Find number of deuterium exchanges in the following compound.



0 D-exchanges.


I have been taught that hydrogen attached to the carbon which is directly or indirectly in conjugation with the ketone group will show deuterium exchange with $\ce{D2O}$ or $\ce{NaOD}$. Why does the solution claim there is no H/D exchange in p-quinone? Are the hydrogens attached to the double bonded carbons not considered in the conjugation?

  • $\begingroup$ Would highly appreciate if someone suggested a better tag. For now I just used a general tag. $\endgroup$
    – Haider
    Mar 17 at 19:07
  • 3
    $\begingroup$ Hydrogen attached to the carbon which will be in conjugation with electron withdrawing group after deprotonation, would be deprotonated. Is there one? $\endgroup$
    – Mithoron
    Mar 17 at 19:42
  • $\begingroup$ Still, I wouldn't be surprised if there was deuteration, if concentration of base was high enough. $\endgroup$
    – Mithoron
    Mar 17 at 19:48

1 Answer 1


It requires a lot of energy to remove the hydrogen atoms from the $sp^2$ hybdridised carbon atoms (involved in double bonding). So there's not really any hydrogen atom which is readily available for removal and deuterium exchange.

  • $\begingroup$ Okay, that makes sense. But, let's say if there were that much energy, then nothing is stopping those hydrogens from being exchanged, right? They are in conjugation with the ketone group, so I dont think there would be any problem. $\endgroup$
    – Haider
    Mar 19 at 16:17
  • $\begingroup$ Yeah well if you provide a sufficient amount of energy, they can potentially participate in deuterium exchange. It's just that usually people don't consider that as a normal answer to problems like this. $\endgroup$
    – afrin
    Mar 20 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.