- Nucleic acids, lipids, and carbohydrates questions
- Nucleic acid structure 1
- Antiparallel structure of DNA strands
- Saponification - Base promoted ester hydrolysis
- Lipids - Structure in cell membranes
- Lipids as cofactors and signaling molecules
- Carbohydrates - Naming and classification
- Fischer projections
- Carbohydrates - Epimers, common names
- Carbohydrates - Cyclic structures and anomers
- Carbohydrate - Glycoside formation hydrolysis
- Keto-enol tautomerization (by Sal)
- Disaccharides and polysaccharides
Created by Ryan Scott Patton.
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- Cl- doesn't really accept the H+ proton because it is extremely stable. I would assume that H2O would as it is in aqueous environment?(7 votes)
- correct. This is just the lazy way of writing it out. HCL donates all of its protons to H20 which acts as a base to make Hydronium ions (H30+). CL- is an extremely weak base that doesn't reaccept H+ ions. So really they should've just written it as H20 Becoming H30+ instead.(5 votes)
- I do not understand how the H2O changes the OH or I am just not sure about this concept at all! I am struggling in my Chem class, and I truly want to get these concepts. But struggling. Where does the below and above come in? Can you please help me understand how the glycoside is hydrolysis change the molecule? Or how do I think of it, and how it works and whys? Thank you so much !(1 vote)
- I hope you finished your class off strong! I know it is late but hopefully I can help you conceptualize these things. Chemistry is a difficult subject, but very satisfying once understood.
OH is converted to H2O in the presence of an acid and is a wonderful leaving group.
When a carbocation is formed (+ charge on carbon), there are three bonds to carbon which results in trigonal planar geometry (due to electron repulsion). If you imagine the carbocation like a piece of paper, a nucleophile can come from above or below to bond to that carbon (SN1). This results in both R and S configuration (roughly 50% of each).
Glycoside hydrolysis is important because this is what the enzymes in our body do (specifically amylase)! Hydrolysis is simply using water to break a bond. For instance, people who are lactose intolerant lack the enzyme to convert lactose into galactose and glucose. In other words, glycoside hydrolysis is how our body digests long carbohydrate chains into monosaccharides. These monosaccharides (glucose) can then be used to obtain energy.
Hope it helps!(8 votes)
- Can the Methanol used in this reaction be a saccharide too since its the same concept in a glycosidic linkage?(2 votes)
- Yeah but instead of a glycosidic bond between two carbohydrate monomers you will just form an acetal, or a glycoside.(1 vote)
- How can you tell if the group added is cis or trans? I can remember it by the way it's drawn in these videos but if it's drawn in another way, how would I know what I'm looking at?(1 vote)
- I think that you would automatically form both the alpha and beta anomer, so you form 2 products. Because both configurations exist in solution in equilibrium, so the alcohol will react with both of them, yielding 2 products. That is the way I look at it.(2 votes)
- why does the carbocation being resonance stabilized mean it's planar?(1 vote)
- Since the carbocation is resonance stabilized, it has partial double bond character which means that the carbocation is sp2 hybridized which has a molecular geometry of trigonal planar which is planar. This just indicates that there is no preference in the mehtod by which a nucleophile will attack. The nucleophile (or water in this case) can attack the carbocation from the top or bottom face.(1 vote)
- In an organic perspective, can you have sugar ring opening in the presence of catalytic acid, heat, and water?(1 vote)
- If acid is used to break down as well as create these linkages, how would we maintain one direction over the other? And what are the enzymes that facilitate this?(1 vote)
- You maintain one direction by the use of enzymes, but there are no enzymes that "cause" this. All you need is acid. The things enzymes do is allow us to form these glycosidic bonds in our body at very low acid levels and the specific enzyme pushes formation in one direction. Similarly we can break down glycogen to produce glucose using a different enzyme. The specific enzyme names are not really important.(1 vote)
- [Voiceover] So, the natural progression, at least in my mind, when I'm thinking about the cyclic nature of monosaccharides and their hemiacetal structure is how do these molecules progress to complete acetals? Remember that one OH and one OR group is a hemiacetal, so if we have a carbon chain and we have one OH group and one OR group, that's a hemiacetal, so, hemiacetal, and if we have two OR groups on that carbon chain, it's a full-blown acetal. So, two OR groups, or two alkoxy groups is an acetal. And this carbohydrate right here, is a hemiacetal. You can see, this is the carbon that would be in the center, and we've got one OH group and one OR group over here, and we want to know how it progresses to an acetal, and well, it actually progresses just like it would if this wasn't a cyclic carbohydrate, so the anomeric carbon, remember, that's the carbon right here, that's the carbon that was formerly the carbonyl carbon, before the ring closed. Now it's the anomeric carbon, and it's attacked by another nucleophilic alcohol. So, another nucleophilic alcohol comes in, and the carbon is oxidized, and loses this hydroxyl group right here, in the form of water, in order to make room for the OR group, or the alkoxy group. So, we have loss of water, and we've made room now for the alkoxy group, and the product then, is an acetal, and kind of looks like this. So now the carbon has two OR groups, and when this happens in the context of a carbohydrate, we call it a glycoside, so that's kind of a special kind of name, a glycoside, and what happens, is we're able to add a functional group onto the carbohydrate. This would be the functional group, and we say that it's bonded by a glycosidic linkage. So that linkage right there would be the glycosidic linkage. Really, that's the big picture idea, but what I want to do in this video, is I want to show you the actual mechanism. Of course, this is an organic chemistry-type idea, so we have to go into the mechanisms, you know, but I want to show you how this actually plays out. So, let me kind of clear some room here, and I'll tell you, I went ahead and I pre-drew the backbone for this mechanism just to save you the endurance of having me draw out all these little diagrams, but I'll try to walk us through, and I'll show us the functional steps for what's going on here. So, right here we have beta-D-glucose. So we've got the beta anomer of D-glucose, and so, this is a hemiacetal. We've got the OR group and the OH group, and what happens in the presence of a mild acid, is this OH group right here, on the anomeric carbon, gets protonated, so gets protonated, and the chloride atom is good at accepting those extra electrons from its bond, and what happens is, so that OH group lost electrons to the proton, and now we have a cation right here, and what happens is we have water loss, so, H2O loss, so this group right here will leave, we've got H2O loss, and then we have, from the previous step, this additional chloride anion, so we've got that chloride anion, and you can see that when the water, when we lose that water, we end up with a carbocation right here, 'cause these electrons left, we've got this cation right here, and it happens to be, as you can see in this image, resonance stabilized, and that's important. This product right here is resonance stabilized. So it's resonance stabilized, and we have, the planar nature of that is gonna be important in steps three and four, so let me pull those in, and we've got steps three and four here, so we start with this planar carbocation, and it's the planar nature of this carbocation that's gonna allow the next alkoxy group, the nucleophiic alcohol, to attack from above or below, so we've got the above attack on this top shelf, and then we've got the below attack on the bottom shelf of the mechanism, and so the alcohol group has nucleophilically attacked, and now the oxygen atom right here has a positive charge on it, but those chloride anions that we lost in the previous step come back into play here, and accept that extra hydrogen back, allowing the electrons from that bond to go to the oxygen. Same thing down here, and that oxygen loses its positive charge, and we end up with our products. Now again, the planar carbocation nature of that intermediate product allowed the alcohol group to attack from above or below, and what that means is that our glycoside, just like our carbohydrate formed two different anomers, the beta anomer and the alpha anomer, our glycoside does the same thing, so if the additional OR group right here is cis with respect to that last carbon, we end up with a beta-glycoside, the beta-glycoside, and again, if it's trans, we end up with the alpha-glycoside, and these glycosides, they actually occur naturally, and a few of them are quite significant. The one that really stands out to me is digoxin. It's a medication that's given for different cardiac, or heart complications. Let me draw a little heart here, and it's actually still extracted from the foxglove plant in the Netherlands, for use as a medication. Stands out to me because, as a nurse, I've given this medication several times, but anyway, it may have occurred to you at some point over the last few minutes, as we've gone through this mechanism, that if an alcohol can join onto a monosaccharide, and remember, that's essentially what we're doing in the formation of this glycoside, that the additional alcohol COULD be another monosaccharide. Remember that monosaccharides have plenty of hydroxyl groups, and that this reaction COULD be a means of joining carbohydrates together, and if that's the observation that you made, that's a really, really, a super-astute observation. This is indeed the method by which carbohydrates join together, and we call these linkages glycosidic linkages, and of course, we'll talk at length in the future about the polysaccharides that are produced. I'm gonna do a whole video on polysaccharides, but this is the means, this glycosidic formation, by which these chains are produced. But before we jump away from the actual glycosidic linkage subject, I want to point out that this whole process of forming glycosides can happen in reverse. So, these linkages can actually be broken down as well, and remember that with this bond formation, we lost a water molecule, so this was a dehydration reaction, so in reverse, what's gonna happen is, we're gonna add water back in a process called hydration. And so, because we see a hydration reaction that breaks down these poly-unit molecules, we call it a hydrolysis reaction. This is glycoside hydrolysis. The lysis for the breaking down, and then the hydro as a reference to the water molecule coming in, so what happens at the beginning is that the alkoxy group that we added before that OR group is protonated, and after protonation of that acetal OR group happens, we lose it in the form of an alcohol, so we lose that acetal alkoxy group, we've formed some water right here, as we lost this extra proton on that hydronium anion, and again, what happens, just like in that previous step, we form a resonance stabilized carbocation again. It's like we're forming that middle product again, so this right here is resonance stabilized, so again, the beginning of steps three through four, with glycoside hydrolysis, is gonna be a planar carbocation. So again, we've got a planar carbocation right here, and again, what that means is that the water that comes in now is gonna be able to nucleophilically attack above or below this molecule, so again, we've got the above and the below, and we'll see what happens. We've got an H2O that comes in right here, and the extra electrons on that water molecule's oxygen are gonna come in and they're gonna nab that proton, and these extra electrons from this bond are gonna come back in, and they're gonna stabilize that hydroxyl group, and the same thing would happen in the below situation. We've got H2O, the extra electrons here nab the hydrogen, nab that proton. The electrons from the bond shift down and stabilize that hydroxyl group, and what we've done, because of that above or below attack, is again, we've formed the beta anomer of D-glucose right here, because this hydroxyl group is cis with respect to that last carbon, so beta-D-glucose, and we've formed the alpha anomer, so alpha-D-glucose, and if the terms beta and alpha and anomer are confusing, I covered those in a previous video on the cyclic formation of monosaccharides, but that's where you'll find that terminology, and I guess, again right here, we've reformed that hydronium, because now the water molecule has picked up that extra proton.