Question

An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a specific rotation of −146°. If this solution were mixed with 500 mL of a solution containing 3 g of a racemic mixture of the compound, what would the specific rotation of the resulting mixture of the compound? What would be its optical purity?

Answer 1

Total mass of sample after mixing both the solutions = 10 + 3 = 13 g

The racemic mixture is a mixtrue containing equal amounts of both the isomers. Thus mass of (-) iomer = mass of (+) isomer in racemci mixture = 3/2 g

and the mass of (-) isomer = 10 g from pure sample + 3/2 from racemic mixture

= 11.5g

and the mass of (+) isomer i= 1.5 g

Thus optical purity = [excess isomer - other isomer ] x100/ total mass

= [11.5-1.5]x100 /13

= 76.92% of (-) isomer

Thus the optical rotationof resulting mixture of solution = optical purity x specific rotation of eexcess isomer

= 76.92 x (-146) /100

= -112.3^o

Thus optical purity = 76.92% and

specific rotation of mixture = -112.3