Specific rotations of freshly prepared aqueous solutions of I and II are + 112° and +18.7° respectively. On standing, the optical rotation of aqueous solution of I slowly decreases to give a final value of +52.7° due to equilibration with II. Under this state of equilibrium, the ratio of moles of II:I is nearly

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1.75

0.57

5.9

answer is C.

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Detailed Solution

$Specific rotation of α - glucose (I) = + 112$

$β - glucose (II) = + 18.7$

Let, at equilibrium

$(I) = x, (II) = (1 - x)$

$∴ (112 \times x) + (18.7 \times (1 - x)) = 52.7$

$\Rightarrow 112 x + 18.7 - 18.7 x = 52.7$

$\Rightarrow 93.3 x = 34$

$∴ x = \frac{34}{93.3} = 0.36 = α$

$\Rightarrow (1 - x) = (1 - 0.36) = 0.63 = β$

$∴ \frac{β}{α} = \frac{0.63}{0.36} = 1.75$

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