Why does cyclization of D-glucose give two isomers of α- and β-D-glucose?
The open chain form of glucose exists in small quantities and under physiological conditions glucose exists in a ring form. Due to cyclization of glucose molecule the C-1 carbonyl group reacts with C-5 hydroxyl group resulting in an intramolecular hemiacetal. Due to this hemiacetal formation glucose exists in α and β isomers.
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