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Question

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Which form of 1-phenylbutane-1,3-dione is more stable?

Answer

3-hydroxy-1-phenylbut-2-en-1-one

My question

I've been told that between the above two, 3-hydroxy-1-phenylbut-2-en-1-one is more stable because the double bond is in the middle which is unclear to me.

Could someone explain the above statement?

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In opposition to the line of thought presented, I argue that the 4-hydroxy isomer is more stable than the 3-hydroxyl form for that the conjugated $\pi$-electron system is larger, than in the former.

This is based on multiple assumptions.

  • For one, given a molecular framework which may have multiple $\ce{C=C}$ bonds, if other parameters are not altered, the system with double bonds in conjugation is more stable than the one with isolated double bonds. As an example, the computed free enthalpy of formation, $\Delta{}_RG$ about 1,5-hexadiene (two isolated $\ce{C=C}$ bonds) is $\pu{92.70 kJ/mol}$ (link to computation on MolCalc.org), the about 1,3-pentadiene (two $\ce{C=C}$ bonds in conjugation) $\pu{86.57 kJ/mol}$ (link to computation). And the less endothermic a compound, the more thermodynamically stable.

  • For the other, I said the 3-hydroxyl isomer would contain a less extended conjugated $\pi$-system, than the 4HP isomer. Elements to participate in this system are the benzene ring, the carbonyl group, and the simple $\ce{C=C}$ bond, present in both. If you imagine a water wave running from one to the other other end of the $\pi$-system, $\ce{C=O}$ only is tangent in 3HP, while in 4HP the wave runs all across these bonds. Thus, equally a reason why the 4HP isomer is more favourable, than 3HP.

    For a raw simplification, you may consider the benzene ring as "super atom" and count the longest pathway accessible. 3HP: 4 atoms, 4HP: 5 atoms participate in the $\pi$ system; again energetically more favourable for 4HP, than for 3HP.

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Sadly, the engines to run MolCalc.org's web site constrain the computations to molecules with 10 non-H atoms maximum. Thus, contrasting to earlier posts by others (e.g., this by Ben Norris), I'm not able to support my arguments by a similar computation for your substrates (12 non-H atoms).

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  • $\begingroup$ I understand that tangent concept you presented, but due to the presence of the carbonyl group the C attached to the group in 3HP is polarized and has a partial positive charge on it, which I think will let it participate in resonance with the benzene ring. $\endgroup$
    – Have a Good day
    Commented May 10, 2021 at 14:48
  • $\begingroup$ You can find a similar resonating structure in Benzoic acid too $\endgroup$
    – Have a Good day
    Commented May 10, 2021 at 14:50
  • $\begingroup$ Kindly elaborate :( $\endgroup$
    – Have a Good day
    Commented May 10, 2021 at 14:50
  • $\begingroup$ The carbonyl group will polarize the $\pi$ system (EWG), as does the phenyl ring (EDG). In 3HP, these two effects partially annihilate each other with lesser interaction of the C=C bond than in 4HP; in the later, this push and pull is right across the C=C bond. The resonance the mesomeric formulae conceptualize for the benzoate anion do not extend the dimension of the $\pi$ system; here, isomerism however affects this dimension. $\endgroup$
    – Buttonwood
    Commented May 10, 2021 at 15:09
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    $\begingroup$ Buttonwood, your argument makes sense. This is a case of cross-conjugation (sometimes called Y-conjugation) vs. linear extended conjugation, the latter is generally energetically preferred. Perhaps you could run MolCalc with a double bond replacing the aromatic ring? $\endgroup$
    – ron
    Commented May 10, 2021 at 17:16

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