139

I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$...


53

To understand the difference between kinetic and thermodynamic stability, you first have to understand potential energy surfaces, and how they are related to the state of a system. A potential energy surface is a representation of the potential energy of a system as a function of one or more of the other dimensions of a system. Most commonly, the other ...


35

It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the ...


31

Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested. However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding trans isomer. trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-...


29

I am using a very simplistic quantum chemical approach of the following isodesmic reaction: I have used Gaussian 16 Rev. A.03 and the DF-B97D3/def2-TZVPP level of theory. The summaries of the calculations are included below. On this level of theory the depicted reaction has an energy change of $\Delta G = \pu{- 37.1 kJ mol-1}.$ Therefore one could assume ...


28

There are two things we need to understand before we can answer the question. 1) More highly substituted double bonds are generally more stable than less substituted double bonds. This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$ hybridized carbon in the double bond. Electron density likes ...


26

We are discussing the following equilibrium We can make the acid a stronger acid by pushing the equilibrium to the right. To push the equilibrium to the right we can destabilize the starting acid pictured on the left side of the equation, and \ or stabilize the carboxylate anion pictured on the right side of the equation. Comparing acetic acid ($\ce{R~ =...


24

This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison. For the answer I assume that you know about hyperconjugation and the various contributing structures it involves. This gives ...


24

$\ce{SF6}$ is extremely stable for purely steric reasons, because S is completely blocked by fluorine atoms from all directions, so the reactions starting with an attack on S that otherwise would readily occur (hydrolysis, etc.) never have the chance to occur. This has nothing to do with electronegativity. For another similar example, look at $\ce{CCl4}$. ...


23

Is the trimethyl carbocation more stable than the benzylic carbocation? There are a number of approaches we can take to try and answer this question. We'll start by first comparing solvolysis rate data to see which carbocation is more stable in solution, and then we can look at thermochemical data to see how the carbocation stabilities compare in the gas ...


22

The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen. \begin{array}{|c|c|c|} \hline \text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline \ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline \ce{CHCl3} & ...


22

It's all about the 3D structure of double bonds. If we look at the tub form we see that all dihedral C-C=C-C angles are 0°. the C=C-C angles are 125°, also pretty close to the optimal value. Everything works out fine, there's basically no strain on the whole molecule. I tried to build the chair form of all cis, but every optimization ended up in either the ...


19

B-strain Both B- and F-strains are the terms I found were originally used to describe steric hindrance in complexes [1, pp. 66-67]: The concept that steric hindrance in a monodentate ligand can affect the affinity of that ligand for a metal was first introduced by Brown and Bartholomay (1944) who suggested that co-ordination of a tertiary amine with ...


18

This is a tough question. I think it might even be unfair to ask such a question on a test in non-advanced classes. In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known. What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl ...


18

Short Answer The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital. Detailed Explanation In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial ...


17

Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying ...


17

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of ...


16

Actually it is not necessary to dig deep into quantum mechanics. There are several reasons why noble gasses are stable (as gasses at room temperature). First of all, there is the obvious full valence shell. Trend in the periodic table make clear that the charge of the nucleus grows from left to right in every period. The attractive force towards the ...


16

What makes both of those molecules so stable? This is a fundamental question. Let's start by looking at the following diagram; a chemist would call it a "potential energy diagram." The y-axis shows the energy of a molecule as it travels along the reaction coordinate (x-axis; transforms, reacts) from the starting material on the left to the product on the ...


16

Can the lone pair on benzene carbanion participate in resonance? As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring. Therefore it cannot participate in resonance. Here is a drawing that illustrates this point. how is it more stable than vinylic carbanion? The $\mathrm{p}K_\mathrm{a}$ approach discussed ...


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

There are two factors that need to be considered to answer your question - entropy and enthalpy. Entropy favors the formation of smaller rings. If we consider a hydrocarbon chain, the number of possible conformations increases dramatically as we go from 3 carbons to, for example, 10 carbons. Only a few of these conformations will be correctly arranged so ...


16

In chemistry ask "why" only after you ask "if". Given a sufficiently strong superacidic medium, $\ce{H3O^+}$ can be protonated to $\ce{H4O^{2+}}$. Evidence for this reaction, by studying isotopic exchange in a $\ce{HF + SbF5 +SO2}$ solvent, is given here. $\ce{H4O^{2+}}$ is, of course, a powerful protic acid, and it would be leveled to ...


15

Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. ...


15

Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\...


15

Vinyl cations can also be generated by the protonation of acetylenes in strong acid. The vinyl cations are captured by a nucleophile to produce a stable alkene with the nucleophile attached to the double bond. Which of the following is a more stable carbocation? This route to vinyl cations has been well studied. As shown in the following diagram, ...


15

Let me start by saying that aromatic compounds are not exceedingly stable, they are just much more stable than antiaromatic compounds. However, they can still be hydrogenated exothermically to the corresponding cycloalkanes. $$\ce{C6H6 + 3 H2 -> C6H12}$$ But why are they more stable than antiaromatic ones? Let’s take a look at the orbital scheme for ...


15

I can only really speak for bromine(VII). Bromine(V) is pretty common and I'm not entirely sure what's the deal with Br(I) readily disproportionating to Br(V) + Br(-1). (The tendency of Br(I) to undergo disproportionation explains why we don't see it very commonly, but why it has this tendency I'm not so sure.) This is an example of what is sometimes called ...


15

Preamble As trivial as this question might seem, it is not. It should certainly never be asked in at an exam level lower than basic quantum chemistry. Any of the tempting explanation schemes will certainly fail. A bonding picture in the Lewis formalism is almost impossible; hybridisation descriptions in these compounds is extremely complex. As a general tip: ...


14

Boron pentachloride is likely not stable except perhaps in extreme conditions, such as under very high pressures. Even then it may be possible that a description such as $\ce{[BCl4^{-}]Cl^+}$ containing a tetrahedral boron anion could turn out to be more accurate than any hypercoordinate structure (a boron atom surrounded by more than four ligand atoms). ...


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