Question

Be sure to answer all parts. An aqueous solution containing 10g of an optically pure compound...

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 = ___________

Homework Answers

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%

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