We have learnt that the energy evel of both R and S optical isomer of a compoud is same, just as well the entropy of each isomer is also same. If we consider the conversion of R form of a compound to it's S isomer we observe that the change in gibbs free energy is "0". This implies that the two isomers must exist in equilibrium. But we also know that any compound does not change from its R configuration to its S configuration for example inversion of configuration in Sn1 mechanisms lead to only one configuration and not both are obtained after the reaction.

So the question is that why don't any compounds R isomer convert into its S isomer on its own so as to maintain equilibrium ?


1 Answer 1


Because even though neither the R nor the S isomer is thermodynamically favored (in an achiral environment), the free energy barrier (or activation energy) between those two states is huge. This is the most common case, when R/S interconversion requires breaking and reforming a chemical bond. Then, even though the thermodynamic equilibrium would be to have a racemic mixture, kinetics prevent the solution from reaching this equilibrium state.

  • $\begingroup$ But isnt there an energy barrier in all equilibriums ? $\endgroup$ Commented Oct 22, 2013 at 14:56
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    $\begingroup$ Yes (well, almost yes: there are barrierless reactions), but in that particular case the barrier is very high, meaning it will take exponentially large time to pass (thus it does not happen on the experimental time scale) $\endgroup$
    – F'x
    Commented Oct 22, 2013 at 15:05
  • $\begingroup$ Great, this is exactly what me and my friend came onto ourselves. We said that maybe the reaction takes too much time and is not observable. Very happy to see that our findings were correct. $\endgroup$ Commented Oct 22, 2013 at 15:09

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