Under structure and functions of biomolecules
The specific rotations of the α- and β-anomers of D-glucose are +112 degrees and +18.7
degrees, respectively. When a crystalline sample of α-dglucopyranose is dissolved in water, the
specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the
basis of this result, what are the proportions of the α-and β-anomers at equilibrium? Assume that
the concentration of the open-chain form is negligible.
degrees, respectively. When a crystalline sample of α-dglucopyranose is dissolved in water, the
specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the
basis of this result, what are the proportions of the α-and β-anomers at equilibrium? Assume that
the concentration of the open-chain form is negligible.
Original post by Padmore98
The specific rotations of the α- and β-anomers of D-glucose are +112 degrees and +18.7
degrees, respectively. When a crystalline sample of α-dglucopyranose is dissolved in water, the
specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the
basis of this result, what are the proportions of the α-and β-anomers at equilibrium? Assume that
the concentration of the open-chain form is negligible.
degrees, respectively. When a crystalline sample of α-dglucopyranose is dissolved in water, the
specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the
basis of this result, what are the proportions of the α-and β-anomers at equilibrium? Assume that
the concentration of the open-chain form is negligible.
The key word is mutarotation: both D-glucose molecules changing their degrees, beta-glucoses increases in turn from +18.7 to +52.7 degree. Thus +52.7 is the degree for an equilibrium for both anomers. In equilibrium the amount of beta is about 64%, the one of alpha is respectively about 36%.
(edited 3 years ago)
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