I cannot understand if carbonyl adjacent to alkene stabilises or destabilises it. Generally, we say that more substituted alkene is stable. In those cases attached groups increase electron density of the alkene. In the case of carbonyl, it decreases electron density of the alkene so it should destabilize the alkene.

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But I read about the stability order of the above compounds to be 1>2>3. The reason being that the first is conjungated and second's enol is conjungated.

So, how can both increasing and decreasing the electron density of alkene stabilise it?

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    $\begingroup$ Ketones and enols are completely different molecules. Sure, they can be converted into one another, but it's absolute rubbish to say that a ketone is more stable because its enol is more stable. You should get a better book. $\endgroup$ Commented May 10, 2021 at 22:50
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    $\begingroup$ On top of that, it doesn't make sense to compare stability of different molecules with different chemical formulae in absolute terms. Absolute thermodynamic stability can be measured using things like heats of combustion, but it's pointless comparing two (or three) non-isomeric compounds because they will have vastly different heats of combustion (for example, the bottom compound will produce a lot more CO2 than the top compound). $\endgroup$ Commented May 10, 2021 at 22:56
  • $\begingroup$ I concluded this from a question on rate of dehydration. OH group was attached on the left end of each of the above mentioned molecules. As their carbocations were almost of same stability so I thought that I should compare the rate based on the stability of products. Thus, I came up with these explanations. Is the order stillcorrect? $\endgroup$
    – Satya
    Commented May 10, 2021 at 22:57
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    $\begingroup$ Well, that's an entirely different question. "Which of these undergo dehydration the fastest and why" can definitely be answered. "Which of these products is the most stable" cannot be answered. $\endgroup$ Commented May 10, 2021 at 22:59
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    $\begingroup$ You should start by considering which type of elimination mechanism (E1, E2, E1cb) is likely to prevail for each of the individual compounds. In order to do this, you also should specify reaction conditions so that any ambiguity is eliminated. However, the comment section isn't the right place to go into this. You can ask it as a separate question (better not edit this since you already have one attempt at an answer). $\endgroup$ Commented May 10, 2021 at 23:04

1 Answer 1


I think the key of your question is when does an oxygen atom donate electron density and when does it withdraw thereof.

According to Lumen Learning How Delocalized Electrons Affect pKa Values, "the oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect."

My professor tells me that oxygen has "dual personality." When its electron density cannot be donated through resonance, its electronegativity withdraws electron density through the sigma framework.

The key to explain the correct stability ranking is perhaps the 1,3-butadiene-like p orbital alignment through two adjacent double bonds. When 4 p orbitals on two double bonds align with each other, butadiene gains a resonance energy of 17.28 kcal. See more here. When two double bonds are not adjacent, p orbitals cannot overlap to gain more stability. I used ChemDraw to draw the tautomerization and the p orbital alignments. Hopefully it is not too bad.

I just finished Orgo 1 this semester, so my Orgo knowledge is at the surface level. Please point out any problems.Quick sketch using ChemDraw

  • $\begingroup$ Thanks alot for the efforts but you seem to be unfamiliar with the concept of positive and negative resonance. I am confused if alkene is stabilised by both positive and negative resonance. $\endgroup$
    – Satya
    Commented May 10, 2021 at 22:50
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    $\begingroup$ The drawings are fine, but please see my comments on the question. On a side note, I recommend using the "ACS Document 1996" style in ChemDraw (File > New from Stationery...). The default is... pretty ugly, let's say. $\endgroup$ Commented May 10, 2021 at 23:01
  • $\begingroup$ I will look into positive and negative resonance, and update the answer as needed. $\endgroup$
    – L. Cang
    Commented May 10, 2021 at 23:48

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