How would you distinguish 3,5-diiodotyrosine, 3-iodotyrosne, and tyrosine by H-NMR

How would you distinguish 3,5-diiodotyrosine, 3-iodotyrosne, and tyrosine by H-NMR?

Does it have something to do with the increasing number of hydrgoens?

Please help! 10 pts. for best answer!

tryrosine would give an almost symmetrical AA'BB' (two sets of doublets) which would integrate to 4 H (aromatic). Symmetrical benzenes are easily recognized by this substitution pattern. (Remember, an amino acid is chiral so the molecule is not strictly symmetrical)

When you introduce I in the 3- position, you destroy the symmetry and you have 3 sets of H's with 1 singlet and 2 doublets. The H's should integrate to 3 H (aromatic).

When you introduce 2 I's in the 3 and 5 position, the "symmetry" is restored and now all you get is 2 singlets which should integrate to 2 H (aromatic)

Once again, it is not really symmetrical because of the chiral center

The iodines replace hydrogens in the tyrosine, so you could distinguish them by integrating the aromatic peaks, the integral corresponding to the number of hydrogens in that chemical shift range. Tyrosine has 4 aromatic hydrogens, 3-iodotyrosine has 3 aromatic hydrogens, and 3,5-diiodotyrosine has 2 aromatic hydrogens.

You would also see differences in the chemical shifts, and in the splitting pattern of the aromatic hydrogens. In particular, the two aromatic hydrogens in 3,5-diiodotyrosine are chemically equivalent, and should appear as a singlet. 3-iodotyrosine has 3 nonequivalent hydrogens giving a complex splitting pattern, and tyrosine has 2 pairs of equivalent hydrogens so you should get a pair of doublets.

Iodine is a very electronegative atom and when it's in a compound, it causes the hydrogens to be pulled down toward the left side of the spectra more. 3,5-diiodotyrosine should be shifted much further down than tyrosine is, and 3-iodotyrosine would be in the middle somewhere. I hope that makes sense if you're looking at the spectras.