pick out the
12:24
beta hydrogen, the carbon bonded to the leaving group is alpha and right next door is the beta. We also have another beta don't we, we have this one here, but we have an unhindered base so I'm not even going to worry about that one because
12:43
I'm going to draw you the major product here. All right, so what's going to happen is I'm going to show it both ways, ethoxide comes in, ethoxide is going to grab the beta hydrogen,
13:07
we're going to break the carbon hydrogen bond and then the leaving group leaves. Since the leaving group is leaving, I want to see lone pairs on it.
13:24
Yeah, so that's going to look like this. Let me just draw it this way, it's a little strange way to draw it, but it works nicely. I'm going to keep the methyl going back, I'm going to keep the phenyl coming forward and the methyl on the right hand side is
13:44
going back and this hydrogen is coming forward. Now we usually don't draw alkenes that way, we usually draw them flat on the page, but in this case I'm just drawing it tilted so
14:03
it would be perpendicular to your page and that's okay to do that. So again, rather than flat on your page, it's just tilted like this, so that's what it would look like. So that's from the dash wedge. If we do it the other way here, we're going to lose that, so this
14:27
is after loss of HBr.
14:47
All right, so I'm just going to remove the HBr, I'm going to keep the Newman projection here, but I just want to make a point here. So this methyl here, still here, this phenyl is here.
15:02
On the back, here's our methyl here, I haven't moved anything. On the back, this is the hydrogen. Well that's a little strange to see it that way, but remember the carbon-carbon bond, here's the front carbon right here, and this round thing is the back carbon, so that still stays,
15:21
this is the front carbon, and this round thing is the back carbon. And so as you can see, our alkene is like this. So in order to figure out what that looks like, we're just going to connect the two lines.
15:41
A little hard to picture, but can you just see that it's going like that. I'm going to redraw it without the circle because it looks a little strange with the circle.
16:07
So notice, if we draw the double bond in here, right here, the methyls are on the same side of the double bond, so they're on the same side here. I've just kind of rotated it this way. These two structures are equivalent.
16:28
It's just that one is in the plane of your page and one is perpendicular, but you can draw that either way on an exam. Questions on how I did that? Anybody?
16:41
Yes. Well you don't want it to, the circle's a little strange. I just left it there so you could see where things are coming from. You don't want to have the circle there. So I would write it like, let's circle what you can do here. So you can write it like this,
17:02
or you can write it like this. Don't want to draw it with a circle because that's a little strange. Like what does that mean, right? Synel elimination of the same alkyl halide does not occur, why? Let's draw it on the next page.
17:22
Of course you guys have this at the bottom of that same page. We're going to fix this all in chapter nine. I know what this program's about so I know how to fix it. So all groups are eclipsed here.
17:44
So what I've done in going from here to here, can I get them on the same page at the same time? No. Too bad. Let's see what I've rotated. I kept the front the same and rotated the back.
18:02
Methyl, phenyl, hydrogen. The front is stayed the same and the back is rotated. That would give you synel elimination. So back is rotated. So number one, the reason we don't get synel elimination
18:22
is because in order to get synel elimination, all groups have to be eclipsed and that's disfavorable. That's going to raise the energy of the translation state. And the other thing is that if we have the base here coming in on the same side,
18:46
if we draw all our lone pairs in, this is what a synel elimination would look like. So not only are all the groups eclipsed,
19:02
but we've got this nucleophile coming in, this nucleophile is surrounded by, it's got three sets of lone pairs coming in in close vicinity to the leaving group which has lone pairs. That's not going to favor the reaction. So that's why we don't get syn. And so let me draw the product that we don't get.
19:25
And again, depending on how you do it, you can draw it like this. Methyl still going back, phenyl still coming forward. Hydrogen going back. Methyl coming forward.
19:52
That's equivalent to flat. We can draw it flat also.
20:08
And it turns out that you don't get any of that. So I'm going to put a big X through all that thing I spent time drawing. Is it going to let me do that?
20:21
Let's try that again. Big X. So two reasons why this doesn't work. Transition state has eclipsed conformation.
20:53
Those are the things we want to look out for. And right here, as you can see, that the base and the leaving group are on the same side.
21:12
So it's also sterically disfavored.
21:31
You think about the lone pairs on the base and the lone pairs on the bromide. Those are going to repel each other as soon as that base gets close to that bromide. So that's the second reason why that's disfavored.
21:43
All right, so that's on the next page. Talks about why we don't get that. Everything we just mentioned.
22:06
Antiperiplanar has a staggered conformation. This one has an eclipse. That's not going to favor it. The base and the leaving group are on the same side of the molecule, which is sterically disfavored. In the anti-elimination, they're on opposite sides. Okay, we want those guys as far apart from each other as possible.
22:25
All right, questions so far? Yes, I'm not going to ask you that. I don't want you to draw it. So you're only going to get antiperiplanar in this class.
22:40
So I don't, you know, I have students who, I guess it's possible to have an NR, no reaction for one of my predicting products, but historically, I don't do many of them. I don't usually have reactions that are no reaction. I usually have an answer, okay? So I have had students before that were not doing well in the class write NR, NR, NR
23:01
for every reaction for predicting products. That's not a good idea in my class because they don't often have reactions that are no NR. I just won't give you one that has to have an eclipse conformation. All right. Okay, so let's see what happens when we're in a six-member ring. Now it's going to look more complicated here. This is mentyl chloride.
23:22
And first we want to see a couple of different things. First, let's find the beta hydrogens. First things first, let's find the beta hydrogens. So we zero in on the leaving group. So this is alpha. The right next door is beta.
23:56
So now, let's look here. This is the most stable conformation
24:13
where the halogen and the isopropyl group are equatorial, right? All the groups are equatorial.
24:21
So this is the most stable conformation. But the beta hydrogen and the leaving group are not antiperiplanar.
25:02
In a cyclohexane ring, the beta hydrogen and the leaving group can assume an antiperiplanar conformation only when they are both axial. So the chlorine's not axial. We have some beta hydrogens that are axial, but the chlorine also has to be axial. So in a six-membered ring,
25:21
antiperiplanar equals trans-diaxial. In a six-membered ring, antiperiplanar equals trans-diaxial.
25:47
So both the leaving group and the beta hydrogen have to be axial, and they have to be trans to each other. If they are not trans to each other, you will never get a reaction in a million years.
26:01
But that's okay, because this ring is flipping rapidly back and forth, right? So we can actually flip this ring and put it in the conformation where the chlorine is in the axial position. You know how to do that, right? In chapter six, you thought you were done with that. I mean, in chapter five, you thought you were done with that.
26:20
You're not done with that. You still have to know how to do that. So we're gonna flip this ring. So now remember, we're looking for the headrest. Here's the, this is all going up here, headrest. That's the footrest. So when we flip that, this carbon right here
26:44
is going to be down, and this carbon right here is gonna be up. So need to know how to do that. So that means that this isopropyl group, which is down, is now gonna be axial.
27:04
It's looking a little weird. Let me fix that. Let's try that again. So that isopropyl is down. This hydrogen is up.
27:21
Remember, we travel around the ring. We're gonna go clockwise. So the next carbon here is chlorine. Chlorine is going to be up. Now normally we want to break the back of that ring, but I can't because it's already drawn there. But the chlorine's gonna be up. If you draw that on your test, you want to break, oh, why'd it do that?
27:41
You want to break the back of that ring. Okay, and then we follow it around. There's chlorine. Next carbon has two hydrogens, so we're skipping that one, and then we have this one here. This methyl is up. All right, so that's not the most stable conformation,
28:03
but it still is flipping back and forth, and that's the conformation we need to be able to eliminate a beta hydrogen. So let me draw in these hydrogens here.
28:21
So can you see that there's really only one hydrogen that we can remove? They have to be trans-diaxial. So this hydrogen right here is not, is cis to the chlorine. So we cannot in a million years eliminate that hydrogen.
28:40
We can, however, eliminate this hydrogen. If we eliminate this hydrogen, that will give us the most stable alkene. But we can't eliminate that hydrogen because that would be syn. We're not gonna see syn. So we're gonna eliminate this hydrogen. So in order to get syn elimination, anti-elimination can't be possible.
29:03
Okay, so this is not, that's not gonna happen, so this is the only hydrogen that we can eliminate. And we know one thing is that these interconvert freely at room temperature, so the fact that we can't eliminate from that conformation is not a problem.
29:23
So no problem. Chair flip is rapid. Even at room temperature, it's rapid. All right, so, so here's our alpha,
29:44
here's our beta, here's our beta. Can't eliminate this hydrogen
30:02
because it's not trans. It is cis, and you can't eliminate the cis hydrogen. This is the only hydrogen that you can eliminate.
30:33
All right, so I'm gonna eliminate that hydrogen. Let's use blue, let's see.
30:42
It doesn't, me too. What's going on? What did I just do? Okay, we'll use blue. Okay, so ethoxide comes in. Boom, removes that hydrogen.
31:04
This goes here. Leaving group leaves. I'm gonna draw it the wrong way and then I'm gonna draw it the right way. So let's draw the product the wrong way.
31:25
And then we're gonna put an X through it. We're gonna spend all this nice time drawing and then we're gonna put an X through it. Okay, I'm gonna draw a drawing with the mouse, guys. I know, right? Well, that's terrible.
31:41
I'm so embarrassed here. Look at that chair. I've never drawn such a bad chair. We're sure taking a long time to make this point, aren't we? Okay. What's wrong with that drawing? You can't see it?
32:02
Besides that it looks like terrible? What is wrong with that drawing? Can you have a trans double bond in a six-membered ring? You can't. It's geometrically impossible. Try to do it at home. You can't be able to do it with those models that you have
32:22
that are in the green box. My models are a little bit more flexible. I'll bring one next time so you can see it or you can come to my office next week to see it. It is absolutely geometrically impossible to have a trans double bond in a six-membered ring. So really the best way to draw this is actually flat and I will let you draw it flat because as soon as we put a double bond in that six-membered ring
32:43
it's not a chair anymore anyway, is it? It's not a chair anymore. Okay, so. So we want to put a big- can I change colors? Ah, look at that. So we're going to put an X through this.
33:07
And what we want to do is we want to draw the right way. Let's draw it flat.
33:22
So I'm going to just draw a flat cyclohexane ring. Oh my gosh. Some people do draw like this on exams, believe me.
33:44
Okay, so there's an isopropyl here that's down. Does everybody see that? This is so funny. That's down. What about this group over here? That's a methyl, believe it or not.
34:01
That's up, isn't it? So that's the best way to draw this. Okay? That's the right way to do it. So, um, this is correct. So on your page, right,
34:20
cannot have a trans double bond in a six-membered ring. I'm not going to draw it on here because it's going to look like a kindergarten wrote it. This is the correct way to draw it.
34:43
Questions, anybody? Should we keep going on this? This is a lot of drawing in this drawing right here. So even though conformation B is less stable, elimination occurs through the conformation because the beta hydrogen and the leaving group are trans-diaxial, yes. Just turn the tablet off and turn it back on again.
35:01
How about that? Okay. Alright, with the neonental chloride, there are two beta hydrogens that can be eliminated. This is neonental chloride. Let's look at it. First of all, let's find our hydrogens.
35:22
Alright, so alpha is the carbon bonded to the leaving group. We have two beta hydrogens, and a beautiful thing is happening here. Both of those hydrogens are anti and in the same plane. They're trans-diaxial, which is the same exact thing in a six-membered ring.
35:41
Antiperiplanar is trans-diaxial. So we've got two possible protons that can be eliminated. So let's do one of them. I'm going to do one of them in red, like that. We're going to eliminate this hydrogen here.
36:01
The electrons are going to come down, kick off the leaving group. Alright, that's one possible product. Let's draw that product. We'll call that pathway A.
36:21
Oh come on, he's doing it again, he's doing it again. Ah, jeez Louise. We'll just won't change colors. Okay, that's A. Let's draw the product. We're going to draw it flat. We're going to save ourselves some grief here and draw these things flat.
36:46
So this methyl is up. Here's our double bond right here. Don't draw this isopropyl group down. Why don't I want to draw this isopropyl group down?
37:01
It's down here. So see how it's down right here? Why don't I want to draw that with a wedge? Hmm? It's trigonal plane or carbon, so that means this carbon, that carbon, that carbon, that carbon, and that carbon are all in the same plane. So there's a sapling problem like that,
37:21
that asks you to pick all of the atoms that are in the same plane. And so we have a double bond here so that means this carbon's in the same plane, that one, and the first atom attached. So those carbons I'm going to put, I'm going to put an X, let me see if it'll let me. This carbon right here, this carbon,
37:43
that carbon, that carbon, and this hydrogen. So the two atoms of the double bond and the first atom attached are all in the same plane so you don't want to draw that wedged.
38:02
So that's one product. Let's do B, see if it's going to let me do it. I'm risking things here. So we'll call this B ethoxide.
38:28
Now we're going to remove the other beta hydrogen. I'll just put this arrow on the other side here.
38:47
Okay, let's draw product B.
39:08
Okay, we're back to the mouse. Now we're going to have to give them a little call
39:20
about this tablet here. That is a double bond on the bottom there. That's a methyl. This is an isopropyl, let's draw it this way.
39:43
Alright, now which one is going to be the major product? We've got two possible products, which one's going to be the major? Top one or the bottom one? Top one is more substituted. So what do we actually get?
40:13
78% here.
40:23
78% and 22% here. So this top one is favored because it's the most substituted.
40:41
I'll let you write that and not me write that. Favored because it's the most substituted. We still have two more minutes of this. I'm going to turn this off and turn it on again. Do we just have to keep doing that? I don't know.
41:08
Alright. Okay. Okay, so definitely let's write this for posterity here.
41:24
Most stable. Favored because more substituted. Alright, we've got one more minute.
41:41
When a mixture of stereoisomers is possible from the E2 reaction, the reactant has two hydrogens bonded to the same beta carbon, both the cis and trans isomers, will be formed because there are two conformers in which the groups to be eliminated are anti. The alkene with the bulkiest groups
42:01
on the opposite sides of the double bond will be formed in greater yield because of the more stable alkene. Alright, so that's the last part of this. We've got two possible products here. I still, it still hasn't, the clock has not turned yet, so we have this and trans.
42:32
Well, is it cis or is it trans? Hard to say because we have more than two groups attached. Which one of these has the bulkiest groups on opposite sides?
42:49
First one or the second one? First one and that's the major. 41% and this is 14%. And then very minor is the least substituted.
43:03
I'll draw that real quick and then we'll end here. This one is minor. Alright, that's a perfect stopping point. We'll stop right there and we'll continue this next time.
43:22
Hope you have a great weekend.
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