H NMR spectrum

Mar 2014
Hey fellow chemistry forumers
I just need a little bit of help regarding an H-NMR question:

THe first part of the question asked me to draw the 5 constitutional isomers of a particular carboxylic acid. It has a molecular formula of C10H12O2.
I was able to come up with the 5 constitutional isomers which are
2-phenylbutanoic acid
3-phenylbutanoic acid
4-phenylbutanoic acid
2-methyl-2-phenylpropanoic acid
and 2-benzyl propanoic acid

The second part of the question is where Im having trouble:
It says that the HNMR for this particular carboxylic acid has 3 absorption signals. 1 of the signals has a delta value of 7.3 which corresponds to 5H.

Can anyone assist me in identifying which of the 5 constitutional isomers is the one that is likely to fit the description of the 2nd part of the question, and

ALso could you please explain how i might go about finding the answer this question. Tanks :):)
May 2014
Genoa (Italy)
Hi Jarrod,

looking at your proposed structures, "2-methyl-2-phenylpropanoic acid" is the molecule which matches with the spectrum better than other ones.
Drawing the structure, we see the exchanging proton of carboxylic group, the two methyls (they're equivalent and have the same chemical shift). The last third signal is generated by aromatic protons.
Effectively, it could be a signal with relative area of 5. But in this last point i've a doubt: I'm not completely sure you can look this 5 protons like an only signal, it's not said...because the alkyl group is able to apply a positive inductive effect and change the shielding on ortho,ortho' position. If it's true, you'll observe a signal with relative area 2 and another signal (very close to previous one) with relative area 3.