The specific rotation of L-alanine in water (at 25°C) is - 2.8. A chemist prepared a mixture of L-alanine and its enantiomer, and 3.50 g of the mixture was dissolved in 10.0 mL of water. This solution was then placed in a sample cell with a pathlength of 10.0 cm and the observed rotation was - 0.14. Calculate the % ee of the mixture.

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Chemistry

Dirty Harry

14 April, 08:54

The specific rotation of L-alanine in water (at 25°C) is - 2.8. A chemist prepared a mixture of L-alanine and its enantiomer, and 3.50 g of the mixture was dissolved in 10.0 mL of water. This solution was then placed in a sample cell with a pathlength of 10.0 cm and the observed rotation was - 0.14. Calculate the % ee of the mixture.

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Crane · Answer 1

14.3%

Explanation:

The compounds that have a chiral carbon (a carbon which has 4 different elements or structures bonded to it) have optical isomers. These isomeres are optically active, and they're called enantiomers. The rotation of then is the opposite.

If the rotation of L-alanine is - 2.8, de rotation of D-alanine is + 2.8. Because the solution has rotation equal to - 0.14, the L-alanine is in excess (if the concentration was equal, it would be 0).

The formula for observed specific rotation is:

αobs = αsol*c*l

Where αsol is the rotation of the solution, c is the concentration of it (3.50g/10.0 mL = 0.35 g/mL), and l is the length of the tube (10.0 cm = 1.0 dm).

αsol = αobs/c*l

αsol = - 0.14 / (0.35*1)

αsol = - 0.4

The enantiomer excess is:

%ee = (solution specific rotation/maximum specific rotation) * 100%

%ee = (-0.4/-2.8) * 100%

%ee = 14.3%