Question

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

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

Answer 1

The correct option is A $35.7$

$\alpha -D$ glucose $\rightleftharpoons \beta -D$ glucose
Initial $1$ $0$
At eqilibrium $1-\alpha$ $\alpha$
$K=\frac{\alpha }{1-\alpha }=1.8$
Solving $\alpha =0.642$
$1-\alpha =0.358$
Percent of $\alpha -D$ glucose remaining at equilibrium
$=35.8$%