Glycosidic Linkage - Video Tutorials & Practice Problems

Glycosidic Linkage Formation Concept 1

In this video, we're gonna talk about the Glyco cytic linkage. Now, our Glyco cytic linkage formation, this is just an acetyl or acetyl bond. So depending on how you want to pronounce it between a sugars anomic carbon, which remember is carbon number one and another mono sac ride. We're gonna say it's formed via dehydration, which is the loss of water. And when we're looking at the equation for a gly acidic linkage formation, we can just say it as monosaccharide one plus monosaccharide two will give me what we call a di saccharide. So remember, disaccharide is two monosaccharide together. And what links them together is the Glyco cytic linkage. So if we take a look here, we're gonna say we have alpha D glucose, alpha D glucose. The anomic carbon, which is carbon number one for the monosaccharide. On the left is gonna interact with the hydroxyl group. The oh group of carbon. Number four of the other monosaccharide unit by their interactions, we're gonna lose water. So if we come over here, we've lost water. So this oxygen now needs to make up the bond that it just lost. This anomic carbon needs to replace the bond it just lost. So what they do is they connect together. So here, this represents our glyco cytic linkage. This is the bond that connects together my two monosaccharide to one another, transforming my individual monosaccharide into a disaccharide.

Glycosidic Linkage Formation Example 1

Here, it says provide the structure of the disaccharide formed when the hydroxyl groups of the highlighted carbons undergo a dehydration reaction. All right. So here we'll treat the monosaccharide on the left. This is our anomic carbon here. And what we're gonna say here is for these two ohs to interact, they need to be next to each other. So we're gonna need to actually flip this second mono sac ride over so that the ohs are next to each other. So we're gonna have to do a little bit of redrawing here. So we still have this year and here we're just gonna worry about just how the rings interact and then we'll come over here and redraw it as our final answer when it's cleaned up. So again, I'm flipping this. So now this position, which is also position one, I'm gonna draw it over here. OK. So there goes the oh All right. So 12345 and six. Again, I'm flipping this. So I'll put the oxygen here, right? So 12345 and six. So I kind of like just flipped it over like that. We have the interaction which causes a loss of water. So at the end, we're gonna form a Glyco Cytic linkage. So we're gonna come over here and clean it up here goes the Glyco Cytic linkage, right? So this is 123456 here, which is 123456 here. So we just gotta draw things in the right positions on carbon number two. The oh is down, then the oh is up, then the oh is down. And then we have C H2O H on the other ring which we have numbered here. Let's redraw this. So we have this oxygen. OK? So then let's see. Two, which is this to here. The oh is down and again numbering this 123456 on three. The oh is up on four, the oh is down and on five, the siege 20 is up. So this is how we would show these two mono sac rides. And in this one, both anomic carbons, both position ones are water connecting together to make the Glyco cytic linkage doesn't always have to be one and four. And this one we're seeing it's between one and one, but the same process still is followed. We have a dehydration reaction. So we have the loss of water to make the Glyco linkage.

Hydrolysis of Glycosidic Linkage Concept 2

In this section, we talk about the hydrolysis of a glyco cytic linkage. Now, we're gonna say under this reaction, a glyco cytic linkage is hydrolyzed into two monosaccharide. And we're gonna say, recall that hydrolysis itself is a reaction that breaks down a molecule through the addition of water. And we're gonna say here, both sugar carbons regain their oh or hydroxyl groups as a result of this hydrolysis. So here we have our disaccharide. So it's two monosaccharide connected to one another through this glyco cytic linkage. Here we're gonna have the addition of water. Water is gonna come in and actually help separate these from each other. As a result. We're gonna have this atomic carbon carbon. Number one, regaining its oh group, the oxygen that was already there is still there. We'll give it to this ca carbon and then it gains an H we're adding water, one gets the 01 gets the H but the end result is that both carbons at the end regain their oh group. So splitting this disaccharide has created for us alpha D glucose and beta D glucose as our two monosaccharide units, right? So just remember hydrolysis is undoing the Glyco cytic linkage to the addition of water.

Hydrolysis of Glycosidic Linkage Example 2

Here, it says provide the M saccharide units produced by hydrolysis of the following disaccharide. So here is our anomic carbon carbon number one. And the Glyco cytic linkage is this right here. We're gonna break it apart. We'll break these two free of one another through the addition of water with a little bit of acid catalyst. H plus what happens here is we're gonna get this monosaccharide being recreated by itself. These old waits are in the same position, this siege to a waits in the same position. Now we have the atomic carbon having its oh group back. Plus we're gonna draw this monosaccharide C h2o H oh oh which C h2o H this oxygen is still there. So it gets to keep it and then it gains a hydrogen so that we have an oh at the end. So through the hydrolysis of this disaccharide, we've created two separate monosaccharide products. Both of the carbons that were involved in the Glyco cytic linkage each have regained in oh group

Alpha vs Beta Linkages Concept 3

We know that a Glyco cytic linkage is what connects two different mono sac rides to one another. But there's also an alpha version and a beta version. Now, here we're going to say this type of linkage created is always defined by the linked anomic hydroxyl group. Remember your atomic hydroxyl group is connected to the atomic carbon, which is carbon number one of one of your monosaccharide. We're gonna say here, alpha and beta linkages are defined in the same way as cyclic monosaccharide. And we're gonna say the exception here is sucrose, it possesses two linked anomic hydroxyl groups. And so we're gonna say both must be named if we take a look here at the different types of Glyco cytic linkages. Let's take a look here at maltose in maltose. If we look, we're gonna say that alpha and beta is based in the same way as cyclic monosaccharide. We would say here that this C H2O H is up. But then the bond here of the an American carbon is pointing down, it's on the opposite side. And based on what we know about cyclic monosaccharide, opposite sides would mean that it is alpha. So this would be an alpha 14 glyco cytic linkage alpha because it's pointing down and 14 because it's the anomic carbon carbon one and on the other monosaccharide carbon number four that are connecting together to fake make my linkage. Next, we have C clo bios. If we look here, C H2O H is pointing up and the bond that is con connect is part of the linked anomic hydroxyl group. It's also pointing up. We said that if your oh and your C H2o H are on the same side with each other, that would be beta, same applies here. So this is a beta 14 linkage for this particular dyak ride. Next, we don't have a 14 here we have a 16. So same rules apply. We look at the atomic carbon, we looked at its linked hydroxyl group. It is pointing down siege to oh is pointing up, they're on opposite sides. That would mean that it is alpha. So here we're gonna say that Amylopsin is going to be alpha 16 linkage because the carbons that are connecting the Glyco bond or Glyco linkage are carbons number one and carbon number six. Finally, we have sucrose. Remember sucrose is different. We have two anomic carbons here. So this is the anomic carbon of this six membered ring. And technically here this is the anomic carbon of this five membered ring. If we take a look here, we would say that this linked hydroxyl group is pointing down and the CH two H is pointing up, they are, they are on opposite sides of each other. So that would be alpha. And then we say that this an American carbon would say this is carbon two. In this case, if we look at it this way where it's linked hydroxyl group, an American hydroxyl group is pointing up and this C H2O H group is also pointing up, they're both pointing in the same direction, same side that would be beta. So here we'd say sucrose is an alpha one, beta two glyco cytic linkage, right? So that's his advances is going to get in terms of these different types of linkages here. So just remember for the most part, we're following the same rules that we know of in terms of cyclic monosaccharide sucrose two. It's just that now there's two an American carbons. So you gotta do it twice the alpha one. And then we'd say beta two Glycolic linkage for sucrose.

Alpha vs Beta Linkages Example 3

In this example question, it says Mela bios represents a di saccharide. That is several magnitudes sweeter than table sugar, determine the type of glyco cytic linkage connecting its two monosaccharide units. All right. So here is our glyco cytic linkage. This is carbon number one. If we look at the other ring, this is the anomic carbon for this ring one and then we count towards 2345 and six. So the connection is between the anomic carbon of this top ring and carbon number six of the second ring. So we know that this is a 16 linkage. It is not sucrose. So we don't have to pay attention to both of the carbons connected it, it's not gonna be a situation of alpha one and beta two and none of that. Now, we just have to determine if this is alpha or beta in nature. If we look, we would say that this linked anomic hydroxyl group is pointing down the C H2O H group that's on the same ring is pointing up, they're on opposite sides. And based on what we talked about when it comes to MOOC cyclic um sugars or cyclic monosaccharide, if C H2O H and the linked anomic hydra groups are on opposite sides, that means it's alpha. So we'd say that it is an alpha 16 linkage. That is the type of glyco cytic linkage for this particular disaccharide.

Lactulose represents a man-made disaccharide that possesses a b-1,4 glycosidic linkage. Determine the two monosaccharide units created from its hydrolysis.

• A.
• B.
• C.