Enantiomeric excess question

[uorganic]

If a natural product was isolated and was found to have an enantiomeric excess (ee) of 84% what would be the percentages of each enatiomer in the mixture?

 

[movies]

Well, do you know the equation (or definition) of enantiomeric excess?

 

[quetzalcoatl9]

If a natural product was isolated and was found to have an enantiomeric excess (ee) of 84% what would be the percentages of each enatiomer in the mixture?

think about it...

how many possible enantiomers are there. hint: the number of possible enantiomers is the same for any compound.

so...let's say that the enantiomers were 50:50 in a mixture. then, the excess would be 0. if it were a 40:60 mixture, then one of the enantiomers would be in 10% excess. notice that (60 - (100 - 60) ) / 2 = 10

see the pattern here?

 

[kandasani]

Enantiomeric excess (ee) of 84%

Then major isomer % = (100+84)/2 = 92 %

%minor isomer = 100-92 = 8%

regards
kandasani

 

[LudusRex]

Enantiomeric excess (ee) refers to the percentage difference between the two enantiomers in a mixture, with a maximum value of 100%. In this case, the natural product has an ee of 84%, indicating that one enantiomer is present in significantly higher quantities than the other.

To determine the percentages of each enantiomer in the mixture, we can use the following formula:

Percentage of one enantiomer = ee + (100-ee)/2
Percentage of other enantiomer = (100-ee)/2

In this case, the percentage of the enantiomer with higher quantity would be 84% + (100-84)/2 = 92%. The percentage of the other enantiomer would be (100-84)/2 = 8%.

It is important to note that the actual percentages of each enantiomer may vary depending on the specific compound and isolation method used. However, the ee value provides a useful indicator of the relative amounts of each enantiomer in the mixture.