Author Topic: 2c-b hcl insolubility  (Read 2427 times)

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Vaaguh

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2c-b hcl insolubility
« on: September 18, 2002, 08:30:00 AM »
Does anyone have an explaination why 2c-b hcl does not dissolve in h2o?

Could the reason be that 2c-b hcl forms a intramolecular bond between the -NH3+ on one 2c-b-molecule and the electronegative -Br on another 2c-b-molecule? (hence the water won't break that bond).

Chromic

  • Guest
Electronegative group on aromatic ring
« Reply #1 on: September 18, 2002, 09:04:00 AM »
Hypothesis: If you've got an electronegative group on an aromatic ring that does not ionize, it'll likely be insoluble. (eg -F, -Br, -OH, but not -O(-), etc)

moo

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The question is: why should an ionic compound ...
« Reply #2 on: September 18, 2002, 09:50:00 AM »
The question is: why should an ionic compound dissolve in water? That is because the water molecules attract the ions with greater force than the ions attract each other. For example NaCl is soluble in water but HgS isn't, even though they both are ionic compounds. The bromine makes the 4-bromo-2,5-dimethoxyphenylethylammonium ion more hydrophobic than 2,5-dimethoxyphenylethylammonium ion, lowering the ability of water to solvate it.

Chromic: the OH-group would enhance solubility in water because of the great polarity of the O-H bond, resulting in hydrogen bonding. Of course, O- should be even better: that's why vanillin dissolves better in basic solutions.

Osmium

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Solubility of compounds
« Reply #3 on: September 18, 2002, 10:05:00 AM »
Solids are usually arranged in a crystal lattice, where the ions are held in place by electrostatic attraction. To dissolve a substance you have to break this lattice. This breaking usually requires energy, which can come from several sources like:
* entropy (meaning the resulting solution has a lower degree of order, is more 'chaotic' than the crystal structure; this usually releases energy!)
* the hydration/solubilisation of the ions (energy released by the solvent molecules arranging themselves around the ions)
* using thermal energy from the surrounding, this means spontanous cooling of the resulting solution
* and probably one or two more energies I don't remember  :P

Depending on the energy budget, derived by simply adding all these energies (which can of course be negative!)  mentioned above, dissolution of a compound might be exothermic (causing the solution to heat up; example would be NaOH), endothermic (solution cools down considerably; example NH4SCN), or pretty much neutral (e.g. NaCl). Some solid compounds simply have such a big lattice energy that the amount of energy produced by dissolution is insufficient to destroy the crystal lattice: they do not dissolve readily or only to a small extend (like Al2O3).


I hope that made some sense. Anyone feel free to correct me.


I'm not fat just horizontally disproportionate.

yellium

  • Guest
Very good. Now, can anyone explain why 2cb.
« Reply #4 on: September 18, 2002, 10:47:00 AM »
Very good. Now, can anyone explain why 2cb.hbr is supposedly  more water soluble than 2cb.hcl ?

hermanroempp

  • Guest
Reason could be that...
« Reply #5 on: September 18, 2002, 11:35:00 AM »
Cl- has a smaller ion radius than Br- => the bond between Br- and the corresponding cation is more easily broken because
a) the electric charge is more delocalized on the bromide ion
b) the water has a much larger surface to "dock on" the bromide ion
than on the chloride ion.
c) all the other things already mentioned by osmium... :)

Quidquid agis, prudenter agas et respice finem!

Osmium

  • Guest
d) because the hydrobromide crystallizes into a ...
« Reply #6 on: September 18, 2002, 03:36:00 PM »
d) because the hydrobromide crystallizes into a completely different lattice geometry.

Meaning the unit cell dimensions might differ enough that one of the salts might for example produce a cubic structure, while the other one produces a hexagonal structure.

There are many possible explanations, it's impossible to give a definite answer without conducting a whole lot of experiments.

I'm not fat just horizontally disproportionate.