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What kind of aromaticity does 1,2-dihydro-1,2-diborete have?

1,2-dihydro-1,2-diborete

I read somewhere that 1,2-dihydro-1,2-diborete is an aromatic compound, but I believe that in order to be aromatic, it should have a continuous resonance. I also think that there should be a break at B that causes no π-electrons to be shared among the B—B atoms. So, I think this is non-aromatic. Is that correct?

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    $\begingroup$ pubs.acs.org/doi/pdf/10.1021/ja00295a033 $\endgroup$
    – Mithoron
    Dec 31, 2016 at 20:05
  • $\begingroup$ FYI this paper^ says it's not aromatic, or even flat. $\endgroup$
    – Mithoron
    Dec 3, 2023 at 21:38
  • $\begingroup$ Annually the calculations are quite old and the level of theory quite crude, but most of it should hold. However, it should be easy to update these findings (if it has not been done already). $\endgroup$ Dec 5, 2023 at 10:13

2 Answers 2

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That molecule is indeed aromatic.

It satisfies Hückel's rule $(4n+2)$ with $n=0$. It's also planar.

Lastly, the boron atom is bonded to three other atoms, meaning it can exist with $\mathrm{sp^2}$ hybridization. The leftover p-orbital, whilst not actively forming π-bonds, can still complete a stable π-molecular orbital.

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    $\begingroup$ There is some delocalisation but it doesn't necessarily mean it's really aromatic. As for answer, there are orbitals not domains, and satisfying Huckel doesn't automagically mean the compound is truly aromatic. $\endgroup$
    – Mithoron
    Dec 31, 2016 at 20:07
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    $\begingroup$ Hückel's rules only work for the compounds that it is derived from: monocyclic hydrocarbons. $\endgroup$ Jan 28, 2017 at 10:42
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The compound you are asking for is Aromatic.

There are 3 conditions for aromaticity:

  1. Compound must be cyclic and planar {Given compound is cyclic and planar since boron is $\pu{sp2}$ hybridised}
  2. Compound must be conjugated {While checking for conjugation there is no need of π electrons. For conjugation the sole condition is there must be parallel p orbitals on adjacent atoms: whether they're participating in π bond formation, have lone pair or just empty p orbitals - (as in this case)}
  3. There must (4n+2) π electrons in the largest conjugated periphery {There are 2 π electrons in this case which satisfies 4n+2 condition}.

PS:

  1. Corrections would be highly appreciated.

  2. MOT explanation for huckle rule: Huckle Rule & MOT

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    $\begingroup$ The same criticism as for the other answer applies here. Hence I have down voted here, too. Arguments towards an answer of this question should be based on observation of physical properties or rigorous theory. Hückel's rules are just a generalisation of an observation and a good guide for cyclic hydrocarbons. $\endgroup$ Dec 3, 2023 at 18:41
  • $\begingroup$ @Martin-マーチン Hückel's rule can also be applied to molecules containing other atoms such as nitrogen or oxygen. As like in pyridine $\endgroup$
    – Amit
    Dec 5, 2023 at 7:28
  • $\begingroup$ It's not been derived from these cases but I'm aware of it's existence. It works coincidentally and probably accidentally for these cases, but it cannot be a substitute for a rigorous theory. Hückel's rules are a good guess estimate, but nothing more. Aromaticity is something that's still not fully understood and there is plenty of discussion still possible. The molecule here is simply to complicated to be treated with such a simple model. $\endgroup$ Dec 5, 2023 at 10:01
  • $\begingroup$ I agree that aromaticity is yet something of a mystery in itself. But at an elementary theoretical level, huckle rules are maybe the best guess we have got. Yeah, experimentally we can do proton NMR spectrum to check the aromaticity, I'm not sure but i read in some research paper that given diborete resonates at about 7 ppm ,,,lil less aromatic then benzene ,,, but yeah, it is in aromatic category. $\endgroup$
    – Amit
    Dec 7, 2023 at 12:47

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