Assume that -D-glucose and -D-glucose are levorotatory and dextrorotatory respec

Question

Assume that -D-glucose and -D-glucose are levorotatory and dextrorotatory respectively. The specific rotation Z of an optically active compound in a set of conditions of temperature and wavelength is give by the equation:

Z = observed optical rotation (°)/optical length (dm) and concentration (g/mL)

A freshly prepared solution of -D-glucose shows a specific rotation of =122°. Over time, the rotation of the solution gradually decreases and reaches equilibrium of 52.5°. In contrast a freshly prepared solution of -D-glucose has a specific rotation of =19°. Th toration of this solution increases over time to the same equilibrium value as that shown by the -D-glucose anomer. Answer the following questions:

(1) Calculate the % of each of the two forms of D-glucose present at equilibrium.

(2) Why does the specific rotation of the two solutions change and reach the same value at equilibrium?

Explanation / Answer

1) At equilibrium, specific rotation =52.5°

Lets assume at equilibrium, % of -D-glucose = a and % of -D-glucose = b

So, we have the equation for equilibrium 122 * (a) + 19 * (b) = 52.5 also at equilibrium % b = (1 - % a)

122 (a) + 19 (1-a) = 52.5,

122 a +19-19a =52.5

103a = 33.5

a= 0.325 or 32.5%

So % of -D-glucose at equilibrium = 32.5 %

and % of -D-glucose at equilibrium = 100-32.5%= 67.5 %

(2) This is due to a phenomenon called "Mutarotation"

So, a Fresh solution of -D-glucose undergoes Mutarotation to an equilibrium mixture containing both and forms of D-glucose. Same applies on a Fresh solution of -D-glucose.

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