Question

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

Answer 1

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

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

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

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

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

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

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

1. `"pH" < 2.2`
2. `2.2 < "pH" < 9.21`
3. `9.21 < "pH" < 10.46`
4. `"pH" > 10.46`

Now let's see how that turns out.

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

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