Fundamentals of Organic ChemistryChemistryBiomolecules
When octa-O-methyl-D-cellobiose is hydrolysed by an aqueous acid, two O-methylated glucose derivatives are formed. One is a tetramethyl derivative, and the other is a trimethyl derivative. Why is a single methyl substituent lost in this process?
Options
- AOne methoxy group is an ester, while the others are all ethers.
- BOne glucose unit is an -methyl glycoside, while the other is a -methyl glycoside.
- COne methoxy group is lost by -elimination.
- DOne methoxy group is part of an acetal, while the others are all ethers.
Correct answer
D. One methoxy group is part of an acetal, while the others are all ethers.
Step-by-step solution
Cellobiose is a disaccharide composed of two glucose units linked by a (1 4) glycosidic bond. When it is fully methylated to form octa-O-methyl-D-cellobiose, all eight hydroxyl groups are converted into methoxy groups. The distribution of these eight methyl groups is as follows: 1. Four methyl groups are on the non-reducing glucose unit at positions C2, C3, C4, and C6. These are all methyl ethers. 2. Four methyl groups are on the reducing glucose unit at positions C1, C2, C3, and C6. The methoxy group at the anomer