The Trend Between Boiling Point and Solubility in Organic Chemistry 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!
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:

$$\mu^{solid}_c(p,T,x_s)=\mu^{liquid}_c(p,T,x_l)$$

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:

$$\mu^{*,\:solid}_c(p,T)=\mu^{liquid}_c(p,T,x_l)$$

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

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

and

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

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!