Organic Chemistry Ethers

Welcome back to Educator.0000

Next we are going to talk about a functional group called ethers.0002

We will discuss their physical properties; we will discuss how to synthesize them; then we will talk about the reactions they undergo.0004

An ether is defined as a functional group that has an oxygen with a carbon group on either side.0013

Because this has no OH group like an alcohol, it cannot undergo hydrogen bonding between ether molecules; that is going to be a very big difference.0020

What it is really going to be doing, because we have no hydrogen bonding, is we are going to see a decrease in boiling points of ethers when compared to alcohols.0029

However it is a polar molecule; when we think about the hybridization of that oxygen, it is tetrahedral; it is sp3 hybridized.0039

These are both polar bonds; there is a net dipole moment; so ethers are polar molecules; we will see that they make very good solvents for that reason.0050

Because of the oxygen that they contain, they can accept hydrogen bonds from water.0062

Although an ether cannot hydrogen bond with another ether molecule, it can be a hydrogen bond acceptor.0067

Meaning if there was another molecule with an O-H bond like water, then we can have a hydrogen bond forming between the partial negative oxygen and the partial positive hydrogen.0078

That means the interaction between water and an ether molecule is very good; we expect to have some water solubility as a result.0091

Let's take a look at some boiling points; again we have seen some of these molecules in the past when we were looking at alcohols.0098

I think it is still very useful to compare alcohols and ethers because they are both oxygen containing functional groups.0103

The simplest ether we could have here is dimethyl ether; you can see that there is very little attraction between dimethyl ether molecules .0110

So that the boiling point is -24 degrees C; that means at room temperature, this is a gas; the simplest ether is not a liquid.0119

It is only when we get to diethyl ether which has about the same molecular weight as pentane that we get to a liquid form.0128

Just like alkanes, you have to have a significant molecular weight before you have strong enough intermolecular attractions because of those van der Waals forces.0136

That you are going to be maintained as a liquid at room temperature; you can see these have about the same molecular weight and they have about the same boiling point.0144

Even though this is a slightly more polar molecule, you can see there is not a huge attraction between ether molecules.0154

Because it has about the same boiling point as if it didn't have an oxygen at all, just a plain old alkane.0161

However when we go from a diethyl ether to this one, this is called MTBE for short; we have a methyl over here; we have a tert-butyl over here.0168

That is a way that we can have a common name for an ether--is we simply list the two alkyl groups on either side of the ether.0180

Just like we call this diethyl ether, this is commonly known as methyl tert-butyl ether; you can see that we have a jump in our boiling point.0188

The reason for that now is we are increasing our molecular weight; as usual as we increase our molecular weight, we increase our boiling point.0199

But you can see that all these are so much lower boiling than something like ethanol even though this is a very small molecule.0207

The fact that it has an OH group means that it can undergo hydrogen bonding; that is really the number one thing you want to look for for things that affect the boiling point.0213

Because that is such a strong intermolecular attraction; it is going to have huge impacts on the boiling point temperature.0229

I mention MTBE because this is something... these letters are something you might see on a gas dispenser at a gas station; this is a fuel additive.0236

We look for oxygenated fuel additives to affect the efficiency of the burning and how cleanly it burns and whether or not it knocks and all those sorts of things.0254

There are a variety of things that go into gasoline mixtures other than just the plain old hydrocarbons--that is really all you need to do the combustion part.0262

But what we can see on the next slide when we look at the water solubility of these ethers, we find that because it can accept hydrogen bonding, we do have some water solubility.0273

Again comparing our pentane to our diethyl ether, we go from something that is completely nonpolar to something that is polar and can accept hydrogen bonds.0285

Suddenly we have a decent interaction with water; we do get some water solubility; about 8 grams of diethyl ether can dissolve in about 100 milliters of water.0298

I discussed this a little before when we looked at this molecule for looking at the alcohol boiling points.0312

Just a little note here that anytime we do an extractive workup in a reaction, our ether layer, when we say we are using ether in a reaction or ether as a solvent.0318

This is the ether we are talking about--is diethyl ether; this is the one we most commonly use; it is referred to as just plain ether.0328

When we do an extractive workup like in a sep funnel where we have both ether and water, we will note that the ether layer is wet, meaning it contains water.0337

It must be dried as part of your workup procedure in order to isolate your product from that organic layer, that ether layer.0348

We expect alcohols to be fairly soluble in water; but again depending on the length of our polar and nonpolar, we see increased solubility when we have a smaller nonpolar group.0359

This is because it has a large nonpolar region, that is going to decrease the water solubility; but a smaller alcohol like ethanol is going to be miscible.0376

We are back to showing MTBE; what is interesting about showing MTBE is that because it is an ether, it has some water solubility.0389

Even though it has this tert-butyl group, it is compact and won't disrupt the water molecules too much.0397

What is interesting about this is if you have a component in your gasoline that is water soluble, then if that gasoline were ever to spill onto the ground.0405

Or if it were to leak from a container, then this component is something that can make its way into the ground water because it is soluble in water.0415

The reason, MTBE, you might find some of that in the news or an issue of discussion is because MTBE is very easily detected by humans at very very low concentrations.0426

If you get some of this in your water supply, the water is going to taste a little funny or smell a little funny; of course that is not acceptable to consumers.0441

It is not that this is necessarily something that is going to be real bad for you.0448

But if it is something that gives your water a bad taste, then this is something that we need to find replacements for.0452

There is a lot research going into just the perfect kind of oxygenated structure that is going to improve our gasoline performance but not be something that has significant water solubility.0460

How would we synthesize an ether structure?--there are a few options that we have for us; one of the most common ways is something known as the Williamson ether synthesis.0475

Here is an example of it; if we take an alkoxide, in other words, we have an RO^-, an alkoxide, and we react it with an alkyl halide.0485

It looks like we have a perfect situation for a nucleophile, electron rich negative charge, coming together with an electrophile; this carbon bearing the leaving group is partially positive.0498

Electrophile and nucleophile coming together, they are going to react; what reaction mechanism would you expect to happen for these pair of compounds?0510

I would expect that my lone pair on my oxygen can attack that partially positive carbon, that electrophilic carbon, and kick off the leaving group.0518

It looks like backside attack; it looks like an Sn2 mechanism; that can happen quite nicely.0526

What product do we get?--our methoxy group has replaced our bromine; how would you describe the structure we just made?--we just made an ether.0532

This is one thing that is very useful to note--is that if you take an alkoxide and an alkyl halide, you can make an ether by doing an Sn2 mechanism.0548

That is very good to keep in mind; it is going to be very useful to us when we are trying to make a variety of ethers.0559

An example of this strategy would be to start with this alcohol; step one, we would react the alcohol with NaH, sodium hydride; that looks like we have H^-.0565

Where have we seen sodium hydride before?--this is very good base; this is a very good base; we have seen it as a way to convert an alcohol to an alkoxide.0576

In other words, this can be used to deprotonate the oxygen and form an alkoxide.0590

Once we have an alkoxide, now in step two, we could add a phenyl CH₂I; that is benzyl iodide; there is our electrophile.0598

If we react our alkoxide with our alkyl halide, we can do an Sn2; we have just created an ether.0609

This would be a way of going from an alcohol to an ether--make the alkoxide and react it with an alkyl halide.0619

Let's look at another example; how about if we had not just an alcohol but we had this structure that has both a chlorine, a leaving group and an alcohol on the same structure.0629

We react this with something that can act as a base; that can act as a base; one thing you might imagine is that hydroxide can also act as a nucleophile.0639

The hydroxide might come right out and want to attack the carbon and kick off the leaving group.0652

While that is a reaction that can happen, there is another reaction that is going to be happening faster; that is the acid-base reaction.0659

Acid-base reactions are the fastest reactions we can have; if there is a possibility of that, that is the first path we want to take.0666

In other words, what is going to happen is it is going to act as a base; it is going to deprotonate; this is not a strong enough base like sodium hydride to 100% deprotonate the alcohol.0674

But we will set up an equilibrium; we will form some of this O^- to generate some of this alkoxide.0686

What that does for us is that makes an excellent nucleophile; this oxygen is now negatively charged.0696

Guess what?--if we see in the same structure, we see a partially positive carbon, we see an electrophile.0704

Then we can have an intramolecular reaction take place if those two are properly separated, properly situated.0710

Let's see what size ring we would form; anytime we have an intramolecular reaction between a nucleophile and an electrophile, that means we are going to form a ring.0719

With this oxygen, we would be forming a one, two, three, four, five-membered ring; it turns out that is a great size ring to form; so we do expect this reaction to happen.0727

As usual, an Sn2 like this, a backside attack, kicking off a good leaving group is not going to be a reversible reaction.0739

Kicking off that better leaving group is going to be what drives this reaction in the forward direction.0749

Even though this deprotonation is reversible and maybe we will just form a very small concentration of this alkoxide, the fact that as soon as the alkoxide is formed.0754

It can do an Sn2 to give this product; then that is something that forces the reaction in the forward direction; it does make this a favorable reaction.0766

We can number our atoms one, two, three, four, five, so you can follow along to see how we just made that ring.0776

We can describe this as an intramolecular Sn2; really it is better to describe this as an intramolecular backside attack; but it is similar to our Sn2 mechanism.0785

When it is intramolecular, it is going to give cyclic ethers; when can we have such cyclization reactions?--remember we can have it when we have a very very small ring like a three-membered ring.0792

That is where our nucleophile and our electrophile are just so close to each other that they will react.0803

But then we jump to five or six-membered rings; those are best ones because those have very little ring strain, little to no ring strain.0809

It is going to be a very favorable transition state, very low energy, and will be a fast reaction.0818

When you consider that this intermolecular Sn2 could have taken place, what we will find is that the intramolecular reaction is always going to be favored.0825

If there is a possibility for an intramolecular reaction, that is going to be the favorable reaction; that will be our major product; that is simply a factor of entropy.0835

The fact that the nucleophile and the electrophile are tethered together, they are already connected, means that they no longer have to come together and collide randomly.0843

Because it is so well favored entropically, when we see an intramolecular reaction is an option, that is the one we should go for as the major product.0854

How might I be suspicious here?--look we have a single starting material here; it has two functional groups.0862

It has something that is potentially nucleophilic and something that is potentially electrophilic.0868

That is something that you would really want to tip off to--let's consider an intramolecular reaction.0872

When we have a haloalcohol and we treat it with any kind of base, we can get a cyclic ether being formed.0877

If we wanted to plan a synthesis, how do we go about that?0888

Knowing that we can do a Williamson ether synthesis, the disconnection that we would make to form an ether would be either one of these carbon-oxygen bonds.0893

We would disconnect on either side of the oxygen; we make a disconnection; remember we are asking what starting materials do I need in this retrosynthesis?0905

When we consider these two atoms that we are trying to bring together in the reaction, we are trying to react an oxygen with a carbon.0921

The oxygen of course is electron rich; that makes a good nucleophile; this was my nucleophile; this was my electrophile.0928

What we should go backwards to is to make this oxygen a good nucleophile, we use the alkoxide; to make this carbon a good electrophile, we use an alkyl halide.0937

An alkoxide plus an alkyl halide gives an ether; an alkoxide plus an alkyl halide gives an ether.0955

That means if we have an ether target molecule, one possible retrosynthesis would be go backwards to the corresponding alkoxide and alkyl halide.0961

Let's see a sample ether; because this is not a symmetrical ether, there is two possible disconnections; I could disconnect this carbon-oxygen bond; let's call that disconnection A.0978

We are going back to some kind of alkoxide and alkyl halide; the group on the right is obviously my oxygen containing group.0995

That would be my alkoxide; that would be my nucleophile; what would that nucleophile be reacting with?1006

It would be this one carbon alkyl halide, CH₃... let's pick any halogen; I think the iodide is the best choice because there is just the methyl; that is a liquid reagent.1014

This combination of alkoxide and alkyl halide would combine to give this ether target molecule.1026

But there is another disconnection; we can go on this side; as our synthesis formed, this is our new carbon-oxygen bond; let's consider that synthesis as well.1033

If we disconnected here then it would be the methoxy group that would come in as my nucleophile; that would be my alkoxide; what would it react with?1044

We need an alkyl halide; we need this three-carbon alkyl halide; again your choice--bromide, chlorine, iodine, any halogen that you want is fine.1056

There is two possible disconnections; are they both equally good?--in order to answer that question, we need to imagine doing the synthesis then and taking this reaction.1071

Taking isoproproxide and methyl iodide and combining the two, what reaction do you expect to have happen?--we know we want to do an Sn2 mechanism.1083

Is this a good Sn2 mechanism?--backside attack is very sensitive to steric hindrance; we want to make sure there is very little steric hindrance.1094

We look at the carbon bearing the leaving group; that is the carbon that needs to get approached by the nucleophile; this is a methyl leaving group.1103

Is that good for an Sn2?--it is the best fastest Sn2 we can have; this is a great Sn2; I would expect this to work very nicely to give the target molecule.1115

How about the second case?--again we want to do an Sn2; our alkoxide needs to attack the alkyl halide; how would you describe the carbon bearing the leaving group in synthesis B?1129

Here we have a secondary leaving group; clearly it has more steric hindrance; now we have these two methyl groups we need to get around.1141

Remember that an alkoxide is a nucleophile but it is also a very good base; it is a very strong base.1149

What we have with methoxide, just like we would have for hydroxide, we have a competition between the Sn2 and the E2.1159

If it acts as a base, in other words, instead of attacking the carbon bearing the leaving group.1167

If it attacked one of these β hydrogens and formed a π bond and kicked out the leaving group that would be the E2; in fact the E2 for secondary is major.1172

What would my product look like?--I would get a CH₂CHCH₃; I would get an elimination reaction instead.1185

Which is my better synthesis?--A is the better retrosynthesis because it leads to the better Sn2 mechanism, less sterics.1194

Anytime we come across an ether that is not symmetrical, we need to consider both possible disconnections; then choose the one that gives the alkyl halide that is less sterically hindered.1218

We have done our plan; we said this is the better plan up here as a way to make our target molecule; let's do the synthesis then because our goal was to synthesize this target molecule.1232

What we need is methyl iodide and propoxide, isopropoxide; where does isopropoxide come from do you think?--how could I make isopropoxide if that was not commercially available?1246

I think I would start with isopropanol; I would need to deprotonate that to make the propoxide, to make the alkoxide.1256

How do I deprotonate?--we need some kind of strong base; how about sodium hydroxide?--is that a nice strong base that would completely deprotonate my alcohol?1267

No, that is not strong enough because we can't make this RO^- from this HO^-; what would be a better base?1276

We saw sodium hydride being used to make alkoxide; remember we also saw sodium metal as an option to do a redox reaction; you can make an alkoxide that way.1283

We need some super strong base that is not going to be reversible; that makes my alkoxide.1294

What did I do with that alkoxide?--I am going to add the methyl iodide to do my Sn2; that is going to give me my target molecule.1300

So ether synthesis is a very nice illustration of how to do different retrosyntheses and evaluate your choices before carrying on.1315

Let's try another one; starting with propanol as the only source of carbon; what is propanol?--three carbons with an OH--as the only source of carbon, synthesize dipropyl ether.1326

Dipropyl ether means I have a propyl group on one side and a propyl group on the other side; one, two, three; this is dipropyl ether; that is my target molecule, TM.1339

It asks us to synthesize it, but it gives us a restriction on where the sources of carbon can come from; we have to consider that when we are doing our synthesis.1351

A synthesis problem really should start the same way every time; that is by doing a retrosynthesis, planning your synthesis first.1364

What starting materials do I need to make this target molecule?--I recognize that this is an ether; what starting materials could I use to make an ether?1373

What are the two ingredients--alkoxide plus an alkyl halide; I can disconnect this one on either side; it would give the same set of alkoxide and alkyl halide.1384

I need this propoxide as my alkoxide, in this case n-propoxide; my alkyl halide would be one, two, three carbons; bromine, chlorine, iodine, your choice.1395

The planning is very useful because it tells me where I have to go; I know that I need this alkoxide; I know that I need this alkyl halide.1413

If I had free range to a stockroom, maybe I could just go and ask for both of those; but since I know I have to start from propanol, I need to synthesize both of them.1421

Here is my synthesis; I will start with propanol; let's first make the alkyl halide; how do I go from propanol to bromopropane?1432

Propanol to bromopropane--it looks like a substitution reaction; this in fact is a conversion we have seen for the reactions of alcohols.1448

We saw a few different strategies for this; the most direct one would be to use one of the reagents that does it just as a one pot transformation, something like PBr₃.1457

Of course if you remembered SOCl₂ and you wanted to use that, thionyl chloride, then you could use that as well.1469

We just picked a halide here; whichever halide you want to use would be great.1473

This is another case where HBr, although tempting because we are probably more familiar with that reagent, HBr would probably not be a very good choice because we have this primary alcohol.1480

It is very likely that after we protonate the OH, it could leave, lose water along with rearrangement to give a secondary carbocation.1490

We could get a mixture; we could get some rearranged product here; we wouldn't necessarily get this as the only bromide.1500

So PBr₃ is a way to get this bromopropane; how do we get the alkoxide?--we need to start with to start alcohol again; we need to start with the propanol.1508

The way we go from the propanol to the propoxide is we need that strong base again, something like sodium hydride.1518

Your choice--sodium hydride, sodium metal, just pick one and go with it; a lot times there is going to be more than one choice for reagents.1527

Then we can combine these two and make our target molecule or we could just follow this one and use the bromopropane that we already showed how to make up here.1535

We do our Sn2 to do our target molecule; great Sn2 because it is a primary alkyl halide; we would expect this synthesis to work pretty well.1547

Here is one more; let's transform the following; we start with this alcohol; we go to this ether.1563

What is very tempting when you see a problem like this is a transform-type problem--is to start with your starting material and just move forward and say I have the OH.1572

How about if I just make the alkoxide with my sodium hydride?--I know how to make an ether; all I need to do is have an alkoxide.1585

Then I add in my alkyl halide; here it is a three carbon chain with another methyl here; chloride, bromide, iodide, my choice; add my alkoxide and my alkyl halide.1594

And I am done; what do you think?--is this synthesis going to work?--we have a good nucleophile; how would you describe the carbon bearing the leaving group?1608

It is a tertiary RX; how good is that Sn2?--impossible Sn2; no Sn2 on a tertiary center; this reaction would not work.1619

What would happen instead is you would form the alkene from the tert-butyl bromide; you would make the alkene because you would do an E2 instead of an Sn2.1630

I think the reason that students are most likely to make a mistake on a problem like this is because they forgot to plan.1646

If you are doing any kind of a synthesis problem or transformation problem, you need to start by making a plan and think about where you are going.1655

If you had made a plan, then you would look at this target molecule which is an ether and you would have asked how could I make this ether?1662

You would realize that there is two possible disconnections; the better disconnection is the one here because that is going to lead to a more sterically available, less sterically crowded alkyl halide.1669

If I put the leaving group on this carbon and I use the oxygen as part of my tert-butyl group, this combination of alkoxide and alkyl halide would work great.1691

This is a great Sn2; we have a benzylic primary alkyl halide, fantastic for the Sn2; even though this is bulky, there is no elimination that can take place.1704

This is going to be something that will work very well to make our target molecule.1720

What do we need to do with this benzyl alcohol then?--what we need to do is we need to convert it to benzyl chloride; we need to convert this to benzyl chloride.1725

How could we do that?--again SOCl₂ is our best choice because that is going to work more often than the others.1739

But in this case if we used HCl, that would work pretty well because there is no other rearrangements that can occur; this is the only product that you can form.1752

Remember sometimes we need the zinc chloride in here though as a catalyst; maybe we would just make the bromide instead of the chloride.1764

But yes, we can convert the alcohol to our alkyl halide; now we can add in the alkoxide; we needed tert-butoxide.1771

Potassium tert-butoxide is going to be the reagent that we use most commonly when we need that alkoxide; we would expect to form the target molecule very nicely.1781

Please do consider planning and thinking about a retrosynthesis because you might have a choice of which bonds to form in your reaction and which sets of nucleophiles and electrophiles that you can use.1794

If we take a look at epoxides, these are a special unique class of ethers--that is the cyclic ether when we have a three-membered ring; this is known as an epoxide.1807

We are going to talk specifically about the synthesis of epoxides; later we are going to see some reactions that are unique to expoxides.1819

If we wanted to make a three-membered ring ether, we could do it the same way we have seen our other ether formations--Williamson ether synthesis; but this would have to a intramolecular case.1827

The way we would get the ingredients in place, the nucleophile and the electrophile on the same carbon chain.1841

We could do that very nicely by starting with an alkene and reacting with bromine and water; what does this do to an alkene?1851

Bromine reacts with the alkene to form the bromonium ion; then water as our solvent is going to come in and open up that bromonium ion; this adds a bromine and an OH across the π bond.1859

Where does that water go?--let's just do a quick review here of this mechanism; we have this bromine, this bromonium ion intermediate.1876

We have two choices here on where the water can attack, the nucleophile can attack; where is it going to want to go?1885

We saw that because of this positive charge, it goes to the more partially positive carbon, which means the more substituted carbon.1891

We saw the regiochemistry of this reaction was very similar to Markovnikov's regiochemistry where instead of H⁺, we are dealing with a Br⁺.1901

That goes to the carbon with more hydrogens; the nucleophile, in this case water, goes to the more substituted internal carbon; this adds a Br and an OH.1910

What would happen if I took this molecule and react it with some kind of base like sodium hydroxide?1920

I would expect to deprotonate that alcohol; then I would expect to do an intramolecular displacement, backside attack.1925

Remember three-membered rings were okay for doing this intramolecular attack even though there is ring strain in this epoxide.1935

The fact that that oxygen and the carbon bearing the leaving group are so close; they are overlapping; there is no way for that reaction to not happen.1944

This would be a suitable way to make an epoxide; probably we would be starting with an alkene as a way to get the OH and the Br into the structure.1952

But we have also seen another reaction that gives epoxides specifically as products; that is via oxidation of the alkene.1963

We learned about mCPBA as a peroxide reagent; it is a peroxy acid; we learned that when an alkene sees a peroxy acid, it gets converted to an epoxide.1974

Which synthesis would we use?--it depends on the complexity of the rest of our molecule.1989

If the rest of the molecule can tolerate mCPBA or oxidative conditions, then this might be the simpler path.1996

But there are other paths that are more acidic, acid-base type conditions rather than oxidative conditions; either of these would be useful for synthesizing an epoxide ring.2003

Now that we know how to make ethers, let's think about the reactions that ethers can undergo; there is not a long list to talk about here.2017

Ethers are generally very very stable; they are not too reactive; that is because they have no leaving group.2025

If you compare them to something like an alkyl halide which can undergo substitutions and eliminations, there is no leaving group here.2031

There is also no acidic protons like we might have in an alcohol where that can be the source of some of our reactions.2037

They are really quite unreactive which means that they make very good solvents because they don't do a lot of reactions themselves; because they are polar, that helps for properties of a solvent.2044

In other words, if we take an ether, we try and react it with a nucleophile or we try and react it with a base.2059

Or maybe some of the oxidation conditions we have seen for alcohols like PCC or Swern or Jones oxidation; nothing is going to happen here; no reaction.2067

There is very few reactions that ethers can undergo; one of the only reactions that we are going to be studying is this one; that is the reaction with either HBr or HI.2080

What is the difference between all the reagents that I was just suggesting on the previous slide?--the difference here is that HBr and HI are both strong acids.2093

That is the one vulnerability of an ether--is that it is subject to reaction with a strong acid; the reaction that happens is we see that we cleave our ether apart.2106

We break both of these bonds; we replace the oxygen bond on both sides with a bromine on both sides or an iodine if we are using HI.2118

Let's see if we can come up with a mechanism for this; what will happen if we take an ether and we expose it to a strong acid?2127

Same thing that happens with anything exposed to a strong acid--we are going to protonate it.2135

The ether reacts as a base; HBr is going to act as an acid; so step one of our mechanism is protonate.2141

We have seen this happening for alcohols; let's use that as our analogy; what happened when we protonated an alcohol?2155

When we protonated an alcohol, it turned the OH group in a water molecule that is attached which made it a very good leaving group; guess what?--we have the same thing here.2165

This is also a good leaving group because, if it left, it would leave as an ethanol molecule, again a very stable neutral molecule; by protonating, we turned this into a good leaving group.2178

Because HBr is a source of not only the acid but also Br^-, that Br^- is going to see a carbon with a good leaving group attached to it; it is going to do a substitution.2192

The substitution... I jumped the gun here... it can be either an Sn1 or Sn2 substitution with a good leaving group; that is simply going to depend on the substrate we are given.2207

Because this is a methyl group, it has very low steric hindrance; backside attack is great; carbocation is awful; in this case, it is going to be an Sn2 mechanism.2221

We are going to be forming one of our products; we just brominated this methyl group; we have bromomethane as a product.2232

What else did we form?--we just kicked off a molecule of ethanol, again very stable great leaving group; this is how we can cleave an ether with HBr and HI.2245

But if we have an excess of this, if we have plenty of this to go around, then this alcohol that is formed in this reaction does not stick around.2258

Because we have seen a reaction of alcohols with HBr, what happens?--we replace the OH with that halide; what is going to happen is we are going to repeat process.2266

Our alcohol under these conditions is also going to get protonated; we are going to protonate the alcohol; now the rest of the mechanism, this is simply a reaction we have seen for alcohols.2278

We protonate the OH; we make it a good leaving group; because we have bromide around, we have something that can substitute for that leaving group.2294

Again this second part of the mechanism can be Sn1; it could be Sn2; because in this case we have a primary, it is going to be Sn2.2304

Our bromide can come in and attack and kick off our water molecule; that is how we get our other alkyl halide.2316

Here is one of our products is bromoethane; one of our products is bromomethane; what was our other product here?--look what we also formed; we formed water.2331

If you would like to balance your reaction here, you can see that not only do we get these two organic products, but we also form a molecule of water.2344

This oxygen get kicked out eventually completely for both carbon groups; these two protons from the HBr will combine with that to give an equivalent of water as well.2352

Ethers in general are going to go just that one reaction as the most common reaction we will see--ether cleavage with HBr or HI.2369

But when we look at an example of an epoxide as a class of compounds, we will find that epoxides can undergo a lot of reactions.2377

They are much more reactive than an ordinary ether; that is what we will spend the rest of our time talking about.2387

This is an epoxide; if we think about the reactivity we might have for an epoxide, I know this is a polar bond and I know this is a polar bond.2394

That puts a lot of partial positive character on the carbons of the epoxide ring; what kind of reactivity do you expect for a partially positive carbon?2405

I think it is going to be an electrophile; this is what we are going to find for an epoxide--is it is going to be an electrophile.2420

In other words, it is going to be E⁺; it is going to be something that wants to react with nucleophiles.2432

The reaction that occurs is called a ring opening reaction; we are going to see lots of examples of those.2438

The general mechanism that we have, the general thing is we have an epoxide; we react it with some nucleophile; that nucleophile is going to attack one of the carbons of the epoxide ring.2444

As usual, if we have a nucleophile attacking a carbon that already has four bonds, we have to break a bond as well; one of these C-O bonds is going to break.2455

If I attack this carbon, this C-O bond is going to break; the product we get is we have an O^- in this case, two carbon chain, and a nucleophile is now attached.2465

The nucleophile is attached to one carbon; the next carbon over still has the oxygen from the epoxide attached.2479

If you take a look at this mechanism, how would you describe the mechanism?--nucleophile attacks, kicks off a leaving group; have we seen that before?--yes, it looks like Sn2 mechanism, backside attack.2487

But this is a strange example of an Sn2 because normally who is your leaving group in an Sn2 mechanism?0--if you have to name a leaving group.2497

Maybe we have a halide--bromide, chloride, iodide, or a tosylate as a good leaving group; we need a good leaving group.2505

Our leaving group in this case is an alkoxide; we just kicked off an O^- that has no resonance stabilization; we have never seen that before.2513

Why is it happening okay?--we said that wouldn't happen on an ordinary ether; why is it happening here?2521

Because in the course of this substitution reaction, in this Sn2, we are also opening up the epoxide ring; that three-membered ring has a lot of ring strain.2529

By breaking that bond and opening the ring, we relieve the ring strain; that is what makes epoxides very reactive and readily undergoing reactions with nucleophiles.2541

It is because of the ring strain they have in that three-membered ring; any other ether is lacking that; that is why we don't expect Sn2s in those cases.2554

Let's see an example; if we take our simplest epoxide--it is called ethylene oxide; that is the epoxide made from ethylene, the two carbon alkene.2563

If we take ethylene oxide and we react it with methanol, we could react it with methanol and acid or methanol with base, methoxide.2573

This little or means that it can be either acid or base catalyzed, the mechanism; we are going to see two different mechanisms for this; but either way, we have a methoxy group acting as our nucleophile.2581

The product we are going to get will have a methoxy group attached to one carbon and a hydroxyl group, an alcohol, on the other carbon; it follows that pattern of our nucleophilic ring opening.2603

Here the methoxy group is our nucleophile; as usual we are going to get an alcohol; epoxide ring opening reactions are always going to give an alcohol product out.2617

The oxygen that used to be part of the epoxide ring will end up as a hydroxyl group on the carbon chain; it will end up as an OH group.2629

Let's one by one take a look at these two different reaction conditions and see what the mechanisms look like for this epoxide ring opening.2638

You can see in this case we get the exact same product out whether it is acid or base catalyzed; in some other cases, we will see how we might see some differences in those products.2645

What does the mechanism look like for an acid catalyzed ring opening?--again here is the product that we are expecting; we are going to get this methoxy group in.2660

Here our reaction conditions, we have the epoxide, we have methanol, we have an acid; what is going to happen as our first step in the reaction?2668

It is going to be reaction with the acid; remember as soon as you see a strong acid present, your first step of your mechanism is going to be to protonate something.2677

Where do we protonate?--we can protonate this oxygen; but that doesn't really lead us anywhere; if we protonate the epoxide oxygen, that is going to be a better step to take.2686

Which means now we have a hydrogen attached to the oxygen; there is now just one lone pair here because the other one was used; remember proton transfer is always two arrows.2702

What does this oxygen look like now?--it has one, two, three, four, five electrons; we know oxygen wants six; it is missing an electron; so we can protonate and give an O⁺.2712

That is our first step--is to protonate; how does that help us?--why is that a good move to make?--our epoxide remember is an electrophile.2721

How about after protonating it?--do you think this structure now is something that is going to be more of an electrophile or less of an electrophile?2734

We have a positive charge now on our structure; is that something that is good for electrophiles, that we associate with electrophilicity, being electron poor?2742

Absolutely; what we did now is we just made a great electrophile by protonating the epoxide; we have a great super electrophile; let's look around for a nucleophile.2750

We look back to our reaction conditions; what nucleophile do we have?--we have the alcohol; we have methanol as our nucleophile; what is that going to do?2762

Every time a nucleophile sees an epoxide ring, same thing--it attacks the carbon and breaks the C-O bond; Sn2--attack the carbon, kick off the leaving group.2777

But because this leaving group is still attached to the carbon chain, it doesn't just disappear from our structure; it stays connected to our structure.2789

This oxygen is now an OH; it has its two lone pairs back; now it is neutral again; let's take a look at this oxygen; this was our methanol oxygen; what does this oxygen still have attached to it?2801

It has the CH₃ and the OH... I'm sorry, the CH₃ and the H; it has just one lone pair still; this looks like another charged oxygen.2813

One, two, three, four, five; oxygen wants six; this is another O⁺; we describe this second step as attack of the nucleophile.2829

We are almost there; we are getting towards our product; we still have an oxygen with a positive charge so I know this can't be my final step; I know I can't be done with my mechanism.2841

I need to get rid of that positive charge; how can I get rid of it and end up with an oxygen with just two bonds?--it is this proton that is most easily removed.2849

What we could show is A-; we formed A- in this first step when we used our strong acid; we can show that A- coming back and deprotonating the oxygen; our third step of our mechanism is deprotonate.2859

There we are; we have our product, our substitution product, our ring opening product where our nucleophile has been attached and we have an alcohol where the epoxide used to be.2875

It is a three step mechanism; we protonate, we attack, we deprotonate; we are going to see that pattern for an acid catalyzed mechanism; we are going to see that pattern again and again and again.2888

First step is protonate; then that makes it possible to attack; then we need to deprotonate to finish things up; protonate, attack, deprotonate.2902

A couple other things I want to point out about this mechanism--notice there aren't any O^- charges in acid.2912

This would be a strong base like hydroxide; there is no hydroxide; there is no alkoxide; those are very strong bases; we can't have those.2917

What kinds of charges to you see in this mechanism?--you see that each structure is either neutral or it is positively charged; only neutral or + charges.2929

That is very much consistent with acid promoted or acid catalyzed mechanisms; I do see a negative charge here; I see this A-; what does A- represent?2944

Remember if HA is a strong acid, A- represents some very stable weak conjugate base; it is around, but it is not something that would be a strong base.2956

It would be unstable; that is a reasonable species to have around; but not hydroxide, not alkoxide.2968

Another thing I want to point out is this is an acid catalyst; you just need a drop of this acid.2977

Because for every step where you use the acid, for every protonation step, there is also a step where you get that acid back; you regenerate it; there is a deprotonation step.2983

It is used and regenerated; it is not consumed; that is our definition of a catalyst, something we don't need a stoichiometric amount.2994

We don't need a full equivalent of this; because all we need is a little bit of it to get the reaction going.3001

Every time we use the acid, we will have another step somewhere in the mechanism that recreates the acid to be used again.3006

Let's compare this with the base catalyzed epoxide ring opening reaction; again we get the same product where we have this methoxy group attached and the alcohol here.3015

But our reaction conditions are different; now we have methanol still but we use sodium methoxide as our reagent.3027

If we want to do our mechanism, we have our epoxide as an electrophile; we have sodium methoxide as a very strong nucleophile; this is a great nucleophile, strong nucleophile.3033

The very first step in this mechanism is going to be attack of that strong nucleophile onto the electrophile; methoxide is something that would great at doing an Sn2; it doesn't need a push at all.3053

We are simply going to attack the carbon and kick off the leaving group; the methoxy group is now attached; this was the carbon that the nucleophile attacked.3066

It is the next carbon over that will have my epoxide oxygen; this oxygen now has three lone pairs on it; let's check the formal charge here.3081

We have one, two, three, four, five, six, seven; oxygen wants only six; we have an extra electron; we will get an O^- on that oxygen.3091

It looks like our reaction isn't done; we need to neutralize this; how do we take care of this O^-?--we need to protonate to get to the OH.3102

Where do we have a proton source?--we have this methanol here as our solvent; our methanol can come in and protonate the O^-; our final step here is protonate; and we are done.3113

It is not uncommon for a base promoted or a base catalyzed mechanism to be shorter than an acid catalyzed mechanism; we are going to see lots of examples of this down the road.3134

But there is our mechanism; we simply attack; then we protonate; and we are done; let's take a look at this overall reaction and ask what kind of charges do you identify in this base catalyzed reaction?3145

There is no positive charges; that is going to be true in all of our base catalyzed reactions; we never want to have an O⁺ species; that is very strongly acidic.3160

Everything is either neutral or it has a negative charge; neutral or negative charge.3169

Once again, just like the acid situation, this is a catalytic reaction in terms of base; it is not consumed.3179

Even though we used our methoxide here in the first step, this last step, the second step, regenerated it because the methanol acted as the proton source; we remade the methoxide.3187

Again we just need a catalytic amount of this base; it doesn't even have to be methoxide; it could be any base.3201

As long as methanol is our solvent, we are going to get this methoxy group down here because that is what we will have in the largest proportion.3206

We have seen an acid catalyzed mechanism; we have seen a base catalyzed mechanism.3216

It is important to point out that you must have one of those conditions in order for the epoxide ring opening to take place; it must be either strongly acidic or strong basic.3221

If we tried to do this reaction without the H₂SO₄ or without the sodium methoxide; what we are doing is we are taking a neutral epoxide which is a weak electrophile.3231

There is nothing super about this electrophile; we are mixing with methanol; methanol itself is a weak nucleophile.3243

If we are asking to bring these two reagents together, if we tried this mechanism to just have the neutral epoxide... this is a neutral epoxide.3252

We have it being attacked by a neutral nucleophile; let's see what product we would get, what intermediate we would get.3264

This oxygen is now an O^-; this oxygen is now an O⁺; what is the problem with this mechanism?3272

If we tried that and you ended up with this structure, what alarm should be going off in your head as how this is not consistent with the mechanisms we have seen so far?3285

You have an O^- which is a strong base, very strong reactive base, and an O⁺ which is an extremely unstable strong acid.3296

These cannot coexist in the same reaction medium; we will never see a mechanism with an O⁺ and O^- in the same mechanism.3309

Our mechanisms are either going to have negative charges throughout or positive charges throughout; we can't have an O⁺ and an O^-.3325

This is a key that you have made a mistake; you need to back up; you need to look at your reaction conditions and see do we actually have a strongly nucleophilic, strongly basic conditions?3332

Or do we have strong acid conditions?--but the neutral epoxide is too weak of an electrophile to be attacked by a neutral nucleophile, weak nucleophile like methanol.3342

Let's look closer at this epoxide ring opening and think about the stereochemistry and the regiochemistry of the reaction.3357

The stereochemistry, we know it is an Sn2 mechanism; Sn2 means backside attack; what does that typically result in for the stereochemistry if this is happening on a chiral center?3364

It leads to inversion of stereochemistry; that is going to be true in an epoxide ring opening, just like it is in a straight chain Sn2.3379

Let's take a look at these reaction conditions; I see I have an epoxide; now I have NaSH and methanol.3392

Every time I have an epoxide, I know have an electrophile; that part is not going to change; I have to look now at my reaction conditions to decide where is my nucleophile?3401

Who is my nucleophile?--I have NaSH and I have methanol; which component is going to be my strongest nucleophile; who is my best nucleophile in this case?3412

NaSH means I have an Na⁺SH^-; we have a negatively charged nucleophile; here we have a neutral nucleophile; who is the best one?--the SH^-.3423

Not only is it negatively charged, remember sulfur is bigger than oxygen; it is polarizable; so for a couple reasons, this is the better nucleophile.3438

That is the nucleophile we are going to have; how can we predict our product?--that nucleophile is going to attack the carbon.3447

In this case, either one is fine; attack the carbon, open up the ring; tell me about the stereochemistry of that attack.3454

If the epoxide ring is coming out toward you as a wedge, that means the sulfur nucleophile has to come in from behind the plane; it will end up as a dash; so I have SH.3465

What I have on this carbon is I still have this oxygen attached as a wedge; there is no chemistry that happened at this carbon so there is no reason to invert that center.3478

This is still a wedge as an O^-; the purpose of this methanol here is our protic solvent; this is going to be our source of H⁺.3487

We needed a protic solvent here so that we can get a neutral product out; we get inversion of stereochemistry at the carbon undergoing any Sn2 mechanism.3499

I showed here that this is attacking right away without really thinking about it; but which mechanism would you expect?3514

Is this going to follow the acid catalyzed mechanism or the base catalyzed mechanism?--the question is: is there any strong acid?3521

We know how to recognize strong acids; it is a pretty short list; things like HX or H₂SO₄; what is another strong acid?3532

Maybe nitric acid, phosphoric acid, those sorts of thing, tosic acid is an acid we have seen before; these are the things that tell us we were under acidic conditions3540

Because there is no strong acid, we are going to be following the base mechanism here; the base catalyzed mechanism.3552

Even though there is no strong base in this case either, it is the lack of a strong acid that causes us to follow a base mechanism.3561

In other words, we do step one; we are just going to come out and attack like I have shown here; then step two, we are going to protonate.3569

You could try that mechanism to convince yourself that that would be how you get to this product.3576

The other thing I want to point out is in this case, we can attack either of these carbons; if I attack the bottom carbon, what would my other structure look like?3583

That is where the sulfur is attached down here as a dash; then the OH would be attached to the other carbon, the top carbon, again still as a wedge.3596

We are always going to get this anti relationship of the incoming nucleophile and the epoxide oxygen; we get this trans anti type product depending on how you are looking at your product.3606

In this case we are seeing, because we started with an achiral starting material, we have a meso compound here.3629

We can't form just a single chiral product; we are going to get a mixture of enantiomers here; this is an example of a racemate.3637

Anytime we are dealing with stereochemistry, we have to be very careful about thinking do we need to say +enantiomer?--are we expecting an enantiomer or are we not?3644

But in terms of the carbon undergoing the Sn2, we will observe an inversion of stereochemistry like we have always seen.3654

How about if we wanted to do this transformation?--we are starting with an alkene and we are going to a diol.3665

The relationship of those two OHs are anti to one another, trans to one another; we have trans OHs; it is a trans-1,2-diol.3674

We have seen a reaction that converts an alkene to a trans... I'm sorry, to a diol; that was one of our oxidation reactions.3686

Remember we saw KMnO₄ or OsO₄; that would give us two OHs; but what would the stereochemistry be for that transformation?3695

Remember KMnO₄ or OsO₄ did a syn dihydroxylation; both of those oxygens got added at the same time; that gave cis OHs; this is an example of syn dihydroxylation.3706

We need another approach to get the trans out; but we just saw a reaction that does give a trans relationship between an OH and something else.3722

That would be if this had come from an epoxide; if we had an epoxide and we opened up that epoxide, the position of the nucleophile and the OH would be trans to one another.3735

That might be a little harder to recognize here because they are both OHs; but you can open up an epoxide ring with an OH; so that would be a possible synthesis.3754

Let's imagine this; if we converted it first to the epoxide and then we opened up that epoxide ring, that would give us a trans diol.3767

Let's think about those reaction conditions; how do we make an epoxide?--we just talked about the synthesis of epoxides.3776

One of the simplest ways is to do just an oxidation with something like mCPBA; a peroxy acid like mCPBA gives us the epoxide; then how do we open up an epoxide ring?3782

We can either have a strong nucleophile like sodium hydroxide; if we have water as our solvent, then we could attack and then protonate to give the trans diol.3798

Or remember we could have base catalyzed reactions; we could also have acid catalyzed reactions; if we just had H₂O and acid, H₂SO₄.3810

Or you could just write H₃O⁺ sometimes to represent that you have water in acidic conditions.3820

That would also work; you could protonate the epoxide ring and open up with water; this would also give the trans diols.3827

This is a nice synthetic trick to know is that, if we need this trans relationship, we could go through the epoxide; that is very useful.3834

In fact instead of doing this in this two-step procedure where first we make the epoxide and then we open it up.3845

We can actually do this transformation in a single step, in a single pot by using peracetic acid or peroxy acetic acid and water as our solvent.3852

What is cool about this is it makes the epoxide; it also gives acidic reaction conditions; it does so in the presence of the nucleophile.3872

What this does is it forms the epoxide in situ, meaning in the reaction conditions; in situ is the Latin term you see italicized.3886

We are making the epoxide in the reaction and we are reacting that epoxide at the same time.3897

So this is a handy set of reagents to be familiar with as well anytime we want to do an anti dihydroxylation.3903

A syn dihydroxylation can happen with osmium tetraoxide or KMnO₄; anti dihydroxylation can happen with peroxy acid in the presence of water.3917

Finally let's talk about the regiochemistry of this epoxide ring opening; let's say we have an epoxide like this which is not symmetrical; we have just been looking at symmetrical epoxides so far.3931

If we have a nucleophile attacking this epoxide, where does it go?--in other words, it can attack this carbon or it can attack this carbon; that would give two different products.3943

Anytime we are deciding what region or what site of a substrate to react, we call that regiochemistry.3953

The regiochemistry of these epoxide ring openings actually depends on the mechanism of the reaction; we saw base catalyzed; we saw acid catalyzed; let's look at them one by one.3962

If we are in base, in other words, there is no strong acid present; it is not necessarily that we are looking for the presence of a base, but we are looking for the absence of an acid.3975

In that case, remember the very first step of our mechanism is right here; we have some strong nucleophile, negative charge; strong nucleophile is attacking the neutral epoxide.3988

That was our very step; a nucleophile attacks a neutral epoxide; in that case, what we are looking at is just a straight out Sn2 mechanism; how is an Sn2 mechanism going to be controlled?3999

It is going to be governed by sterics; in other words... there is a typo there, sorry... it is going to attack the less hindered carbon.4011

We come back here; we see that this is a secondary carbon; it has two carbon groups attached; this is a primary carbon; it has just one carbon group attached.4021

Which is going to be the faster Sn2, the better Sn2?--the one where it goes after the primary carbon; the nucleophile goes to the less substituted carbon.4033

It is governed by sterics; it is the plain old Sn2; we do what we would normally expect for our regiochemistry.4047

When it is an acid, in other words, there is a strong acid present, remember the very first step of our mechanism is going to be protonate the epoxide.4057

The attack then happens on that protonated epoxide; we are going to have this species, a protonated epoxide, getting attacked by a weaker nucleophile, some kind of neutral nucleophile.4068

We are asking the same question: which site is going to be preferred?--we have a nucleophile attacking a protonated epoxide; the presence of that charge means that it is not a simple Sn2 any longer.4081

The way we describe it is we say it is an Sn2, but it has some Sn1 character; an Sn1, we describe as having a carbocation that gets attacked by a nucleophile.4097

There is no carbocation here; both of these carbons are just partial positive; let's keep that in mind; it is partial positive, not a full positive.4109

That is why we are still describing it as an Sn2; it is still backside attack; but the presence of that partial positive gives it some Sn1 character.4121

Which means we are no longer being governed simply by the Sn2 sterics, the backside attack sterics.4131

The presence of that charge means we are concerned a little bit more with electronics, electron density.4138

Now what we look at is we see this partial positive is on a secondary carbon; we have a secondary partial positive; this partial positive is a primary partial positive.4146

Which of those partial positives, those δ+'s has more electrophilic character?--which of those carbons better handles the partial positive?4161

Just like carbocation stability goes, the more carbon groups we have, the better; this secondary partial positive, this is better.4173

It is the better electrophile; it is the more partial positive of the two carbons; in this case, the nucleophile doesn't want to go to the primary center.4184

It wants to go to the secondary center, the more substituted; in this case, because we are being governed by electronics and not sterics, the nucleophile goes to the more substituted carbon.4197

We end up getting exactly opposite regiochemistries; if it is protonated, it goes to the more substituted carbon; if it is neutral, it goes the less substituted carbon.4213

Let's take a look at a few examples; here we have a not symmetrical epoxide; in other words, attack at either carbon would give a different product; let's look at this.4224

What if we reacted this with sodium cyanide and water?--sodium cyanide and water; what we have here is a very strong nucleophile; we have water as our solvent; we have no strong acid.4240

That means our mechanism is going to be simply the cyanide attacking the epoxide right off the bat; which of these two carbons is it going to prefer to attack?4260

We are looking at a plain old Sn2 mechanism; it is going to be governed by sterics; we have a secondary carbon over here; we have a tertiary carbon over here.4271

It is going to prefer to go to the less sterically hindered carbon; what does our product look like?4282

Here is our carbon chain; how do we draw our epoxide ring opening products without going through the complete mechanism?4290

This is the carbon that got attacked by the cyano group; it is attached here; the other carbon didn't get attacked; it still has the oxygen here.4297

Because we have a protic solvent here, we have something to work this up; we are going to get this alcohol.4309

Notice the pattern of epoxide ring opening; we always have the nucleophile added to the carbon next to the carbon that has the OH remaining.4314

This is going to be like our base catalyzed one; it is controlled by sterics; take a look at my stereochemistry; why did I draw by cyano group up here and not down here?4324

Shouldn't I have drawn it down here because it is backside attack?--yes, except no stereochemistry has been shown for this starting material.4340

Even though the cyanide is attacking at the bottom here, whether I draw the CN up here or down here, it doesn't matter.4349

This chiral carbon has not been shown any stereochemistry; so we could just draw it anywhere we want.4355

Stereochemistry is only going to be relevant when we are dealing with a chiral carbon and the stereochemistry of that chiral carbon is indicated at the beginning.4362

How about the next case?--if we react this with HI, the same epoxide with HI; now I see that we have a strong acid.4371

Do I have a nucleophile?--remember an epoxide is always an electrophile; our goal in looking at our reaction conditions is to find our nucleophile.4384

There it is; we have HI so I^- will be our nucleophile; where is that nucleophile going to prefer to add?--remember we are going to protonate our epoxide first.4394

Where will that nucleophile prefer to add?--it is going to prefer this tertiary partial positive because this is more electrophilic; what does our product look like?4407

Again here is our carbon chain; our iodide is going to be attached to the carbon on the right, the more substituted carbon; the alcohol, the OH, is going to be attached to the other carbon.4418

Again no worry about stereochemistry here in this case because no stereochemistry has been shown for our starting epoxide; these are acid conditions; it is controlled by the better partial positive.4431

A little note here, a little note, this is not about the more stable carbocation; we have seen that argument before.4453

When we looked at Markovnikov's rule in addition to alkenes when we had a carbocation in the mechanism, absolutely we always want to go for the more stable carbocation.4461

Why is it not correct here to say that the reason I get this regiochemistry is because it is the more stable carbocation?--because the mechanism has no carbocation.4470

It has a protonated epoxide; the I^- attacks this; in other words, the ring doesn't open first and then have the iodide attack; the iodide attacks; that is what forces open the ring.4481

We want to be very careful not to mention anything about a carbocation when we are discussing epoxide ring opening reactions.4495

But instead refer to the partial positive in the intermediate in the transition state; that is what is governing it.4502

How do I know it is Sn2 here?--how do I know it is backside attack and not a stepwise where the ring opens and then the nucleophile adds in?4511

The stereochemistry of the mechanism is what gives evidence that it is an Sn2; remember we observe backside attack; we observe inversion of stereochemistry.4520

We know it must be a concerted substitution and not a stepwise substitution.4529

What other nucleophiles can we have?--we have seen Grignards and hydrides attacking carbonyls; these would be great for attacking epoxides as well.4537

Let's see an example of that; we have an epoxide; we know that is our electrophile; we look at our reaction conditions; step one here, we look for a nucleophile.4546

LiAlD₄; I have seen LiAlH₄ before; that is lithium aluminum hydride; what do you think this reagent would be?4558

We could call this lithium aluminum deuteride; just like hydride is a source of H^-, deuteride is a source of D^-.4569

D represents a deuterium; it is simply an isotope of hydrogen; it is a hydrogen that has a neutron in there; but we use it so commonly that it has its own name.4582

Anytime you see a D in a structure, you are going to treat it just like you would a hydrogen; it is going to do the same reactions.4597

Just like this was an H^-, this is going to be a D^-; we will keep those quotes around it because it will always be coordinated with the aluminum.4602

It is the aluminum that is delivering it; it is not ionic; but that is going to be our nucleophile; we have a D^-.4614

Now we need to decide... we know what we are adding; we have to decide where are we adding?--we have to decide the stereochemistry.4620

We want to consider both the regiochemistry--where we are adding it?--we want to consider the stereochemistry.4628

What is the stereochemistry of that addition?--because we are clearly dealing with some chiral centers here; we have some dashes and wedges in our starting material.4635

What we need to decide for the regiochemistry, we need to decide is this an acid catalyzed mechanism or a base catalyzed mechanism?4644

We take a look at our reaction; it is tempting to think it might be acid catalyzed because I certainly see some acid here; I see some H₃O⁺.4653

But see these numbers here; this tells us that the first thing we are doing is reaction with lithium aluminum deuteride; after that is done is when we add in the acid.4662

The important thing to note is there is no acid in step one, in the first step; in fact the reason this is separated and we don't have them combined like we did before.4671

You can't have acid in the presence of a really strong nucleophile like a Grignard or anhydride; you can't mix water with that because those would quench those.4683

With those really strong nucleophiles, what we do is we do it in a two-step procedure; first we react it with that nucleophile.4692

Then we do an aqueous workup to protonate anything that needs protonating; this is going to be our base type mechanism.4699

In other words, we have nothing in step one to protonate this epoxide; it is just being attacked right off by the nucleophile; it is the neutral epoxide being attacked.4708

How are we going to decide where it goes?--it is just a plain old Sn2; it is going to be governed by sterics; let's look at the two carbons that are being attacked.4718

How would you describe this first carbon or the carbon on the right?--this is a secondary carbon; it has two carbons attached; how about this carbon?4732

Be careful; it might be tempting to say it is tertiary because we see this tert-butyl group; but if you look carefully, this is also secondary.4743

But is there a difference in their sterics?--absolutely, the one group attached here is just a methyl group; the other group attached here is this tert-butyl group offering a lot of sterics.4751

Most definitely the less sterically hindered one will be the carbon on the right; there is our mechanism; but you want to be careful in your explanation.4762

If you were asked to describe or explain why this regiochemistry was observed, you want to be careful not to say it prefers the secondary over the tertiary site.4772

You want to say it prefers the carbon without the tert-butyl group because it is going to be less sterically crowded.4780

Let's draw our carbon chain; we can keep our carbon chain in the plane here; because our epoxide was drawn as a wedge, we are going to add the deuterium here.4788

Tell me about the stereochemistry of that group; it has to be backside attack; if the epoxide is sticking out, that means the deuterium comes from behind.4797

We could keep our carbon chain the same and show the deuterium coming in from the back here; what is the stereochemistry?--what is on this other carbon?4806

It still is a wedge; we have an O^-; if we just show our mechanism, we could show it is stepwise just as a little review here.4816

We would get the O^- first as a wedge; but then after step two when we did our work up, that is what we would use to protonate to get to our final product.4827

You wouldn't need to show this whole mechanism for a predict-the-product; but I just want to add a little detail here so we are clear on the purpose of that aqueous workup.4837

Step one, we still are doing our attack and our protonate; but in this case, it required two separate steps for our reagents.4848

Because our nucleophile was too strong of a reagent to tolerate a protic acid; we would have to do it stepwise.4855

Any enantiomer here, +enantiomer?--I know we have that reaction every time to say yes I want to get the enantiomer as well because I have drawn a chiral product here; this is clearly chiral.4865

Where would the enantiomer come from?--what does the enantiomer look like?--that would have a wedged deuterium and a dashed oxygen; how could that come from this starting material?4877

It couldn't; because we were given just a single enantiomer as our starting material, it is okay to have just a single enantiomer product here; that is going to be common for many of these products.4887

This is the only product; this is the only regiochemistry; this is the only stereochemistry; we do get this single enantiomer as our major product.4901

Let's take a look at another example; here we have an alcohol and a bromine; in our first step, we are reacting it with sodium.4908

Sodium all by itself means that it is not Na⁺; it is actually sodium metal; sodium metal has that electron it is dying to get rid of; that makes it a good reducing agent.4919

Where have we seen that as a reagent?--we have seen it reduce an alkyne to a trans alkene; that is a possibility.4930

We have also seen it reduce any acidic proton, reduce that and give off hydrogen gas; as a result then we are going to get the alkoxide.4940

To make an alkoxide, we could use a base or we could use a metal like sodium lithium potassium; to do a redox reaction, we do that same process.4958

Step one here, we make the alkoxide; is that our final product then after the first step or is there something that alkoxide can do before we get to our second step?4968

I just made an alkoxide in the presence of an alkyl halide; I think we are going to have a reaction take place here.4981

Furthermore the O^- is a wedge and the bromine is a dash; which means when we want to do a backside attack, we are perfectly set up stereochemistry to do that anti attack.4988

Sure enough, we are going to get attack of the alkoxide to kick off the leaving group; we are going to make this epoxide.5000

Step one after we deprotonate the OH, we are going to do an intramolecular Sn2, an intramolecular Williamson ether synthesis; we would get an epoxide after step one.5012

An epoxide is an electrophile; I am hoping in step two we have some kind of nucleophile; who is that nucleophile?5023

I see that I have NaOH; that is Na⁺ which means I have hydroxide, HO^-; I also have water.5033

This 18 is simply an isoptopic label; it is useful to be able to track the incoming OH and track him as a different oxygen than the one that is already here.5042

Just like we used deuterium in the last example, that was useful so we could see where the new hydrogen, the deuterium, ended up even though we already had a hydrogen in the structure.5057

This labeled oxygen is going to be doing the same thing; we have hydroxide, labeled hydroxide, as our nucleophile; that is going to open up the epoxide ring.5066

When we do that, we have to think about two things; we have to think about the regiochemistry because in this case, our two carbons are differently substituted.5081

And we have to think about the stereochemistry; how do we decide the regiochemistry?--we need to decide whether it is an acid catalyzed or a base catalyzed mechanism; what do you think in this case?5089

Surely I see hydroxide in here, clearly base catalyzed; there is clearly no strong acid here; this is going to be our base mechanism.5100

Which means our hydroxide is simply going to attack the epoxide as shown, the neutral epoxide.5110

It is going to be our ordinary Sn2; there is no Sn1 character, ordinary Sn2; it is going to be about sterics.5117

When I compare my two carbons, I have a secondary carbon over here; I have a tertiary carbon over here.5126

My hydroxide is going to attack the carbon on my left and open up the ring; tell me about the stereochemistry of that Sn2.5134

Since the epoxide is a wedge, that means the new oxygen has to come in from behind to do backside attack; it must come in as a dash.5147

Here is my 18, my labeled oxygen is on the carbon on the left, is a dash; what do we have on the right?5156

This stereochemistry remains unchanged; there is no reaction here; so my methyl group is a dash and my oxygen is a wedge.5165

Because I have a protic solvent here, then I can expect to protonate this after I open up the ring; I would have an O^-; then I can protonate it to get this OH.5174

We have formed an epoxide; then we have opened it up; we considered the regiochemistry; we considered the stereochemistry.5185

It is okay to have this single enantiomer as my product because I started with a single enantiomer as a starting material.5192

It looks interesting; the nucleophile has the same stereochemistry as this leaving group; it looks like we had retention of stereochemistry.5200

How did we do that?--really there were two Sn2s; there was an inversion in this first Sn2; then there was an inversion in the second Sn2.5207

A double inversion ends up giving us retention of stereochemistry; that makes this problem an interesting one.5215

Finally let's look at an example of a synthesis; if I asked you to synthesize this target molecule and I gave you a hint; I said you should start with some kind of epoxide and some kind of nucleophile.5223

That will help you maybe see the pattern in the target molecule that makes it look like a product that might have come from an epoxide ring opening.5236

What do those ring opening products look like?--they always have an OH on one carbon; the next carbon over, we have a nucleophile that has been added.5246

It is possible to add a methyl or an ethyl via a Grignard reagent; but this oxygen group most definitely he couldn't have been there in the epoxide form.5256

So this was my nucleophile; this was the group that was added; so that is the disconnection we are making here.5266

If you want to imagine doing in your retrosynthesis, we are asking what starting materials do I need?5276

You can almost doing that backwards reaction where the epoxide ring closes back up and kicks the leaving group out.5283

Instead of just doing a disconnection, you could think about that reverse mechanism, that imaginary reverse mechanism.5293

The stereochemistry on this carbon stays the same; I still have a methyl back and a hydrogen out.5302

But I want you to think carefully about the stereochemistry of this; what did the stereochemistry used to be over here?--we have a wedge and a dash that we need to fill in.5310

What did these groups used to look like so that after the nucleophile added in, we ended up with this stereochemistry at the carbon?5319

Think about what it means to be an Sn2 and do an inversion of stereochemistry; I want to point this out.5326

Because when our leaving group is in the plane, it is a little different than some of the examples we have seen.5333

Just by having this oxygen come in from in the plane, it causes the dash and wedge bonds to be pointing in the up direction; then the inversion now points them in the down direction.5340

Because my ethyl group was a wedge and my methyl group was a dash, there is still going to be a wedge and a dash when this nucleophile adds in.5355

My nucleophile... because my leaving group in this case... let's think about it in the forward direction.5368

My nucleophile in this case is in the plane... I'm sorry... my leaving group is in the plane; that means my nucleophile must come in from in the plane, approach from in the plane.5373

It is not going to come in as a wedge or a dash because my leaving group is a straight line; that means the nucleophile comes in, ends up as a straight line.5386

So the inversion... you have already seen the 180 degrees; it used to be up here; now it is down here; there is your inversion.5395

The other two groups are simply like your umbrella flip; the one group that was pointing out towards you is still pointing out towards you; but instead of going down, it is going up a little bit.5402

There is no way for these groups to do this; that would again be a double inversion; you can check your stereochemistries here.5410

If you want to assign R and S before and assign R and S afterwards, you will be able to confirm that they have indeed inverted.5416

If you are not convinced by this or you are having some trouble with this; of course working with a model can help as well.5423

This is my epoxide; that is my electrophile; I know I need a nucleophile; who is my nucleophile?--it needs to be this ethoxy group; maybe I can have sodium ethoxide.5431

That would certainly be a nucleophile; this might be a good way to do this synthesis; but what we need to do after we do our planning is we need to check to make sure that would work.5444

If I took these two, this epoxide and ethoxide and I went to predict the product, would this give the target molecule?--how do we decide that?5458

We take a look at our reaction; we think about the regiochemistry; we think about the stereochemistry; we need to confirm it as if it were a predict-the-product.5469

Do we have acid catalyzed conditions or base catalyzed conditions here?--clearly this is a strong base; there is no acid here.5479

Which means the ethoxy group would go where?--it would go to the less sterically hindered carbon; what product would I get?5488

I would get the product where my ethoxy group adds to the carbon on the left; the alcohol would end up on the carbon on the right.5498

Let's think about that stereochemistry here; my ethoxy group would come in from down here; that kicked off the oxygen.5514

What happens with this methyl and this hydrogen?--the hydrogen is still a wedge but it is up here a little bit; the methyl is still a dash but it is up here a little bit.5520

This would be the product formed by this synthesis; is that the product we want?--no, that is the wrong regiochemistry; I don't want the OH on this tertiary carbon.5528

I want the OH over here on the secondary carbon; this is not going to work; this gives the wrong product; it gives the wrong regiochemistry.5539

How do I force my nucleophile into this carbon?--I want the nucleophile to attack this carbon; how do we force it over there?5552

How do we get the nucleophile to the more substituted carbon?--we do it by using a strong acid so that it will be the protonated epoxide that gets attacked.5562

How do I do that?--I don't use methoxide; I use methanol... I'm sorry... I don't use ethoxide; I use ethanol and a strong acid, H₂SO₄.5576

I can't use ethoxide anymore; what would happen to ethoxide if I tried to mix that with acid?--it would simply get protonated.5591

Remember it is okay to have a weak nucleophile; we need to have a weak nucleophile; we are in acidic conditions.5599

It is impossible to have a strong nucleophile in acidic conditions; so I would use ethanol and H₂SO₄.5604

Now when I go to predict the product, I protonate the oxygen; my ethanol goes to the more substituted carbon; I get this target molecule out instead.5611

This is an interesting thing to point out; I have a tertiary carbon here; I am doing an Sn2 on a tertiary carbon; this is the first time we have ever seen that.5622

This is the only time we are ever going to see that; what is unique about this mechanism is that it is not an ordinary Sn2; we have always said Sn2s don't happen on tertiary carbons.5630

What is different about this mechanism?--because of the strong acid, because we are protonating the oxygen, it now is an Sn2 with some Sn1 character.5640

It is that Sn1 character, it is that presence of that positive charge that makes it different from an ordinary Sn2.5650

Now we go to the more substituted carbon, even if it is tertiary, and we do our backside attack.5656

We saw lots examples here about ethers, how to synthesize ethers, what reactions do ethers undergo, what are some physical properties.5663

Of course the most interesting ether that we have is the epoxide; we saw lots of examples of epoxide ring opening reactions and stereochemistry and regiochemistry implications of those.5671

Hope to see you next time at Educator.com; thank you.5682