What happens to the NMR of an alcohol when D2O is added?
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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