Name each monosaccharide unit & the glycosidic linkage in ...

Question

Name each monosaccharide unit & the glycosidic linkage in the following disaccharide (gentiobiulose).

Chemical structure of gentiobiulose showing monosaccharide units and glycosidic linkage.

Accepted Answer

Alright, so this practice problem wants us to name each monosaccharide unit and the glycosidic linkage in the following disaccharide down below which is genteobiolose. So really all we need to do is name this sugar unit right here and name this sugar unit over here and then name the glycosidic linkage that's linking together these 2 sugar units. And so, of course, there are multiple different approaches you can take to this problem and once you guys start getting really, really good at recognizing these Haworth projections, you'll realize that this sugar unit over here is galactose and this sugar unit over here is fructose. And of course, this is just naming these 2, monosaccharide units. So this first part here is complete. So we can move on to, naming the glycosidic linkage here in the middle. And so of course we're gonna be looking at the anomeric carbon here which is gonna be carbon number 1 and then all these other carbons we can sequentially number. And so, notice that the glycosidic bond here, the o glycosidic bond is going in an upwards direction in the same direction as the highest numbered carbon and so that means that it's gonna be in the beta configuration and so, this will be beta so up above here we can write in beta configuration and of course, the first carbon here is, being linked. So we need to figure out which, carbon this is what number it is and so for fructose over here it's anomeric carbon is over here. So this will be, carbon. The lowest possible number for this anomeric carbon is actually gonna be carbon 2. So this will be carbon 1. So we have 1, 2. This will be carbon 3. This will be carbon 4. This will be carbon 5 and this will be carbon 6. And so what you'll see is that this is a beta 16 glycosidic linkage. And so really this is the quick answer to this practice problem. Now, if you guys are good on this practice problem, you guys can end the video here. But if you're looking for a way that you could have recognized this as galactose and recognize this as fructose, then you can stick around for a little bit longer. And so notice that, if we want to convert cyclic sugars into, linear Fisher projections, then all we need to remember are the two words up, lefting, and down right. And so we can highlight chemical groups that are going up in green and highlight chemical groups going down in blue. And we can do the same here for furoctose. And so, the anomeric carbons, we are not going to be, caring about their positions because recall the anomeric carbons, are the reactive carbons and so, we only care about the ones that are not going to be reactive. The chiral carbons that are not going to be reactive. And so, of course, what we can see is that this sugar, we can see that it has 6 carbons on it. And so if we draw our Fisher projection, we know that we wanna have 6 carbons. So, let's go ahead and put in our first carbon here and we can put in our last carbon and then we want to make sure that there's a total of 6. So, 1, 2, 3, 4, 5, and then this is 6. And of course, what we can see here is that this anomeric carbon is part of a hemiacetal group and we can tell that this is part of a hemiacetal because it has 2 OR groups and coming off of this anomeric carbon going down we just have a hydrogen atom, So that makes it a hemiacetal. And hemiacetals we know are derived from aldehyde groups, which means that this carbon up here we would expect to be an aldehyde group. So we can go ahead and draw that in. There's our aldehyde group. And so that means that this is gonna be our carbon 1 and we can sequentially number all of the other carbon atoms going from top to bottom. So then this is where we can see that carbon number 2 has its hydroxyl group going down which means that it should be pointing to the right here. So, we can go ahead and fill that in and of course, the other group is going to be a hydrogen over here and then notice that carbon 3 and carbon 4 both have their hydroxyl groups pointing to, pointing upwards which means that they're going to be pointing to the left in the Fischer projection. So pointing to the left and so we have an Oh here and an Oh here. And of course the other groups are going to be Hydrogens. And then normally we are working life, prefers the deconfiguration of monosaccharides. And of course, the deconfiguration is just gonna be the configuration that is of the chiral carbon that's furthest from the carbonyl group. And the chiral carbon furthest from the carbonyl group is carbon 5. 5 and so the deconfiguration would be if it were going to the right. And so, we can go ahead and fill in on carbon 6. The other, the CHtwoOH group that is left And so here, what we have is the structure, the linear form of this, sugar. And so we were able to derive that just by using up, lefting, and down right. And so what you might notice here is that this kind of looks like, glucose a little bit, except the c four position is, has a different configuration than glucose. So this is the c 4 epimer of glucose. And you might recall that the c 4 epimer of glucose is galactose. And so that's another way we could have identified this as galactose. And recall that the 4 vowels that are in, galactose remind us that it's going to be the c 4 epimer of glucose. And, we could have done a similar process over here with defructose. Again, what we can see is that it has, defructose has a total of 6 carbon atoms. And so, of course, we would want to, make sure that we had 6 carbon atoms. So, 2, 3, 4, 5, and then 6 carbon atoms. And notice that the anomeric carbon here on fructose, on the other hand, notice is part of a ketal group. I'm sorry, a hemiketal group. Notice that it only has one OR group. It still has its hydroxyl, but notice going down it has another R group. And so if this were a hemiacetal, then this group here would be a hydrogen, but it's not a hydrogen, it's another R group going down. And so that's what makes this a Hemiketal and recall that Hemiketals are derived from ketones and, this ketone is gonna be on the C 2 carbon. And so, of course, what we're saying is that, this is going to be carbon 1. The c 2 carbon here must, be a ketone. So we can go ahead and draw in our ketone right here. And then of course looking at the c 3, hydroxyl group here, you can see that it is going up. And of course, if it's going up, it should be going to the left. And so, if we finish numbering off our carbons here, you can see that the C3 hydroxyl should be going to the left. So we can fill in an o h here. And of course the hydrogen is gonna be on this side. For the hydroxyl group going down on the C4, notice it's going down so it should be going right on the Fischer projection, so an Oh. And then of course we know that life prefers those D amino, those D monosaccharides. And so, of course that would mean that it would have to be going to the right here. And so we can go ahead and fill in our hydrogens And then, of course, this c 6 here is just going to be a a a CH two o h group. And, also the c one is also going to be a CH 2OH group. And so what you'll notice here is that from C3 all the way down, it's looking a lot like glucose and really the only difference is this top part up here. So this is really the ketose form of glucose and of course the ketose form of glucose is going to be fructose. And so, this is another way we could have identified this sugar unit as being fructose. And so, this here again concludes our practice problem. The answers are galactose, fructose, and beta 16 glycosidic linkage. And so I'll see you guys in our next video.

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Name each monosaccharide unit & the glycosidic linkage in the following disaccharide (gentiobiulose).

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