Question

When $\left(\alpha \right)d-$glucose is dissolved in water, it undergoes a partial conversion to $\left(\beta \right)d-$glucose to exhibit mutarotation. This conversion stops when $63.6\%$ of glucose is in $i-$form. Assuming that equilibrium has been attained, calculate $K_c$ for mutarotation.

Answer 1

The equilibrium for the mutarotation, the initial percentage and the equilibrium percentage of $\alpha$-D-glucose and $\beta$ -D-glucose are as shown below.
$\alpha$-D-glucose $\rightleftharpoons \beta$-D-glucose

	$\alpha$-D-glucose	$\beta$-D-glucose
Initial percentage	100	0
Equilibrium percentage	$100-63.6=36.4$	63.6

The equilibrium constant for the mutarotation is as given below.
$\frac{[\beta -D-glucose]}{[\alpha -D-glucose]}$.
Substitute values in the above expression.
$K=\frac{63.6}{36.4}=1.747$
Hence, the value of the equilibrium constant for mutarotation is 1.747.