An unknown disaccharide gives a positive Tollens' test. A glyc...

Question

An unknown disaccharide gives a positive Tollens' test. A glycosidase hydrolyzes it to D-galactose and D-mannose. When the disaccharide is treated with methyl iodide and Ag2O and then hydrolyzed with dilute HCl, the products are 2,3,4,6-tetra-O-methylgalactose and 2,3,4-tri-O-methylmannose. Propose a structure for the disaccharide.

Accepted Answer

All right. Hi, everyone. So this question says that an unknown dissect ride with the formula C 12 H 22 0 11 reduces tall and reagent and meter rotates it undergoes complete hydrolysis with beta galactosidase to give D galactose and D glucose when it is treated with dimethyl sulfate and base followed by hydrolysis. 2346 tetra omey galactose and 236 trom methyl glucose are obtained, determine the complete structure of the sty sacra. So throughout the reading of this question, I've highlighted some key bits of information here starting off with the molecular formula C 12 H 22 0 11. Because the number of carbons here in this case, 12 can be divided evenly by two. To let us know that our dy saccharide is going to compose of two hexes, right? Two monosaccharide each with six carbons. And this confirms or rather this is confirmed by the next sentence because after hydrolysis of the unknown disaccharide, we are given d galactose and oops D glucose, both of which are hexes. So now the question becomes how these two monosaccharide are connected together. And because the second sentence here mentions that hydrolysis took place with a beta galactosidase. That means that our disaccharide here must have been linked by a beta galacto seal linkage. Meaning that the anomic carbon of galactose must have been involved in the glyco cytic bond between our two monosaccharide. And also the question makes a point of mentioning in the first sentence that our disaccharide reduces to reagent and because it reduces tollins reagent, it's therefore considered to be a reducing sugar with a free hemi acetal group on the anomic carbon. Now, because dec galactose is already participating in the actual glyco cytic bond. Then what that means is that glucose d glucose must be present as a hemi AOP in order to make it a reducing sugar. So with all this in mind, let's go ahead and discuss the methylation procedure, start it off with that of galactose. So if I scroll down here a bit, then in galactose, right, the hydroxy groups that were methylated were the ones on carbon two, carbon three, carbon four and carbon six. Now recall the methylation only occurs here with free hydroxy groups on a rink. So the only hydroxy groups that were not actually meth laden in this case were the ones on carbon one and carbon five. Now, carbon one is the anomic carbon. So because carbon one is participating in the glyco cytic linkage, it was not going to be methylated because it did not actually have a free hydroxy group. And similarly carbon five or rather the oxygen on carbon five is the one that is involved in the actual ring structure. And then for glucose, right, the hydroxy groups that have been methylated were the ones on carbon two, carbon three and carbon six. Now, notably, this leaves out carbon one, carbon four and carbon five. Now, because glucose is a hexose just like galactose, the oxygen on carbon five is participating in the ring structure, meaning that could not have been methylated. Now, additionally, right, carbon one of glucose in this case must be left as a he acetyl in order to make our disaccharide a reducing sugar. So its hydroxy group must not have been methylated, which means that the oxygen on carbon four must have been involved in the glyco cytic linkage because it was not methylated, right? It must not have been free, meaning it must have been involved in the glyco cytic linkage. So now we have all the pieces we actually need to start drawing our dy sac, right, starting off with galactose, right? So first, I'll be drawing my six number ring to which I ended to which I added, excuse me, some shading on the bottom half here. So now for galactose, I'm going to be adding my sixth carbon outside of the ring as well as all of my hydroxy groups and hydrogens. Now here because we need a beta galactose. The hydroxy oxygen of the anomic carbon is going to be in the equatorial position in this case, pointing upwards. So now I'm just adding in my hydrogens before I go ahead and add in glucose. So now my glyco cytic linkage is going to connect to carbon four of the glucose. So here is my second six member ring like so right, so here I'm going to go ahead and add my sixth carbon as well as all of my hydroxy groups first. Now when it comes to carbon one of D glucose, the orientation of the hydrogen and the hydroxy group aren't necessarily specified. So that's why I'm drawing these kind of curved lines to showcase that they are connected, but that their configuration is not specified. And there you have it, here is the complete structure of our di saccharide. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Key Concepts

Tollens' Test

Glycosidic Bond and Hydrolysis

Methylation and Structural Analysis

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