Lecture 07. Alkenes, Part 2
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00:00
VancomycinAlterungGenMannoseHydroxybuttersäure <gamma->BohriumTuffAlkeneArgininPökelfleischChemische VerbindungenVerdampfungswärmeWursthülleResveratrolChlorMolekülAtomGenWasserReaktionsmechanismusDerivateAlkeneBromChemische ReaktionFülle <Speise>Chemische ForschungEsterStereochemieAcetateAktivität <Konzentration>SäureGesundheitsstörungBenzolringDoppelbindungEssigsäureChemisches ElementSirtuinOrganische ChemieChemische StrukturEliminierungsreaktionAlkoholateAlkohole <tertiär->StoffwechselwegSetzen <Verfahrenstechnik>BiochemikerinPharmazieFunktionelle GruppeAllmendeBukett <Wein>ProteineIonenbindungEtomidatSchweflige SäureStammzelleBaseFluorkohlenwasserstoffeHydroxylgruppeElektronische ZigaretteOrdnungszahlMethanolOrganisches LösungsmittelEssigsäureethylesterGenregulationEthanolZündholzNeotenieBrillenglasChlorwasserstoffAlterungPeriodateQuerprofilAktivierung <Chemie>Computeranimation
08:52
MannoseBohriumAlkeneVancomycinMonoaminoxidaseKirVerhungernChemische ReaktionVerdampfungswärmeVerhungernAlkeneHydroxylgruppeDoppelbindungNatriumhydroxidAbfüllverfahrenKonformationsisomerieWasserstoffperoxidGummi arabicumBromSchweflige SäureBiogasanlageStereochemieBorGangart <Erzlagerstätte>SubstituentMischenChemische FormelChlorWassertropfenOrganische ChemieWasserAlkohole <tertiär->OrdnungszahlKomplikationBoraneFunktionelle GruppeWursthülleIodNatriumMolekülGesättigte KohlenwasserstoffePferdefleischPenning-KäfigAtomMolekülbibliothekGezeitenstromAlterungSubstitutionsreaktionElektronische ZigaretteSubstrat <Chemie>IonenbindungChemische VerbindungenPeriodateQuellgebietKrankengeschichteHope <Diamant>Chemischer ProzessChemische ForschungMultiproteinkomplexPeroxideTumorHydroxyethylcellulosenMischanlageComputeranimationVorlesung/Konferenz
17:44
MannoseVancomycinBohriumInsulinMagnetometerCalciumhydroxidInsulinkomabehandlungSymptomatologieIonenbindungKohlenstofffaserOrdnungszahlMischenReplikationsursprungAzokupplungChemische ReaktionBorRingspannungFülle <Speise>AlkeneFunktionelle GruppeHydrateMolekülReaktionsmechanismusPrimärelementCarbokationVerdampfungswärmeWerkzeugstahlEinsames ElektronenpaarTerminations-CodonVerhungernReaktivitätBaseKohlenstoff-14SäureElektron <Legierung>SalzsäureDoppelbindungOrganische ChemieValenzelektronBromideUmlagerungMethylgruppeBromChlorWasserstoffbrückenbindungBiosyntheseBoraneAlterungAdditionsreaktionStockfischKatalaseSingle electron transferSekundärstrukturPipetteSynergistSandChemischer ProzessStoffwechselwegAllmendeMetallWursthülleKluftflächeSubstrat <Chemie>QuerprofilWasserstoffOktanzahlAbfüllverfahrenComputeranimationVorlesung/Konferenz
26:36
MannoseCycloalkaneBohriumAltbierVancomycinMethylmalonyl-CoA-MutaseProlinSpätburgunderWursthülleAlkoholfreies GetränkKohlenstoff-14KohlenstofffaserDeprotonierungUmlagerungReaktionsmechanismusIonenbindungKonzentratChemische ReaktionBiochemikerinRingspannungZuchtzielElektronische ZigaretteWerkzeugstahlSubstitutionsreaktionAzokupplungElektrolytische DissoziationSekundärstrukturCarbonateBromideGesundheitsstörungAlterungAusgangsgesteinChlorwasserstoffDoppelbindungPrimärelementElektron <Legierung>AktionspotenzialCarbokationFunktionelle GruppeAlkylierungAlkeneSetzen <Verfahrenstechnik>ComputeranimationVorlesung/Konferenz
35:29
MannoseVancomycinEnoleDifferentielle elektrochemische MassenspektrometrieBaseMühleAtomspektroskopieAlkeneKohlenstofffaserMolekülAtomMassendichteChemische ReaktionCarbokationReaktionsmechanismusOrganische ChemieDeprotonierungBromideVerdampfungswärmeStoffwechselwegPenning-KäfigChlorideEthylgruppeStereochemieDoppelbindungKohlenstoff-14Single electron transferGesättigte KohlenwasserstoffeAllmendeInitiator <Chemie>Substrat <Chemie>SekundärstrukturEnantiomereElektronische ZigaretteMethylgruppeSubstituentIonenbindungFülle <Speise>AmrinonWeinfehlerThermoformenBiogasanlageGezeitenstromAlterungKatalaseFunktionelle GruppePhasengleichgewichtStockfischAzokupplungWursthülleTumorStrahlenschadenAdamantanIsomerBodenschutzChemische ForschungWassertropfenQuellgebietAusgangsgesteinMeeresspiegelComputeranimationVorlesung/Konferenz
44:21
ZigarreAtomspektroskopieBohriumMannoseVancomycinGelöster organischer StoffEnantiomereBromideCarbokationChemische ReaktionAlkeneWasserHalogenideDeprotonierungChemische StrukturFunktionelle GruppeMethylgruppeKohlenstofffaserEthylgruppeReaktivitätAlkohole <tertiär->ZündholzAtomorbitalStoffwechselwegMischenReaktionsmechanismusChemischer ProzessMolekülFrischfleischIsotopenmarkierungAlterungAdvanced glycosylation end productsKonkrement <Innere Medizin>AzokupplungBiogasanlageWassertropfenLactitolChemische ForschungChemische VerbindungenTumorVerletzungBaseSystemische Therapie <Pharmakologie>CadmiumsulfidBukett <Wein>WursthüllePhasengleichgewichtGeochemiePeriodateComputeranimationVorlesung/Konferenz
53:13
Chemische ReaktionFülle <Speise>Chemische ForschungStereochemieChemisches ExperimentVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
00:07
So a reference recently to this, some news article on, of course I don't use this kind of stuff myself, maybe my wife would, about some new product that I guess is supposed to rejuvenate your skin,
00:23
make you young, and apparently it contains a multi-patented gene activating 4AR molecule. And when I see the word molecule in the news story, of course it immediately turns me on and makes me think, gee, there's something cool there.
00:40
And so I wanted to know what is it that you could be smearing on your face that might be gene activating? And these mysterious letters 4AR, right, that must be better than 3AR. And so what is this about? So I wouldn't have looked up with this 4AR molecule as I said it was developed by somebody in Utah.
01:03
And who was it in Utah that patented? Actually it's an organic chemist just like me there that made a very simple derivative of a compound that got a huge amount of press within the past decade. And that's resveratrol. It's this molecule on the bottom. I'm not sure if you can see my laser pointer here.
01:22
But at the bottom here I've got resveratrol. And it's a red wine compound that many people were touting as the cure to diabetes. The cure to diabetes, the cure to aging, the cure to every ailment known to man. And it's very active. It turns out that this molecule resveratrol,
01:42
well I wouldn't say it's super active in terms of potency. But it acts on a lot of things through this target SIRT1. So this is a compound that cleaves amide bonds. It's a protease. Specifically it's a histone deacetylase. And what that does is when this protein acts, it releases all kinds of genes.
02:01
That's the function of histone deacetylases. So they were correct. This is a gene activating molecule. The trick here behind this 4AR molecule is that a chemist went in and said, well gee, you know, everybody knows that resveratrol isn't very useful. Let me turn down the sound here. Because you have to take huge quantities.
02:21
The body clears resveratrol so fast that you have to sit there and eat it like breakfast cereal in order to keep enough in your body to do anything. And so what this chemist did is went in and made a very simple ester derivative. That's the ester right here on top. That's called an acetate ester.
02:42
It's an ester of acetic acid. And we'll teach you how to do that in Chem 51C. You may do that in the laboratory and make simple esters. So somebody going and making a simple ester here gives this molecule enough hang time so that you don't need to eat huge quantities of this for it to have some sort of an effect. Now I don't know if this is actually going to make your skin seem younger and to regenerate it.
03:05
But it is a technically trivial idea that you'll be able to accomplish just by taking a class like this. I hope in your Bio 98 and related classes, you're learning cool stuff like this about biology.
03:20
These are the kinds of things that biologists care about these types of pathways and these types of activating molecules. Okay, so let's go ahead and get back to our, I'm seeing that there's a double bond right in between these two benzene rings. And it makes me want to add something to that double bond. And so let's go ahead and talk about the chemistry of double bonds.
03:41
That's the focus of Chapter 10 for this class. And we spent the lecture on Friday, which seems forever because there was a, we just had this holiday, this university holiday. Let me try to get this document camera aligned here.
04:10
So it seems like forever ago since we just had a Monday holiday. And this is what we're going to cover in Chapter 10.
04:22
We spent our last lecture really just covering some basic concepts that aren't the kinds of useful concepts that you'll use throughout the rest of the course. But this is. These five reactions here are really Chapter 10 in a nutshell. If you can grasp these 10 reactions, you've got it. And moreover, if you can grasp these 10 reactions, you're set
04:43
up for the rest of organic chemistry, especially Chapter 18, which is a big focus for this class towards the end of the quarter. When I teach Chem 51C, when people are behind and feel like they're totally lost, I tell them go back to Chapter 10 and learn those reactions again. So these are absolutely fundamental
05:01
to your understanding of organic chemistry. Okay, so here's the reactions that we're going to learn. We're going to learn how to put a bromine atom on an alkene. And there's a very simple reagent for doing this. I can't imagine reagents that are that much more simple. And it's HBR.
05:20
And if you wanted to add chlorine in the same way, you'd add HCL. Now, this may look very mysterious to you because you can't see the H that got added there. But I added not just a bromine, I also added an H. And just in case it's not clear that there's an H there, I'll draw it for you. But I expect you to be very good by this point in time seeing the H's that aren't drawn.
05:41
There's the H that was drawn that was added with HBR. Okay, so very easy. You could do the same reaction with HCL. I've never seen anybody do a reaction like this with HI, but hypothetically, that's possible. It may depend on the structure of the alkene. Okay, so if you also want to add a hydroxyl group
06:02
to an alkene, you can add the elements of water to an alkene. But we need a trick here. We have to have some sort of an acid in order to add the elements of water. And that is to add catalytic sulfuric acid. And I'm very picky about which acid you choose. I want you to choose an acid
06:20
that has a non-nucleophilic counter ion. So don't pick HBR, don't pick HCL. Use catalytic sulfuric acid. And in order to make this work, you have to add water. And so I, and you have to add a lot of water. What makes this reaction go is that water is the solvent and it's present in big excess.
06:42
So whenever I draw a reagent below the arrow, if those reagents are common solvents in organic chemistry like water or methanol or ethanol, you can assume that that's present in a large excess. Okay, so again, I've added the elements of OH and H. Here's the OH from the water and here's the H. And if that's not clear to you,
07:03
let me just remind you, there was one H here before and now in the product there's two H's. So I did add an extra H and an OH to make this product. Now you have seen both of these reactions before. Let me draw a dividing line. You've already seen these reactions. These aren't new. And you may think, well, I didn't see that reaction before.
07:23
What you've seen is you've seen that reaction in reverse. You've seen elimination of HBR to make alkenes. We showed you strong bases like t-butoxide to do that. Now what I'm showing you is you can take that same alkene and go back again to the alkyl halide.
07:40
We've shown you how to dehydrate alcohols using sulfuric acid as a catalyst. We just left out the water was the trick. Now we're driving the equilibrium backwards simply by running this reaction in water. Okay, so you'll see the mechanisms for those two reactions that I show you look very much like the reverse of mechanisms you already know.
08:02
Okay, so in a way these are kind of new, these first two reactions, but really they're not that new. Okay, the reactions three, four, and five here, those are new, and you haven't seen anything like these before. And there's going to be a lot of stereochemical issues that come along with these.
08:20
So if I want to add two bromine atoms here to an alkene I'm going to use bromine, molecular bromine, that's Br2. You don't have to draw the bond there. I'm just showing you that Br2 has a bond in it. And this is a reagent we haven't used yet in this class, and you're going to use this a lot. And you could also do this with chlorine, Cl2, chlorine gas.
08:42
Bromine is a liquid. It's just a dark brown liquid, whereas chlorine is a gas. That's why it would be very rare for people to do this with chlorine in an academic lab. We usually use bromine because it's a convenient liquid. Okay, we can take the same reagent, bromine, Br2, and this time I'm going to draw it as a molecular formula
09:01
because that's more convenient. And if I simply run the reaction in water, same bromine, same bottle of bromine, I take out a dropper and drop a few drops in there. And if I run the reaction in either an alcoholic solvent or in water, that water or alcohol acts as a nucleophile.
09:22
And so I don't get two bromines. I get one bromine and a hydroxyl group in this case. Now I want you to notice something particular about the stereochemistry. These two bromines on top here end up anti to each other. And the bromine and the hydroxyl group also end up anti to each other, on opposite sides of where the double bond was.
09:42
So that's one stereochemical issue I'm going to harp on and make a big deal about. And the other thing, of course, here is that we've got, I'm symbolizing this with plus E to mean plus enantiomer. So you also get the mirror image. And the same down here. On exams, I always front end my questions by saying,
10:02
show any stereochemistry unless otherwise indicated. And this is one simple way for you to indicate that there's also a mirror image that's produced. You can just write plus E, and I will understand that to mean that you've generated a racemic mixture of two enantiomers. Okay, so that's two new reactions. Addition of bromine across a double bond or addition
10:22
of bromine in water, and you could do both of those reactions with CL2, with chlorine if you wanted. You cannot do those reactions with iodine. Maybe the second one in special cases, but I would never ask you to do that. Okay, so the last one here looks just
10:44
like the sulfuric acid in water. And what I'll show you is that it's complementary. It's like, if it gives the same product, why would I use this complicated bottom reaction? And here's the reagents for this complicated bottom reaction. The first reagent is a reagent you've never seen before.
11:02
It's called borane. Yes, that's a boron atom. And you haven't seen anything yet with boron atoms. So we're going to talk a lot about that when we get to our discussion of these reactions. And there's a second step here. Notice how I enumerate the steps. If I leave out the numbers, it's totally wrong. If I don't enumerate these steps as step one and step two,
11:24
then that means I've simply mixed everything together. And if I mixed everything together, I'd have a dangerously explosive mixture. And I would get none of the product that's shown. So you have to enumerate these. I have enumerated these with Roman numeral I and Roman numeral II.
11:41
You could use Arabic numerals, the book, just to show you the alternative. You can enumerate things the way the book does. The book uses brackets. That's totally inconsequential. And they use Arabic numerals. Generally, in organic chemistry, if you use these little Roman numerals, it means you didn't isolate
12:01
and purify the intermediate after the first step. Whereas if you use the Arabic numerals, it means you isolated and purified the intermediates. And it doesn't matter. Throughout this quarter, you may use Arabic numerals. Wherever I use Roman numerals, and that's fine. Because you don't have to worry about the details of workups.
12:21
Okay, step number two is to add hydroperoxide anion. And we make that in the lab by mixing hydrogen peroxide. So you can buy hydrogen peroxide. And the book shows HO minus. That's a minus sign there. If you wrote, I would prefer if you wrote sodium hydro,
12:41
in fact, let me write sodium hydroxide because it's bothering me. There is no reagent OH minus. So that's sodium hydroxide. So you mix these two things together. Sodium hydroxide and hydrogen peroxide. And I know it looks kind of complex there. Look at all those steps and those weird reagents. You don't know anything about. We'll get to that.
13:00
Just hold on and we'll show you why you would want to do this other way for adding an OH group. So again, let me just remind you that I've added two things across this alkene. There was one H here down at the bottom in the beginning. And my pen is cooperating. There you go. There was one H there in the beginning. And now in this product there's two H's.
13:21
So you can see this fifth reaction here. I've added H and OH across that double bond. So I urge you to practice until you are an expert seeing the H's that aren't drawn for you. I expect you to already be able to see that there was an H added. It'll hurt you if you can't see that.
13:41
And it will help you immensely if you're fast at it. Okay, so those are the five reactions. Let's talk about them. Let's just talk about them over and over and over again. You'll see in some ways they're very similar to each other until we get to the last one. And then things got kind of crazy. Okay, so let's talk about stereochemistry.
14:00
It's a big issue in this chapter. And stereochemical relationships. So I'm going to start off by drawing a substrate in the middle of my page here. And I'm going to try to pretend that I'm standing and looking
14:23
at this alkene edge on. If I were standing and looking at this edge on kind of like looking at a piece of paper so that the paper looks thin I would see two substituents going back and two substituents going forward. So I could see sort of the edge.
14:42
What I'm trying to emphasize is that I want to try to imagine looking at an alkene so that I can see it has a top face and a bottom face. Every alkene has a top face and a bottom face. And I want you to imagine two of those five reactions. Let's take a look at the stereochemical consequences for two of those five reactions so that we can see
15:02
that sometimes things add on the top face or the bottom face and there are two modes in which the two atoms can add. So I'll show you two reactions from that list. Okay, the first reaction that I'm going to show you is going to be the addition of bromine, Br2. So when I add Br2 across this alkene let's take a look
15:25
at what happens to those two bromine atoms. And we're going to have a special name for this that allows us to discuss it and keep things straight. So what happens is I'm going to get, there will be no double bond in the product
15:43
and I'm really trying hard to see anti-relationships by drawing this in this saw horse conformation. It looks kind of like this, this staggered conformation here. Okay, so here's the product and in the product one bromine adds from the top face and the other bromine adds
16:02
from the bottom face. And this is always true with bromine addition, with Br2 addition, one always adds from one face and then the other one adds from the other face. And you'll also get a mirror image product, I'm not going to draw on where this Br is down and the bromine on the other side is up. The important point here is that we refer to this relationship
16:22
in the product, the relationship of these two groups as anti. That's an anti-relationship and we call this mode of addition anti. So that's the description for the relative orientations of those two bromine groups in this reaction. Okay, now let's go back and look at this boron reagent. That was the fifth reaction because it's different.
16:42
If I'm going to invent this whole sort of nomenclature here, it must be because I have something, some alternative to anti. And so here's, it doesn't matter what R is, it can be H's or alkyl groups. So if I take a borane molecule that has a B-H bond,
17:00
the B and the H will add across the double bond. But in this case, you don't get an anti-relationship between the way the B and the H add, those add with a syn orientation.
17:22
And so let me draw this B and this H. So here's the H and here's this weird boron atom. I hope you're going, gee, what's with that boron thing? We've never seen that before. It's this whole group with the two R groups attached. Don't worry about those R groups, doesn't really matter. Okay, so notice clearly and obviously that the boron
17:43
and the H are now on the same face. If the boron adds from the top face, the H will also add from the top face. And if the boron adds from the bottom face, then the H will also add from the bottom face. You'll get usually a one-to-one mixture of addition from top face and bottom face. We call this relationship between these two groups syn.
18:04
And we call this mode of addition syn. So some of the reagents that I showed you on that last page, I showed you five reactions that we're going to talk about over and over and over again. Some reagents add anti, like bromine and chlorine. Some reagents add syn, like this borane with the BH bond,
18:23
the boron hydrogen bond. And then some give you mixtures of syn and anti. And so you're going to have to keep that straight. For every one of those five reactions, I want you to know, I expect you to know whether the mode of addition is syn or anti, and I expect you to understand the mechanistic origin
18:41
of that stereoselectivity. Okay, so generally what you'll find is that reactions that involve carbocations give you mixtures of syn and anti. And that'll be things like hydration or HBR addition. Okay, and we'll talk about that. Okay, so note those syn and anti relationships.
19:05
Okay, let's talk about our first mechanism for that first reaction that I showed you, the addition of HBR across double bonds.
19:20
So I'm going to start off by drawing a substrate. It'll be a simple trisubstituted alkene. And I've chosen this alkene because I want to make it clear that in this particular case, the two carbon atoms are different. One carbon atom has two substituents, two methyls. That's the one on the bottom. Whereas the carbon atom on top in the alkene has only one methyl group.
19:43
And because of that, the reaction will exhibit some regioselectivity. So if I go ahead and draw up my reaction with HBR, and I'll draw out the HBR bond because I'm drawing the mechanism for you. So I can add H and BR across that double bond.
20:01
H adds to one carbon atom in the alkene. BR adds to the other carbon atom in the alkene. And the mechanism looks something like this. We've got this double bond here. And this is what you haven't seen before. And this is really, this arrow I'm about to draw is really this chapter in a nutshell.
20:21
This chapter in a nutshell is alkenes attacking stuff. Alkenes acting as bases and to nucleophiles. And there you go. It's like, wow, I've seen three chapters already where lone pairs attack things. And now you're telling me that pi bonds can attack things? Yeah. In fact, that's, we're going
20:42
to spend both this chapter and chapter 11 and chapter 18 talking about over and over and over again. And that's the heart of modern organic synthesis. Okay, so you have to break the bond to the H, the HBR bond if we attack. So really we're just doing an acid base reaction.
21:00
And it's like, who would have thought that an alkene could act as a base? Well, if you've got super strong acids, it can act as a base. Okay, so now we're going to take this intermediate. What you can't tell, and this is what's frustrating about arrow pushing when you use alkenes is, I've just drawn an arrow here where the double bond attacks the H. But you can't tell
21:21
from my pink arrow which carbon atom picks up the H. You won't know until I draw the intermediate which of those two carbon atoms in the alkene picks up the H. Some people get really frustrated by that. And they want to, they try to make their arrows start from one carbon atom. No, you make the arrow start from the pi bond.
21:41
That's what's acting as the base. That's the pair of electrons that's engaging this HBR. And so let me go ahead and draw the intermediate for you. It's this tertiary carbocation. And let's draw the H's that are here. There's two H's on this alkene. There's one H. And here's the other H. So I started off
22:03
with one H sticking off of here. And now I have two H's. So I didn't have to draw those in for you. But just to help you see, where did that H go? Okay, that leaves, if I pop off the bromide anion, that means that the bromide anion is still floating around right here. There's four pairs of valence electrons.
22:21
I'm just going to draw one of those lone pairs. And that sets me up to see what's going to happen next. So next, that bromide anion is going to attack that carbocation, that poor, lonely carbocation that doesn't have an octet. And that BR minus will satisfy that. And so that's what gives you this final product.
22:40
We've now added HBR across the double bond. I'll typically draw it like this. I typically won't draw the H that I just added. You have to get over this. Stop. We're not going to draw H's for you anymore. You have to envision where those H's are. Okay, so we just showed you that's addition of HBR across double bonds. You can do the same mechanism with HCL, hydrochloric acid.
23:02
That's a very cheap reagent that's common in the laboratory. You could do it with HI, very uncommon. And typically, alkaliodides are so unstable. You just don't see people doing that very often. So HCL and HBR are the most common. Okay, so we're going to see that kind of reactivity
23:21
where double bonds attack things over and over and over and over again. Yes? The question is how do you know the carbocation will be on the bottom carbon?
23:42
How do you know the carbocation will not be on the top? I will talk about that. I will in just a couple's overheads show you something called Markovnikov's rule, which tells you that. Okay. Okay, that reaction generated a carbocation,
24:00
and we've been trying to teach you in this course, whenever you have reactions that generate carbocation intermediates, you have to watch out for rearrangements. So here's an example of a reaction that generates a carbocation, and you have to watch out for rearrangements. So let me go ahead and draw out what you get. Here's a substrate, and what do I see
24:21
when I see this reaction? Okay, you're going to add HBR across a double bond, but I immediately am drawn to that strained four-membered ring. Look at all that strain in that four-membered ring. It's like a square, right? Bonds don't want to be square in organic chemistry. They want to be 109 degrees.
24:41
So maybe it shouldn't be that surprising to me that somewhere in this reaction pathway, this molecule found the capacity to rearrange and get rid of the 26 kcals per mole of strain in that four-membered ring. And so let's try to follow this through. So if I ask you to explain how you generate this final product,
25:02
it's like all kinds of crazy stuff is happening there. One trick that I usually emphasize that I want you guys to try to hone in on is if you have a mechanism, if I ask you a mechanism, and you're sitting there, gee, I can't see what happens, try enumerating the carbon atoms. Try to track which carbon atoms end up where
25:22
as just a basic tool. So here's some of the things that I see here. I see in the starting material I've got an alkene, and there is no alkene here in the product. So if I arbitrarily start to number my carbon atoms, I might call this carbon atom one, two.
25:40
That's a CH2 group. Notice how I'm not just looking at the atoms. I'm looking at whether they're CHs or CH2s or CH3. So carbon atoms is a CH2 group. Carbon atom three is a CH2 group. Here I'll enumerate carbon four as a CH2 group. And the symbols you use are arbitrary. I could use smiley faces or flowers.
26:00
It doesn't matter. I'm just going to use numbers because they're easy. And what I'm trying to do here is just come up with a hypothesis. Okay, let me try numbering some other atoms. Maybe carbon one is one of these two carbons between the rings. I don't know which one. But one of the things that I immediately see is three CH2 groups in a row. I'll bet the three CH2 groups on this side
26:22
or on the other side are carbons two, three, and four. Right? In the mechanism, I don't break every single bond. There's just a few bonds that break. And so I don't really know which of these is carbon atoms two, three, or four. I might arbitrarily do this. One, four, three, two, or one, two, three, four.
26:41
You know, I don't really know what's attached to what. You start off by making a hypothesis. And if anywhere along your mechanism the numbers stop matching, then you back up and start over. And number it a different way. So let me encourage you to number carbon atoms. And you'll oftentimes pick the wrong numbering at first. But at least it helps keep you focused
27:01
on is my mechanism heading in the right direction. Okay, so let's go ahead and draw out what happens if I protonate this alkene. There's two different places you could protonate this. You could put the proton on here on the right-hand side or on the left-hand side. And I'm going to try protonating over here on the right-hand side, which will leave the positive charge
27:22
right next to this strained ring. And you wouldn't necessarily know that this is going to work out. You might have to try protonating the other place first and then if you couldn't get to the right place. So if I draw that proton adding, here I'll just draw that proton just in case it bothers you that you can't see where it went to. There's the proton that I just added.
27:41
And that leaves me with a secondary carbocation. It doesn't matter which of those carbons I protonate. I'm going to get a secondary carbocation. When I add a proton to the double bond. And now, this particular intermediate, I have to be very worried. Well, I don't know if worry is the correct emotion here.
28:02
It's what I can see is that this has a capacity to undergo a 1, 2 alkyl shift to make a more stable carbocation. Those reactions tend to be fast. 1, 2 alkyl shifts that generate more stable carbocations, particularly tertiary carbocations, tend to be very fast. And if they relieve ring strain when they do that in a 3
28:23
or 4 membered ring, they're unstoppable. So in other words, let me draw this other counter ion that's floating around. You may wonder, well, gee, how could I have guessed that the bromide doesn't simply attack the carbocation? And it's because rearrangements that relieve ring strain are always faster under conditions
28:41
that you see in this class are always faster than the, this kind of attack of bromide on the counter, on the carbocation. Okay, so let me go ahead and draw that rearrangement. If I'm claiming it so fast, there we go, zoop. See how fast that was?
29:01
Okay, now, here's a place where you might get mixed up really easily. In other words, how do I make sure I draw all of these carbon atoms faithfully? This is where numbering can often help you. If you're worried that you're not going to redraw this intermediate correctly, numbering your carbon atoms can help you.
29:20
So in this case you might want, you might consider finishing and enumerating all of your atoms, 5. There's a CH2 group number 5. Here's a CH2 group number 6. Here's a CH with the alkene, number 7. And here's a CH number 8. And right or wrong, I have to match these up.
29:40
I'm looking for a CH2 number 5. Gee, maybe it's still attached to carbon 1. So I'll draw that as CH2 group number 5. Here's CH2 group number 6. Maybe that matches. Oh, here's a CH2 group. If 6 is still attached to 7, I have to explain in my mechanism somehow how carbon 7 picked up an extra proton.
30:01
And if this is right, now I've got this very weird, gee, carbon 8 now has all these carbon atoms attached. So, again, your numbering could be wrong at first. Be just, be ready to back up with your mechanism and start over. So let me keep numbering. The numbering should still correlate here.
30:25
And now in this rearrangement, this is where I really have to keep track to make sure I don't mess up during my intermediate. So I'm going to try to make this 5 member ring, this one over here where I have carbon 1, 5, 6, 7, and 8. And you don't have to number.
30:41
If you're good enough to figure things out without numbering, great. It's just a tool to help you when things start to get really confusing. So if I move, if I break this carbon 1, 2 bond, so carbon 4, I'm not breaking. So carbon 4 has to still be attached. And carbon 3 is still attached to carbon 4. And carbon 2, 3, I'm not breaking that bond.
31:03
So let me go ahead and enumerate those so you can see. No, I'm, my arrow pushing didn't break the bond between carbon 1 and 4. My arrow pushing didn't break the bond between carbon 3 and 4. And my arrow pushing didn't break the bond between 2 and 3. But according to my arrow pushing, I broke the bond
31:21
between 1 and 2, and I made that bond move over to carbon 8. So let me draw that bond here. There it is. And if I take electrons away from carbon 1, I have to leave a carbocation there. There's that carbocation right there on carbon 1. And just to be clear, there was an H
31:40
out of attached here on carbon 8. I always see those H's, and it's still attached there on carbon 8 in this intermediate, and it is still attached in the product. So that's a relief. At least I'm not veering off in some weird direction. And now I'm going to draw this bromide counterion that's just floating around, and it's ready to attack that tertiary carbocation.
32:00
Maybe it's not drawn in a particularly beautiful way, but what I've ended up with is the right ring system with two five-membered rings and a tertiary carbocation. And so now my bromide counterion can come in and attack that tertiary carbocation. Okay, so what's the important thing? This is not new, the idea of 1, 2 rearrangements.
32:23
This is just a different reaction that leads to carbocation intermediates. The addition of HBR and HCL across double bonds leads to carbocations, just like SN1 reactions lead to carbocations. Watch out for rearrangements of carbocations, especially if they lead to more stable carbocations,
32:43
and they're unstoppable if they relieve ring strain in a three- and four-membered ring. In a three- or four-membered ring. Yeah, yep, yep, you have to assume
33:08
in this class that that's true. It's concentration dependent, so for the reactions that we show you in this class, you have to assume that at the concentrations we're showing you that the rearrangements of three- and four-membered rings
33:21
to make tertiary carbocations are faster than attack. And it's, I'm kind of doing a bait-and-switch because it's also, there's a potential reversibility, but I'm not going to, you don't need to worry about that. Just assume it's faster for three- and four-membered rings.
33:42
Okay, so how do you know which carbon you're going to add the proton to? And that's answered by, gee, I don't even know how old this rule is. It's a rule from the, I'm pretty sure the 1800s, Markovnikov's rule, maybe that's wrong,
34:01
maybe it's early 1900s, but it's about that time frame. So let's take a look at some unsymmetrical alkenes. And I'll show you a simple, really empirical rule. This was developed without an understanding of the mechanistic basis for it. But how do you predict when you've got an alkene
34:22
that has two different types of ends, and you add HCL or HBR to that? How do you know which end of the alkene gets the proton and which end of the alkene gets the halide? And that's what Markovnikov's rule is good for. What Markovnikov said is, well, look at the alkene,
34:42
and then you have to decide if the two ends are non-identical, if one is more substituted and one is less substituted. So I'll write less substituted. So the top end of this alkene is less substituted, and the bottom end of this alkene is more substituted.
35:03
I'm just going to abbreviate substituted there. So I've got two different ends. And Markovnikov's rule is very simple. You add the proton to the less substituted carbon, and you just need to remember that. So I'll refer back to Markovnikov's rule later in the class. It's an empirical rule, and you just have to remember, oh,
35:23
that means the H adds to the less substituted carbon. So let's draw up the product for this. Here's the product for that reaction. Well, wait, hold on, what, where's that H?
35:42
Well, I didn't draw it. The H is there. Here it is. There's three H's on this top methyl group that I didn't draw for you. There were two in the beginning. And when I say less substituted, I don't mean H is not considered a substituent for Markovnikov's rule. I mean substituted by alkyl groups.
36:02
Those two H's are still there, and here's that third H. And I emphasize over and over that you have to be very good by this point in time at seeing H's that aren't drawn for you. So you can see I've added H and CL across the double bond. The H goes on the less substituted carbon, and the nucleophile chloride or bromide goes
36:22
on the more substituted carbon. Okay, let's try that again. Let's see if we've really mastered this Markovnikov's rule. Okay, there's a differentially substituted alkene.
36:40
And I'm tired of HCL, so I'll use HBR now. And I add HBR across that double bond. So you should be able to figure out which of the two carbons in the alkene, one of them gets the H, one of them gets the CL. So if we apply Markovnikov's rule, then, again, it's just an empirical observation that, gee, the less substituted carbon gets the H. That's this one
37:04
down here at the bottom. So now the one on the bottom is less substituted, and now the one on the top is more substituted.
37:20
And by the end of this chapter, you're going to be very fast at recognizing, okay, that carbon is more substituted, that carbon is less substituted. And so when I draw this product out, the bromide is going to be up here, and that H that we usually don't draw,
37:45
we usually don't draw the H, I'll draw it for you, is down there. Again, I didn't draw the other Hs, so it just seems weird to me that I'm drawing that one H in. Okay, so the H goes on the less substituted carbon.
38:01
So you're going to be using this over and over. I expect you to be good with that rule. That's a common kind of question for MCATs and stuff like that. Okay, why does that work? Why is it that this, how is it that Markovnikov figured this out, and he didn't know anything about the mechanism? You could have invented this rule if you knew the mechanism.
38:21
Actually, you don't need the rule if you understand the mechanism for the reaction. So let's take a look at what's going on mechanistically that makes the H add to the less substituted carbon. I'm going to come back to that initial example, and I'm going to try to draw it kind of here vertically in the middle of my page, because I want to show
38:42
that this substrate has two possible pathways, and it chooses one. So here's an alkene. The top carbon in my alkene is less substituted, and the bottom alkene is more substituted, and this has two possibilities when it adds to HBR. Here, I'll draw my HBR molecule just seductively
39:02
hovering over that alkene, waiting to do something, and it's got two pathways. It likes the bottom pathway that I'll show you in a moment, and then we'll try to rationalize why it doesn't take this top pathway, and so, again, let's just note for ourselves
39:20
that this carbon atom over here on the bottom is more substituted, because this is what we're paying attention to with Markovnikov's rule, and the top alkene, top carbon is less substituted. There we go. Okay, so when this double bond attacks, when this pi bond,
39:42
this CC pi bond attacks that H, the proton. I'll draw my arrow starting from the middle of the pi bond. That's the way we do it, and we're going to pop out the bromide leaving group, and so the question is, does the proton go to the bottom carbon or the top carbon?
40:00
And what I showed you when I drew up that mechanism was that that proton likes to go to the less substituted carbon, that top carbon, to give you this tertiary carbocation. I'll just, in my tiny pen here, sketch out that H atom that I just added. That's not our focal point. The point is that this is a tertiary carbocation.
40:23
I'll just write tertiary R plus. That's tertiary. If I protonate the bottom carbon, if I did stick the proton on the more substituted carbon, it starts to look very painful.
40:41
There's that H, and now what I'm left over with, left with over here, is a secondary carbocation. More stable carbocations form faster.
41:01
This is a rule you already know in organic chemistry, and I'm not telling you that more stable products always form faster in organic chemistry. That's only true half the time, but for carbocations, more stable carbocations form faster, and so given a choice to protonate on the top carbon and form a less stable, sorry, on the top carbon
41:21
and form a more stable tertiary carbocation, that's the choice, and the alkene does not pick up the proton on the bottom carbon because that would leave an unstable secondary carbocation. That's why Markovnikov's rule works. Markovnikov's rule works because it basically, when you stick the proton on the less substituted carbon,
41:40
you leave a carbocation on the more substituted carbon, and so that's why Markovnikov's addition correctly predicts that the nucleophile will be attached where there was a more stable carbocation. So there's the bromide after it attacks, and there's the H atom that you typically don't draw,
42:00
but I'll draw it in there so we don't get lost there. Okay, so Markovnikov's rule works. You don't generate less substituted carbocations. When it's a race, it's more substituted and more stable carbocations that form faster. Okay, so that's why Markovnikov, so again, you don't have
42:21
to remember Markovnikov's rule if you know the mechanism for the reaction, there you can just fall back on, oh gee, more stable carbocations form faster in organic chemistry, and I'm not going to draw this other possible pathway because I've already shown you the top pathway is not the favorable pathway there.
42:43
Okay, so let's talk a little bit more about issues of stereochemistry because this whole idea that we can add to top faces or bottom faces of double bonds, that's some very real implications for how you should draw your answers for my exams and for your understanding of organic chemistry.
43:03
So let's go ahead and plot out how you get these two possible products in this reaction. So if you had HBR across this alkene, and generally on my discussion section problem sets on my exams I say clearly indicate stereochemistry
43:22
and the presence of stereoisomers if any, so you'll, if I ask you to draw the products you need to draw something like this or draw a plus E. These are enantiomers of each other and so why do you get to enantiomers, to help you see that I'm going to try to draw the substrate edge on again because I want to try to envision this idea that there's a top face
43:41
and a bottom face to this molecule. So that bottom end of the alkene has an ethyl group on there. I'll just draw that ethyl group like this and then if I'm trying to envision this so I can see those two faces then some of the substituents will be sort of angled back and some
44:00
of them will be angled forward. So I'm just trying to draw this edge on so I can see, okay, alkene has a top face and a bottom face. Every alkene has a top face and a bottom face. There's electron density on both faces of that alkene and so when this HBR adds it's equally possible and we'll get to this in a minute that it adds from the top face
44:20
to the bottom face. That's not the important point yet. We'll talk about that in just a moment. So when I take my HBR molecule and I use that alkene as a base, which is weird to think about, the alkene is a base, to grab a proton from the HBR. I want us to look at the carbocation that we generate.
44:44
Just to remind ourselves that when we look at that carbocation, which I'm also going to try to draw an edge on here. So now I've got a CH3 group. If I pick up the proton and I follow Markovnikov's rule, Markovnikov's empirical rule says I want to put the proton
45:00
on the carbon that is least substituted. So that's now going to be a CH3 group and it's kind of weird. It's like all of a sudden I'm drawing it as a CH3, whereas here I just symbolize it as normal. You have to try to keep that straight. So now I've got this carbocation intermediate.
45:21
It's got a big, fat, empty P orbital there and it is equally possible for the bromide counterion to attack from either the top face or the bottom face. So if I make the bromide add from the top face of the carbocation, that will give me this enantiomer
45:44
over here and if you assign Coningle, Pralog, stereochemistry, you'll find that's the S enantiomer. But it's just as possible that the bromide could add from the bottom face. So I'll write or the bromide adds from the bottom face.
46:00
You know, there's no steric difference between these two faces. I could spin this little ethyl group around any way I want. If I add, let me make that arrow go a little closer here. Sorry, I didn't do a good job of drawing that. So if I make my bromide attack the carbon from the bottom face, that's going to give me the other enantiomer and I'll get a one to one mixture of those enantiomers
46:20
because it's equally possible for the bromide to attack from the top and the bottom face. So that's something you should already know from SN1 reactions. This is just like the products you get when you take tertiary alcohols and you add HBR. We're just generating the carbocation intermediate in a different way.
46:46
Okay, so the last stereochemical, what I'm doing here is I'm spending a lot of time to harp on stereochemical issues and all these things that I'm saying for HBR and HCL, they're true for HCL and they're also true
47:00
for the addition of water which we'll talk about when we come back on Friday. So really I'm going to spend, of those five reactions that I spent that I said were so important, we've basically spent this whole lecture just talking about reaction number one. Okay, so let's talk about a different issue and again, I'm going to have two different pathways, one going toward the top and one going toward the bottom.
47:23
And I'm going to take an alkene, in this case, I'm not worried about Markovnikov's rule, I just want to come back to this idea that there's two faces to every alkene. And so when I add HBR to an alkene that has two different faces, the proton can add
47:40
from the top or the bottom face. And so let me go ahead and draw, let's just imagine what will happen if I protonate this alkene from either the top face or the bottom face. And let's not, I'm not going to worry about the fact
48:01
that there's two different carbons, top and bottom, because those are, those have equal reactivity. So if I protonate the bottom carbon, there's two different faces where I could add the proton from. And so on the top pathway here, I'll draw what happens
48:21
if I protonate that alkene from the top face. So the proton is on the top face and that methyl group is now pointing downward if I've protonated on the top face. And now on the bottom pathway here, I'll show what happens if I protonate that alkene on that same carbon, except I protonate from the bottom face.
48:44
And there's my proton on the bottom and there's my carbocation. It is equally likely, let's ignore regiochemistry, I don't want to worry about protonating the top carbon, I just want to worry about faces of the alkene now. So it's equally likely to protonate from top and bottom face.
49:01
I'll have a one-to-one mixture of these intermediates floating around. Now when I do that, this is just to show you how hairy things can get and why you have to know, understand the mechanism to appreciate the product distributions. So now if I have bromide counter ion floating around along with that carbocation, remember what I just told you
49:22
on that last overhead, I just told you that with carbocations the bromide can attack from both the top face and the bottom face of the carbocation. So this carbocation in the top pathway can give two different products. If bromide adds from the top, then the bromide ends up on the same face as the H. That would be syn addition.
49:45
So if bromide adds from the top, I'm tempted not to draw this proton here, but I'll draw it in anyways. We usually don't draw protons on our structures.
50:03
So if the bromide adds from the top, then it ends up on the same face as the H. And again, that's syn addition. I'm not going to label every one of these. But it's equally, you know, it's also likely that the bromide can attack from the bottom face. And so in that case, the bromide would be down
50:24
with a dash, and that H will still be up, and now the H and the Br are on opposite faces. That would be the anti-product. So I'll get both of those products in this reaction,
50:41
both syn and anti-addition. And that's because you can protonate and generate a carbocation, and the halide can attack that carbocation from either face. Let's follow through with this bottom pathway, and we'll see how the product distribution relates to that from the pathway above. So I'm going to avoid attacking that positive charge.
51:02
I'll use my arrow to attack the carbon atom. Okay, so once again, the bottom pathway, the difference is that I was drawing out the initial protonation occurring from the bottom face. So both of the products I generate here will have that proton going down.
51:21
Look at all these products here. So if the bromide adds from the top face, I should have drawn this other line going down. If the bromide adds from the top face, now it's going up,
51:41
and that methyl group is going down. And so now that gives me the anti-product. Sorry, the, you know what, I didn't mean to do, I'm just, it's going to become very difficult for me to, I'm going to scratch that out, and I'm going to put the bromide. What if the, let's consider what happens
52:01
if the bromide attacks from the bottom? I just want this to try to match. If the bromide attacks the carbocation from the bottom, then this will be the syn product, H on the bottom, bromide on the bottom. And if I take that same carbocation and I add the bromide from the top face, that would give me the anti-product.
52:31
So there's four products that will arise from this really simple reaction. I take an alkene and I add HBR across this.
52:40
Now, there's a simple relationship here. These two products are enantiomers. In other words, as long as you drew one of them, you could just write plus E. And these two anti-products, those are also enantiomers. As long as you drew one of them, you could just write plus E if you were asked to draw.
53:01
So if you can see these enantiomeric relationships, you wouldn't have to draw all four of these for products in your reaction. You could just write two of them and write plus E, as long as you pick the right two. Okay, so that's it for the first reaction of those five reactions. I'm not going to go through all this stuff for those other reactions because the concepts
53:24
of stereochemistry and Arkonikoff's rule are similar. ------------------------------d4a2b5b9a4b5a0