Why is solubility phe > tyr ?

[Eric Lucas]

Davy Kalonia wrote:

> The free energy of solubilization (del G) depends on three factors.
...
> The energy of crystallization for Tyr is much larger than that of Phe,
> ie. the O..H--O interaction  and stacking of the hydrophobic rings
> results in stronger interaction in the solid state.  Thus more energy
> must be provided to move the molecules from a solid_solid state to a
> solvent_solid state.  Thus poor solubility for tyr.

Quite correct, but it's not just hydroxy groups that have this effect; methoxy 
groups can have the same effect.  There are several synthetic tricks that 
structural chemists can use to make compounds that are easier to crystallize for 
X-ray structural analysis.  One of the tricks is to use phenyl rings; they stack 
beautifully.  In this case, both phe and tyr have benzene rings, so that isn't it.  
However, beyond this, another common trick is to attach hydroxy or methoxy groups 
to the phenyl ring.  Very often they really make for excellent crystallization, 
because they really increase the lattice energy.  In grad school, I tried to get 
two rather different reactions to give crystals for quite some time.  When I 
changed the ligands in both (very different ligands) from having a phenyl to a 
having a dimethoxyphenyl, I got x-ray quality, analytically pure crystals in the 
NMR tubes I did the reactions in.  Without any effort at all.  Just left em there 
and came in the next day, and handed the tubes to the crystallographer. 

The argument about hydroxy groups and hydrogen bonding makes a lot of sense.  
However, methoxy groups obviously don't have the same option.  It's also hard for 
me to believe it's simple polarity--typical crystal stacking with phenyl rings is 
in a direction orthogonal to the benzene ring, and the dipole of a methoxyphenyl 
is in the plane of the ring.  Any crystallographers out there have a good 
explanation as to why methoxy groups seem unique in making phenyl-containing 
compounds much more crystalline?

	Eric