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Here is what I tried: enter image description here

These are the two enolic forms that I could make. So according to me, the total number of distinct enol forms of acetylacetone is $2$. But the answer is 3, what am I missing?

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    $\begingroup$ Write the formula of acetylacetone, this time including all hydrogen atoms bound to the molecule. Then, write down reaction equations for the equilibria between acetylacetone and its tautomeric enols; again with all hydrogen atoms bound to the molecule. $\endgroup$
    – Buttonwood
    Sep 28 at 6:38
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    $\begingroup$ In practice, you're not missing anything. Dienols aren't really viable in practice, for this compound. Even your unconjugated tautomer is hardly important. Then again, if you push into dienols, then there is even fourth one, with two double unconjugated bonds. Either ways this answer is wrong. $\endgroup$
    – Mithoron
    Sep 28 at 13:31
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enter image description here

I have drawn the enols and keto forms of acetyl acetone . The ones with equality are the same structure just flipped along the line I have shown on right . I don't see how your first structure had a chiral carbon atom because the carbon you marked is bonded to two hydrogen atom . Regardless I am also getting 3 stable forms . One could say we would have another pair of enantiomers of cummulated alkene . But that is generally not considered stable . So I didn't include that .

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  • $\begingroup$ Understood, thanks a lot. I accidentally took it as a chiral carbon, my bad. My mind is off today. $\endgroup$
    – Vega
    Sep 28 at 7:11
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    $\begingroup$ @Vega Suggestion: Take a brisk walk around the block and do not engage active thinking about anything. (But pay attention to the traffic, of course.) Somewhere between 5 to $\pu{10 min}$ is a break which helps me to clear my mind, leave a rut of thinking too intensely. This may be way better than a chat in the coffee/tea kitchen (a potential trap, since colleagues of yours feed you there with new ideas [possibly unrelated, but eventually scattering your thoughts further]). $\endgroup$
    – Buttonwood
    Sep 28 at 7:54
  • $\begingroup$ @Buttonwood yes I think that would help, too much work nowadays. $\endgroup$
    – Vega
    Sep 28 at 7:59

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