A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent
2 answers:
Answer: The proportion of the R-enantiomer is 62 %, the S-enantiomer is at 38 %, and the enantiomeric excess is 24 %
Explanation:
We know that:
A chiral molecule limonene is recorded as 62 % enantiopure.
This compound has two enantiomeric forms, R-limonene, and S-limonene.
Let's state that 62 % of the sample is R-limonene.
Consequently, the remaining S-limonene will amount to = (100 - 62) = 38 %
Enantiomeric excess is determined by the difference between the percentage of the major and the minor enantiomer.
In mathematical terms,
% major enantiomer = 62 %
% minor enantiomer = 38 %
Substituting the values into the equation, we have:
Thus, the R-enantiomer is 62 %, the S-enantiomer is 38 %, and the enantiomeric excess is 24 %.
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