Is it true in assuming that a higher boiling/melting point means that an organic compound will be more soluble in water?

I'm trying to distinguish between the solubility of aldehydes vs. ketones based on boiling point. I received a question in a test today about it, comparing propanal with propanone, based solely on boiling points given, and my structural understanding of the two. I thought propanone had lower solubility of the two, as the ketone group is surrounded by carbons, while the propanal is exposed. Scouring through the internet, I'm not so sure my answer was correct!

  • $\begingroup$ You mean propanone as in acetone? It is miscible in water, whereas propanal has limited solubility. Both compounds have similar bp and mp (particularly bp), so solubility would be very difficult (imho) to decide on the basis of this alone. $\endgroup$
    – Buck Thorn
    Jun 26 '19 at 18:13

As andselsik pointed out, definetley no in the general sense. Just think of some practical examples like PTFE (Teflon) - which melts at about 330 °C but is definetly not water soluble.

However there is one connection I'd like to point out that allows to relate the enthalpy of melting to the solubility:

Consider phase equilibrium between a soluble compound C and water. At equilibrium, we have equality of chemical potentials:


Assume that $x_s$ at equilibrium is 1 (there would be a tiny fraction of water in the solid, but its fair to neglect that). Then the LHS is just the chemical potential of the pure solid:


Now using the solvent convention for the chemical potential of the liquid mixture:

$$\mu^{*,\:solid}_c(p,T)=\mu^{*,\: liquid}_c(p,T)+RT\mathrm{ln}(x_c^l \: \gamma_c)$$

Rearranging this and recognising that a chemical potential difference of a pure substance across a phase boundary is the gibbs energy of that phase transition gives

$$\Delta g_{c;\: melt} = \Delta g_{c;\: S \rightarrow L}=\Delta \mu_{c;\: S \rightarrow L} = RT\mathrm{ln}(x_c^{l} \gamma_c)$$


$$\Delta g = \Delta h + T \Delta s$$

using the limiting behaviour of the activity we find that for $x_c , \gamma_c \rightarrow 1$, the RHS tends to zero; we can reexpress the entropy of the phase change by setting $\Delta g$ in the above equation zero, and substitute the result into the equation obtained from equality of chemical potential:

$$\Delta h_c^{*,\:melt} \Big(\frac{T}{T^{melt}}-1\Big) = RT\mathrm{ln}(x_c^l \gamma_c)$$ ... $$x_c^l\gamma_c=\mathrm{exp} \Bigg( \frac{\Delta h_c^{*,melt}}{R} \Big(\frac{1}{T^{*,melt}}-\frac{1}{T}\Big)\Bigg)$$

The activity coefficient itself is obviously also a function of the composition, so this does not give us a straightforward way to calculate the solubility of compound C (it does however provide us with a route to $\gamma$, which is normally harder to measure than a solubility) but it shows that phase change properties have all sorts of implications!

  • $\begingroup$ What do you mean with " phase equilibrium between a soluble compound C and water"? $\endgroup$
    – Buck Thorn
    Jun 26 '19 at 14:40
  • $\begingroup$ Between solid solute and solvated solute. There is mechanical, thermal and chemical equilibrium and two distinct phases, so this is also a phase equilibrium $\endgroup$ Jun 26 '19 at 15:20
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    $\begingroup$ By the way, the OP asks about the connection between the boiling point and solubility, not the melting point. $\endgroup$
    – Buck Thorn
    Jun 26 '19 at 18:06
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    $\begingroup$ I interpreted "higher boiling/melting point" as referring to both boiling and melting point $\endgroup$ Jun 26 '19 at 19:32
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    $\begingroup$ I think its hard / next to impossible to deduct a very general trend. However a high bp can point to hydrogen bonding, which would mean favourable interactions with water. For propanon / propanal your approach was right but I'd argue that the keton is so miscible because the oxygen is in a symmetrical central position, whereas the aldehyde has a hydrophilic and a hydrophobic end - maybe there is more, surprised it would have such a strong effect with a 3C chain length (that two-tail effect is used in tensides / soaps, where the chain length is around 12-18 Cs) $\endgroup$ Jun 27 '19 at 10:46

The boiling point is directly dependent on the various forces of attraction that the chemical species can exert. However, for solubility, it is the relative strength of these forces of attraction.

Note that higher molecular weight species, in general, boil at higher temperatures due to greater magnitudes of van der Waals interactions.

Boiling Points

The boiling point of organic compounds of comparable molecular weight increases as the types of forces of attraction increase. As an example, methane exhibits only van der Waals interactions. Formaldehyde shows polar interactions as well. Ethanol can also hydrogen-bond. Note that this is with respect to a pure component, i.e. no solvent has been introduced yet.

Propanal and propanone (acetone) show van der Waals interactions and polar interactions.


Imagine methane in water. Methane molecules can have only van der Waals interactions with water. Moreover, by just being present inside the bulk of water, methane molecules disrupt the hydrogen bonding between surrounding water molecules. Water is "unhappy" with this loss of stabilization and methane, as expected, has a very low solubility in water. In an aqueous solution of formaldehyde, some hydrogen bonding between water molecules is disrupted. But, formaldehyde can hydrogen-bond with water due to its $\ce{O}$. In this case, the overall energy of the system is better reduced when formaldehyde is completely mixed with water. The case for ethanol is similar.

In a propanal-water and acetone-water systems, there are van der Waals interactions, polar interactions and hydrogen bonding (as with formaldehyde). Their magnitudes, however, differ, resulting in different solubilities.


Boiling point and solubility depend on similar properties. Solubility also depends heavily on the solvent properties. There is no direct correlation between boiling point and solubility (for non-homologoues)


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