# Why is the deprotonated form of dopamine more reactive than the protonated/neutral form of dopamine?

See this (from the Wikipedia article on dopamine):

Dopamine, like most amines, is an organic base. At neutral or acidic pH levels it is generally protonated. The protonated form is highly water-soluble and relatively stable, though it is capable of oxidizing if exposed to oxygen or other oxidants. At basic pH levels, dopamine becomes deprotonated. In this free base form it is less soluble and also highly reactive and easily oxidized

I'm very curious about this since I'm curious about dopamine auto-oxidation, as it is often a cause of neurodegeneration observed in those who take amphetamines (and seems more likely in the cytosol, than the vesicles it would otherwise be in). Though I'm also curious how much auto-oxidation would happen for extracellular dopamine as well.

• Just to give some concrete numbers (which appear to be lacking in the Wikipedia article), one can calculate the concentrations of the various species. I crunched the numbers in Mathematica. Assuming aqueous solution, physiologic $pH \approx 7.35$, and using $pK_a$ values from this article, the mole fraction of the fully protonated form is roughly $0.98$. – Greg E. Aug 2 '14 at 15:13
• Thanks for the calculation! I'm curious: Even if the deprotonated form was a very tiny minority of all the forms of dopamine (which would be < .02 because most of the non-protonated form would be the neutral form of dopamine), would that tiny minority be enough to explain the auto-oxidative damage of dopamine if it were reactive enough? – InquilineKea Aug 2 '14 at 20:18
• I can't answer that with any certainty, sorry. I just felt it might be useful to actually quantify some of the relevant parameters; the chemistry section of the Wikipedia article is sparse in that respect. – Greg E. Aug 2 '14 at 20:34
• @IK: You are partly correct. Even a 0.02 fraction of active compound could induce activity. Imagine that the active 0.02 would be continuously removed from the equilibrium and new active "0.02" would take its place until all the inactive form is consumed. Obviously the active concentration would also be 500 times weaker. But then the oxidation should not be directly connected to the $-NH_2$ group (see my post below). "Autooxidation" is a set of interconnected reactions with multiple paths and products and $-NH_2$ could play some role downstream. Only way to answer is experiment (or literature) – K_P Aug 2 '14 at 21:30