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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 ...

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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 $\ce{N}$ ...

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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 ...

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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 (Ref.1) reported that deuterated bicyclo[3.2.1]octa-2,6-diene 1b, below, undergoes base-catalysed de-deuteration (potassium tert-butoxide in DMSO) to give 1a much faster (ca. $3 \times 10^4$) than the ...

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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 $2\pi_\mathrm ... 19 General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding. General Rule #2: The more s-character in a bond the shorter the bond (reference). For example a$\ce{C(sp^3)-C(sp^3)}$single bond length is ~ 1.54$\mathrm{\mathring{A}}$a$\ce{C(sp^2)-C(sp^3)}$... 19 I approach this question from the opposite direction. Benzene is commonly drawn as a ‘cyclohexatriene’ corresponding to the Kekulé structure, i.e. with three single bonds and three double bonds, despite the fact that the six bonds of benzene are actually indistinguishable from each other. This graphical representation of benzene is in accordance with ... 18 I would appreciate learning whether such reasoning, for these two type of substituted phenols is plausible or whether furthermore solid arguments should be invoked. Your reasoning seems generally well thought out. You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series. Structural effects don't ... 17 The nitrogen in aniline is somewhere between$\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}$hybridized). However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, ... 16 It is only the electrons from atoms in the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system. The reason for ... 16 Interesting question, more subtle than I realized! Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must ... 15 To append Ringo's good answer and to add some more insight into the bonding situation, I performed a calculation on the DF-BP86/def2-SVP level of theory. Since this is a linear molecule, there are symmetry restrictions. Its point group is$C_{\infty\mathrm{v}}$, which means, that there are degenerated orbitals. As we will see, these correspond to$\pi$bonds.... 15 In addition to the species mentioned in the answer above, I found another one in Organic Chemistry by Morrison and Boyd(Seventh Edition) 14 There definitely is an easy way to do this. One of the easiest examples that comes to mind is benzene. If we look at the structure: we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction. If we draw the other resonance structure these bonds shift by 1 atom to look like the ... 14 I agree absolutely with Max, and while I was still running the calculations, he already provided the answer. The goldbook provides a nice definition for aromatic compounds: In the traditional sense, 'having a chemistry typified by benzene'. A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than ... 14 Disclaimer: This is not a full answer. I was just too long for a comment. I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein et. al. ... 14 one might expect that the bond lengths would be exactly intermediate between a single and a double bond They actually are. The relevant reference structure, that can provide us lengths of single and double bond for$sp^2$carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths ... 14 When it comes to resonance, you should not take the notion of superposition of different states literally. Feynman mentioned it explicitly right from the get go, just read carefully what he says: We could imagine that [...] See, we could imagine. The superposition of states in a description of what is called resonance in chemistry is not the ... 14 It seems likely that the tricyclopropylcarbinyl carbocation would be more stable than the tropylium carbocation for the reasons I'll outline below, but if you have a reference proving this point, it would be nice to add it to your question. Background The cyclopropyl group is similar to an olefinic double bond in that it is very effective at stabilizing ... 14 Bon has a very nice example of motion potentially restricted because of a barrier, and although it turns out that this is not the case for the norbornyl cation, there are simpler examples that do show inversion through a barrier, such as ammonia inversion or cyclopentene ring puckering. The two structures either side of the barrier would be not be called ... 14 The correct anion is stabilized by the mighty resonance. That's about it. Hydrogen bonding is irrelevant, since it affects both variants the same way. 13 Which of the contributing structures of the resonance below is more stable? Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities. What the question ... 13 Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide. The carbon is obviously$\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 OK, let's step through these terms and let's add "cross-conjugation" and "homoconjugation" for good measure to round out the series. Backbonding: This term is most often used to describe a type of bonding that occurs in inorganic compounds. Metals bonding with carbon monoxide provide good examples of backbonding. The carbon atom in ... 13 Compare the$\ce{OH}$bonds in Vitamin C (ascorbic acid) and decide which one is the most acidic. The most acidic proton in ascorbic acid is the one whose conjugate base is most resonance stabilized. Removal of either of these$\ce{H}$'s at hydroxyl group A or B does not give a resonance stabilized anion:$\bf{Scheme \ 1}$The proton at D is less ... 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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