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Q.

Specific rotation for $α$ - anomer of a given mono saccharide is 100° and for $β$ - anomer is +20°. Specific rotation of equilibrium mixture is 68°, if % of $α$ - anomer is x × 10% then ‘x’ is

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a

7

b

6

c

4

d

5

answer is C.

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Detailed Solution

Percentage of α - anomer = 10 x %
Percentage of $β$ - anomer = (100 – 10x) %
$∴ \frac{(100 \times 10 x) + (20 \times (100 - 10 x))}{100} = 68 x = 6$

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Specific rotation for α - anomer of a given mono saccharide is 100° and for β- anomer is +20°. Specific rotation of equilibrium mixture is 68°, if % of α - anomer is x × 10% then ‘x’ is