Lecture 22. Conjugation, Resonance, Diels-Alder Reactions, Part 2
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SecretionGuar gumChemical reactionGalactoseKohlenhydratchemieOrganische VerbindungenÖlRock (geology)Chemical compoundSlatePharmacyAddition reactionDeposition (phase transition)BenzeneÖlschieferGasPressureResonance (chemistry)LactitolLegierenNatural gasFarmerFluidOil fieldFoodAqueous solutionFeinchemikalieChemistryHexamereOligosaccharideIonenbindungIce creamPolymerExplosionMannoseGlykolsäureMan pageMedical historyBiomolecular structureDiamondQuantum chemistryIceSauerrahmHardnessComputer animationLecture/Conference
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BiosynthesisSurface finishingIonenbindungChemical reactionDiamondCarbon (fiber)DoppelbindungFunctional groupMaterials scienceSodiumWursthülleCycloalkaneUranhexafluoridMoleculeHeck-ReaktionWalkingBreed standardHexaneResonance (chemistry)Film grainProcess (computing)Grading (tumors)CarbonylverbindungenRiver sourcePotenz <Homöopathie>StuffingAgeingLactitolSystemic therapyDyeingController (control theory)MethylgruppeOrganische VerbindungenHydrocarboxylierungConformational isomerismDieneCyclohexenAlkeneKetoneSubstituentDieneEthylgruppeWattLecture/Conference
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Chemical reactionWursthülleCyclopentanEnantiomereFunctional groupCarcinoma in situAlkeneCarbon (fiber)SubstituentDoppelbindungCarbonylverbindungenHydrocarboxylierungHuman body temperatureISO-Komplex-HeilweiseAtomic numberStereochemistryChemical propertyVeresterungIonenbindungCyclohexenDeterrence (legal)Kohlenstoff-14StuffingConformational isomerismElectronic cigarettePleuramesotheliomKonstitutionsisomerieDieneMixtureDieneChemical compoundTumorPipetteActive siteChemistryMoleculeThermoformingSetzen <Verfahrenstechnik>Click chemistryDiamondPentaneSubstitutionsreaktionRing strainSystemic therapySample (material)Growth mediumMultiprotein complexBoyle-Mariotte-GesetzSynthetic rubberPotenz <Homöopathie>IceComputer animationLecture/Conference
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SubstituentDoppelbindungChemical reactionIonenbindungNorbornaneHydrocarboxylierungIce frontSetzen <Verfahrenstechnik>MethylgruppeStarvation responseAlkeneMoleculeChemical compoundFunctional groupSystemic therapyCarbon (fiber)Wine tasting descriptorsWursthülleCyclohexenConnective tissueDieneReactivity (chemistry)AlkaneStuffingThermoformingRiver sourcePleuramesotheliomSubstitutionsreaktionTopicityElectronic cigaretteKatalaseConformational isomerismAgricultureWattPeriodateHexaneAldehydeAageActivity (UML)MixtureAlp (name)Tidal raceLactitolDiketoneAgateMedical historyComputer animationLecture/Conference
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Chemical reactionBenzeneChemistryElectronic cigaretteHydrogenDoppelbindungFunctional groupSetzen <Verfahrenstechnik>Reactivity (chemistry)Chemical propertyChemische SyntheseMoleculeSecretionSingulettzustandDeathChemical compoundCommon landSubstitutionsreaktionDerivative (chemistry)DiamondThermoformingRiver sourceCell disruptionTolueneColourantStuffingSlateCalcium ammonium nitrateWursthülleIonenbindungMethylgruppeBromideHalogenationHydro TasmaniaMalerfarbeHydrateAusgangsgesteinResonance (chemistry)IslandAnilineAromatasePipetteSchweflige SäureChemical reactorElektrolytische DissoziationAromaticityAngiotensin-converting enzymeAddition reactionAlkeneDieneHydrocarboxylierungInternationale Union für Reine und Angewandte ChemieCarbon (fiber)CyclohexenSchwefelblüteLactitolComputer animation
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EthylbenzolAusgangsgesteinKonstitutionsisomerieProtonationBenzeneChloroformAgeingChlorineCarbon (fiber)MethylgruppeSubstituentChemical compoundDerivative (chemistry)Chemische VerschiebungMarch (territory)Hydrophobic effectFunctional groupVeresterungPharmacyEmission spectrumSystemic therapyAreaTolueneMissouri Farmers AssociationStuffingSpectroscopyProcess (computing)Chemical propertyBenzoic acidGasolineHydroxylCombined oral contraceptive pillKohlenstoff-14Baton (law enforcement)DoppelbindungPhenolAnilineISO-Komplex-HeilweiseSet (abstract data type)Phenol formaldehyde resinAtomChlorphenoleConstitutive equationAcidEthylgruppeController (control theory)Internationale Union für Reine und Angewandte ChemieEstradiolHeterocyclic compoundComputer animationLecture/Conference
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Lecture/Conference
Transcript: English(auto-generated)
00:08
So I just saw this news article about chemists doing some sort of work related to the oil industry. This looks like some kind of a picture from the 1960s
00:21
with some old dude working alone in the lab. Most labs aren't like that anymore. And so you might wonder what does organic chemistry have to do with underground rocks and the oil field industry? And so let's talk about that. Why are they showing this picture of some chemist trying to have some sort of an impact on the oil
00:41
and gas field industry? And it has to do with fracking. This is a term you may have heard of before. This is revolutionizing energy production in the United States. So for my entire lifetime the United States has been an importer, a net importer of oil and gas. And that could change. It could reverse to the point where we have so much oil
01:01
and mainly natural gas that we're able to export that to other countries. And it's based on this concept of fracking. You take an oil drilling rig trying to access natural gas where there isn't much. And if you drill into shale deposits you can get it to go sideways. And you initiate small explosions here.
01:21
Then you can use pressurized fluids to break up this shale rock so that the gases can escape and come back out. That's fracking. So breaking up the shale rock using high pressure liquids in order to allow natural gas to seep through and escape where you can collect it.
01:41
So what does it take to take fluids that you can pressurize so much that they can break up shale rock? Well, you need to add special chemical additives to that. And those recipes are secret. No chemical, no oil company wants to tell you what secret recipe they're using that allows them to pressurize basically aqueous solutions
02:01
so much that they will break up shale rock. But here's one of the additives. It's now well known within the industry that one of the secret additives that's no longer secret is a compound called guar gum. And this has changed the economics certainly of guar gum production. Seventy-five percent of the guar gum that's produced in the world comes from India.
02:21
It's grown in India. And it's used as a thickener in ice creams and in foods. You can buy it on the internet. But of course the oil and gas industry uses lots of this. In fact, Halliburton last year issued a profit warning. This is a $38 billion company that said, we're going to have problems with our profits next year because the price of guar gum has shot up so much.
02:40
So these Indian farmers are, I guess they're in the money because they're producing something that everybody wants. Here's the structure of guar gum. It's an oligosaccharide. And later on in Chem 51C, probably the last chapter that you cover, you're going to cover glycobiology and oligosaccharides like this one. This is guaran. It's the oligosaccharide you find in guar gum.
03:01
And it's an oligomer of the sugar mannose. So I hope you can see there's kind of a polymer here with the sugar mannose. And attached to every other sugar, there's a different sugar called galactose. So that's the structure of guaran or guar gum. And it's changing oil production in the United States and around the world, giving us access to gas deposits
03:22
that previously we couldn't have extracted out of the ground. So let me remind you that there's two new assignments that are now up and running on the Sapling homework set. That's Chapter 16. That's the dines chapter that we're doing right now. And Chapter 17, which we're going to start today. And I'm making that assignment due on Sunday
03:42
because I don't want you to fall behind or put this off. The last chapter, Chapter 18, is huge. And I want you to have a clean slate by Monday when we start on Chapter 18. I want you to be done with the problems for Chapter 16 and 17. So I'm not letting you fall behind. That's not going to be a choice for you. So our goal today is to finish Chapter 16 on dines
04:01
and resonance and something called the Diels-Alder reaction. And then we'll have just a little bit of a time to start on Chapter 17 and talk about benzene rings. Okay, so on the problem set that you've been doing
04:21
in discussion sections this week, you've been introduced to something called the Diels-Alder reaction. And let me show you the Diels-Alder reaction. And I'll preface it by saying I guarantee you I will have a Diels-Alder reaction on the final. Who knows? I'm not going to have 20 Diels-Alder reactions. But every organic chemist loves Diels-Alder reactions. Professor King loves Diels-Alder reactions.
04:42
Professor Guan, Professor Freeman. And it's because it makes two carbon-carbon bonds simultaneously. So one of the reactants in a Diels-Alder reaction is a diene. That's why it's in this chapter. Sorry, there's a question? Oh, you can't see anything. Sorry about that. Let me fix that so you can see what I'm drawing. Okay, so I'm talking about the Diels-Alder reaction.
05:02
And again, here's the reaction that I love so much. And it's a reaction that uses diene. So here I'm showing a diene as a reactant. And then there's some sort of another reactant in here. It's an alkene. And we call that the dienophile. I'm not a Latin expert, but I think you can get what
05:22
that means is it just loves doing Diels-Alder reactions with dienes. And in 90% of cases, there's a carbonyl attached to the dienophile. That makes the reactions work faster so you don't have to heat them to 1,000 degrees. So here's my recipe for a Diels-Alder reaction. And there's no extra reagent that you add.
05:40
Typically, even with the carbonyl group there, you do have to heat these. So if you just see a diene and an alkene and heat, that's usually a sign that you're going to undergo a Diels-Alder reaction. And the product of this Diels-Alder reaction is a six-membered ring, and specifically a cyclohexene. You will generate a cyclohexene ring
06:01
with a double bond somewhere in there. Every single time I see a six-membered ring with a double bond in it, I'm thinking, I could make that through a Diels-Alder reaction. And I would like you to think the same thing. I'd like you, every time you see a six-membered ring with a double bond in there, I'd like you to think, gee, I could make that through a Diels-Alder reaction. Okay, let me draw out the arrow pushing.
06:22
This is a concerted process. I don't know if you can see this, but you form two new bonds. Here's one of those new bonds. Here's the other new bond. So I'll just write new. You form two new bonds, CC bonds in this reaction. That's spectacular. And so let's push the arrows to understand this. You could imagine that this pi bond down here could be a nucleophile
06:42
and attack this double bond. That means I have to break this pi bond on the dienophile, and I'll make that attack this carbon. And I don't want to have too many bonds to that carbon, so I'll swing this bond down here. So there's three arrows, and the correct arrow pushing for a Diels-Alder reaction. You make two new carbon-carbon bonds, and it's very powerful.
07:02
It's very powerful to be able to make two bonds in a single reaction. Okay, so let me just remind you of these names, because I'm going to refer to these, and I'll point out why I hate the names so much. So this fragment we call the diene. I hope that's obvious why it's called the diene, because it's always a diene. It's always got two double bonds conjugated to each other.
07:23
This other fragment that reacts in a Diels-Alder reaction, if I want to describe the other partner in a Diels-Alder reaction, I would call this the dienophile. O-phile. And the reason I always hated this is because I felt like the smaller component had the longer name.
07:40
That always seemed opposite of logic to me. But I didn't invent the system. That's the naming system. So the diene, the one with two double bonds, that's the diene, and then the dienophile, where just one double CC double bond reacts. That's the smaller fragment that tends to get the larger name. Now, I expect you to practice this reaction.
08:02
There are lots of problems in the back of the chapter, and I expect you to do those. There's problems on the Sapling homework, and I expect you to be good with this reaction backward and forward. I expect you, if I give you the two components, to recognize that a Diels-Alder reaction can happen and correctly draw the product. And let me finish this product here because I'm missing that carbonyl group.
08:22
That carbonyl group would be attached right there. There we go. Sorry, that looks kind of small, but I forgot to draw that. I was so excited about the cyclohexene ring. Okay, so you need to practice the Diels-Alder reaction backwards and forwards. If I give you two starting materials, I expect you
08:41
to be able to draw the product. If I give you the product, I expect you to be able to draw the two starting materials. This is a very important reaction in organic synthesis, and in those kinds of problems on the last page of my exams where I say, here's some starting materials. Here's the product. Now show me what reagents you would use
09:01
to convert all those starting materials into products and steps. So the key thing for me here, if I want to figure out what my starting materials would look like, would be to recognize that cyclohexene ring and recognize the two new bonds that would form in a Diels-Alder reaction. Here's one of the new bonds. There's the other new bond.
09:21
So by seeing those two new bonds that would form in a Diels-Alder reaction, by recognizing this pattern, you can back out what the two starting materials will be. The two starting materials will be a diene. Maybe I didn't draw that as straight as I could. And then the other starting material will be the dienophile. And I just have to figure out what the substituents are on there.
09:41
Sometimes the dienophile could be an alkyne, which would leave a second double bond on there. That makes it very confusing. Okay, so if I look at this, there's an H attached here at this position, and then there's this ketone group. So I'm not going to draw the H, but I'm going to draw that ketone group with an ethyl hanging off the other side.
10:00
And there I go. I'm done with the dienophile. And now I have to look at what's attached over here on the diene, and it looks like there's a methyl group hanging off of this carbon. And so I'll just draw that methyl group there. And that's it. That's how easy it is. These are the two components, a diene and a dienophile that will lead to that six-membered ring through this concerted bond-forming reaction
10:21
where two new carbon-carbon bonds form simultaneously. So again, this is also the kind of question they'll ask you on standardized exams. If you're applying for medical school or the DAT, they expect you to know this reaction. Very powerful stuff. Okay, so we have a requirement for the diene.
10:42
You can't just take any old diene and have it engage in a Diels-Alder reaction. And in the last lecture on Monday, I introduced to you this nomenclature called S-cis and S-trans to describe two different flat conformations of a diene.
11:00
So let me draw out a diene here. Here's a molecule that has two, and in fact, I'll draw out a conjugated diene. And there's two different flat conformations of this. I would refer to this flat conformation as the S-trans conformation. You may recall that. I've drawn it flat in the plane of the paper.
11:20
And that can equilibrate. This is not a resonant structure because I move atoms. I have to move, swing this double bond around. I have to rotate around this bond one way or the other. And if you move atoms, it's not a resonant structure as soon as you start moving atoms. So when I draw this diene and the alternative flat conformation,
11:42
the S-cis conformation, now I can more readily see how these two carbons on the right-hand side can position themselves so that they are close to a potential dienophile. And so let me draw this dienophile here. And just to be different, I'll draw the carbonyl group pointing downward.
12:02
You don't want to get stuck always drawing it in the same, maybe you do, but I try not to get stuck drawing it in the same conformation. I'm not going to draw the arrow pushing. I'm just going to try to show you here that when a diene, a conjugated diene is in the S-cis conformation, these two carbons over here are positioned and ready to form a six-membered ring.
12:23
So you can see which carbons will be attached there. Okay, so you can only engage in a Diels-Alder reaction through the S-cis conformation. So let's take a look at these four dienes on the bottom. They're all conjugated dienes. Which ones of those would we expect to be able to engage
12:41
in a Diels-Alder reaction? And if I look at this diene over here on the bottom, this is an S-trans diene. And unfortunately, it cannot, I'll just use a dash line here, it cannot rotate about that bond because this bond, double bond on the bottom is locked into a six-membered ring.
13:01
And so because this cannot change to the S-cis conformation, that cannot engage in a Diels-Alder reaction, right? There's no way for me to flip around this bond to position this double bond so that it's S-cis in an S-cis conformation. When I look at this diene down below, here's the diene,
13:22
and there's a sigma bond in between the two alkene moieties here. And this could rotate around that bond. So I've drawn it in the S-trans conformation, but it's capable of rotating to put this double bond into an S-cis conformation. So that could act in a Diels-Alder reaction.
13:41
Over here, here's a cyclohexadiene, and it's already got the two double bonds in an S-cis conformation. In fact, this spends 100% of its time in the S-cis conformation. This would be a marvelous diene for doing a Diels-Alder reaction. So it's always S-cis and can't, there's no way to rotate around this bond here to become S-trans.
14:04
And then this fourth example is another example where you can't really rotate around this central sigma bond between the two alkenes, and so that's locked S-trans. Can't ever do anything different. And so that could never engage in a Diels-Alder reaction. So you need to be able to distinguish dienes
14:22
that are either locked in S-cis conformations. That's great. S-cis is great for Diels-Alder reactions. Or, and to contrast those with dienes that can never adopt S-cis conformations, things that are stuck in an S-trans conformation. Those are no good for Diels-Alder. Okay, so two different conformations, S-cis and S-trans,
14:44
and only the S-cis is good. One of the reasons why the Diels-Alder reaction is so powerful, and it's a shame we don't have more time to spend on the Diels-Alder reaction, because there are lots
15:01
of added nuances to this reaction that make it powerful, but would also require you to learn some new stuff. Okay, so here's an example of two Diels-Alder reactions, and I've got the same diene, butadiene, in this reaction. But what I've done is I've changed the dienophile, and all I've done is I've changed the stereochemistry
15:21
of two substituents. These are ester functional groups, so in case it's not obvious to you, that CO2CH3 here is, has a carbonyl group on there. In 90% of cases, the dienophile will have some sort of a carbonyl substituent, one, two carbonyl substituents. You could have three.
15:41
The more carbonyls you have on the alkene, the lower the temperature you can use, and if you had no carbonyl groups, you'd have to really heat these to high temperature. So again, I'm showing delta. There's no reagents. Sometimes the book will be so polite to you. It's really nice of them. They write Diels-Alder so that you know it's going to be a Diels-Alder, and I'm not going to do that for you.
16:02
I expect you, when you see heat, to say, gee, it's heat, and there's no reagents there, and I see a cyclohexene ring in the product. Maybe that's a Diels-Alder reaction. I expect you to do that kind of an analysis. Okay, so the Diels-Alder reaction is stereo specific, and stereo specific means that if I start with one isomer
16:24
or one geometric isomer of a starting material, like trans, I'll get one product, whereas if I start up with the other geometric isomer, like cis, I'll get a different product. That's what stereo specific means. Like E, the E isomer gives one type of product. The Z isomer gives another product.
16:42
So let me start off by drawing the product of this Diels-Alder reaction. It's going to be a cyclohexene ring. Maybe it will help you if you, some, you know, you could do arrow pushing to help yourself see the new bonds it forms. Sometimes I like to just draw dashed lines to help me see what's going to be connected in the products. Those little dashed lines help me see.
17:02
And if you are at all worried that you're going to mess up on the positions of the substituents, then do something as simple as numbering the carbon atoms. Oftentimes that will save you so that you can see where things are going to be attached. So here if I number these substituents, one, two, three, or number the carbon atoms that are going to end
17:21
up in my six-membered ring, it will help me to keep track. So here's the four carbons in my diene, and here's five and six. And just by looking at those numbers, I can tell, well, gee, I'm going to form a bond between carbons four and five and carbons one and six. And let me try to number those over here in the product.
17:40
One, two, three, four, five, six. That's just a little trick that can help you sometimes keep track. And what I know now from looking at this is I need to have those two ester groups attached to carbons five and six. Now the key thing to note here was that there were hydrogen atoms attached. This helps me to see stereochemistry is
18:03
when I see hydrogen atoms. So let me point out to you that there were two hydrogen atoms on carbons five and six. And they were trans to each other just like the esters were trans to each other. And so the rule here is if you have two substituents on your dienophile that are trans to each other, they will end up trans to each other in the product.
18:24
So let me draw those trans to each other. So here I'll draw the ester functional group. And maybe I'll draw the whole functional group so we can see the ester in all its glory there. Okay, there's one ester. It's sticking up. And if the other ester is truly trans in the product,
18:41
it needs to be going downward. And there we go. And the hydrogen atoms are going the opposite direction. So there was a hydrogen atom also attached to this carbon. Sometimes on Sapling they ask you to draw stereochemistry. And on the online Sapling problems, it's if you draw a stereocenter, you have to draw both the bold and the dash substituents.
19:02
Now there's no reason why this particular enantiomer would be favored over the other. You get the other enantiomer out of this. There's a mirror image that's non-superimposable. And you get equal amounts of those. But both enantiomers, in both of the enantiomers, those two groups are trans. Okay, so this is part of being stereospecific.
19:22
The trans isomer of the dienophile gives you a trans relationship between those substituents in the product. And maybe this is not going to be such a big surprise. If we have a cis relationship between these substituents in the starting material, those groups will end up cis in the product.
19:52
And just to make it, I feel like it makes it more clear. I'll draw these H atoms, the H atoms substituents on there. Because seeing those oftentimes helps me resolve confusing
20:03
looking relationships. Okay, so that's what stereospecific means. Trans gives one product. Cis gives a different product. And usually those relationships are preserved. Excuse me. Okay, notice I'm not drawing plus E here. It just so happens through pure chance that this product is meso.
20:22
And so the mirror image is identical. So I'm not going to write plus E. So that means meso means that there is no different mirror image. The mirror image is the same molecule. That's true for meso compounds. Okay, but look for that issue. That's not the typical case. The typical case is you get mixtures of enantiomers
20:42
in the Diels-Alder reaction. Because you generate, usually in most cases, new stereogenic centers. And always look for that. Okay, so that's stereospecificity. Very powerful. It means if I wanted to make one or the other, I would know exactly which diene profile I should be choosing
21:02
in order to get those relationships. So in more than half of cases with the Diels-Alder reaction, we use dienes that are cyclic.
21:20
And the prototype of all cyclic dienes that are used in the Diels-Alder reaction is cyclopentadiene. So it's a cyclopentane ring with two alkenes in there. I'm not sure if you can get smaller than that. You can't get smaller than that. We'll address that in the next chapter. You can't have a four-membered ring with two double bonds. It has a special property called anti-aromaticity.
21:42
But you'll often see five-membered rings with two conjugated alkenes and six-membered rings with two conjugated alkenes. So in other words, cyclic. There's a little bridge here that joins two ends of that diene. So when this engages in a Diels-Alder reaction, look how complex that structure looks. It's not always easy for you
22:00
to tell exactly what was going on in this reaction. I could use the little dash lines, but I'll use the arrow pushing to help me see what gets connected with what. Okay, I'll use this pi bond to attack that carbon. That means I have to break this cc pi bond. So I'll use that cc pi bond to attack this carbon.
22:20
And look how precise I am with drawing my arrows. This first arrow ends on a carbon atom. This second arrow starts with a pi bond. In fact, all three of these arrows I'm going to do is start with the pi bond. And then this third arrow starts with a pi bond and swings over and makes the cc double bond that you see in the product. So that's the correct arrow pushing
22:41
for the Diels-Alder reaction. And I've highlighted these two new bonds in red here. I hope you can see those. And I've arbitrarily drawn this little bridge head flipped up. There are two new stereogenic centers in this molecule. That's this over here. And this down at the bottom. So I'm going to write plus e just to note that, yes,
23:03
there are stereogenic centers on there where you have sp3 carbons that have four different substituents. So if you generate new stereogenic centers, you have to watch for whether you're going to have either a mirror image or the far less usual case, you're going to have a meso compound.
23:21
Okay, so because it's so common to use cyclic dienes, you have to practice drawing these types of structures. And what's important is that you can draw them in a conformational drawing. And so I'm going to draw for you the most common conformational drawing for this type of system, this type of bridged system.
23:42
And so first I'll draw it for you. And then I'll explain what I'm doing to draw this. This is called a norbornyl ring system, which I don't see any reason for you to know that. So I hope you can see when I draw this kind of conformational drawing of a norbornyl ring system, there's a bond in back that I've drawn in a broken form.
24:02
And then there's this bond up front here. Let me try to bold the bond up front just so it's more clear. And the bond up front is passing in front of the bond that's vertical and in back. So that's the relationship we're conveying. We've got two bonds crossing over each other. And the bond that's in back is drawn in a broken fashion.
24:21
Now, you may not realize this, but I was being super-duper careful when I drew this. What I was doing is I was drawing basically a chair with two axial substituents. That's how I drew this.
24:40
You didn't notice what I was doing there, but I drew this by drawing a chair with two axial substituents. And then to turn this into this type of bicyclic system I simply drew a bond here, here, and then I erased or never really drew that stuff on the end. Why is it so important to draw so carefully these types
25:00
of norbornyl bicyclic ring systems? It's because when you do that, you can easily see stereochemical relationships, in particular anti-relationships between substituents. But here there's a different relationship I want to point out. And that's the relationship between these two groups here and this double bond.
25:22
Wow, look at that. So that's the double bond you see over here. And that's the two methyl groups that were on the dienophile. And the two new bonds that we formed here, let me highlight those new bonds. Here's one new bond. I'll just write new. And here's the other new bond right there.
25:40
And I expect you to be able to recognize norbornyl ring systems. I don't think it's very easy to draw these in using the sapling online homeworks, but there will be problems in the back of the chapter where they draw things in this conformational drawing. Now, I hope you can see just how difficult it is to see the cyclohexene ring when I draw it
26:02
in that conformational drawing. There's a cyclohexene ring in there. Right? This is exactly the same molecule as above. And it's really difficult to recognize that cyclohexene ring. And I expect you to be good at that. Even when I draw it in that conformational drawing, I expect you to see that there's six carbons here in a ring, and there's an alkene in there.
26:21
And that that can be made through a Diels-Alder reaction. Now, when I draw it like this, we call this a bridged bicyclic system. There's two types of bridges here that connect this. And I'm going to give those names. Every Diels-Alder reaction with a cyclic diene generates a bridged bicyclic system.
26:42
And one of those bridges is going to have an alkene. Right? You generate a new six-membered ring with an alkene. And so this bridge at the bottom down here, that's called the alkene bridge. And then there's another bridge up here. I'm just going to circle, and I'm going to call that the alkene bridge, so two different bridges.
27:04
And why is it, why am I taking the time to tell you the names of these two bridges, to distinguish those two types of bridges? It's because I want to distinguish the relative positions of these two methyl substituents. And here's why. We have a name for those. So now that I've shown you how
27:21
to draw a norbornyl bicyclic ring system, let's go ahead and practice over here. I'm going to start drawing my, part of a chair. I won't finish up that last one. And you're not as good at drawing chairs as I am. Maybe you are. Maybe you've been practicing over and over like I used to do when I was an undergrad.
27:41
So here's my two axial substituent. And I'm not going to draw this last carbon. I'm just going to draw a cross here. So here's my norbornyl ring system. And I oftentimes like to bold the bonds that are closer to me because it makes me feel like it's more three-dimensional. It just gives me this satisfying feeling of three-dimensionality.
28:01
Okay, that's my norbornyl ring system. There's my alkene bridge on the bottom. And here's my alkene bridge on the top. And we have names, special names for these two substituents. They're not the same. One of those substituents is on the same side as the alkene bridge. The other substituent is on the same side as the alkene bridge.
28:24
And Diehl's Alder experts have given these two substituents different names. They call this substituent down here that's on the same side as the alkene bridge, they call that the endo substituent. And then this substituent on top that's on the same side as this alkene bridge, they call that the exo substituent.
28:43
And that's obviously going to be important. And I'll explain shortly why that's important. Okay, so notice this relationship here. The endo substituent and the alkene bridge are on the same side. And that's not so easy, I feel like that's not so easy
29:01
to see when you draw it in that flat drawing. Whereas it's very easy to see that relationship when you draw it with this conformational drawing, the bridged bicyclic system. Okay, so the substituents can have two different orientations, exo or endo. When I was a postdoc at Berkeley, the guy I worked next to was doing Diehl's Alder reactions.
29:22
And he wanted the exo product out of those reactions. And he ended up naming his cat exo. It was such an obsession for him. It was fun to go to his house and he went, exo come over here. Okay, so why am I making a big deal out of endo versus exo?
29:43
So I told you that in 90% of cases, and maybe for Chem 51 it's 100% of cases. But in 90% of cases, there's a carbonyl substituent on your dienophile. 90% of the time, maybe 100% of the time. And so let's try to imagine what the product
30:01
of this reaction would look like if I did a Diehl's Alder. I'm not going to draw the arrow pushing here. I'm just going to use dash lines so I can see the connectivity. Here's the two new carbon-carbon bonds I'm going to form between there. And I'm going to draw this. Now my bridge doesn't have just one carbon. My bridge is going to have two carbons. And so I'm very good at drawing these.
30:21
You're not going to be quite so good at drawing these yet. But I'm going to try to draw this in this conformational drawing that allows me to see the alkane bridge and the alkene bridge. Okay, there's my alkene bridge on the bottom. And here's my alkene bridge on the top. And then on the other side, I had the two carbons
30:41
from my dienophile. And when you get, when you look at the products of Diehl's Alder reactions with cyclic dienes, the carbonyl substituent ends up in the endo position. The carbonyl substituent ends up in the endo position.
31:01
And so that's the preferred mode of reactivity. This is called the Alder-Endo rule, or sometimes just called the Endo rule. And it's pretty simple. The carbonyl ends up on the dienophile, ends up on the endo position. So let me just write that here so I can see that.
31:22
Endo. And what happens if there's two carbonyl substituents? Well, you'll get a mixture, one carbonyl on the bottom and the other carbonyl on the bottom, if they're not, if they're not identical in some way. So you need to look for this, let me just draw it down here. You need to look to make sure that you draw the product so that the carbonyl is on the same side as the alkene bridge.
31:49
Now, of course, there's a new stereogenic center, actually multiple new stereogenic centers in this molecule. So, of course, I'm going to write plus E here to recognize that there's a mirror image that's formed.
32:03
Okay, so the carbonyls end up in endo positions. It's going to take some practice for you to recognize. And, again, imagine when I draw it like this how hard it is to recognize these six carbons that there's a, it's hard to see that there's a cyclohexene in this product. Let me just remind you of where those carbons are.
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I'll number these carbons, 1, 2, 3, 4 is going to combine with carbons 5 and 6 to make a six-membered ring. Where is that six-membered ring? It's here, 1 forms a bond with 6. Six is attached to 5. Five was attached to 4 in the new, because there's a new bond form there.
32:41
Four was attached to 3. And then there's two right there. It's not easy unless you've practiced to see the cyclohexene rings in these bicyclic drawings. So you have to practice with that. So just be prepared to practice. Okay, so there's other sections in the book here. And I'm just simply going to point
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out that what does section 16.14 say? You'll go through and it'll give you all kinds of Diels-Alder reactions to practice with. And I expect you to work lots of problems in the back of the chapter to practice the Diels-Alder reaction and incorporate it into synthetic roots. Section 16.15 tells you
33:21
that when you have conjugated dienes, they absorb light at longer wavelengths than when the dienes are isolated, when the alkenes are isolated. And I'm not going to test you on that. So generally, the more double bonds you have in a molecule, the longer wavelength light it's going to absorb. Most organic molecules don't absorb visible light
33:40
or don't have colors, but when you have lots of double bonds, they tend to be colored. Okay, so that's it for chapter 16. There's lots of problems in the Sapling homework. I do those by Sunday night. Let's go ahead and start on chapter 17 so we can try to finish that by tomorrow so that when we get started
34:01
on Monday, we have a totally clean slate to start with. Chapter 17 is very short. It's basically teaching you, okay, this is a benzene ring. And here's some compounds that have similar properties with benzene rings, so recognize them.
34:21
And I feel like a lot of this chapter is review. So there's not very many problems in the Sapling online homework problem set. And that's why I'm making it due on Sunday. So you have less than a full week to do that. And I don't want you to still be working on that stuff next week because next week we're moving on and that chapter 18 moves very quickly.
34:43
Okay, let me remind you, so this is not new stuff. I'm just going to point out to you that when you take an alkene, what do we do with alkenes in Chem 51? We add things to them, and if you add bromine across a, whoops, that's not the right, forget that. So when you add bromine across a double bond,
35:04
it reacts very easily and efficiently to give you trans-dibromides. And there's an anti-mer, I'm not really worried about that. The point is that alkenes are reactive. And in chapter 10, we showed you reaction after reaction after reaction that you could do with double bonds. Addition of HCL, hydration, hydroboration,
35:23
double bonds are reactive. But if you take a double bond that's part of a benzene ring, we didn't make a big deal about this, you get no reaction. Now this is exactly at odds with what I told you in the last chapter.
35:41
I told you that when you have conjugated double bonds, the double bonds are more reactive than regular isolated double bonds. So what's going on here? Now I've got three double bonds all conjugated. Why is it that in this particular arrangement, these pi bonds are less reactive than a typical alkene? There is something special about three double bonds,
36:02
not four, not eight double bonds, but three when they're in a ring like this, in benzene. And this is not just some bromine thing. These double bonds in benzene aren't reactive with any of those reagents from chapter 10, HBR, hydroboration,
36:21
sulfuric acid, and what, it's not, they don't react. Okay, so let's talk about this. That's what this chapter is about. What's so peculiar about the double bonds in a benzene ring that they don't react like regular double bonds? Okay, let's, what's making them so special is that these double bonds in benzene have a very special form of resonance
36:43
that gives them a special property. And I'm going to draw that other resonance structure. So here's one resonance form of benzene. And if I move those double bonds, I can use arrow pushing to see that other resonance structure. So I'll use a double-headed arrow here. And I'll draw this other benzene resonance form.
37:02
So I've moved this other pi bond here. Here and here. And I could use arrows to push back. So this special type of resonance with three double bonds in a six-membered ring confers a property called aromaticity. And aromaticity generally means that those double bonds,
37:23
the pi bonds, are unreactive. That's what you should associate aromaticity with. If you want to do any sort of a chemistry with those double bonds, you're going to need some screaming white-hot reagents that are more aggressive than anything you've ever seen before. Because those double bonds are totally happy
37:41
with this special type of resonance. And in the next chapter, Chapter 18, I'm going to show you those screaming hot reagents. I'm going to give you five recipes for super white-hot screaming reagents that are so reactive that they'll even react with these unreactive double bonds in benzene. And what's particular is that in the end, even when I do show you these super secret reactions,
38:04
they're too secret for you to see right now. Your head will explode. But if I show you these reactions, they don't ever add to the double bond and disrupt double bonds or disrupt aromaticity. Those double bonds are still there in the products. What I'll show you in Chapter 18 is when you finally do learn about these special reagents,
38:24
the end products still have the double bond there. What has happened is that they replace a hydrogen. That's what we're going to see. When benzene ring double bonds react in Chapter 18, we're going to replace hydrogen atoms. It's a different type of substitution reaction than you've ever seen before.
38:41
It's not SN1. It's not SN2. We substitute hydrogen atoms, things that don't look at all like leaving groups. Okay, so benzene rings are clearly special. This aromaticity property of three double bonds in a six-membered ring is super special. It makes the overall molecules very stable and it makes the double bonds
39:02
in the benzene ring unreactive. Those double bonds in benzene rings are super happy like that. Don't try to take away that arrangement, that aromaticity. You're going to be going uphill in energy. Okay, let's talk about some nomenclature.
39:21
And as you know, I don't test on IUPAC nomenclature, but I've got sapling problems on the online problem sets that do because the standardized exams seem to think that's important at this stage. So I'm going to show you three common parent benzene ring derivatives.
39:41
These are considered parent ring systems that have their own special name. So if you have a methyl group on benzene, we refer to this as toluene. We don't call it methylbenzene. The IUPAC people decided that's such a common molecule that we're not going to call that methylbenzene.
40:01
We're going to call it toluene. So if you had a substituent to that, you'd call it chlorotoluene. That's a parent ring system. And it's one of three. You can go to Ace Hardware and buy gallon cans of toluene. It's a solvent that's good for stripping paint, doing who knows what else.
40:22
Okay, here's another parent kind of ring system. We put an NH2 group on there. And we don't call that aminobenzene. Again, they've got a special name for this. We call this aniline. And you simply need to know these two words. You've got to memorize that. Because I'm going to point to something and say, see that aniline over there?
40:41
And I want you to know what I'm talking about. So we don't call that aminobenzene. And then the last one is to put a hydroxy group. And I feel like maybe you guys might have used this as a nucleophile or something like this as a nucleophile in the Chem 51 LB lab. We don't call this hydroxybenzene. We call this phenol.
41:03
So that's a phenol or derivatives you might refer to as phenolic compounds. So you need to know these special names, right? I'm never going to say hydroxybenzene or aminobenzene or methylbenzene. I'm going to use one of these names. Okay, so let's talk about how to name things that have substituents on there.
41:20
So if I have a benzene ring with a hydroxy group and a chlorine on there, the parent compound is the phenol, not the, I mean there is no special name for chlorine on a benzene ring. So the parent compound would be phenol. So if I wanted to name this, I would refer to this as a chlorophenol. This other one over here I'd call a chlorophenol.
41:40
And this one over here I'd call the chlorophenol. And since the phenol is the parent ring structure here, the hydroxy group is assumed to be at the one position. So I now number all the other substituents relative to that hydroxy group because the hydroxy group is part of the phenol parent compound. So in other words, the way I number this is I go in a way
42:01
that makes the chlorine have the lowest numbering. This would be 2-chlorophenol. I don't need to say 2-chloro-1-phenol because the hydroxy is always at position one. So that's the way we number that compound. If I move the chlorine over by one more atom, now that's 3-chlorophenol. And this is IUPAC nomenclature.
42:22
And then if I move it over by one more atom so that it's farther away, 2, 3, 4, that's 4-chlorophenol. These are isomers of each other. They have different boiling points, different chemical properties. They move differently on TLC plates. They're all distinctive.
42:40
They're constitutional isomers. Now there's a different naming way to describe the relationship between these two substituents on this benzene ring. When I have two substituents that are directly next to each other on a benzene ring attached to carbons that are directly next to each other, I would describe the relationship between those two positions as ortho.
43:01
I would call that the orthoisomer and expect you to know what I'm talking about. When the two substituents are directly next to each other, that's ortho. And if I called that ortho-chlorophenol, every chemist would know what I'm talking about. When the two substituents are separated by, I guess I would call that by three carbon atoms, then I would call
43:21
that not ortho, but we have a different name. We call that meta-chlorophenol. And the relationship between those two groups is that they are meta to each other. That's the way I use that language. The two substituents are ortho or they're meta. And then the last possible relationship is when they're on opposite sides of the benzene ring. And we call that relationship para.
43:40
So if I said para-chlorophenol, every chemist will understand that that means they're on opposite sides. Now this ortho, meta, and para is super commonly used. And I'm going to use it and we are going to use it throughout the next chapter. So you have to get used to that naming system. And the top one where we use the numbers, that's IUPAC naming.
44:02
That will be less common for us. Okay, now these names are used so frequently, ortho, meta, and para, that chemists usually abbreviate that O. And it's in italics. You can't tell that I'm writing in italics. I'm not doing a good job. If I just write O-chlorophenol, that's understood to mean ortho-chlorophenol.
44:20
If I write M-chlorophenol, that's understood to mean meta. MCPBA, that little M, that's meta. Meta-chloroperoxybenzoic acid. And MCPBA, the little M means meta. Here's para-chlorophenol. If we write P-chlorophenol, every chemist will understand that that means para.
44:42
So earlier you learned about something called peritoluenesulfonic acid. That P was para. So we've been using this without you knowing what the M and the P and the O meant. And so I expect you to practice that M and the P and the O. Okay, so let's, we already covered spectroscopy.
45:01
But let's cover spectroscopy again. All right, I expect you to understand spectroscopy and benzene rings. But let me just remind you of what we already said when we covered NMR spectroscopy back in chapter 14. So what did we say about benzene rings and carbon NMR spectroscopy?
45:21
It's super easy stuff. So where do you find benzene ring carbons in the C13 NMR? You find them in this range from 160 to 110. And it's in the same range that you find regular C-C double bonds. And there's a tiny bit of aggravating overlap here with ester derivatives from about 155 to 160.
45:44
I hate that overlap region because I feel like my bad luck is every time I've got some carbon in that overlap region and I can't tell immediately whether there's an ester or a benzene ring carbon. But when I look at ethyl estradiol, right, I expect you to freak out initially when you look at this.
46:02
But after you freak out, just stop for a minute and there's all kinds of useful information in this C13 NMR spectrum for, this is part of the birth control pill, ethyl estradiol. But we can immediately see these carbons, 1, 2, 3, 4, 5, 6 in the C13.
46:21
Take a look right here from about 110 to 160. Let's count the peaks, 1, 2, 3, 4, 5, 6. Isn't that amazing? Six peaks that correspond to those six carbons in the benzene ring of ethyl estradiol. And then I've got all these other carbons there and who really cares about those? But I'm just showing you how easy it is and how distinctive this region
46:41
of the carbon 13 NMR spectrum is. At a very minimum, I would expect you to look at the structure of ethyl estradiol and predict how many peaks you would expect to see in the C13 spectrum. And I would expect you to know that those carbons appear somewhere in this region from 110 to 160. So we've already done that kind of stuff.
47:00
I hope that's review for you. I'm just trying to show you this wildly complex spectrum. So you can see that it's not too complex for you to say anything. You can say something about that. Okay, so C13 is pretty easy. Let's review proton NMR spectroscopy and benzene ring derivatives.
47:22
So again, this is review. I expect you to know that way downfield here, somewhere between about 6 1 1 2 and, you know, you can have things as far out as 8 1 2 parts per million, but typically to about 8. That is the region for protons directly attached
47:43
to benzene rings. So if I look at the NMR spectrum for toluene, these protons on toluene will be somewhere over here around 7 parts per million. That's the benzene ring area of the spectrum. And the protons that are not directly attached to benzene, but 1 carbon away, those are typically
48:01
around here 2.2 parts per million. So there's this region for protons directly attached to benzene rings and there's this region for things that are 1 carbon away, benzylic protons in other words. So let's just confirm that we get that. Let's take a look at ethyl benzene. And I see 10 protons just by looking
48:22
at the integrations 5H, 2H and 3H. I hope it's very easy for you to identify these 5 protons that are all on top of each other at 7.2 parts per million. Here's those protons right here. These are all these protons all jumbled on top of each other
48:41
in the aromatic region of the proton NMR spectrum. So I hope you would know that 5H is all those protons jumbled on top of each other. Then down here around 2.6 parts per million, that's the protons in the benzylic region. That's these protons attached to the carbon that's attached to benzene. There you go.
49:01
And then the last set of protons here, this 3H signal, that's a methyl group. And just by looking at the chemical shift, there's nothing special about those. That's these protons over here. So I expect you guys to already know something about proton NMR spectroscopy. That was just a little review. There's two other sections in Chapter 17 that talk
49:24
about cool, interesting compounds. I won't test you on that, but it's fun reading, so read that. Okay, get started on those Sapling online problem sets. I want you to be done with those by Sunday night. So we're ready to cruise into Chapter 18, which is very long.
49:41
And I'll see you on Friday.
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