Question

1. The amount of branching (number of (alpha 1->6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample of amylopectin is exhaustively methylated - treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyl with a methyl group, converting -OH to - OCH3. All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of 2,3-di-O-methylglucose so formed is determined.

(a) Explain the basis of this procedure for determining the number of (alpha1->6) branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure?

(b) A 258 mg sample of amylopectin treated as described above yielded 12.4 mg of 2,3-di-O-methylglucose. Determine what percentage of the glucose residues in amylopectin contain an (alpha1->6) branch. Assume the average molecular weight of a glucose residue in amylopectin is 162 g/mol (180 -18 for loss of water). The molecular weight of 2,3-di-O-methylglucose is 208 g/mol.

Answer for B should be: % of (alpha1->6) branch = 3.74%. Can someone show me how this is obtained?

2. A polysaccharide of unknown structure was isolated, subjected to exhaustive methylation by treatment with excess methyl iodide, and then hydrolyzed. Analysis of the products revealed three methylated sugars: 2,3,4-triO-methyl-D-glucose, 2,4-di-O-methyl-D-glucose, and 2,3,4,6-tetra-O-methyl-D-glucose, in the ratio 20:1:1. What is the structure of the polysaccharide? What is the percentage of monosaccharide units that are at nonreducing ends?

Answer 1

1. a) In glucose residues at branch points, the hydroxyl of C-6 is protected from methylation because it is involved in a glycosidic linkage. During complete methylation and subsequent hydrolysis, the branch-point residues yield 2,3-di-O-methylglucose (that is -OCH₃ group at 2nd and 3rd carbon) and the unbranched residues yield 2,3,6-tri-O-methylglucose.

b)Given the average molecular weight of a glucose residue = 162 g/mol

then 258 mg of amylopectin contains = (258 X 10^-3 g) / (162 g/mol) =1.59 X10^-3 mol of glucose

The 12.4 mg yield of 2,3-di-O-methylglucose (M_r = 208) is equivalent to (12.4 X 10^-3 g) / (208 g/mol) = 5.96 X10^-5 mol of glucose.

Thus, the percentage of glucose residues in amylopectin that yield 2,3-di-O-methylglucose is

[(5.96 X 10^-5 ) / (1.59 X 10^-3 )] X100 = 3.75%

2. The polysaccharide is a branched glucose polymer. As exhaustive methylation gives 2,3,4-tri-O-methyl-D-glucose as the major product, the predominant glycosidic linkage must be (16). When branch point occurs through C3, 2,4-di-O-methyl-D-is formed. The ratio of these two indicate that a branch occurs at an average frequency of once every 20 residues. The non reducing chain ends, which compose about 5%, of the residues, give 2,3,4,6-tetra-O-methyl-D-glucose. Thus, the polysaccharide has chains of (16)-linked D-glucose residues with(13)-linked branches, about one branch every 20 residues.