[element41skater]

For organic chemistry we are learning Ir spectroscopy in lab.  With this la we are given a Ir spectrum of a compound.  Mine is C12H10O.  And i am assuming from what i have been able to understand from the Ir that it is 2-Phenylphenol.  I came to this from these conclutions:

There is is a alcohol/Phenol group: 3200 to 3400 wavenumber-1

There are C=C: 1600 wavenumber-1
(which makes me assume the benzene rings)

Now is where i am really getting stuck.   I not sure where to find the Benzene rings. I used both http://en.wikipedia.org/wiki/Infrared_spectroscopy and
http://en.wikipedia.org/wiki/Infrared_spectroscopy_correlation_table

and i am just getting confused at this point.

Am i missing and thing else?

Of and i calculated my degree of unsaturation to be:8
this just makes it more obvious that the structure is 2-Phenylphenol.  There must be 2 benezene rings attached together by a single bond with a OH coming off one of them.


Here is my spectrum.
Thank you

[Chepsy]

The benzene rings are implied by the absorption at 1600 cm^-1 (benzene has three conjugated double bonds, which absorb at a lower frequency than any other double bond due to overlap between six pi orbitals, which makes the double bonds have significant single-bond character).  You can look for unsaturated (sp²) carbon-hydrogen stretches from 3000 to 3100 cm^-1.

I hope I understood your question right.

[ATMyller]

Look for benzene C-H out of plane bending at 900-700 cm-1 and C-H deformation summary bands at 2000-1650 cm-1 for determining substitution. Also you are right about it being 2 benezene rings attached together by a single bond and a hydroxyl group, but those are called biphenylols.