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:
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asked Feb 2, 2017 at 4:08
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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 ChoyFeb 2, 2017 at 10:14
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$\begingroup$ 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. $\endgroup$– stoic-santiagoJan 14, 2018 at 15:13
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2 Answers
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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:
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. $\endgroup$ Feb 2, 2017 at 10:26
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$\begingroup$ 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 NgoFeb 2, 2017 at 11:24
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Starting from the aldehyde,
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. $\endgroup$– MithoronFeb 2, 2017 at 12:09
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$\begingroup$ 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. $\endgroup$ Feb 2, 2017 at 12:17
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$\begingroup$ "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... $\endgroup$– MithoronFeb 2, 2017 at 12:24