Aspartic acid conjugate base acidity

Question

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 H_b and H_c

I know H_d 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 H_c 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 H_c 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.

H_b may be acidic but no special effect may be attributed to it, similar to H_a.

I think the order will be H_b > H_c > H_a > H_d

Is my assumption correct?

Answer 1

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, H_d is is less acidic than H_a or H_b.

As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before H_c is removed. This means that H_d is more acidic than H_c.

Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to H_b than H_a, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of H_b would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of H_a.

Therefore the order of stability would be:

H_b > H_a > H_d > H_c

Answer 2

I highlighted the some of your attempts:

$\ce{H_d}$ has $\mathrm pK_\mathrm{a} \approx 9$, so it will not be the most acidic.

This assumption is correct. For aspartic acid, $\mathrm pK_\mathrm{a}(\ce{H3N+}) = 9.6$.

$\ce{NH3+}$ has very high -I effect but no mesomeric effect, so $\ce{H_c}$ 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.

This assumptions are correct as well. The $\alpha$-hydrogens in acids and esters are not as acidic as those in aldehydes and ketones because carbonyl groups of them already involved woth attached $\ce{OH}$ or $\ce{OR}$ groups, respectively. However, $\alpha$-hydrogen in amino acids fairly acidic than those in aldehydes or ketones due to strong -I effect from $\ce{H3N+\!-}$ group.

Here can the electron pair if formed by loss of $\ce{H_c}$ be conjugated on the $\ce{C=O}$ bond?

Yes, that is the main reason it has being acidic (compared to $\beta$-hydrogen or other $\ce{C-H}$ in the same alkyl chain). However, it won't leave to make a carboanion until it has reached the correct $\mathrm{pH}$. The $\mathrm pK_\mathrm{a}$ of $\alpha$-hydrogen in an $\alpha$-amino acid is in the range of $16$ to $17$ (Ref.1). The exception is been D-phenylglycine, $\ce{Ph-CH(NH2)-CO2H}$ $(\mathrm pK_\mathrm{a} = 14.9)$, which is not a natural essential amino acid. The reason for lower $(\mathrm pK_\mathrm{a}$ value is the additional resonance contribution from neighboring $\ce{Ph}$ group. The few examples are given below:

$$ \alpha\text{-Hydrogen $\mathrm pK_\mathrm{a}$ values (Ref.1)}\\ \begin{array}{c|ccc} \hline \text{Amino acid} & \mathrm pK_\mathrm{a} \\ \hline \text{Alanine} & 16.5 \\ \text{Valine} & 17.0 \\ \text{Isoleucine} & 16.9 \\ \text{Leucine} & 16.7 \\ \text{Phenylalanine} & 16.2 \\ \text{D-Phynylglicine} & 14.9 \\ \hline \end{array} $$

Therefore, it is safe to assume that $\alpha$-hydrogen of acid is the least acidic among other acidic hydrogen in the molecule, even lower than that of $\ce{H3N+\!-}$ group.

$\ce{H_b}$ may be acidic but no special effect may be attributed to it, similar to $\ce{H_a}$.

This assumption is incorrect. As a matter of fact, $\ce{H_a}$ and $\ce{H_b}$ are the most acidic hydrogens among other acidic hydrogens in the molecule (this was already proved in the other answer, so that I'm not going to repeat it again). Between the two, $\ce{H_b}$ is more acidic than $\ce{H_a}$ since it is closer to $\ce{H3N+\!-}$ group and feel more -I effect (through less numbers of $\sigma$-bonds). The actual $\mathrm pK_\mathrm{a}$ of $\ce{H_a}$ and $\ce{H_b}$ are $3.65$ and $1.88$, respectively.

Thus, the correct order should be $\ce{H_b} \gt \ce{H_a} \gt \ce{H_d} \gt \ce{H_c}$.

Reference:

1. Eric D. Stroud, Dennis J. Fife, Grant Gill Smith, "A method for the determination of the $\mathrm pK_\mathrm{a}$ of the $\alpha$-hydrogen in amino acids using racemization and exchange studies," J. Org. Chem. 1983, 48(26), 5368–5369 (https://doi.org/10.1021/jo00174a045).