Question

If pure (R)-2-bromobutane rotates light 100 degrees to the right, how much rotation would occur for a solution with 80% (R)-2-bromobutane and 20% (S)-2-bromobutane?

Answer 1

Specific rotaiton of pure  (R)-2-bromobutane = +100 Degrees (Rotates right)

Enantiomeric excess of sample = 80 -20 = 60 %

% Enantiomeric excess  = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

60 % = 100 x specific rotation of sample / 100

specific rotation of sample = + 60 Degrees (Rotates right)

Hence solution with 80% (R)-2-bromobutane and 20% (S)-2-bromobutane solution will have the rotation 60 degrees to the right.