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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

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  • $\begingroup$ 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 '17 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. $\endgroup$ – epsilon-emperor Jan 14 '18 at 15:13
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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.

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  • $\begingroup$ 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 '17 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 '17 at 11:24
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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.

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  • $\begingroup$ OP is asking specifically about ring creation, so this doesn't answer the question. $\endgroup$ – Mithoron Feb 2 '17 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 '17 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 '17 at 12:24

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