[uri.koffman]

hey i dont get the trick in this one.
things that i do know are:
1. i have a tert butyl group
2. methyl group
3. oh group
4.dou= 4 so maybe benzene but im not sure
how is it possible 3h multiplets?

thanks!

[criss]

My guess is that you have indeed a t-Bu, a Me one the molecule. The signal at about 7 can be a OH from a phenol and yes a aromatic ring, therefore when you do the math you still have 3 position blocked by t-Bu, Me and OH and still 3 H on the aromatic ring, that's why 3H as multiplet (looks like an 1,2,3-substitution)

[wildfyr]

I agree with Criss. The multiplet is not referring to the splitting on a 3 identical protons, but is referring to 3 protons with complex splitting sitting at about the same ppm. In real life a higher field NMR could possibly resolve the situation, but for a practice problem its just "a mess of similar but not identical protons on top of each other."

Its very difficult to say whether its 1,2,3 or 1,2,4 substitution. From the provided info either would be correct in my eyes.