# Can tyrosine form dianionic form since it has acidic carboxyl group and also hydroxyl group at the phenyl group?

Mar 16, 2016

This is tyrosine at physiological $\text{pH}$ ($7.4$):

Recall that a $\text{pH}$ lower than the pKa indicates greater acidity and thus $\left[\text{BH"^(+)] > ["B}\right]$ and $\left[{\text{HA"] > ["A}}^{-}\right]$, and vice versa.

"pH" = "pKa" + log \frac(["B"])(["BH"^(+)])

= "pKa" + log \frac(["A"^(-)])(["HA"])

From the Henderson-Hasselbalch equation, we could prove that:

• If $\text{pH}$ $<$ $\text{pKa}$, $\log \setminus \frac{\left[{\text{B"])(["BH}}^{+}\right]}{<} 0$ and thus $\left[\text{BH"^(+)] > ["B}\right]$, or $\log \setminus \frac{\left[\text{A"^(-)])(["HA}\right]}{<} 0$ and thus $\left[{\text{HA"] > ["A}}^{-}\right]$, because $\log b$ where $0 < b < 1$ is negative.
• If $\text{pH}$ $>$ $\text{pKa}$, $\log \setminus \frac{\left[{\text{B"])(["BH}}^{+}\right]}{>} 0$ and thus $\left[\text{BH"^(+)] < ["B}\right]$, or $\log \setminus \frac{\left[\text{A"^(-)])(["HA}\right]}{>} 0$ and thus $\left[{\text{HA"] < ["A}}^{-}\right]$, because $\log b$ where $b > 1$ is positive.

In examining the possible $\text{pH"-"pKa}$ relationships, $\text{pH}$ can ONLY be any of the following:

1. $\text{pH} < 2.2$
2. $2.2 < \text{pH} < 9.21$
3. $9.21 < \text{pH} < 10.46$
4. $\text{pH} > 10.46$

Now let's see how that turns out.

1. The carboxyl is neutral, the ${\text{NH}}_{2}$ is ${\text{NH}}_{3}^{+}$, and the phenol is neutral. This has a net $+ 1$ charge.
2. The carboxyl is ${\text{COO}}^{-}$, the ${\text{NH}}_{2}$ is ${\text{NH}}_{3}^{+}$, and the phenol is neutral. This is net neutral. The isoelectric point, $\text{pI}$, is at about $\text{pH} = 5.65$.
3. The carboxyl is ${\text{COO}}^{-}$, the ${\text{NH}}_{2}$ is neutral, and the phenol is neutral. This has a net $- 1$ charge.
4. The carboxyl is ${\text{COO}}^{-}$, the ${\text{NH}}_{2}$ is neutral, and the phenol is ${\text{PhO}}^{-}$. This has a net $- 2$ charge.

So yes, that can happen, but only at $\text{pH} > 10.46$, which is technically harmful to the body. Best to stay at blood $\text{pH}$...