Organic Chemistry Alkenes

Welcome to Educator.0000

Next we are going to talk about alkenes; there is actually a couple lectures we are going to have on alkenes.0002

We will start by looking at the structure of alkenes and the synthesis of alkenes; an alkene is defined as a molecule containing a carbon-carbon double bond.0006

When we look at the structure of a carbon-carbon double bond, we remember that it is a planar structure; each of these carbons has sp2 hybridization; that requires a trigonal planar geometry.0018

Connecting the two carbons are two bonds; we would describe these bonds as one σ bond and a π bond.0041

The way a σ bond is made is we join the hybrid orbitals on each carbon; this is an sp2 overlapping with an sp2 hybrid orbital.0055

For the π bond, the way we form a π bond is have a p orbital on each of these carbons and that is overlapping to form the π bond.0070

This model has a very nice picture of what an alkene looks like; here in the gray, we see the σ bonds; they are all planar.0077

On each carbon, each sp2 hybridized carbon, we have a p orbital; remember that is the dumbbell shape type orbital.0089

It is overlap of the top half of that orbital and the bottom half of that orbital that comprises a π bond; a π bond is a cloud of electrons above and below the plane of the molecule.0095

We could describe the π bond as being the overlap of a p orbital and another p orbital.0107

If we want to do a 3D sketch of this molecule, one way to draw it is to draw all the σ bonds in the plane; that is certainly an easy way to draw all the σ bonds.0114

Trigonal planar means these bond angles are about 120 degrees; but then our 3D sketch should also show the π bond and where the π bond is.0127

If the molecule is in the plane, then the π bond is perpendicular to that plane, sticking straight out and straight back.0135

We can sketch that as maybe a wedged and a dashed lobe, a wedged lobe and a dashed lobe, one on each carbon; then we can show some overlap at the front and back.0142

What we need to show is that the p orbitals are perpendicular to the sp2 plane; or orthogonal; we need to show that they are perpendicular.0153

Actually another way that we can draw this molecule is to draw it so that the p orbitals are in the plane, the π bond is in the plane.0164

We could draw our p orbitals nicely shaped and our π bond, again, top and bottom shows overlap.0174

But when we do that, what happens to the hydrogens?--if there are hydrogens here or the σ bonds, whatever they are attached to?0181

The σ bonds are now--these two are projecting out toward you and the others are pointing back; the rest of the molecule is now perpendicular to the plane.0186

We need to show them as a dash and a wedge; when we go to draw this, what we do is we tilt it just a little bit so you can see the wedged bonds and the dashed bonds as well.0197

You want to make sure you use the same angle though; either tilt it up a little or tilt it down a little; you don't want to try and twist the molecule and draw it like that.0206

Here I drew both wedged bonds pointing down toward the bottom of the page; that would be an acceptable drawing.0215

Either of these 3D sketches look really good as a way to draw an alkene, a carbon-carbon double bond.0222

What else do we know about π bonds?--we've seen a little picture of them; we should also know that π bonds are higher in energy than σ bonds.0229

The p orbitals are higher energy, that are coming together to form those π bonds; that is going to make them more reactive.0238

We are going to see lots and lots of reactions that break π bonds and all sorts of reactions of alkenes that we will talk about in our next lecture.0245

Also, we want to note that π bonds are electron rich; they are a good source of electrons; those two electrons are fairly loosely held; they are fairly available.0255

What are some things we associate with electron rich species?--they could be a nucleophile; what does it mean to be a nucleophile?0263

That means you can react with an electrophile; we will see a lot of reactions of alkenes with electrophiles.0269

It could also be a base; what does it mean to be a base?--an acid is something that donates a proton; a base is something that accepts a proton; in other words, it can be protonated.0278

Many of our mechanisms for alkenes, we are going to start with reaction with an acid and our very first step is protonation of that π bond.0290

We should also review some things that make alkenes stable; some of the things we've already seen before; but some other things that we want to know.0299

One feature is that an alkene is more stable if it has more alkyl groups attached to it; more carbon groups, the better.0309

If we take a look at this carbon-carbon double bond, it has a possibility of having four alkyl groups attached.0318

In this case, we have one, two, three of those positions filled with alkyl groups; we describe this as being tri-substituted.0328

We know that is going to be more stable than... for this alkene, our carbon-carbon double bond is here; how many alkyl groups do we have attached?0340

We have just one and two carbon groups attached here; this we could describe as di-substituted; it is less stable; a di-substituted alkene is less stable than a tri-substituted alkene.0348

Another way you could describe these two particular alkenes is you could say that this one is internal to the carbon chain while this one is terminal.0360

The second one is terminal; it is at the end of a carbon chain; that is also not a good place for a double bond to be.0369

We can look for features like that to determine whether or not our alkene is more or less stable; we can also look for the relationship of the groups that are attached on a double bond.0374

For example, if we have a di-substituted double bond, we know that the... we could have a cis arrangement or a trans relationship.0388

This cis one is when they are both facing on the same side of the double bond; trans is what we call it when they are on opposite sides of the double bond.0397

The trans is more stable; that is because by forcing these two alkyl groups in the same direction, they have some steric strain, some steric hindrance there.0404

There is some more crowding; cis is going to be less stable than trans.0417

The exception is when we have a double bond within a ring; if we take a look at something like this--cyclopentene or cyclohexene.0421

In order for these two groups on the double bond to connect in a ring, they actually must be pointing in the same direction on the double bond; it has to be the cis conformation.0432

For small rings like these, it is cis only; it is possible to have trans if you get to a larger ring.0442

It is possible if it is seven; that is still pretty unstable; but as soon as you get to an eight-membered ring, you can have...0450

For example, cyclooctene can either be cis-cyclooctene where we draw our eight-membered ring and we put a double bond in there; now we see that those two alkyl groups are cis to each other.0456

Or it could be trans-cyclooctene; when we draw this, it gets a little more complicated because the molecule ends up being twisted because we put one alkyl group on the opposite side of the other.0475

But if the ring is large enough, then those two alkyl groups can still come around and be connected; so it is possible to have cis or trans.0488

Let's see, we have one, two, three, four, five, six, seven, eight carbons there; this is kind of passing behind the double bond; this is the 3D sketch.0495

Try building these models and you can get a feel for what cis and trans looks like.0505

But if it is a smaller ring, if we just have cyclopenetene or cyclohexene, we do not put the word cis as part of that name because it is impossible to have trans there.0509

Cis is assumed; and cis is going to be much more stable because it is a more stable ring that way.0517

If we happen to have more than one π bond, something we can look to for stabilizing the π bond is conjugation.0524

Conjugation is when we have a π bond; then a σ bond; and then another π bond; if our π bonds are separated by just one single bond, that is going to be a good relationship.0531

Because what happens... this is an example where we have a double bond, single bond, double bond; this is conjugated.0544

And this is not conjugated because separating these two π bonds, we have an sp3 hybridized carbon.0556

We take a look at the p orbitals here; we know that this π bond is a p orbital and a p orbital; and this π bond is a p orbital and a p orbital.0565

Look what happens when they are conjugated; we have p orbital, p orbital, p orbital, p orbital.0573

We end up getting delocalization of those π electrons over all four of these atoms; what we get is resonance stabilization.0578

We will be looking more at these conjugated systems down the road; but for now, we want to see it.0588

If we ever see it, we want to recognize that that is something is being stable; and that is more stable than having this π bond totally isolated from this π bond.0594

And there is no relationship between the p orbitals from one to the other; this is going to be more stable and the non-conjugated is going to be less stable.0603

That is something else we can look to stabilize a double bond if we can.0618

Finally, if we have something like this; this is an example of a bicyclic compound; it is a bridged compound; we have two rings in this molecule.0624

When you have a double bond in a bicyclic compound like this, we are going to look for the bridgehead carbons.0636

The bridgehead carbons are defined as those that are shared between multiple rings, where the two rings come together and there is a juncture; this is bridgehead carbon; this is a bridgehead carbon.0642

When you look at this molecule from the perspective of the bridgehead carbons, you see that these carbons are connected by three bridges.0652

Here is a one carbon bridge; here is a two carbon bridge; here is another two carbon bridge; we call these bridgehead carbons.0660

It turns out that having a double bond attached to one of those bridgehead carbons is very unstable; this molecule is not going to want to exist; it is going to be highly unstable.0666

It is okay to put a double bond somewhere else in the bicyclic compound; it is okay to have a bicyclic compound; over here would be okay because the double bond avoids the bridgehead position.0678

This is known as Bredt's rule; you might come across some examples where you are looking for where to put a double bond as a result of a reaction; we want to avoid bridgehead carbons.0689

This alkene stability can be measured; this is something that we can find some evidence for, these various rules that we are looking at.0701

By doing a reaction called a hydrogenation reaction and measuring the heat of that hydrogenation.0709

We are going to be talking about that in our next lecture when we look at some of the... that is one of the many reactions that we will be studying that alkenes can undergo.0713

How would we make an alkene?--if we want to put a carbon-carbon into a structure, there is really two major approaches that we would want to take.0723

One of them is going to be an E2 reaction on an alkyl halide; let's see an example of that.0731

How about if we wanted to transform this given starting material into the desired product; what a transform problem looks like is it means provide the necessary reagents to convert one to the other.0737

More than one step might be possible; it might not just be one reagent that you are putting in there; it might be multiple steps to convert one to the other.0750

If we compare our starting material to our product, we see that an elimination has taken place.0759

It must be an elimination because we've formed a double bond; we've lost the bromine--the leaving group; but we've also lost a hydrogen.0766

What we would need to consider is what mechanism do we want to employ in doing this elimination?--we have two choices; it could be either an E1 elimination or an E2 elimination.0780

There is some benefits to some; some might be more suitable; let's think about an E1 elimination; that was the mechanism that was the multistep elimination mechanism.0790

It started by loss of a leaving group to form a carbocation; carbocations are involved in the E1 mechanism; would this be a good substrate to form our carbocation?0803

We have our leaving group on a primary carbon; that would give us a primary carbocation; that is not a very good carbocation; that would be highly unstable.0814

I don't think an E1 is a good idea here because we have a primary leaving group; that would give a primary carbocation; that most certainly would rearrange.0826

I think we should avoid trying to do some kind of unimolecular elimination and count on a carbocation; instead we are going to do an E2.0841

E2 elimination was where we had some strong base attack in a single step mechanism; attack the β hydrogen, form the π bond, kick off the leaving group; that is our E2.0850

What we need here is we need a strong base; we need a strong base; if you had to think about a strong base, maybe sodium hydroxide comes to mind; that would definitely be a strong base.0861

But let's take a look at this starting material, this n-bromobutane or 1-bromobutane, and think about its reaction with sodium hydroxide.0882

Yes, it could do the E2; but is there another reaction that hydroxide can do?--remember that hydroxide can be both a base and a nucleophile; we have our competition here between Sn2 and E2.0892

Because we have a primary alkyl halide, what would be favored?--there is no steric hindrance; Sn2 is going to be favored.0906

Sodium hydroxide wouldn't work because that would give us the substitution product; how can we force the elimination then?--how can we suppress the substitution, the backside attack of the Sn2?0915

If there is some way we can increase the sterics of that backside attack, then that would help us; that would force the E2 be favored over the Sn2.0926

How about if we just used a different base?--what would be a bulkier base than sodium hydroxide?--how about t-butoxide as a base?0936

T-butoxide has a tert-butyl group; now it is very bulky; this is bulky; so E2 is major; even with a primary halide, the tert-butoxide favors the elimination.0945

All we need to do, in this case, our reagents... it can be done just in a single step... is we use t-butoxide.0960

Maybe we can throw in some heat if you want; that usually favors the elimination as well; but all we need in this case is a strong bulky base.0970

The E2 is a very good method for forming alkenes; very reliable single step; let's review some of the features of the E2.0978

Remember the stereochemistry of the E2; there was a relationship between our leaving group and our β hydrogen.0986

They had to be anti to one another; we call that anti elimination... anti elimination; we need to be able to achieve that stereochemistry in order to do the E2.0991

Our regiochemistry, this was governed by Zaitsev's rule; we wanted to get the most stable alkene ;in this case, we didn't form a very stable alkene; it is terminal; it is mono-substituted.1011

But because there was only one β hydrogen that was... one type of β hydrogen that is possible, this is the only E2 elimination product that is possible; this would be in fact our major product.1031

Zaitsev's rule comes into play when we have more than one β hydrogen from which to choose; the one we select is based on the one that would lead to the most stable alkene product.1043

Finally, how about the Sn2 versus E2?--because we are going to have that competition with our strong bases; they can also be nucleophiles.1056

As usual, as we increase our sterics, we know that is bad news for the backside attack for the Sn2; that is going to increase the proportion of the E2 elimination product.1065

We need to... if we want to increase our sterics, we need to have a nice repertoire of bulky bases from which to choose.1078

Of course, the t-butoxide is the one we are most familiar with; the t-butoxide; but there is some of other ones you can do as well.1087

There is some nice amine bases; amines have a nitrogen in them and they usually have some groups attached on that nitrogen.1096

For example, triethylamine, a nitrogen with three ethyl groups attached to it, like the tert-butyl group, very big and bulky; that is very good for doing eliminations.1105

Or diisopropylamine has a nitrogen with two isopropyl groups on it; look at all that steric hindrance, all that bulk; diisopropylamine is another example of a base that is good.1114

If you put that in there with an alkyl halide and warm it up, then you can have a good bet that E2 is going to be your major product there.1129

A second approach, besides doing the E2 elimination, is to start with an alcohol and do a dehydration reaction; here is an example of an alcohol.1140

In order to do a dehydration, we are going to react this either with H₂SO₄ and heat or maybe H₃PO₄; these are the ones we will probably see most often.1149

Some strong concentrated acid and heat--what happens is we get an elimination reaction to take place; we form an alkene product; we start with an alcohol and we form an alkene.1160

What did we just lose?--what has been eliminated in this elimination reaction?--we lost the OH, of course; but remember we also lost a β hydrogen; we lost an OH and an H.1175

It means we have a loss of H₂O; that is why we call this reaction the dehydration reaction.1188

Because just like when you are out in the desert or you are running and you are dehydrated, you are low on water, dehydration means you are losing water; the other product in this reaction is water.1196

Let's think about the mechanism for this; how can this alcohol get converted to an alkene?--how can it lose water?1211

Clearly, we have very strongly acidic reaction conditions; what do you think our first step should be?--in a strong acid, we need to protonate something.1217

Let's look at our alcohol and think about where to protonate; we really only have one choice--on the oxygen; that in fact is going to be our first step of the reaction.1225

Let's just use HA to represent our strong acid; two arrows to do a proton transfer; and we can protonate the alcohol OH group.1235

Great idea for our first step because we are in strongly acidic reaction conditions; so protonate; what does that do for us though?--why would that be something that might move us toward our product?1251

Remember, once we protonate an alcohol, we turn that OH into a very good leaving group... very good leaving group; it is going to be... it would be water once it leaves, very stable molecule.1265

We now have a leaving group to do our elimination; now we should think, do you think it is going to be an E1 elimination mechanism?--an E2 elimination mechanism?1281

E2 means we have a strong base come in and attack the β hydrogen to kick out this leaving group; is that what happens?--do we have any strong bases in these reaction conditions?1288

No, of course not; we have very strong acid conditions; it can't be an E2 elimination; it can't be an E2 because there is no strong base.1299

Our only other choice is to be an E1; what does it mean to be an E1?--it means our leaving group just leaves on its own.1312

In acidic conditions, that is what we would get; our leaving group leaves on its own; this next step is loss of leaving group to give a carbocation intermediate.1320

This carbocation can go on to be a carbon-carbon double bond, to be an alkene; this how the E1 continues; how do we do that final step?1337

Remember we are losing water; we are losing the OH leaving group; but we are also losing a β hydrogen; what we do is we look to one of the hydrogens on the neighboring carbon.1349

I'm going to choose over here because that is going to give me a double bond that is more highly substituted, that is more stable; you could see that is the major product that was shown here.1359

I could just use A- as a good base; that was formed in that first step; but I could use A- to come back in here and deprotonate; carbocations can be deprotonated to form double bonds to form alkenes.1367

This last step we could describe as deprotonate or loss of the β hydrogen; we know that has to be part of an elimination; we want to put in that context.1385

We are going to protonate, lose our leaving group, and deprotonate; that is our E1 mechanism for dehydration.1398

What are some other things that we want to know about this dehydration mechanism?--because it is the E1 mechanism, it is going to give the most stable alkene; this will also follow Zaitsev's rule.1409

It involves a carbocation; we know carbocations can rearrange; if there is a possibility that we can have a hydride shift or an alkyl shift to go to a more stable position.1422

A more substituted carbocation, that can happen; that will happen; taking a look at our reaction conditions, remember that we are dealing with a very very strong acid.1433

We need that to make OH a good leaving group; we really have to have something like sulfuric acid as our catalyst in this reaction.1442

In my mechanism, I just used HA to represent sulfuric acid; that is a safe species to use in a mechanism to represent a strong acid.1451

But let's just remind ourselves what H₂SO₄ looks like; this is H₂SO₄.1460

After it protonates something, after it is used as an acid, we are left with A-; that is this species right here.1468

Why is sulfuric acid such a great acid?--why is it such a strong acid and so willing to donate its proton?--we could take a look at its conjugate base.1478

This is the conjugate base of sulfuric acid; do you think this is a pretty stable molecule or is this pretty reactive?--what do you think about it?1490

It has an O^-; sometimes we consider that to be a strong base, but this is not a strong base; that is because this O^- is resonance stabilized.1499

We can draw several resonance forms for this; I will just draw one here; this negative charge can also be delocalized on the third oxygen, the other oxygen down here.1510

That negative charge is equally delocalized over all three of these oxygens; which means it is very very stable and is resonance stabilized.1524

We have a very weak conjugate base; this is very stable, very unreactive, very weak; we would expect that for something that is the conjugate of a strong acid.1537

Because it is so stable, that makes it non-nucleophilic; in other words, there is going to be no Sn1 competition.1552

Normally up till now, every time we've seen a carbocation, we've used that carbocation; it had a nucleophile add to it and we've done a substitution reaction.1565

The only time we've seen elimination, we said that was just usually a side product for our substitutions; substitution is usually favored.1573

The dehydration reaction is the one example that deviates from that norm; it is simply because, in these reaction conditions, we have no nucleophile to add to the carbonyl, to the carbocation.1580

All we have is a strong acid and its weak conjugate base; once you form the carbocation, your only choice is to dehydrate, to eliminate and form the alkene double bond.1591

You can think of the A- like a spectator ion; you can use it as a mild base to deprotonate something that needs deprotonating.1603

But it is not going to be something that is going to be nucleophilic; that is not going to be participating in the reaction.1614

What I wanted to do is I wanted to compare a strong acid like H₂SO₄ with another strong acid like HCl.1621

What we just said about H₂SO₄ and heat, this looks like what kind of reaction conditions?--this looks like you are going to lose water; it is going to be dehydration.1629

Strong acid and heat on an alcohol, what would that product look like?--we have four carbons; we are going to lose water.1639

We would form the double bond between the middle two carbons here--that is more stable than being at the end; we can keep this trans relationship; that would be more stable.1649

Remember we could also form the cis-butene product; but that is going to be less stable; that would not be our major product.1658

Our major product is the most stable alkene we can have after rearrangements if they are necessary; in other words, we would expect E1 elimination to take place with H₂SO₄.1667

How does that differ when we react it with HCl?--HCl is unique because, not only is it a strong acid and a source of proton, it is also a source of Cl^-.1679

It is a strong acid, plus it is a nucleophile, it is a source of a nucleophile; what happens when we react an alcohol with a haloacid like HCl?1691

We protonate to form a good leaving group; then the halogen replaces that leaving group; we get a substitution mechanism taking place either by the Sn1 mechanism or the Sn2 mechanism.1705

Both of those can happen with the halide; we can form a carbocation or the halide can do a backside attack.1717

But with an acid like HCl, we get substitution because, if there is a nucleophile present, that nucleophile will want to attack the carbocation.1724

With H₂SO₄ and heat, there is no nucleophile present; once we form the carbocation, our only choice is to eliminate and get the alkene product.1732

Let's try another example here; here we are given a reaction and we are asked to provide a mechanism; it looks like we have these two methyl groups here.1742

Remember our line drawing means these are methyl groups; it looks like those methyl groups used to be on the same carbon; now they are on different carbons.1757

I think we are going to have to do some kind of rearrangement in our mechanism to account for that.1766

How do you think we should get started?--I see that strong acid; I am going to protonate as my first step; that gives you HA to protonate.1771

Every step in this mechanism, the dehydration mechanism, is reversible; this oxygen now has just one lone pair; it has a positive charge; I protonate it.1783

After protonating, now I have a great leaving group that can leave; I show this bond breaking and leaving with water; this is where I just kicked out my water molecule; now I have a carbocation.1796

Let's put those CH₃s back in again so we can see them a little more clearly; we are at the carbocation stage; now we can rearrange.1811

Remember a carbocation rearrangement is where we look over to one side or another and we try and find and steal some electrons, steal a bond that would result in a more stable carbocation.1825

What this carbocation sees is, if one of these methyl groups picked up and moved over to the next carbon over, we call that a methyl shift... a 1,2-methyl shift.1836

We now have one methyl up here and one methyl down here; where does that place our carbocation?--this carbon now has four bonds; it had only three before.1846

But this carbon just lost one of its bonds; the carbocation is now down here; why is that rearrangement favorable?1855

It is favorable because we have gone from a secondary carbocation to a tertiary carbocation which is more stable.1865

We always want to be on the lookout for any mechanism involving a carbocation; we want to be on the lookout for rearrangements; because if it can find a way to stabilize, it will.1876

How do we get to our alkene?--how do we do our last step here?--we need to form the double bond; now we need to deprotonate.1885

Where is the proton that we want to go for, we want to deprotonate?--it is not going to be at the carbon of the carbocation; it is going to be one of the adjacent carbons, one of the β positions.1893

Why did we not deprotonate in this direction?--because that would give us a double bond that is less highly substituted.1907

If we go for this hydrogen, that is going to give us the double bond between the two methyl groups; it is going to give us a tetra-substituted, which is the most stable we can have.1915

In this case, because it gave us the major product, we didn't have to make this decision; I just want to keep that in mind.1930

If you had to predict the product here, this is the product; this is why you would get the product you did.1936

We need to deprotonate; what can we use to deprotonate?--A- is the safest bet here; we just formed A- in the first step; you can also see that this is catalytic in acid.1941

For every protonation step, there is a deprotonation step; we grab that proton and use those electrons to forms a double bond.1950

We protonate, lose water, and then, in this case, we rearranged; then we deprotonated to form the double bond.1959

One more example, let's think about a transform; if I had an alcohol here; I am starting with an alcohol; I want to go to an alkene.1971

It is important to consider the functional groups in our starting material and our product to decide how we are going to convert one to the other.1983

We are back to our original problem; we want to form an alkene; the way we form an alkene is by eliminating to form the double bond.1991

You know our choices are E1, E2; those are the only elimination mechanisms we know so far; those are the only ways we know how to make alkenes so far; we will learn about some other down the road.2000

We have an alcohol; what if we took this and we treated it with H₂SO₄ and heat; we just learned that alcohols can be dehydrated; certainly this alcohol could be dehydrated.2010

Would that give this product that is shown as the major product?--let's think about that mechanism; we know this is going to be an E1 mechanism; it is carbocation conditions.2023

We are talking about a primary carbocation, highly unstable; but with immediate rearrangement, we can get to a secondary carbocation, maybe even a tertiary carbocation.2040

But most definitely, we would rearrange this carbocation; what do you think that major product would look like if we had to predict this?2050

I think we would end up forming the tri-substituted double bond, the internal double bond, as the major product.2060

Because we don't want that as our target molecule, as the thing we are trying to synthesize, then this would not be a good synthesis.2069

What is another way that we can put in that double bond?--we need to do an E2 elimination to form the double bond at the end here.2077

Why don't we treat this with t-butoxide?--we talked about t-butoxide as our bulky base... throw in a little heat... as our bulky base.2088

What would happen if you take an alcohol and you react it with t-butoxide?--would you expect the E2 elimination?--what is wrong with trying to do an E2 here?2098

You lose a β hydrogen and what?--a leaving group; we do not have a leaving group here with hydroxide.2107

This is no reaction because there is no leaving group; we wanted to try and do the E2; but that is not going to happen here.2113

How can we possibly do the E2 elimination?--what do we have to do?--we have to make the OH a good leaving group; then we can do the E2.2122

An ideal way to do that is to convert the alcohol to a tosylate; we can use some tosyl chloride in pyridine; that keeps the alcohol right where it is.2130

But turns the OH into an OTs which is now a great leaving group; now that you have this great leaving group, now you can treat it with t-butoxide and heat and do the E2 elimination.2145

If you don't want the tosylate, you can maybe use something like thionyl chloride, SoCl₂, or PBr₃ to convert it to a bromide or chloride.2163

Then we have a good leaving group and we can do an E2 elimination.2173

We want to keep in mind, in synthesizing alkenes, we need to know the mechanisms very well for the E1 and the E2 and decide which is appropriate in each condition, in each situation.2176

Next we will be talking about the reactions that we can do for alkenes once we have a carbon-carbon double bond in our structure.2188

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