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Properties of Anomers - Mutarotation

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Properties of Anomers - Mutarotation
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25
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27
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This is the third (and final) quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. Index of Topics: 01:05 - Ways to Draw Glucose 02:10 - Chair Conformation 09:01 - Chair to Haworth Projection of Glucose 11:35 - The 5-Membered Ring Cyclic Hemiacetal Form of Fructose 14:54 - Drawing 5-Membered Ring Cyclic Monosaccharides 20:43 - Properties of Anomers: Mutarotation 33:42 - Formation of Glycosides 44:16 - Hydrolysis of Glycosides 47:02 - Reactions of Monosaccharides as Alcohols: Ether Formation
Molecular geometryMannoseConformational isomerismCarbon (fiber)Process (computing)Ice frontIonenbindungFunctional groupGlucoseChain (unit)CarbohydrateHydroxylThermoformingHydrogenWine tasting descriptorsIsotopenmarkierungMethylgruppeElectronic cigaretteComputer animation
Carbon (fiber)FructoseDeterrence (legal)GlucoseAlpha particleBeta sheetCarbohydrateHydroxylHydrogenColourantAldoseFuranosenAldosteroneLeft-wing politicsFunctional groupMonosaccharideIsotopenmarkierungKetonePentoseIonenbindungHybridisierung <Chemie>Wine tasting descriptorsKetosenThermoformingCobaltoxideHexoseCrevasseComputer animation
Beta sheetAlpha particleHydrogenWaterFunctional groupQuartzSolutionGlucoseSeleniteCarbon (fiber)Chain (unit)MixturePyranoseSyrupCarbohydrateHydroxylSecretionOptische AktivitätThermoformingCobaltoxideMixing (process engineering)Wine tasting descriptorsIsotopenmarkierungHardnessCyclische VerbindungenFilm grainHydrocarboxylierungNitrogen fixationComputer animation
Alpha particleRiver sourceOptische AktivitätGlucoseThermoformingMixtureBeta sheetCarbohydrateChain (unit)Sense DistrictHydroxylChemical structureCell membraneCalculus (medicine)WursthülleDeterrence (legal)Computer animationLecture/Conference
Reaction mechanismAcidHydrocarboxylierungBiochemistrySolutionGlykosideHydrolysatBase (chemistry)Alpha particleAssetRiver sourceAcetateWine tasting descriptorsAldehydeOptische AktivitätThermoformingFunctional groupIonenbindungEsterActive siteMethylgruppeCarbohydratePenning trapGlucoseCarbonylverbindungenMonosaccharideMethanolMixtureSetzen <Verfahrenstechnik>Carbon (fiber)Beta sheetElectronAlcoholHydroxylHydrogen chlorideProtonationGlucosideMethoxygruppeLecture/Conference
EtherWine tasting descriptorsAcidVerdünnerChlorideBase (chemistry)MethanolHydrogenReaction mechanismMethylgruppeWalkingCobaltoxideAlpha particleElectronHydroxylSeparator (milk)SolventIodwasserstoffHydrolysatCarbohydrateConcentrateFunctional groupAcetateGlykosideEthylgruppeResonance (chemistry)Lone pairOxideSilverAlcoholBiosynthesisGemstoneSodium hydrideMethyl iodideDimethyletherBeta sheetMoleculeHydrocarboxylierungMonosaccharideSodiumComputer animation
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Transcript: English(auto-generated)
We're going to get started, we're going to talk some more about carbohydrates today, talk about how to draw carbohydrates and then reactions of carbohydrates.
Does anybody have questions before we get started? I did not get the file back from the midterm, I anticipated getting that by about, usually I get it about 6 o'clock at night, so maybe tonight at about 6 o'clock at night. It's late because of the holiday. Questions anybody? Okay, so we left off last time talking about how to, yes, you have a question?
Sometime today, or tomorrow morning, early, okay, alright.
So we want to know how to draw Fischer projections of sugars, we want to know how to draw epimers of sugars, we want to know how to draw anomers of sugars, we want to be able to Draw a Heyworth projection and a chair, and a chair conformation for sugars.
And what you're going to do is you're going to memorize glucose, you're going to memorize the open chain form of glucose, and then you're going to memorize the chair of glucose which is really easy because all the groups are equatorial. And then you're going to make changes from there and that's the easiest way to do it.
You can also certainly, if you want to, go through this whole process for drawing a six-member grain, you can certainly do that, but I think it's a little easier just to remember what glucose looks like. And then for the Heyworth projection, you can either memorize the Heyworth projection of glucose or if you do it like I do, I draw the chair, and then I kind of mentally
flatten the chair in my mind. And that's what I'm doing on the next page. So we left off last time doing that, so let's go to the next page so I can show you what I mean. So we're doing D-mannose, the C2 epimer of glucose, I have the chair of glucose memorized,
and this is beta D-mannose and so this is beta, that's the wrong amount here, okay so this is the beta anomer, you will recognize this as the beta anomer.
And so since mannose is the C2 epimer of glucose, I'm just going to draw the chair of glucose, and then you got to know how the numbering goes. So the anomeric carbon is number one. So the anomeric carbon is carbon number one for glucose.
Alright so it's beta so that means that this hydroxyl is up, that's carbon number one, this is carbon number two, I'm not going to put the two there because that's where my group needs to go. And what you're going to find is that depending on how you draw your chairs, like I didn't
do a very good job on that, this is the chair that I have trouble with drawing, and so I don't do it as well as I do the other chair. So I'm going to just kind of move this over. So carbon number two, rather than the hydroxyl being equatorial, it's going to be axial.
I'm drawing in all of the other hydroxyls. The CH2OH group, depending on how you draw your chair, classically you're supposed to tilt it this way and what I usually do is draw it like that because it gives you a little more space. Depending on how you draw your chair you might have trouble fitting it, what you really
want to be careful of is not to make that group look axial. So it has to be bent as much as you can, bend it right here. Because if you draw it straight up and down, parallel with the edge of the paper, you'll get it marked wrong. No matter what else you've drawn, you get it marked wrong. Because on this bond, on this carbon right here, axial is down.
So your equatorials need to look equatorial, your axials need to look axial, and if you just draw this coming straight up you'll get it marked wrong. Because it's not. Now I left hydrogens off because I think it's pretty clear what's equatorial here and what's axial, but you may want to draw some of your hydrogens in if you want.
Some of you draw really bad chairs so you're going to have to work on that before the final. So, and it has to be really clear. So what I sometimes get people doing is they're not really sure whether it should
be axial or equatorial, they kind of draw it halfway in between. You know, hoping that they'll get a sympathy point. So if the grader has to look to the person next to him and say do you think this is axial or equatorial, it's wrong. So in other words, if this bond is going out at this angle right here,
everybody following that angle, automatically it's wrong. No matter what else you've drawn. So this needs to be equatorial. So here's what we're going to look for. This bond right here should parallel this bond. This bond right here should parallel that bond.
Okay? This bond right here, hard to do it, but that should parallel that bond also. The other way to look at it is if I put a plane of symmetry right here, this bond angle should be, this should be the mirror of this bond angle.
The mirror image. So however you want to draw it has to be clear. But you need to be able to draw a chair clearly. So some people, they'll just draw this out at an angle and this out at an angle and they'll label this as equatorial and this as axial. You don't need to label it, you need to be able to draw it so it looks that way. I know it's been a long time since we've drawn chairs but you need to be able to
do that. And the other thing that I see students doing is putting a bond here, let me draw it, and not putting hydrogens on. So like this, okay? Okay, that's wrong. Because those are all methyls now.
That is not, that is not beta-d-manos, okay? Those are methyls. So please make sure that if you're drawing bonds and you don't put anything on the end it's a methyl. So if you want to add hydrogens you can certainly add hydrogens. Where you're going to get into trouble is these middle carbons here when you're drawing.
So these middle carbons it's sometimes hard to fit things in, okay? The other thing you want to do is sometimes you'll see them drawn in books, let me erase this one and then we'll re-fix it. So like for example, if I go up like that, see that?
Okay, if you draw it like that that means that now we have, if you do it like that that means that this carbon now is bonded to four different things. Do you see that? If you, the intersection of any two lines is a carbon.
So if you're going to draw it like that then you need to break the back bond. So I'll do that, I'll erase this back bond here and I'll move that a little bit. So I think it might be easy, you break the back bond, right?
Just like that. That means that this is coming, this is in front of that but they're not bonded together. You don't want to break the front bond because then you're kind of doing like a little M.C. Escher thing going on there, it can't possibly be that way. So you can either make this bond really short like I did, that works, or you can
break the bond behind it. Okay? So it's your points on the test so you want to make sure that that's clear. And if you want to draw it and come show me and see if it's clear, I'll certainly let you know. Alright so now we also have to draw the Haworth projection and what I like to do is I like
to just flatten the chair. So now either memorize the Haworth of glucose and make changes or flatten the chair.
Alright so I'm going to add some more hydrogens here so you can see how I'm flattening the chair. So in a Haworth projection the ring oxygen is in the top right so our chair is going to exactly line up.
We wedge this guy right here and now all the bonds are straight up and down. This is carbon number one and so it's going to look like that. And so what you're looking for here is when you flatten the chair, notice that this hydroxyl is up above the hydrogen.
It's on the same carbon so it's up. This hydroxyl is also up. That sounds like, that's my ring but it doesn't sound like it's coming from here. Nope, somebody else is. Okay so that hydroxyl's up, this is up.
Above the hydrogen it's on the same carbon. This hydroxyl's also up above the hydrogen on the same carbon. This hydroxyl is down below the hydrogen that's on the same carbon and this CH2OH is up. And this CH2OH is always up for D sugars so you're going to make sure you get that right. So that's all I'm doing here is I'm looking at this, this hydroxyl is up.
What is hydroxyl is up? So that's carbon number one, carbon number two, carbon number three, the hydroxyl's also up. Carbon number four, the hydroxyl is down. So these are all straight up and down. And then carbon number five, the CH2OH is coming up and of course we want to make sure
and I'm just going to change colors so you can see we want to make sure of course that we put our hydrogens here. So just like that. So guaranteed you will have a problem on the exam like this where you'll have to draw
the chair and the hayworth and the fissure. So make sure you know how to do that. Questions? Anybody? Alright. Alright let's talk about five-membered ring form of fructose.
This is your least favorite. So not all sugars form six-membered rings in their hemiacetal form, many aldopentosis and ketohexosis form five-membered rings instead. Alright so this is, who recognizes this?
What's this, what is this? What is the fissure projection of? It's C-fructose right? I know it was right on the tip of your tongue. D-fructose. Okay so how do we know it's D-fructose? So we have CH2OH on the bottom, it's right, right, left, if it's glucose this is a hydroxyl
and it's right. So it's right, right, left, right. But this is a ketose and so instead of a hydroxyl here, it's a ketone. That's fructose. So fructose, remember, the whole bottom part of fructose, the bottom one, two, three, four carbons are identical to glucose.
So need to know that. Alright so we can cyclize here. So notice it looks a little different here. Now we gotta find our anomeric carbon. And we're looking for a carbon that's bonded to a hydroxyl and an OR. So that's gonna be right here.
Now remember with glucose in the six-membered ring, this was a hydrogen. But this is an aldose so instead of the hydrogen, we have a CH2OH. So we actually have two CH2OHs here. So there's our anomeric carbon and so the hydroxyl can be up or the hydroxyl can be down. If the hydroxyl is up that's beta, if it's down it's alpha.
So this is beta-D, we could call it fructose or we can call it, be more specific, beta-D fructofuranose. And all the furanose tells us is that it's fructose
that's cyclized into a five-membered ring. And then we have here alpha-D fructofuranose. Alright so let's label the anomeric carbon here
so we know what that looks like. So remember the anomeric carbon is the hemiacetal carbon.
It's the hemiacetal carbon. Alright and so this CH2OH right here is always up for the D sugars.
Okay and so we can draw the five-membered ring cyclic monosaccharides using the same strategy that we used with glucose. It just looks a little bit different but this is what it would look like here. And so let's draw it as we go. Mentally lay the Fischer projection over on its right hand side.
Groups that were on the right are down and groups that are on the left are up. So let's draw it right here.
And I'm gonna throw all my hydrogens on here. Okay so we're gonna mentally lay that down on its side. I'm gonna number first before I do that. One, two, three, four, five, six.
So six carbons. And I'm gonna lay that down on its side. And I'm gonna kind of curl it up a little bit like it's gonna cyclize. And I will wedge this so it's kind of sticking out here.
So when I lay that on its right hand side I have the ketone here, I have the CH2OH. So let me number that for you again. This is one, two, three, four, five.
And then let's draw carbon number six. So that's carbon number six. Okay so when I lay this down on its side I'm just like, it's almost like I'm knocking it over.
If this is going down on its right hand side which is what I've done here. If I look at carbon number three the hydrogen will be down. The hydroxyl will be down on carbon number four. The hydroxyl will be down on carbon number five. So let's draw that in.
So carbon number three, hydroxyl's down. Again I'm going down this direction laying it on its right hand side. Hydroxyl's down, hydroxyl here is down on carbon number four, on carbon number five the hydroxyl is down. And then of course we wanna draw our hydrogens in.
So that's what it would look like. And we need to rotate because if we keep it like this then what's gonna cyclize is this oxygen that will make a six membered ring. But for this particular ring system we're making a five membered ring. And so this oxygen has to rotate
so that it's where the CH2OH is. Okay, hard to do that rotation but if we're drawing D sugars the CH2OH will always be up, yes. Should be an OH where?
Oh yeah, yeah, yeah, oh gosh. I'm drawing deoxyribose, oh that's bad. Okay so yeah we wanna fix that. Okay, is that better?
All right, so we're gonna rotate this. If you don't know how to do the rotation very well again, doesn't matter because we know that the CH2OH is always going to be up. So when we do this rotation
everything down here will stay the same.
All right so that CH2OH group is now up
and it just made my carbon disappear. Okay, so that looks like that. Hydrogen down and we're ready to cyclize. Hydrogen here, hydrogen here and now we're directly in line to cyclize.
All right and that can detect the carbonyl from the bottom it can also detect the carbonyl from the top. Okay, so that's how we get our two anomers. And so we're going to get beta-D-fructofuranose
and then we're gonna get also alpha-D-fructofuranose.
All right, questions anybody about how we do a five member grain cyclization. All right, so the next thing I wanna talk about is mutarotation.
Something completely different. Well, not really completely different but. All right, so if we have a solution of D-glucose and we have it in water and we allow it to reach equilibrium so it doesn't matter which glucose,
it's a solution of D-glucose. It's gonna contain an equilibrium mixture of alpha-D-glucopyranose, beta-D-glucopyranose and their intermediate open chain form. That's the equilibrium mixture. And so here we have the alpha anomer.
This is open chain. I don't have the open chain drawn in a Fischer projection because it's gonna be cyclizing so I want it kind of in position to cyclize. And then we have the beta anomer. So what's gonna happen, this is a reversible reaction
and so this is going to, this is gonna cyclize to make either alpha or beta and then it's gonna open back up. So it's opening, closing, opening, closing, opening, closing. Okay, back and forth. So you could start with this and you'll still get the same equilibrium mixture. You could start with open chain completely.
You'd still get the same equilibrium mixture. You could start with the alpha anomer and you could, if you allow that to sit in water you're gonna still get the same equilibrium mixture and I'm gonna show you that right now. All right, so there's our equilibrium mixture and let's label that in equilibrium.
All right, now sugars are notoriously difficult to work with and crystallize. I don't know for science fair projects if you've ever tried to grow crystals of sugar.
Has anybody had any luck? Can I see a show of hands of people who've had luck doing that? It's kind of hard, isn't it? Okay, so what happens when you heat sugars up? So when you recrystallize something you put it in solution, you heat it up, you get everything to dissolve and then you allow a crystal to grow.
And if the crystal won't grow you put in a seed crystal and then the crystal grows on the seed crystal. The problem with sugars is they like to form syrups and crystals don't like to grow very well out of syrups and so maybe you ran into that when you were trying it. Anyway, so but if you're very good at this
and you crystallize, if you've got some good seed crystals and you crystallize below 98 degrees centigrade, I'm pretty sure those of you who have grown the seed, that have done it with the seed crystals haven't grown pure alpha anomer or pure beta anomer because that's pretty tough to do.
You can also crystallize above 98 degrees centigrade. 98 degrees. And if you do very careful crystallization below 98 degrees you can isolate the pure alpha anomer as crystals. If you very carefully crystallize above 98 degrees
you can isolate the pure beta anomer. So after all of that hard work of isolating those pure anomers, if you take this and you put it back into water,
so by the way the pure alpha anomer has an optical rotation of plus 112.2 degrees and the pure beta anomer has an optical rotation of plus 18.7 degrees. That's how you know you have pure alpha and how you know you have pure anomer. But when you add that back into water after all that careful work, what you're gonna find is the optical rotation is gonna gradually decrease upon standing
and you're going to, the optical rotation, once it gets to 52.7 it's going to stabilize at 52.7.
If you take the pure beta anomer and you do the same thing, what you're gonna find is the optical rotation now is going to gradually increase upon standing until it stabilizes at optical rotation of what?
52.7, okay? So the 52.7 is this equilibrium mixture up here. That's the optical rotation of the equilibrium mixture of the alpha, the beta, and the open chain form. Okay, so that's what's going on there.
So this is an equilibrium mixture of the alpha, open chain, which incidentally is a very small amount in the open chain, and the beta anomers.
All right, and so this change into, in specific rotations to a mutual value is called meter rotation. And if we have these numbers here, we can actually calculate how much alpha, how much beta we have in the mixture.
So something you want to know how to do, I haven't decided whether I will put it on the final or not. If I do put it on the final, you will need a non-graphing calculator and I will let you know. So you'll know even before you take the final whether you're going to have to do one of these calculations, okay? But right now I'm going to show you what it looks like.
It's not a hard calculation. So if we have any non-math people out there, it's not a hard calculation. So how much alpha and beta is there in the equilibrium mixture? This can be calculated as follows. I'm going to let A be the fraction of glucose
as the alpha anomer, I'm going to let B equal the fraction of glucose as the beta anomer. And so the fraction of glucose as the alpha times its optical rotation, which is plus 112.2 degrees, plus the fraction of glucose as its beta anomer,
which has an optical rotation of plus 18.7 degrees, and that's equal to 52. We are ignoring the open chain because it's such a small amount. We are ignoring the open chain form. So ignoring open chain form
because there's only a tiny amount. And I'm going to tell you in a second how much open chain we have and you'll see that that was a good approximation to completely ignore it.
All right, so we have two unknowns, one equation. We do know since we're ignoring the open chain form that A plus B equals one, right? Fraction of alpha, fraction of beta equals one. So A plus B equals one. And I'm going to solve for B. B equals one minus A.
And so now I have two equations, two unknowns, I'm going to be able to solve it. A times 112 point, whoops, 112.2 degrees,
plus, and for B I'm going to plug in one minus A, times its optical rotation, 18.7 degrees. And that's equal to 52.7 degrees. And then I'm going to solve for A. And what I'm going to get is A equals 0.36.
So that's the fraction that is in the alpha, of the fraction of the alpha anomer. So in order to get the percentage, it's the fraction times 100%.
So the alpha anomer is 36%. And so the beta anomer is 100 minus 36, 64%. All right, so does that make sense?
Does it make sense if, let's look at the, go look at the structures again, if this would scroll up for me. Does it make sense that the beta oh, the scroll bar's terrible on this. Does it make sense that the beta would be favored?
What does the beta anomer have that the alpha anomer doesn't have? Yeah, it has equatorial hydroxyl. So it's more stable. The beta anomer is more stable than the alpha anomer. It's not necessarily going to be the case for all of the sugars. You're going to get different percentages for the different sugars because in the different sugars,
some of these hydroxyls are going to be axial, so it kind of changes things, doesn't it? But for glucose, definitely beta is favored. All right, so that makes good sense.
And so for the beta, the anomeric hydroxyl is equatorial. And so you expect the beta anomer to predominate.
All right, so let's talk about some of these other forms of glucose. De-glucose can also cyclize into two furanose forms, so five member ring forms.
But furanose forms constitute only 0.14% at equilibrium. So alpha, this would be alpha-D-glucose furanose, 0%. So we didn't have to worry about that. Open chain form.
It looks like I don't have the number here, but it's less than 1%. I know it's less than 1%. And then we have beta-D-glucose furanose, 0.14%.
So for D-glucose, it overwhelmingly likes to be in the beta. The beta anomer of a six-member ring, okay? Questions on need of rotation, anybody? Yeah, 0.14%, it's really low.
So six-member rings are more stable than five, so I'm not surprised about that. Yeah, so yeah, this is 0.14%, not 14%. Yeah, that's the optical rotation of the pure anomer, which I got from right here, see that?
Yeah, does that make sense now? And the 52.7 is this value right here. That's gonna be different for different sugars, yep. You would be given, oh yeah, of course.
Of course that would be given. More questions on mean to rotation, yes. No, the open chain is less than 1%.
Yeah, it's less than 1%. It's not less than 0.14, but it's very small, I don't remember off the top of my head. And I know that my two TAs don't remember off the top of their head. It's less than 1%. So it was okay for us to ignore that value. All right, so let's talk about,
start talking about some reactions of monosaccharides as carbonyl compounds. So maybe if you looked at and learned glucose in bio, you probably thought, oh, how's it acting as a carbonyl compound? It's not a carbonyl compound, but it certainly is, right? So even though glucose is mostly in a hemiacetal form,
so it's all cyclized. Remember, it's opening and closing and opening and closing. So when it opens up, it can react as a carbonyl, right? It's an aldehyde. So it doesn't matter that most of it's in the ring form and not going to react. As soon as it opens up, it can react as a carbonyl.
If it can't open up, then it won't react as a carbonyl and that's what we have when we have a glycoside. So that's the first thing I want to talk about. So here we have this, and I'm just going to draw a squiggly line here. That means that I have a mixture of alpha and beta.
Mixture of alpha and beta is the squiggly line. And so this is a hemiacetal. And of course the hemiacetal carbon is right here.
It is the carbon that's bound to a hydroxyl and an OR group simultaneously. All right, if we take that and we add methanol and catalytic acid, we'll just say HCl here, or H3O+, or you could say H2SO4, catalytic.
What happens to a hemiacetal when we put it in methanol with a catalytic acid, what happens to it? You can make an acetal from it, right? Okay, so that's what we have here. And so we can make an acetal here.
Again, I'm going to just show this squiggly line for a mixture of alpha and beta. So this is now an acetal. All right, and so now that it's an acetal, we're going to name this. So we're going to name this kind of like you do with esters,
those of you who looked at the podcast and decided to actually learn how to name carbonyls, which was not everybody. But we named this kind of like an ester. So this group is named first, and then the rest of it is named as an, rather than an eight for an ester, it's an oside for a sugar.
So this is methyl, alpha, D. Well, since it's alpha, let's draw it as alpha. Let me go erase that. Let's fix that. Sorry if you wrote that in pen. I do apologize.
OCH3. That's alpha, right? So I'll give you a second to fix that. So methyl, alpha, D,
glucopyranoside. So it's kind of cool, because these names are really descriptive. So we have a methyl, it's a methyl glycoside. It's alpha, so we know that at the anomeric carbon, it's alpha, and we know it's D-glucose,
and we know it's in a six-membered ring. And so there's a lot of information there. And it turns out that you get about 66% of the methyl, and you'll also get about 33% of the beta anomer. All right, so this bond right here is the,
well, not that one. This bond right here is the glycosidic bond.
And the group that you put on here, this right here, that's called an A-glycone. And an A-glycone is a group that is bonded
to the anomeric carbon of a glycoside.
So methylglucopyranosides do not undergo meter rotation, and they are stable to base, but hydrolyze easily in acid solution, just like acetals, same as acetals.
So there is no meter rotation. We don't get the ring opening and closing and opening and closing. So once this forms methyl alpha D-glucopyranoside, it's stuck as the alpha anomer. It cannot go back to the beta. They don't go back and forth. There's no mechanism by which they can open up.
You can hydrolyze a glycoside, but then it's not a glycoside anymore. All right, so they're stable to base, just like acetals, and they hydrolyze easily in acid. So let's look at the mechanism here. And if you know the mechanism for acetal formation, which most of you got on midterm one, and assuming that's still in your head,
then you know this, okay? So this is like half of the mechanism for acetals. And I'll just start with alpha here just for fun. So we protonate, if I have catalytic acid,
I'm gonna have some of my alcohol protonated. If we use H3O+, you could also do this with H3O+. Not very many mechanisms in this chapter. I think this might be the only one, which means that's a likely candidate for the final,
just saying. Not to mention, it's also, you know, cumulative. I'm trying to make the final as cumulative as possible. Okay, all right.
So you're gonna encounter the same intermediate when you go back and look at midterm one and you look at your acetal mechanism and you say to yourself, how did I even do that? This is the same thing. And so what's gonna happen is electrons are gonna come down and we're gonna kick this off
as a leaving group. And we get something that looks like a protonated carbonyl, it's not a protonated carbonyl, but now we're gonna have methanol attack.
All right, so and I'm just gonna go ahead
and draw again the alpha here, but I could get beta also. And so this is the alpha, I'm gonna get beta also.
And then in the last step, I will deprotonate. So depending on what the acid here is, I'm just gonna use methanol. If it was hydrochloric, you could deprotonate with the chloride.
Now, notice I have left off all of the extra ethyl hydroxy groups. I'm just trying to focus on the group that's changing here. So I left off all of the other ones and if I have this mechanism on the final,
it would be the same thing. You're just gonna draw the ring like this without all the extra hydroxyls. No reason to take the time to do that, yeah. You can, it just depends on what the acid here is, what the acid is given. All right. So that's from the methanol coming in from the bottom. If you attack from the top face of this ring,
so if the top of the face of the carbonyl, you get the beta anomer.
And so we're not really gonna talk too much about why alpha's actually favored here, but you might wanna think about, if you think about when, if this is coming in from the top,
it's approaching this direction, we have lone pairs on this oxygen, that may affect how this methanol wants to come and attack. It's preferring to attack from the bottom rather than coming in on this side. So just something to think about, but I will not be testing you on that. All right, questions? Anybody on that mechanism?
So you should also be able to draw the reverse. Take a glycoside and hydrolyze it to get back your sugar, right? It's just the opposite. So I want you to be able to do that also. And look at that, I added this in.
I didn't use it to cover this, but I added it in. So we can do this though. And again, what we're gonna do is we're just gonna, we're gonna just draw in the necessary things. I'm gonna take the alpha, and I think these are the only two mechanisms in this chapter if I'm not mistaken.
Here I'm gonna use hydronium ions. So one of these two, huh? You can handle that. You guys usually do really do well on mechanisms too. It's synthesis that's the trick, right? That's the thing. Okay, so we're gonna protonate that.
That group's gonna leave because we need to replace that with a hydroxyl. All right, so we're gonna have electrons on oxygen come down, kick that off.
And I will tell you just from my experience, if I put this mechanism, one of the two of these, which I usually do, probably about 90% of the students will get it right. Okay, because you guys do really well on mechanisms.
Maybe you don't believe me, but I see it so. All right, then this is gonna attack. So in the previous one we had the methanol attack as we're trying to make a glycoside. Now we're trying to hydrolyze a glycoside.
And we're almost there.
Now we can undergo mutarotation because now it's a hemiacetose. So in order to undergo mutarotation, it has to be hemiacetal. It can't when it's an acetal. All right, questions on that one? Anybody? Yeah.
Oh, have that come off? Yeah, if you just wanna have this come off instead and form the carbocation, it's also okay to do that. It's just a resonance structure of what I've drawn here. All right. All right, so that's hydrolysis of glycosides.
Let's talk about reactions of monosaccharides as alcohols. So we know from chapter nine, I know that was a while ago, but when we did Williamson ether synthesis, what did we do? We deprotonated the alcohol with sodium hydride, and then we did, like if we wanted to make the methyl ether, we would use methyl iodide.
So this is the same idea. It's just that they use, with sugars, silver two oxide, silver, not silver two oxide, silver oxide is used instead of sodium hydride. Okay? But as you can see, this is now an acetal,
but all of these other hydroxyls we can make ethers out of those by deprotonating and attacking with methyl iodide. So let's draw the product of that.
All right, so this is now gonna be O-O, we'll just do methyl to save space here. O-methyl, O-methyl, O-methyl.
So notice I have a little bend here for that CH2 there. And let's draw some hydrogens in to make this look a little better. All right, so these guys, all of these,
this one, this one, this one, and this one, those are now ethers, methyl ether.
They are stable to base and dilute acid. What did we have to do to cleave a regular ether? I'm not talking about an epoxide, but a regular ether back in chapter nine, what did we have to use?
Do you remember that? Hot concentrated hydroiodic acid. Hydroiodic acid has a pKa of like minus nine or eight, nine, or 10. So they're pretty stable, that's why we use them as solvents all the time. That's why you guys use them in the lab when you're doing separations in the step funnel.
This one on the other hand is not an ether, is it? What is it? What is it part of? It's an acetal. This is part of an acetal. Therefore sensitive to dilute acid.
So if I treat this molecule with dilute acid, the acetal is gonna, the glycoside's gonna get hydrolyzed, all the rest of the ethers are gonna stay exactly the way they are, and we'll talk more about that next time.