[Roxo]

I am studying proton NMR at school and my textbook says that adding a few drops of D₂O to ethanol will cause the hydrogen in the OH group to be displaced by deuterium and thus remove the peak from the spectrum enabling identification of the OH peak in the original sample.  This has raised a question in my mind as to why the D should replace the H. I could understand this if there was a high concentration of D₂O and if an equilibrium would be formed of OH and OD but the implication is that all the OH is converted to OD otherwise why would the OH peak be eliminated completely rather than still appearing but with lower intensity. I have looked at various reference sources and discovered that hydrogen / deuterium exchange is very important in all kinds of fields in chemistry and I have also come across a page from Liverpool University which suggests a mechanism for the displacement in ethanol and also a reference elsewhere to an enolate type of mechanism but none of them address my main questions which are:
Is there in fact an equilibrium established between the OH and OD versions of the ethanol or is it effectively a one way reaction and deuterium does indeed replace more or less all the hydrogen: in which case why does it do that?  If the deuterium does not displace all the hydrogen from the OH group then why does the OH peak disappear from the proton NMR spectrum rather than just be weaker?

[Enthalpy]

It would be a miracle if ethanol did catch deuterium selectively.

Industrial deuterium production uses far less direct and efficient processes, with many cycles to gain some purity at each step.

[Roxo]

@Enthalpy...thanks for your reply. You seem to be confirming that a dynamic equilibrium is indeed set up and thus molecules of CH₃CH₂OH and CH₃CH₂OD will be present. So why does a proton NMR spectrum not pick up the OH hydrogen? My text book says it completely disappears from the spectrum.

Thinking about this some more I have a further question: if D and H are exchanging then presumably there will be some DHO molecules and even potentially H₂O molecules in the equilibrium mixture. Why don't these show up on a proton NMR spectrum? This is very confusing - someone on here with expertise in NMR spectroscopy must know the answer. My textbook makes a point of saying that this deuterium exchange is an important technique for confirming OH groups (by their absence after adding D₂O) and there have been exam questions about it (Oxford and Cambridge UK A-level chemistry) but the book and nothing I can find online says what happens to the displaced hydrogens nor why they don't show up somewhere on a spectrum. They can't just disappear. I'm confused.

[Vidya]

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Structural_Assignment/High_Resolution_Proton_NMR_Spectra

Check above link for great explanation for missing OH peak.

[Roxo]

@Vidya...thanks for your reply. I have looked at that link and don't find the explanation very satisfying. I am familiar with Jim Clark's website and generally as you say it is full of great explanations but in this case it seems less than rigorous.
If the H ion is displaced from the ethanol in the way he describes then it must go somewhere. Jim's explanation is that it gets lost somewhere in the mixture. Well that's not very satisfactory. It surely must still exist either attached to deuterium oxide as an acidic ion or to form HDO as a neutral molecule or to exchange with two D atoms and form H₂O. If there was enough hydrogen in the original OH groups present in the sample to give a resonance on the NMR spectrum then when that hydrogen is displaced (where ever it goes to) there should surely be enough to produce a resonance at some other chemical shift value, to say it just disappears is not an explanation that I find acceptable. To say it appears somewhere else on the spectrum (for example as water) would make sense to me but to just disappear doesn't. I appreciate that deuterium does not normally appear as a peak in spite of having integer spin but that is because it has such a different chemical shift .   So at this point I feel no wiser as to why the displaced hydrogen simply disappears from the spectrum. However it may well be that I am misunderstanding something ...I often do 

[wildfyr]

It basically gets lost in the baseline as it is incorporated into a wide and weak HDO peak is what I think.

[Enthalpy]

[...] a dynamic equilibrium [...]

That's what I expect, with very little selectivity. That is, nearly the same D/H proportion in water and alcohol. Processes that offer a bit of selectivity are rare
https://en.wikipedia.org/wiki/Girdler_sulfide_process

Said equilibrium may take very long to achieve. Acidic conditions are known to speed it up.

I have no knowledge hence opinion about NMR.

[wildfyr]

D and H exchange equilirbium of alkyl alcohol will be on the order of a seconds or less.

[Babcock_Hall]

My recollection of the relevant theory is that under the conditions of fast exchange (which almost certainly apply here), that a new peak appears at the weighted average of the two individual peaks.