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Looking at prop-2-yn-1-ol spectrum that was taken from:

Spectrometric identification of organic compounds. Robert M. Silverstein and G. Clayton Bassler. J. Chem. Educ., 1962, 39 (11), p 546. DOI: 10.1021/ed039p546.

Why OH peak is so sharp? And why it is located around 1 ppm? The book does not specify the solvent used.

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At the same time if one takes a look at 1H NMR spectrum of the same compound taken on 90 MHz machine in CDCl3. Source: Spectral Database AIST with SDBS No. 74HSP-00-252

  1. Why the OH peak shifted so much? But it is still as sharp as in other spectrum. Please someone explain such pattern.

enter image description here

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  • $\begingroup$ There is no way that top spectrum comes from the reference indicated. It is a FT spectrum; FT NMR was not developed commercially until several years after that article was published. $\endgroup$ – long Dec 18 '17 at 21:52
  • $\begingroup$ @long Well spotted! $\endgroup$ – Karl Dec 20 '17 at 18:43
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The chemical shift of protons in polar groups can vary a lot between solvents of different polarity, because the average electron density around the proton changes. Hydrogen bonding etc. A nice overview can be gotten from the article NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist.

For example water shifts from 4.79ppm in water($D_2O$) down to 0.4ppm in d6-benzene. Or the OH group of ethanol from 0.5 in benzene to 1.3 in chloroform to 4.6(!) in DMSO. The methyl and methene groups otoh just shift by 0.3 ppm max. between solvents.

The signals in your spectra are sharp, because there is either a fast exchange, meaning the peak comes at the average CS of both positions of the proton, or no exchange. Slow exchange (either statistically or because the reaction is slow) leads to broadening.

(Slow = exchange happens on the NMR timescale, a millisecond plus/minus a few decades, depending on your $T_{2}$ etc.)

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The chemical shift and linewidth of the hydroxy signal varies considerably, as has been discussed. Exchange rates will influence both linewidth and position, but in the case of the sharp signal shown you most likely have very fast exchange, otherwise coupling to the -OH signal would be apparent.

The shift of -OH signals will vary considerably, and is influenced by solvent, temperature and concentration. More specifically, the shift will be influenced by conditions that affect H-bonding. H-bonding leads to greater downfield shifts. Usually, you see that as you increase the concentration of the alcohol, the shift of the -OH peak moves downfield.

So, all else being equal (same solvent, same temperature), it is likely that the bottom spectrum is recorded at a higher concentration than the top spectrum.

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  • $\begingroup$ +1 Also you don't see the chloroform peak in the lower spectrum. That being said, why is it invisible? 99.99% CDCl3? Bit strange, that. $\endgroup$ – Karl Dec 20 '17 at 18:52
  • $\begingroup$ Alcohol + chloroform don't make a fast exchange. I wager the sample also contained a rather large amount of water, with which the exchange takes place. $\endgroup$ – Karl Dec 20 '17 at 18:54
  • $\begingroup$ @Karl - No suggestion that alcohol exchanges with chloroform. All untreated commercially available CDCl3 contains at least trace levels of water. Apologies if that was not clear. The amount of water does not have to be large. In both cases, the -OH peak appears to give very good integration. And as you say, the sharpness of the peaks indicate either fast or no exchange. No exchange is not an option for a simple primary alcohol. $\endgroup$ – long Dec 20 '17 at 20:35
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    $\begingroup$ @Karl - AINST-SDBS remove solvent signals from the spectrum prior to public release. This is stated as per their experimental conditions "Solvent, reference, and impurity peaks were removed prior to open to the public[sic]". $\endgroup$ – long Dec 20 '17 at 20:35

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