Organic Chemistry Reactions of Alkenes

Hi; welcome to Educator.0000

Next we are going to talk about reactions that alkenes can undergo; many of the reactions we are going to be seeing are described as electrophilic addition reactions.0002

One such reaction is the addition of HX across the double bond; we call this a hydrohalogenation reaction since we are adding a hydrogen and a halogen.0011

For example, if we react an alkene with HBr, what is going to happen in these electrophilic addition reactions is we are going to break the π bond.0021

Then we are going to be able to add a group to each carbon; we are going to see the same pattern again and again throughout this lecture.0032

When we are reacting with HBr, those are the two groups we are going to be adding--a hydrogen and a bromine; you can see we have a choice on which carbon gets the hydrogen and which gets the bromine.0039

It turns out that the hydrogen prefers to add to this end carbon and the bromine goes to the middle carbon; this is the only product that is formed.0048

Just so we are clear on what is not happening, the other choice we could have made is that the hydrogen goes to the middle carbon and the bromine goes to the end carbon; that product is not formed.0057

We describe this reaction as being a regioselective reaction since it reacts with just one region, one site, preferentially.0072

This is a question of regiochemistry--when we see that there is more than one place that we can react and we need to make that decision.0080

The regiochemistry that is shown is described as Markovnikov's addition; or we can call it Markovnikov's rule.0086

What Markovnikov's rule states is that the carbon with more hydrogens gets the hydrogen; the carbon with more hydrogens originally, initially, gets the hydrogen in the electrophilic addition.0093

We go back to our alkene; we see this carbon has just one hydrogen; the end carbon has two hydrogens; the hydrogen goes to the carbon with more hydrogens; that is exactly what is observed.0109

If we want to understand why we follow Markovnikov's rule, we need to look at the mechanism; it is a two-step mechanism; it is going to start with our alkene reacting with HBr.0124

What do we know about the reactivity of HBr?--we know it is a very strong acid; in fact, that is going to be our first step of our reactiom--is our alkene is going to act as a base.0137

HBr is going to act as an acid; we are going to protonate that alkene as the very first step; usually when we have done a protonation, it has been a lone pair on a base.0147

A π bond can act as a base as well; we just start with that π bond, those two electrons attack the proton, kick off the Br^-.0157

This is a proton transfer as usual--two arrows; but we are starting from the π bond in this case.0166

Right now in this very first step... let's call this step one... in this very first step, we have to decide which carbon gets the hydrogen.0173

We know based on the product observed that the hydrogen goes to the carbon with more hydrogens, the end carbon in this case.0181

What happens to this carbon?--it just lost one of its bonds; this carbon now has one, two, three bonds; if we do its electron count, we get one, two, three; we know carbon wants four.0192

We end up with a positive charge on this carbon; we are going to get a carbocation intermediate; that is what happens when you protonate an alkene with acid--you get a carbocation.0203

Let's think about that other possible regiochemistry; if I instead added the proton to the middle carbon, that would give me a carbocation on this end carbon.0218

Take a look at those two possible carbocation intermediates and think about why it is that this first carbocation is favored and leads to product and the second carbocation is disfavored.0231

Do we know anything about carbocation stability?--how would you describe this one?--it has one, two carbon groups attached to the carbocation carbon; we call that a secondary carbocation.0243

This is a primary carbocation; we know the more alkyl groups donating into that carbocation, the better; so this is more stable.0254

This primary is less stable; primary carbocation is really awful looking; that is why this again is not formed; we get this carbocation; let's redraw it down here so we can do something else with it.0263

What is going to happen to that carbocation?--it is now going to react with the Br^- that was just formed in that first step.0278

The carbocation acts as an electrophile; it looks like a great electrophile; it has a full positive charge; it is electron deficient.0292

The Br^- acts as a nucleophile; we simply have the bromine bonding with the carbocation; our reaction is done; we now have our addition product as we were expecting.0300

We added H and Br across the π bond; it is a two-step mechanism--first we protonate and then our nucleophile attacks, adds to the carbocation.0321

Considering that mechanism now, let's see if we can get a good understanding of why that Markovnikov's rule is followed.0332

It turns out Markovnikov addition is favored because that is the one that favors the more stable carbocation intermediate; it is all about the stability of the intermediate.0338

If you have a more stable intermediate, that means it is lower in energy; if the intermediate is lower in energy, that means the transition state leading to that intermediate is lower in energy.0348

That means the energy of activation is lower leading to that transition state; there is a smaller hill to climb; simply the reaction is faster.0361

What we are looking at here in the observed regiochemistry is kinetics in play; it is just simply which reaction is faster; in other words, it is not about product stability.0369

We don't look at the two possible regiochemistries and think: where would the bromine like to be?--is that interesting to us?--no, it is simply about the stability of the carbocation intermediate.0384

If we think about our reaction here, our energy diagram, our starting materials are at some combined starting energy; our products are typically going to be at a lower energy.0395

We have a pathway to get here; this is a two-step mechanism going through an intermediate; our carbocation is going to be a high energy intermediate; we could put that here.0412

If this represents our secondary carbocation, the other path leading to the primary carbocation would be higher in energy; these are the two possible paths that we can take.0423

Our starting material can go to the secondary carbocation and then onto product; or the starting material can go up all the way to the primary carbocation and then down to the product.0435

If we look at these two paths and we ask which one is faster, it is going to be going from the starting material energy up to that transition state energy... my curve missed that a little bit.0450

Here is our transition state; this is our energy of activation; trying to go up to this transition state is going to be a larger hill to climb; this is the faster reaction.0462

In this part, formation of that carbocation is going to be our difficult step; this is our rate determining step; it is going to be very difficult to protonate and form the carbocation.0477

That is an uphill battle; because this is a faster step, that means overall, it is a faster reaction since it is our rate determining step.0494

Let's see an example--if you want to predict the product here; I also pointed out that, rather than just predicting the product, let's also think about the stereochemistry of that product.0511

We have an alkene; we are reacting with HCl; we are going to be breaking the π bond; we are going to be adding a group to each carbon; we have a hydrogen and a chlorine.0521

Where do you think the hydrogen should go?--it wants to go to the carbon with more hydrogens; we could think about our line drawing; we have one hydrogen here; we have two down here.0533

Once again, the end carbon is going to be the place that we would want to put that hydrogen; so this is our addition product; that is our regiochemistry.0544

What about the stereochemistry?--what do you think... we have a chiral center here on this carbon; what do you think the orientation of that chlorine is?--is it a wedge or a dash?0559

In fact it is going to be a mixture of those; we could just draw it flat; but what we really mean is we are going to get some with the chlorine as a wedge and some with the chlorine as a dash.0571

In fact we are going to get exactly 1:1 of this mixture; what is the relationship of these two products?--they are enantiomers of each other; in other words, we are going to get the racemate.0584

Stereochemistry is going to be a big issue for us in looking at reactions of alkenes; because for alkenes, we are starting our planar; we are starting out flat; these are achiral starting materials.0598

But when we break the π bond and add something to it, we have the potential for creating new chiral centers in the new tetrahedral centers, sp3 hybridized centers that we are making.0611

Stereochemistry is going to be an issue constantly; if we think about that protonation, we can protonate to form this carbocation intermediate; there is a hydrogen here.0620

This carbocation intermediate is sp2;--it is still planar just like the alkene starting material.0637

But now when the chlorine attacks this planar species, it can attack from the top face... attack from the top; or it can come in and Cl^- from the bottom.0646

This is how we can get to either enantiomer; you can imagine your planar intermediate or your planar starting material.0664

Attacking from one from one face or the other face is going to lead to your two possible stereocenters.0673

What we need to keep in mind here, what this is demonstrating, is that any new chiral center or chiral carbon is generated with both R and S configurations.0679

We have an achiral starting material; so if we come up with a chiral product, we must get a racemic mixture; if we come up with a product that is chiral, it must be formed as a racemic mixture.0704

How do we normally draw this product?--we normally draw... it is assumed we know it is a racemic mixture because that is always the case.0720

So we usually just draw it flat; we don't need to draw any stereochemistry here; but technically what this is representing is an equal mixture of both enantiomers.0727

In some other examples where we create two new chiral centers in the addition to an alkene, we will find that we are going to have to talk more about the relative stereochemistries of those two groups.0736

We will see those in some of our next examples; what else can we add across a double bond?--we can also add water across the double bond; we call this the hydration of alkenes.0746

We are going to break the π bond; what are the two groups that we are going to add to that π bond if we are adding the components of water?--we are going to be adding an H and an OH.0762

We are going to break this π bond; we are going to add these two groups; if you had to take a guess on where the hydrogen goes and where the OH goes.0775

What if this followed Markovnikov's rule just like addition of HBr did or HCl?--then we would expect the hydrogen to go to this end carbon; guess what?--that is exactly what happens.0786

We add H and OH; we get Markovnikov addition of the H and OH; we started with an alkene; we end up forming an alcohol product; this is one way we can make alcohols--is by hydrating an alkene.0802

Our reaction conditions for this is we need water of course if we want to add water; we also need acid; this is acid catalyzed mechanism.0824

I just want to reflect for a moment on what happens when you take a strong acid like H₂SO₄ and you put it in water.0832

What you end up with is H₃O⁺; sometimes your reaction conditions for this hydration might just show H₃O⁺.0840

Sometimes it will show H₂O and a strong acid; it can have a few different ways that it will be presented.0846

But if you ever see H₃O⁺, it means that you have both HA, some strong acid, and H₂O present; it implies that you have both of these.0852

The reason we can assume we have H₃O⁺ is by definition a strong acid is something which fully dissociates in water.0862

Dissociates in water... in other words, it is going to protonate water and give A-.0873

A strong acid, when we represent HA in aqueous conditions, if we ever we use HA, the strongest acid we have is going to be H₃O⁺, the hydronium ion.0888

You no longer have H₂SO₄ molecular in an aqueous solution; you have H₃O⁺ and you have the conjugate base of that strong acid.0897

When you see these reaction conditions, you can just go ahead and use H₃O⁺ in your mechanisms knowing that this is the mechanism by which that H₃O⁺ was formed.0909

Let's take a look at that mechanism; what do you think our first step should be in these clearly acidic reaction conditions?--I think we need to protonate; we can use our π bond for that.0922

I am going to go ahead and draw out H₃O⁺ since I know that is the acid in these reaction conditions; I am going to have my alkene act as a base.0936

I am going to protonate my π bond; just like our mechanism for addition of HBr, it starts the exact same way--it starts by protonating.0947

Remember this protonation step is the point at which we determine our regiochemistry--which carbon is going to get the H⁺.0963

Just as a quick reminder then again, the other possibility is that we had the proton to the middle carbon; if we do that, our carbocation ends up on the end carbon.0971

Exactly like we saw for addition of HBr, we want to form the more stable, in this case, secondary carbocation.0984

Always going to form the more stable carbocation; every time we go to protonate an alkene, we will keep that in mind.0995

We have a carbocation; where do we go from here?--we want to form this alcohol; we need an oxygen to attack.1002

How about using hydroxide?--how about if we wanted to use hydroxide to add to the carbocation?--our mechanism would be finished very quickly there.1010

But there is a problem with that; what is the problem with using hydroxide?--let's take a look at our reaction conditions.1021

Do you expect to have a lot of hydroxide present in these strongly acidic conditions?--no way; there is no hydroxide around.1027

Who is our nucleophile?--who is the strongest nucleophile we have in these conditions?--it is going to be water.1036

Water is going to be the strongest nucleophile that we have and that we can use; that is what is going to attack the carbocation.1043

Let's think about this oxygen now that has just attacked; it still has two hydrogens on it; what else does it have?--it still has one lone pair and a positive charge.1055

My water can attack; we can call this second step--attack of our nucleophile onto the carbocation.1068

We have added a group to each carbon; but we are not done yet because this oxygen has a positive charge; it is very unstable; we need to get rid of that positive charge.1077

How can we do that?--we know oxygen just wants to have two bonds and two lone pairs; it is one of these protons that can be removed.1085

Our last step is going to be deprotonate--that is what we call it when we remove a proton; what could we use to deprotonate?1093

We can come back and use this water that we just formed in our first step; and we are done; a couple things we can note--that it is catalytic in acid; we need that acid there.1103

But for every step where we protonate, there is step where we deprotonate and get that H₃O⁺ back; we just need a tiny bit of acid to go along with our water.1129

We could take a look at this pattern; it is going to be something we are going to see again and again with acid catalyzed reactions where we protonate and then we attack and then we deprotonate.1141

We are going to see that pattern again and again; something else to keep in mind is that if this mechanism seems at all familiar or these species that we have.1152

It is because we have actually seen this mechanism before; but we have seen it in the reverse direction; we have seen it in the reverse direction.1163

What do we call it when we take an alcohol and we lose water to go to an alkene?--that is the dehydration of an alcohol; how did we do that?1170

We protonated to make it a good leaving group; then it left; then we deprotonated the carbocation; same exact intermediates.1181

When we compare hydration for the dehydration, we will see that the hydration mechanism is the exact reverse of the dehydration; same number of steps.1193

We just saw this part when we want to hydrate a π bond, we use water and H₂SO₄ to go from an alkene to an alcohol.1202

How do we dehydrate?--what conditions do we do to get rid of water?--we use H₂SO₄ and heat; very similar conditions in that they both have a strong acid; they both need acid.1214

But when we have, in the presence of water, we do hydration; we add water; these reaction conditions are aqueous with some added acid.1230

Where these conditions--look we have H₂SO₄ and heat; these are strong concentrated acid... concentrated acid.1246

We have an absence of water; there is very little water; that is how we do a dehydration; of course the heat also helps in doing the elimination.1255

But take a look at your reaction conditions; they are very similar; but the key here is that we are adding water, we are removing water.1266

But this is truly an equilibrium; you can force a reaction in either way by your reaction conditions; also by having an excess of one or the other or removing your product as it is formed.1276

Let's try an example of this hydration reaction; we have an alkene; we have H₂O, H₂SO₄; this looks like we are going to add water.1289

We are going to add an H and an OH across the double bond; this looks like hydration conditions; here is where we have our alkene.1304

Notice that these are not going to be reactions of benzene π bonds; benzene π bonds are not alkene π bonds.1314

All the reactions we are going to be studying in this lecture are not relevant to benzene rings; just isolated double bonds, separated double bonds.1320

What we have to decide is which carbon gets the hydrogen, which gets the OH; we could think about Markovnikov's rule.1333

Markovnikov's rule tells us that the more stable... I'm sorry, the carbon with more hydrogens, gets the hydrogen; that doesn't help us here; each of these carbons has just a single hydrogen.1341

You can see very quickly that Markovnikov's rule is only going to get us so far; it won't help us predict every case.1354

We need to have a better understanding of the mechanism and the foundation underneath Markovnikov's rule.1360

That foundation is the notion that in our mechanism we want to have the most stable carbocation intermediate.1367

In our first step, we are going to protonate this π bond; that will give us a carbocation either here or here; where is the better carbocation?1374

How would you describe this carbocation in this position?--let's imagine the carbocation here; this is a secondary carbon; it would be a secondary carbocation; that is not a bad one.1384

How about if we had the carbocation in this position?--it is also secondary; again we are seeing that we have a very similar condition, very similar situation.1395

But this not just secondary; we have this benzene ring here; what do we call the position that is next to a benzene ring?--we describe that as benzylic.1405

A carbocation in this position would be secondary and benzylic; that is better than just being secondary; we want to think about the stable carbocation.1419

We want to know the mechanism and think about the mechanism, what is important to it; you might want to think: are sterics important in this case?1436

Sterics have nothing to do with carbocation formation; that is if we are trying to do a backside attack maybe.1445

But we really want to think about what is the governing thing in this situation; it is really about the carbocation stability.1452

What we want to form is the benzylic carbocation; how do we get to the benzylic carbocation?--what does that mean?1460

That means we protonate in the other position; the hydrogen goes here so that the carbocation goes in the position we want.1471

This is resonance stabilized; try it; that is a good exercise; can you show the resonance stabilization of this benzylic carbocation?1479

Once we know the carbocation that is going to be formed, now we can go to the product hopefully without doing a complete mechanism because where the carbocation was is where our OH group will be added.1494

Remember we are adding an H and an OH; in this case, we would want to put the OH on the benzylic carbon because that would be the more stable carbocation.1509

Notice I am not showing any stereochemistry here; we did form a new chiral center; but we just know that is going to be racemic; we could just draw them all as straight lines.1517

Addition of H₃O⁺ is one way to hydrate a π bond and add the components of water across the π bond.1529

But there is two other alternative hydration methods that we are going to consider; we will see some benefits of that down the road.1535

This first one is called oxymercuration-demercuration; as the name implies, it is a two-step process.1548

It is good to pay attention to the name of this because that is going to help us understand more about the mechanism and remember some of the details of this.1555

Oxymercuration-demercuration looks something like this; we start with an alkene; the first thing we do is we treat it with Hg(OAc)₂; that is called mercuric acetate.1563

OAc represents the acetate group; this is called mercuric acetate; and water; we also add water in this step; this adds a mercury group and an OH group.1580

A little reminder--the Ac group, every time we see Ac, it means we have COCH₃; we have a carbonyl and CH₃; this is like an ester group here.1596

It breaks the π bond and it adds a mercury and an OH; that is our first step; the second step, we treat this with NaBH₄; that is called sodium borohydride.1608

It replaces the mercury part; it replaces the mercury with a hydrogen; we call this a reduction reaction because we are introducing hydrogen; we are increasing the number of C-H bonds.1626

Sometimes actually this process is called oxymercuration reduction; some textbooks might have it described as that process because the second step is both.1640

It is a demercuration; we are getting rid of the mercury; because we are replacing it with a hydrogen, it is a reduction.1648

What happens here?--we look at our structure; this mercuric acetate group gets replaced and it is just a hydrogen.1654

What happens overall?--let's look at the big picture comparing our starting material to our product; overall we broke the π bond and we added two groups; we added an H and an OH.1662

This is a way of hydrating an alkene; how would you describe the regiochemistry of that addition?--the hydrogen added to the carbon; it already had a hydrogen.1675

That is going to be described as Markovnikov addition; this is an alternate way of doing Markovnikov addition; why do we need an alternate way because we already have H₃O⁺?1686

It turns out that this is going to be higher yielding; in the laboratory, this just works better than H₃O⁺.1700

There are a lot of other functional groups that react with H₃O⁺ or might not be stable to those reaction conditions; so these are alternative reaction conditions.1705

The mechanism does not involve a carbocation; we saw that H₃O⁺ does; that is going to be a benefit; it is not going to have any rearrangements in the reaction.1715

Experimentally this is a superior way to do the same thing--add water across the π bond with Markovnikov regiochemistry.1724

Let me just briefly show you this mechanism here for this first step, the oxymercuration step; we have our mercuric acetate.1732

What happens is the π bond attacks the mercury and kicks off one of the acetate groups; but then the mercury comes back and adds back into the π bond to form a three-membered ring.1743

That gives a positive charge on the mercury; I am showing you this because we are going to see some intermediates later in this chapter and in future chapters that look a lot like this.1767

After we get a more thorough explanation of those species, you can come back and maybe have a better understanding of this species.1779

But it turns out that this now gets attacked by the water; it chooses to attack on the carbon that is more substituted; again we are going to look at that in detail more later.1787

That opens up this three-membered ring to form the OH here and the HgOAc group here; the other reason I want to point this out is because instead of doing this reaction in water.1800

If we did this reaction in the presence of an alcohol as our solvent, then it could be the alcohol oxygen that opens up the mercurium ion.1816

We would end up forming an OR group; what would happen if we did step two here, NaBH₄, and replaced our mercury?1826

We could get a product where instead of adding the components of water across the π bond, we add the components of an alcohol across the π bond; this could be a way that we make some ethers.1838

Another possibility for adding water across the π bond, our second one, is called a hydroboration-oxidation; again with that big name, that tells us that it is another two-step process.1853

We start with an alkene; the first thing we do is we do a hydroboration; we add BH₃; BH₃ is in quotes because BH₃ is a very unstable molecule.1865

What we are going to use are various reagents that are the equivalent of BH₃; we could use B₂H₆; it is called diborane; this dimerizes; that is a little more stable.1876

Or we could have BH₃-THF where the BH₃ is complexed with a solvent molecule, the THF molecule.1890

Pretty much any boron reagent that you might see is all going to be doing the same reaction--that is a hydroboration; we are going to be adding a hydrogen and a boron across the π bond.1900

Our second step is an oxidation step; that means we are adding some oxidizing agent; in this case, what we typically use is hydrogen peroxide.1915

In our oxidation, we are going to replace the boron group with an OH; our boron used to be here; now there is an OH in that place; hydroboration-oxidation.1924

What did we do overall?--we started with a π bond; now we have an alcohol; what did we add across the π bond?--we added water across the π bond; we added the H and the OH.1948

How would you describe the regiochemistry of the H and the OH?--the hydrogen went to this carbon; the hydrogen went to the carbon with less hydrogens; would you describe that as Markovnikov addition?1960

No, that would expect the hydrogen to add to the end carbon; we describe this as anti-Markovnikov addition, anti-Markovnikov regiochemistry.1973

This is interesting because now it gives us a method we can use to get the opposite regiochemistry of the other methods.1994

Let's take a look at the mechanism to see again a little bit of this mechanism so we understand where it goes and some details of the product that unfolds.2003

This first step, the hydroboration, is our first step; we add a hydrogen and a boron; this is a one-step mechanism; we describe it as a concerted reaction.2016

We are going to form a bond from the hydrogen to one end of the double bond and the boron to the other end of the double bond; both of these bonds are going to get at it simultaneously.2025

Our mechanism--our alkene acts as a nucleophile; that is electron rich; this boron acts as an electrophile; boron, again we talked about why it is so unstable.2037

It is because a BH₃ with just three σ bonds has no lone pairs; it is missing an octet; it is highly electrophilic, highly reactive.2049

What happens is the π bond attacks the boron and forms a bond here; but at the same time, the boron gives back one of its hydrogens and adds that back into the double bond.2059

We have formed these two... these two arrows are moving these four electrons all at one time; our product then is we have a boron attached and we have a hydrogen attached.2072

This is going to explain two things; this is going to explain the regiochemistry; this is a regioselective reaction where the boron goes to the end carbon and the hydrogen goes to the middle carbon.2086

Remember we call this anti-Markovnikov; there is a few explanations for this; one of them is just simply about sterics.2097

Because the boron is bigger than the hydrogen, it prefers to go to the less hindered carbon.2106

There is also an electronic argument saying that again, because this is the electrophile, as this bond breaks it is better to have some positive charge building up on this carbon.2114

There is more than just a sterics argument here; but we can keep it simple by pointing to that; it turns out this reaction is also stereoselective.2125

We describe it as being syn addition; that is because, since the boron and the hydrogen both added at the same time, they had to add to the same face of the alkene.2136

We call that syn addition when they come from the same side; they add at the same time to the same face; this is called syn addition.2148

Let's see an example then; if we wanted to do a hydroboration-oxidation on this alkene; the first thing we are going to do is BH₃-THF; how do I interpret that reagent?2160

I know I am going to add a boron and a hydrogen; that is what all of our boron reagents are going to do; they are going to deliver a boron and a hydrogen.2174

We are going to break the π bond; we are going to add a group to each carbon; now we have to decide where to add what.2183

Is it going to follow Markovnikov or anti-Markovnikov?--it is going to be anti-Markovnikov; the hydrogen is going to go the carbon with less hydrogens.2191

That is because this big boron, remember that is going to be larger; sterics plays a role here; the boron is going to prefer to go here; that is the regiochemistry.2201

Remember we are thinking about that; how about the stereochemistry?--because this is a concerted mechanism, they have to add to the same face.2213

When we draw our product here, the way we can show them adding to the same face is, for example, we can draw them as both wedges.2223

If I add the hydrogen to the top face, what happens to this methyl group?--this CH₃ can no longer be in the plane anymore; this is a tetrahedral carbon.2235

That methyl group gets pushed back, gets pushed down; it can add to the top face or it can add to the bottom face; it is your choice whether you want to show them both as wedges or both as dashes.2242

Let's take a look at that other product where they are both dashes and see if they are the same thing.2254

Sometimes they are going to be the same thing; sometimes they are not; we will have to evaluate each case.2260

If they added to the bottom face, would this be the same intermediate product or are they different?--it looks like they are mirror images of each other; it looks like they are enantiomers.2265

We could either draw both of them or we could just say +enantiomer when we see that we have made a chiral product.2279

This is because this is chiral and we came from an achiral starting material, we know we can't just form this single chiral molecule; we always have to form its enantiomer as well.2290

We can either just say +enantiomer or we have to draw them both; drawing both usually takes a little too much time; there is our hydroboration reaction--stereoselective, regioselective.2300

Our next step is oxidation; what we need to know is that this oxidation step retains the stereochemistry.2312

I am not going to go through the mechanism for this oxidation on how it is that the boron gets replaced by the oxygen; but you could think of it as just inserting itself into the carbon-boron bond.2324

If it was a wedge, the oxygen will end up as a wedge; if it was a dash, it will end up as a dash; we will draw that for our product here.2334

What used to be a BH₂ as a wedge is now an OH as a wedge; the other carbon stays the same; and +enentiomer of course.2344

If you want to see what that enantiomer looks like, that is where both of our groups have been added as dashes; an OH here and a hydrogen here.2358

The methyl group must be pushed into the wedged position, must be pushed up, as the hydrogen added from the bottom face.2374

A couple things to point out--a racemate is formed; as always now we have to be very careful that is a right +enantiomer.2381

Because we have added dashes and wedges to our product and we have drawn one specific enantiomer, now we have to point out that the other enantiomer is also present.2389

We are going to keep that in mind and see lots of those examples in this unit; it is going to be anti-Markovnikov regiochemistry.2399

The hydrogen went to the more substituted carbon, the one with less hydrogens; and it needs to reflect syn addition of H and OH.2409

With the hydroboration-oxidation, it is one of the most complicated reactions we will be studying for alkenes.2421

Because not only does it have complicated reagents that we have to take a look at--this BH₃-THF and this H₂O₂.2428

But we also need to know what does it add, what is the stereochemistry, and what is the regiochemistry?2435

Here is an example; it is coming back to just what I pointed out; these are the questions we need to ask when it comes time to predicting a product.2446

We have to interpret the reaction conditions and decide what the groups are adding; then we have to decide where are they going and what is the stereochemistry.2453

Where are we adding it is another way of asking what is the regiochemistry; then what is the stereochemistry--dashes and wedges.2461

Let's take a look at this two-step process; we are almost always going to be seeing those two steps back to back.2472

We are not so concerned about that intermediate product, what it looks like; we want to get in the habit of being able to draw a final product.2478

I see a boron here; this looks to me like hydroboration-oxidation; hydroboration-oxidation; we are going to be adding an H and an OH; this is one of our hydration reactions.2486

Here is our two... we are going to break the π bond; we are going to add a group to each carbon, going to add an H and an OH.2501

Where should the hydrogen go?--because hydroboration-oxidation is anti-Markovnikov, we want the hydrogen to go to the carbon with less hydrogens.2508

This carbon has one hydrogen; this carbon has none; that is a good place to do it; again if you draw that exaggerated boron here, then that helps you remember that you want to avoid some sterics with it.2518

What is the stereochemistry?--we need to do syn addition; that means we can add them from the same face, either from the top or the bottom; that is your choice.2530

But we need to show that in our product; we can do that over here... where the boron adds, in our final product, we are not going to have a boron there; what are we going to have?2543

We are going to have the OH; where the boron goes is where the OH ends up; we can draw that as a wedge right here.2559

The tricky part I think is right here--the hydrogen; where do we draw that hydrogen as a wedge?--it is not correct to draw it down here because this is not what a tetrahedral carbon looks like.2566

Let's try that again; this is our OH as a wedge; where do we put a wedge on this carbon?--the dash and wedge are pointing up in this direction.2578

Remember the H and the OH both came from the top face; but because we have a zigzag structure, they are coming in from slightly different directions; that is what our product should look like.2590

What happens to this methyl group?--he is no longer in the plane; he is being pushed back because we have a dash; so we must have a dash in that position.2599

Being able to predict these products, we are going to draw on all that experience we have in drawing 3D structures and understanding tetrahedral carbons.2608

And being able to draw something that represents the three-dimensional shape; let's double check; we added an H and an OH.2616

We added it with anti-Markovnikov regiochemistry; we added them syn; they are both wedges; very nice; is that my only product that I formed?2625

That is going to be hard to do because this is a chiral product and I started with an achiral starting material; what is missing?2635

We have to write +enantiomer; or we could say that it is racemic; we could say racemic if we want; we could either say +enantiomer or we could say racemic.2641

Some instructors do something like they say +/- indicating that we are getting a mixture of both isomers, both enantiomers; so we will have no optical rotation.2651

Either one of these is fine; but you have to do something; it is improper to draw a single chiral carbon as a product.2664

A lot of textbooks get a little sloppy with this; try not to pick up on those bad habits.2671

Again why do we need to learn about all these different hydration methods?--it really comes to being synthetically useful; there is a very good reason that we need to be able to use these.2679

One thing is that we can add water with either Markovnikov or anti-Markovnikov regiochemistry.2693

These are wonderful techniques that are complementary to each other and let us go in different directions synthetically.2698

These alternative methods have no carbocation in the mechanism; therefore we have no rearrangements; carbocations can do all sorts of weird things.2705

Here is a perfect example; if we wanted to hydrate this alkene and we just used plain old H₃O⁺, you would not get the alcohol that you would normally expect.2712

Because it will rearrange; try this; try and do this; try protonating, looking at the carbocation, and see how we could rearrange.2728

What we have is all these methyl groups over here right next to where you would have the carbocation; one of those methyl groups is going to shift over to give a more stable carbocation.2737

This is the case where if you wanted to hydrate this π bond with Markovnikov regiochemistry.2746

Instead of using H₃O⁺, you would do oxymercuration-demercuration or oxymercuration-reduction.2751

In doing that, because there is no carbocation, there is no possibility for rearrangement, you would get exactly the product you expect.2757

Hydrogen goes to the end carbon; OH goes to the middle carbon; perfect.2766

Or if you took the same alkene and you did hydroboration-oxidation, then you would break the π bond and you would add an H and an OH.2772

But in this case, you would now do it with anti-Markovnikov regiochemistry, where the hydrogen goes to the inside carbon and the OH goes to the outside carbon.2782

Why did I not... we know this is syn addition of the H and the OH; why did I not show any dashes and wedges in this particular case?2794

I didn't show any dashes and wedges because neither of these is a chiral center; so there is only one stereoisomer that is possible here.2803

It is not a chiral molecule; there are no chiral centers; in that case, there is no stereochemistry that needs to be shown.2812

It is only when both carbons involved are going to be forming its new chiral centers that we need to show the dashes and wedges and the configuration of those chiral centers.2820

If we are not dealing with chiral centers, then stereochemistry is not so relevant; how do we keep all of these reagents straight in our minds?2831

Because this is just the start and there is a lot more reagents to come even in the reactions of alkenes.2841

Let me give you just a little suggestion on how to use flashcards; I think flashcards are very useful for studying and learning organic chemistry.2846

Because it is a great way not only to organize your information, but to store it somewhere that is handy that you can refer to very quickly.2855

You could test yourself on the various information we are going to have for these chemical reactions; we are going to be seeing many many many reactions of this nature.2861

Where we start with some kind of starting material and we react it with some sort of reagent or set of reagents and it gives one or more products; we have these three possibilities.2871

One way that you can test the reaction is you can have on one side your starting material and your reagent; in other words, set up the reaction and make it like a predict-the-product.2885

Then of course, on the back, you have the answer; that is a way that you can test yourself to see whether or not you can do that problem; but that is not the only way you can or should test yourself.2900

Another way that you can test yourself is to say, if I had this starting material and I wanted to go to this product, how could I do that?--that is a transform-type problem.2911

Those are a little more challenging because now you have to plan what reagents and what reaction conditions are going to be necessary to do the mechanism that you want to employ.2922

But that is still not the only possibility; you can also say, given the following reagents and given this is the product, what starting material did we start with in order to go there?2932

This kind of problem is going to be very useful as a way to test yourself and to organize, to remember, as these number of reagents and reactions pile up and accumulate.2947

You need to remember which functional group is associated with which reagent; we just learned about hydroboration; we saw B₂H₆.2960

You need to be able to associate B₂H₆ as a reaction of alkenes going to alcohols; seeing it from all those directions is going to be very useful.2970

I think managing information and developing those good study habits is one of the most important things that you get out of organic chemistry.2980

This lecture is really good place to start thinking about those things; what else can we add to double bonds?--we saw addition of HBr; we saw addition of water.2987

How about the addition of a halogen like Br₂?--this is called the bromination reaction; we could also do chlorination.2999

It is described as anti-addition; that is because when we react an alkene with Br₂, it is common to see carbon tetrachloride as a reaction condition.3006

That is simply a solvent that is used in this reaction; try not to let that distract you too much; it is the bromine that is doing the chemistry here.3019

What happens as you might expect is you break the π bond; you add a group to each carbon; for Br₂, the two groups you are going to be adding is a bromine and a bromine.3026

What happens is the bromines end up adding to give the trans product only; we get one up and one down in the case of a cyclic compound; you can see that very nicely--trans product.3035

We call the mechanism anti-addition; we call the product trans; the opposite, let's see what is happening; we are not getting syn addition; we just saw an example of syn addition.3049

Syn addition is what we called it when the groups add to the same face, if they were both wedges for example; that would give cis bromines; that does not happen.3064

Just getting to know some of this terminology, some of this vocabulary--syn and anti describe the mechanism; cis and trans describe the products that result from those mechanisms.3081

I showed it trans; this bromine can be the wedge and this could be the dash or the top bromine can be the dash and this can be the wedge; aren't these the same product?3092

We are really going to have keep making these decisions about stereochemistry again and again; what do you think?3102

Is there any way you can flip one over or rotate it, do anything to get them to superimpose?--no, indeed; because this is a chiral compound, it has no plane of symmetry.3107

Then these in fact are enantiomers; we are going to get, as usual, we are going to get the racemate formed here.3122

You don't have to draw both; you could just draw one and say racemic or +enantiomer; just for argument sake, let's think about this syn addition product.3129

If I added these bromines to the same face, added them to the top let's say, would I also have to add them to the bottom?--would that be a unique product?3137

Is it right to say +enantiomer here?--actually not in this case because if I added the bromines to the bottom face, it would be the exact same product.3147

The difference here is because this is a meso compound, it is symmetrical; it is achiral; it does not have an enantiomer; we are really going to have to keep this in mind.3158

It is not enough to just go through as you are doing predict-the-product and say +enantiomer, +enantiomer, +enantiomer; you need to be thoughtful about that and think about it.3170

Back to bromination, what we observe is anti addition; we will end up with these trans bromines; we describe this as a stereospecific reaction; we get one specific stereochemistry.3178

If we want to understand why we get this anti addition, we are as usual going to have to look at the mechanism; let's do that.3190

The bromination mechanism starts with our two ingredients; we have an alkene and we have bromine.3201

What we can imagine happening is the alkene acting as a nucleophile, the bromine acting as an electrophile.3211

You could have the double bond attack one bromine and kick the other bromine off; let's see what that what that would give us.3221

That would attach a bromine to one carbon; that would form a carbocation on the other carbon.3232

Like protonating a double bond gives a carbocation, adding a Br⁺ to an alkene also can give a carbocation.3238

However this is not formed because this is less stable than another arrangement this can have.3249

Because this bromine has these lone pairs, those lone pairs can come back in and can instead form a three-membered ring.3260

This bromine has two bonds and still has two lone pairs; it is an unusual looking bromine; we are used to it just having one bond.3274

Let's think about the formal charge for this bromine; we have one, two, three, four, five, six; bromine wants seven; it is missing an electron; this is a Br⁺.3282

This is called a bromonium ion; that is actually the intermediate that we get in the bromination reaction; we don't form this carbocation.3295

Instead we add the bromine, the Br⁺, to both carbons of the double bond to form this bromonium ion.3309

It turns out that the reactivity for the bromonium ion involves each of these carbons; they are partially positive; that makes them electrophilic.3317

Those are electrophiles; those are electrophilic carbons; we will see what happens to those next.3328

This is an introduction to the bromonium ion; let's step back and look at the complete... when we want to start the bromination mechanism from the beginning again.3339

We have our alkene; we have our bromine; we know that every time we have bromine in the presence of an alkene, we are going to go to this bromonium ion.3348

I am actually drawing this as a wedge; they are both bonds pointing in the same direction; we know we are going to get this bromonium ion--how do we get there?3357

The mechanism involves both the double bond coming out and attacking the bromine and kicking off the bromine; like we thought was going to happen in the first place.3371

But at the same time, one of the lone pairs of this bromine adds back in to the second carbon of the double bond.3381

We end up forming two bonds to the double bond in one step of the mechanism; that is why they end up adding to the same face; that is why we are showing them both in this case as wedges.3388

This is an electrophile; do we have any nucleophiles around in the reaction?--sure, we just formed in this very first step, we formed Br^-.3399

This Br^- is going to attack the bromonium ion; it will attack either one of these carbons; we can show it attack this one.3414

If it attacks that carbon, what happens is it treats this carbon-bromine bond as a very good leaving group; it opens up that bromonium ion ring.3427

What does this mechanism look like?--have we ever seen a mechanism where a nucleophile attacks a carbon and kicks off a leaving group?3440

In a single step, attacks a carbon and kicks off a leaving group?--absolutely, this is the Sn2; we describe that as backside attack.3448

If my bromonium ion is a wedge, if it is sticking out at you, where does that second bromine have to come in?--it has to come in from the opposite side; it has to come in from the back.3459

If it attacks this carbon on the right, then that new bromine is going to be a wedge and that original bromine is still here now with its three lone pairs; it is back to being neutral.3468

We can call that path A if you want, product A; or we could have it attack the other carbon and do an Sn2 on that carbon, path B; that is how you can see we can get to our other enantiomer.3481

If we attack that carbon, then this bromine would be the one that is back and this one would be out; we get our 1:1 mixture of enantiomers.3500

It explains the backside attack is why we get the anti addition; because whatever face the first bromine added to to form the bromonium ion, the second bromine has to come in from the opposite face.3513

We always get anti addition, one up and one down; it is possible to add more than Br₂ in this fashion.3529

We can have other nucleophiles add to the bromonium ion in something that is known as forming a halohydrin.3540

The example here would be if we had an alkene reacting not just with bromine but bromine in water or maybe bromine in an alcohol; what happens in that case?3547

We form the bromonium ion; but instead of the Br^- attacking it, it is the water that is going to attack it.3561

The two groups we end up adding will be a Br and an OH instead of two Br's; this is called a bromohydrin; when you have a Br and an OH, it is called a bromohydrin or a halohydrin as a general term.3568

We are going to get trans stereochemistry again; it will still be anti addition.3593

Of course, we have to say +enantiomer because it was totally random but I chose to make the bromine a wedge and the OH a dash; the other one will also be formed.3599

Let's think about a mechanism; how do we combine these three ingredients to make this bromohydrin?3609

The first thing that is going to be happening is this bromine is going to be reacting with the alkene; like I said, anytime you see bromine and alkene, we know we are going to form the bromonium ion.3616

I think it really helps to draw the bromonium ion that we are expecting; a lot of times we draw it as a wedge; you could draw it as a dash if you want; but this is often how we see it.3628

I know this is the bromonium ion I am going to get; I am going to be adding a bromine, Br⁺, to one face of the alkene; that helps me draw my mechanism.3638

This π bond is going to attack one bromine, kick off the other bromine; then that first bromine takes its lone pair and adds back in to form the second bond.3647

You can see there is still two lone pairs left; one lone pair has gone to form this bond; the π bond has gone to form this bond; here is our bromonium ion.3658

We have an electrophile with a bromonium ion; we look around for a nucleophile; what nucleophiles do we see?--we have water as a nucleophile; and we have Br^- as a nucleophile.3669

Who do you think would be a better nucleophile?--I see a negative charge with this bromine; I think he would be a very good nucleophile.3684

However what we are looking at here is a matter of statistics because water is our solvent.3693

Water is surrounding all these molecules including the bromine atom; it is solvated; it is totally surrounded by water.3698

So statistically, as is this bromonium ion, statistically the nucleophile that is going to be most likely to attack and therefore leading to our major product is going to be the water instead.3711

Water does in fact, when you have water present, water will be the most significant nucleophile to attack; again Sn2 mechanism; we have our backside attack; it will come in as a dash then.3722

The bromine on this left-hand carbon is still here; he is still a wedge with three lone pairs now; he is neutral.3738

What do we have on the other carbon?--we have an oxygen bonded in the back position, in the dash position.3744

What else is it on this oxygen?--we still have our two hydrogens; and we still have one lone pair; and we have a + charge.3753

We formed a bromonium ion and we attacked the bromonium ion; is our mechanism done yet?3768

Once again, anytime we use water as a nucleophile, our mechanism isn't done until we get rid of that positive charge; how do we get rid of that positive charge?3774

We can deprotonate; you can use anything you would like; some books use Br^- to be nice and tidy.3782

But really again, because water is our solvent, water is surrounding everything, that is going to be the most prevalent base that we have; we can deprotonate; then we are done.3792

In our mechanism, we usually show the mechanism just to get one of the enantiomers, knowing that we would have the same mechanism with the opposite stereochemistry to get the other enantiomer.3807

We can alter our reaction conditions a little bit, put in a solvent that could be nucleophilic; this is a new nucleophile that we are adding in here.3819

Therefore when we form a bromonium ion, it is going to be that solvent that attacks to give a halohydrin.3828

Because we are adding two different groups across the double bond, we need to think about regiochemistry again.3837

In other words, which carbon gets the bromine?--which ones gets the OH?--the regiochemistry that is observed is shown here.3843

The major product formed is the one on the top here where the OH goes on the middle carbon and the bromine goes on the end carbon; the other product is not formed.3852

We need to explore this a little more to understand why this would be the major product that is formed.3868

In order to do that, we need to look at our bromonium ion intermediate; both of these products would come from the same bromonium ion.3874

Here is our bromonium ion; our only question is: when water attacks this bromonium ion, where does it prefer to go?--in order to answer this question, let's take a look at the transition state.3882

Let's imagine that water attacking here and starting to form a bond, starting to open up the bromonium ion ring; let's show those as partial bonds.3896

As that carbon that carbon-bromine bond stretches, the positive charge on the bromine is now being shared with the carbon.3906

We have a partial plus on the bromine and a partial plus on the carbon that is being attacked by the nucleophile.3912

Or we could have the nucleophile attack this carbon; our partial plus will be on the bromine and this carbon.3920

If you take a look at these two transition states and we try and decide which one is more favorable, which one is going to be a faster reaction again?3927

We take a look at this location of this partial positive charge; how would you describe the carbon bearing this partial positive charge?3937

It is attached to two other carbons; this is a secondary partial positive; in this case, on the end carbon, it is a primary partial positive.3946

Just like we know for carbocation stability, we are going to extend that same logic to our partial positives here.3957

We will acknowledge that the secondary partial positive is better; it is more stable and therefore it is a lower energy transition state.3964

When the water goes to attack this bromonium ion and has to decide between these two positions, it is actually going to attack the carbon that is more highly substituted.3984

Because this carbon, let me write this down here, it is more partial positive; it is the better electrophile; that means the nucleophile adds here.3995

Water does not add to this end carbon; it is drawn to the more substituted carbon because that has more of a partial positive.4012

What is interesting about this is our mechanism is an Sn2; it is attacking this carbon and breaking open the ring.4022

But it is an Sn2; but it is going to the more substituted, more sterically crowded site; it is not just a simple Sn2; we describe this as an Sn2 with some Sn1 character.4031

The reason it has some Sn1 character is because of the positive charge; because we do not have just a neutral electrophile being attacked, it is not a plain old Sn2.4047

Because there is a positive charge on the leaving group, electronics beat the sterics; electronics win over the sterics.4061

Even though it is more sterically crowded, the partial positive character, the electron deficient character, is significant enough that that is where the nucleophile goes in spite of increased the sterics.4073

That explains our regiochemistry then; the nucleophile goes to the more substituted carbon; we can't really describe this as Markovnikov or anti-Markovnikov because we are not adding a hydrogen.4086

But this is similar to Markovnikov's rule that we have seen; although we are not adding an H⁺ in our mechanism, we are effectively adding a Br⁺.4099

Just like the H⁺ prefers to go to the carbon with more hydrogens, the Br⁺ prefers to ultimately end up there as well.4117

Br⁺ goes to carbon with more hydrogens just like H⁺ did.4125

Hopefully you can see, you can tie it in a little bit with the addition of HBr or the addition of water; those mechanisms had carbocation intermediates; there we were looking at competing carbocations.4136

Here, because both of these have the same bromonium ion intermediate, there is only one intermediate; what we are looking at is the path from that intermediate to the product.4148

Who is going through the lower energy transition state--therefore that is the faster reaction; slightly different explanations; but still they have some similarities.4157

Let's try and predict the product here; how about if we reacted this alkene with bromine and methanol as our reaction conditions, Br₂ and CH₃OH?4169

Again what are the questions we need to ask?--what are we adding?--if all we had was Br₂, then we add a bromine and a bromine.4182

But because we have a solvent that can be a nucleophile, that can be a nucleophile, it will be a nucleophile.4194

The two groups you are adding are going to be Br and... what do we get if CH₃OH was our nucleophile?--would we add the OH group?--no, we get an OH if we use water.4202

We get the OCH₃ group when we have an alcohol; we add the OR group; in this case, with methanol, we would get the methoxy group.4215

Where are we adding it?--that is our regiochemistry; that is our regiochemistry; we are breaking the π bond; we are adding a group to each carbon.4226

Which one gets the bromine?--which gets the OCH₃?--we need to consider that; we need to think about the stereochemistry.4237

What is the relative stereochemistry of those two things?--for that, we need to understand the mechanism.4243

We need to understand the mechanism; that is going to tell us the relationship of those two; let's look at some choices for our products and see if we can decide.4250

We said we are going to add a Br and an OCH₃ to these two positions; let's take a look at those choices; all these are chiral so they all should say +enantiomer.4263

Assuming these are all racemic, which one has the proper combination of the regiochemistry and the stereochemistry?--first of all, the stereochemistry; let's think about our mechanism.4276

We go through a bromonium ion; the bromine adds to one face; then our nucleophile, our methanol, comes in from the opposite face; remember it is anti addition.4290

Anytime we have bromonium ion, we have to get anti addition; which are the ones that have anti addition?--here our OCH₃ and our bromine added to the same face; that is syn addition.4303

Here the bromine and the OCH₃ added to the same face; that is also syn addition; I know those answers are wrong.4316

But here the bromine is a dash and the OCH₃ is a wedge; that is good; here same thing--bromine is a wedge and the OCH₃ is a dash; perfect.4324

These are both anti addition; they have the stereochemistry right; how about the regiochemistry?4334

When we make this bromine ion... let's imagine that; let's imagine that bromonium ion; let's do a little work here to think about that.4340

The methanol is coming in and attacking; where is it going to go?--is it going to be governed by sterics?--no, it is going to be governed by the positive charge in this case.4350

Because this carbon is tertiary, it has a higher partial positive; so the nucleophile goes to the more substituted carbon of the bromonium ion.4362

Which product is that?--it looks like D is our best product because it has the proper regiochemistry as well; the methoxy went to the carbon that was tertiary.4373

Interesting again, an Sn2 happening on a tertiary center; this is not an ordinary Sn2.4385

It is definitely because of that charge that we are seeing that Sn1 character and we are allowing this to happen; very good.4390

Another reaction we can have for alkenes is one called catalytic hydrogenation; as the name implies, hydrogenation, we are adding hydrogen across the double bond.4402

We will break the π bond and we add two hydrogens; what you can see here from this example is that the hydrogens have both added from the top face.4413

We describe that as syn addition; we describe the hydrogens in the product as being cis to one another.4424

The reaction is a catalyst; that is why we usually call it catalytic hydrogenation; we will talk about catalysts in just a second.4432

It is considered a reduction reaction; that is because we are increasing the number of C-H bonds; we are increasing the number of C-H bonds; that is going to be a reduction.4438

Overall it is an exothermic reaction; like most of the addition reactions we are seeing for alkenes, this is exothermic; it is releasing energy.4453

That is because overall we are breaking a weak π bond and forming a stronger σ bond.4461

All these addition reactions are going to be favorable because overall the net change is breaking of a π bond and forming stronger σ bonds; that is usually good news.4475

Let's talk about the catalyst a little bit; what is going on here?--we have some kind of metal catalyst.4490

It is going to provide a surface upon which this reaction can happen; it is going to speed up the reaction.4495

What happens is the metal adsorbs the hydrogen gas; the metal also grabs onto the alkene; then the metal delivers the hydrogens to that alkene; in doing so, it delivers them to the same face.4502

It delivers them to just a single face; that is why we get syn addition; the hydrogens end up on the same face of whichever face of the alkene it goes to.4519

Our catalyst can be described as either heterogenous catalyst or homogenous catalyst.4530

The heterogenous catalyst, the ones we see most often--you will see palladium nine times out of ten as the complete reaction conditions for catalytic hydrogenation.4535

These are metals that are added into the mixture; they don't dissolve in the solvent; so we just stir them very rapidly or we shake the bottle maybe.4546

Because they don't dissolve, we need some way to mix them in with the gas and with the alkenes that is dissolved in the solution.4556

At the end of the reaction, you just filter the catalyst off and you can isolate your product.4564

The other class of compounds are called homogenous catalysts; they are called homogenous because the reaction mixture in these cases are homogenous.4569

They are a solution; they don't have any solids in there because they do dissolve in organic solvents.4577

That is because these are organometallic species; they have both organic components attached to the metal.4583

We have a variety of transition metals like rhodium or ruthenium or iridium; they have various organic ligands attached to them; this guy is called rhodium tris(triphenylphosphine) chloride.4590

Each of these phosphorus has three phenyl groups, has three benzene rings attached to it; then there is three of those coordinated around the rhodium.4607

A lot of benzene rings in there; that makes it organic, soluble; so we can use organic solvents; they are called homogenous that way.4614

By varying the ligands that we have here, we can have a huge variety of catalysts; you can do some cool things with that; you can have some increased selectivity.4626

For example, if you have a diene, if you have two alkenes in a structure, you could use this catalyst to just react with the less hindered alkene; it would be sensitive to sterics in that case.4634

You can't really do that with catalytic hydrogenation, with just ordinary catalysts; that just reduces every π bond it sees.4651

But we can do some more interesting things with the organometallic catalyst; we might see some of those too.4658

The heat of hydrogenation of this reaction is going to be useful to us; it can tell us something about the stability of the alkene starting material that we had.4667

For example, if we take a look at this series of alkenes, 1-butene, cis-2-butene, and trans-2-butene, what is interesting about this series of alkenes?4679

If we take any of these alkenes and you react it with hydrogen and palladium, you will break the π bond; you will add hydrogen to each carbon.4690

The product you are going to get is the same; in every case, you are going to get butane.4701

This is an interesting series of compounds to study because they all give the same product with the same energy at the end of the reaction.4706

We can measure the heat of that reaction; we see it here given at kcals per mole: -30, -28.6, -27.6; we see a difference in the energy given off.4718

This is going to be related to the stability of the alkenes; we already know something about the stability of alkenes; which of these do you think is going to be the least stable?4733

We know that the less substituted an alkene is the less stable it is; so we know that the terminal mono-substituted is going to be the least stable.4742

These are both di-substituted; of the di-substituted, who is the most stable?--the trans is the most stable; the fact that these energies are going in this trend is a reflection of that.4753

We can show a diagram here to show how we could relate the heat of hydrogenation data.4768

If we take a look at our alkene A, we know it is higher in energy than alkene B which is higher in energy than alkene C.4777

But when they get hydrogenated, they all go to the same place; they all go to butane as the final product; they end up with the same final energy.4788

To go from A to butene is a longer path than to go to B or to go to C.4797

Because A starts out as the least stable, the highest energy, it gives off the most heat when it is hydrogenated; that is why this number is the biggest number.4806

Where C, because it is the most stable, it starts out with the lowest energy, it does not have as much heat to give off when the hydrogen takes place.4819

If given some heat of hydrogenation data, we should be able to convert that, interpret that into something about the stability of our starting alkenes, of the alkenes that we are comparing.4830

This is also useful when we are looking at dienes; if we had two double bonds, what are the relationships we can have for those two double bonds?4841

There is three possible relationships they can have; they can have one where they share a carbon; they are right on top of each other, right next to each other; we call these cumulated systems.4849

The p orbitals on that middle carbon, because it is sp hybridized, the p orbitals on that carbon are orthogonal.4863

The p orbitals for the first double bond are perpendicular or orthogonal to the p orbitals for the second double bond.4870

We can have double bonds that are unrelated to each other; we call those isolated π bonds because we have these π bonds here and these p orbitals and these p orbitals.4877

Because of this sp3 center in between, there is no relationship to them; or we could have conjugated π bonds.4887

That means that the p orbitals on the first double bond nicely align with the p orbitals on the second double bond.4894

When we have conjugated, it means we have a π and then a σ and then a π like that; that allows for overlap; it allows for resonance.4901

Can we see that in our heat of hydrogenation data?--absolutely; when we have these isolated π bonds, we get about 60kcals of energy taken off.4916

That is approximately the heat of hydrogenation for an alkene, for an ordinary alkene times two.4928

An alkene was somewhere around 30kcals; an isolated π bond, we get about 60kcals because we have twice as many π bonds to hydrogenate, to react.4938

But when we force those double bonds to be cumulated to each other, it actually increases the amount of energy given off for hydrogenation.4949

That tell us that that arrangement now is less stable; this is the least stable of all three of these.4957

It destabilizes; putting those two π bonds perpendicular to each other destabilizes the π bonds.4963

Where putting them in a conjugated relationship actually stabilizes it; this is the most stable; that is because of that resonance.4970

We can use the heat of hydrogenation data to get some evidence for which arrangement of double bonds is more or less stable; what I forgot to point out is these are all pentadienes.4979

Again if we have excess hydrogen and palladium, if you have as much hydrogen as you need, you will reduce every π bond there is and convert it to an alkane.4992

We will get just pentane as our final product; it is useful to study this series because they all go to the same product.5004

We can do the same cartoon again for our energy diagram; A--the cumulated π bonds are unstable, are less stable than an ordinary diene; having them conjugated is more stable than an ordinary diene.5012

They all go to pentane as their final product; A has the highest distance to go; B and C has the lowest distance to go.5027

This -54.1, the smaller the number, the smaller the magnitude of it... they are all going to be negative; remember this is always going to be an exothermic reaction.5037

But the smaller that magnitude tells us the lower in energy our alkene or diene started before we did the hydrogenation.5045

Let's see if we can predict a product here; if we reacted this alkene with H₂ and catalyst.5056

One thing I wanted to point out with this example is here we have a carbonyl; a carbonyl has a π bond but it is not a carbon-carbon double bond; so there is no reaction here.5066

Catalytic hydrogenation is a reaction for alkenes, carbon-carbon π bonds; we will see it for alkynes also when there is a triple bond.5078

We will leave that carbonyl there; we can draw our backbone; this is our carbon chain; we are adding hydrogens; do we know anything about the stereochemistry of that addition?5085

Regiochemistry is not an issue of course because we are adding a hydrogen and a hydrogen; there is no decision to make there; but how about the stereochemistry?5099

Let's think about that mechanism of the catalytic hydrogenation; the hydrogen is bound to the metal; it is going to be delivered to the same face of the alkene; we get syn addition.5106

How do we draw those hydrogens coming from the same face?--we could draw them both as wedges or we could draw them both as dashes.5117

If they are wedges, that means for the zigzag down, our wedge is pointing down; but for the zigzag up, our wedge is pointing up; here they are coming from the same face, all from the top.5125

What does that do to these two methyl groups?--that pushes them back into the plane; it is your choice on which carbon part of the chain you want to keep in the plane.5139

But you want to make sure that the other one gets pushed from being in the plane; it is no longer in the plane.5152

This is a chiral product; we want to make sure we say +enantiomer or we say racemic so that we know that having the hydrogens come from the bottom face would give a different product.5161

That product is also formed--the enantiomer; another way we can draw this... there is a lot of different ways to draw this.5171

It is important to keep that in mind when you compare your answer that you have worked on to the answer in the textbook or the solution manual.5179

That if theirs is slightly different, you need to be able to evaluate whether your answer is still right or not or if you have done something wrong.5185

Another possibility is we can say I want them to both come from the top face and pointing down here for example; we could do that; that would keep this first carbon the same.5191

But the second carbon, if the wedge is down here, then the bond that is in the plane is this methyl group is still in the plane.5209

It is the ethyl group that got pushed down to behind the board or behind the screen; a lot of different ways that you can draw this; but get into something that you are comfortable with.5217

And that again, with a lot of practice, you are used to drawing these chiral products and these new stereocenters that are going to be formed when we form new chiral centers in addition reactions.5230

The next group of reactions we are going to look at are called oxidation reactions; there is several ways that we can oxidize an alkene.5243

First though let's take a look at the word--oxidation; we know it comes hand in hand with reduction; we have learned about redox reactions in the past.5251

But in organic chemistry, we view it a little differently; it is a little easier to see than some of the redox reactions that you may have seen in general chemistry.5259

Let's just take a look at organic examples of oxidation and reduction reactions; let's just look at a one carbon species.5268

Every carbon has four bonds; we have this example with methane; all four bonds are to hydrogen; then we replace one of those bonds with an oxygen--this is methanol.5278

Then we replace two of those bonds with oxygen--this is formaldehyde; then we could have three bonds to oxygen--this is formic acid.5288

Then finally all four bonds can be to oxygen--this is carbon dioxide; we have gone from four bonds to hydrogen and one by one we are replacing a hydrogen bond with an oxygen bond.5299

This process in going from left to right is described as an oxidation; all these are oxidations.5313

For example, you could say that methanol could be oxidized to formaldehyde or even it could be oxidized to formic acid.5325

The way that we can identify it as an oxidation is we are increasing the number of C-O bonds while decreasing the number of C-H bonds.5336

Any process that trades a C-H bond for a C-O bond, we know is an oxidation; as a result, the oxidation number will increase.5348

But we don't have to calculate the oxidation state of each carbon in order to make that determination; it really can be done visually by looking at the oxygens.5362

This reverse process in going from right to left, these are all described as reductions; what happens in a reduction is we have an increase in the number of C-H bonds.5372

At the same time, there is a few different things that can happen; in this case, we are seeing a decrease in the number of C-O bonds; other bonds can be lost instead.5387

In general, as soon as we are seeing we are increasing in the number of C-H bonds, then we are going to see it as a reduction reaction; like catalytic hydrogenation, we call this reduction reaction.5396

As a result, by increasing those CHs, our oxidation number of the carbon will go down, will be reduced just like in a reduction reaction.5406

It is not changing the definition of a redox reaction; we are still having a change in our oxidation state; but we are just going to look at it more simply in this case.5414

There is going to be three... as you can see, there is going to be three types of oxidation reactions we are going to be studying for alkenes.5427

This first one will give as a product--we break the carbon-carbon double bond and we put in its place a three-membered ring with an oxygen in it; this is known as an epoxide.5433

It is called an epoxide when you have a three-membered ring with an oxygen; that is one possible oxidation.5448

If you react a double bond with KMnO₄, what happens is you break the π bond and you add an OH to both carbons; that forms a diol or a glycol.5455

It could be called a glycol if you have two OHs right next to each other or a vicinal diol.5467

Or you can react it with ozone, a reaction called ozonolysis; in that reaction, we actually break the carbon-carbon double bond and we replace it with carbon-oxygen double bonds.5474

We get these carbonyl products; we are going to get either ketones or aldehydes or some combinations of those depending on what your substitution patterns were on the original alkene.5489

Handful of new reagents; very different looking products; but do you see why these are all grouped together and all described as oxidation reactions?5505

In each case, you have lost carbon-carbon bonds; what have you replaced them with?--carbon-oxygen bonds, carbon-oxygen bonds, carbon-oxygen bonds.5514

You are increasing the content of oxygen in your structure; we call that an oxidation; let's look at our first one--the epoxidation reaction.5521

If we take an alkene like we said and we react it with this guy... we will define that in minute; it is called a peroxy acid.5532

The product we get is an epoxide; we are going to add both of those C-O bonds from the same face; we call that syn addition; I have drawn both of those bonds as wedges.5539

I added the oxygen from the top face; what if I added it from the bottom face?--is that another product?--do I say +enantiomer here?5556

Would it be a unique product if I drew both of those bonds as dashes?--or would it be just the same thing flipped over?--same thing flipped over.5564

Because this is a meso compound, it has no enantiomer; this is an achiral product; this would be a case where we just draw the stereochemistry as either wedges or dashes.5573

Then we are done; we don't have to say +enantiomer; it would be wrong to say +enantiomer.5588

Let's take a look at this formula RCO₃H; it looks very similar to a formula we may have seen before, RCO₂H.5593

RCO₂H, when you expand that, you have a carbonyl and an OH; that is a carboxylic acid.5600

That is an ordinary compound; we will study its reactions down the road; but as the name implies, it is just an acid; it is not an oxidizing agent.5610

But now the formula we are looking at is RCO₃H; just by looking at the formula and recognizing that it deviates from the carboxylic acid, it has extra oxygen in it.5619

That should tip you off to the fact that it must be an oxidizing agent; it must be something that causes oxidations.5632

Let's look at that structure; it still has a carbonyl; but now it has two oxygens; this is described as a peroxy acid.5641

This oxygen-oxygen bond is called a peroxide; this is very unstable; this is dying to get rid of an oxygen--very easy to cleave that bond.5657

That is what leads to its reactivity and the fact that it is possible to oxidize things.5672

Another example of a peroxide is this guy, H₂O₂; what is that called?--that is called hydrogen peroxide.5680

This is another oxidizing agent; this in fact would also work to do this epoxidation reaction; that would work just fine because you still have the peroxy group.5694

Look at the formula, H₂O₂; it has an extra oxygen compared to water which is nice and stable; this is now instead a strong oxidizing agent.5702

We might just draw a peroxide; we might draw a generic formula like RCO₃H to know that we are dealing with a peroxide or we could pick a specific peroxide reagent.5715

One that is very commonly used is this guy; he is called meta-chloro; meta describes the relationship between these two groups--that they are 1,3.5727

We have a chloro and we have a peroxybenzoic acid; without this double oxygen, this would be benzoic acid; but because it has the extra oxygen, we call it peroxybenzoic acid.5736

Luckily we can abbreviate this guy as mCPBA; mCPBA stands for meta-chloroperoxybenzoic acid; if we saw the formula, what part of the formula would clue you in that this is an oxidizing agent?5746

Right there--that extra oxygen; if we saw the name, we need to just recognize the name, very common to just use the acronym here; again another great flashcard to get to know mCPBA.5764

Just a general mechanism; for the most part, we are not going to be studying detailed mechanisms for these oxidation reactions because they are outside the scope of our organic chemistry mechanisms.5776

But I want to give you a little bit of an idea of what the mechanism is so it is not just this total magical black box thing.5786

What happens, by having these two oxygens attached to one another and having this acetate group here, is we have a leaving group attached to an oxygen.5793

Very unusual; that is again what makes it such a great oxidizing agent; what happens is the π bond of the alkene attacks the carbon and kicks out the leaving group.5804

But at the same time, this lone pair adds back in and forms the three-membered ring; we saw this similar mechanism for the oxymercuration, forming the mercurium ion.5817

We saw this with Br₂ to form the bromonium ion; in this case, we are seeing that this would be a way to form a three-membered ring with oxygen, not as an intermediate but as a final product.5828

This mechanism happens again; because it is concerted, because both of those bonds happen in a single step, that is why they must come from the same face; we get syn addition.5839

It is concerted so we have syn addition; let's look at another way that we can make an epoxide; what would happen if we took an alkene and we react it with bromine and water?5850

I know that bromine always makes that bromonium ion; would this add a bromine and a bromine?--or do we have to pay attention to this water here as our solvent?5870

We do in fact because that it going to win over the bromine, the Br^-, as what gets to attack the bromonium ion.5883

This is going to add a Br and an OH across the π bond; what about the stereochemistry of those two groups?5894

Again because we know the bromonium ion is the intermediate, that means if the bromine comes from the top face.5902

Then the water must come from the bottom face due to that backside attack for the Sn2 to open up the bromonium ion.5908

We can show... anti addition is what we call it; we are going to get the OH and the Br trans to each other; +enantiomer; it doesn't matter who you draw as a wedge and who you draw as a dash.5916

We saw this reaction as a way to make bromohydrins; what I want to show now is there is a nice application for bromohydrins.5934

If I took this structure and I treat it with sodium hydroxide, what can happen?--sodium hydroxide is a base; it is a nucleophile; you might think maybe we can do some substitution, elimination.5942

Anytime you see something that is a base, we want to look first to see is there any place that it can act as a base?--is there any proton that is unusually acidic?5958

Sure enough, we have an OH group here; in the presence of base, we are going to deprotonate some of that alcohol group to make an O^-.5967

What can happen here is now we have a great nucleophile, very strong nucleophile, in a structure where we also have a leaving group.5979

What can happen is that nucleophile can attack the carbon and kick off the leaving group; this is an intramolecular Sn2 if you want to call it that; it is a backside attack.5989

Those are possible; we have usually seen it just for five or six-membered rings; but three-membered rings are possible too.6015

Now we are seeing that if we did that here, we would get a product we call an epoxide.6021

What is nice about this synthetics scheme is it relies on some newer reactions we have seen with addition reactions to alkenes.6025

And some older reactions we have seen where we are doing a substitution of a good nucleophile on a carbon bearing a leaving group.6035

Why might I use this instead of just using mCPBA of some other oxidizing agent?--maybe I have some other functional groups on this molecule that are sensitive to oxidizing agents.6045

It might not be suitable to do an oxidation reaction; but I could still convert the alkene to an epoxide by doing electrophilic addition and then a base-promoted substitution reaction.6055

The next oxidation reaction we will study is dihydroxylation; dihydroxylation means that we are adding two hydroxyl groups to the alkene.6074

The way we do that is with either KMnO₄ as a reagent or OsO₄ as our reagent, either of these.6086

You can see the similarity in their structures; we have a metal with four oxygens on it.6092

In fact, our reaction conditions are usually more complicated; they have some other additives in there; some other oxidizing agents.6097

But these are the key ingredients we need to see that will tell us we are doing a dihydroxylation.6103

What we do is we break the π bond; we add an OH to each group; this turns out to also be a syn addition; we get syn addition of two OHs.6109

We call this a dihydroxlyation, a syn dihydroxylation, meaning the OHs end up cis to each other.6130

Any enantiomer in this case?--I drew the OHs both as wedges; how about if I drew them both as dashes, would that be the same compound or would that be a different one?6138

I don't see any plane of symmetry in this molecule; this is a chiral molecule; that means it must have an enantiomer; this is a case where we would want to say +enantiomer.6148

How do those two OHs get added in a syn fashion?--again just a little bit of a mechanism to give you some background on that, not a detailed mechanism.6160

But if you think of this potassium permanganate or the osmium tetraoxide, if you think of either of these reagents as some metal with four oxygens.6170

What is going to happen is that metal is going to take two of those oxygens and add them at the same time to the double bond.6179

It is going to be another concerted mechanism; we are going to form this bond and this bond all in one step.6185

Our mechanism looks something like this--we form a bond here; this π bond moves; this π bond moves; something like that.6192

Because they are adding in the same step, the oxygens have to add to the same face.6204

It is concerted; it goes to be a cyclic intermediate; when we follow our electrons around, we get something like this.6213

Again the exact details of the metal species is not important because it is going to vary whether it is manganese or osmium.6220

But the idea is we have added them in one step; that is why we must get syn addition; like the epoxidation, any concerted mechanism is going to give you that syn addition.6228

How does this get to this?--how do we end up with the two hydroxyl groups?--again we do some kind of workup; we cleave off the metal and we free those oxygens to be OHs.6245

But again carbon-oxygen bond is now set and the stereochemistry is fixed.6255

Finally let's talk about ozonolysis; this is our third oxidation reaction; as the name implies, it uses ozone; let's take a look at this ozonolysis.6264

We have two words here; the ozone tells us that we are using O₃ as our reagent; what does lysis mean?--what does it mean to lyse something or to cleave it?6277

It means we are breaking apart; we are going to be breaking or cleaving the molecule with ozone.6289

Just knowing the word ozonolysis, knowing the name of this reaction is going to help you predict the product properly.6299

But first let's talk about ozone just for a second because we have all heard of ozone before; it is interesting that we are using it as a reagent in the lab.6306

The question I want to ask is do you think that ozone is something that is good or something that is bad?--it kind of has a dual nature.6313

We know that there is an ozone layer and we are very concerned that there is a hole in the ozone layer; that let's us know that ozone is a good thing and the destruction of ozone is a bad thing.6323

That is true that ozone can be good; that is when it is in the stratosphere--that is the layer of the atmosphere, the outermost layer that is protecting us from outerspace.6336

That is because what the ozone does is it absorbs UV radiation; the sun is coming down with a ton of this ultraviolet radiation.6356

The ozone layer is doing a good job of absorbing a lot of that so only a fraction of that makes it down to earth.6370

That is a very good thing because this radiation is... meaning it is coming from sunlight I should say... that is a good thing because this UV radiation is damaging.6376

It can do damage to your skin; that is what gives you a sunburn; that is what causes the simplest cases; it is a sunburn.6391

But you can also damage DNA; if it penetrates, you can have damage there, cause skin cancer, and so on... put that here too.6402

The more UV we have coming down to earth, the worse shape we are; so the ozone is very good for that; it is protecting the earth; we don't want to put anything in there that might destroy the ozone.6413

But it is bad, ozone is something that is bad when it is in the atmosphere; in other words, if that is something that is surrounding the earth.6427

If it is something that I am interacting with, then ozone is not a good thing; it acts as an irritant to your mucus membranes and so on.6436

It is a smog component; it is one of the parts of smog that is tracked for health alerts and that sort of thing because it can cause irritation in your nose and your eyes and that sort of thing.6448

It also... we are seeing it is an oxidizing agent; it is a very strong oxidizing agent; if we have a lot of ozone around.6463

Then that can do a lot of damage to just materials that we have that might have double bonds in them; we are seeing that it reacts with double bonds; obviously ozone will do that rapidly.6469

Just the oxygen in the air is also an oxidizing agent; it could cause some of these damaging reactions; but ozone is even more reactive and will do it more rapidly.6481

For example, if you have ever taken a rubber band and tied up some bills or something and thrown it in the attic or thrown it in a box and stored it for a while in your closet.6490

What happens, when you go back to that box and you are down the road and you take that thing that was strapped with the rubber band, what happens to the rubber band sometimes?6499

Have you ever seen it just crumble or crack or fall apart or become stiff, something like that?6507

That is an example of oxidation that is happening to it where it is breaking down the structure of the rubber.6513

We can have oxidation reactions like rubber bands for example; or your tires on your car, they will crack; they will wear because of oxidation reactions; and so on.6521

What we have is we have a little ozonizer; we have a little instrument that we can bring in the lab and it generates a stream of ozone gas.6535

You stick that into your reaction; you can bubble some ozone through it and we can cause this oxidation reaction to occur.6542

Let's see what happens at the molecular level when it reacts to the double bond.6548

What is going to happen when it reacts with ozone... again not doing a detailed mechanism here but just giving you some idea... is we are going to completely cleave all bonds between the two carbons.6553

The two carbons, you still have two bonds connecting them; the ozone inserts three oxygens in between the carbons so the two carbons are no longer connected.6564

This is known as an ozonide intermediate; that is always formed as your first step in an ozonolysis; that is why there is always a second step in an ozonolysis reaction.6574

It is always followed up by some other set of reagents; these are typical ones we might see like zinc or dimethyl sulfide; these are reducing agents; this is known as a reductive workup.6583

What happens is we simply take these carbons; this carbon has two separate bonds to oxygen; after our workup, we are just going to put both those bonds to a single oxygen.6596

This carbon has two bonds to oxygen; after workup, we are going to have two bonds to a single oxygen; that is all.6610

With this reductive workup, there was a hydrogen here; there was a hydrogen that started out all along; that hydrogen is going to remain in our product.6618

Depending on the type of workup you have that might change; sometimes we can do an oxidative workup where we get rid of that hydrogen.6627

But this is a typical procedure for ozonolysis where we react step one with ozone and step two with some kind of reducing agent.6634

Our product, how do we predict our products?--we completely break the carbon-carbon double bond; we replace the carbon-carbon double bond with a carbonyl at each position.6644

We insert a C-O double bond called a carbonyl; in this case, we got one ketone and one aldehyde; we describe it as an aldehyde if it has a hydrogen attached to it.6654

But we will get a mixture of ketones and aldehydes depending on what alkene we started with.6665

Let's look at a few examples of predicting the products of ozonolysis; again looking at our reaction conditions and trying to decide what to expect here.6671

I see ozone; there is exactly one reaction we know that uses ozone; that is ozonolysis; right there in the name, we find out what we are doing to this molecule.6681

We are finding our double bond and we are completely breaking that double bond; when we go to draw the products then, we will now have two separate products.6695

We have broken our molecule into two pieces; we are going to have a carbonyl on the five-membered ring and we are going to have a carbonyl...6704

You can leave it upside down; you don't have to flip it over; it is fine to leave it in the orientation it had; we are going to have a carbonyl with this three carbon chain.6712

One great check that you can do when you are doing an ozonolysis is to take a look at your starting material; I had eight carbons in my starting material.6719

When I am done, I should still have eight carbons; I do--I broke it up into a five carbon ketone and a three carbon ketone.6729

Because ozonolysis will take every double bond in a structure and break it; if you have multiple double bonds you will end up with lots of fragments for your product mixture.6739

In fact ozonolysis is a nice way to study the structure of a compound--is by breaking all the double bonds and studying the fragments that you get.6752

You can help determine where the double bonds were located in your structure; those are some good exercises too--is looking at ozonolysis products and putting them back together.6761

How would we do that in this case?--that is a good problem; if you said that these are the ozonolysis products and you were asked what is the starting material that gave these products?6771

How would you approach that reaction?--you know that ozonolysis forms carbonyls from carbon-carbon double bonds; we are going to take those two carbonyls and we are going to reattach them.6786

We are going to reattach them as double bonds to each other; take a look for some of those problems; those are good exercises as well.6797

How about this one where we have a double bond within a ring?--how is that going to affect our product?--again ozonolysis and zinc reductive workup; looks pretty typical.6806

Here is the double bond that we are going to be breaking; but now because we are cleaving a ring, we are going to cause the ring to open up.6818

This is a really good idea to number our carbons so we don't get lost.6826

What I recommend actually as a good way to handle some of these problems is to redraw your starting material with your carbons numbered--one, two, three, four, five, six.6831

Then literally erasing the bond that you are breaking; that way I know I haven't lost any carbons or gained any carbons.6843

I am breaking that double bond; what am I replacing it with?--a carbonyl at carbon 1 and a carbonyl at carbon 6.6850

You don't have to straighten it out and draw it as a nice zigzag line; you can leave it in whatever conformation you want.6859

The only thing I will point out here though is because I made an aldehyde, it is okay to leave your line drawing like this.6866

But by convention, we usually draw in the hydrogen of an aldehyde; that is the only hydrogen attached to a carbon that we normally leave in a line drawing.6873

I am just going to feel a little more comfortable adding those hydrogens in because that looks a little better to me rather than having a carbonyl dangling at the end of the chain.6885

But that is optional; even though they weren't shown on carbons 1 and 6, of course there is a hydrogen here and a hydrogen here; those are still preserved with our reductive workup for the ozonolysis.6892

Finally let's look at some additions to alkenes that follow a completely different mechanism; that is a radical mechanism; let's think back to what we know about radical reactions.6908

The one reaction that we have seen before is the free radical halogenation of alkanes; we take an alkane and we react it with bromine and light for example; we would form an alkyl halide.6920

What we are doing is we are taking our alkane, we are plucking off a hydrogen, and we are replacing it with a bromine; we call that free radical halogenation.6933

Let's review the mechanism for that so we can look at a new radical mechanism; the very first step in any radical mechanism is some kind of initiation--something that forms radicals.6941

In this case, the reaction, the exposure of Br₂ to light, to hν--we represent it that way; it causes homolytic cleavage of that bromine-bromine bond where we split it.6952

One electron goes to each bromine; we are going to form a radical; we are going to form two of these; we are going to form two bromine atoms; that is our initiation.6964

Then what that bromine does is a hydrogen atom abstraction; it is going to look for a hydrogen on the alkane to remove.6975

It is going to remove the one that results in the most stable radical possible; I am going to go for one of these internal ones rather than one of these end ones.6985

What an atom abstraction looks like is you break the carbon-hydrogen bond; one electron pairs up with the radical; the other electron stays behind.6995

We just formed a molecule of HBr; in fact I forgot to write up here--that is the other product you get in free radical halogenation.7006

I usually am more concerned with the organic product; but you are also forming HBr here.7014

The other product we get is this secondary radical; what does this radical do?--same thing--atom abstraction; that is what radicals do.7020

In this case, it is going to go after bromine; it is going to take a molecule of bromine and it is going to do a bromine atom abstraction.7031

We are going to, just like we did before, break this bromine-bromine bond; one electron pairs up with the radical; the other electron stays behind.7041

That is how we get our product; that is our bromination product; the other thing we formed here was Br·.7052

What was nice about this radical reaction is it also reformed the Br· which can come back and do another hydrogen atom abstraction and so on and so on; we could say et cetera here.7061

We get this chain reaction mechanism that started with the initiation; that is because the majority of the steps that are taken in a radical mechanism are described as propagations steps.7072

We start with one radical; it reacts with some stable species; then as a result, we get a new radical; here we had a Br· reacting with an alkane; that gave an alkyl radical.7087

Then we had an alkyl radical reacting with Br₂ to give a bromine radical; it is propagating because one radical begets a new radical.7101

There is two general types of propagation steps we can have; in free radical halogenation, the only one we saw was this first one--for example, atom abstractions.7112

The radical comes after an atom from a stable molecule such as a hydrogen atom and it removes that hydrogen; we saw it can pluck off a hydrogen; it can pluck off a bromine from Br₂.7121

We are going to make a molecule here of a hydrogen attached to the radical; we are going to form a carbon radical in this case.7136

Free radical halogenation relies on only atom abstraction as our propagation steps.7145

But if you happen to have an alkene around, a carbon-carbon double bond, there is another propagation step that can happen; that is called addition to an alkene.7152

What does it look like?--a radical comes in and it is the π bond that breaks; one electron pairs up with the radical; the other electron stays behind.7160

Let's take a look at that product; what has happened now is this radical has bonded to the alkene and formed a radical where the π bond used to be.7173

We call this an addition to an alkene; since we are studying reactions of alkenes, let's take a look at some examples of radical additions to alkenes and see what those products look like.7183

The most common reaction to study in the first place is the addition of HBr; we have already seen... let's review the reaction we have already seen.7195

That is addition of HBr; just normally what happens is we break the π bond, we add an H and a Br, and we do it with Markovnikov regiochemistry.7209

This is our electrophilic addition; we saw that mechanism; that mechanism was protonation with our acid and then addition of Br^-.7218

But if we do this reaction in the presence of ROOR, which is the formula for some kind of peroxide, a different mechanism occurs which is a radical mechanism.7228

As a result of that mechanism, we get this product which has the bromine on the end carbon and the hydrogen on the middle carbon; we describe that as anti-Markovnikov.7242

It is possible to add HBr with Markovnikov regiochemistry or anti-Markovnikov regiochemistry; we control that by whether or not we have added in some peroxides.7252

Let's see how this mechanism works, the radical mechanism works, and why is it that we get the anti-Markovnikov product.7264

As usual, our reaction has to start somewhere; we have to initiate a radical; that is the job of the peroxide; this is a radical initiator.7272

It is this oxygen-oxygen bond that is so weak and just loves to cleave homolytically and form two radicals, alkoxy radicals in this case.7280

We have a source of radicals now; our reaction conditions are--we have an alkene and we have HBr.7295

What this radical is going to do is it is going to do a hydrogen atom abstraction on HBr and pluck off a hydrogen to give Br·.7302

Ultimately the result of this reaction of HBr with peroxides is we are forming bromine radical; we are forming some Br·.7312

How is that Br· going to interact with the alkene?--if we have an alkene around, it can add to that alkene; we have a four carbon alkene here.7325

We are going to break the π bond; one electron joins up with the radical; the other electron stays behind; the bromine adds and forms a radical.7338

This is the step at which the regiochemistry has been determined; the first group you add has to go to one carbon or the other; that is when you decide the regiochemistry.7356

Let's look at the choice we have to make; the bromine can add to the end carbon or the bromine can add to the middle carbon; that would give the radical over here.7364

Let's look at our two possible intermediates, our competing intermediates, and decide which one is better and wide.7375

Radical stability turns out has the same general order as carbocation stability; if we take a look at the secondary one and this primary one, we will see that the secondary is going to be favored.7385

Just like we have seen before, we are going to follow the path that gives the more stable radical intermediate; that is important.7401

It turns out because radicals are not quite as unstable as carbocations... this is not the only issue that is going on here.7421

It turns out that it is also nice that the Br· adds to the less hindered carbon.7428

For this radical reaction, actually steric hindrance does play a little bit more than it would for just the addition of HBr with our other electrophilic addition mechanism.7439

But either way, we can see that the bromine wants to add to the end carbon; we now have this radical; how do we get to our product?--we wanted to add HBr.7448

Let's come back and redraw our radical; what can that radical do to get to the final product?--we only have two choices; it can either do an atom abstraction or it can add to an alkene.7456

Because we have plenty of HBr around, it is going to choose to do an atom abstraction; it is going to pluck a hydrogen atom off of HBr; we have just formed our product.7471

What else do we generate in this step?--we just generated some more Br·; he can continue to come back and add to a new radical and so on.7489

Once again, these propagation steps go on and on and on; this is the way that we make our product--is we start with a radical and then we make a new radical and we keep going and keep going and keep going.7500

We saw that we have some techniques to add water across a π bond with Markovnikov or anti-Markovnikov regiochemistry; that is very important.7515

Now we see we can do the same thing with the addition of HBr; we can add it with Markovnikov or anti-Markovnikov by adding some peroxides to our reaction mixture.7523

Let's take a look at one more interesting reaction that has a radical mechanism involving alkenes; that is the idea of doing a polymerization starting with an alkene.7536

This molecule is called vinyl chloride because it has a chlorine attached to a vinyl group; it is attached directly to a double bond; this is described as a monomer because it is just a single unit.7550

If you added some kind of radical initiator, then we would expect that initiator to add into the alkene and form a radical.7562

Let's suppose that in these reaction conditions all we have is this monomer; all we have is a ton of vinyl chloride, just a tiny bit of radical initiator to get a radical mechanism going.7573

What product do we expect to have happen here, to observe here?--we have this radical; if this has plenty of vinyl chloride around, then this new radical is going to do another addition to an alkene.7586

Now we have two units of vinyl chloride here; we call this a dimer because we have two of them; if we still had vinyl chloride around, then this could continue to add, break the π bond.7600

We could get a trimer and a tetramer and so on; it can continue going until we get a polymer--many many many repeating units.7615

This is one of the ways that polymers can be built--is by counting on this radical chain reaction; we can see here that we have a repeating unit.7624

This is what we see; we see our vinyl chloride components at two carbons with a chlorine on there.7638

Depending on the monomer that we start with, we can get a different polymer each time; depending on what groups are attached, then we can get very different properties.7645

This is PVC--this is polyvinylchloride; it is used for water piping, for example, for sprinkler lines, that kind of thing is PVC piping.7655

But also a wide variety of fabrics and plastics and materials that we can get from this polymerization reaction.7665

It is also possible to do the same sort of polymerization reaction with an acid catalyst.7671

Because if we form a carbocation in the presence of lots of other alkenes, then that alkene can act as a nucleophile.7682

And so on, giving a new carbocation which then can add to another carbocation and so on.7702

That would be another way of doing a polymerization; but radical mechanisms are very well suited for that as well.7709

That sums it up for the many reactions we can have for alkenes; we will talk to you later; thanks for coming to Educator.com.7716