Which of the following sugars are reducing sugars? Which ones ...

Question

Which of the following sugars are reducing sugars? Which ones would undergo mutarotation?

(a) 4-O-(α-D-glucopyranosyl)-D-galactopyranose

Accepted Answer

Hello everybody. Let's take a look at our next problem. It has three parts. Question one says draw the structure of 40 and then in parentheses, Alpha D Manop Perano closed parentheses DT Oyo and question two says, is it a reducing sugar? And question three, can it undergo muta rotation? I note that questions two and three are linked because all reducing sugars can undergo meter rotation. Uh Reducing sugar essentially has an available carbonel group to be oxidized. So in the cyclic form, it means there's no Glyco linkage to the anomic carbon. They have a hemi aal group that an aqueous solution can be hydrolyzed back to the straight chain form of the sugar with a carbon group. Now a shortcut to this question before we even draw the structure is just looking at our name. So we know that looking at our Alpha D Manno Pera seal, this syl ending tells us that the Glyco link is through its anomic carbon. While in the second part of our name Tallo Pyin Nose, that OSE ending tells us that the Glyco link does not go through the anomic carbon. If the link were through both animate carbons, the ending here would be Perano side rather than pinos. So we know that we have this avail, we should have this available he acetyl group. And so just looking at the name, we know that we have a reducing sugar that can undergo muto rotation. So we already know that our answers to these two questions will be yes and yes. So just in the interest of if you were on test and maybe didn't have time to get through all of it, you could at least answer those two parts. But now we need to get the structure, which is a little more laborious note, this slightly different naming system. Um We often are used to seeing the links specified by number in the middle and here instead, we have 40 in the front. We know that our man Ayana is linked through the anomic carbon and that 40 tells us that the Glyco link is through carbon four of Talo Perino. So put C four. So let's get to drawing some structures. So we need to look at our straight chain structure first to see what the cyclic structure looks like. So we need to look up the straight chain structure of D manos. So we'll draw that it's an Aldo heos six carbons with the cho at the top, draw that vertical line with C H2O H at the bottom draw for middle carbons. And in terms of the orientation of the oh groups, carbons, two and three have the, oh, on the left and carbons four and five have the, oh, on the right. Not going to bother to put in the hydrogens because we know they're there. So this is our structure. So, what is our cyclic form of this? Well, let's draw six membered ring and then we'll leave our, an americ carbon until we look at other carbons. We know it's alpha D Manana seal. So we need to compare it to that group on the other side of the oxygen. So we'll leave carbon one for now and start with carbon two. So carbons two and three, we know that that oh group is going up, put that on there. Carbon four goes down and then carbon five would go down. But of course, that's what reacted with the aldehyde. So we would have rotated around which puts our carbon six up above the plane of the ring. So there's carbon six and the oh group. So if our carbon six is above the plane of the ring and we have alpha, an alpha arrangement, then our oh group on their anomic carbon must go down for the trans arrangement. So put alpha in parentheses. So that's our D manos and we know that our gly acidic link, let's put pink highlight here is going to go through the anomic carbon. So that O group on carbon one is going to be used to link the two and then we just need our D Talos, which is the other saccharide in there. And this is also and Aldo Heo. So a very similar structure. So starting with cho at the top C H2O H at the bottom and then four carbons in the middle, except the only difference is that we have the, the hydroxyl group at C four is on the left instead of the right. So we've got the hydroxyl groups on carbons 12 or excuse me, 23 and four on the left. And then the hydroxyl group on carbon five on the right. So that's the only difference between Taus and Manus. So looking at our cyclical structure, we have, I'm going to draw below just to save a little space, draw our six membered ring there. And then we know that the only difference will be that C 40 So I'm just going to fill this in except that we don't know, we're not told because we don't have it indicated in the name whether art Telop Peros is alpha or beta. So we draw a squiggly line to the oh on carbon one because it could be either way. So then go through the oh group of carbon two is going up, the oh group of carbon three is going up. And then again, the one difference there, in this case, the oh group at carbon four is going up rather than down. And then we know from our name that the glyco cytic linkage is going to go through carbon four. So find carbon four, that's the one that's going up here and that's where that linkage is going to be. So now we just need to draw the linked molecules. So we'll start with our manus. So put in the six membered ring and we know it's C one, the oxygen is going to be going down since it's alpha. That's where the glycolytic bond is going to be, it's no longer a hydroxyl group. C two that oh group going up C three, going up C four going down. This is the manos and then C five, we've got C six and that hydroxyl group. So that's the first part of our molecule. And then our gly acidic bond is going to the carbon four. We know that on our Talos, uh the hydroxyl group on carbon four was above the plane of the ring. So our glyco cytic bond will also be above the plane of the ring. So we have line down to the oxygen then down from that to carbon four. And then let's draw our second ring here. Oops didn't drop quite in the right place being sort of a weird structure there. So we've got our linkage to carbon four and then put on our carbon six and hydroxyl group gan our wiggly line from carbon one since we could be alpha or beta configuration going up to the hydroxyl group on carbon two up to the hydroxyl group on carbon number three. So there we have it and we'll highlight that as our structure. We have our 40 alpha Di Manno Perano di Tayo. It's like acidic linkage from see one of our manus to see four of our Talos. And then our question asks us and we, we know it's alpha Manno Perano seal so that um like acidic bond is going down below the plane of the ring of the manos. And then it's a reducing sugar because it has the Hammy acetyl group. I'll highlight down at the bottom in yellow that oxygen within the ring, then carbon one and the O group. So I just label that, that means it is a reducing sugar and due to that it can undergo muter rotation to get the accessible carbonyl group that will allow it, that can be oxidized, making it into reducing sugar. So that is our answer. We'll see you in the next video.

Which of the following sugars are reducing sugars? Which ones would undergo mutarotation?
(a) 4-O-(α-D-glucopyranosyl)-D-galactopyranose

Key Concepts

Reducing Sugars

Mutarotation

Glycosidic Bonds

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