What happens to the NMR of an alcohol when D2O is added? What difference you expect in the 1H NMR spectrum of phenol in $\ce{CDCl3}$ after a drop of $\ce{D2O}$ is added to the test tube?


1) Since the $\ce{-OH}$ of the phenol is quite acidic, it undergoes rapid $\ce{H+/D+}$ exchange with the $\ce{D2O}$. My expectation is that the $\ce{-OH}$ signal from the original phenol will get reduced and broadened (maybe even disappear, but I am not sure) because it is replaced with an $\ce{-OD}$ group.

2) Aside from this, will the aromatic regions of the NMR spectrum change at all? I expect that the ortho and para $H$ can slowly turn into $D$ through EAS, but I am not sure if that is possible when only "one drop of $\ce{D2O}$ is added", especially without acid and base present.

My book does not contain an answer key, but I would like hear if my thought process is correct.
Thank you for the question. This is a very basic physical chemistry question but the way it has been asked is interesting.



In order to understand the probability of displacement of a H from any X-H bond (X is any atom), you need to know its pKa. The pKa values are as follows (ref.):


pKa of water = 14
pKa of OH (phenol) = 9.95
pKa of CH (benzene) = 43


If the pKa of a bond is $x$, it means that for one $\ce{X^- + H^+}$, there has to be $10^x$ no of $\ce{X-H}$ compound. Thus, it clear that for one dissociation of OH bond of phenol, ~$10^{10}$ phenol molecules are required, while for one dissociated CH bond of benzene, $10^{43}$ molecules of benzene is reuqired. To put it in perspective, in 1 mole of phenol, there will be ~$10^{13}$ dissociated OH bonds but only ~$10^{-20}$ dissociated CH bonds.

This hereby implies that dissociation of CH bond and exchanging with $\ce{D_2O}$ is highly improbable, but dissociation of OH bond and its exchange with $\ce{D_2O}$ is highly probable. That's why you get the OH peak to be lowered due to exchange but no similar phenomenon occurs for CH bonds in the system.



Please reply in comments if you have doubts.

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