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The answer written in book is 3 but no solution is given. After referring to some other question I think that hydrogen bonding can take place in compounds like 3 but am not sure about it. Am I right that it is the only reason why it has the highest enol content?

I would like to ask one more thing that, while comparing the keto and enol content, if in a compound more than one ketone is present, than do we enolize any one of them only to check?

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    $\begingroup$ It's a 1,3-dicarbonyl... $\endgroup$ – Zhe Sep 10 '18 at 11:40
  • $\begingroup$ 3 is the only 1,3 diketone so there is a delocalisation effect between the enol and the adjacent ketone $\endgroup$ – Waylander Sep 10 '18 at 11:41
  • $\begingroup$ What do you mean by delocalisation effect .@Waylander . Does it means what is shown here -> chemistry.stackexchange.com/questions/42738/… $\endgroup$ – Shresth Jain Sep 10 '18 at 18:00
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If you can form an enol where the carbon-carbon double bond is conjugated to one or more other double bonds, the enol is more stable and the equilubrium mixture will therefore have more. With that in mind, try seeing what happens to (3) if you tautomerize with the proton on the bridging carbon between the two carbonyl groups.

For a further exercise: render the structure for cyclohexa-2,4-dien-1-one where the carbonyl group is conjugated to two carbon-carbon double bonds in the ring. Now tautomerize to the enol and see what "enol" you get. When the conjugation of the enolic double bond forms an aromatic ring, the stabilization effect is so strong that the enol takes over completely.

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(This is too elaborate to fit into a comment.)

Although one would, at first blush, immediately think of the β-diketone (3) as quite acidic, a moment's pause of what the putative enol should look like will confront you with a violation of Bredt's rule. Thus, it would prefer to enolize at the non-bridgehead hydrogens, and the acidity would likely not be quite different from the other three choices.

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    $\begingroup$ (-1) Bredt's rule is irrelevant here. $\endgroup$ – orthocresol Sep 10 '18 at 13:55
  • $\begingroup$ Why is it so? @orthocresol $\endgroup$ – Shresth Jain Sep 10 '18 at 17:55
  • $\begingroup$ I also thought of your point @derek correa and hence made a guess of (2) because in that case we get a conjugated system also but then i saw this chemistry.stackexchange.com/questions/18904/… ..please tell me that can hydrogen bonding effect solve this problem? $\endgroup$ – Shresth Jain Sep 10 '18 at 17:59

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