Given the following information, draw a Haworth projection for...

Question

Given the following information, draw a Haworth projection for the disaccharide gentibiose:

1. Gentibiose is a dimer of glucopyranoses.

2. The glycosidic linkage is β(1 → 6).

3. The anomeric carbon not involved in the glycosidic linkage is in the α configuration.

Accepted Answer

Alright. So this practice problem says, given the following information listed down below, draw a Haworth projection for the disaccharide, GENTI Bios. And so notice that the first piece of information that we're given says, Genti Bios is a dimer of glucopyranosis. And dimer is just telling us that again, it's going to be a disaccharide consisting of 2 identical sugar units. And those sugar units are gonna be glucopyranosis. And so we know that pyranosis are cyclic sugars with a 6 membered ring and the glucose tells us that this sugar is gonna be derived from glucose. And so we need to recall the structure of glucose and recall that the structure of glucose looks like this. And so if we want to convert the linear form of D glucose into the cyclic pyronose form glucopyranose, then, we're going to need to remember the 2 words, up, lefting, and down right. And so, we can go ahead and start to draw the base of our pyranose, our 6 membered ring. So, we know that the 6 membered ring is going to have an oxygen atom in it. And so we're just drawing our, hexagon here. So, there it is right there. And of course, we know that the carbon atom furthest to the right, we're gonna want to be our anomeric carbon. So, this will be our carbon number 1 and then we can sequentially number all of these other carbons in order here. So, this is carbon number 5 and of course, we know that glucose has a total of 6 carbon atoms. So, we need to add our CH two o h group here. So, we can go ahead and add that right here. Our CH two O H and that will be our 6 carbon atom. And then of course looking at all of these chiral carbons here, we want to draw in all of these hydroxyl groups. So notice that the C2 hydroxyl is going to the right and so we think about the word down right and so in the Haworth projection we expect this to be going down, so the hydroxyl will be going down. And then notice that the c 3 hydroxyl is going to the left So, we think about the word uplifting and so here we can draw the hydroxyl group going up and then again, the C4 hydroxyl is going to the right. So, we think about down right and the hydroxyl will be going down. And so now, we've got the base of our gluco pyronose. And so looking at, the second piece of information that we're given, it tells us that the glycosidic linkage that's gonna be linking together the 2 glucopyranoses is gonna be a beta 16 glycosidic linkage. And so, of course, the beta is referring to the configuration of the anomeric carbon and the two bumps of the beta being on the same side tells us that the glycosidic linkage here on the anomeric carbon has to be going to the same side here as the highest numbered carbon. And so it's gotta be going in an upwards, fashion here. So we can go ahead and draw that in. And, we know that disaccharides are linked by o glycosidic linkages. So we know that this is going to be an oxygen atom here. And so, of course, it tells us that, in the third piece of information here that the anomeric carbon that's not involved in the glycosidic linkage is going to be in the alpha configuration. And so this is referring to the anomeric carbon of the second glucopyranose. So we have our first glucopyranose drawn. Now we just need to draw our second one. So essentially, what we can do is take this structure right here, and we can redraw it. And, again, we're told that this is going to be a beta 16 glycosidic linkage. So here's, the one. It's already in beta configuration. So we now just need this to be linked to our 6 carbon. And so we wanna make sure that we redraw our second Glucopyranose here so that it can be linked, here like so. And so, essentially what's going to happen is we're going to get rid of this and we will have a connection right here. And so we'll have our 6 carbon and, we'll still have, the 2 hydrogens here. And so, ultimately what you can see is that we've got one glucopyranose, our second glucopyranose, and they are linked by a beta 16 glycosidic linkage. So we can actually write beta here to remind us that this is already in the beta configuration. And then, again, this last piece of information is telling us that the anomeric carbon not involved in the glycosidic linkage, which would be this anomeric carbon over here, is gonna have the alpha configuration. And alpha, of course, means that it's gonna be going the opposite direction of its highest numbered carbon, which means that the hydroxyl group here is gonna be going down. And this is the alpha configuration here. And so really this is it. This is the answer to our practice problem. And so this is the structure of GENTIBIOS. And even though we hadn't discussed GENTIBIOS before in our previous, videos, Notice how we were able to use these clues to draw the structure of Genty Bios. And so we have developed a lot of critical skills here. And, again, this is the answer to our practice problem and I'll see you guys in our next video.

Struggling with Biochemistry?

Given the following information, draw a Haworth projection for the disaccharide gentibiose:
1. Gentibiose is a dimer of glucopyranoses.
2. The glycosidic linkage is β(1 → 6).
3. The anomeric carbon not involved in the glycosidic linkage is in the α configuration.

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Disaccharides practice set

Struggling with Biochemistry?

Given the following information, draw a Haworth projection for the disaccharide gentibiose:1. Gentibiose is a dimer of glucopyranoses.2. The glycosidic linkage is β(1 → 6).3. The anomeric carbon not involved in the glycosidic linkage is in the α configuration.

Watch next

Disaccharides practice set

Given the following information, draw a Haworth projection for the disaccharide gentibiose:
1. Gentibiose is a dimer of glucopyranoses.
2. The glycosidic linkage is β(1 → 6).
3. The anomeric carbon not involved in the glycosidic linkage is in the α configuration.