7
$\begingroup$

My chemistry textbook says that hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane are ring chain tautomers, but it did not mention the steps for conversion of one into the other. I tried to figure them out but got stuck.

My attempt:

tautomerisation

$\endgroup$
2
  • $\begingroup$ I disagree that you start with the ketone as the aldehyde is more unstable than the ketone is, since it is far more electrophilic than the ketone. The ketone has a methyl group attached to it which makes it far more stable. I suggest you approach this question the other way around. $\endgroup$
    – Linus Choy
    Feb 2, 2017 at 10:14
  • $\begingroup$ Hello. Which textbook is that? I hadn't read about such an interesting example of ring chain tautomers before, so just curious. Thanks in advance. $\endgroup$ Jan 14, 2018 at 15:13

2 Answers 2

7
$\begingroup$

At first, even though this is not related to the question, I want to say that you're confused between double-headed arrow between the two resonance forms and double harpoon in reversible reactions. (The harpoons should be written as below.)

Your incomplete mechanism has two errors:

  • The hydrogen atom bonded to carbon is far less reactive than the one bonded to oxygen.

  • The positive carbon, which is bonded to oxygen, a very electronegative element, is very, very unstable, which is not likely to happen.

This is the right mechanism for the reaction:

enter image description here

Step 1: Oxygen is a good nucleophile, so it is most likely to take the most acidic hydrogen.
Step 2: Then, the negative charged oxygen will attack the electrophile carbon atom, forming epoxy ring.

$\endgroup$
2
  • $\begingroup$ While I think your answer is slightly better than the author's, it's still unlikely. I think that tautomerization should happen only in the aldehyde form since it is more reactive. Besides, the formation of the geminal hemi-epoxide is quite unstable as well. $\endgroup$
    – Linus Choy
    Feb 2, 2017 at 10:26
  • $\begingroup$ Yeah, I agree that hemi-epoxide is quite unstable, but isn't that the tautomers are the two substances, that include the hemi-epoxide? $\endgroup$
    – Huy Ngo
    Feb 2, 2017 at 11:24
-1
$\begingroup$

Starting from the aldehyde,

enter image description here

The double bond of the aldehyde breaks, allowing the oxygen to bond with the H in the alcohol. The alcohol subsequently breaks the O-H and forms a ketone.

Note that this molecule is stabilised due to hydrogen bonding; therefore it may be prudent to add a base or acid to encourage tautomerization.

$\endgroup$
3
  • $\begingroup$ OP is asking specifically about ring creation, so this doesn't answer the question. $\endgroup$
    – Mithoron
    Feb 2, 2017 at 12:09
  • $\begingroup$ If you see my diagram there is a six-membered ring formed due to the hydrogen bond formed between the aldehyde oxygen and the alcohol hydrogen. Thus this reaction is more likely to happen then the above. Besides, ring-chain tautomerization happens more for hemi-epoxides which can form five or six membered rings so I doubt that this is likely to happen in the first place. $\endgroup$
    – Linus Choy
    Feb 2, 2017 at 12:17
  • $\begingroup$ "hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane" - does acetylacetone and its enol look like them? No? OP wrote normal tautomerisation instead of this cyclisation and you changed compound to have a ring... $\endgroup$
    – Mithoron
    Feb 2, 2017 at 12:24

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.