Question

A sample of enantiomeric 2-bromobutane is observed to have a specific rotation of +8.9. If the specific rotation of S-2-bromobutane is +23.1, what is the enantiomeric excess of S-2-bromobutane? What is the percentage of R-2-bromobutane and S-2-bromobutane within the sample mixture?

Answer 1

we know that

enantiomeric excess is given by

ee = observed rotation x 100 / specific rotation of pure enantiomer

in this case

observed rotation = 8.9

specific rotation of pure enantiomer = 23.1

so

ee = 8.9 x 100 /21

ee = 42.38

now

we know that

enantiomeric excess is given by

ee = ( %R - %S ) x 100 / ( % R + % S)

42.38 ( % R + % S) = 100 ( % R - %S)

0.4238 % R + 0.4238 %S = %R - % S

1.4238 %S = 0.5762 % R

% R = 2.471 %S

now

we know that

% R + % S = 100

2.471 %S + %S = 100

%S = 28.81

now

%R = 100 - 28.81 = 71.19

so

% R is 71.19

% S is 28.81