Enantiomeric excess: Difference between revisions

From Wiki
Jump to navigation Jump to search
No edit summary
Cgoehry (talk | contribs)
No edit summary
Line 1: Line 1:
go back to [[Main Page]], [[Chemistry & More]], [[Chemical & Physical concepts]]
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 so 30% of B in solution
so also 40%A + 30%A +30%B
                [––––––––]
                 


==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:
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.
* the pre-exponential factors are equivalent for both pathways leading to enantiomer A and B.

Revision as of 15:36, 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 so 30% of B in solution
so also 40%A + 30%A +30%B 
               [––––––––] 
                 

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