[chmjmhi]

I have noticed that sometimes a phenyl group shows up in 13C NMR as three unique peaks rather than four as I would expect (ipso, ortho, meta and para expected). Could someone please point me towards a good explanation for this?

I have also been tearing my hair out trying to characterize an oxindole with a (4-trifluoromethylphenyl)-substituent and I can swear that the four hydrogens on the aromatic substituent appear at identical shifts. That's what the integrals tell me at any rate (integrates to 4; coupling constants 23.9 and 8.8 Hz), although I think it sounds insane since the ortho- and meta-hydrogens should experience different amounts of CF coupling. Or am I wrong there? Do I have reason to doubt my integrals?

[movies]

I don't recall observing H-F coupling beyond three bonds (you would have 4) and I wouldn't expect it to be that large.  Typical H-F coupling through 3 bonds is around 24 Hz.  A CF₃ on a phenyl ring will give C-F coupling to each of your carbons though!

As for your first question, you ought to see 4 peaks.  Are you sure that you have enough scans and delay time to see the ipso carbon?  Since it isn't bonded to any H it takes longer to relax.

[chmjmhi]

Yep, the 13C spectrum is much easier (or rather, would be if I had enough material to be able to see the %$*#& quaternary fluorine-split peaks...).

As a sidenote to the first question, I obtain my spectra in CDCl3 typically. Could it be so simple that it is a solvent effect (I know vaguely where to read up on that) and I could resolve the peaks by changing to, for example, methanol-d4?

[movies]

Sometimes solvents can help, but usually the difference in signal isn't much.  Where solvent changes really help are when you are trying to see two different, but very similar compounds.  For example, I have encountered some mxtures of diastereomers that look like essentially one compound in CDCl₃, but separate quite nicely in C₆D₆.