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An optically active compound enriched with R - enantiomer (60%enantiomeric excess ) [alpha]d= 90degree .if the[alpha]d value of the sample (mixture) is-135 the ratio of R and S enantiomers would be?
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An optically active compound enriched with R - enantiomer (60%enantiom...
Calculation of Enantiomeric Ratio:
- Given: [α]d = -135
- Specific rotation of a mixture = (60% * 90) + (40% * x) = -135
- Solving for x gives x = -225

Enantiomeric ratio:
- Let the ratio of R:S enantiomers be x:y
- According to the formula for specific rotation:
xαR + yαS = -225
(0.6 * 90) + (0.4 * (-225)) = -135
54 - 90 = -135
-36 = -135
x/y = 135/36
x/y = 15/4

Ratio of R and S enantiomers:
- The ratio of R to S enantiomers in the mixture is 15:4. This means that for every 15 molecules of the R enantiomer, there are 4 molecules of the S enantiomer present.
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An optically active compound enriched with R - enantiomer (60%enantiomeric excess ) [alpha]d= 90degree .if the[alpha]d value of the sample (mixture) is-135 the ratio of R and S enantiomers would be?
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