NMR Analysis of Ethylene glycol I have the spectrum obtained by an NMR analysis of Ethylene glycol :


I would like to explain the number of peaks in this spectrum,
and assign them to the proton spins present in the molecule.
I would say that one of the picks is due to the OH groups, and the other one to the CH$_2$, but I don't understand very well the physics behind this.
Could anyone try to explain me ?
Thanks for your help (and sorry for my bad english)
 A: You are right when you say that one peak is due to $\mathrm{OH}$ groups and the other one to $\mathrm{CH_2}$ groups. The protons in each groups are chemically equivalent and contribute to the same peak.
You should compute the area below each peak.
Since there are 4 $\mathrm{CH_2}$ protons and only 2 $\mathrm{OH}$ protons, my guess is that one peak's area is two times greater.
The chemical shift is in general greater for protons near oxygen. Then the left peak may be assigned to $\mathrm{OH}$ group.
To verify this you should check peak's areas.
A: Fast rotation keeps the paper-symmetry protons equivalent.  High concentration makes for fast proton exchange in the alcohols (sharp line, no coupling to methylene protons).  Integration identifies populations.  The difference in chemical shifts tells you the temperature of the sample,
http://www.emerypharmaservices.com/blog/calibration-variable-temperature-nmr-thermocouple 
http://staff.ustc.edu.cn/~liuyz/methods/NMR_VT_calibration.htm 
   with calibration equation.
Are there C-13 satellites?
