# Formation of zwitterion from amino acids

Why it is easier for the carbonyl group to lose a proton to become negatively charged and the amino group to accept a proton to become positively charged?

I know it has to do with the very polar bond due to the oxygen. And the $$\ce{H+}$$ ions attracted to the lone pair of electrons on $$\ce{N}$$ in the amino group.

But I don't understand why would an electronegative atom such as oxygen wants to lose a proton.

I was told that they will attract more electrons towards towards them. But that does not help in my understanding of why protons will leave.

• It's not that the proton is "unhappy" on the oxygen, it's just happier somewhere else :) – Jan Jensen Jun 27 '17 at 11:04

The reason amino acids exist largely in their zwitterionic form at biological $$\mathrm{pH}$$ (usually around 7) is due to the $$\mathrm{p}K_\mathrm{a}$$ of the constituent groups. It's not that the oxygen 'wants' to lose a proton, but more that at that $$\mathrm{pH}$$ the equilibrium lies towards the deprotonated state (things are rarely 100% protonated/deprotonated).
At a $$\mathrm{pH}$$ of 7, the medium is basic enough to cause the carboxylic acid to be deprotonated, and the amine to be protonated. If you look at the diagram below, you can see that increasing or decreasing the pecan shift the equilibrium towards fully protonated (at low $$\mathrm{pH}$$) or fully deprotonated (at high $$\mathrm{pH}$$) amino acid.
Ionisation state of amino acids over 0-14 $$\mathrm{pH}$$ range, taken from Biochemistry, Berg (7 ed.)