Question

The specific rotation of two glucose anomers are $\alpha =+{110}^{\circ }$ and $\beta =+{19}^{\circ }$ and for the constant equilibrium mixture is $+{52.7}^{\circ }$. Calculate the percent composition of the anomers in the equilibrium mixture.

Answer 1

Let a and b be the mole fraction of $\alpha$ and $\beta -$ anomers in
equilibrium mixture.

Thus,
$a+b=1$
${110}^{\circ }a+{19}^{\circ }b={52.7}^{\circ }$

Solving both equations, we get $a=0.362$ and $b=0.638$