An unknown reducing disaccharide is found to be unaffected by ...

Question

An unknown reducing disaccharide is found to be unaffected by invertase enzymes. Treatment with an α-galactosidase cleaves the disaccharide to give one molecule of D-fructose and one molecule of D-galactose. When the disaccharide is treated with excess iodomethane and silver oxide and then hydrolyzed in dilute acid, the products are 2,3,4,6-tetra-O-methylgalactose and 1,3,4-tri-O-methylfructose. Propose a structure for this disaccharide, and give its complete systematic name.

Accepted Answer

Hello everyone. And welcome back. Our next problem says honey contains a trace amount of an unknown disaccharide with reducing properties when subjected to alpha glucosidase. This disaccharide produces one molecule of D fructose and one molecule of D glucose upon subjecting it to an excess of iota methane and silver oxide followed by hydrolysis. It yields 2346 tetra o methyl glucose and 3465 tetra o methyl fructose. From these findings, what is the molecular structure of the unknown disaccharide? And what is its systematic nomenclature? So let's work through what we know about our disaccharide based on what we're told in the problem. So first, it tells us it has reducing properties. So this means that in the gly acidic bond, only one of our sugars is linked through its anomic carbon because for it to have reducing properties, at least one ketone or aldehyde needs to be available for oxidation. And when the glyco cytic bond is linked to the anomic carbon, that particular monomer can't go back out until it's into its straight chain form, it remains in its cyclic form and there's no carbon group there. So we know that only one is linked through its carbon one or is anomic carbon, excuse me. And then we're told that it breaks into two molecules when subjected to alpha glucosidase. So we know that this is an alpha glyco cytic link. And then of course, we're told that our two sugar monomers are D fructose and D glucose. So then last of all, and perhaps more most informative, we look at this methyl methylation reaction and it's important because this one's a little bit more complicated to note that we first do this exhaustive methylation. So first ex exhaustive methylation when all of the exposed hydroxyl groups are converted to och and then the second step is hydrolysis which breaks the gly acidic bond. So we need to recall that the methylation is going to tell us which hydroxyl groups are exposed in its disaccharide form. So that is going to tell us for our, which one is linked by its anomic carbon and which one is linked by a different carbon and which carbon links it because the carbon in the glyco bond doesn't have a hydroxyl group and therefore won't react in step one. So glucose is a little bit easier. So let's look at glucose first. So we know that it yields 2346 tetra omey glucose. So I'm going to start by looking at the cyclic form of glucose. And of course, I've drawn um the alpha form since we're told we have an alpha gly acidic bond. So if this is the one linked through its glyco through its anomic carbon, it will be in the alpha form. So let's just check whether this makes sense for the glucose to be linked in this way. So we have 234 and six are, these are exposed hydroxyl groups in this form. So we have carbon two, carbon, three, carbon four and carbon six, carbon five would not get methylated because carbon five has no hydroxyl group since it has the bond to carbon six oops accidentally put the six on top of the hydroxyl group instead of the carbon. So we can tell sort of looking at this right away. We don't see that number one and we don't expect to see number five because of the structure of glucose, glucose is, you know, one that's used quite a bit and it's good to just, you know, know this structure really well. Now fructose is a bit more complicated. So we're going to approach this slightly differently and look at it in a little bit more uh step by step kind of format. So first we're going to talk about, we'll start with our product of our 3465 tetra o methyl fructose. And we're going to look at it in its straight chain form. So let's think about fructose. And the reason I say it's a little more complicated is because it's a keto sugar. So its cyclic form, the anomic carbon is not carbon one, but carbon two. And it makes it a little more. We have to think through this a little more carefully. So let's draw our structure. So it's a six carbon sugar. So we start with carbon one C h2o H then carbon two, we have the carbon group and then carbon three, we have a hydroxyl group. On the left side, we're drawing a fissure projection here. Carbon four, our hydroxyl group is on the right side. Carbon five is on the right side. And then at the end, we have carbon six C h2o H. Now I'm going to convert this to this is the form of fru of fructose. Now I'm going to draw it to demonstrate what our tetra methyl fructose looks like. So the groups that have been converted two methyl groups, the hydroxyl groups that have been converted to methyl groups, we have 346 and five. So I'm going to go ahead and erase the hydroxyl groups, carbon three, carbon four, carbon five and carbon six. And we will replace those with methyl groups. So och three, och three and och three and then lastly one on carbon six. So what does this mean for what this looks like in our dy saccharide? So, recall that we're given this molecule we methylated first. And so the methyl groups represent exposed hydroxyl groups in a dy saccharide form. And we know that fructose can become either a purose or a pin nose depending on whether the hydroxyl group on carbon six or on carbon five reacts with the carbonel group. But what do we see here? Neither carbon five, both carbon five and carbon six had exposed hydroxyl groups. That's very odd. What does that mean? Well, that means that in our disaccharide, the fructose was in the straight chain form, it wasn't in its cyclic form. And again, we know that our ketone must be available. So what's the only hydroxyl group that would have been able to be involved in a gly acidic bond? And that's going to be the hydroxyl group on carbon one. Now note, we often think of like in glucose, carbon one is the anomic carbon but keto sugars, it's not carbon one, that's the anomic carbon. So our gly acidic bond is from the anomic carbon one on glucose to the non anomic carbon one on fructose. So we'll highlight both of those as our galic acidic bond there. So what then is the structure of our dy saccharide? And I think I'm going to scroll down. So we're not trying to squeeze it in and can see it clearly. So, scroll down here. So we know we have our glucose ring. So we'll put that in its cyclic form. So there's our oxygen in the ring and we have an alpha linkage. So since our carbon six projects up above the plane of the ring, we know that our gly acidic bond, the oxygen is gonna be going down from our anomic carbon. Then hydrox scoop on carbon two going down on carbon three, going up on carbon four going down. So that's the glucose portion. And now again, we have a gly acidic bond to carbon one of our fructose, but again, in its straight chain form. So the bond goes from oxygen to carbon one. And then now we're destroying it as a chain. So carbon two, which is a carbon group, then we have carbon three, carbon four, carbon five and carbon six. And then carbon six has that hydroxyl group. On the end. Now, we just need to think about the stereo chemistry of our hydroxyl groups. And that's always a bit tricky when going from official projection. But the best way to remember it is that in a Fisher projection, if we kind of envision all of the horizontal bonds as projecting up out of the plane and then the vertical ones as projecting down and then remember as we convert to a zigzag, we have to flip them around. So look at our, our structure that we've just drawn and see where our ketone is. So we're starting with, we'll number them carbon 12345 and six. And I'm going to number 34 and five on the Fisher projection to help me think about which way groups are going. So if I turn the whole thing, if I rotate my whole structure 90 degrees to the left that puts my ketone pointing up, that would put my hydroxyl group on carbon three pointing back. So put it on, eh sorry, I need to switch to blue there on a dashed wedge. So there's the oh then carbon four, we know that the hydroxyl group is going in the opposite direction. But because we've converted to a zig zag structure, it's as if we've spun the whole thing around. So it will also be a dashed wedge for that hydroxyl group. Carbon five is my zigzags pointing down and that hydroxyl group is the opposite side from the one on carbon three. So it will be a solid wedge projecting forward on carbon five. So now we have our structure, we'll go ahead and a little yellow box around that. We just need to get our systematic name. So let's go through the rules of naming a sugar. The first sugar is defined by the non reducing end. So glucose is going to be first, we label it by the configuration of its anomic carbon. So in this case, this is alpha. So we start with alpha and then we know we have D glucose. So alpha D and then it's a cyclic form. So if it's a six membered ring, it's a Perano seal if I remember doing Furano seal. So this is a Perano seal. So it's glucose is the root name of our sugar Perano seal. Now we define our linkage. We don't need an alpha or beta because we don't have a link with the anomic carbon. So we have the final linkage. It's from carbon one to carbon one. So we have hyphen and then open parentheses. One comma one closed parentheses. And then again, no alpha Arita, our carbon six hydroxyl group here. This is a straight chain form. There's no alpha or beta here. And then our second sugar, it's not a cyclic form. So it's just the sugar's name D fructose. So that's the systematic name, alpha D gluco pericle, 11 D fructose. And again, our structure here, we have our cyclic form of glucose. It's in the alpha configuration. So the anomic carbon which has the gly acidic bond has the oxygen going downward below the plane of the ring that's linked to the first carbon on fructose, which is in its straight chain form. So we have the ketone available since we know it's a reducing sugar. And then the rest of the chain of six carbons in our fructose. And the systematic name again is alpha D Gluco parental 11 D fructose. See you in the next video.

Key Concepts

Disaccharides

Enzymatic Hydrolysis

Methylation and Hydrolysis

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