Question

Be sure to answer all parts.

An aqueous solution containing 10g of an optically pure compound was dilluted to 500mL with water and was found to have a specific rotation of -123 degrees. If this solution were mixed with 500mL of a solution containing 3g of a racemic mixture of the compound, what would be the specific rotation of the resulting mixture of the compound? What would be its optical purity?

Specific rotation = ____________

Optical purity = ___________

Answer 1

ee % = 100 * ( [ R ] - [ S ] ) / ( [ R ] + [ S ] )

% Optical purity of sample = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

Enantiomeric excess and optical purity are same.

Amount of racemic mixture = 3 g

Amount of pure R compound = 10 g

So total amount of R compound after mixing = 10 g + (3 g/2) = 11.5 g

Amount of S isomer = 1.5 g

[ R ] = 11.5 g/L

[ S ] = 1.5 g/L

ee % = 100 * ( [ R ] - [ S ] ) / ( [ R ] + [ S ] )

= 100 * 11.5-1.5/11.5+1.5

= 77%

% Optical purity of sample = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

or, 77 = 100 * (specific rotation of sample)/ -123⁰

or, specific rotation of sample = - 94.71 ⁰

optical purity = 77%