Lecture 03. Alcohols, Ethers, and Epoxides Part 2
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VancomycinChemistryCanadian National RailwayInflammationFunctional groupNana (manga)MeatAuxinInclusion (mineral)HypobromiteAmpicillinElastizitätsmodulPICTHighway Addressable Remote Transducer ProtocolLymphangioleiomyomatosisAcidHydro TasmaniaHydroxylMechanically separated meatInternational Nonproprietary NameMoleculeCobaltoxideElectronic cigaretteAlcoholFunctional groupPeriodateMalerfarbeCarbon (fiber)HydroxylWursthülleAcidStuffingSetzen <Verfahrenstechnik>Breed standardZigarettenschachtelMorphineBranntweinHeroinSubstitutionsreaktionGasController (control theory)GemstoneScreening (medicine)CarboxylierungLactitolWaterChemical propertySpeciesChain (unit)HexaneAgeingChemistryDimethyletherHuman body temperatureHydrogenMixturePentaneDyeingMetalContainment buildingRecreational drug useChemical structureFaserplatteOrganische VerbindungenStorage tankWasserwelle <Haarbehandlung>Diethyl etherHydrogen bondFlüchtigkeitSubstituentButanolEtherCarboxylateAlkaneHexanoic acidChemical compoundEthylgruppeVerdampfungswärmeHardnessInternationale Union für Reine und Angewandte ChemieLeft-wing politicsMethylgruppeComputer animation
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Sodium dodecyl sulfateSodiumHydrideVancomycinSodium hydrideTitanium carbideDigital elevation modelAluminium nitrideSpawn (biology)Rye breadMashingCross-flow filtrationAlcoholMill (grinding)EtherHamInternational Nonproprietary NameCaliforniumAlkaneHalideAlkoxideAngular milKalisalzeSodium hydrideMethyl iodideBromomethaneStuffingSetzen <Verfahrenstechnik>Sense DistrictMethylgruppeSodium hydroxideAlcoholEpidermal growth factorProcess (computing)PharmacyMethylchloridElimination reactionLibrary (computing)OperonLeadYield (engineering)EthanolProtonationCobaltoxideBromideBenzeneIonenbindungOrganische VerbindungenProteinMedical historyHydroxylHydroxideLone pairGolgi apparatusWursthülleCarbon (fiber)Cardiac arrestChemical reactionFireSodiumAlkaneHaloalkaneBase (chemistry)IodidePotassiumAlkoxideAlkylationHydrogenHalideEtherPhenyl groupBromoethaneSprayerComputer animation
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Ring strainCycloalkaneKalisalzeGesundheitsstörungAlcoholAcidCarbokationDehydration reactionMatchCan (band)HamSemioticsChemical structureProcess (computing)WalkingCarbon (fiber)Functional groupKohlenstoff-14ThermoformingSchweflige SäureWursthülleAcidChemistryUric acidAlcoholBy-productProtonationHydroxylSubstrat <Chemie>Reaction mechanismPotenz <Homöopathie>ColourantSulfateElektrolytische DissoziationÜbergangszustandAgeingGesundheitsstörungDoppelbindungBase (chemistry)Ring strainElimination reactionSkarnIonenbindungSpeciesCHARGE syndromeDrop (liquid)WaterIce frontCommon landDehydration reactionMedical historySulfateConcentrateBeta sheetAlkoxideButylEpoxideMoleculeCarbokationLeadSchwefelblüteEtherVerdampfungswärmePenning trapComputer animationLecture/Conference
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AlcoholEtherFormaldehydeComputer animationLecture/Conference
Transcript: English(auto-generated)
00:11
Yeah, I normally, I might have to worry about feedback here. I'm going to turn this down just a notch. I was walking in Albertson's yesterday and over
00:23
up those automatic, those checkout stands where you do your own checking out of the groceries. I noticed on the floor there was this poster sticker thing that had organic molecules on there and I said, gee, that's cool, I can walk into my grocery store and they're showing me the structure of organic molecules. And the sticker says, student's choice.
00:42
It's sort of rubbed off now. I think it says items selected by your classmates. I wasn't aware you were in the habit of selecting hard narcotics because that's the structures that they have on the board. You know, the molecule on the board is called dextrophan. It's a morphine alkaloid related to heroin and morphine.
01:02
When you go buy cough syrups, they contain a methyl ether called dextromethorphan, an o-methyl, but when you take the methyl off, there's no accepted use for that molecule other than to abuse it as a drug like morphine or heroin. So I get the spirit of their advertisement. I think it's cool they had molecules
01:20
on there even though maybe that wasn't the actual molecule that you guys are selecting. Okay, so let's get back to some issues of business here. We did not quite have a chance to finish up talking about IUPAC nomenclature of alcohols.
01:44
And you might recall that I said I'm not going to ask you IUPAC nomenclature questions on my exams, but I will ask you those types of questions on Sapling because you will get nomenclature questions on standardized exams like DAT, MCAT, PCAT, stuff like that.
02:03
And so the big point that I was making when we left was when you name a molecule using IUPAC nomenclature, you have to establish what's the highest priority functional group on your molecule. So if the highest priority functional group is a carboxylic acid like this one right here,
02:22
the name of that molecule is going to end in oic acid. If the highest priority functional group in your molecule is an alcohol, the name of that will end in all, cyclopentanol, butanol, hexanol, there will be an o-l ending, all at the end of your molecule. But if the highest priority functional group
02:41
in your molecule is one of these other molecules like oic acid and you have a hydroxy group, you're going to, you won't end the name of the molecule in all, you're going to use the word hydroxy to show it as a substituent. So let me show you this molecule that has a higher priority function group than the alcohol here. This is going to be some sort of a carboxylic acid.
03:01
And so the way that you would name this is you start off as before counting the carbons connected to the carboxylic acid. There's one, there's one, two, three, four, five, six. And with a carboxylic acid, as with an alcohol, you always number so that the carboxylic acid carbon has the
03:22
lowest number possible, which will always be one because it's always at the end of the chain. Therefore, this six carbon species is some kind of hexanoic acid. The hexane tells me there's six carbons. The oic acid tells me that it's a carboxylic acid. And now, all I have to do is find
03:42
out where those substituents are attached to name and number those. So in this case, it would be five hydroxy hexanoic acid. So you notice, and you don't need the dash there. Let me, I'll just write omit the dash.
04:02
So you don't need a dash in that position. So you can see the importance of noting what's the highest functional group, priority functional group. The highest priority functional group is an alcohol. The name ends in all. But if there's a different and even higher priority functional group, you'll just simply describe it as a hydroxy something else. Okay, so again, what do I want you to do?
04:23
I'm going to point over on the screen and say, oh, that carboxylic acid over there or that alcohol over there. I just want you to recognize those functional groups when I talk about them. When I ask you questions on the exam about compounds that have functional groups, I don't want there to be any confusion about, oh, he's talking about the alcohol molecule.
04:43
Okay, so let's go ahead and skip a little bit forward past this nomenclature business and talk about some properties of alcohols. I can't really think of that many types of questions that I would ask you, but let's just start off by, what would I like you to know? And I want to start off by comparing three different molecules.
05:03
There's pentane, five carbons, all in a row. And what I'm going to do is I'm going to take one of those carbons out and substitute it with an oxygen atom. And I'm going to put that oxygen atom in two different places. First I'll put it in the middle, and then I'll put it over here at the end.
05:21
Okay, so here's three molecules. They have very, very similar molecular weights. If you replace a CH2 group with an oxygen, that's only a difference of two mass units. Essentially, those weigh about the same. Let's go ahead and look at the boiling points for these. If I look at the boiling point for pentane, it's a very volatile liquid.
05:42
So in the lab, we use pentane. It boils almost at room temperature, not that far off from, you know, close to human body temperature. You know, we use that in the lab when we want a solvent that's very volatile and very greasy. So if I come over here and I look at diethyl ether, that is also very volatile.
06:02
We use diethyl ether, we'll see some reactions in chapter 20 where you have to use an ether as the solvent. You don't have a choice. And if you want a solvent that's very volatile, diethyl ether is good for that. Simply by moving that oxygen over to the end, we double the boiling point. That's no longer what I would consider to be extremely volatile
06:22
and maybe if I compare it to water, which boils at 100, I might consider that volatile. So the point is alcohols are vastly more polar than either ethers or alkanes. Alcohols, and it's that H at the end attached to the oxygen that makes the huge difference there.
06:40
This is what accounts for the higher boiling point of the alcohol. But there's one more point that I want to make. It's not just the fact that the boiling point is higher. Let's look at the enthalpies of vaporization. In other words, once you heat one of these liquids up to the boiling point, you then have to add extra heat
07:02
to lift those molecules into the gas phase. So not just to get them to the boiling point, but once they're at the boiling point, you have to add even more heat. And so this would be called enthalpy of vaporization. It's one of these things I think you probably learn about in freshman chemistry. If you look at the enthalpy of vaporization,
07:20
and these are in kilojoules per mole, but the units don't really matter. So 26 kilojoules per mole for pentane. Let's compare that to diethyl ether. It's about the same. Once you raise the temperature of those two liquids up to the boiling point, it takes about the same amount of energy.
07:40
You have to heat, you have to raise the temperature even hotter to get to the boiling point of butanol up above here. And once you do, you have to put even more energy in to lift those molecules out of the liquid phase and into the vapor phase. It costs you in two different ways. You have to heat things hotter. They're alcohols. You have to heat them hotter. And once you get to the boiling point,
08:01
you have to add even more heat to get them to vaporize. The impact of that is if you take a mixture of butanol and ether and put it on your skin and just sort of get it to evaporate, the alkanes and the ethers will evaporate faster. Anything with a hydroxyl group, whether it's water or an alcohol, will evaporate more slowly.
08:21
And so that's the big point. Okay, so why is it that alcohols are harder, require more heat to get them into the vapor phase to get them to evaporate? And that's all hydrogen bonding. So any time you have an alcohol, there's something they can do that alkanes and ethers cannot do. And that's that they can form hydrogen bonds.
08:40
So a typical hydrogen bond looks something like this. And unless you have an OH, there's no capacity to form those hydrogen bonds. A typical hydrogen bond is about two. That's not a good drawing. But a typical hydrogen bond is about two angstroms. Whereas a typical OH or NH bond is about one angstrom. So that bond down there between the OH is about a one. Sorry, I didn't write that very clearly.
09:01
But it's, that says two angstroms. It's between this sort of measurement bar. And then that's a one angstrom OH bond. So hydrogen bonds are about twice the length of a typical bond to a proton. And that OH is just sitting there vibrating back and forth between these two structures. And it's that stickiness that holds your proteins together.
09:22
It's that stickiness that holds the, holds your DNA strands together. It's that stickiness that holds water on the planet Earth. It's that stickiness that keeps Kevlar vests from allowing bullets to pass through. Hydrogen bonds are immensely powerful. So look for bonds between N and H and O and H
09:41
because those have the capacity for hydrogen bonding. You know, before each lecture, I'll try to tell you,
10:00
show you the structures of some of the, sorry for my misspelling here. Just, I apologize. It should be, of course, you are. I don't know that these are cool structures. Here's structures of ethers that I expect you to know. And you don't have to memorize. In the book, there's a section 9.5 that shows you some cool molecules that are ethers and alcohols.
10:21
You don't have to memorize those structures. Here's a structure I expect you to memorize. This is tetrahydrofuran. It's a five-member ring furan. And I expect you to use that because it is one of the most common solvents in organic chemistry. So when I say THF, and this will be probably much more important in the next quarter, I expect you to know that that's an ether solvent with lone pairs on the oxygen.
10:44
And I would expect you to contrast that with diethyl ether. This is the single most common ether solvent, more common than THF even. So to the point where we just call it ether. And everybody knows you're talking about diethyl ether. No chemist that I know would confuse,
11:02
when I say ether, would think of, would think that I'm talking about THF, even though they're both, officially they're both ethers. Okay, so when I say ether, I really mean diethyl ether. That's just the accepted lingo. Okay, so I expect you to know those structures even though I'm
11:22
not big on memorization. What are some alcohols that I expect you to know the name of? Ethanol. Not just because you drink it, because a lot of medicines come as something called a tincture. Tincture of iodine means iodine dissolved in ethanol. So if you ever hear somebody call, say, tincture,
11:41
that means it's a medicine that's dissolved in ethyl alcohol or ethanol. I'll just write tinctures here. Maybe that's sort of old school pharmacology type nomenclature or pharmaceutical nomenclature. I'm not sure of that. Okay, let's take an example of another alcohol
12:01
that you'll see on a regular basis. This is just to give you a sense that you can have many hydroxy groups on it. You can have compounds where every single carbon has a hydroxy group on there. That's a carbohydrate. You're going to have an entire chapter on carbohydrates later on in Chem 51C. This is called glycerol.
12:20
It's important in lipid biosynthesis and lipid signaling. When you make soap by hydrolyzing fat, glycerol, sorry that my writing is not so good there. I'll try to pay attention to that. When you hydrolyze fat to make soap, one byproduct is actual soap, the hard stuff
12:41
that you make bars out of. The other byproduct is glycerol. So when you buy fancy boutique soaps, they leave this in there as a contaminant because it makes it feel really moist. It's actually just less pure. There's all kinds of natural products that contain. You don't have to draw this one out because I would never expect you to know this. But this weekend when you're toking up a doobie in order
13:01
to celebrate these new relaxed laws they have, you're going to be ingesting this compound. This is tetrahydrocannabinol. Tetrahydro, that's not the important part. Cannabinol. It's actually the delta 9 isomer.
13:20
We don't use this nomenclature, the delta. That means there's a double bond at the 9 position. You can see the double bond up there. But it's got a hydroxy group on there. I think in the lab you're going to be doing some SM2 reactions with hydroxy, with benzene compounds. We call those phenols. There's an ether in there. And the key point that I want you to note is even
13:41
if I didn't know anything about the structure, as soon as I hear all, somebody said blah, blah, blah, all, I know there's an alcohol somewhere in that molecule. And that's kind of the level I'd like you guys to be operating at. So when you hear those kinds of endings, you know something about what's going on in that reaction or in that medicine
14:02
or in that package of food that you're buying. Okay, let's talk about some chemistry. We had kind of a review on Wednesday for the kinds
14:25
of molecules I was trying to sort of spark your memory for stuff from Chapter 7 and 8. And so a lot of Chapter 9 is basically just reworking stuff from Chapters 7 and 8, SN2 reactions, SN1 reactions, E2 reactions. And so let's just remind ourselves again, it doesn't hurt.
14:43
It's always good to get this stuff down. So let's go ahead and talk about how can you use oxygen as a nucleophile and how can you make ether molecules by using them as nucleophiles. Let's go ahead and go through a couple of examples where we use alkoxides as nucleophiles to attack things.
15:03
So here's a tertiary alkoxide. It's kind of hindered. It depends on what I make R out to be. If R is an H, that would be an isopropyl alcohol anion called isopropoxide. If R is a methyl group, this would be t-butoxide, which you guys ought to be familiar with.
15:20
Very basic, a little bit hindered because the hindrance is not actually attached to the oxygen. It's one atom away. All the steric hindrance is one atom away over here. So I would classify this kind of hindered. If I react that with a completely unhindered alkyl halide, like methyl iodide, all I'm going to get is SN2,
15:42
right? I'm not going to pop off the iodide and make a methyl cation. Well, what could be worse than that? I'm not going to make a methyl cation, and I'm not going to do an elimination. There's only one carbon in methyl iodide. How can I make a CC double bond if there's only one carbon? The only thing this could do is SN2. This would be a good example of an SN2 reaction
16:01
where the alkoxide, no matter how hindered it is, basically only has one choice. And so when you mix those two together, you'll get an SN2 reaction going on. And nothing else can happen, really. That, to me, looks like the kind of SN2 reaction I would run in the lab,
16:20
rather than these lousy things they have with secondary substrates in the book. I'm going to turn the sound down just a notch. I apologize, but I'm getting some feedback here. You can get SN2 reactions with alkoxides to work.
16:41
And let me just make a note here. So, of course, I'm careful to put my charge. I'm only drawing one lone pair on oxygen. Of course, there's three lone pairs on oxygen. When I was an undergraduate in your position taking a class just like this, I wrote every single lone pair out. It doesn't take that much extra time, but it'll help you
17:01
to reinforce your knowledge of bonding. Yeah, do you have a question? It would occur very quickly because methyl iodide is unhindered. So the extra hindrance on the nucleophile won't slow that down.
17:23
It's the methyl iodide part that makes that super fast with any alkoxide. Okay, so quite often I won't draw all the lone pairs on my nucleophiles. I'll just draw one because all I need is one for arrow pushing. I encourage you to draw every lone pair on your molecules because you never know when you're going
17:41
to need those in your mechanism. So I can have a more hindered. I could have a primary alkyl halide. This would be a very common one called allyl bromide. And in this case, the alkoxide could still attack and give me very good yields of SN2
18:03
in that sort of a system. And so in this case, it would attach two different allyl groups together. I find one of the biggest, this is one of the most common alkylating agents for making ethers, the allyl group. There's three carbons in here, and I find that a lot
18:20
of people seem to forget to draw this carbon right here. So you'll see this electrophile a lot more. I'll just write common here for SN2 reactions. It's extremely good and unhindered, and it undergoes much slower rates of E2 elimination. That's why I like to draw it as an example. But again, I find that lots of people forget to draw
18:41
that carbon right there. So I'll just write don't forget. And the way not to forget is to say, oh I had three carbons here. Let me make sure I draw three carbons there. Okay, so what happens if we go from methyl or primary alkyl halides? Those are unhindered. What happens if we add a little bit of hindrance to the alkyl halide?
19:01
Well, it kind of wrecks your reaction, or at least the expectation of good results. If I switch to a secondary alkyl halide like this, all I'm thinking when I draw this out is, oh God, I'm going to get this mixture. I'm going to get this horrible mixture because now it's more hindered.
19:21
Look at all that hindrance, and I'm going to take the least hindered alkoxide I can have, hydroxide. Nothing is less hindered than that as far as an O minus is concerned. But even with this unhindered thing, all that hindrance in this secondary alkyl halide, I'm going to get competing E2 and SN2 because there's all, I'm not going to draw the beta hydrogens on the other side,
19:42
but as this hydroxide is trying to worm its way in here and attack that bromide like that, it's going to find that it's a lot easier to just stop short and pluck off that hydrogen, that proton right there. So what you'll see is elimination reactions occurring in competition with SN2, two different pathways that lead
20:03
to two different products, and you should fully expect that you're going to get mixtures in this reaction. And so the two products you would get is you would get the substitution product over here. You'd get SN2, and then you'd get E2 elimination.
20:23
And in this particular case the ratios are not perfectly one-to-one, but pretty darn close. You get 53% of the alcohol through SN2, and you get 47% of the elimination product. You get one-to-one, right? So this typifies the problem of trying to do nucleophilic substitution on secondary alkyl halides,
20:44
and that's why I would never go into the lab and try to do because I don't like getting one-to-one ratios. At least with primary, at least in the cases above, you'll get very good yields of the SN2 reaction. I say no chemist would ever do this reaction.
21:01
Here's why no chemist would ever do this reaction. Because when you get to Chapter 20, we're not there yet. You won't get to there until you take Chem 51C. We're going to show you a top secret method. This is too secret for me to show you. It will work in 100% yield with no E2 or any of that.
21:23
You're going to totally forget about all this E2 and SN2 business when you get to 50. By halfway through this course, we won't be talking about it anymore. You'll never talk about that again when you get to Chem 51C because we're going to show you secret methods for doing things that are vastly better
21:40
and don't have problems with elimination associated with them. Yeah, so this Chapter 20 method, you don't get a 53%. You get 100% yield, and the reaction is done in three seconds. Okay, so I guess what I'm saying is don't fret if you found the SN2, E2 business confusing because, yeah, it's really confusing, especially when you talk
22:02
about secondary alkyl halides. And I don't know why, but the book just has this fetish for secondary alkyl halides. It's unreasonable. Okay, cool new reagent, sodium hydride.
22:21
I guess I didn't write it anywhere here. That is a base, and it is very specifically a base, not for E2 eliminations ever, but for alcohols. It gives you very fast and very efficient deprotonation of alcohols. Don't use it for SN2.
22:41
And so let me show you how this sodium hydride works. So if I take any alcohol under the sun, it doesn't matter how hindered it is. It doesn't matter how much other stuff is sitting around. If you take sodium hydride, and I'm going to draw this ion-dissociated form because that's what the book does,
23:01
and I want to be consistent with the book. So here's one representation of sodium hydride. You can think of it, even though it's not, you can think of it as sodium plus and hydride minus. And if that hydride minus looks kind of uncomfortable to you, good, because it is. And it's going to yank a hydrogen off of,
23:20
a proton off of something and make hydrogen gas. So when you expose things to sodium hydride, we just draw it NAH, and here I'm just, for the purposes of arrow pushing, I'm drawing it as sodium plus H minus. This will very rapidly pull a proton off of any alcohol. So if you ever want to make an alkoxide base or nucleophile,
23:45
this is how you do it. I'm going to put a little descriptor here or a qualifier here. That is a money reaction. If you ever hear me call something a money reaction,
24:02
baby, that is money. That means it's going to work every single time, and there won't be some weird side reaction with this, like E2, that's, this is a reaction you will use over and over, even when you get into Chem 51C, you'll be using that. So remember that. So what's the key to this? It's fast, and it's, most importantly, it is irreversible.
24:24
That's the key. Once it pulls that proton off, it doesn't go back, and here's why it doesn't go back, because when you look at the byproduct of this reaction, hydrogen gas, that just floats out of the reaction mixture. That hydrogen back, hydrogen gas is never going to float back
24:42
in and redeliver a proton to your alkoxide. It's game over. Okay, so how do you use this? Typical way you use sodium hydride, at least in this chapter, is to make alkoxide so that you can use them as nucleophiles. And so let's go ahead and take an example.
25:01
Here I'm going to take, as an example, a tertiary alcohol, and I'm tired of drawing t-butanol, so I'll try to jazz this up, because I'm tired of t-butanol. Just showing you that hindrance has nothing to do with this reaction. It's great. The reaction's great. Then I have sodium hydride as my reagent, and I'm not going to push arrows, so I'm not going to draw it as H minus or NA.
25:22
Chemists draw the formula for that as NAH. So whenever I take an alcohol and I mix that with sodium hydride, I will now get an alkoxide anion, and here's my sodium counterion.
25:42
And now I can use that for whatever I want to use it for. I'm not going to use it as a base for E2 elimination. I could buy t-butoxide base from the store, but if I took the trouble to make this, I'm probably going to use it to make an oxygen carbon bond. So if I want to use this as a nucleophile for SN2, let's go back to my friend here, methyl iodide.
26:04
Could I use methyl bromide for, yeah, I could use methyl bromide, but no chemists do. Methyl bromide is a gas. So if you wonder why I never draw methyl bromide or methyl chloride, it's, well, I like to work with liquids. I don't like to haul this huge gas tank out and spray it in. So, of course, when you're answering questions
26:21
on an exam, it doesn't matter to you. You could draw methyl bromide, but if you're wondering why I always choose iodide on this, on methyl, it's just because in the lab, we only use methyl iodide for convenience, and there you go. Look how easy, of course, notice how I was careful to choose an alkyl iodide
26:43
that can't undergo competitive E2 elimination. So, of course, this is a great synthesis of an ether. So this is called the Williamson ether synthesis, this two-step process where you start off with an alcohol, you throw in your money reagent sodium hydride, and you make the sodium alkoxide, and then you use
27:00
that as a nucleophile, an SN2 reaction. And so the book will refer to Williamson ether synthesis, the sapling problems will refer to Williamson ether synthesis, and it's this two-step process that they're talking about. Okay, let me urge you not to make a mistake, and your tendency will be to say, oh,
27:21
I need to make an alkoxide. Oh, but I can't remember this new money reagent, so I'm going to use sodium hydroxide, because that's the only base I can remember, and you're in for a disappointment. And the problem is that if you take some sort of an alcohol, and it doesn't ethanol or T-butane, any alcohol, and you think you're going to deprotonate it
27:42
with sodium hydroxide, unlike sodium hydride, this is reversible. It is fast, but the problem is that it is reversible. I just like drawing the lone pairs on there. So yes, the hydroxide anion will deprotonate the alcohol,
28:06
but the problem is you will very quickly reach in equilibrium where you have a ratio of these that is one to one. And nobody should be running reactions with a one-to-one mixture of nucleophiles.
28:20
You're not going to get a good result. There is no way you can get an efficient reaction of this nucleophile when there's hydroxide anion floating around. Hydroxide anion is less hindered. It's just as nucleophilic. You're never going to get a good result by deprotonating. Now, in the lab, you're going to take a very different alcohol.
28:43
So let me say a 51LB lab. You're going to use this method on an alcohol that's a million times more acidic than a regular alcohol. Here it can work, and the reason is this is a million times more acidic than a regular alcohol. Now it can work. Now you don't get a one-to-one ratio.
29:02
So I'm sure Professor Link will tell you about that stuff when you take the 51LB lab. But for regular alcohols where the hydroxy group is not attached to a benzene ring, sodium hydride. So why don't you use sodium hydride in the lab for your alcohol?
29:22
It has a tendency to create fires, and they don't want you to catch on to fire. It's that good. It's that reactive. She wouldn't catch on fire, but I don't know why they don't want you to do that.
29:41
Okay, so let's talk about some strategy here. If somebody gives you an ether, so let me highlight the problem here. Suppose I ask you this. On my exams, I can't think of any recent exam where I didn't ask you, show me how to make this thing. Right? That's the question that I always ask those questions
30:02
to make sure you're learning reactions. So if I ask you how to make this, you know, it's an ether. The only reactions you know for how to make ethers so far are SN2 type reactions. And if I tell you to make this using a Williamson ether synthesis, you have two different carbon oxygen bonds to choose from. Which of these two carbon oxygen bonds would you make
30:22
using a Williamson ether synthesis? So here's what, here's one strategy for making this. Let's try taking this left-handed portion here, ethanol, and what if we could deprotonate that ethanol with sodium hydride to make sodium ethoxide?
30:54
That would certainly work. There's a 100% chance that that would work. And then I throw into that reaction mixture an
31:01
alkyl halide. Sodium alkoxides are very strong nucleophiles. Not the strongest, but they're strong. So if I throw in an alkyl bromide with that, that could make an oxygen carbon bond. Not, no it couldn't. I hope you can see the problem immediately. The problem is that there are beta hydrogens all
31:21
over the place. The problem is that tertiary alkyl halides are sterically hindered, and it is extremely hard to get any nucleophile. No nucleophile is going to attack this particular tertiary alkyl halide at the central carbon. The only thing you're going to get out of this is E2 elimination. This would be a bad strategy for making the ether
31:40
that I just showed you above. You can fully expect that you're going to get 100% something called isobutylene or 2-methylpropene, and then your OH back. All that work you did to pull the proton off the ethanol is now for nothing because now it's got a proton back on there. So this would be a bad approach to making this.
32:00
If you really wanted to get good yields out of this, out of this ether, what you should do is you should start with this alcohol that kind of looks like the right-hand side of this molecule and treat that with sodium hydride. And again, that reaction will work great.
32:25
So you get sodium t-butoxide. Actually, I wouldn't do that because that's a cheap reagent you can buy, but any more complex alcohol you can buy potassium t-butoxide and sodium t-butoxide. They're very cheap. And so now, I could throw in with this,
32:44
the two-carbon alkylating agent, bromoethane. Now, there is still potential for elimination here, but because that's a primary alkyl halide, bromoethane is a primary alkyl halide, I'll get pretty good yields of SN2 substitution and probably less than 10%, well,
33:04
less than 20% E2 elimination out of this. So this will give me good yields when this O minus comes in and attacks without too much competing E2 elimination. So this would be, of the two different approaches to make this particular ether,
33:21
this is the approach that will work well. So in other words, you always want to, when you do SN2 reactions, choose some strategy that allows you to attack the least hindered alkyl halide. And so that's important. Okay, so I'll show you a type of product that, or a type of reaction that I don't think you
33:42
in a million years could ever have guessed. I just feel like this is extraordinary based on the stuff that you know at this point. PH stands for benzene ring, stands for phenyl. I'll often use that because I'm too, I feel too constrained that I can't draw in a benzene ring,
34:00
but all organic chemists write PH to mean phenyl. Let me just write that abbreviation here. If you'd like, you can draw later, go back and draw a full benzene ring on there, but you'll find that I often use abbreviations. I'll write ME for methyl, ET for ethyl, PH for phenyl,
34:21
and I will expect you to kind of make sense out of those. So in case you see me doing this, it's out of habit. ME is an abbreviation for a methyl group, ET is an abbreviation for an ethyl group, and then PH is an abbreviation for a benzene ring, a phenyl group, so that's what I mean by phenyl.
34:48
Okay, if you take this particular alcohol, the thing to notice is that, gee, there's an alcohol that could be a nucleophile if I treat it right,
35:00
and right next door there's a leaving group, and who would have thought this, but if you treat this with sodium hydride, you'll make the alkoxide anion and it will swing over and push out that bromide leaving group. Wow. What's amazing to me is I just happen
35:23
to know there's 26 kilocalories per mole of ring strain in epoxide. That's amazing to me that that could swing over and create something that is so strained, and you would never have known that that's possible until I just now told you, or I guess by reading the book you would know. So, of course, the sodium hydride, I'll draw the intermediate, is deprotonating this alcohol
35:43
to make an alkoxide, and what you couldn't have known is that all the alkoxides floating around in this mixture, hypothetically they could do an E2 elimination, there's a proton here at the beta position I'm not drawing,
36:02
but this reaction is fast enough, this is what you couldn't have known, is that this intramolecular SN2 to create a strained ring is fast enough that it outpaces all the E2 elimination reactions. So that's an intramolecular, it's like an SN2 reaction.
36:21
Actually, SN2 requires two molecules to collide. Oh, sorry about that. Thank you for, here, let me, okay, so, yeah, who would have thought that this would be fast enough, that strained ring forming reaction would be fast enough
36:40
to outpace all the possible E2 eliminations. So this is a great way to make epoxides. We'll show you another way to make epoxides in, I think, chapter 12 on oxidation reactions or else in the next chapter, chapter 10, I can't remember where that gets covered, but we'll show you a different way to take alkenes and make epoxides. But until then, this is your only way you've got
37:01
for making epoxides, epoxide functional groups. And epoxides are valuable because you can use those to, as electrophiles, to do more SN2 reactions on.
37:28
Okay, I'm going to give you a rule that you're going to have a tendency to abuse. And I'm going to draw out some potential substrates where I see oxygen. Oxygen's electronegative, and, gee,
37:42
aren't electronegative atoms leaving groups? I mean, chloride can act as a leaving group. It's electronegative. You know, one of the strongest nucleophiles that you're really spending time working with in this chapter is a hydroxide anion. And there's no hindrance with a hydroxide anion. But no matter how good you think this hydroxide anion is,
38:01
it is still not good enough. In fact, there is no nucleophile that's good enough to do an SN2, even with this unhindered methyl group. Sorry, I meant to draw dashed lines. I was just going, no, no, not ever. You will never do an SN2 reaction and, well, asterisk.
38:24
Well, I'll give you an exception. You'll never do this kind of SN2 reaction and displace and this as a leaving group. So, in other words, the problem with this is I'm trying to use this as a leaving group. I'll just write bad leaving group.
38:40
That's not to say it can't ever act as a leaving group. But in SN2, that's a no-no. In other words, if I take another alkoxide here, and it looks so seductive, and, gee, why couldn't I do this and just displace a hydroxide anion? How bad is hydroxide anion? I can buy hydroxide.
39:00
It can't be that bad. You'll never do that kind of SN2 reaction where you displace. And that's not to say that hydroxide can't act as a leaving group. You're going to see all kinds of that stuff in Chapter 21 when you take Chem 51C, but not SN2 reactions. And there's not a fundamental problem with hydroxide or alkoxide leaving groups. The fundamental problem is
39:20
that SN2 reactions are crappy, crappy reactions that you will forget all about by the time we reach the end of the quarter. Well, no. I mean, when you get to Chapter 17 and 18, you'll never use those. You're not going to use SN2 reactions when you get to 51C. We've got better reactions for you. It's because the SN2 reaction is lousy.
39:42
But that's all you've got right now, so sorry about that. I wish we could show you better secret stuff. You have to wait for that. Here's a four-membered ring. It's got good amounts of ring strain in there, and it is still not strained enough to allow an SN2 reaction to open that up.
40:02
Okay, so what is the exception? What is the one example of an SN2 reaction where you can pop out an alkoxide leaving group? And I've already mentioned that to you before. It's an epoxide. Four-membered rings don't have enough ring strain, but three-membered rings have enough ring strain.
40:22
To allow you to open those up by SN2 reactions. So if I take hydroxide anion, that now has enough ring strain. Just barely to allow you to do that. Even though SN2 reactions are lousy reactions, they're not very fast, this has enough ring strain.
40:43
And it's only three-membered rings, not four-membered rings. Now when you open an epoxide, it's just like any other with hydroxide anion or alkoxide anion. This is just like any other SN2 reaction. You attack at the least hindered center. And so the intermediate you initially get
41:03
in this reaction would be this. And if you didn't have something that could protonate that, that could go and open up another epoxide. But if you're clever and you run this in a ton of water, so if this is roaming around in a sea of water, so you've got aqueous sodium hydroxide,
41:21
the water could protonate that before the alkoxide does something bad. Okay, so you can open up, you can do SN2 reactions on ethers if and only if the ether is an epoxide. So that's the only case. So again, why does that work?
41:43
That works because of ring strain. And there's more ring strain in a three-membered ring than in a four-membered ring. The three-membered ring just barely works and the four-membered ring doesn't work. I just want to remind you of why it's so fast to do SN2 reactions of alkyl halides.
42:02
Why is it so good to do an SN2 displacement, for example, of methyl iodide? And here's the reason. That's why it's fast. If you look over at the periodic table, I don't even know why I used to have a, there's a laser pointer here that doesn't quite work.
42:22
But if you look over at the periodic table at iodine, which my laser pointer doesn't quite reach, here. Here's the halogens, fluoride, chloride, bromide, iodide, and the bonds are long. Once you drop below the second row, below fluorine,
42:41
those bonds get very long and it is much easier. It is always easier to break long bonds. I mean, this iodide is practically all the way off in outer space here. That's why it's fast to displace halides. This is why you spent two entire chapters talking about alkyl halides and halide leaving groups. The bonds to chlorine, bromine, and especially iodine are long.
43:03
And those bonds are easy to break. If you take graduate classes, we'll explain that in terms of HOMOs and LUMOs and frontier molecular orbitals. But you just need to know that long bonds break more easily as far as leaving groups go.
43:24
Yeah, so, yeah, just to contrast that, epoxide bonds, those aren't long. It's all the ring strain that's making that work. And a regular ether, those bonds are not long. That's why it's so hard to pop out alkoxides as leaving groups.
43:51
Okay. We're going to return back to a concept from chapter eight, and that's dehydration.
44:01
And I'm going to show you a new reaction of kind of a scary reagent, sulfuric acid. There's very few reagents in the lab that scare me, but concentrated sulfuric acid is one of those scary reagents. You take just a paper towel and you drip concentrated sulfuric acid on there. It chars the towel.
44:20
It turns black and chars right in front of your eyes. If you got that on your hands, that would be not good business. So sulfuric acid is a super powerful acid, super powerful. Whoops, that's not what I intended to draw. And when you treat alcohols with sulfuric acid, they will tend to eliminate to make CC double bonds. This is an important theme of this chapter is that alcohols
44:42
and ethers can also be induced to eliminate to give CC double bonds. So let's go ahead and draw out the mechanism for this reaction. Whenever I see a reagent that's a super powerful acid like sulfuric acid, I'm kind of expecting that in step one I'm going to protonate something. And for this alcohol, the OH group is the only game in town.
45:04
So I see that super strong acid, super duper scary powerful acid there, and I'm thinking, oh, I want to protonate that. Now, I'm going to symbolize my acid here. I'm not going to draw sulfuric acid. I expect you to know the Lewis structure for sulfuric acid.
45:20
But I'm going to represent my acid, and I urge you to do this. Whenever you draw arrow pushing mechanisms, and I'm making a point here at the bottom, when you draw curved arrow mechanisms, I encourage you to symbolize acids and bases when you do curved arrow mechanisms by drawing HA for the acid and A minus for the base, for the conjugate base.
45:43
If you're doing a reaction under basic conditions, then use the symbols B minus for the base and HB. I don't want to, it's not just that I'm too lazy to draw the structure. When I need to, I'm totally happy drawing big complex structures. I'll explain in just a minute why it's better
46:02
to use these kinds of symbols like HA. If you want, you can draw sulfuric acid. You're certainly welcome to do that. So step number one here is that that super powerful acid is going to protonate my alcohol. It's way stronger than HCL as far as protonating power.
46:21
Well, significantly stronger. Okay, so now I've converted my alcohol into a good leaving group. This is an equilibrium process, but the equilibrium very heavily favors sticking the proton on. This will reach equilibrium in a fraction of a second, and at equilibrium, probably about only one
46:42
out of every hundred million of these has no proton. All the others, the other 99,999,999, those all have protons on them, so almost every single alcohol in there greater than 99.999% are protonated. So that's why you use sulfuric acid is it's powerful
47:02
enough to put protons on almost all of the alcohol molecules. And now that I've done this, I now have a good leaving group, and in this particular case, I chose a substrate that is very capable of forming a t-butyl carbocation, a stabilized tertiary carbocation.
47:22
I don't have any good nucleophiles anywhere in my reaction. The byproduct of this would be a sulfate anion, which is not nucleophilic, and so let me go ahead and draw out that t-butyl carbocation, and I'm going to choose one of the beta protons to draw.
47:42
Of course, there's beta protons everywhere. I'm just going to draw one of them, and this is one of the biggest problems that you'll find when you transition from Chapter 7 to Chapter 8. In Chapter 8, we stopped drawing all the Cs and Hs for you. We switched over to these line drawing structures, and I expect you instantaneously to know that this has three Hs, this has three Hs, this has three Hs,
48:03
even if I don't draw them for you, because here we've reached a step, and the book won't have that drawn for you. You have to know that there was an H there and draw it yourself, and you need that H right now because what's going to happen is this A minus, the counter ion from the sulfuric acid, can now come back in and pull that proton off
48:22
to complete an E1 elimination mechanism, and so that's the mechanism by which you form that isobutylene product there. Now, the reason I symbolize with A minus is there's another product that's floating around in your reaction mixture, and that's water. So as soon as you pop the water out,
48:40
you've got water floating around. You've got sulfate anion. This is called bisulfate because there's two, well, it doesn't matter. It's why it's called bisulfate. Don't worry about that. The point is that you don't know which of these two species is pulling that proton off. You couldn't possibly know. I couldn't know, and I know a lot of chemistry,
49:03
and that's the advantage of drawing A minus instead of specifying that, oh, it's the sulfate anion that's pulling that. You don't know that. Water is actually more basic than this sulfate anion, so that's why it's better to symbolize using HA and A minus because initially, you know, and as soon as I put this in,
49:22
it starts to get protonated. You don't know what's the acid that's transferring the proton there, so I like those symbols. It's not just because I'm too lazy to draw the sulfate anion. Let's draw that sulfate anion down here because I want to make a point. When you see this, this, yeah, the third, this one?
49:51
Oh, that's any one of these three carbon atoms has an H, so let me draw dots on the carbon atoms, dot, dot, dot. So there's three carbon atoms here. When I pop this bond off, I'm going to leave a carbocation
50:04
at the central carbon. There's that carbocation, and this carbon is one of those, this C here is one of these three carbons, so let me draw this out. There's, this is a CH3, this is a CH3, this is a CH2 plus 1 is 3. This is a CH3, this is a CH3, and this has three H's on it,
50:27
and like I said, this is one of the problems when you transition from chapter 7 to 8 is that you've got to practice to see all the hydrogen atoms. Okay, so there is, down below here, I'm going to draw the fact that there's a sulfate counter ion associated with this cation,
50:43
and you might wonder, well, how come the sulfate anion with its negative charge doesn't come back in and attack the T-butyl carbocation? I see a minus, I see a plus, and in fact, that is happening. I didn't ask you to show that this is happening. I just asked you to draw the mechanism here, but in fact,
51:02
this is coming in and doing attack, and let me try to use a different pen color here so we don't get all confused, but once that attacks, it will give you something that simply re-dissociates again, and the book doesn't show you this, and it leaves you wondering, gee, how come sulfate anion doesn't attack?
51:21
Well, the sulfate anion is attacking, but once that sulfate anion attacks, this is such a good leaving group, it just pops right back out again, and that's not really part of the E1 elimination mechanism. You don't need to show that, even though it's happening, so in case you're wondering why it's not happening, it is,
51:40
but it's just not important. Who cares, right? It doesn't lead to product. You don't have to show that, so don't be worried by the fact that we didn't draw that. We only draw the things that lead to the product. Okay, so when we come back, we'll learn some cool new reactions with alcohols and ethers, and maybe we can home
52:02
in on what we can do with three-membered rings.