Lecture 20. Radicals
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VancomycinMan pageEthanolEnolPhenolsAltbierSodium hydrideMagnetometerIce shelfWaterStratotypProanthocyanidinHydroxylBreed standardGoldPotato chipPhase (waves)Klinische ChemieExtractPreservativeActivity (UML)Chemical compoundKlinisches ExperimentChemische NomenklaturAcidCatechinChemical plantPharmacopoeiaBottling lineNaturstoffBET theoryGesundheitsstörungCuring (food preservation)FerulasäureRecreational drug useStatinCirculation (fluid dynamics)IsotopenmarkierungBenzeneStuffingGasolineGasAlcoholHeadacheInflammationLecture/ConferenceComputer animation
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Man pageProton-pump inhibitorSetzen <Verfahrenstechnik>VancomycinArginineMagnetometerInsulinNitrosamineMixtureBohriumLone pairAltbierTopicityProtonationFunctional groupVerdampfungswärmeOrganische ChemieAminationLactitolProcess (computing)Chemical structureStuffingOptische AktivitätSetzen <Verfahrenstechnik>Emission spectrumIonenbindungZunderbeständigkeitAzo couplingMoleculeElectronic cigaretteOrigin of replicationSense DistrictCyclohexanonBromineButylChemistryHeadacheAction potentialNuclear magnetic resonanceSingulettzustandGrading (tumors)Carbon (fiber)AmineChemical compoundMethylgruppeDoppelbindungLone pairRadical (chemistry)Library (computing)AgeingLambicEthyleneAmmoniaGezeitenküstePEEKWaterMultiprotein complexComputer animationLecture/Conference
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Man pageSemioticsBis (band)Optische AnalyseBohriumMagnetometerVancomycinCalcium hydroxideInternational Nonproprietary NamePan (magazine)Hydroxybuttersäure <gamma->LavaRadical (chemistry)StuffingChlorinePondSetzen <Verfahrenstechnik>Chemical reactionHydrogenWursthülleBromineAlkylationHydrogen bondFunctional groupAlkaneCarbon (fiber)AlkaneBiomolecular structureIonenbindungBy-productReaction mechanismStereoselectivityNuclear magnetic resonanceGrignard-ReaktionElectronic cigaretteElimination reactionAfterburnerRiver sourceWine tasting descriptorsElectronGesundheitsstörungClick chemistryCobaltoxideSemioticsProtonationSolutionWeaknessSpaltflächeCarbokationAtomic numberWalkingBiosynthesisAgeingCalculus (medicine)PeriodateStiffnessSeleniteThermoformingAreaStress (mechanics)Spawn (biology)MixtureKatalaseAnaerobic digestionElectronegativitySubstrat <Chemie>Coordination numberLecture/Conference
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Man pageSodium hydrideInternational Nonproprietary NameKoch (surname)Dominican RepublicMagnetometerVancomycinMixtureArginineKohlenstoff-14MethylgruppeRadical (chemistry)Elimination reactionEthylgruppePropylgruppeCarbon (fiber)Spawn (biology)BromineIonenbindungChemical reactionElectronBiomolecular structureRekombinante DNSMixtureElectronic cigaretteChain (unit)Process (computing)AlkaneAtomic orbitalSpeciesDeposition (phase transition)HydrogenProtonationElektronentransferAtomWalkingSubstrat <Chemie>Hydrogen bondLibrary (computing)Chemical compoundAtomic numberMoleculePeriodateCobaltoxideAgeingOperonPhase (waves)Plant breedingPondHaloalkaneDiet foodBattery (electricity)GesundheitsstörungKatalaseFunctional groupLecture/Conference
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Man pageHydroxybuttersäure <gamma->Nana (manga)BohriumEnolCalcium hydroxideVancomycinAtomic absorption spectroscopySetzen <Verfahrenstechnik>MagnetometerArginineMedroxyprogesteroneBenzoylS-Adenosyl methionineMashingHausmannitAnomalie <Medizin>Carbon (fiber)CobaltoxideGesundheitsstörungPeroxideAgeingBromineLactitolReaction mechanismChemical reactionIonenbindungMedical historyProcess (computing)NobeliumBiomolecular structureCalcium ammonium nitrateSprayerAerosolRadical (chemistry)SemioticsSetzen <Verfahrenstechnik>DoppelbindungElectronic cigaretteSubstrat <Chemie>Addition reactionWine tasting descriptorsAtomConcentrateFunctional groupSpawn (biology)Hydro TasmaniaWursthülleActive siteElectronDeposition (phase transition)PipetteHydrogen peroxideChemical structureStereoselectivityKohlenstoff-14HydrogenSauerrahmAcetonitrileSubstitutionsreaktionDrop (liquid)Hydrogen bondOxycodonInitiation (chemistry)BromideAlkeneCatalytic converterAllylStuffingInternationale Union für Reine und Angewandte ChemieResonance (chemistry)Paste (rheology)Mixing (process engineering)Potenz <Homöopathie>Computer animationLecture/Conference
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FirnChemical reactionChemical reactor
Transcript: English(auto-generated)
00:07
I stopped by the local Trader Joe's supermarket yesterday morning to pick up some water and turned around. And I saw on the shelf behind the checkout stand there were some bottles of something labeled pycnogenol,
00:22
which I've never heard of, but that to me sounds like a chemical name. And if it ends in all, it's some sort of an alcohol. And so I immediately got interested in what is pycnogenol that they're selling to people. It says here, may support healthy circulation, exclamation point.
00:40
So what is this stuff? It turns out the pycnogenol is an extract of the bark of these pine trees called the maritime pine. And the Trader Joe version of this pycnogenol is isolated from the French version of the maritime pine pinus pinaster. So I got very excited about the idea that, gee,
01:01
maybe they're taking pine bark and extracting some useful thing out of that. Maybe I could make tea out of my own pine bark at home, and it'll have some therapeutic benefit to that. To my disappointment, it turns out that pycnogenol is not actually one chemical compound that has an IUPAC-like name to it.
01:21
It's actually, the extract is actually a lot of different compounds, dozens of different compounds. In order to label anything pycnogenol, so pycnogenol is this crude extract. The U.S. pharmacopeia has laid down the law saying it has
01:40
to contain these four natural products that you'll find in lots of different plants and extracts. You'll notice that two of these are very similar, catechin and taxofolin. These two are not quite so similar, cathaic acid and ferulic acid. And the common feature of all of these is they are phenolic. They have hydroxy groups attached to benzene rings.
02:02
Any time you have a hydroxy group directly attached to a benzene ring, you can expect that to have antioxidant activity. The kinds of preservatives that they put in potato chips have hydroxy groups attached to benzene rings. That's not that spectacular. There are a lot of companies that now sell pycnogenol,
02:20
these extracts that come from this French maritime bark. And so the question is does pycnogenol, this bark extract, really support healthy circulation? I don't know. There's actually clinical trials that have gone on, many different clinical trials, nothing beyond phase two, which would be the gold standard of approval. Maybe it does.
02:40
Looks like it probably does. Of course, so does aspirin. So does steroids. There are lots of compounds that can improve circulation and would be related to things like heart disease, Alzheimer's disease, anything involving the brain, headaches, inflammation, menopause. So it may be active towards that, but there's lots
03:03
of things that are active for circulation. Is this going to be the next big hit, like the statin drugs that have gone a long way to cure heart disease in the United States? I'm not going to bet money on that. But very cool. The name is very cool. Okay, so we have an exam coming up on Friday.
03:23
So remember to be here in your seat at 7.55 so we can start at 8 a.m. because we have a very limited amount of time. You know there's a class that comes in here after us. And bring your ID for that. I've already posted your seat assignments on EEE grade book. And so be ready to sit in your assigned seats.
03:42
The problem set for NMR is due this evening. And I've got one last little snippet to cover about, or a couple last little snippets here to cover about NMR before we start on this next chapter, which is radicals, that I didn't quite have time to finish. And so here's one important feature
04:01
that I was leading up to. And let me just succinctly state here that this splitting really complicates NMR. And it can make NMRs of even simple, seductively simple looking compounds will have complex looking NMRs if you have multiple different types of coupling going on. So let's take an example of this dichloro-stiral compound,
04:27
don't worry about the name, that has three protons on here. And what happens if they're coupling to each other with different magnitudes of coupling constants? This will create huge headaches in the NMR. And we refer to these types of effects
04:40
as higher order splitting. And basically I'm not going to have this on my exams, higher order splitting. If you're taking the Chem 52 lab, then you need to know about higher order splitting. If you made cyclohexanone or some sort
05:01
of t-butyl cyclohexanone in the Chem 51 lab, well, you weren't able to interpret all of that NMR because of this type of higher order splitting. So just to give you a sense for what's going on here, imagine taking one peak, one proton NMR peak, and then having it split by some neighboring proton to give two peaks.
05:21
And so if I split that peak into two peaks, now it will look like this. And there's going to be some coupling constant associated with that. So I'll just measure that coupling constant, J equals 7 hertz. As it doesn't matter what the coupling constant is, I'm just trying to help you imagine what happens when a peak gets split into two peaks by some magnitude.
05:43
And now let's imagine there's a second proton that is splitting this. And so now this peak could get further split. So now there's more splitting. And so now this can further be split again by an even larger magnitude. So imagine taking this pair of peaks.
06:01
I don't know if I can do this with my fingers here and splitting that pair of peaks even further to make four peaks. And so what you'll end up with is a pattern of four peaks. Here it is. And now here it's split again. And I'm really exaggerating this so that you can see. So now I can go from one peak, but with two different types
06:21
of splitting, I can create four peaks out of that. So you can still see there's that original coupling in here. But in addition to that, there's a bigger coupling, which I've grossly exaggerated. So that new bigger coupling there, it's more than twice as big, I can tell by my eye. But just imagine it's only twice as big, 15 hertz.
06:40
So this is what happens when you have two different types of coupling that aren't of the same magnitude. Right? You can go from a singlet to a set of four peaks. And what is important is that those four peaks are not a quartet. That is not a quartet. We call that a doublet of doublets.
07:00
Here's a doublet of doublets down in my spectrum. Here's a doublet of doublets. And the reason I know that that's not a quartet is because a quartet has peak ratios of one to three to three to one, and I don't need to see any exact integration there to tell that is not one to three to three to one.
07:20
Those sets of four peaks here look closer to one to one to one to one. So that's not a quartet. So every time you see four peaks, even if they look like they're equally spaced, it has to have that one to three to three to one ratio for you to call it a quartet Q. These are doublet of doublets. Okay, so splitting can become very complex when there's more
07:41
than one type of proton, and I'm not going to put examples like that on the exam. It's important stuff for interpreting proton NMRs of more complex structures, and enough said about that. And again, it makes proton very complex, but once you learn proton NMR, it makes it very powerful
08:03
to help you deduce the structure of organic molecules. Okay, so an important feature that I already mentioned of NMRs has to do with the equivalence of protons that is created by bond rotation. So one of the most common situations where you encounter that is a methyl group.
08:22
So in other words, all of these protons on a methyl group and on every methyl group are indistinguishable by an NMR. The H's on a methyl group are indistinguishable because I can rotate about that bond, and once I rotate, you can't tell which proton was which before I rotate it.
08:40
And so that causes all the protons on any methyl group to be indistinguishable. But remember, you cannot rotate about a double bond. On the NMR time scale, on the laboratory time scale, on the time scale of your four-year degree here at UCI, that double bond will not rotate. That means the H that is next to cis to the bromine or syn
09:03
to the bromine is never going to experience the same environment as the H that's anti to the bromine. So you have to look for double bonds. There's a potential for H's attached to double bonds to be distinguishable and to give, unless there's just some weird symmetry, they'll be different if there are the two protons
09:22
on one carbon of a CC double bond. So bond rotation can make things identical. Not all processes on the molecular scale are fast relative to the NMR experiment. And so this is the last thing that we'll say about NMR.
09:42
So bond rotation makes things equivalent, makes the H's of methyl groups equivalent on the NMR time scale. Let me give you an example of another process that's not bond rotation that makes things equivalent. And it's kind of not important here so much for this chapter.
10:04
But coming up in Chapter 25 at the end of Chem 51C, this is going to be sprung on you as though, oh, you knew this, but you will have forgotten it by then. And so I'm just going to tell you now and I'm telling you you're going to forget it so that when you were told you were supposed
10:21
to remember it you'll, it'll jog some neurons in there. Okay, so when you take amines, like ammonia, ethylamine, that's the whole Chapter 25 is all about amines. There's a lone pair that makes that pyramidal. And so when you look at that you might think that these protons could be diastereotopic if there was some chiral center nearby
10:42
or some other chiral center in the molecule. But the important thing about amines is that they rapidly invert. And again, you'll cover this in Chapter 25, the idea that amines rapidly invert. And this inversion process, kind of like an umbrella flipping inside out, is so fast that those two protons will be able
11:03
to effectively switch their environments. So that fast, rapid process there of having protons, of that flipping through amines, makes protons on most amino groups indistinguishable. If you have an NH2 group the H's
11:21
on an NH2 group are indistinguishable even if they look like they're diastereotopic. So I'll just write never diastereotopic. No matter how many other stereocenters or other features there are in the molecule, diastereotopic. And it's because of that fast flipping.
11:43
Let's take a look at another type of process that is not bond rotation that can allow protons to switch places very rapidly. So you need to imagine this picture where we take a molecule, and I'm going to do my best job here to draw a chair. And I've had years of practice, but I hope you've been practicing drawing chairs.
12:02
So here's my example of a chair. And when I draw the two protons here on the right-hand side, the right-hand carbon of this molecule, they're not identical. One of those protons is axial and the other proton is equatorial and that's not the same.
12:21
So I have an axial proton and an equatorial proton that are distinctive. They experience different environments. They'll couple with each other, with other protons in a different way, axial and equatorial, with different coupling constants. But because chairs flip very rapidly on the NMR time scale,
12:40
in the NMR experiment what happens is you pulse this with energy and the molecule sits there. And while it's sitting there waiting to emit its energy back, it will flip many hundreds of times. So I'm drawing a chair flipped version here. So I don't know if you're good at drawing two different types of chairs.
13:00
Chair flipped, but you need to flip that chair. I'm good at that because I've had lots of experience, but you should get good at that. I remember when I was an undergraduate I used to just sit and draw chair after chair after chair practicing to see how good I could draw chairs. You may think, what a dork.
13:22
I was into that. And what I noticed is that most of my colleagues who ended up going into organic chemistry were into that kind of stuff. It turns out that well-drawn chairs help you see anti-periplanar relationships. That's a really important skill. So now the top proton that was axial is now equatorial. And this proton that was going down that was equatorial is now axial.
13:42
So that fast ring flipping causes axial and equatorial protons to rapidly exchange. So unless you have some big bulky group on here that prevents ring flipping, like a t-butyl group, you can expect cyclohexanes to rapidly flip and cause the CH2, the H's, the CH2's to exchange positions.
14:04
So in other words, when I go take the spectrum of cyclohexane I don't see the difference between the axial and equatorial protons. They all look identical. Okay, the last issue of things that might prevent splitting is something I already talked
14:20
about before, and that is hydrogen bonding. So as we mentioned before, you can't see any typically in most cases, so in 95% of cases you won't see any splitting between OH protons and the protons that are on adjacent carbons.
14:41
And that is simply because hydrogen bonding starts, there's so many zillions of different ways that this can hydrogen bond back and forth between acceptor atoms. And so I'll just put a water molecule here. There's so many different ways that that proton can bounce back and forth with water molecules on the NMR time scale that you don't see any coupling between protons
15:02
and the neighboring protons. So those are things that are fast on the NMR time scale. This proton transfer of this bond vibration of a hydrogen bond is fast on the NMR time scale. So you don't see splitting. Okay, so that's it for NMR. Again, we're not going to have high order splitting
15:20
on my exams, but you should know what that is, and when you run across that, for example, when you run across a doublet of doublet, you should know, gee, that's not a quartet. There's something else going on there. Okay, so let's move on to chapter 15. It's actually, there's not a lot of, it looks like there's a lot of stuff in that chapter,
15:41
but I'll show you why there's not a lot of stuff in that chapter. So we're going to talk about radicals. I mean, you can have people that are radicals that are radically different, but we're going to talk about radicals in the context of organic chemistry,
16:00
and what I'm going to do is I'm going to give you a series of neutral radicals that will be common when we discuss radicals in organic chemistry so we can talk about stability, and I'm going to show you that if you want to have an unstable radical, let me start up by drawing this radical. This is what I would consider to be a reactive radical. It's an oxygen atom.
16:22
And it's an oxygen atom that has only seven electrons, right? It's not an alkoxide minus. There's no H on that oxygen or no R group. It's missing something, and what it's missing is the eighth electron, and oxygen is electronegative. You know that.
16:40
So you can imagine how angry that oxygen atom is, that it only has seven electrons. Here's an electronegative oxygen atom, and I promise you that thing is going to pluck some sort of an atom from somewhere in order to make itself happy. Very selfish, very angry. It's going to pluck an H off of something, and we'll see some stuff like that.
17:02
Okay, let's take a less electronegative atom, something that's not. What I'm going to do is I'm going to draw some radicals in terms of increasing stability, and at the high end here, this is the most reactive and least stable.
17:20
So an alkoxy radical is the least stable in this series, but it is the most reactive and the most unhappy. So let's compare this to a halogen. I could draw chloride. It's the same idea, but I'm going to draw bromide, which is more common. It does more useful chemistry. Bromine is not as electronegative as oxygen,
17:41
so you can guess. It's not happy, but it's not going to be as reactive as an oxygen-based radical, and so, you know, if I compare these, oxygen-based radicals are more reactive than Cl dot, and then once again, there's only seven valence electrons with bromine. I don't want to use the term octet rule, because bromine is not in the second row.
18:02
The octet rule is only for second row atoms, so you can have more than eight electrons on a bromine atom, but in this case, it wants eight valence electrons. I guess that's the important idea here. Okay, so a bromine radical is more reactive than a radical that's centered on carbon, so if I come over here
18:24
and I draw a series of carbon-based radicals, let me try to draw these in black so I can be consistent here. I'm going to start by drawing three different types of carbon-based radicals, and the difference here will be whether they're primary, secondary, or tertiary. A primary radical is less stable than a secondary radical,
18:42
and that's less stable than a tertiary radical, so as I go through this series of primary to secondary to tertiary, they get more and more stable. Again, the carbon doesn't have an octet. It's only got seven electrons. That can't possibly be happy,
19:00
but the more alkyl groups you have, the happier that that carbon radical is going to be, the more alkyl substituents you have, and that's just like carbocations. I don't think that's that weird. Carbocations aren't happy because they only have six electrons.
19:22
Attaching alkyl groups makes them happier, so it's the same idea with radicals. Tertiary radicals, there's this change in stability as we walk through this series that primary radicals are more reactive and less stable. Secondary radicals are in between, and tertiary radicals tend
19:42
to look rather stable to me, not stable enough for you to isolate that would require some weird structure, but of those three types, tertiary radicals are the most stable, and primary radicals are actually very hard to form. Okay, the last structure I'm going to draw in this series is a rather special type of radical.
20:03
It's a radical that's primary, but it's a special type of primary radical, and that would be one that has a carbon-carbon double bond next door. In other words, not where the carbon is part of a CC double bond. That would be just insane and crazy unstable,
20:21
but if this CH2 group is attached to a double bond, that is going to have extraordinary stability, and we'll talk about that. So I would consider that to be a really stable radical, and we call that an allyl radical. Okay, so there's this trend. Electronegative atoms, they want eight electrons, not seven.
20:41
Oxygen radicals are angry and reactive. Carbon-based radicals are more stable, but it depends on how many alkyl groups are attached, and if you're going to form a carbon-based radical, then you really want to form a tertiary carbon-based radical or one that's allylic.
21:02
Okay, so we're going to introduce a very different type of arrow pushing for talking about radical reactions. I hate this, because it's so hard to get regular arrow pushing down. The idea that just for one chapter I'm going to introduce a single type of arrow pushing that we only use
21:22
in this chapter and we never see again, that kind of kills me because I feel like pedagogically that's not a good idea. So let's take our angry oxygen radical. It really wants to have eight electrons, and let's just imagine that it bounces into a piece of your DNA, and look at that proton,
21:44
and not proton, that hydrogen atom there. That hydrogen atom has an electron. When oxygen-based radicals collide with your DNA, they will pluck hydrogen atoms off the DNA backbone. This is why everybody talks about free radicals and, oh,
22:00
you should take antioxidants, and it's not oxidants they're worried about. It's free radicals on oxygen that they're worried about. What will happen is that this oxygen will pluck off this hydrogen atom, and it will leave behind a carbon-based radical that's more stable than the original oxygen radical. The oxygen-based radical was not stable.
22:22
Now that oxygen is happy, and we need some way to depict this process where the H moves from carbon to oxygen and brings with it a single electron, and we can't use the regular arrow pushing that we've been using throughout the book, and we'll continue to use throughout the book. We need a new type of arrow pushing, and the types of arrows
22:43
that we use are called fishhook arrows. So here's the way we use fishhook arrows. I'm going to take this bond here. We know that the HC bond has to break. One of the electrons in the HC bond goes back to carbon, and so I'll depict that with a half-headed arrow, and there it is.
23:00
Notice that little half head? It looks kind of like a fishhook upside down. So we call these fishhook arrows or half-headed arrows. You may not have noticed this, but when you do the sapling problem sets and you draw arrows for arrow pushing mechanisms, if you click on the arrow, it toggles back and forth between double-headed and single-headed. And you'll work some sapling problems where you use
23:23
that feature where you click on the, you just tap, whatever, mouse click on the arrow, and suddenly it toggles back and forth between double-headed and single-headed. Okay, so there's a second electron in this bond. When I break that bond, I have to do something with it. What I'm going to do is I'm going to depict using a second arrow that the second electron
23:41
in that bond recombines with this single electron, this unpaired electron on oxygen in order to make a new bond. So notice what happens. I don't show some arrow going from electrons and attacking proton. None of these arrows ended on a proton. This is totally different from the regular arrow pushing
24:03
that I've been using before. It's an accounting device. It just is an accounting device to help me see where do the electrons go. This O dot electron and one of the electrons from this bond combine to make a new OH bond. There it is. I'll just darken it in blue here, so I don't know if that's darkening it in blue, so you can see.
24:21
There's two electrons, and here's the third electron. It's still sitting here on carbon. Okay, so that's fishhook arrow pushing. And we won't use this ever again after this chapter, so I'm not a big fan of forcing you to stop and learn this new type of arrow pushing, but we're going to have to do that.
24:41
And you may end up on some standardized exam at some point staring at this stuff, and you'll need to know how it works. Okay, so new type of arrow pushing just for this chapter and not beyond this chapter. Yes?
25:00
No, there's nothing in this chapter 15 will be on the exam. This is all for the final exam, will be wrapped into the final exam somehow. Yeah, the exam on Friday we'll just cover through chapter 14, the NMR chapter. So what's that, 11, 12, 13, and 14.
25:21
But that's not to say you still have to invest in our lecture today. Okay, so how can you use this reaction? The reactions of radicals turn out to be sometimes not obvious that they're occurring, but when you see mechanisms that you don't understand many times those involve radicals.
25:43
Let's take an example of a reaction that involves radicals. And it looks like it involves reagents that you've seen before. We're going to take bromine. You've seen that reagent before, simple alkanes. I don't think you've seen reagents like simple alkanes before.
26:01
Those generally don't react. But if we choose our conditions carefully, look at that. I just added a leaving group. Think of all the things that you can do with that now. You can do E2 eliminations, SN1 reactions,
26:21
E1 elimination reactions. When we get to chapter 20 we'll show you how to make a super powerful nucleophile called a Grignard reagent out of that. I can start with something that has no functional groups on it and using this simple reaction. And there's one secret key that I need to remember here. And that is we need something to make this work.
26:41
Basically we need something to generate radicals. And that is energy. Here's one type of energy that you'll use. Light. You need some sort of photon source in order to get some radicals going in this reaction. And the book gives you another way to do that, although this is less common, and that is heat. And I'm less of a fan of showing you heat as a way
27:02
to generate radicals because it looks so much like the non-radical reheating stuff we've been doing in other chapters. So I'll generally choose, you know, the book says, oh, use either one, I'm not sure why they tell you two different things. Why not just tell you light? Okay, so yeah, if you heated it hot enough you could get
27:22
some radical stuff going on here. Okay, so notice nowhere above the arrow does it say radical. Right? You can't just look at the reagents and, well, as soon as I see light I'm thinking radical. Okay, there's another byproduct here. You may see that I formed a new carbon bromine bond because I'm trying so hard to make a big deal out of this,
27:41
but I want you to notice that something else happened here, and that is that I broke this carbon hydrogen bond. And in some ways that's as important as seeing the fact that I formed a new carbon bromine bond. So where did that H go? It's over here. It's in the byproduct for the reaction. The byproduct of the reaction that we generally don't care about is HBR. So there were two bromines in BR2.
28:02
One went to the alkane. One went to HBR. Now hypothetically you could do the same thing with chlorine. The problem is that chlorine ends up everywhere. Right? If you take chlorine and apply the same conditions, you'll end up chlorine all over the place. In this case above,
28:20
what I showed you is the reaction is selective for tertiary. And there's a phenomenal degree of selectivity here, and it goes like this. When you do this reaction, tertiary RHs, alkyl carbon hydrogen bonds, are much, much faster to react than secondary alkyl hydrogen bonds.
28:42
And that's much, much faster. In fact, I've never seen a case where this reaction can apply to a primary alkyl hydrogen bond. It's very selective. And in general, in almost all cases, you'll only use this for tertiary carbon hydrogen bonds. So you notice of all the different types of CHs
29:01
in the substrate, it only went after the tertiary CH. The problem with chlorine, when you try to do chlorine in this reaction, it reacts with every type of CH pond. Right? And since that's useless, I'm not going to put it on my exams. I'm not going to discuss that. I'm not going to ask you questions about chlorination.
29:20
There's no use to that because it gives you horrible mixtures of stuff. Right? Why should we spend our time? What's ironic to me is the book loves that. I don't see why it's good to ask you questions about stuff that's not useful. Okay, so bromination is useful because it's so selective for tertiary CHs.
29:41
And it's going to be useful to you. Now that I've taught you how to add functional groups where there was no functional group before, you can think of all the synthesis questions I'm now going to ask you where you start with simple alkanes and then add leaving groups to them out of nowhere. Okay, let's talk about the mechanism for that reaction. Now what's going on there when you add bromine and light
30:02
or heat, so I'm going to draw three steps here, the three steps involved in that bromination reaction. And I'm going to start off by drawing out bromine. And the key thing is that whenever I see two electronegative atoms, oxygen, oxygen, chlorine, chlorine, bromine, bromine,
30:23
with two electronegative atoms tugging on each other, those bonds are weak. And they'll have a tendency to pop apart to give two radicals, not to give a carbocation or a cation and an anion, but to give two radicals. And let's use Fisch-Hugh-Arrow pushing to depict that bond cleavage.
30:42
And we have a certain name for this. We call this homolysis. The bond is cleaving homolytically in an even way so that each bromine gets a single electron. And that's how we depict it with that Fisch-Hugh-Arrow pushing. And so the products that we get out of that are two bromine radicals.
31:02
Now, I call the bromine-bromine bond weak. It's not like every bromine-bromine bond is falling apart. It's only a few. You only get a tiny amount of these things in solution. So I'll just write small amounts. You only get tiny, tiny amounts of bromine radicals floating around in your reaction mixture.
31:24
Okay, so what's the energy? Let me just remind you of the energy that we're using in order to create that. Light is one, and we always symbolize that with H-new. So Planck's constant times the frequency symbol new
31:41
at E equals H-new. Or the book puts or heat. I favor the light. So you'll find that when I write questions, I always write the light version because that's distinctive from anything you've seen in any other chapter. Okay, so what are these tiny amounts of Br dot doing as they float around in your reaction mixture?
32:02
You've got Br dot floating around, and it's not happy. Let me just draw the seven electrons. That's not happy. Br minus is happy, but Br dot with just that unpaired electron isn't happy. And so when it comes along and it collides with an alkane CH bond, it's going
32:24
to abstract a hydrogen atom. It is going to pluck off a hydrogen atom from that carbon. And as I've drawn it, it's a tertiary CH. Those are particularly easy to pluck off. And I need to use this arrow pushing to depict that. And I already showed you how that works.
32:41
I need to show how this carbon hydrogen bond breaks. So I'm going to form a new BrH bond. I'll do that by taking one electron from here, and I'll take one electron from here. That's going to form a BrH bond, and it's going to leave behind one electron on that carbon atom.
33:01
And again, it's this very weird fishhook arrow pushing that you have to get used to for this chapter. There's my BrH bond. That's happy now. Oh, no. Now the carbon is unhappy.
33:23
Maybe it's not quite as angry as the bromine radical was, but now we're in this situation where the carbon is not so happy. You know carbon wants to have eight electrons, and so let's take a look at the fate of this carbon-based radical. I'll redraw it again down here. And now I'll watch what happens.
33:41
Now hypothetically, it could bounce into another tertiary carbon H, but that would be a very slow hydrogen atom transfer. And notice how I'm avoiding calling this a proton. A proton has no electrons with it. When Br dot pulls off the H, it's bringing an electron
34:00
with it, and that's a hydrogen atom. H with an electron is a hydrogen atom. Okay, so when this carbon radical runs into a Br2 molecule, right, not all of the Br2 homolyzed, only small amounts of it homolyzed, the rest of the bromine is just floating around as Br2. And as soon as that carbon radical bounces into a bromine,
34:23
it abstracts, it pulls off a bromine atom, and I'm going to use arrow pushing, this fishhook arrow pushing, once again to depict that. So in the first reaction, we abstracted a hydrogen atom, and in the second reaction, this carbon radical abstracts a bromine from Br2. Most of the bromine is just floating around as Br2
34:43
and never got homolyzed. And the end result of this process is that it makes a carbon-bromine bond. And now the carbon is happy. And what's the leftovers here? Ah, now we're back to the Br dot.
35:03
So you can see what's going to happen. The Br dot is going to come back and find another alkane. So you start with Br dot, and after two reactions you've regenerated another Br dot. And this process will continue in a chain. It's called a chain reaction over and over and over and over again.
35:21
And the end result is that you're converting CH bonds into carbon-bromine bonds. That's the end result of this process. And at any point in time, there's only small amounts of radicals floating around in your reaction. So notice the language that I'm using. And so you should, I'm not going to test you on this, but the language that I'm using is I call this
35:42
homolysis, where I cleave a bond, where a bond just simply flops apart. That's only weak bonds that do that. Oxygen, oxygen, bromine, bromine. These two steps down here, I'm using a different term that we've never, that we haven't used before, and that's abstract. I abstract an atom.
36:01
I abstract a hydrogen atom up above here. Here I'm abstracting another atom. The carbon radical is abstracting a bromine atom. In other words, it's taking the atom and one of the electrons from the bond. So those are very classical radical reaction verbs that we use, or lingo, or descriptions that we use.
36:26
Okay, these radicals that we generate, there's some stereochemical consequences that arise when we talk about making carbon-based radicals.
36:40
So when I, let's just imagine, I'm going to take a substrate here. Let me try to draw this out in a way that gives me a little bit of room. What I'm going to do is I'm going to draw a chiral center. And here's my proton. There's only one tertiary proton here in this substrate.
37:00
So I have this carbon atom in the center, and on it I have a methyl group going back. I have an ethyl group coming forward, and I have a propyl group. So methyl, ethyl, and propyl. So there's only one tertiary CH bond in this substrate. If I take that and I apply it to my conditions,
37:20
I put in some bromine. That's a reagent, you've used lots before. And then I shine a light on that. And in our lab we have, you know, it's not a fancy light. It's a light you buy from Home Depot. So you don't have to use super fancy light with most of the common reagents. And the product of this reaction is
37:41
that we replace the CH. And here's my new carbon-bromine bond. And immediately I'm thinking, gosh, I really want to do some SN1 reactions with that, or an E2 elimination. All the things you can do with a tertiary carbon-bromine bond. Okay, let's take a look at what's going
38:01
on in this reaction. I'm not going to show how you homolyze the bromine-bromine bond to generate Br dot. And I'm not going to draw, I guess I could draw, why don't I draw out this hydrogen atom abstraction just to give us even more practice with this fishhook arrow pushing. Here's our Br dot. It's floating around.
38:21
It's looking for some choice easy hydrogen atom that it can pluck off. Here's the bond. So I'll take this electron, use a fishhook arrow. I'll take one of the electrons from the CH bond. And there I go. I've depicted how I'm going to form a new BRH bond. So what do I do with the other electron that was in this carbon-hydrogen bond?
38:42
Well, I have to do something with that. So why don't I take that remaining electron from that CH bond and I'll deposit it on the carbon. So once again, that's the use of fishhook arrows. The point is that we make a radical here that is essentially planar. It's not perfectly planar, but it's essentially planar.
39:02
So I'm going to do my best to draw this resulting radical as a planar species edge on. And so here's my radical. And here's my orbital, my empty, not empty, it's half-filled.
39:20
Here's this half-filled orbital. And it's very P-like. It's almost like there's no significant difference between the top and the bottom face for this radical. And let me draw that one unpaired electron. I can draw it in either the top lobe or the bottom lobe. It spends 50% of its time in each lobe. I'm not going to even phase the lobe there.
39:40
So the important point here is that this is essentially planar. What that means is, as I showed you before, this carbon radical isn't happy. It's going to react with Br2. So when Br2 comes along, the Br2 can collide with either the top face or the bottom face of this molecule.
40:01
50% of the time it will collide with the top face. The other 50% of the time the bromine will collide with the bottom face. And so in the end what you end up with is a mixture of products where 50% of the time the carbon bromine bond is
40:21
on the top and the other 50% of the time the carbon bromine is on the bottom. And so I'll draw my fish hook arrow pushing here. So you get an equal mixture. You'll lose the, even if you started off with 100% of chirality, one stereogenic center there, you'll end up with a mixture. So here's the way, again,
40:41
I'm going to show my fish hook arrow pushing. I'll take one electron from this carbon radical and I'll recombine it with one of the electrons from this bromine-bromine bond. That will make a carbon-bromine bond. Notice I never showed anything attacking bromine. That's not the way fish hook arrows work. And then I'll take the second electron from that bromine-bromine bond and I'll give it to the other bromine.
41:02
And that simply takes experience and practice drawing these fish hook arrows to get a feel for how they work. Okay, so because these carbon-bromine radicals are essentially planar, you get this effect where it adds top
41:20
or bottom with equal facility. So top or bottom of the, of that carbon radical that's nearly planar. Now there are other reactions that involve radicals that, and in particular the book has an entire section about fluorocarbons.
41:41
In fact, this is the basis of the ozone layer being very sensitive to fluorocarbons that they used to use when I was a kid in all kinds of aerosol spray cans. Our former colleague, Sherry Roland, won the Nobel Prize for showing that this type of radical abstraction process is a problem
42:00
and destroys the ozone layer. And I, we just don't have time to put that on my exams. Read about it, it's cool, it puts you in touch with the history of our department, but I won't test you on that. It's more stuff that's cool reading. Okay, so replacing tertiary CH bonds with bromine. Super useful, right? You can start with no functional groups whatsoever
42:22
and introduce functional groups. That's one of three important reactions in this chapter. Let me show you important reaction, well, semi-important reaction number two. So you can replace tertiary CH bonds with bromine. Wow, that's powerful. Okay, here's a reaction that's not quite as powerful.
42:40
And I don't know why, but the book loves this reaction. What I'm going to do is I'm going to take an alkene. Well, you've seen, you've had an entire chapter on alkenes. And I'm going to show you a reagent that we mentioned sometime in the past, but we didn't go into in much detail. It's called N-bromosuccinimide.
43:01
Let me remind you of what N-bromosuccinimide is, and you don't need to remember it. I'm not going to draw the mechanisms for you. And I don't want you to try to draw the mechanism for this because you won't be able to. Here's N-bromosuccinimide.
43:20
Usually the N is in italics. IUPAC rules say that the N is in italics, if you ever have to write that out. So here's our recipe. We take N-bromosuccinimide, and we take light. It's supposed to be H nu. I'm not doing a good, it's not HV, it's H nu.
43:41
That's the Greek letter nu equals Greek nu. And we can replace, we can convert alkenes into allyl bromides. Effectively, what you're doing is you're replacing one
44:01
of the CHs that's next to the double bond with a carbon bromine. And so let me show you that this reaction is amazingly complex. What is this NBS for? The only reason they're using N-bromosuccinimide in this reaction is that under these conditions, under the conditions of this reaction,
44:20
NBS generates very small amounts of bromine at low concentrations. So really, this is a reaction of Br2. Just plucking off a hydrogen atom, just like the tertiary CH substitution reaction. So why don't you just draw Br2 here? Because you know what high concentrations of Br2 do.
44:42
You already know what happens if you dump Br2 in high concentrations into alkenes. They add bromine across double bonds. It would not work if you simply wrote Br2 above the arrow because you would get the dibromide. So NBS is a super secret reagent that under these conditions generates very small amounts
45:01
of bromine at any point in time in the reaction mixture. And I'm not going to talk about how that happens. It's not important. The one thing I will show you is that there's all kinds of allylic carbon hydrogen bonds here. And I'm just going to draw one of them. Well, I'll draw two of them here. And so you can imagine what would happen if a Br dot abstracted a hydrogen atom from one
45:25
of these two positions here. What you would generate is a carbon-based radical that is stabilized by resonance. And so here I'm going to draw this carbon-based radical. There it is. If I pluck off one of those two hydrogen atoms,
45:41
and I'll draw the other hydrogen atom there just so we can see that one of them is still there. And that radical is stabilized by resonance. Let me draw a resonance structure. And I will use arrow pushing, fishhook arrows, to depict how I get from one resonance structure to the other. So I'll take this electron and one of the electrons from this pi bond, and I'll create a new pi bond.
46:03
There it is. And then with the remaining electron that was in this pi bond, I'll deposit that electron back on this carbon atom on the end. So you can see the resonance structure I'm going to create is still another allyl radical.
46:24
So that's why allyl radicals are more stable than tertiary. And there's one H over on this side, so I'll draw it. Now, this is a very carefully chosen substrate. It's carefully chosen because on the starting material when you look, there was another set of CHs
46:42
on the other side that are indistinguishable from, right, those because the substrate is so symmetrical, the CH2s on each side of the double bond are indistinguishable. And moreover, it generates a radical that is also by symmetry symmetrical. So what happens is this carbon radical will abstract a bromine
47:06
from Br2 that's floating around in small concentrations in the reaction. And it doesn't matter which end you abstract a bromine atom on because by symmetry, if I attach a bromine here, it's the same as if I attach it to bromine on the other side.
47:22
You just flip the molecule over and they will look the same. This reaction, unless you use it on super symmetrical substrates, will give you lots and lots of products. And the book is not totally clear about that. It only gives you useful product mixtures if you use it on super symmetrical substrates. So I'm not a big fan of reactions
47:41
that require super special substrates. So I consider this to be less useful than just replacing tertiary CHs with bromine. So that's not a very, so not quite as useful
48:01
as replacing tertiary CHs. Okay, so that's the second reaction, allylic bromination. And here's the last reaction that's important in this chapter. There's three reactions that are important. Replacing tertiary CHs. Second reaction is replacing allylic CHs using N-bromo 6 centimeter, NBS. And here's the last reaction.
48:21
So the last reaction is adding HBR across double bonds. And that may surprise you because you've done that before. Let's take an alkene and let's take HBR. Well, you've seen that before.
48:42
And the key here that should clue you in that something is different is that they write this. This is a peroxide. It has an oxygen-oxygen bond. You can see it has an oxygen-oxygen bond in there. So ROOR is the symbol for peroxide. As soon as you see ROOR, you know this is not going
49:02
to be a regular HBR addition across a double bond. So you can recall from chapter 10, we did this over and over and over again. You took alkenes, you added HBR, and the bromine ended up substituted at the more, or ended up attached to the more substituted carbon of that double bond. What's going to happen up above
49:20
when we add a peroxide in this reaction is the bromine ends up at the less substituted double bond. And there's an H that gets attached here. I'll draw it for you, but just to make sure you're clear where that H went. The H ends up at the least substituted. So in the framework of Markovnikov
49:42
versus anti-Markovnikov selectivity, we refer to this as anti-Markovnikov selectivity, regioselectivity. And it all has to do with this addition of peroxides. And I think the book mentions or tries to tell you that you can just use heat or, sorry, just light for this, but really you need a peroxide initiator.
50:02
So let me demystify this peroxide business for you. This is the same peroxide that they put in acne creams. It's benzoyl peroxide, oxy ten. You could take a smidgen of this paste and put it into your reaction and you would reverse the selectivity and get anti-Markovnikov. That's very powerful, right?
50:21
It's very powerful. You know how to hydroborate and put oxygen at that least substituted double bond, but just by including peroxides in your mix here, you can reverse the selectivity of an HBR addition. Okay, so let me summarize what we did in this chapter. Again, we didn't do a whole lot, but it's this whole fishhook arrow pushing
50:42
that underlies everything. I just showed you three reactions. Here they are. The first reaction I showed you is that you can replace CHs that are tertiary with bromine. It's powerful. It adds a leaving group where there wasn't a leaving group before. I showed you that you could replace allylic CHs using n-bromosuccinimide, and that's a secret reagent
51:03
that generates small concentrations of bromine. And then last, I showed you if you add peroxides to your HBR additions, the bromine will add to the least substituted carbon. And those are your three reactions you need to know for this chapter, and that's pretty much it. The rest of it you can just sort of skim through.