I’m not a chemist (or even a well-informed layman), but I ran across an entry in Wikipedia that looks fishy to me. From a article on lactic acid I found this:
Emphasis added by me.
Now, isn’t it more or less impossible for an atom to lose a proton without becoming a different element? My understanding is that ions are formed by the gain or loss of electrons, and that protons pretty much stay where they’re put unless they happen to be in a nuclear reactor or particle accelerator.
But I before I go correcting my betters in Wikipedia, I thought I’d post this question here. I am having a lapse of reason. or should the above reference to a proton read “electron”?
Well, what it looks like is happening to me is not losing a proton from an atom that still stayed part of the molecule. Instead, they’re losing one of the hydrogen atoms from the molecule… or losing most of it, because they’re KEEPING that hydrogen atom’s electron. That, I think, might be possible. Notice that we’re going from a formula with H6 to a more strung-out formula with a total of 5 H’s, and the - indicator.
Shows what I know. I blame the Bohr model of the atom I learned in junior high. I hear that the “miniature solar system” analogy isn’t very close to reality, but I admit I’m lost when I see the more accurate diagrams where the atom looks like a balloon animal made by a party clown.
wheel It’s a long time since I did any chemistry but I don’t ever recall chemical processes described in terms of protons. It’s always about electron sharing, valences and such.
That’s to say, that in chemical terms a proton is described as a hydrogen ion.
Nope, it actually does lose a proton (H[sup]+[/sup] ion). The problem is that this is not a free proton. The proton is bound to an H[sub]2[/sub]O molecule to make an H[sub]3[/sub]O[sup]+[/sup] ion.
If you go into even more depth, it’s technically not even an H[sub]3[/sub]O[sup]+[/sup] ion, because that in turn is bound to a bunch of other water molecules. So basically you’re transferring a proton from the orginal C[sub]3[/sub]H[sub]6[/sub]O[sub]3[/sub] to a bunch of water molecules. Even that is still a lose approximation of what’s really going on, but real detail would mean studying Physical Chemistry for a while
To go into a little more detail (but still very cartoonish detail):
When the water molecule approaches the lactic acid molecule, The electronegative oxygen in the water molecule loosens the bond between one of the carbon and hydrogen atoms in the lactic acid while itself forming a bond. It in effect “steals” on of the hydrogen atoms, leaving behind it’s electron. But what is a hydrogen atom with a missing electron? It’s a proton!
Well… keeping track of the differences between atoms and molecules is a start. (I do know that in some of those heisenberg diagrams the hydrogen atom can look pretty funky shaped, but it’s when we start getting into the molecules that the shapes really look weird.)
Exactly. All a hydrogen atom (assuming the major isotope) is is a proton and an electron. “Proton” is often used when talking about reactions where all that is moving is just the bare proton or in other things involving hydrogen, such as calling it proton NMR instead of 1H NMR.
As for lactic acid, pictures are always better than strings of symbols. As the article says, it’s a carboxylic acid with a hydroxy group alpha to the carbonyl. The proton is coming off the carboxylic acid. I think you were having a problem with your definitions. Yes, loss of a proton from the nucleus of an element results in transmutation to a different element, depending on the way the loss occurred (and thus the type of radiation given off.) But lactic acid isn’t an atom, it’s a molecule.
It is of course loosening the bond between the hydrogen and oxygen in the -OH group attached to one of the carbon atoms. NOT between the carbon and hydrogen!
The OH bond in the carboxylic group is weaker than the OH bond in the water molecule.
That’s absolutely correct. What’s being discussed, however, is a molecule losing a proton, not an atom losing a proton. And “losing a proton” is just loose chemist jargon for losing a hydrogen atom. The tendency to lose protons is one way to define acids, and the strength of the acid depends on how easily it gives up protons.
Just to amuse yourself, here’s a really cool gallery of orbital images (artist impressions of course) called The Orbitron . I think my favorite is the 7f, which is what I’ve linked.
WROOOOONG!
“losing a proton” is just loose chemist jargon for, get this, “losing a proton”!
The lactic acid does not lose a hydrogen atom. If it did that it would have a neutral charge afterwards, and it would be seriously hungering to fill up that missing valence in the oxygen atom that suddenly only has a single bond to the carbon atom. I’d stay out of it’s way if I was you.
It actually loses a proton. However, this is not a proton that is suddenly flying around loosely in our vat of lactic acid. It is attached to a water molecule leaving us with an H[sub]3[/sub]O[sup]+[/sup] ion and a CH[sub]3[/sub]CHOHCOO[sup]-[/sup] ion in solution.
Yep, but whenever chemists talk about losing a proton, it’s always a group containing a loosely bonded hydrogen atom that’s losing it’s protons. Now, when hydrogen loses a proton, what do you end up with? An electron! And what’s a hydrogen ion with a missing electron? A proton!
The extra electron is taken up in a valence bond (You need to know that the bonds in the carboxylic group are not O=C-O or O-C=O, but somewhere in between, with the extra electron also “smeared” across the O-C-O (angled in reality) bonds.
The proton is always bound to another molecule (usually water, but can be other stuff, see Organic Chemistry), and not just flying around loosely, like say in a particle accelerator.
However, it is true that when we have H[sub]3[/sub]O[sup]+[/sup] ions in solution, the extra protons are constantly being passed back and forth between individual water molecules, sort of like a hot potato.
This is where oddities in terminology cause problems.
Acids are referred to as proton donors and bases as proton acceptors, but this has absolutely nothing to do with nuclear physics.
Rather, the idea is that an ionized hydrogen atom, i.e., one from which the electron has been stripped, is ordinarily (99.1+% of the time) a naked proton. HCl dissociates in ionization into H+ and Cl- , and H+ is an ionized hydrogen atom, AKA hydrogen ion, AKA a proton.
So then, an atom that has “lost a proton” still has the same number of hydrogen atoms. Right. :rolleyes: This is just a terminology difference, there is no need to get snarky.
I wasn’t getting snarky. Just trying to fight ignorance.
Well, if it’s deuterium, and we lose a proton, we still have hydrogen.
That sentence makes no sense. I think you meant: a molecule that has “lost a proton” still has the same number of hydrogen atoms. The answer is that it does not, because hydrogen is just a proton and an electron, so if a molecule loses a proton, it is missing an H in its chemical formula, and has gained an electron.
The problem is that we are talking about chemical formulas, which are just a simple symbolic way of showing more or less what a molecule is made of, and in many ways have very little to do with the physical structure of the molecule. In reality, the carboxylic acid has lost a proton! That’s why I said that “losing a proton” means just that, and nothing else.
It would be nice if it were actually possible to diagram or draw what is happening when the lactic acid and water molecules meet (in the correct way). At one point druing the reaction you can actually imagine a loose bond between the carboxylic group and the -OH of the water molecule. At that point the proton from the H in the carboxylic group is actually kind of in between and bonded to both molecules, with the electron from the H “smeared” across the O-H-O “bridge”.
In the end the water ends up winning, and “steals” the proton, while the electron is left over in the carboxylic group. Of course, the whole time we also have -COO[sup]-[/sup] groups meeting up with water molecules, and “stealing” protons from them leaving OH[sup]-[/sup] ions, OH[sup]-[/sup] ions meeting H[sub]3[/sub]O[sup]+[/sup] ions and leaving two water molecules behind, and so on and so on. Pretty quickly we reach an equilibrium, which is best symbolized by the formula:
