(+)ibuprofein has a specific rotation of 57(degrees). Supposed a mixture of (+)-

Question

(+)ibuprofein has a specific rotation of 57(degrees). Supposed a mixture of (+)- and (-)-ibuprofein is analzyes by polarimetry and the observed rotation is +48 degrees. What is the % enantiomeric excess (%ee) and the + isomer in the mixture. Assume 1g dissolved in 1 mL of water, and the sample cell was 1dm long.

(+-lbuprofen has a specific rotation of +57. Suppose a mixture of (+)- and (-)-ibuprofen is analyzed by polarimetry and the observed rotation is +48°. What is the percentage enantiomeric excess (%ee) of the (+) isomer in the mixture? Express your answer in a whole number and use no units. Assume a 1 g sample was dissolved in 1 mL water and the sample cell was 1 dm long. Number

Explanation / Answer

Specific rotation = Rotationobs / (c *l)

where "Rotationobs" is the experimentally observed rotation, "c" is the concentration in g/ml and "l" is the pathlength of the cell used expressed in dm (10 cm).

Specific rotation = 48 / (1g/ml * 1dm) = +48°

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

Remember, the positive signs tell us that S is the dominant enantiomer over R.

% Optical purity of sample = 100 * (+48/+57) = 84.2 % ie there is a 84.2% excess of S over R.

This corresponds to a mixture of 92.1% S and 7.9% R. How do you get this quickly?

Well, if there is a 84.2% excess of S, then the 15.8% leftover must be equal amounts of both R and S ie. 7.9% of each. So the total amount of S is 7.9% + 84.2% excess = 92.1%.

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