This is what I have thus far. C11H14 has an IHD of 5. The multiplet way downfield is a pretty good indication of a mono-substituted benzene ring so let's start with that (5 hydrogens, 6 carbons used = C5H9 left). The benzene ring consists of three pi bonds and, obviously, one ring therefore occupying 4 of the 5 pi bonds/rings = 1 pi bond must be made. This is probably the double bond required for the acid catalyzed hydration. With respect to the spectrum, specifically integration number (#H's at each signal), # of signals, and splitting pattern I believe this concludes:
Four separate signals must be present (excluding the multiplet):
A). Quartet ≈5.4ppm (1H): 3 neighboring hydrogens
B). Doublet ≈2ppm (3H): 1 neighboring hydrogen....(I'm thinking this signal to be a deshielded methyl group due to benzene ring?)
C). Quartet ≈2ppm (2H): 2 equivalent H's with 3 neighboring hydrogens
D). Triplet ≈1ppm (3H): another possible methyl group considering the expected ppm...2 neighboring hydrogens
So I'm just having trouble trying to piece these carbons and hydrogens together while maintaining a double bond somewhere in the midst of that. Any hints/help appreciated. Thanks in advance.