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

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"#...