The equilibrium rotation of an aqueous solution of $\alpha$- and $\beta$-D-glucose is +50°, and the rotations of the pure $\alpha$ and $\beta$ forms are +102° and +24°, respectively. Assuming no other forms are present, the percent $\alpha$ form at equilibrium is:

Question

Quizplus · Accepted Answer

The equilibrium rotation is a weighted average of the rotations of the $\alpha$ and $\beta$ forms. Using the equation: $50 = 102x + 24(1-x)$, solving for $x$ gives the fraction of the $\alpha$ form, which is 33.3%.

The Equilibrium Rotation of an Aqueous Solution Of α\alphaα - And β\betaβ

The Equilibrium Rotation of an Aqueous Solution Of $\alpha$ - And $\beta$