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39

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 ...


25

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 ...


24

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 ...


23

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 ...


22

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 $\ce{N}$ ...


21

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 ...


20

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 $\ce{C(sp^2)-C(sp^3)}$ single ...


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 ...


18

In addition to the species mentioned in the answer above, I found another one in Organic Chemistry by Morrison and Boyd(Seventh Edition)


17

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 ...


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, ...


17

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 ...


16

Preamble One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle; -merism from Gk. merismos &...


16

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 ...


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

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 ...


15

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 ...


15

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.


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

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

The acidity of nitrophenols (or any acid for that matter) is determined by the stability of the conjugate base. In the case of m-nitrophenol and p-nitrophenol, the relative stability can be determined by looking at the resonance structures. You can see that p-nitrophenol has an additional resonance structure where the negative charge is delocalised onto the ...


14

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 ...


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

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity. Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol. Resonance structure B requires that the nitro group be planar with the aromatic ring. If methyl groups are placed ...


13

That is really not a fair question and even with a computational approach the differences are more or less marginal. The approach by electron pusher is certainly the way to go, and here is the computations to back it up. See if you can find the carbon with the highest electron density in the following plot showing the value of the electron density for the ...


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