Question: [![IR spec and C NMR](https://i.stack.imgur.com/WiZ5D.jpg)](https://i.stack.imgur.com/WiZ5D.jpg)
[![H NMR](https://i.stack.imgur.com/Ef6cm.jpg)](https://i.stack.imgur.com/Ef6cm.jpg)
I have the following IR spec, C NMR and H NMR data. The boiling points determined for the unknown compound were 116.8 and 119.1°C (two trials). Given the possible compounds, the unknown can be one of: 
```

3-methyl-2-butanol
1-butanol
3-pentanol
2-pentanol
2-methyl-1-butanol. (These are based on boiling point)
```

I have eliminated some of these choices based on the number of different types of carbons (4, as shown on C NMR) but am not sure where the H of the OH is indicated in H NMR. There is no singlet present among the 3 groups of peaks, so where would it be?

Answer: 
Peaks of acidic protons in an NMR spectrum can change (in position, form and intensity) due to exchange with other volatile protons.

* A very fast exchange makes the peaks coalesce, you get one singlet between the two "original" (expected) chemical shifts, the exact position depends on the respective amount of both.
* A slower exchange, one "on the NMR timescale", broadens that peak considerably, typical case is residual water and a COH proton. The peak might in some cases be so broad it becomes invisible.
* Even slower, this peak splits into two, which become sharper and further removed from each other.
* A very slow exchange makes no change to the positions, but an exchange with deuterium (i.e. from the solvent) changes the relative intensity of the respective peaks over time.
The proton on O or N atoms can be acidic enough to exchange with the deuterium atoms of the NMR solvent, thus "hydrogen-bonding protons" are often absent from H NMR spectra, or may show up "[anywhere](https://www.google.com/amp/s/www.researchgate.net/post/How_to_interpret-NH_and-OH_peaks_in_proton_NMR_and_how_to_report_them_in_a_publication/amp)".