[romanwahoo]

You should have 4 protons in the aromatic region between 6.5 and 8, which include two triples and two doublets. If you have the 1h nmr and you are just assigning the peaks, then simply figure out which carbons are more positively charged and which are more negatively charged. Since the two groups' effects are additive and not interfering, just go with the carboxylic acid, which is an electron withdrawing group. The carbon to which the -COOH is attached therefore is more negative, the next more positive, then more neg, and so forth. The protons attached to carbons that are more positive are more deshielded, so they have a higher shift. Using this method, you should be able to assign all four aromatic protons.The -COOH and the -OH are generally much harder to see, and they are concentration dependent. A broad peak around 6 though is usually indicative of the -OH peak. Because the oxygens in the -COOH can resonate charges, the carboxylic acid proton is usually shifted waaay downfield, around 14 or so, I believe, and which is usually not shown in a normal proton spectra.