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Standard XII

Chemistry

Mutarotation

The optical r...

Question

The optical rotation of the $α$ - form of a pyranose is $+ {150.7}^{\circ}$ , that of the $β$ -form is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of these anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α$ - form in equilibrium mixture is:

28%

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32%

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68%

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72%

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Solution

The correct option is A 28%
If we assume the percentage concentration of $α - f o r m$ to be $x$ , we know the $β - f o r m$ will be $100 - x$ % in solution.
From there, we can write

$\frac{x \times 150.7 + (100 - x) \times 52.8}{100} = 80.2$

$\frac{97.9 x + 5200.8}{100} = 80.2$

$97.9 x + 5200.8 = 8000.2$

$97.9 x = 8000.2 - 5200.8$

$⟹ 97.9 x = 2799.4$

$⟹ x = \frac{2799.4}{97.9}$

$⟹ x = 28.59$

$⟹ \approx 28$ %

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The optical rotation of the α- form of a pyranose is +150.7∘, that of the β-form is +52.8∘. In solution an equilibrium mixture of these anomers has an optical rotation of +80.2∘. The percentage of the α- form in equilibrium mixture is:

The optical rotation of the $α$ - form of a pyranose is $+ {150.7}^{\circ}$ , that of the $β$ -form is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of these anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α$ - form in equilibrium mixture is: