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(+)-2-Butanol has [alpha]20D=+13.90. A sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.50 under similar conditions. The percentages of the(+) and (-) enantiomers present in the mixture are, respectively. |
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Answer» Given (+)-2-Butanol has [alpha]20D=+13.90 So its enantiomer (-)-2-Butanol has [alpha]20D=-13.90 Now a sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.50 under similar conditions. So itis due to the presence excess (-)-2-Butanol. Hence the excess percentage of (-)-2-Butanol in the mixture will be = (3.5x100)/13.9 ≈ 25.18%. The remainder of the sample is a racemic mixture of the enantiomers and in the racemic mixture each is present in (100-25.18)/2%=37.41% Hence we get the percentages of the(+) and (-) enantiomers present in the mixture are, respectively.as (+)-2-Butanol=37.41% and (-)-2-Butanol=37.41%+25.18%=62.59% |
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