The equilibrium constant for the mutarotation, α-D–glucose$\rightleftharpoons$β–D–glucose is 1.8. Wh

Question

The equilibrium constant for the mutarotation, α-D–glucose$\rightleftharpoons$β–D–glucose is 1.8. What percent of the α-form remains under equilibrium?

Accepted Answer

α – D – Glucose $ightleftharpoons \beta$– D – Glucosea 0Equilibrium a – x x$K = \frac{x}{a - x} = 1.8 \Rightarrow x = \frac{9}{14}a$∴% of α – D – Glucose remained = $\frac{a - x}{a} 	imes 100 = 35.7\%$

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