Ch_322a_5.08 - Enantiomeric Excess(ee The enantiomeric...

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Optical Purity Racemic Form or Racemate Because the rotatory power is balanced by the equal numbers of the enantiomers, a racemic form shows no net rotation of plane-polarized light. Optical Purity and Enantiomeric Excess (ee) If the specific rotation of a single enantiomer of a chiral compound is known, it is possible to determine the enantiomeric mixture of samples of the compound from polarimetric experiments. A sample of a chiral compound that contains only a single enantiomer is enantiomerically pure, and is said to have an enantiomeric excess (ee) of 100% . A mixture of equal amounts of the two enantiomers of a chiral compound is called a racemic form (or mixture) or simply racemate .
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Example: (S)-(+)-1-chloro-2-methylbutane C H CH 2 Cl CH 3 CH 2 CH 3 As the levorotatory (R) enantiomer is added to the sample, it cancels the dextrorotatory power of an equal number of (S) molecules. The rotatory power of the sample is due only to the enantiomers that are in excess.
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Unformatted text preview: Enantiomeric Excess (ee) The enantiomeric excess of a sample is (ee) = (moles of one enantiomer -moles of other enantiomer) (total moles of both enantiomers) % x 100 and is directly calculated from the specific rotations by (ee) = % (observed specific rotation) (specific rotation of pure enantiomer) x 100 A sample that is 100% of this enantiomer has a specific rotation of [ ! ] 25 = +1.64 o . D Optical Purity of (R)- and (S)-1-Chloro-2-methylbutane observed specific rotation in degrees optical purity (ee) in % % S % R +1.64 100 100 -1.64 100 0 100 +0.82 50 75 25 +0.41 25 62.5 37.5 0 0 50 50-0.41 25 37.5 62.5- 0.82 50 25 75 Quiz 5.08 A sample of 100% ee (R)-2-butanol shows [ ! ] = -13.5 o . What is the enantiomeric mixture of a sample of 2-butanol that shows [ ! ] = +1.35 o ? %ee = +1.35 o +13.5 o x 100 = 10% excess of S This sample of 2-butanol is 90% racemic form and 10% excess S, or S = (45 + 10) = 55% and R = 45%....
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