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    Specific rotation of glucose molecule (anomers)

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    Q1. ?-D-Galactopyranose has [?]D = +150.7o, and ?-D-galactopyranose has [?]D = +52.8o. When either of them is dissolved in water and allowed to reach equilibrium, the specific rotation of the solution will be +80.2o. What are the percentages of each anomer at equilibrium? (Original formatted question attached.)

    I really have no ideal how to attempt this question but I will provide a solution:

    150.7 + 80.2 = 230.9

    150.7/ 230.9 * 100 % = 65.3%

    52.8 + 80.2 = 133

    52.8/133 * 100% = 40%

    Right away, I know this answer is wrong because the percentages of the two anomers do not add up to 100%.

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    https://brainmass.com/chemistry/chemical-equilibrium/specific-rotation-glucose-molecule-anomers-14310

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    Let first anomer have f fraction and the other one 1 -f fraction.
    hence,
    150.7*f + ...

    Solution Summary

    The solution is given step-by-step equationally. Specific rotation of glucose molecules such as anomers are examined. The percentages of each anomers are determined. D-Galactopyranose is examined.

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