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# Calculate how much of the $\alpha$ and $\beta$ anomers of glucose are present in equlibrium mixture with a specific rotation of 52.6 degrees. (pure $\alpha$ D- glucose has specific rotation = + 112.2 degrees and that of $\beta$- D-glucose has specific rotation + 19 degrees)

$\begin{array}{1 1}(a)\;\text{36 and 64 percentage}\\(b)\;\text{30 and 70 percentage}\\(c)\;\text{40 and 60 percentage}\\(d)\;\text{20 and 74 percentage}\end{array}$

If the fraction of glucose present as the alphaanomer is a the fraction present as the $\beta$ anomer is b, and the rotation of the mixture is +52.6 degree, we have
a(+112.2degree) + b(18.7) =52.6 degree
There is a very little of the open chain form present, so the fraction present as the $\alpha$ anomer (a) plus the fraction present as the $\beta$ anomer (b) should account for all the glucose a+b = 1 or b = 1-a.
Thus a(+112.2degree) + (1-a)(18.7) =52.6 degree
solving this equation for (a), we have a = .36 or 36 percentage. Thus (b) must be (1-0.36) = .64 or 64 percentage
Hence (a) is the correct answer.

edited Nov 28