How Water Freezes Lower on a Negatively Charged Surface
by orgopete on Feb 10 2010 (4228 Views)I first heard this on National Public Radio and then I searched for it. In short, David Ehre, Etay Lavert, Meir Lahav, and Igor Lubomirsky report in Science, (Water Freezes Differently on Positively and Negatively Charged Surfaces of Pyroelectric Materials) water freezes at a lower temperature (-18°C) on the negatively charged side of a lithium tantalate plate with a strontium titanate film than on the positive side (-7°C, and -12°C uncharged).
Is this unique or is this a manifestation of something in our standard introductory organic chemistry textbooks? I thought it was the latter. Let me explain how.
For the purpose of thinking about this problem, let us assume the metal surface is simply a flat charged surface, without contour. If the surface has a negative charge, then the water should be attracted like a flagpole. One hydrogen should be anchored to the surface of the metal at right angles and the other hydrogen could spin about that axis with the flag hydrogen at 105°. It should not be surprising that this configuration should not be as good of a surface as one with greater rigidity.
If we compare with the positively charged surface, then both pairs of non-bonded electrons should be anchored to the surface and locking the hydrogens in a fixed position. This should limit the degrees of freedom and enable crystal growth.
For those that may be wondering where this is found in your textbook, it may not be there. The negatively charged surface is the one that seemingly will have the most important stereochemical constraints and information in a textbook. The analogy I was comparing is the stereochemical restrictions of proton transfer reactions. In that context, the angle between a proton and donor-acceptor electron pairs in a hydrogen bond is usually 180°. One can find smaller bond angles in intramolecular proton transfer reactions, such as the decarboxylation of a beta-ketoacid or a Cope elimination reaction of an amine-oxide as six and five-membered ring examples.
You may also encounter a … transition state which transfers a proton via a four-membered ring. While this mechanism is present in some textbooks, I am troubled by a lack of precedent for this proton transfer. In a normal hydrogen bond, the preferred bond angle is 180°. Variations from 180° are commonly found in six and five-membered rings …
While the four-membered ring is expedient and avoids a zwitterionic intermediate, I am skeptical sufficient experimental data exists to support it. In the normal hydrogen bond, the electron-electron repulsion forces the nuclei to be linear. While smaller angles are present in six and five-membered rings, a continued decrease in bond angle increases the electron-electron repulsion exponentially as predicted by Coulomb’s Law. This could be compensated for with a large nucleus…. A larger nucleus can attract electrons and mitigate their repulsion. However, I have resisted writing any examples of proton transfers via four-membered ring intermediates. [A Handbook of Organic Chemistry Mechanisms, p 65]
I could have drawn a model with two attachments points for water. That would probably look better if a plane charged surface is present rather than several pairs of electrons. If a two point model were to be present, then another model for the melting point difference is needed.
P.S. this is my first post here. As I often seem to think of something bleeding edge, not obvious, heretical, or downright wrong, I hope if there were any comments, this is just an idea. I may change my mind tomorrow.
Although the flagpole looks appealing, I don’t see why both protons couldn’t be on the surface as these are just coulomb interactions (charge with partial charge) not proton dissociations. More opposite charge touching the stronger the interaction.
If I were to look at the electron density map of water:
It would seem easier to stack water molecules with two protons on the charge surface, than stacking water molecules on the rounded edge of the lone-pair electrons. Molecules that stack well have lower freezing points than molecules that don’t stack well.
Nice write up. You certainly put a lot of effort into it (more than I ever do…). Now, is there a practical use for this nifty result?
Re: one or two protons attracted? My thinking was the differences in freezing points must be essentially an entropic effect. After all, both do freeze. I could not think of why water should back in more slowly to a fixed matrix than forward. It seemed to me, the effect is more a question of which atoms are moving on a micro scale. In terms of movement, it shouldn’t matter if a proton from an attacking water moves more than an electron pair. Therefore, if both protons were attached, there shouldn’t be a difference in the freezing points. I agree with the stacking argument. I thought it must be something else.
I agree with your electron density map making it appear little difference is present in the molecule and that a general dipole exists as drawn from the oxygen toward both hydrogens. However, I believe the strongest hydrogen bonding occurs along the H-O axis. (The best example of a variance I can find is the enol form of a diketone, such as acetylacetone.) Accordingly, I thought water would most likely bond similarly to an electron pair on the surface. If that were so, I thought the free rotation of the other hydrogen could inhibit productive collisions for freezing.
Re: practical?
Well, that is putting the screws to this. I have taken some time to think about it. Could this be related to rain making? If you wanted to make it rain, you would seed clouds with silver ions. We know that silver nitrate impregnated TLC plates can coordinate with alkenes to alter their elution. Therefore, silver crystals may present a surface to coordinate with the non-bonded electrons of water and enable condensation. If that were true, then we could consider searching for a similar molecular property. (I am only guessing that seeding clouds was done by trial and error and may not have undergone a more modern re-examination. If someone knows better, they may correct me. I have not researched this.)