How to rationalise the protonation state of amino acids

Question

I am learning about amino acids in various $\ce{pH}$ environments and how their groups are protonated or deprotonated in these environments. I don't need to mention a specific acid for my question, I will just use the generic form of one: $$\ce{H2N-CH(R)-COOH}$$

The $\ce{H2N}$ group has a $\mathrm{p}K_a$ of $8$, I believe, and the acid has a $\mathrm{p}K_a$ of $3.1$.

If I drop these into a solution with $\ce{pH} = 12$, the $\ce{COOH}$ will donate it's hydrogen ion, but the $\ce{H2N}$ will not. It will only add another hydrogen ion if it is basic compared to the solution to become $\ce{H3N+}$. Why is it, that it won't donate its hydrogen ion like $\ce{COOH}$, when it is acidic compared to the solution, to become $\ce{HN-}$?

Answer 1

Let's assume you drop the amino acid in water at $\ce{pH} = 7$. You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$ Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.

You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$ This means $\ce{NH3+ -> NH2}$.

You could define another $\ce{p}K_a$ for $\ce{NH2 -> NH^{-}}$. The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35. You may look at this site: How to use a $\mathrm{p}K_a$ table You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.