Trehalose is a nonreducing disaccharide (C₁₂

Question

Trehalose is a nonreducing disaccharide (C₁₂H₂₂O₁₁) isolated from the poisonous mushroom Amanita muscaria. Treatment with an α-glucosidase converts trehalose to two molecules of glucose, but no reaction occurs when trehalose is treated with a β-glucosidase. When trehalose is methylated by dimethyl sulfate in mild base and then hydrolyzed, the only product is 2,3,4,6-tetra-O-methylglucose. Propose a complete structure and systematic name for trehalose.

Accepted Answer

Hello everybody. Welcome back. Our next question says, isom maltose is a reducing disaccharide used in a variety of foods, including baked goods and baking mixes. When isomaltase is treated with alpha glucosidase, it produces two molecules of glucose. However, when treated with beta glucose, no reaction occurs when isomaltase is methylated with dimethyl sulfate in a mild base and then hydrolyzed, it yields two products, 2346 tetra o methyl glucose and 1234 tetra o methyl glucose based on these observations, draw the structure of isomaltase and give its systematic name. So let's work through what information is provided in this problem. So dy saccharide, we've got two molecules and we're told that when treated with alpha glucosidase, it produces two molecules of glucose. So we know we have a disaccharide composed of two molecules of glucose and they are linked by an alpha gly acidic bond. Since we know that we produce these two sugars when it's treated with alpha glucosidase, but beta glucose has no effect. And then in terms of the carbons that are linked, we're told it's a reducing sugar, which means we know that while the first molecule is linked by its anomic carbon or at least usually is the there must be a free anomic carbon or excuse me, free an americ carbon oh group to convert to the open chain form and have an aldehyde group available. So it means our second glucose is not linked by the anna maric carbon or I should say an americ carbon hydroxyl group. So which carbon is it linked by? Well, now we turn to our methylation reaction. So me uh this methylation reaction will methylate every exposed hydroxyl group on the sugar. So for our first molecule, we know that the result is 234 and six methylated groups. So as we, you know, assumed the anomic carbon is the one involved in the linkage. It doesn't have an exposed oh group. And then of course, we don't expect to see C five in there since that oh group was involved uh when the ring was formed. So we have C 10 group in linkage. And then what we really need to determine is which of the oh groups is involved from the second molecule. And there are methylation occurred at carbons, 123 and four. So that way we see that it is the hydroxyl group on carbon six that is in the linkage. So let's go ahead and think about drawing our structure. We'll draw that, you know, chair confirmation there. So we're going to skip C one just for a second to go around and do the others to think about our glyco cytic bond. So we know that because we've got glucose here, we know that at C two, the hydroxyl group points down at C three, it points up at C four, it points down. And of course, at C five, we have the link to C six projecting up above our ring. And then there is the C six hydroxyl group. Now to go back to our anomic carbon, this is going to be our alpha glycolytic linkage. So since the C 60 is above the plane of the ring, the alpha arrangement means that the C one oxygen is in the opposite location. So it must point down. So pointing down, we've got the oxygen and then that's going to be linked to C six on our second glucose molecule. So we've got our bond to C six and now we need to draw our second glucose molecule nor my kind of weird looking structure there. I always find it a little challenging to draw in a nice looking way. So we have our C six on here. We'll go around to C one now because this anomic carbon has its hydroxyl group exposed. It's going to be a mix of alpha and beta forms. So what we do is we draw this little wiggle to the oh group to signify that it's this equal mix. And then of course, C two, the oh group points down C three, it points up C four down and then C five of course, is linked to C six, which is then in the Glyco cytic bond. So that is our structure here. And now we just need to give it its official name. So let's think about our recall, our naming rules for Dy sac rides. So the first sugar is the non reducing end. So the one where our anomic carbon is involved in the link and we start with the configuration of our anomic carbon, which is, as we said is the alpha configuration. So we start with alpha and then the name of our first mono sac ride, which is its root name. And then we follow that with pise seal. If it's a six membered ring or Purana seal, if it's five membered, this is of course six membered. So our sugar is D Gluco parana and then we have the atoms that are linked together. So we have open parentheses, one comma six closed parentheses. And then the name of our second sugar, which is followed by a pinos if its six ring or pinos, if its five ring. So that would be, now we'd add an alpha beta if we had a linkage that was specific that way. However, we're going to have a mix of alpha and beta forms. So we're just going to leave off the alpha or beta for the second sugar. And then, so we have D Gluco Pinus and the one sort of exception to that is if both anomic carbons are involved in the glyco cytic bond, then instead of having purose or furino, you would have pranic or furan aside. But in this case, it's C six, not the anomic carbon in the second sugar molecule. So that is our name. And again, the structure, we have the alpha gly acidic bond between the two between the carbon one and the carbon six of the second molecule. And then our systematic name is alpha D Gluco Perano 16 D Gluco Perino. So you can see why Isom Altos is an easier name to use. But that is the answer to this problem. We'll see you in the next video.

Key Concepts

Disaccharides

Glycosidic Bonds

Methylation and Hydrolysis

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