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Thiols and Thioethers

Predict the product formed from this reaction:
Draw the product formed from this reaction:
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Thiols and Thioethers

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  • Intro 0:00
  • Thiols and Thioethers 0:10
    • Physical Properties
    • Reactions Can Be Oxidized
  • Acidity of Thiols 3:11
    • Thiols Are More Acidic Than Alcohols
  • Synthesis of Thioethers 6:44
    • Synthesis of Thioethers
  • Example 8:43
    • Example: Synthesize the Following Target Molecule
  • Example 14:18
    • Example: Predict

Transcription: Thiols and Thioethers

Hi and welcome to

Now that we know what alcohols and ethers look like and how they behave, let's take a look at the sulfur analogs of those; they are called thiols and thioethers.0002

When we have a compound with an SH group, we call that a thiol; when we have a sulfur with an R group on either side...0012

Just like if this were an oxygen, we would call that an ether; with a sulfur version of that, we call that a thioether.0021

If we take a look at some of their physical properties, the most notable thing about thiols and thioethers is their odor.0028

This molecule, this is called dimethyl sulfide; it is a common name; dimethyl sulfides are an older name for thioethers.0036

Dimethyl sulfide is one of the molecules that gives rise to the odor of cabbage when you cook it; that foul odor, rotten kind of smell is from dimethyl sulfide.0047

The combination of these two molecules, these two thiols are primary constituents of skunk spray.0062

Many of the foul odors that we experience are going to be sulfur related compounds, might be sulfur related compounds.0072

We could take advantage of that fact when we look at this molecule; this is ethanethiol--is the IUPAC name for this molecule.0080

Ethanethiol has a very distinct odor, very easy to detect in very very small quantities; a use for this is it is added to natural gas.0087

This is the odorant that is added methane, to natural gas; natural gas is an odorless gas; but that is very dangerous because we use that to heat our homes and so on.0099

If there were a leak in a pipe, we would not be able to detect that; what is added to the natural gas reserves is a little bit of ethanethiol.0114

If there is ever a leak of the methane mixture, we would be able to detect it; that is a smell that is familiar to us.0122

Many people think that is what methane smells like; but in fact it is the thiol has been added to the methane.0131

Some of the reactions that they can undergo, one that is very special is that it can be oxidized very easily; remember oxidation is increasing the oxygen content.0137

We could use any good oxidizing agent like KMnO4 or hydrogen peroxide or sodium hypochloride; this is bleach; this is the oxidant that is in bleach.0146

What happens is we increase our number of S-O bonds; in fact sulfur can have more than eight electrons around it; we can even have this two, four, six bonds to sulfur.0155

Really readily oxidized; we put lots of oxygen on there; these are things now that are going to be odorless.0170

This is very useful to us in case let's say we have a dog that gets sprayed by a skunk; how do you get rid of that odor?0177

You can treat the dog with peroxide or a dilute bleach solution; that is going to be very effective at getting rid of that sulfur smell.0184

Another feature of thiols we want to consider is their acidity; remember if you have an OH group like in an alcohol, you could deprotonate that, making it a reasonable acid.0193

The pKa of an alcohol is somewhere around 16; the pKa of a thiol, having an SH group, is significantly lower; it is about a million times more acidic having a hydrogen on a sulfur than an oxygen.0203

Let's take a quick look at why that is; as usual we need to consider the conjugate bases in order to explain that phenomenon.0216

If we were to deprotonate the thiol, we would end up with this S-; if we were to deprotonate the alcohol, we would end up with the alkoxide and O-.0224

We take a look at those conjugate bases and we try to find a difference in their stability; what is the difference between oxygen and sulfur?0235

Where are they located on the periodic table?--sulfur is actually right underneath oxygen on the periodic table; they are in the same group, but it is lower down.0241

As you know, as you move down a group in the periodic table, you get bigger and bigger and bigger; sulfur is larger.0251

It turns out that for the negative charge, that negative charge then has more surface area to be distributed on; we could say since sulfur is larger, it better handles the negative charge.0259

We could describe the extra electrons, the extra electron density, being better delocalized over the large surface area; the S- is more delocalized.0280

Delocalization of the charge is always a good thing; maybe we could do it by resonance; maybe we could do it by inductive effects; here is a case where we are seeing it by having a larger atom.0296

What does that do for us?--if the negative charge on sulfur is more stable, that means that this S- is more stable.0305

More stable means that it is less reactive and therefore the weaker conjugate base; this is a much weaker conjugate base.0317

Of course because it is the weaker conjugate base, it has the stronger parent acid which was the thiol with the lower pKa of about 10.5.0331

In comparing alcohols to thiols, it is important to know that thiols are more acidic; when we might have an application for this or need to know this is when we go to choose a base.0347

Let's say we wanted to deprotonate an alcohol; if we wanted to go from an alcohol to an alkoxide, what kind of base do we typically use to do that?0359

We need a pretty strong base; we need something like sodium hydride--is a very good choice because it is an irreversible reaction; or maybe sodium metal to do a redox reaction and form the alkoxide.0366

But we can't use something like hydroxide; sodium hydroxide is not an effective base to make an alkoxide because their pKa's are too similar.0379

However if we wanted to make an S- from a thiol, from an SH, we could use sodium hydroxide.0388

That would be plenty basic enough and very readily completely deprotonate the SH to give us the S-.0395

Let's see an example; how could we use this to make a thioether?--we could start off with a thiol; if we were to treat this with sodium hydroxide--this is a base.0406

This is base; the thiol is an acid; what happens in this first step?--we can deprotonate to make it completely deprotonate; very little if any reverse reaction there.0418

We could show that as a one way direction; we could deprotonate; this would give us an S-.0434

An S- would be a great nucleophile; in fact not only because it is negatively charged, but once again because of its size.0441

That makes it polarizable; those electrons can stretch out and attack an electrophile; that makes it an exceptionally good nucleophile.0448

It is large and it is polarizable; of course it is negatively charged; that makes it electron rich as well; it is a great nucleophile.0458

Here in our second step, what are we adding?--we are adding benzyl chloride; this would be a good electrophile; anything with a leaving group looks like a good electrophile.0469

What is going to happen here?--I think we are just going to attack the carbon, kick off the leaving group; what is that mechanism?0479

Attack the carbon, kick off a leaving group?--Sn2; perfect Sn2; nucleophile loves to do Sn2s; it is not as strong a base as hydroxide so we don't have that E2 competition working against us.0487

We can use this approach very nicely to add an alkyl group onto a thiol to make a thioether; we had one R group on the sulfur; now we have two R groups on the sulfur; we can make a thioether like this.0505

Let's take a look at another example; if I asked you to synthesize the following target molecule, how would we make this?0525

We are going to use what we know about regular ethers with oxygen and use that same approach; remember for a regular ether, the disconnection you make is on either side of the oxygen.0531

The same is true; we can disconnect on either side of the sulfur; those would be good disconnections; we want to make sure that one of these carbons...0542

Let's take a look at the one on the right here and consider that as disconnection A; then we will compare the two after.0555

For disconnection A, we have a sulfur and a carbon that we are trying to bring together; one of them started out as a nucleophile; one of them started out as an electrophile.0562

Anytime we see a heteroatom like a sulfur, nitrogen, oxygen, those are going to be naturally great nucleophiles; I know this was my nucleophile; this means that this carbon was my electrophile.0572

When I ask myself with my retrosynthesis, what starting materials do I need?--I know that my nucleophile is going to be this sulfur group; we can make that as an -.0585

How do I turn this four carbon chain into an electrophile?--I put on a halide, your choice, chloride, bromide, iodide, to make the alkyl now partially positive.0601

This would be a good electrophile; now I have a good nucleophile and I have a good electrophile.0616

Let's take a look at disconnection B; then we can compare the two; disconnection B would be disconnecting on the other side of the thioether.0622

Who would my nucleophile be in that case?--now this would be this four carbon chain with the sulfur; that would be my nucleophile.0633

What electrophile would it be reacting with?--we need this one, two, three carbon chain; we want the middle carbon to be electrophilic.0640

What we do is we add on chloride, bromide, iodide, your choice, any one you want; we can make the halide there.0650

Normally what we would do is we would compare these two combinations and imagine them coming together and doing the reaction; what we want them to do is we want them to come together and do an Sn2.0657

We know that for backside attack, steric hindrance is very important; we want to consider those; in terms of steric hindrance, which of these syntheses would be better?0674

Remember it is the carbon bearing the leaving group that has to be approached from the backside; it is that carbon bearing the leaving group we want to be as unhindered as possible.0684

What we have in this first case, our electrophile has a leaving group on a primary carbon; this is a primary alkyl halide.0692

In the second one, we have a leaving group on a secondary carbon; this is a secondary alkyl halide; which one is the better Sn2?0700

This top route, route A is the better Sn2; but guess what?--because we are dealing with an S- as our nucleophile and this is a weaker base compared to an alkoxide.0707

Once again because that sulfur is so large, the charge is delocalized; that makes him an excellent nucleophile and decreases its basicity.0726

Even on a secondary alkyl halide, this Sn2 is okay as well; in other words, there is going to be no E2 like we would have with RO-.0734

If this seems familiar; it is because this is very much like the Williamson ether synthesis that we did; the Williamson ether synthesis is a way to make ordinary ethers.0749

Oxygen ethers involves an alkyl halide and an alkoxide; an alkoxide and an alkyl halide gave an ether; now we are using a S- and an alkyl halide to give a thioether.0759

Because sulfur is not as strong a base, we don't have to worry as much as we did about the E2 occurring; in this case, both of these answers would be acceptable.0773

But let's go with the better Sn2; this is a faster Sn2; maybe it is going to be higher yielding; the reaction is going to go faster; so let's do that synthesis instead.0785

How would we make this?--how would we make this anion?--we would start with the thiol; then we would convert it to the S-; that means we need to deprotonate it.0798

Let's pick a base; you could use sodium hydride here but you don't need to; in general we use the weakest base we can because it is going to be easier to use, easier to handle, less expensive.0815

All we need is sodium hydroxide here; that will deprotonate; that made our nucleophile; now we want to add in an appropriate electrophile; that is why we did our planning.0827

We need n-butyl chloride or n-butylbromide or something like that; we would expect this to do the Sn2 mechanism and give us our target molecule.0837

Williamson ether synthesis is going to work well to make thioethers just like it did regular ethers.0852

Let's look at one last example because this is a pretty interesting reagent that we have, Na2S.0859

We know each of these sodiums is positively charged; what does this reagent give us?--we have S2- as an anion.0866

Once again we have never seen that on an oxygen; that would be impossible with an oxygen; but because this sulfur is larger, it could accommodate those extra electrons; it can have a -2 charge.0876

What do you think would happen with this reagent if we treat it with this dihalide?--how do you think this reagent behaves--acid, base, electrophile, nucleophile?--what do you think it does?0887

It looks like a pretty good nucleophile to me, a pretty great nucleophile because it is negatively charged; what are we going to do with it?0896

Here let's look at our Lewis structure; we have two, four, six, eight electrons; but sulfur, just like oxygen, wants only six electrons; that is why we have our -2 charge.0904

With this alkyl halide, we certainly have an electrophile; what I expect to happen is attack the carbon, kick off the leaving group.0915

Now the sulfur has just three lone pairs and a bond; what kind of charge do we have there?--one, two, three, four, five, six, seven; sulfur wants six; this is still an S-.0927

Which means after it does this first Sn2, it is still reactive; it is still a great nucleophile; what do we have over here within the same molecule?0941

We have an electrophile; we have an electrophilic carbon bearing the leaving group; what can happen is we can get an intramolecular Sn2 to take place; we have a second Sn2 here.0950

What size ring would this form?--one, two, three, four, five; is a five-membered ring acceptable for doing an intramolecular backside attack?--it is very good; five and six are both great.0961

What we are going to form here, just like we could make a cyclic ether by doing an intramolecular Williamson, we can make a cyclic thioether here; we have a five-membered ring this way.0976

The more you know about alcohols and ethers, the easier it is to think about thiols and thioethers because they do many of the same things that oxygen does.0990

But it is important to remember that the sulfur is larger so it has some interesting consequences in terms of acidity and in terms of nucleophilicity.0998

Thanks for visiting; hope to see you again soon.1007