In an aqueous solutions of `D-` glucose the percentages of `alpha` and `beta` anomers at the equilibrium condition are respectively

Options

A. `64%` and `36%`
B. `20%` and `80%`
C. `36%` and `64%`
D. `80%` and `20%`

Answer

C

Explanation

The `alpha` and `beta - D -` glucose have different specific rotations. When either anomer is dissolved in water their rotation change until the same fixed value results. This rotation change is called mutarotation.
The `alpha` and `beta-D-` glucose are each in equilibrium with the open chain aldehyde form and therefore with each other.
`alpha-D-"Glucose"hArr underset("form")("aldehyde")hArr beta-D-"Glucose"`
As each anomer begins to establish this equilibrium, its specific rotation changes. When equilibrium is reached, the experimentaly determined rotation remains constant. A base such as `NaOH` catalyzes the attainment of the equilibrium. Assuming that the concentration of the open-chain form is negligible, one can, by use of the specific rotations, calculate the precentage of the `alpha` and `beta` anomers, respectively Let `a` and `b` be the mole fractions of the `alpha -` and `beta -` anomers, respectively. Solving the simultaneous equations
`a + b = 1`
`112^(@) a + 18.7^(@) b = 52.7^(@)`
gives
`a xx 100% = 36%`
`b xx 100% = 64%`
Note that these percentages, `36% alpha` anomer and `64% beta -` another are in accord with a greater stability for `beta - D - (+) -` glucopyranose. The preference is (what we might expect) on the basis of its having only equatorial groups.
However the `beta`- anomer of a pyranose is not always the more stabel. With `D`- mannose, the equilibrium favours the `alpha-` anomer, and this result is called an anomeric effect.