# Tag Info

44

The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true. It does not switch; ...

32

Aromaticity is not binary, but rather there are degrees of aromaticity. The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene. Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be ...

28

To clarify @ron's point, the general "cheminformatics" rule for deducing aromatic fused-ring systems is whether there is one path (usually the peripheral one) that satisfies $4n+2$: So for naphthalene: The bold bonds give you a ring path of $\mathrm{sp^2}$ carbons with 10 π electrons, so it's aromatic. There are some arguments about whether the central ...

26

Pyrene is aromatic. The Hückel $4n+2$ rule works best with monocyclic ring systems. If you look at the following resonance structure for pyrene with a central double bond, the monocyclic periphery has 14 π electrons (ignoring the greyed-out central double bond), but that is a rationalization. Nonetheless, pyrene undergoes reactions characteristic of ...

23

Introduction It is fairly obvious that the statement given by Wikipedia cannot be entirely correct. There is some evidence that the anion (which obeys the 4​n+2 Hückel rule) is aromatic. The trinitromethane anion cannot obey the Hückel (4​n+2) rule because it is neither monocyclic nor planar. And assigning a number of π electrons is fuzzy at best, so ...

21

TL;DR: We don't count $\pi$ electrons here, but the $\pi$ conjugated circuits (1). [7]-сirculene has all conjugated circles of size $4n + 2$, and none of $2n$, what makes it an aromatic compound. Strictly speaking, the aromaticity $4n + 2$ and antiaromaticity $4n$ rules in their original interpretation are very limited and cannot be universally applied: ...

20

The molecule actually optimizes to a planar, delocalized $\pi$ system Using B3LYP/6-31G(d) optimizations, we see that the molecule optimizes to become planar. (The initial configuration was intentionally bent.) Moreover, if I plot the HOMO, it's highly delocalized: Similarly, the LUMO is also delocalized across the entire molecule: Consequently, I don't ...

20

The modern definition of aromaticity from deep theoreticians is that the π-system needs to support aromatic ring currents. Borazine can support it, so it is technically aromatic. Aromatic systems that do not contain carbon are not really all that common, but they are known. Pentazole was detected, for example. The problem is, that for aromaticity to be a ...

19

Very interesting question, and it kept me up despite daylight saving time cheating me of one hour of sleep last night... A good reference is Albright, Burdett and Whangbo, Orbital Interactions in Chemistry 2nd ed. pp 282ff. which explains this in much greater detail than I can. (In general, that book is fantastic.) I will try my best to summarise what they ...

18

1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic ...

17

If we go back to your earlier question on Frost diagrams (I've reproduced the key figure below), we see why simple molecular orbital theory or the "$4n+2$" rule suggests that benzene is aromatic while cyclobutadiene is antiaromatic. Forming a planar, conjugated, 6-membered ring and placing 6 π-electrons in it creates a π-system that is energetically more ...

17

Graphite is definitely aromatic and boron nitride is at least partially aromatic. For instance, in this paper, the authors calculate the percent resonance energy (%RE) of graphite as a comparison to other known aromatic systems. The %RE is a measure of the total resonance stabilization per carbon atom relative to some reference system. They find the %RE of ...

16

It is only the electrons from atoms in the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system. The reason for ...

15

There are quite a few conjugated $\pi$-systems out there; some of them are pretty stable, some are less so. The ultimate way to find out is to go and solve the eigenvalues problem for the corresponding matrix; in some cases this can be done manually, with pen and paper. There are simple cases with analytical solutions. One of them is the case of cyclic ...

15

When you have a ring that has $4n$ π electrons it's anti-aromatic if you force it into a fully conjugated cycle. Often such rings find a way to break the unfavorable (or at least, less favored) anti-aromatic conjugation. Large ones like cyclooctatetraene are likely to bend so all ring atoms are no longer in the same plane. Small ones like the cyclopropenyl ...

15

Huckel's rule can be applied to "inorganic" cyclic molecules but only if they have a delocalised pi system. They also have to have flat (or close to flat) rings. Recently some N5 rings which 6 pi electrons have been made, these are "inorganic" aromatic rings. Also borazine has been the subject of considerable debate, it has been discussed on stack exchange ...

14

I agree absolutely with Max, and while I was still running the calculations, he already provided the answer. The goldbook provides a nice definition for aromatic compounds: In the traditional sense, 'having a chemistry typified by benzene'. A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than ...

14

The two rings in biphenyl are inclined at an angle of 44° to each other because of the large steric repulsion between the ortho hydrogens on each ring. For Hückel's rule to be applied, the molecule must be able to be planar and it must be monocyclic. Due to the molecule not being planar, the two rings are not strongly conjugated to each other and you ...

14

TL;DR Yes. For cyclic conjugated systems in the $T_1$ state (i.e. lowest energy triplet excited state), $4n$ electrons means aromatic and $4n+2$ means anti-aromatic. The $T_1$ state of the cyclopentadienyl cation, which has 4 pi-electrons, is therefore aromatic. Aromaticity The (quite understandable) confusion here arises over the interpretation of the ...

14

The confusion arises because of a simplistic definition of aromaticity. What matters for aromaticity is the planarity of the relevant part of the molecule. In triptycene there are three planar benzene groups joined together. Each is individually aromatic, but the whole molecule is not. Similarly, triphenyl-methane has three individually planar benzene ...

14

In general, we can describe a quasi aromatic compound as a compound, which is ionic in nature with a counter ion, and the $\pi$ electrons in such compounds follow Huckel's rule ($4n+2$). In other words, quasi aromatic compounds are those in which the charges present on the molecule are a part of aromaticity of the compound. A few examples of such compounds ...

13

The justification comes from molecular orbital theory. I'm going to analyze the MO diagrams for benzene (a molecule with $4n+2$ π electrons) and an MO diagram for cyclobutadiene (a molecule with $4n$ π electrons). These results generalize for all planar cyclic molecules with π orbitals on every atom. The above π MO diagram of benzene may be obtained through ...

13

Essentially it's a case of aromaticity vs number of resonance structures. Well put! Aromaticity is a very strong driving force so aromaticity wins out; Huckel's rule is more important than the number of resonance structures. Therefore cyclopentadiene is more acidic than cycloheptatriene. Even though we can draw 7 resonance structures (one with a ...

13

Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon. If you look at the resonance structures in the drawing below you'll see that structure B depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons. Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is ...

13

Number 2 is not aromatic because transannular C-H interactions cause the molecule to be non-planar although it satisfies 4n+2. This problem was solved by the synthesis of number 4, which, while a little less than planar but rigid, displays aromatic properties (NMR, etc.). E. Vogel, H. D. Roth, Angew. Chem. Int'l. Ed. 1964, 228.

13

Oxepin is certainly not aromatic. If it were flat and the oxygen atom was sp2 hybridized, then it would be an 8π-electron anti-aromatic system, but to avoid being anti-aromatic, it exists in a boat-like conformation, with incomplete conjugation of the π-system.[1] Interestingly, in solution, oxepin is in equilibrium with benzene oxide - the two ...

13

Phosphorine (IUPAC: phosphinine) actually has aromatic character nearly as great (88%) as that of benzene. According to the reference, phosphorine is sufficiently stable to be handled without air-free techniques; and it undergoes electrophilic substitution reactions similar to those of benzene.

12

Algebraically, the levels of cyclic polyenes may be derived using simple Hückel theory (see also: Pi molecular orbitals of polyenes). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is: $$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$ where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ ...

12

The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any ...

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