The specific rotation of two glucose anomers are α=+110∘ and β=+19∘ and for the constant equilibrium mixture is +52.7∘. Calculate the percent composition of the anomers in the equilibrium mixture.
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Solution
Let a and b be the mole fraction of α and β− anomers in equilibrium mixture.
Thus, a+b=1 110∘a+19∘b=52.7∘
Solving both equations, we get a=0.362 and b=0.638