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I was on the Wikipedia page of Tartaric Acid, and I noticed that its solubility in water varies depending on whether the solution is racemic, meso, or optically active: $$\begin{array}{|c|c|} \hline & \text{solubility (kg/L)} \\ \hline \text{D or L (active)} & 1.33 \\ \hline \text{DL (racemic)} & 0.21 \\ \hline \text{meso} & 1.25 \\ \hline \end{array}$$

After some pondering, I can see why the D-solution and the meso solution may have different solubilities; I would guess it is because of varying degrees of intramolecular H-bonding that can occur between the -OH and the -COOH groups. I would be grateful if anyone could formally justify this.

tartaric acid

What surprised me, however, was the difference in solubility of the D-solution versus the racemic mixture. I would imagine that due to symmetry, every individual molecule of the acid in a racemic mixture would have an equal chance of being ionized/solvated as any other, as well as equal to the chances of molecules in a pure D-solution. Ergo, their solubilities should be identical. However, it appears my intuition is wrong here.

What is the reason, then, for these differences in solubility between the three types of solutions?

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    $\begingroup$ Solubility is represented by something like A(s) <=> A(aq). You’re essentially only looking at the stability of A(aq), measured by inter/intramolecular interactions, but ignoring that of A(s). I don’t know what the exact interactions in the solid state are but this is probably the fundamental issue $\endgroup$
    – orthocresol
    Commented Mar 26, 2018 at 10:02
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    $\begingroup$ I understand that, but the solid state aside, why would the A(-)(s) <=> A(-)(aq) vary depending on whether there are more A(-) or A(+) molecules in the solution? unless some of the solvation energy is due to other molecules of A rather than only those of water... $\endgroup$
    – Scripter1000
    Commented Mar 26, 2018 at 10:45
  • $\begingroup$ I was ready to comment by using @orthocresol argument as well as your last lines in your comment. The first is correct because is the general way to think. The second makes sense as for just based on similes cum similicum, a solvent containing a certain amount of X can be a better solvent for a compound Y as compared to itself alone. For the interaction between between X and Y we are back to how they pack in solid state. In the Tartaric ac. case, there is a stronger interaction between a D and a L molecule. When perfectly balanced in rac. mix. this lead to a higher melting point. This. ... $\endgroup$
    – Alchimista
    Commented Mar 26, 2018 at 13:09
  • $\begingroup$ .... means that the behaviour in your question is not as surprising as it is at a first glance. I have no time to make a nice looking answer out of this. Perhaps @orthocresol can or I will later. But this is the answer. $\endgroup$
    – Alchimista
    Commented Mar 26, 2018 at 13:12

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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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  • $\begingroup$ Interesting. I realize that until now, for some reason, I'd 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? $\endgroup$
    – Scripter1000
    Commented Mar 27, 2018 at 13:55
  • $\begingroup$ It is ast really. Because the reaction will be already between solved $\endgroup$
    – Alchimista
    Commented Mar 27, 2018 at 14:14
  • $\begingroup$ 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. $\endgroup$
    – Alchimista
    Commented Mar 27, 2018 at 14:23

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