14

A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane. For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole): Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the ...


9

This is a rather unusual and interesting case of aromaticity, which has been given a special name: homoaromaticity. The Wikipedia page does a quite nice job of explaining what's going on. As you state, protonation of cyclooctatetraene generates a $\ce{C8H9^{+}}$ cation containing an $\mathrm{sp^3}$-hybridised carbon atom between all the other $\mathrm{sp^2}$-...


9

Simply for the beginners in organic chemistry, this explanation of aromaticity is good enough: An aromatic (or aryl) compound contains a set of covalently bound atoms with specific characteristics: A delocalized conjugated $\pi$ system, most commonly an arrangement of alternating single and double bonds. Coplanar structure, with all the contributing atoms ...


5

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$). Because of various reasons, ...


5

In keto-enol tautomerism, the equilibrium is mostly towards the keto form, as it should be, since $\ce{C=O}$ is a very strong and stable bond. In some cases however we find that the equilibrium sometimes shifts to enol form. This happens in cases where some additional benefit is done by the enol structure. here are two examples: A) Removing anti-aromaticity ...


4

The rules in determining the aromaticity of a compound are as follows: The system must have 4n+2 π electrons The system must be planar The system must be completely conjugated The molecule must be cyclic. The cycloheptatriene anion does not follow two out of these four rules. Namely, The anion system has 4n electrons (whereas it should have had 4n+2 ...


4

Maybe the original structure contained the motif of 2-hydroxypyridine. If this guess is correct then a 2-pyridone is one reasonable tautomer, as shown in the first line of the illustration below. And it seems plausible to draw such an isomer for a molecule like 1H-pyrazolo[3,4-b]pyridin-6-ol, too.


4

According to Huckel's rule Four Criteria for Aromaticity The molecule needs to be (1) planar, (2) cyclic, (3) fully conjugated, and has (4) 4n+2 electrons. Your molecule does not seem to be planar (b/c of N and O atoms) but assuming that it has a nearly planar conformation it satisfies all other criteria. You only need to count one of Oxygen's lone pair and ...


4

Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced $\pi$-electron bonding difference. Less than meets the eye Suppose you were to use ...


3

Setting aside the arguments over umlauts, yes you can add the nitrogen atom to the Hückel matrix; in fact, theoretically you should do so and allow the matrix eigenfunctions and eigenvalues to decide whether the nitrogen atom in this case is really conjugated. The carbonyl oxygen can also be coupled in. But there is a catch. Since the nitrogen or oxygen atom ...


3

The given dihydroborinine is indeed anti-aromatic if is it fully conjugated and planar, but counting pi electrons in a conjugated ring is not as simple as it seems when you have boron in the ring. If the boron is not joined to a ligand, as in (mono)boronine or unhydrogenated diborinine, it could pass its electron deficiency into the sigma bonding framework ...


2

Yup, they aren't. With the saturated carbon atom in the ring, you can't get a delocalized ring of pi electrons that you would need to apply Huckel's Rule. A more rigorous analysis in Ref. [1] give us a curious conflict between theoretical and experimental results for this molecule. The theoretical calculations suggest that delocalization of the pi bond ...


2

Simpler explanations, can by their nature be less accurate than a detailed description, but are helpful in beginning to understand things. Getting to the heart of your question: “How is the aromaticity in graphene different from the aromaticity in benzene? . . . but is it possible to find a simpler if less accurate way to explain the difference,” Perhaps a ...


2

G3 is generally a term used to describe a particular computational model used in calculating molecular bonding in quantum chemistry. The G refers to Gaussian and 3 is just a third iteration of including additionally complex terms to the molecular orbital calculations being used. There's a brief discussion on wikipedia, but you'll find more extensive ...


1

It is correct that the $\ce{N}$ with the double bond has a localized lone pair, because of its $sp^2$ hybridization, due to which the lone pairs are in the same plane as ring, and hence not conjugated. However, the other $\ce{N}$ is clearly in conjugation, hence, if you draw the $5$ possible resonance structures, you would notice that each bond in the ring ...


1

You can think of it as: polarity causes slight concentration of pi electrons on nitrogen atoms in contrast to boron in the resonance hybrid due to which cyclic delocalization does not occur "completely"; hence the aromatic character is hindered Further reference: https://www.researchgate.net/publication/...


1

Let us understand and try to answer your question with a few simple points. First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons. According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for ...


1

Antiaromaticity doesn't exist in states that are observable. Its extended presence is a contradiction to its definition. Instead the double bond system exists as an independent system instead of being conjugated. This effect is known as the Jahn-Teller Distortion. A very famous example of said effect is cyclo-butadiene which exists in the form of a ...


1

As Ivan Neretin pointed out in the comments, $\pi$ bonds will remain same whether or not in resonance. However, if you have trouble counting the double bond equvalent, you can directly use the formula for it given as $$\ce{DU=\dfrac{2C+2-H+N-X}{2}}$$ so here number of carbons are C=23, H=21, N=1 which also gives 14 as the answer.


1

Plants like fully decomposed (composted) organic materials just like humans prefer cooked food. The solution which you are preparing, is going to hurt the plants rather badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in the ground or soil, mix it with old leaves and let it ...


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