A sample of limonene has a specific rotation of +81.5. (Given that R-(+)-limonene = +125.5 and, S-(-)-limonene = -122.1).
What is the %ee?
______%ee
What is the molecular composition of this same sample?
S isomer = ______%
R isomer = ______%
This is how I solved for the R and S isomer:
Let x = % of R
Let 100-x = % of S
125.6x + [(-122.1)(100-x)/100] = 81.5
125.6x - 12210 + 122.1x = 8150
247.7x = 20360
x = 82% = R isomer
100-82 = 18% = S isomer
That's how I solved for the % of R and S isomer. I'm not even sure if I solved correctly.
And for the %ee, I have no idea how to do it! I know that you divide measured specific rotation of mixture by specific rotation of pure enantiomer and then multiply by 100, but in this case, the the values for the R and S are different, so I'm not sure where to go from there.
Please help!
%ee = (+81.5/+125.5) x 100%
it was in the techniques book! :)
R-S= %ee I think
or +125.6. idk which one is the one in your problem.
Ah, the mysterious world of chirality and specific rotations! Don't worry, I'm here to help with a humorous explanation!
First, let's tackle the %ee conundrum. %ee stands for enantiomeric excess, or the excess of one enantiomer over the other. Since we have the specific rotations for both R-(+)-limonene and S-(-)-limonene, we can calculate the %ee using this formula:
%ee = (Measured Specific Rotation - Specific Rotation of Other Enantiomer) / Specific Rotation of One Enantiomer * 100
In this case, let's use R-(+)-limonene as our reference. So, the %ee would be:
%ee = (81.5 - (-122.1)) / 125.5 * 100
Now, you can do the math and calculate the %ee. But remember, whatever the answer is, make sure to add a little extra humor to it!
As for the molecular composition of the sample, you've already done an excellent job! By solving the equation, you found that the sample is 82% R isomer (which is quite "Rrrr-emarkable") and 18% S isomer (which is not too "S-serious"). Bravo!
Now, go forth and impress your chemistry pals with your stellar solving skills and humorous chemical insights! You've got this!
To calculate the %ee (enantiomeric excess), you need to compare the specific rotation of the sample to the specific rotations of each pure enantiomer and determine the difference in their rotations.
%ee = (specific rotation of sample - specific rotation of opposite enantiomer) / specific rotation of pure enantiomer * 100
In this case, the specific rotation of the sample is given as +81.5. You already know the specific rotations of the pure R-(+)-limonene (+125.5) and S-(-)-limonene (-122.1).
To calculate %ee for this sample, you can follow these steps:
1. Determine the enantiomer with the opposite rotation to the sample's rotation:
The sample has a positive rotation (+81.5), so the opposite enantiomer would be the negative rotation (-81.5).
2. Calculate the %ee for the sample:
%ee = (+81.5 - (-81.5)) / +125.5 * 100 = (163 / 125.5) * 100 ≈ 130.0% ee
The %ee represents the excess of one enantiomer over the other. In this case, the +81.5 rotation indicates a higher proportion of the R isomer (R-(+)-limonene) compared to the S isomer (S-(-)-limonene).
Now, let's solve for the molecular composition of the sample.
To determine the molecular composition, you need to calculate the percentage of each enantiomer in the sample.
Using your previous work, where x = % of R isomer and (100 - x) = % of S isomer, you found that x = 82% and (100 - x) = 18%.
So, the molecular composition of the sample would be:
R isomer = 82%
S isomer = 18%
Therefore, the molecular composition of the sample is:
R isomer = 82%
S isomer = 18%