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Effect of stereoisomerism and racemization on solubility
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Mark Cox
Effect of stereoisomerism and racemization on solubility
Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.
Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.
As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.
However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.
In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.
However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.
This is right the case for tartaric acid,
Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.
Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower
according to the experimental data that you have posted.
(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).
Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.
Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.
As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.
However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.
In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.
However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.
This is right the case for tartaric acid,
Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.
Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower
according to the experimental data that you have posted.
(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).
Interesting. I realize that until now, for some reason, Id always assumed that the only manner in which these solutions differ is their rotation of plane polarized light. Am I correct in imagining, then, that the rates of different reactions that any given compound participates in would differ slightly, depending on what kind of mixture (D or racemic or meso) it is, because the activation energies for some of the steps would differ slightly?More
It is as you said concerning the attained solution optical activity. No you cannot expect different reaction activation energies unless the reaction is taking place in the solid state . Again, is rather the solid side of the story that matters. Think of it as a "third compound" unless dissolved, In the latter case just as a mixture in solution. For sake of completeness: not all 1:1 mixtures of enantiomrrs give a raceme with higher melting point.More
Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.
Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.
As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.
However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.
In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.
However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.
This is right the case for tartaric acid,
Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.
Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower
according to the experimental data that you have posted.
(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).
Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.
Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.
As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.
However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.
In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.
However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.
This is right the case for tartaric acid,
Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.
Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower
according to the experimental data that you have posted.
(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).
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