This is a postmortem of a topic I read and did some research upon, but I don’t recall the details of some of the web pages I had followed. The original topic can be found in the link, Why do halogen substituents make molecules more lipophilic?
My interest began because I thought the actual reason was halogen substituents made molecules more hydrophilic, the opposite of what was asked. It actually is a little more complicated than that. Let us first look at the water solubility of some small hydrocarbon analogs. The change we should note is replacing a CH3 group with a fluorine or other halogen. A fluorine atom contains the same number of protons and electrons as a methyl group. You can see there is a great increase in water solubility. Replacing a fluorine with chlorine, bromine, and iodine increases the water solubility up to bromine. Then it begins to decrease again.
The molecular changes dynamically alter the interactions with water. Replacing a methyl group with a fluorine dramatically reverses the polarity. A methyl group has positive charges on its perimeter while fluorine negative charges. A polarization of the carbon-halogen bond also is present, but it should be the least with fluorine and the greatest with iodine.
We should conclude from this data that the interactions of these compounds with water are as hydrogen bond acceptors. Thus, fluorine should be greater than a methyl group. Also consistent with this model are the solubilities of methanol and methylamine. They are infinitely miscible. The non-bonded electrons of methanol and methylamine should be much greater hydrogen bond donors and are consistent with this observation.
Why is methyl bromide the most soluble methyl halide? If we were comparing the nuclear proton field, we would find it also increasing with the increase in molecular weight. This increase in the proton field effectively pulls the electron shells toward the nucleus. If we compare the bond lengths of the halo acids we would find that as the electrons are pulled toward the nucleus, they begin to encounter an increasing nuclear proton field. This proton field repulsive force would be the largest for iodine than the other halogens. This effect is consistent with the increased acidity of hydrogen iodide. A proton repulsive force should also decrease the ability of iodine to act as a hydrogen bond acceptor and increase the carbon-iodine bond length. An increase in the bond length should expose the carbon to interact with the electrons of water.
I did a quick calculation  of the lipophilicity of heptane (3.42), fluorohexane (2.51), chlorohexane (3.03), bromohexane (3.15), and iodohexane (3.54). The numbers in parentheses are the calculated Log P values for the partition between octanol and water. Therefore replacing a methyl group with a halogen results in a compound that is less lipophilic. For reference, hexane is less lipophilic (3.0) so adding a chlorine, bromine, or iodine does result in a compound that is more lipophilic.
 Calculated with ChemBioDraw v12.