4
$\begingroup$

The enol and keto forms are said to be tautomers of each other, and the interconversion of the two forms involves the movement of an alpha hydrogen and the shifting of bonding electrons.

At what temperature does tautomerization stop? (E.g I can obtain both the enol and keto form without them interconverting)

To make this simple, let's just assume the molecule of choice is: acetaldehyde

In addition there is no tunneling of the two tautomers. Lastly it is known the the two tautomers of acetaldehyde are stable and do not convert at 11 K with argon matrix trapping. What's the highest temperature that do not interconvert?

$\endgroup$
4
$\begingroup$

Tautomerisation, as an observable process by NMR, is entirely system dependent, and there are many cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. As always, equilibrium is both sterically and electronically controlled. Some systems require very low temperatures (liquid nitrogen and liquid helium temperatures) to slow the exchange processes, and of course, going to higher magnetic fields will also improve the case of observing individual exchanging species.

Acetaldehyde is a very small molecule; this would require extremely low temperatures; as you have said 11K allows for observation without interconversion.

A lot of this sounds like a homework question .... you might want to be more specific about what your own research is contributing towards answering your own questions.

$\endgroup$
  • $\begingroup$ Well absolutely there are cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. Take for example phenol as an example. What I wanted to know is at what temperature does it truly stop. As for why I mentioned acetaldehyde is because it is the most simplest case. To be honest it could have been any aliphatic aldehyde or ketone. It is easy to say make a generalization and say "at low temperatures", but I already knew that. What makes you think this is a homework answer, it is just pure curiosity. $\endgroup$ – Mr Flux Feb 17 '17 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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