The glycosidic bond in a disaccharide was determined to be α(1...

Question

The glycosidic bond in a disaccharide was determined to be α(1→6). Hydrolysis of the disaccharide produced one galactose and one fructose. Draw the structure of the disaccharide.

Accepted Answer

Hello, everyone. Let's look at our next problem. During a laboratory experiment, a student hydrolyzed a dy sac ride and obtained one molecule glucose and one molecule of fructose. The experiment notes indicate that the glyco cytic linkage is alpha and then in parentheses, one right arrow five. So 1 to 5, draw the structure of the dy saccharide. So we know our diacide will have one glucose and one fructose. So let's think about the structures of both of those. So we'll start with our straight chain structures. So glucose is a an aldose, it has six carbons. So we'll draw our aldehyde group at the top cho down to ch 20 for carbon six. Then we add in the four internal carbons by making four lines. And we know in glucose carbons, 24 and five have hydroxyl groups on the right. Well, carbon three has oh on the left. So fill those in and we'll just number our carbons to make drawing our cyclic cyclical molecule a little easier. And then let's just take a look at fructose. So we've got them both here for drawing our disaccharide and fructose. We can remember as a ketose with that carbonel group of carbon two. So we have C H2O H as our carbon one group here. It's also a hexose six carbons there. C h2o H at the bottom. But then carbon two draw that double bond to an oxygen. It's carbon two and then 34 and five lines. And carbon 30 on the left, carbons four and 50 on the right. So we've got our structures here. We'll number our carbons again and we know that our, we'll draw our cyclical structures and then figure out how they link together. So for glucose, let's draw our, we have glucose makes a pin us a six membered ring. So draw the top bar of that with oxygen on the right. Draw the rest of our hexagon there. Our anomic carbon is carbon one and that goes to the right there. It's carbon one and then we'll go around carbon two. The oh goes down. Carbon three, it goes up carbon four down. And then of course, carbon five has RC H2O H on it. Carbon five has carbon six attached to it. And I'm just going to draw a little wiggle for our hydroxyl group on carbon one because we'll think about that alpha and beta in a little bit. And that's our glucose in its cyclical structure. So fructose forms a fewer nose, a five membered ring. So for that, draw the oxygen on top and then draw r rest of our five membered ring, our anomic carbon goes on the right. In this case, that's carbon two. So we have to draw C H2O H with carbon one coming out of there. And then little wiggly line for the hydroxyl group since we don't know which animal we have and then go around. Carbon three has the oh group, carbon three has that. Oh group going up, carbon four and five, going down, put those on there And then carbon six also is attached to carbon five. So we put another ch 20 group up there. So fructose we see has two C h2o H groups on either side of that oxygen within the ring. So we have our two monosaccharide here. Let's think about our glycolytic linkage. So it goes from carbon one on the glucose. So we'll go ahead and highlight that in pink to carbon five of the fructose. So let's label that. So this one's fairly straightforward to draw us a disaccharide because our carbon five on fructose is already on the right side of the ring. So the last thing we need to do is note that our glyco cytic linkage is an alpha linkage and that refers to the aer of glucose. So alpha linkage on glucose. So we'll draw the alpha form of glucose will mean that the hydroxyl group on that anomic carbon needs to be pointing opposite of the carbon six group. So our carbon six is going up. So our alpha hydroxyl group needs to be going down to form that gly acidic linkage. But alpha, yeah. So let's draw glucose again. Again, our oxygen on the top there and then I'm gonna kind of go backwards around the ring so I can get to the glyco cytic linkage last. So at our hydroxyl groups on carbon four, carbon three and carbon two, and then the hydroxyl group on carbon one is going to be now the one that links the two and again, it's alpha. So this bond needs to go down into the right. There's the oxygen then up and to the right. And that's going to be linking to carbon five. So then I will draw my little Pentagon to make that furino structure the oxygen at the top and that's carbon five there linking the two. And so this has AC H2O H group going up and then moving around my ring clockwise. Then there's the oxygen in the ring. Again, I don't know, I know that my glucose had was an alpha ener, but I don't, I'm not told what my fructose monomer is. So I'll draw the C H2O H on there and then I'll draw a little squiggly line to the oh as it could be either version and then going around continuing around carbon three has an oh going up and carbon four and oh going down. So we've drawn our dy sack right here, which is going to be sucrose with one unit of glucose and one of fructose, it has that alpha linkage going down between them from carbon, one of the glucose to carbon five of the fructose. See you in the next video.

The glycosidic bond in a disaccharide was determined to be α(1→6). Hydrolysis of the disaccharide produced one galactose and one fructose. Draw the structure of the disaccharide.

Key Concepts

Glycosidic Bond

Disaccharide Structure

Hydrolysis Reaction

Watch next