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

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

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Answer 1 · 2016-03-16T07:55:00

This is tyrosine at physiological $pH$ $"pH"$ ( $7.4$ $7.4$ ):

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

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

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

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

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

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

$pH < 2.2$ $"pH" < 2.2$
$2.2 < pH < 9.21$ $2.2 < "pH" < 9.21$
$9.21 < pH < 10.46$ $9.21 < "pH" < 10.46$
$pH > 10.46$ $"pH" > 10.46$

Now let's see how that turns out.

The carboxyl is neutral, the ${NH}_{2}$ $"NH"_2$ is ${NH}_{3}^{+}$ $"NH"_3^(+)$ , and the phenol is neutral. This has a net $+ 1$ $+1$ charge.
The carboxyl is ${COO}^{-}$ $"COO"^(-)$ , the ${NH}_{2}$ $"NH"_2$ is ${NH}_{3}^{+}$ $"NH"_3^(+)$ , and the phenol is neutral. This is net neutral. The isoelectric point, $pI$ $"pI"$ , is at about $pH = 5.65$ $"pH" = 5.65$ .
The carboxyl is ${COO}^{-}$ $"COO"^(-)$ , the ${NH}_{2}$ $"NH"_2$ is neutral, and the phenol is neutral. This has a net $- 1$ $-1$ charge.
The carboxyl is ${COO}^{-}$ $"COO"^(-)$ , the ${NH}_{2}$ $"NH"_2$ is neutral, and the phenol is ${PhO}^{-}$ $"PhO"^(-)$ . This has a net $- 2$ $-2$ charge.

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

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

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