Question

The equilibrium constant for mutarotation $\alpha-D \rightleftharpoons \beta-D$ Glucose is $1.8$. What percentage of $\alpha$ form remains at equilibrium?

• A. 35.7
• B. 64.3
• C. 55.6
• D. 44.4

Answer 1

Answer: (A)

35.7

Answer 2

$K=\frac{\alpha}{1-\alpha}=1.8$

Solving $\alpha=0.642$
$1-\alpha=0.358$

Percent of $\alpha -D$ glucose remaining at equilibrium

$=35.8 \%$