The equilibrium constant for mutarotation `alpha-"D Glucose "hArr beta-"D Glucose 1.8"`. What percentage of `alpha` form remains at equilibrium?

Options

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

Answer

A

Explanation

`{:(,alpha-"D glucose ",hArr,beta-"D glucose"),("Initial",1,,0),("At equilibrium",1-alpha,,alpha):}`
`K=(alpha)/(1-alpha)=1.8`
Soving `alpha=0.642`
`1-alpha=0.358`
Percent of `alpha-D` glucose remaining at equilibrium `=35.8%`.