Organic Chemistry Elimination Reactions

Welcome back to Educator.0000

Next we are going to talk about a class of reactions called elimination reactions.0002

The first reaction we will study is called an E2 reaction for an E2 mechanism; and the name of that reaction comes from the fact that it is an elimination reaction; and it is bimolecular.0008

It is a one-step mechanism very similar to the Sn2; the other mechanism we saw that was a single step was also bimolecular.0020

That has the same 2 that the E2, as the Sn2 backside attack substitution reaction.0031

Something that the E2 mechanism needs is a strong base; so what are things that we know of as strong bases?0038

Things like hydroxide or alkoxide; maybe an N^- would also be a very strong base; so pretty short list of species that we would describe as strong bases.0044

Those are the kinds of reagents we will need to do an E2 elimination; and here is an example of one.0059

Let's react... we will start with an alkyl halide; and we will react it with sodium hydroxide; that is a nice strong base.0065

What do bases do?--bases go after protons; the proton we are going to attack is located right here.0072

It is not on the same carbon as the leaving group; remember the halide is going to behave as a leaving group just like it did in the substitution reaction.0080

The hydrogen that gets attacked is not on the same carbon as the leaving group; it is the next carbon over.0089

We call this not the α position but the β; this is called a β hydrogen.0095

The hydroxide or whatever strong base we are using is going to come in; and in a single step, it is going to grab that proton, form a π bond, and kick off the leaving group.0101

The product we are going to get will be an alkene; we are going to get an alkene product; we are going to be forming a carbon-carbon double bond as a result of an elimination.0115

The other products that are being formed... if I used hydroxide, I am also forming water, the conjugate acid of hydroxide; and of course I always lose my leaving group, so I also form bromide as well.0131

This is described as an elimination reaction because you have eliminated both the β hydrogen and the leaving group.0144

We lost HBr so sometimes this reaction is described as a dehydrohalogenation; because you have lost the hydrogen and the halogen.0152

What do you think an energy versus POR diagram would look like?--well, it would be a lot like our Sn2 mechanism because that was also a single step reaction.0162

Our energy versus progress of reaction, we would start at some combined energies for our starting materials; the hydroxide, the alkyl halide.0174

We are going to end at some final product energy; assuming it is a favorable reaction, that it is going to be an exothermic reaction.0184

In order to get from the starting material to the product, as usual for our one-step reaction, we are going to go through just a single transition state to go here.0192

So really, the energy diagram looks very similar to that for the Sn2.0202

What does the transition state look like?--the transition state has a lot of bonding changes taking place all in a single step.0206

First of all, we are forming a new bond between the hydroxide and the hydrogen; so forming bonds, we show as partial bonds in the transition state.0215

We are also breaking this carbon-hydrogen bond; so breaking bonds, we are going to draw as a partial bond in the transition state; let me draw the rest of my carbon chain here.0227

We are also forming a π bond between these two carbons so we will show a partial double bond being formed between those two.0239

Finally our leaving group is leaving so we are breaking the carbon-bromine bond; and we will draw that as a partial bond as well; so four bonds involved in this mechanism, being formed or broken.0248

How about any partial charges that we need to account for in our transition state?0260

In our starting materials, our oxygen is negatively charged, but it is neutral in the product so that charge is disappearing; we have a partial minus on the oxygen.0267

Our bromine starts out neutral but ends up as bromide; so we have a negative charge developing on the bromide; so that is what our transition state looks like for the E2 elimination mechanism.0277

What about the stereochemistry for the E2?--if we are dealing with a chiral molecule and we have some asymmetric centers, what is the ramification of that stereochemistry when we do an E2?0292

It turns out that there is a special relationship we have to have between the β hydrogen and the leaving group that got eliminated; they need to be anti-coplanar.0305

That is the relationship we described when we looked at Newman projections; we said two groups were anti if one was straight up and one was straight down; anytime they have 180 degree dihedral angle.0314

That is the required relationship of the leaving group and the β hydrogen; and for that reason, the E2 is described as anti-elimination.0326

Let's see an example of this; here we have... I had to hand draw some of this... we have an alkyl halide--here is our leaving group that is going to be eliminated.0336

We are not just eliminating a leaving group; we also have to eliminate a hydrogen on the next carbon over.0349

We go this carbon and we see there are no hydrogens; so we go to the next carbon, and here we find there is only one possible β hydrogen.0355

But when we look at this conformation as drawn, the H and the Br are not anti to one another; they are both dashed bonds, dashed lines.0365

If we were to look at this from the side, we see that they would be gauche to one another; they are close to one another.0377

But that is okay; this is a molecule that can be rotated; we could rotate around that carbon-carbon single bond.0382

That is what we will do; we are going to rotate first and rearrange it such that the hydrogen and the bromine are 180 degrees.0391

One way to draw that very easily is to show the hydrogen pointing straight up, the bromine pointing straight down, both in the plane; that must mean they are anti-coplanar.0400

Now the next two drawings we see, the H and Br are anti; and that is the necessary conformation; we have to be able to do this before we can do the E2 elimination.0411

But if we are going to rotate this, that is going to move these other groups; where are they going to go?0429

Let's start with, let's look at this bromine; and I built a model of this; so this bromine right now is pointing away from us; it is a dashed line.0436

What we need to do is we need to rotate it into the plane; then we need to bring it forward.0449

What does that do to this phenyl group?--it was in the plane, but when we rotate it, it is like we grab onto this group and turn all three groups clockwise.0455

That pushes the phenyl group backwards; and so on this carbon now, the phenyl is going be a dashed bond.0463

This hydrogen was a wedge; it is still a wedge; it was pointing down a little bit; and now it is pointing up a little bit; but it still pointing towards you.0472

Let's see if we can do this second carbon; we want the hydrogen to be pointing straight up and in the plane; so what did we have to do?0481

We had to grab onto that hydrogen and bring that forward as well; so we grab onto that molecule and twist that carbon.0490

What is going to happen to this ethyl group?--is it now going to be a wedged bond or a dashed bond?--well let's take a look.0498

Here is our drawing; our ethyl group is in the plane; our methyl group is a wedge; hydrogen is a dash.0507

We want to bring this forward so that it is in the plane; and when doing so, that is going to push our ethyl group back.0515

So the ethyl group will also be pointing backwards, CH₃CH₂; and our methyl group is still a wedge.0524

So wWe are going to need to manipulate our molecule a little bit to get the leaving group and the β hydrogen anti to one another.0533

Another perspective of this drawing is if we stand here and look at it this way.0541

On this front carbon, we see the hydrogen is pointing straight up; on the back carbon, you see the bromine pointing straight down.0548

This is that relationship of anti that we described back when we first learned about these Newman projections.0554

Once it is in the proper conformation, now my base, whatever base I am using, can come in and do the elimination.0562

One, two, three arrows; grab the hydrogen, form the π bond, kick out the leaving group.0568

What happens is because it is a concerted mechanism, all the other groups that are on the carbon chain end up getting frozen in place the relative positions they were in this anti conformation.0574

Notice that the phenyl and the ethyl are on the same side here; after I do the elimination, the phenyl and the ethyl are required to still be on the same side.0587

I could see it in this drawing too here; the ethyl is a dash, the phenyl is a dash; they are both behind the page.0598

So when we flatten it out... remember when we do the elimination, we are going to a planar product.0605

Everything is going to be pushed up or down a little bit; but the phenyl and the ethyl are still going to end up being on the same side.0611

This going to be the only stereochemistry that is observed; in other words, we are not getting this possibility where the ethyl and the hydrogen are on the same side.0618

This product would arrive from a different conformation, one that does not have the hydrogen and the bromine anti originally.0632

Let's try another example; here we have two phenyl groups; let's identify our leaving group; that is always easy to find; it is going to be a halide, maybe a tosylate, but a halide would do great.0639

Where do we have our β hydrogen?--this is another example where there is only one possible β hydrogen right here.0654

Are they anti right now?--they are not anti right now, they are both dashes again; so let's rotate this.0663

The easiest way to rotate this compound would be to bring that bromine slightly forward, bring that hydrogen slightly forward, so that they are both in the plane with one up and one down.0670

Let's fill in our dash and wedge that we know are here and see where those phenyl groups go.0683

What do you think?--this bromine, when you bring that bromine forward, what does that do to the phenyl group?0690

If you bring the bromine forward, that is going to push the phenyl group back; and when you bring this hydrogen forward, that is also going to push this phenyl group back.0697

This is the proper conformation that we have where they are now anti to one another; so what we did here was we just rotated so that they are anti.0710

Now we could have our base come in and do our elimination; three arrows--base attacks the hydrogen, forms a π bond, kicks out the leaving group0722

Since it is just a one-step mechanism, it is a great idea to go ahead and draw the arrows so you have a better chance of drawing the product accurately.0734

We are going to get an alkene product which is planar; and what is the relationship of those two phenyl groups going to be?0741

Because they are both behind the board, one is a little up and one is a little down, they are both going to get flattened out, but they are both still going to be behind the board.0750

So whether you draw them on the top or the bottom, it doesn't matter; but you need to show them on the same side of the double bond.0757

We have two phenyls on the same side; and what are the other two groups?--we have the methyl and a hydrogen; those were both wedges pointing out towards you; and they still will be.0763

Again it doesn't matter if you draw the phenyls up, or you flip it over and draw the arrows down, that is the same exact product.0773

Anti elimination is the way we describe the stereochemistry of the E2 mechanism.0778

What about the regiochemistry of the E2?--regiochemistry is when we ask about what region or site reacts.0786

If we have a choice of more than one site on a molecule to undergo a reaction, how do we make that decision?--and here is such an example.0799

Here we have an alkyl halide; we see our leaving group; we treat it with a strong base; let's say we want to do an E2 elimination.0807

Actually I think there is more than one possibility here so let's consider both possibilities.0816

If we wanted to do an E2, we have our leaving group is on this carbon, where do we have β hydrogens?0823

Well, we have a hydrogen on this carbon; and on this carbon, we have two hydrogens; we have one that is a wedge and one that is a dash.0829

The hydrogen on the far left, we will call that H_A; and the hydrogen that is a dash, we will call that H_B.0840

And let's call this D; I will refer to that hydrogen in just a moment.0848

What is another thing that can happen?--let's think about this reagent that we are using, hydroxide; what do you know about hydroxide's reactivity?0857

What can it do?--is it an acid or a base or an electrophile or a nucleophile?0866

It certainly can be a strong base; then that is what we are considering right now, doing the E2 elimination as a strong base and attacking a proton.0871

But isn't it also a strong nucleophile?--it definitely is a strong nucleophile; and what mechanism do we associate with strong nucleophiles?--backside attack, Sn2 mechanism.0879

So another thing that hydroxide can do is it can attack in this position; let's call that path C; and that would be an Sn2 path; that would be a backside attack path.0892

Let's look at what those three products would look like A, B, and C; A, if I took this hydrogen, would end up with a double bond between these first two carbons.0905

B, if I took the hydrogen from this middle carbon, would give me a double bond between these middle two carbons; now I went ahead and took that hydrogen and left the carbon chains where they are.0916

Because notice this H_B is a dash and the bromine is a wedge so they are already in the anti-conformation; so I don't have to do any rotations.0931

If I wanted to take hydrogen D and do an elimination there, what I would have to do is I would have to twist the molecule to get those two, the hydrogen and the bromine, anti.0941

That would give product D like this; D isn't even formed, but we will draw that up here so you can see what that looks like.0951

What does product C look like if you do the Sn2 mechanism?--backside attack; that means you replace your leaving group, substitute your leaving group for your nucleophile.0960

Tell me about the stereochemistry of that reaction; since the bromine is a wedge, the hydroxide must have come from behind; and you would get a dash here.0968

These are in fact the three major products; 16% of A is isolated; 75% of B and 9% of C; so Sn2, we get a very small amount here.0980

We are going to talk a little later about how we would make that prediction of the E2 versus the Sn2 mechanism.0995

But of these, these are both E2 eliminations, but there is a huge difference on which is the major product.1001

B is favored by far; why is that?--it is the major product because it is the more stable alkene.1012

This is known as Zaitsev's rule; and Zaitsev's rule states that when you have a choice between different alkenes that you can form, you are always going to want to form the more stable alkene.1019

Can you tell between B and D, why B would be favored over D?--why is that the major product?--let me fill these in; looks a little better.1032

Why is B better than D?--well, this is cis versus trans; and the cis is going to have some steric hindrance so that is definitely going to be less stable than the trans.1042

That explains why no D was formed even though it is a possible product.1055

But other than cis and trans, what are some other things we should know about alkene stability?1060

If we want to predict which is the major product, we have to understand which alkene is more stable; let's take a look at some factors that will affect alkene stability.1066

One thing is that if you are ever given conjugate, if it is ever possible to be conjugated, that is going to be a more stable alkene, a more stable diene.1078

If you have multiple π bonds, then it is better to have those double bonds alternating double bond, single bond, double bond, than isolated.1086

Here they have no relationship to each other; here they have one right after another; we have a p orbital here and a p orbital here, and right away we have more p orbitals.1095

This relationship is more stable because we could have resonance delocalization of these π electrons.1104

If you happen to already have a double bond in the structure, when you go to form a second double bond, if it is possible to put that new double bond in conjugation with the original one, that would be a good thing.1110

Another rule is that as you increase the number of alkyl substituents on the double bond, you increase stability.1123

What do I mean by alkyl substituents?--these are the number of carbon groups on the carbon-carbon double bond.1131

What is the maximum number of carbon groups that you can put on a carbon-carbon double bond?1143

Remember each of these carbons wants four bonds; so there is two bonds over here and two bonds over here; that would be a very stable carbon-carbon double bond.1148

Let's take a look at some examples; here is one such example of that; this alkene has four alkyl groups attached; we describe this as being tetra-substituted.1158

This is called a tetra-substituted alkene; that is the most alkyl groups you could possibly have; and this is the most stable alkene that there is; it really likes having those alkyl groups on there.1175

That is going to be more stable than an alkene with just three alkyl groups; we call those tri-substituted.1186

That is more stable than any di-substituted pattern; and look at all these possibilities; these are all di-substituted.1197

But within the di-substituted patterns, we want to know that the trans arrangement where the two alkyl groups are as far apart from each other as possible, that is going to be the best arrangement.1205

That is more stable than cis; cis is not so good because it has some sterics here; remember there is some steric hindrance in cis that makes it less stable, higher energy.1219

That is not such a great arrangement; and this one isn't so great either; this is called geminal or gem for short; having two groups on just one end of the double bond is not a very stable arrangement.1232

I usually don't care too much on trying to compare these two, but for sure you should know that neither of these is as good as the trans relationship; but typically cis is better than geminal.1246

All of these are better than having just a single alkyl group; we would describe that as a mono-substituted alkene.1257

Or you could also describe it as terminal; it is called a terminal alkene because it is at the terminus; it is at the end of a carbon chain if it has just one alkyl group on it.1269

That is not a very good arrangement for alkyl groups as well; this is the least stable.1279

It turns out that this is the same general trend in stability we saw for carbocations; carbocations like to have alkyl groups attached.1286

That one is a little easier to understand or maybe predict because it is an electron deficient species.1293

Why is it that alkyl groups also like to have... I'm sorry, why is it that alkenes also like to have alkyl groups attached?1300

That discussion of that stabilization is a bit beyond our scope here; so we simply need to know that the more alkyl groups we have, the more stable it is.1306

Let's take a look at a few examples and see if we can apply Zaitsev's rule; here is an example--draw both E2 products and select the major.1318

We have our leaving group here; how is it possible to get two different E2 products?--well, let's look for our β hydrogens.1328

There is no hydrogens on this carbon; it is the next carbon over; we could grab one of these hydrogens or one of these hydrogens.1339

Would those give two different products?--actually those would both give the exact same product; so β hydrogen type A is going to be right here or right here.1345

Where is the second type of β hydrogen?--how about going in this direction right here?--going to this methyl group; β hydrogen type B.1359

Product A would form a double bond within the ring; and like I said, whether you go to the right or left, it is the same product.1370

Tell me about this methyl group; is that still a dash?--is that still pointing behind the board?1378

Remember what an sp2 hybridized carbon looks like, an alkene carbon?--this is planar; and so that methyl group now after the elimination is completely flat; and it should be drawn as a straight line.1385

Now you can see that if you just flip it over, taking the hydrogen from either side would give the exact same product.1396

But what is the other possible product we can form?--instead of going to one of the side ones, we can go up to this top carbon.1403

What does he look like now?--he used to be a CH₃; it is now going to be a CH₂ because one of those hydrogens was taken away.1411

These are the two possible products we can get; and now which is the major product?1419

We are going to look to see which one is more stable; and for that, we are going to see which one is more highly substituted.1426

How would you describe the number of alkyl groups attached to this first carbon, to this first alkene?1431

Here is our alkene; we have one, two, three carbons attached to it; in other words, there is only one hydrogen.1438

This is a tri-substituted alkene; and how about this second one?--we have two hydrogens up here and two carbons.1448

We are not counting the carbons in the double bond; we are asking how many carbons are attached to those double bonds; this is a di-substituted alkene.1459

Which would be the major product?--the tri-substituted would be the major; why?--it is usually not such a great idea to just say because Professor Zaitsev told me so or just invoke the rule.1471

The reason it is the major product is because it is the more stable alkene; that is how we determine the regiochemistry of E2 elimination.1486

Let's take a look at another example--if our leaving group is attached to a six-membered ring.1502

Now it is the chair conformation of that compound that needs to be taken into consideration when we look at the stereochemistry and the regiochemistry.1510

Here again we have a strong base; we forgot to mention that up here just noting we were using hydroxide; now here we are using methoxide, also a strong base.1522

When we look at it in this conformation, if you think about where you are going to take your β hydrogen, we have hydrogens over here and we have hydrogens over here.1532

Which one is more tempting to take based on Zaitsev's rule?--for regiochemistry, where do you think you want to go?1542

I am guessing we might want to go to this side because that would give us a more highly substituted double bond; but there is a problem with that.1549

What did we learn about the stereochemistry of the E2?--let's draw in that hydrogen and see if you can find a problem with that hydrogen.1557

That hydrogen is a wedge; that bromine is a wedge; is it possible to rotate that to get them anti to one another?--it is not; so because this is not anti, I can't eliminate in that direction.1567

Instead I have to look over to this side; this is a CH₂; one is a wedge and one is a dash; this is the only anti β hydrogen; so there is only one possible product we can get.1580

When you look at the chair conformation, the most stable chair here has a bromine up; we could put that in an equatorial position.1595

And it has a methyl down; we can also put that in an equatorial position; so the most stable chair has the leaving group in an equatorial position.1603

Unfortunately the hydrogen we want to take next door is not anti; those are not anti to one another; again those are going to be gauche to one another when you take a look at that Newman projection.1614

What we are going to have to do is we are going to have to flip our chair into the other conformation.1628

Let's see; that means we are taking this carbon and bringing it down; so that is down here; this hydrogen is now pointing straight down.1642

This carbon got flipped up; so my bromine up is now axial up instead of equatorial up; and this carbon has my methyl group down; it was equatorial and now my methyl group down is now axial.1650

In this chair conformation, now my H and Br are now anti; we call that anti-diaxial as the necessary conformation.1668

In a chair conformation, the leaving group must be in an axial position in order to be 180 degrees from the β hydrogen.1683

Our base can come in, grab that proton, kick off the leaving group; and what does our final product look like?--we can go back to our line drawing here.1692

We are going to form a double bond on this side; and our original methyl group is still there; and that is the only possible product we can draw.1703

This product was really just a question of regiochemistry only in order to draw the product; but I wanted to point out that the required stereochemistry requires a specific conformation of our chair.1713

That is going to be relevant because depending on the substitution pattern throughout the chair, this chair flip might not be such an easy thing.1727

You will find that some cyclohexyl halides do very fast E2 elimination; some our very very slow; this is an important thing to take a look at when you are considering such cases.1734

Now that we know what an E2 mechanism looks like, let's consider its competition with the Sn2 mechanism.1747

The Sn2 mechanism was described as backside attack; and what is required for the Sn2 is that it needs a strong nucleophile.1754

In an Sn2 substitution, a strong nucleophile attacked the carbon, kicked off the leaving group; that backside attack; we needed a strong nucleophile.1768

Another thing that was important with that backside attack was something about steric hindrance; remember that backside of the carbon had to be very accessible.1778

Let's just write that steric hindrance... so it needs a strong nucleophile; and steric hindrance is bad; it is not a good thing; that slows down the reaction.1787

What do we know about the E2 mechanism?--what does it require?--E2 is when we have something attack a β hydrogen and form a π bond and kick out a leaving group.1802

We are attacking a hydrogen; the E2 needs a strong base... it needs a strong base.1812

It turns out that this deprotonation reaction is not affected by sterics; so the sterics is going to be a major factor in making the decision between an Sn2 and E2, we will see.1821

When do they compete?--when do we have to make this decision?--well, most definitely we are going to see it for reactions that involve hydroxide, that is HO^-, or RO^-, that is known as alkoxide.1835

In other words, we can have methoxide, ethoxide, propoxide, so an alkyl group with an O^-.1853

What is special about these species is they are both strong nucleophiles, meaning they love to do backside attack; they are great for an Sn2 mechanism.1858

And they are strong bases, meaning they can also go after a proton and do an E2 elimination.1868

We really need to be on the lookout when we are using hydroxide or alkoxide; and we will have to make a decision between the E2 and the Sn2.1875

Let's compare some rates; if we took a variety of alkyl halides and treated them with sodium methoxide and ethanol.1885

We could have either a substitution reaction take place; what would the product look like if methoxide decided to act as a nucleophile and do a substitution?1895

That means instead of RBr, we have our ethoxy; so our nucleophile has replaced the leaving group; that is our Sn2.1908

Why don't we draw out the ethoxy here; so I'll do some more abbreviations on this page; it is real good to start getting familiarized and seeing back and forth between these abbreviations.1922

This would be our Sn2 product; what would it look like if we did elimination instead?--well, that of course is going to depend on which alkyl group we started with.1934

But let's just say plus some kind of alkene or mixture of alkenes; in other words, we could do an elimination and form a double bond instead.1942

Let's take a look at these series of alkyl halides and try this reaction and see what product composition we get.1953

The first one we have here is methyl bromide; we could just abbreviate this MeBr for methyl bromide.1961

What kind of carbon is bearing the leaving group in this case?--it is simply just a methyl group; we just describe it as a methyl group.1970

This next one has one, two, three, four carbons; you could call that a butyl bromide or n-butyl bromide since it is the normal butyl group.1977

What kind of carbon is bearing the leaving group here?--well, because we have just one carbon group attached to it, we describe that as a primary carbon.1987

This next one is another common arrangement of four carbons; when we have something that looks like the isopropyl group here, but you have an extra carbon, this is called the isobutyl group.1997

If we had isobutyl bromide, this is still primary; but right next door to that on the β carbon, the next carbon over, we have some branching; so let's call this primary with β branching.2014

The second one, this three-carbon arrangement with the attachment at the middle carbon, that is the isopropyl group.2031

So this is isopropyl bromide; it is an example of the secondary carbon bearing the leaving group.2037

Then once again we come back to good old t-butyl bromide; tert-butyl bromide is just our classic example of a tertiary carbon bearing the leaving group.2043

What have we done?--why have we chosen this series of compounds?--because as we move down this list, we see that we are increasing in our sterics.2054

We have increasing sterics about the carbon bearing the leaving group.2065

What proportion of products do we see as a result?--well, when we have methyl bromide, it turns out that we see 100% Sn2 reaction; no E2 at all.2072

Let's think about what that E2 product would look like; what does E2 mean?--it means you remove the leaving group and a β hydrogen and you form a carbon-carbon double bond.2083

How does that look for methyl bromide?--it is looking pretty impossible because not only does it not have any β hydrogens, it has no β carbons.2095

This is let's just put NA here because there is no β hydrogens; you can't form a carbon-carbon double bond if you only start with one carbon; so of course we expect 100% for Sn2.2103

But when we move to primary, now the E2 is possible; and we do get a mixture of the two; but it is heavily in favor of the Sn2; we get about a 90:10 mixture.2116

Almost all Sn2, but we start to see a little bit of the E2 side product.2125

Adding just the slightest bit of β branching here, slightest bit of sterics, now pushes it in the other direction where the elimination is favored.2132

That is just how sensitive the Sn2 substitution, that backside attack, is to steric hindrance.2140

In fact if we go to a secondary carbon, then it is favored in the opposite direction even more so; where E2 is the major product by far, 4:1.2147

Now we are really shifting; and how about tertiary?--what do you think about the tertiary Sn2 reaction?--does that look like a good reaction?2159

No, we know that there is way too much steric hindrance here; so let's say there is essentially 0 and 100 now in this switch; remember tertiary alkyl halide is no reaction with the Sn2.2166

It is no reaction with the Sn2; but a reaction in fact does take place; instead it is the E2 elimination reaction.2183

Let's continue looking at this comparison and come back to tert-butyl bromide.2196

What we are asking it to do when we react it with ethoxide, or some other strong base that can also be a nucleophile, is it has two choices.2201

It can either attack the β carbon... I'm sorry, attack the carbon bearing the leaving group, do a backside attack; that is the Sn2 path.2210

Or what can it do?--it can attack the β hydrogen; that is the E2 path; attack the β hydrogen, form the π bond, kick out the leaving group.2220

Which of these is possible for the tertiary center like t-butyl bromide?--the Sn2 is impossible; it is impossible to get in here with all this steric hindrance and do backside attack.2231

But look how accessible that β hydrogen is; you can see that sterics is not going to be a problem; anytime we are going after a tiny little hydrogen, eliminations are no problem.2243

Let's just make a note here that the backside attack is too hindered and that the β hydrogen is more accessible.2254

We get 100% E2 with a tertiary leaving group; that is kind of the ultimate in the examples of Sn2 versus E2.2272

Another possibility is to vary the base that you are using; what if you had a primary alkyl halide?2285

Here we have fifteen methylene units; fifteen of these carbons; so it is just a really long carbon chain with a bromine at one end.2292

What if we reacted this with a species that could be either a base or a nucleophile?--what kind of mixture do we get, distribution between Sn2 and E2?2300

If we use a small base like methoxide, what we see, the same thing we saw on the previous slide, is that Sn2 is favored because it is very little steric hindrance for a primary carbon.2311

We get something like 96% of the Sn2 and maybe just a little 1% of the E2.2326

However if we use this guy, what is this structure?--that is a strange looking structure.2331

Remember potassium is just a K⁺; so that means we have O^-; and this is the t-butyl group... t-butyl group.2336

This could be abbreviated as tBuOK; tBuOK is potassium tert-butoxide.2344

When we have a very very bulky base now, that is going to hinder that backside attack; and we see a push in the other direction; now elimination can be favored.2352

This guy is called t-butoxide; he is a very bulky base because he brings along with him that t-butyl group.2363

I brought some models here again just as a reminder about the sterics like we are seeing here.2371

The methyl is so tiny; but the tert-butyl, when you have three methyl groups attached to a carbon, that really inhibits this backside attack.2377

If we are thinking about this coming in and acting as a nucleophile, this is going to have the same problem.2388

This is going to have a hard time getting in to do a backside attack on a carbon bearing a leaving group because it is bringing steric hindrance with it.2395

T-butoxide is a very bulky base so it is never good for an Sn2 unless maybe you are attacking a methyl group that can't do an E2.2402

Really these are classic E2 reaction conditions; if you really want to favor an E2 elimination, then t-butoxide is a great base to choose.2415

By the way, addition of heat also favors eliminations so we should get used to seeing heat as part of the reaction conditions.2424

That is because we are breaking the molecule up into different pieces so that is going to favor the entropy of the reaction; so elimination reactions are also usually done in the present of heat.2430

Let's summarize; what would we say the preference is for E2 over Sn2?--looking at a reaction type, which type of alkyl halide would really like to do the E2 elimination?2443

The tertiary is going to be the best; this is pretty much all E2 and no Sn2 at all; secondary still prefers E2, but you are going to have a little substitution.2455

Here is where we see that big jump when we are looking at primary; this now is essentially all Sn2; so only if it is a primary alkyl halide can you still expect to do an Sn2 with a strong base.2472

As soon as you get to secondary and of course if you are at tertiary, if you take a look at that data from the previous slide, you will see that E2 is now the major product.2487

So only when there is absolutely no steric hindrance, because you have a primary leaving group.2496

Of course, we are leaving off the methyl here intentionally because methyl has no β hydrogens; that would be impossible to do the E2 in that case.2500

Let's take a look a second elimination mechanism; we saw the E2 elimination which is the one-step mechanism; base attacks a β hydrogen, forms a π bond, kicks off the leaving group.2513

Another mechanism that we can have for doing a dehydrohalogenation, losing a hydrogen and a halogen, is an E1 mechanism; that stands for elimination unimolecular.2525

Let's take a look at an example of that; if we take this alkyl halide and we treat it with water, how would we describe water as a reagent?2538

Water is a weak nucleophile; it is weak everything; it is a very stable molecule so it is a weak nucleophile; it is a weak base.2549

But we can still undergo substitution-elimination reactions here; in fact, we do get the substitution product.2559

Would you expect that substitution to occur by Sn2 mechanism, backside attack?--no, because this is such a weak nucleophile and because this is a tertiary leaving group; so this can happen by Sn1.2570

It is also possible to get some elimination product; could this elimination be the one-step E2 mechanism?--no, because we don't have a strong base.2583

Instead a different mechanism must be in operation and it is the E1; so the Sn1 and E1 we will find are always in competition.2598

Just like we saw the E2 and the Sn2 are often in competition, can be in competition, the same is true for Sn1 and E1, these unimolecular reactions; and the Sn1 is usually the major product.2608

In most of the reactions we are going to see, there will be a nucleophile around; it is going to prefer to add; and so we will get this substitution.2627

E1 is maybe a side reaction; so you might see a little bit of elimination and this is how you get it.2635

We will see one example where the E1 is the primary mechanism in action when we do the dehydration of alcohols; but with that exception, Sn1 will be major.2646

The mechanism for an E1 is going to be two steps; and it is going to be via a carbocation; again very similar to Sn1; that is why they compete.2655

Our first step is going to be loss of a leaving group; our chloride is going to leave which results in a carbocation intermediate; notice this is the same slow rate determining step as Sn1.2667

I lose my leaving group and I form a carbocation; but now instead of having that carbocation get attacked by a nucleophile to do a substitution reaction, instead we can do an elimination.2683

What did it mean to eliminate?--you eliminate a leaving group plus a β hydrogen; so what have we done so far?--we've lost our leaving group.2699

What's next?--we lose our β hydrogen; and we have a word for loss of a proton; we call it deprotonate.2707

We are going to take our carbocation, we are going to take a look at one of these β hydrogens, and we are going to eliminate that β hydrogen.2718

Some base comes in; again you can use chloride if you want; I think water is a better choice because that is going to be present in this reaction as well; and it is your solvent so you have more of that around.2731

We are going to do a deprotonation; we grab the proton; why can carbocations get deprotonated so easily?2744

Because when you fill in those electrons as a π bond, you are going to get rid of the carbocation; you are going to add a new bond and get to a stable alkene product.2751

Let's consider the regiochemistry; why did I select this β hydrogen?--we did see that that is the major product; but is that the only β hydrogen that is available?2766

Here is our carbocation; no, this is a β hydrogen; this is a β hydrogen; I could have also made this product.2775

But why was that not listed as one of the products formed here?--this is actually not formed; how would we make that decision?2782

It turns out that this elimination also follows Zaitsev's rule just like the E2 did; and we are going to get the most stable alkene possible.2793

This would be a tri-substituted alkene; that is going to be more stable than a di-substituted alkene.2802

And so as usual we are going to go for the more substituted β hydrogen to give the more substituted alkene.2810

Let's take a look at the kinetics for the E1 mechanism.2820

Because it had the same slow rate-determining step as the Sn1 mechanism, it has the same rate expression, the rate is proportional to simply the concentration of the alkyl halide.2823

That is all you need to lose your leaving group; so that is why it is described as unimolecular like Sn1.2836

The E1 and Sn1 are both referring to the fact that just a single species is present during the rate determining step.2844

It is directly proportional to the rate of the reaction; its concentration is directly proportional.2851

The rate is not dependent on the concentration of water; it doesn't matter how much of that base we have in there; it won't affect the rate.2857

That also reinforces the fact that water is not involved in the rate determining step or whatever base you have; tt is simply the loss of the leaving group from the alkyl halide.2865

How could we speed up this reaction?--well, once again, the more stable the carbocation you have, the faster it is going to be formed.2874

If you bring down the energy of your carbocation intermediate, that is going to have a lower energy transition state as well; and you have a lower energy of activation, faster reaction.2885

Again the faster the carbocation is formed, once you are at that carbocation, you can now do substitution or elimination.2897

The same thing that speeds up the Sn1 reaction is also going to speed up the E1 reaction; and we come back to that competition.2905

So the E1 rate is exactly the same as the Sn1 rate; looking at the type of carbons bearing the leaving group, these are all great carbocations; so they are going to be fast E1 eliminations.2912

Allylic means you have an allylic carbocation; so in other words, an allylic leaving group... let's put a chlorine here.2928

That would be a great substrate to do an E1 elimination just like an Sn1 elimination.2940

Benzyl means it is next to a benzene ring; a leaving group next to a benzene would be great because that would be a resonance stabilized carbocation.2945

And then of course tertiary is what we just looked at because that is also a good carbocation; those are better than secondary.2954

This is where we have our huge jump for carbocations; primary and methyl are so bad; these are poor carbocations so it is going to be a very very slow E1 elimination.2961

Actually let's cross out that methyl again because you can't have an E1 elimination with just a methyl halide because you don't have a second carbon to form the carbon-carbon double bond.2979

How about the stereochemistry of the E1; is there any kind of relationship that we need to have between the leaving group and the β hydrogen?2992

Remember E2 elimination, what did we need?--our leaving group and our β hydrogen had to be anti to one another; one up and one down; they had to be anti-coplanar, 180 degrees.2999

Remember the E1 mechanism is a stepwise mechanism that goes through a carbocation; that carbocation is achiral; it is planar; so this is another case where we are going to see loss of stereochemistry.3011

This is the same story for the Sn1; remember we described it as racemization because we were dealing with chiral carbons in that case; here we just describe it as loss of stereochemistry.3021

Let's see an example; if we take this substrate and we treat it with a strong base, hydroxide; strong base, what does that do?--that is going to come out and attack; this likes to do E2.3031

Let's just define it; hydroxide is a strong base and it is a strong nucleophile; as a strong base, what mechanism can you do?--E2.3045

As a strong nucleophile what mechanism can you do?--backside attack, Sn2; which one is going to be favored in this case?3059

We take a look at the carbon bearing the leaving group; it is a secondary carbon; secondary--E2 is going to be favored.3069

We add in a little heat; that doesn't hurt; but even without that, with secondary, we are going to go with E2; there is just too much steric hindrance to do the backside attack instead.3077

We've conveniently drawn it already so that the hydrogen and the leaving group are anti-coplanar; so we could come right out and we don't have to rotate at all.3088

We could just take our hydroxide, grab the proton, form the π bond, kick off our leaving group; that is going to give us an alkene product.3097

What groups are going to be on the same side?--you have a phenyl that is a dash; and you have a methyl that is a dash; so those are both on the same side of the alkyl halide.3108

They will be trapped on the same side of the alkene; it doesn't matter whether you put them at the top or bottom; the important thing is they just need to be on the same side.3119

What other two groups do we have?--we have a methyl and a hydrogen; those are still on the structure.3128

Remember, because this is anti-elimination, this is our only product; it requires those two groups to be anti; and so this would be the only possible stereochemistry.3136

Now let's compare this same starting material and let's just react it with water; no hydroxide; so now we have a weak base and a weak nucleophile.3152

What kind of reactions are possible there?--is a weak base going to go and attack a β hydrogen?--no.3167

Is a weak nucleophile going to go and do a backside attack and kick off a leaving group?--no way.; so this water does reactions like Sn1 and E1; and of course the E1 is probably going to be major.3174

The Sn1 is probably going to be major, but let's take a look at the E1 to see what its stereochemistry would be.3190

What is going to happen here is because we have no strong base or strong nucleophile, the only other thing that can happen is the leaving group can just leave on its own; so that is exactly what happens.3197

When that chlorine leaves, the resulting carbocation is now planar; it is flat.3212

When we go to deprotonate our β hydrogen, the β hydrogen can be removed now or we can have some rotation; it can rotate here.3223

There is no longer anything trapping these two methyl groups into the trans position; it can rotate around.3237

Because it is a multistep mechanism, we lose that stereochemical relationship; once that leaving group leaves, it is gone.3245

When water comes in and it deprotonates, we now have a mix of relationships between these methyl groups.3254

We are going to get two products; we are going to the one where the phenyl is on the same side as the methyl; and we can also form the one where the methyls are on the same side.3264

You are going to get a mixture; and of course you are also going to get Sn1; and Sn1 is actually going to be your major product probably; but we just want to look at our E1 product composition here.3282

These are both possible; does that mean they are going to be formed in equal amounts?--remember Zaitsev's rule is still holding true; we still want to form the most stable alkene possible.3294

Is there a difference in this case in their stability?--the phenyl remember is a benzene ring; that is going to be larger than a methyl group.3306

This product has more steric hindrance than the other one; so this is going to be our minor E1 because of sterics; and this is going to be our major E1.3315

For anti-elimination, we only have so much say on which alkene we can get; it has to be from anti-elimination.3332

With na E1 elimination coming from a carbocation, we could rotate around and get to the most favorable carbocation conformation.3342

Which leads to the most favorable transition state, leading to the most stable alkene product.3349

Let's talk just a little bit about carbocation rearrangements; because like the Sn1 mechanism, we are going through a carbocation.3362

Carbocations can rearrange; so let's just briefly review what we saw for this Sn1 mechanism.3376

This was a rearrangement when our leaving group left; it gave a secondary carbocation; and that secondary carbocation could have a hydride shift to give a tertiary carbocation.3386

That would be a favorable rearrangement; so that is how we can get this substitution product where the nucleophile has come in to a different position.3405

What is important to remember is that this new carbocation can undergo E1 elimination products.3425

Sometimes our double bond is going to end up in a different place from where our leaving group used to be.3434

Because both of these mechanisms deal with carbocations, and carbocations can rearrange, a lot of times we can get some interesting product mixtures.3441

So these end up becoming excellent practice problems for mechanisms; in other words, how can this combination of reaction conditions lead to all one, two, three, four, five of these products?3450

This is definitely a reaction that looks like it is coming from the carbocation because the carbon chains have rearranged; the positions of the substituents have rearranged.3467

This is another example where our initial secondary carbocation can rearrange, reorganize itself to become a tertiary carbocation.3482

In this case, it is by rearrangement of one of these methyl groups, a shift of a methyl group.3495

If this carbon group moved over here, that carbon has lost its carbocation; it has got four bonds again; but this carbon is where the carbocation is now.3504

Hopefully you can see how that can lead to this substitution product; and it could lead to these two elimination products.3515

This original carbocation can lead to this substitution product and this elimination product down here.3522

Just considering the various carbocations that are possible, and the various Sn1 and E1 products that can result from those, is how we can get to these various products.3530

Of the elimination products, which would you say would be the major elimination product... be the major elimination product?3541

Again the substitution products are going to be the favorable one; and probably this is going to be the major overall because it comes from the most stable carbocation.3550

Because we do have a nucleophile present, that is why we do get a substitution; that nucleophile will attack the carbocation more often than not.3563

This will be our major overall; and of the elimination products, which do you think would be the major elimination product?3570

I am thinking the one where we can have a one, two, three, four, a tetra-substituted alkene.3578

And that comes from the more stable carbocation; so both kinetics and thermodynamics are favoring this as the major E1 product.3588

Let's try a few examples where we are looking at substitution and elimination competition; the ones shown here... let's just focus on the Sn2 and the E2 and see if we can decide which will be major.3600

First of all let's take a look at our starting material, our starting carbon compound, and see what we have; OTs is called a tosylate; that is a good leaving group.3618

What kind of carbon is it on?--is it on a primary carbon?--it's not even on a primary carbon; this carbon has no carbons attached; so we just call him a methyl; so we have a methyl tosylate.3635

What are we reacting it with?--the sodium means that I have a CH₃O^-; and what do you know about methoxide?3648

This methanol, what is this methanol doing here?--well, it is very common that when you are using methoxide as a reagent, you use methanol as your solvent.3657

But because he is neutral and stable and less reactive, we are not going to consider him for our reaction.3664

It is going to be the less stable, more reactive species that is going to dictate the path of this reaction.3670

What do we know about methoxide?--acid, base, electrophile, nucleophile?--well, it is a strong base; and it is a strong nucleophile.3676

So what mechanisms does it have?--these are perfect for E2 or Sn2.3689

We said that alkoxides like this; like methoxide is a perfect substrate to have this competition between E2 and Sn2.3695

Who is going to win in this case?--if you have a leaving group on a methyl carbon, then it is going to be impossible to do the E2 because there is no β hydrogen.3704

I am expecting to do the Sn2; that means backside attack; attack the carbon, kick off the leaving group.3713

My methoxy group is going to now be attached to my methyl group; it is going to be a substitution, backside attack.3722

We have another alkyl halide; this time it is on a secondary carbon; so this is a secondary leaving group.3731

What do you know about ammonia?--what kind of reactivity does ammonia have?--is it a strong acid, base, electrophile, nucleophile?3737

He is a very good nucleophile; which means he would love to do the Sn2; but is he a strong base?3744

He definitely is basic, but he is not one of the strong bases we identified for the E2; he doesn't have a negative charge on it.3751

Not a really really strong base... so in the case of a neutral amine, our favored reaction is going to be the Sn2 which means backside attack.3760

Tell me about the stereochemistry of that nitrogen; since our leaving group is pointing out towards us, the nucleophile has to come in from behind.3772

Let's keep our carbon chain fixed; and so now instead of the wedge pointing down, we have a dash pointing down.3780

What happens to that nitrogen when it gets a fourth bond?--one, two, three, four; nitrogen wants five; it is going to be an N⁺.3790

Which means... a lot of times, we draw that leaving group in there because the bromide is not going to get very far; it is going to have an ionic bond with the ammonium.3799

So a lot of times we draw it as the salt here as this product.3808

Who is DMF?--it is another commonly used solvent in organic chemistry; it is an example of an aprotic solvent.3811

A lot of these abbreviations or names, you should be familiar with and get used to seeing them there.3820

Even if we don't draw a solvent, all of our reaction are run in a solvent; so it is good to get some practice seeing those and recognizing those.3826

In most cases, the solvent is not going to be dictating the mechanism we choose.3836

But of course it is possible, if this solvent is the only variation, that you are going to see a difference in the mechanism that you pick.3840

How about this next one?--we still have our same secondary leaving group; but instead of ammonia, we are using this guy, tBuONa.3848

This is Na⁺ so this is tBuO^-; what do you know about tBuO^-?--it is t-butoxide; strong bulky base.3856

He loves to do the E2; not so good at the Sn2 because he is so bulky; so even if he was primary, he wouldn't be able to do the Sn2; this looks like perfect E2 conditions.3869

We look for our β hydrogens; we have some over here and we have some over here.3880

First we should consider regiochemistry; where would we prefer to deprotonate?--well, we want to get the most substituted alkene possible; so we are not going to go to the end carbon.3890

Then how about stereochemistry?--well, we know stereochemistry needs to be anti; but remember this is an acyclic structure; so we can rotate; it is not locked into any one hydrogen.3902

In fact, if it took this hydrogen, it would keep the carbon chain the same; and if it took this hydrogen, meaning it had to flip first and then do the elimination, it would give this product.3915

Which of those two products do you think is the better product?--how about stability of the alkene as a guide?3932

Sure, the trans alkene is more stable; so that is going to be the one that we get; so don't be fooled by the conformation that happens to be presented.3938

Remember that you can rotate the molecules and manipulate them around; these are in solution; they are mixing around; they can be any way they want.3952

So we are going to do whatever we can to get the most stable alkene possible, assuming that it is still anti-elimination.3959

Here are a few more; this one is sodium methoxide in ethanol so you could just abbreviate it as EtONa and EtOH.3968

What do we know about ethoxide?--acid, base, electrophile, nucleophile?--strong base, meaning it can do the E2; strong nucleophile, meaning it can do the Sn2.3977

How do we decide which is going to happen?--we take a look at our carbon bearing our leaving group; what do you see?--where is our leaving group?3993

Well, there is a problem; what do you expect to find as a leaving group?--chloride, bromide, iodide, tosylate?--is hydroxide a leaving group?--no, not a leaving group.4003

If you don't have a leaving group, you can't do a substitution; you can't do elimination.4018

It doesn't matter which; you can't do any of the four mechanisms that we've learned for Sn2, E2, or Sn1, E1; no reaction.4022

We learned a couple strategies where we could make this reaction happen; if we had a made a tosylate first, then we could do a substitution-elimination.4031

If we had a strong acid present, then it is possible to do a substitution; but that is not the case here.4039

We have strongly basic conditions; so nothing is going to happen with that alcohol.4045

Same thing is going to be true in this next step; we have sodium azide and acetone; azide is a very good nucleophile.4050

It is not a strong base; it is a good nucleophile; another nucleophile we should get used to seeing for the Sn2.4057

But because this is not a leaving group, again no reaction; nothing it can do.4063

How about our next one?--we have hydroxide; hydroxide is a strong base; it is a strong nucleophile; so this is a great example where we have our E2, Sn2 competition.4072

Do we have a leaving group this time?--yes, we have an iodide; we are back to our alkyl halide so we have a leaving group that we can get rid of.4089

What kind of carbon is that leaving group on?--it is a one, two, three; it is a tertiary carbon; a tertiary leaving group.4096

What do you think?--is that backside attack?--is tertiary good for a backside attack?--no way, impossible because of sterics; so this is going to be all E2.4105

Now we have to think about regiochemistry; we have to think about stereochemistry; we've already identified our leaving group; where is our β hydrogen?4117

We have this β hydrogen; we have either one of these β hydrogens; this is very similar to a substrate we worked with earlier.4127

We are going to want to eliminate one of the inside, one of the ones within the ring; so then we end up with this tri-substituted product, more substituted.4136

Just a little reminder, we don't want to eliminate in this direction because that is going to be di-substituted; this is less stable; we always want to get the more stable alkene possible.4145

Finally here we have our same tertiary leaving group; and we have sodium cyanide; what do we know about sodium cyanide?--great nucleophile; great nucleophile; it loves to do the Sn2.4160

Is it a strong base?--we have our list of strong bases; this is not one of them; this is a weak base so it cannot do the E2.4175

Let's take a look at our Sn2, our backside attack; does it look like a good Sn2?--tertiary leaving group?--no way.4190

So guess what?--we have a great nucleophile; but our carbon bearing the leaving group is too hindered; and it is not a strong base; so it can't do E2 elimination.4198

This is another example where no reaction is going to happen.4207

You might ask, well can't we have the leaving group just leave and make a tertiary carbocation in this case and maybe do an Sn1?--and do the substitution that way?4210

Well, here is one case where the solvent is going to come into play; the cases where we did see an Sn1 reaction like in the solvolysis reactions, we always had a protic solvent.4222

That is going to be true for a carbocation formation; it actually needs a protic solvent; like water or an alcohol.4238

Protic solvents have an acidic proton; they are extremely polar; and they are needed to stabilize if it is an extremely unstable polar carbocation.4249

If you have an aprotic solvent, it is not going to be possible to form a carbocation.4259

Most textbooks aren't testing that much detail; but this is an example where really having an understanding of the details of the solvent capabilities is going to help us determine something about a mechanism.4266

In this case, I was specifically asking you to compare E2 versus Sn2; so with those restrictions, for sure we could conclude that it is no reaction.4279

Sn1 would be a good guess in this case; but it turns out that would not be possible either.4288

That wraps it up for looking at elimination reactions and considering their competition with substitution reactions.4293

Hope to see you again soon; thank you.4299