Aspartic acid conjugate base acidity Question


Which hydrogen among the four in the conjugate base of aspartic acid is most acidic?

Answer

 No answer given

My attempt
I am confused between Hb and Hc
I know Hd has $\mathrm pK_\mathrm a \approx  9$, so it will not be the most acidic.
$\ce{NH3+}$ has very high -I effect but no mesomeric effect, so Hc appears to be quite acidic.
But at the same time it is at the alpha position of the carboxylic acid and may not be effectively conjugated in the $\ce{C=O}$ bond.
Here can the electron pair if formed by loss of Hc be conjugated on the $\ce{C=O}$ bond? This will prevent the lone pair on oxygen from being delocalized, however, something that is not characteristic of carboxylic acids.
Hb may be acidic but no special effect may be attributed to it, similar to Ha.
I think the order will be Hb &gt; Hc &gt; Ha &gt; Hd
Is my assumption correct?
A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion.
A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point.



Let's consider this molecule to have reached the isoelectronic point. We know that the value would be less than $7$ since there are two acidic groups (meaning that this is an acidic amino acid). When we reach that point, we see that the both the carboxyl groups are de-protonated and the amine group still exists as $\ce{-NH3+}$.
This means that the $\ce{-NH3+}$ group is less acidic than the the acid groups. Therefore, Hd is is less acidic than Ha or Hb.
As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before Hc is removed. This means that Hd is more acidic than Hc.
Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to Hb than Ha, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of Hb would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of Ha.
Therefore the order of stability would be:

Hb &gt; Ha &gt; Hd &gt; Hc