Enantiomeric excess: Difference between revisions
No edit summary |
No edit summary |
||
| Line 6: | Line 6: | ||
|| stands for absolue value | || stands for absolue value | ||
example: | example: | ||
we have 70% of A | we have '''70%''' of A and '''30%''' of B in solution | ||
so also 40%A + 30%A +30%B | so also '''40%'''A + '''30%'''A +'''30%'''B | ||
[––] [––––––––] | |||
excess racemic part | |||
so the ee in this case is : '''40%''' | |||
Revision as of 15:41, 18 August 2009
go back to Main Page, Chemistry & More, Chemical & Physical concepts
from relative concentrations to ee
A and B are enantiomers, to get an ee from relative concentration, just do :
ee=|[B]-[A]| || stands for absolue value
example:
we have 70% of A and 30% of B in solution
so also 40%A + 30%A +30%B
[––] [––––––––]
excess racemic part
so the ee in this case is : 40%
from ee to relative concentrations
As enantiomeric says, the ee value is an exces, so do if A is in excess: %A=100-(100-ee)/2 %B=(100-ee)/2 Sometimes, it is interesting to access a theoretical value for the enantiomeric excess, written ee. This value can be accessed using the Free Energy of activation, assumed that:
- the pre-exponential factors are equivalent for both pathways leading to enantiomer A and B.
- Starting compound is identical for A and B
exp(-δΔG˜/RT)-1
ee(%)= ––––––––––––––– *100
exp(-δΔG˜/RT)+1
where δΔG˜ is the difference of Free Energy of activation separating the TS leading to A and B
δΔG˜=ΔG˜(favoured)-ΔG˜(unfavoured)
This theoretical value doesn't take into account the Gibbs Free Energy of the compounds, the formula given is derived from reaction rate constants. Thus, after equilibration in solution,depending on the temperature, the observed ee might change in respect to this.
Finally, one can get, under same conditions, a Gibbs Free Energy activation difference :
1+ee
δΔG˜= RT ln ––––
1-ee
/!\ The value of R is often in cal K−1.mol−1 [1], so if your δΔG˜ value is in kcal/mol, choose R=1.9858775*10^-3 Kcal K−1.mol−1