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When $\ce{D2O}$ is added to a sample, the OH and NH peaks in a $\ce{^1H}$ NMR spectrum disappear. This is the basis of (for example) the $\ce{D2O}$ shake test, and is also the reason why peptide/protein NMR is commonly run using $90\%~\ce{H2O}/10\%~\ce{D2O}$ as a solvent (or else the NH peaks would be unobservable).

What causes them to disappear from the spectrum?

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    $\begingroup$ The $\ce{D^{+}}$ exchanges with the proton and the exchange process leads to line broadening; usually the proton signal becomes so broad that it is no longer visible. $\endgroup$
    – ron
    Commented Apr 25, 2021 at 17:04
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    $\begingroup$ Even if there is no broadening you attenuate the signal according to the concentration of protons and deuterons in the solvent, which exchange quantitatively on the OH. $\endgroup$
    – Buck Thorn
    Commented Apr 25, 2021 at 18:14
  • $\begingroup$ @BuckThorn I read the question such that the OP was asking about an OH proton signal from an organic alcohol sample (e.g. hydroxy or hydroxyl, not hydroxide). In such a case the concentration of the hydroxyl proton does not change $relative$ to the protons in the organic backbone of the alcohol. So while the chemical shift of the hydroxyl proton will change with addition of $\ce{D2O}$, the hydroxyl proton signal should still be visible (if the backbone protons are still visible) if exchange could be slowed compared to the timescale of the nmr experiment. $\endgroup$
    – ron
    Commented Apr 25, 2021 at 21:32
  • $\begingroup$ @ron I'm not sure I agree: $\ce{ROH + D2O <=> ROD + HOD}$ the hydroxyl proton concentration is decreased, with a corresponding increase in the concentration of HOD. My interpretation of the question is the same as yours, although the first edit muddled the question a bit, I must admit. I rolled back to the original. $\endgroup$ Commented Apr 26, 2021 at 9:37
  • $\begingroup$ @ron In fact I was doubly unclear in not writing explicitly "deuterium oxide" or deuterated solvent with exchangeable protons/deuterons ("protic"). I also meant that OD is not going to show up in a proton spectrum (although might have an effect through couplings). Exchangeable protons (including alcohol OH) will see their signals attenuated although the mechanisms and extent may differ depending on exchange rates and concentration of exchangeable protons in the solvent. $\endgroup$
    – Buck Thorn
    Commented Apr 26, 2021 at 10:54

1 Answer 1

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R-OH + D20 interact to equilibrium. So you get R-OD + H2O. The more D2O you add the less R-OH you get until this is not longer detectable .

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    $\begingroup$ To be perfectly honest here, this doesn't add much to the topic and it's been discussed in the comments above. $\endgroup$ Commented Mar 3, 2023 at 12:29

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