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Lecture IX Polysaccharides I

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1 Lecture IX Polysaccharides I
Structural Determination Lecture #9 - Polysaccharides I - Determination of structure Will cover Structural Determination of PSs (general principles only)

2 CHO Classification I Monosaccharides Oligosaccharides Polysaccharides

3 CHO Classification V Polysaccharides > 20 monosaccharide units
Homo- and heteroglycans Linear Branched Note: DP = Degree of Polymerization

4 Carbohydrate Structure I
Monosaccharides Glycosides Oligosaccharides Polysaccharides Generalizations I

5 Carbohydrate Structure I
Monosaccharides Glycosides Oligosaccharides Polysaccharides Generalizations II polydisperse = range of MW polymolecular=linkages vary…. A polydisperse collection

6 Polysaccharide Structure
Preparation/Purification Remove foreign material Extraction/Isolation Filtration, dialysis, etc. Precipitation

7 Lignin Wood = cellulose + lignin(20-30%)
Lignin = non-CHO polymer, contain-ing aromatic rings bearing -OCH3 (methoxyl) groups. Methanol derived from wood comes from the lignin component (hence, MeOH = “wood alcohol”)

8 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Ring size(s) Anomeric configurations Linkage types Sequence of sugar residues Substituent groups DP/MW/Higher Orders of Structure

9 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Hydrolysis HPLC and/or GLC Structure of Heparin

10 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues HPLC ( Separation of sugar anomers

11 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Linkage types/ring sizes Methylation analysis

12 Reactions of Carbohydrates III Hydroxyl Groups- Ether Formation
Methylation analysis Methylate polysaccharide Hydrolyze (acid) Analyze hydrolizate Deduce where PS linkages were

13 Methylation Analysis I
CH2OH OH HO A CH2 B C ROH RO- 30% NaOH For example …….. 1. Strong base in DMSO 2. Methyl Iodide (Me I) O CH2OMe OMe MeO A CH2 B C 90% formic acid 100o, 1 hr Note, “Me” = CH3-

14 Methylation Analysis II
MeO O CH2OMe OMe A OH m.p. = 72oC 2,3,4,6-Tetra-O-methyl-D-galactose + O CH2OH OMe MeO HO B OH Cool, concentrate Dil H2SO4 14 hrs, 100o, neutralize A syrup + 2,3-Di-O-methyl-D-mannose O CH2OMe OMe HO C MeO OH A syrup 2,3,6-Tri-O-methyl-D-mannose

15 Methylation Analysis III
MeO O CH2OMe OMe A OH O CH2OH OMe MeO HO B OH O CH2OMe OMe HO C MeO OH + + 2,3-Di-O-methyl-D-mannose 2,3,6-Tri-O-methyl- D-mannose 2,3,4,6-Tetra-O-methyl- D-galactose 1. NaBH4 reduction 2. Acetylation CH2OAc CH2OMe C OAc OMe MeO A < CH2OAc C OAc MeO B < CH2OAc CH2OMe C OAc MeO < + + Note, “Ac” = CH3C- =O Note, “Me” = CH3 Partially methylated alditol acetates

16 Methylation Analysis IV Possible Structures (shorthand)
4)Man(1 Gal 1 6 A 6)Man(1 4)Man(1 Gal 1 4 A 1 2 B C B C Gal 1 4 Man 6 4)Man(1 3 A Gal 1 4 Man 6)Man(1 A C C B B

17 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Linkage types/ring sizes Monosaccharide sequence

18 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Linkage type Monosaccharide sequence Linkage configuration

19 Polysaccharide Structure
Structural Analysis(generalities) Sugar residues Linkage type Monosaccharide sequence Linkage configuration DP/MW/Higher Orders of Structure

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22 H2O H+ bond w/in chain H+ bond between chains H2O K+ ion ligands (6 each)


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