First an overview of ionic solutions and a nomenclature primer.When harmaline.HCl dissolves in water, it forms a dissolved harmalinium (protonated harmaline) cation
harmalinium+(aq) and a dissolved chloride anion
Cl-(aq) . AFAICS all harmaline.HCl is dissociated in aqueous solution.
The same goes for hydrochloric acid. Being a strong acid, it is completely dissociated in aqueous solution as chloride anions
Cl-(aq) and free proton cations
H+(aq) , sometimes also denoted as hydronium
H3O+(aq) .
With acetic acid it is a bit more complicated, because it is a weak acid and only partially dissociates into acetate anion
H3COO-(aq) and proton cation
H+(aq) . A certain percentage of the acetic acid will remain undissociated as
H3COOH(l). I write the latter as a liquid (l), as I assume that it is mixed with water, not dissolved with a solvation shell - real chemists may want to correct me if I am wrong.
When you add zinc dust to an acidified solution of harmaline, the solution initially contains:
harmalinium+(aq);
Cl-(aq);
H+(aq);
Zn(s) and of course
H2O(l):
(For completeness, there is also a relatively minute amount of
OH-(aq) anions as a result of water's spontaneous dissociation, and for the same reason there are slightly more
H+(aq) cations than
Cl-(aq) anions. These details are not important here.)
Now onto the reduction part. This is quite complicated for me. I'm not sure if I really understand it, but this is what I can make of it:
A zinc dust atom
Zn(s) reacts with the
harmalinium+(aq) cation by donating an electron to the harmalinium cation . The result is a dissolved zinc cation
Zn+(aq) and a neutralized harmalinium radical
harmalinium*(aq). The latter is called a radical because it now has an unpaired electron -- in a molecule, electrons always bind as pairs. And perhaps it isn't really neutral, maybe it is a negatively charged cation with both a localized
+ charge (on the nitrogen) and a localized
- charge (on the adjacent carbon). It's complicated.
Anyway. Next, a proton
H+(aq) joins the harmalinium radical, forming a
hydroharmalinium+*(aq) radical cation.
The zinc cation
Zn+(aq) reacts again with the
hydroharmalinium+*(aq) radical cation by donating another electron to it. This results in a dissolved zinc cation
Zn2+(aq) of a higher oxidation state and a
hydroharmalinium(aq) species (another negatively charged cation?)
Finally, another proton
H+(aq) joins the
hydroharmalinium(aq), turning it into a
dihydroharmalinium+(aq) cation. Since dihydroharmaline is the same thing as tetrahydroharmine, we can also call it a
tetrahydroharminium+(aq) cation.
Well, that is the simple version. In actuality, the steps can happen in a different order too. I am not going to expound on that, apart from showing this diagram of a similar reduction scheme:
Perhaps mindlusion, ulim or downwardsfromzero could draw a nice scheme with curly arrows. I don't know if it would make it more intelligible, but it would sure look better.
There are some side reactions to consider too. HCl is a strong acid and will attack the zinc directly, generating hydrogen gas
H2(g). Perhaps using acetic acid is better here because it is less vigorous in its attack on the zinc. That's just a speculation. Sometimes ammonium chloride is suggested as a proton source for zinc reductions. Could be even better.
Another possible side reaction is the linking of two harmaline molecules in various stages of reduction. That is called dimerization. I have no idea how likely that is in practice and how much of what exact sideproducts to expect.