Nucleophile Aromatic Substitution to Aniline Rings
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BenzeneChemische SyntheseStickstoffatomSodium nitriteIsotopenmarkierungCigaretteBeerWine tasting descriptorsDiet foodNitrosamineMeatCancerSmoking (cooking)CyanidionDiazonium compoundCopper(I) cyanideSchweflige SäureAmineHydroxylReaction mechanismChemical reactionAminationAlkylationAcidWaterPhenolsHalogenationBenzodiazepineSodium thiosulfatePhosphorusHydrochloric acidAnilineHydrogenSalpetrige SäureAromaticityKaliumiodidCopper(II) chlorideChemical compoundFunctional groupHypobromiteTolueneRadical (chemistry)DyeSubstituentKernproteineAromatic hydrocarbonSodiumHandelsdüngerMethylgruppeWursthülleSodium nitrateChemistryMolecularityCopperChlorineBromideNucleationSubstitutionsreaktionDiagram
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CryogenicsHypobromiteAmineProcess (computing)Functional groupHalogenationAminationTolueneBenzeneStickstoffatomChemical compoundElectronPalladiumLibrary (computing)Hydrochloric acidMethylgruppeCarboxylateSodium hydroxideDiazonium compoundAcidMoleculeChemistryAageWaterSubstitutionsreaktionReaction mechanismDiazonium compoundNamensreaktionAluminium hydrideChemical reactionChemische SyntheseAldehydeLithiumReducing agentHydrocarboxylierungWalkingBase (chemistry)FloodRiver sourceGrading (tumors)ZincAmalgam (chemistry)Mercury (element)EisenbromideAluminiumRedoxWine tasting descriptorsCommon landComputer animation
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BenzeneFunctional groupElectronChlorineHalideVinylverbindungenMethanolWine tasting descriptorsCarbon (fiber)IodideEliminierungsreaktion <beta->FluorideAddition reactionBromideChemical reactionChlorideHalogenationWhitewaterAtomic numberColourantHydrocarboxylierungStickstoffatomKernproteineMixtureResonance (chemistry)NitroverbindungenAromaticityFood additiveAtomAcylWursthülleWaterAryl halideCyanidionReaction mechanismWalkingOctane ratingKohlenhydratchemieAcidCobaltoxidePhenolCarbonylverbindungenAminationCarbokationPhenyl groupBlock (periodic table)
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Chemical reactionCarbon (fiber)AcidEliminierungsreaktion <beta->AldehydeResonance (chemistry)KetoneFluorideHalogenationAromaticityFunctional groupNitroverbindungenOctane ratingFormaldehydeBeta sheetReaction mechanismEnamineImineAmmoniumChemical compoundIminiumsalzeSemioticsThermoformingCHARGE syndromeStickstoffatomBase (chemistry)EthanolEnolHydrocarboxylierungAldolHydrochloric acidElectronCobaltoxideAddition reactionAminationSteric effectsSunscreenIodideHuman subject researchAmineInsertionselementProtonationIronChlorideWalkingChlorineSubstitutionsreaktionCarbonylverbindungenAldol reactionComputer animation
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WalkingStickstoffatomHydroxylBeta sheetCobaltoxideElectronFormaldehydeKetoneAcidFluoxetineHydrocarboxylierungFunctional groupChain (unit)ThermoformingAldehydeKohlenhydratchemieOrlistatGlucoseSample (material)AldoseKetosenFructoseChemical plantStereochemistryBody weightChemical structurePolymerGlykogenCelluloseAldol reactionSodium hydroxideChemical reactionChemische SyntheseReaction mechanismHydrolysatChemical propertyOrganische VerbindungenCell (biology)MonosaccharideAbundance of the chemical elementsChemical compoundEthanolBiochemistryStarchTopicityWaterCheminformaticsCarbon (fiber)ÜbergangszustandLactitolElectronic cigaretteCarbon dioxideComputer animationLecture/Conference
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Computer animation
Transcript: English(auto-generated)
00:09
So, we left off last time talking about nitrous acid. It's generated in situ by treating NaNO2 sodium nitrite with hydrochloric acid and
00:23
what you can do with this is it decomposes to nitrous sodium ion and these are actually considered to be very very highly carcinogenic compounds. So there is some concern if you look on the label when you eat any processed meat or hot dogs or things like that, it says NaNO2 sodium nitrite, that's what they use to preserve
00:46
the meat and of course you have hydrochloric acid in your stomach so this sort of thing can happen in your stomach and so there's some concern about this. It's been proven to cause things like stomach cancer and bladder cancer in animals
01:01
but maybe not in humans, the problem is that it's ubiquitous, they use nitrogen fertilizers and so even if you have to eat really healthy and you consume lots of fruits and vegetables you still will get these nitrosamines in your diet. It's also a little bit in beer, cigarette smoke, things like that but what they found
01:23
is that people who eat a lot of fruits and vegetables actually have lower risk of these cancers so the mechanisms aren't entirely understood but just to keep you a little aware so you'll see some meats that claim to be healthier instead of using sodium nitrite
01:41
they'll use celery extract, well celery has a lot of sodium nitrite in it already and so they're just really doing the same thing, it just sounds healthier, celery extract rather than sodium nitrite so something to be concerning but certainly we do need to preserve these meats because if you don't preserve the meat you're going to get a
02:02
lot sicker, right away you'll get sicker, later on you'll get sicker if you get cancer right? Okay so we talked about the mechanism, I do not ask this mechanism on the exam whatsoever and it's really not very synthetically useful for simple amines like this.
02:22
So we're not going to be doing this, this is not what we're doing, it actually is synthetically useful when you use aromatic amines. Okay so let's label this, this is a electrophile with a fantastic leaving group
02:46
so can wreak all sorts of havoc in your body when you have a good great nucleate so we call it a really great alkylating agent. Alright so the reaction with primary amines, primary aliphatic amines is not synthetically
03:02
useful that's what I just showed you on the previous page but when you use primary aromatic amines it's actually very useful. When aniline is allowed to react with nitrous acid, diazization occurs the same as we did here on the previous page and we form an arene diazonium salt, this diazonium salt
03:22
has a very good leaving group, molecular nitrogen, it undergoes substitution. And so there's a bunch of different substitution reactions we can do with this, let's first draw the arene diazonium salt, so this is the intermediate. We have a positive charge on the nitrogens that's bonded to the benzene ring
03:45
and then depending on what we use, in this case we use hydrochloric acid, we would have Cl- as our counter ion and this is stable at zero degrees. So it's stable enough at zero degrees, when we do it with an aliphatic amine
04:02
it is not stable at zero degrees. So this is stable at zero degrees so we can form the diazonium salt in situ and then we can treat it with a bunch of different nucleophiles. So here's all the reactions we can do with this. So this reaction here, it's called the Sandmeier reaction and this is why you want to go back and you want to review aromatic chemistry.
04:22
You want to go back and review chapter 18. If you had me last quarter, I talked a lot about problems with aniline and our ways to work around that. If you didn't have me last quarter, that is a handout under the practice link and you definitely want to take a look at that. And even if you had me last quarter, you definitely want to review that.
04:43
So here's all the things that you can do here. We can replace that diazonium salt with a bromine using copper bromide. We can replace it with a chlorine using copper chloride.
05:00
We can replace it with an iodine using potassium iodide. HBF4 and heat will replace it with a fluorine. So this is the first time we've been able to put a fluorine on an aromatic ring. Copper cyanide is a way to replace that diazonium salt with a cyano group.
05:24
So this is our first synthesis of an aromatic, this would be benzonitrile, aromatic nitrile, benzonitrile, very synthetically useful. If you treat this with water and sulfuric acid and heat,
05:40
so basically you just would use sulfuric acid as your acid and then you would just let this warm up. You can now replace that diazonium salt with a hydroxyl group. So this is now our first synthesis of phenols. This is the first time we've been able to synthesize phenols. And then a very kind of different reagent H3PO2,
06:02
it's actually called hypophosphorus acid. These numbers are not wrong, it looks a little strange. So hypophosphorus acid. And if you do that, you can replace the diazonium with a hydrogen. Why would you want to do that?
06:21
Well, there's instances where that would, that's actually a very useful thing to be able to do. All right, so this can be used to remove an amine group. And let's say we put an amine group on the benzene ring
06:42
and we use it to direct incoming substituents. Once it directs the incoming substituents, we can take it right back off again when we're done with it. So to remove an amine group that's used to activate the benzene ring
07:12
or direct an incoming substituent.
07:25
And sometimes you can actually use it to also to block a position on the benzene ring that you don't want substitution on. Now as for the mechanism of this reaction, it is not an SN1, it is not an SN2. I'm not gonna ask you the mechanism
07:41
because it's not completely understood. Probably involves radicals, so not SN1 and not SN2. Mechanism not exactly known
08:03
and probably involves radicals. Which means there's evidence to suggest that there's radicals as part of this reaction. So one less mechanism you need to know.
08:22
But it's very useful in synthesis and guaranteed you will have some synthesis on the final where you use the Sandmeyer reaction. All right, so let me give you an example of how you would use this in synthesis. Apart from just being able to put these groups on a benzene ring.
08:42
And we'll do some more in discussion next week. So here's an example. Let's say we want to synthesize the following compound from toluene. So if you think about synthesizing that, bromine's an ortho paired director, methyl's an ortho paired director, how are we gonna get those things meted to each other?
09:03
Okay, so that's a tricky problem. If you brominate toluene, you will get, and let's say we use two equivalents. Two equivalents of bromine.
09:21
Toluene's an ortho paired director, so our major is going to be one bromine in the ortho position, one bromine in the paired position. That's our major. And minor, because it's a little bit more sterically hindered, would be to have both bromines in the ortho position.
09:43
So that's gonna be a minor product. And that's only because it's more sterically hindered. But not even close to what we want to make.
10:01
All right, so let's see how we can do this and get things going exactly like we would want them to go. So instead, what we can do is nitrate the benzene ring and change that to aniline. Okay, so nitrate, reduce that
10:21
and use the amine group as a directing group. So again, all that aromatic chemistry should come flooding back to you hopefully at this point. We will get ortho and para. So we would isolate the para, nitro toluene.
10:42
We reduce it with H2 palladium. Now what do we know about this compound? If we brominate that, where are the bromines gonna go?
11:07
So that amine, and so again, you want to look at that problems with amine that is very highly activating group. You will get both ortho positions filled. Okay, and you don't want to use FeBr3
11:22
because this is so activated that you just throw bromine in and they're gonna go right on the ring, even at low temperature. So you're just gonna say 2Br2, FeBr3 is not needed.
11:45
So you see where we're going here, we've introduced an amine and that amine is doing a great job for us. It's directing where these bromines are going to go. Notice both bromines are meta to the methyl group, which is exactly where we want them to be.
12:02
And now we can get rid of that amine group. So we're gonna do a diazotization. So NaNO2, H2SO4, and you can use hydrochloric acid, you can use H2SO4, your choice.
12:21
Those are the two common ones, H2SO4 and water. That makes the diazonium salt.
12:42
That makes the diazonium salt. And then we use, again, you just want to get these numbers right for this one. It's really common for students to change this around, H3PO2. I won't tell you the common mistakes, because then you'll do that on the final. But it is H3PO2, so it looks a little strange. And then we get exclusively the product that we want
13:02
with the bromines, the two bromines and the methyl meta to each other. All right, so now I've had students say, oh, I have a better approach that doesn't use that. So let me give you an alternate approach to this molecule
13:22
that also works, that you might have thought of if you hadn't had this problem if I hadn't shown you this. So let's do a different approach.
13:41
All right, so we start with toluene. And this is gonna bring back some old chemistry from chapter 18. So one of the things that we can do is we can convert that into a carboxylic acid. Remember how to do that? KMnO4, our favorite purple reagent.
14:02
KMnO4, sodium hydroxide heat, and followed by, and that should be a two, followed by acid. Why do we have to follow by acid? Because we're making a carboxylic acid in base, so it will be deprotonated. So we need to make sure to throw in acid here. That gives us the carboxylic acid,
14:23
which is a meta director, and now we can brominate. Two equivalents of bromine, FeBr3. We definitely need the FeBr3 here.
14:46
And I honestly don't care which method you use here. Whatever works, if as long as all the steps work, all the steps work. Now we need to convert that carboxylic acid back into a methyl group. So the way to do that would be to reduce it with lithium aluminum hydride first.
15:12
And so the reason I'm showing you this is I do try as much as possible to bring up old chemistry, so that when you see it on exams, you will be ready for it.
15:21
You'll remember what that was. So lithium aluminum hydride, this goes back to chapter 20. And then PCC, we convert this to an aldehyde. And now what would we do from 51B?
15:41
Two possible name reactions here to get rid of that carbonyl. Clemensin or Wolf-Kishner? I kind of favor Wolf-Kishner. I mean Clemensin myself. We'll just do Clemensin reduction. Zinc, mercury, amalgam, HCl.
16:04
So I'm hoping that many light bulbs are going off in your head to remind you of these reactions. So that's another way to do it. Both of those would get equal credit on the exam. Even though the first mechanism, I mean the first synthesis
16:20
has a little bit more pizazz to it. We do like pizazz, especially when we're grading. You know, you grade 370 exams and you like when somebody does something interesting. Okay, so it's delightful to see. All right, questions, anybody?
16:43
We are adding, I'm adding two new reactions onto this chapter, just for fun. And so one of the reactions I'm adding is nucleophilic aromatic substitution, which is in chapter, it's in the newest edition of Smith. I did not have time to cover it last quarter.
17:02
It is in chapter 18, okay? If you have the third edition, how many people have the third edition? I will put some practice problems for you. I will cover as much as you need to know about this reaction, and I will put some practice problems for you under the practice link so that you will not have any disadvantage
17:20
over people who have the fourth edition, okay? And I do believe the fourth edition is on reserve in the library. Can anybody verify that, is it on reserve? It usually is. Okay, should be in the library if you wanna take a look at it. All right, so we covered in chapter 18, we covered electrophilic aromatic substitution, and that is where the benzene ring acts as a nucleophile
17:45
and attacks electrophiles, right? So we made all sorts of souped up electrophiles and the benzene ring attacked. This is the opposite. This is nucleophilic aromatic substitution, and now the benzene is acting as an electrophile and nucleophiles are attacking it.
18:02
So if we want benzene, and so we're used to seeing electron rich benzene in chapter 18, and we'd have electron donating groups to make that benzene really electron rich, if now you want the opposite and you want benzene to be electron poor,
18:20
what would you put on that benzene ring? Electron with your ion groups, right? The more the merrier. The better, the more electron with your ion groups. So back in chapter 18, when benzene was acting as the nucleophile, if you put electron withdrawing groups on it, it slowed down the reaction,
18:41
you got meta substitution and you had to heat it up a lot higher. Here, putting lots of electron withdrawing groups actually enhances the reaction and makes it work better. So let me show you what it looks like. So chapter 18, fourth edition, aromatic rings don't react easily with nucleophiles, nucleophiles can displace aryl halides
19:00
in highly deactivated benzene derivatives via addition elimination mechanism. All right, so let me show you what I mean by highly deactivated. So and our criteria is gonna be a little more strict
19:21
than the general criteria. For highly deactivated, we want two nitro groups or two cyano groups, or and I guess we'll do, or two carbonyls,
19:44
where of course the carbonyl is bonded directly to the aromatic ring. And so we want two of them. Two of these, or you could have a nitro and a cyano group, but two deactivating groups. Ortho or para to a halogen, especially fluorine.
20:12
So I'm gonna make these very, very recognizable to you. Okay, you see a benzene ring, you see two nitro groups, two cyano groups, two carbonyl groups or two of any mixture,
20:23
ortho or para to a fluorine. And that should just jump right out at you because we barely use fluorine at all, right? So those are the best cases, and that's what we're gonna be looking for. All right, so that's our criteria.
20:43
So fluorine's better, this one uses chlorine though, but notice the chlorine is ortho and para to two deactivating groups, two strongly electron withdrawing groups. And so the reason we're introducing this here is this is a way to make aromatic amines, another way to make aromatic amines.
21:06
So our nucleophile is NaNH2, it is displacing the chloride. What do we know about this reaction? Is it an SN1 or an SN2?
21:23
What do you think? SN2, we don't want sp2 hybridized, right? So it's not gonna be SN2 because we can't do backside attack. Is it gonna be SN1?
21:40
Yeah, if it's SN1, we certainly wouldn't want a bunch of electron withdrawing groups on a benzene ring. If it's SN1, we'd have to make a carbocation on the benzene ring, we would have to make a phenyl carbocation. And having a bunch of electron withdrawing groups on a carbocation is gonna not be a good thing.
22:00
So it's not SN1, it is not SN2. So let's look at the characteristics of the reaction. Strong nucleophile required. So absolutely strong nucleophile required. Water wouldn't work, methanol wouldn't work. Can't proceed by an SN1 because the electron withdrawing group would destabilize this intermediate. So let's draw that,
22:21
the intermediate that you would use for this that we're not gonna use. Definitely that's looking really bad. We've got two really powerful electron withdrawing groups. So no SN1, let's put a big X through that. Okay.
22:41
Reaction cannot proceed by an SN2 like vinyl halides. Aerial halides cannot achieve the correct geometry for backside displacement. Aromatic ring blocks approach to the nucleophile to the back of the carbon bearing the leaving group. So let's just draw this out just for a reminder of this because it's been a while since we've had this. So here's benzene on its side.
23:05
So I'll wedge it to show that it's kind of... So it would be floating above your page. So we know we've got overlap on the top and the bottom. And just for a moment here, let's leave off the nitro groups to make this a little easier to see.
23:21
Let's put our chlorine here. Can we do backside attack? I mean, basically what this means is this, this electron rich nucleophile. So we've got this electron rich, we've got these electrons circulating on the top and the bottom.
23:41
And this electron ring, what's gonna happen when this nucleophile approaches that pi electron cloud? It's gonna be repelled, right? So is that gonna be able to sneak in here into the middle of that donut hole and do backside attack and kick off chloride? Not in a million years is that gonna happen. Okay, so impossible to have SN2, impossible to have SN1.
24:03
Let's do this a different color so you can see what I'm talking about here. It's gotta go in here, not gonna happen. So none of this either. So SN2 requires backside attack,
24:26
which is impossible here. And there is no, and we already talked about this multiple times, there's no SN2 on SP2 hybridized atoms.
24:59
All right, so SN1 is out, SN2 is out.
25:02
The other thing that's really kind of wonky here is that the fluoride is a much better leaving group than iodide. So we definitely isn't, we're not looking at SN2. So recall in SN2, in SN2 or SN1 even,
25:22
iodide is a better leaving group than bromide, which is a better leaving group than chloride, which is a better leaving group than fluoride. And here fluoride is better. So definitely we're not talking about the same reaction.
25:42
And this is worse leaving group. And in fact, we've never seen fluoride as a leaving group in this class. So what, if we, but we are doing a substitution. So what kind of substitution do we do when we have an SP2 hybridized atom? What do we do instead of SN1, SN2?
26:01
What do we do, what do we do all throughout chapter 20 when we wanted to do substitution of an acyl acid chloride, for example? What is that? What is that called? Additional elimination, right? Right? You attack the carbonyl, you kick electrons up onto oxygen, electrons come down and you do an elimination.
26:22
So it's the same thing here. It's additional elimination. All right, so let's look at it. Additional elimination.
26:40
Now I have so many choices here. I can, what I'm gonna do is I'm gonna actually attack the carbon that's bonded to the leaving group. I'm not gonna do direct displacement, I'm gonna do additional elimination. So I'm gonna have this nitrogen attack here. That carbon, I'm gonna move electrons over
27:00
and I can go all the way up onto the nitro group. There should be a positive charge on that nitro group, by the way. That's addition. A little bit longer, a little more arrows than attacking carbonyl, but that's essentially what we're doing. And that is the slow rate determining step.
27:22
Now there are a bunch of resonance structures we can draw. I'm not gonna draw all of them. I'm gonna do the two main ones. So I pushed electrons onto one of the nitro groups. I can also push electrons onto the other nitro group. And that's why these two nitro groups
27:42
need to be ortho or para to the leaving group. Because if they're ortho or para to the leaving group, then we're gonna be able to especially stabilize. We've disrupted aromaticity, so we need to do some stabilization of this high energy intermediate here.
28:01
And that's the nitrogen here. Let's put the nitrogen there. All right, and that was so much fun. I'm gonna push onto the other nitro group here, okay? Again, we're leaving out the extra resonance structures. There's a bunch. But what I can do is I can, because of the orientation of these two nitro groups,
28:22
I can go all the way onto the other nitro group. And in fact, I could have gone onto this nitro group at the start. So you see why we need two of them.
28:46
So this is a possible mechanism for the final. And of course, we can push this charge onto the ring, three places in the ring in addition to these two resonance structures.
29:03
So my recommendation to you is know how to draw a nitro group. That's the wrong spot there. Know how to draw a nitro group for the final. And know how to draw resonance structures for a nitro group, okay?
29:21
So there's our slow rate determining step. It's a high energy intermediate because we've disrupted aromaticity. And now we wanna do the fast, now we wanna do the fast step, fast elimination.
29:42
All right, so in addition, we add electrons, we push electrons up onto the oxygen, and then elimination electrons come down and kick. And now they're gonna kick off the fluoride. Let's see, we'll do it from this one. We could do it from either one. There's a negative charge here.
30:01
Let's go here, just like that. So electrons on oxygen come back down and we kick off fluoride as a leaving group. Can you think of a possible reason
30:21
why fluoride would be a better leaving group than say, let's do the other extreme iodide? Why would fluoride be a better leaving group than iodide, for example, here? There's two reasons. So maybe you guys can come up with one of them. Well, let's draw the product first
30:44
while you mull that over. Fluoride ion versus iodide would both be. Yeah, actually that would both be kicking off. So that's not, that wouldn't be the reason.
31:01
So substitution, think about sterics here. Sterics is one of the factors here. Look at where the amine is attacking. Do you see any possible advantage
31:20
for that leaving group to be as small as possible? Because it's attacking on the same side. All right, so it's attacking on the same side and it's also more electronegative. So let's, so carbon with, remember, we want this ring as electron withdrawing as possible.
31:44
We want it to be very activated. Fluoride's more electronegative than iodide so it's gonna make that ring even more electrophilic and that's what we want. So carbon with fluorine attached has larger partial positive charge.
32:08
Therefore, it is more subject to nucleophilic attack in the rate determining step.
32:27
And the other thing is, of course, the size of the fluorine. In SN2, we have backside attack so size doesn't matter here, the leaving group ability.
32:42
In nucleophilic aromatic substitution, NAS, attack is from the same side. So the fact that fluorine is smaller is better.
33:01
Questions on nucleophilic aromatic substitution, anybody? All right, we've got one more reaction to talk about. That's the manic reaction.
33:27
All right, so when an aldehyde or ketone is heated with an acid catalyst in the presence of formaldehyde and an amine, this is the most standard version of the mnemonic. There's other versions of this, but this is the one that we're gonna be using.
33:40
The product is a beta aminocarbonyl compound that should be the symbol beta rather than that. So turn that into a real looking beta there. So I'm gonna go through this mechanism and then we'll look at the product.
34:05
Part one, and I'm not gonna go through all the steps because we've already done this and you can go back and check how to do this. Part one, we form an aminium ion of formaldehyde.
34:30
All right, so here's formaldehyde, and we're gonna form an aminium ion.
34:42
So not an imine and not an enamine. So normally when we use a secondary amine, which we're doing here, we form an enamine. Can we form an enamine from formaldehyde? No, you'd need a carbon here, right?
35:03
We deprotonate that carbon and we form an enamine. So this is what you form. We can't form an imine, we can't form an enamine, so we form an iminium ion. And that's definitely going to be electrophilic. The mechanism for this is page 46 of your notes.
35:24
Okay, so take a look at that. So this would be a possible mechanism on the final also because it's got something from midterm one, it's got something from midterm two,
35:40
and then it's got something from this third section of the class. So part two is acid-catalyzed aldol-like reaction.
36:04
So the first part, part one is from midterm one, the second part is from midterm two material. And so let's go through that, acid-catalyzed aldol. We've done that before, but let's do it again because this is a little bit different.
36:20
So acid-catalyzed reaction, we protonate the carbonyl first. We've got hydrochloric acid. I'm just gonna use hydrochloric acid to do this. So remember in the acid-catalyzed reaction, an enol is our reactive intermediate.
36:41
So we're gonna make the enol from this. All right, so we've got two bases here. We've got chloridine, we have ethanol. You can use either, I'm not gonna be particular about this ethanol, or you can use Cl minus.
37:04
I'll just use ethanol here. We're gonna remove this acidic proton, push electrons up onto oxygen. So remember in the acid-catalyzed reaction,
37:22
enol is our reactive intermediate. All right, so now that rather than in the aldol reaction, we would attack another protonated carbonyl, but instead we have this iminium ion that we just made in the first part. So that's what we're gonna attack.
37:46
So electrons on oxygen come down. We attack this carbonyl. We kick electrons up onto nitrogen.
38:03
So hopefully this is all looking very familiar to you. It's just a little bit of a twist on that reaction. And what you see is gonna happen is we've got this amine,
38:21
and we're gonna do an intramolecular deprotonation. And remember, in order to do an intramolecular deprotonation, we need to have a five or six-membered ring in transition state. So if we count, this is one, two, three, four, five, six. So that's exactly what we're looking for.
38:43
So we're gonna deprotonate. Intramolecular deprotonation.
39:10
So this nitrogen is gonna be protonated. And then you can see step two. In step two of our reagents,
39:21
step two, we add sodium hydroxide. So then we will just deprotonate this. So here's our final product. So remember, in an aldol reaction,
39:40
we get a beta hydroxy carbonyl. Here we get a beta aminocarbonyl. So if you see that functionality, you know we can make that using a manic reaction. So let's go ahead and draw that up here.
40:15
It doesn't like being up there. Questions, anybody, on the manic reaction?
40:30
So possible mechanism for the final. Manic reaction is used as a key step in the synthesis of Prozac. We're not gonna do the whole synthesis of Prozac, but you can see that this is Prozac,
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and maybe might wanna on your own think of how to finish this synthesis. All right, so we've got our ketone, we've got formaldehyde, we have methyl amine here. We have an acid catalyst, and this is ethanol as our solvent.
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So maybe you wanna practice the mechanism on this one.
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And sodium hydroxide. And then think about how we would take that
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and do Prozac. So what you can see here is that we have this part done right here. So basically, but this is a ketone, we need that to be reduced. And we'll just think about how else, how you might finish that synthesis to make Prozac.
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Questions, anybody? All right, we're going to, I'm gonna save this page, and then we're gonna start chapter 27. And we're almost done here. So let me close this.
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Any questions while we're waiting for that to save? Anybody?
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So probably most of you know a little bit about carbohydrates from bio, right? So we're gonna just use that as a jumping off point
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to go a little bit more in depth into carbohydrates. The carbohydrates are the most abundant class of organic compounds in the plant world, so super important. So they're synthesized by nearly all plants and animals. They use them to store energy and deliver to the cells. And then most living organisms take carbohydrates
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for which they extract glucose, they oxidize it to get CO2 and water and energy. And that's not certainly not a topic for this class. That would be a topic for your class, your bio class. Starch, glycogen and cellulose are all polymers of glucose.
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Plants store energy by converting glucose to starch. Starch animals store energy by converting glucose to glycogen. And plants use cellulose as structural material to support the weight of the plant. All of these are polymers of glucose with subtle stereochemical differences which make profound differences
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in the physical properties of these compounds. All right, we're gonna start by classification of carbohydrates. So there's a little bit of, I guess you could call it nomenclature of carbohydrates that we need to make sure we're all on the same page here. So monosaccharides are carbohydrates that cannot be broken down into simpler units
44:41
by hydrolysis. You can take a monosaccharide and convert it into smaller units by other means but not by hydrolysis. So monosaccharides, two examples are glucose and fructose. This is drawn in the open chain form which you're not used to seeing it in but certainly. There is a small amount of the open chain form
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when you have a sample of glucose. And so basically we see an aldehyde here. So this is an aldose. So ose means sugar and ald means it's an aldehyde.
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You won't be able to tell it's an aldehyde when you look at the closed form, the ring form which is the most predominant form but if you have the open chain form you'll see that it's an aldehyde. De-fructose on the other hand is not an aldehyde, it's a ketone. So it is a ketose. All carbohydrates end in ose
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so that means you just have a sugar or a carbohydrate and ketose means that you have a ketone. So that's classification number one that we all need to get down.
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The other thing is that, and I pretty much use Fisher projections exclusively for carbohydrates. Fisher projections are introduced in chapter, what's this stereochemistry, chapter five? So Fisher projections are introduced there
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but I don't talk about it in my class, I wait till now. So let's talk about Fisher projections, we definitely need to know what to do with Fisher projections. So they look like this and there's some rules here. Most highly oxidized end carbon on top. So the most highly oxidized is the aldehyde so that's on the top.
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Here most highly oxidized is as close to the top as we can get it, second from the top. It looks like these are Lewis structures but this is actually conveying stereochemical information. We don't have time to talk about that now, we will continue this on Friday.