Answer
a) All the anomeric carbons in cellulose are involved in glycosidic bond with the subsequent residue. Thus they are not free and cannot be methylated by methanol. Only the terminal anomeric carbon of the cellulose fiber is free (not involved in glycosidic bond) and therefore can be methylated by methanol. Thus by treatment with methanol, only one methyl group can be incorporated per cellulose molecule.
b) The glucose residues when present in cellulose fiber, have three free hydroxyl groups in its ring which will get methylated on treatment with dimethyl sulfate. After hydrolysis, the released glucose monosaccharide units will have two additional hydroxyl groups in its ring, but they won't be methylated. The structure of the monosaccharide is shown below:
Work Step by Step
a) All the anomeric carbons in cellulose are involved in glycosidic bond with the subsequent residue. Thus they are not free and cannot be methylated by methanol. Only the terminal anomeric carbon of the cellulose fiber is free (not involved in glycosidic bond) and therefore can be methylated by methanol. Thus by treatment with methanol, only one methyl group can be incorporated per cellulose molecule.
b) The glucose residues when present in cellulose fiber, have three free hydroxyl groups in its ring which will get methylated on treatment with dimethyl sulfate. After hydrolysis, the released glucose monosaccharide units will have two additional hydroxyl groups in its ring, but they won't be methylated. The structure of the monosaccharide is shown below:
