# Tag Info

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To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) ...

31

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site. Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and ...

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The answer is you are referring to neither of them. That is because resonance structures don't actually exist in reality. We only use them to give us a rough idea what the actually molecule and bonds look like. A common way to explain resonance structures is this: An explorer from a far distant land travels to a new continent and sees a strange animal that ...

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First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity. Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It ...

20

Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ ...

20

The mechanism of the Cannizzaro reaction is illustrated below. The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate. Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon ...

19

Interesting question. It is much less studied and reported on than the case of non-classical carbocations, but I did find a few papers. Brown and Occolowitz (Brown, J. M.; Occolowitz, J. L. Chem. Commun. 1965, 376–377) reported that deuterated bicyclo[3.2.1]octa-2,6-diene 1b, below, undergoes base-catalysed de-deuteration (potassium tert-butoxide in DMSO) to ...

18

Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm ... 18 I approach this question from the opposite direction. Benzene is commonly drawn as a ‘cyclohexatriene’ corresponding to the Kekulé structure, i.e. with three single bonds and three double bonds, despite the fact that the six bonds of benzene are actually indistinguishable from each other. This graphical representation of benzene is in accordance with ... 17 The short answer to your question is that no, they don't all occupy the same orbital but yes, they occupy very similar regions in space. What tends to be confusing in these pictures is it sometimes unclear that the pictures are sometimes of the molecular orbitals and sometimes of the electron density, which is an observable quantity and quite another thing ... 17 I would appreciate to learn whether such a reasoning, for these two type of substituted phenols, is plausible or whether further more solid arguments should be invoked. Your reasoning seems generally well thought out. You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series. Structural effects ... 17 General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding. General Rule #2: The more s-character in a bond the shorter the bond (reference). For example a$\ce{C(sp^3)-C(sp^3)}$single bond length is ~ 1.54$\mathrm{\mathring{A}}$a$\ce{C(sp^2)-C(sp^3)}$... 17 The nitrogen in aniline is somewhere between$\mathrm{sp^3}$and$\mathrm{sp^2}$hybridized, probably closer to the$\mathrm{sp^2}$side. We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$hybridized). However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, ... 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 ... 16 Interesting question, more subtle than I realized! Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must ... 14 I very much like the first answer, especially the superb visuals! However, I will nonetheless add a simpler approach to answering your question, which is this: Pauli exclusion holds as long as there is something in space or in spin (spin being the pairs of electrons) that firmly distinguishes two "stable" states. What that means is that your initial ... 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 Disclaimer: This is not a full answer. I was just too long for a comment. I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein et. al. ... 14 To append Ringo's good answer and to add some more insight into the bonding situation, I performed a calculation on the DF-BP86/def2-SVP level of theory. Since this is a linear molecule, there are symmetry restrictions. Its point group is$C_{\infty\mathrm{v}}$, which means, that there are degenerated orbitals. As we will see, these correspond to$\pi$bonds.... 14 When it comes to resonance, you should not take the notion of superposition of different states literally. Feynman mentioned it explicitly right from the get go, just read carefully what he says: We could imagine that [...] See, we could imagine. The superposition of states in a description of what is called resonance in chemistry is not the ... 14 It seems likely that the tricyclopropylcarbinyl carbocation would be more stable than the tropylium carbocation for the reasons I'll outline below, but if you have a reference proving this point, it would be nice to add it to your question. Background The cyclopropyl group is similar to an olefinic double bond in that it is very effective at stabilizing ... 13 Which of the contributing structures of the resonance below is more stable? Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities. What the question ... 13 Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide. The carbon is obviously$\ce{sp^2}$hybridised (as this concept is very well applicable here), hence organising it's ligands in one plane with roughly$120^\circ$angles. Naturally one would assume the nitrogen to be$\ce{sp^3}$hybridised, which ... 13 one might expect that the bond lengths would be exactly intermediate between a single and a double bond They actually are. The relevant reference structure, that can provide us lengths of single and double bond for$sp^2$carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths ... 12 TL;DR: Lewis$\to$Non-Lewis$\mathbf{E(2)}$values have no direct physical significance, are intrinsically un-measurable, and serve only to quantify the extent to which the "real" wavefunction for a system deviates from the fictional idealized Lewis-structure wavefunction.$E(2)$values do, however, correlate with a variety of trends in chemical bonding ... 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 ... 12 Even when the molecule is present in a neutral form, as a phenol, the oxygen will still be close to a$\ce{sp^2}$shape with the$\ce{p}_z$orbital partially overlapping the$\ce{p}_z\$ orbital of the carbon bound to it. The proton NMR of phenol clearly shows characteristic aromatic peaks, demonstrating that the ring is aromatic. Hückel's rule states that a ...

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I think that a major teaching opportunity was missed. If MO theory had previously been introduced in the course, then there is absolutely no excuse, here is another topic where it was meant to be applied. If MO theory had not been previously introduced, then it still remains a teaching failure, here's why... This was a great opportunity to introduce a new ...

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Why should they be bang in the middle? Or to better phrase the question: What reasoning are you applying to assume an averaged bond length? Bonds don’t work as we laymen like to write them, with either a single line or a double line; and a double line being equivalent to two single lines. Rather, bonds — and most importantly, their lengths — are the result ...

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Yes, it's anionic sigma complex, more precisely Meisenheimer complex. They are much more stable then arenium ions (Wheland intermediates) and can be isolated much more easily. There was actually synthesized a sigma complex which is zwitterionic combination of both Meisenheimer and Wheland complexes: (picture created by Wikipedia User:V8rik) The term "...

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