Question

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a) An aqueous solution containing 10g of optically pure fructose was diluted to 500mL with water and placed in a polarimeter tube 20 cm long.The measured rotation was -5.20 degrees.Calculate the specific rotation of fructose.b) If this solution were mixed with 500 mL of a solution containing 5 g of racemic fructose,what would be the specific rotation of the resulting fructose mixture?What would be its optical purity?

Answer 1

a) Specific rotation of pure fructose = - 130°

b) Specific rotation of racemic fructose = 65°

Specific rotation of the racemic mixture = -65°

Optical purity = 50%

Step-by-step explanation:

specific rotation = observed rotation/ (tube length in decimeters x concentration)

[α] = α/lc

Observed rotation, α = -5.2°

Tube length in decimeters, l = 2 decimeters (1 decimeter = 10cm)

Concentration in g/mL or g/cm³ = 10/500 = 0.02 g/mL

Specific rotation, [α] = -5.2/(2×0.02) = - 130°

b) The racemic fructose would rotate light in the opposite direction. If the two enantiomers have the same concentration, specific rotation of the racemic fructose would be -(-130°) = 130°, but the concentrations are different.

Concentration in g/mL or g/cm³ = 5/500 = 0.01 g/mL

At half of the concentration of optically pure fructose, specific rotation would be 0.5 × 130° = 65°

Total specific rotation of mixture = - 130° + 65° = -65°

Optical purity = (specific rotation of mixture/specific rotation of pure sample) × 100% = [-65/(-130)]× 100 = 50%

Hope this Helps!!!