The specific optical rotations of pure -D-glucopyranose and -D-glucopyranose are and , respectively. — Biomolecules Chemistry Question
Question
The specific optical rotations of pure $\alpha$ -D-glucopyranose and $\beta$ -D-glucopyranose are $+112^\circ$ and $+19^\circ$, respectively. The specific rotation of their constant equilibrium mixture in water is $+52.7^\circ$. If $100\text{ g}$ of pure $\alpha$ -D-glucopyranose is dissolved in water and allowed to reach equilibrium, what is the approximate mass of the $\beta$ -anomer formed? (Assume the mass of the acyclic intermediate is negligible).
💡 Solution & Explanation
Let the fraction of the $\beta$ -anomer be $x$. Therefore, the $\alpha$ -anomer fraction is $(1-x)$. Using specific rotations: $19x + 112(1-x) = 52.7$. Solving this yields $x \approx 0.637$ or $\approx 64\%$. Thus, in $100\text{ g}$ of the mixture, approximately $64\text{ g}$ will be the $\beta$ -anomer.