10

You can find a nice discussion of diethynyl ether and some of its derivatives here: Ooi, I. H., & Smithers, R. H. (1989). Bis [(trimethylsilyl) ethynyl] ether: a moderately stable C4H2O derivative. The Journal of Organic Chemistry, 54(6), 1479-1480. https://pubs.acs.org/doi/pdf/10.1021/jo00267a053 They write that "little or nothing is known about ...


10

As Oscar Lanzi suggested, both $+M$ and $-I$ applies here, but $\ce{Cl}$ stabilizes carbocation, meaning $+M$ is more effective than $-I$. This fact was confirmed by this peer-reviewed paper (Ref.1): The lowering of $\ce{C_\beta–H}$ stretching frequencies in carbocations 1a–d and 2a–c induced by hyperconjugation was tested as a possible probe for ...


7

Have a look at this table from Organic Chemistry by Wade and Simek [1, p. 134]: Every time a $\ce{-CH2}$ group is added, the heat of combustion is increased by almost 650 kJ/mol; whereas the maximum relief in ring strain attained by adding that $\ce{-CH2}$ group is not more than 25 kJ/mol. So there is no competiton between increase in heat of ...


7

Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms. Even so, carbon tetraiodide is definitely established and in use as an iodinating agent. Storage below normal room temperature is recommended. Carbon tetraiodide is also very red....


7

A carbonyl group normally destabilizes carbocations but stabilizes carbanions. In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, ...


7

I think the third option as greatest number of resonance structures can be alluded to it. The following may suffice: In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it. In the second one the lone pair on oxygen, the ...


6

Following is an image of hyperconjugative effect. We can see a $\ce{\sigma}$-orbital delocalising electron pair with an empty $\ce{\pi^\ast}$-orbital. The explanation to your problem can be given by deciding whether the effect is "centralized" or "decentralized". Let me explain it by an example. A quick scene: Suppose, you and your ...


6

Disclaimer: This post might be hard to read for some people. Some may even consider this a rant. You have been warned. This answer will question the validity of the used models, as well as the basis for such exercises. (It will also point to a fundamental flaw of this specific exercise.) If you are a student, you might learn concepts that will later turn out ...


5

At least the trioxide $\ce{CO3}$ exists, and it has even been detected in at least two isomeric forms. Wikipedia reports that it forms as a transient species in reactions between carbon dioxide and atomic oxygen or ozone, in irradiation of solid carbon dioxide, and even in oxidation of carbon monoxide by $\ce{O2}$.


5

This answer is years later than the original question, but I add it for anyone who come across it now or in the future... All acid anhydride and acid chlorides are thermodynamically unstable in water. They will eventually hydrolyze, but that time could be seconds, minutes, hours, days, or years. Moreover, in the absence of a catalyst, the rate of hydrolysis ...


4

Expanding what Tan Yong Boon said, there is a full scientific article detailing why ethylene dione is unstable Chem. Eur. J., 1998, 4, 2550-2557. Long story short: Two CO molecules are far more stable than OCCO. The ground state of ethylene dione is a "diradical". This is, OCCO has two unpaired electrons instead of forming the intuitive chain of double ...


4

According to the authors of Ref. 1 this is a general property. They provide an explanation for the stability of such sheets based on formation of particular buckled geometries: The discovery of a flat two-dimensional crystal known as graphene has contradicted Landau−Peierls−Mermin−Wagner arguments that there is no stable flat form of such crystals. Here, ...


4

There are several things which are needed to be known to be able to deal with this question. Firstly there is the 18 Valence Electron (18VE) rule (description at libretexts.org). The iron pentacarbonyl is an 18VE complex. It is coordinatively saturated. A low valent transition metal complex with strong ligand field ligands is considered to be coordinatively ...


4

For this question, given the lack of initial information, I'm making the following assumptions: Butane-2,3-diol dehydrates to give butan-2-one 2-butanol dehydrates to give but-2-ene Butane-2,3-diol dehydration is nicely described in this paper from Emerson1: Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of ...


4

Your answer is the correct one. Acid-catalysed hydrolysis of epoxides proceeds exactly as shown. In the specific case of styrene oxide, nucleophiles attach themselves to the benzylic carbon1,2, which means that A should be the major product. I'm attaching an image from (1) which gives the products of reaction of methanol with styrene oxide: If you want to, ...


3

According to Palstics International's Chemical Resistance Chart, the A-rated plastics (no solvent attack) towards acetone are: ECTFE (Halar®): transparent films available Fluorosint® PTFE: white HDPE: transparent films available Nylon®, Type 6/6: white PP: clear sheets available PPS: opaque white PTFE: white Among these, polypropylene appears to be the ...


3

The Rules Actually the rules for determining the relative stabilities (more accurately, contributions) of resonance structures do have their own priorities. Here is one version of them, with descending order of importance listed below (as is indicated by the bold words). The rules can be found in common organic chemistry textbooks (like [1], also supported ...


3

Well, I think your book is wrong. Read the extract: A-Values are numerical values used in the determination of the most stable orientation of atoms in a molecule (conformational analysis), as well as a general representation of steric bulk. The utility of A-values can be generalized for use outside of cyclohexane conformations. A-values can help predict ...


3

You get both, but the +M effect wins. See the discussion of the effect of atoms with lone pairs over here.


3

These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question: The order (B > A > C > D) was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the ...


3

Your reasoning is correct, but I think you'd mistakenly compared the ring strain of the rings, while I would have compared the ring strain in the corresponding ketones. Note that, you are asked about the stability of gem-diols, which could convert into the corresponding ketone. The point is, the ring isn't going to change, so the stability of rings don't ...


2

Nobody is able to foresee the solubility of a product. There are some experimental rules, but they all have exceptions, that nobody is able to explain. Just have a look on the Calcium salts made with the halogens (F, Cl, Br, I). There is a nice analogy among Cl, Br and I, but not F. Look ! The Calcium chloride $\ce{CaCl2}$, bromide $\ce{CaBr2}$ and iodide $\...


2

Let's look at following table from a Yale website: $$ \begin{array}{c|ccc} & \ce{(CH2)3} & \ce{(CH2)4} & \ce{(CH2)5} & \ce{(CH2)6} & \ce{(CH2)7} \\ \hline \Delta H_\mathrm{Combusion} \ (\pu{kcal/mol}) & -499.8 & -656 & -793.5 & -944.6 & -1108.3 \\ \Delta H_\mathrm{Combusion} \text{ per } \ce{-CH2}- \ (\pu{kcal/mol}) ...


2

How to decide whether +M effect or -I effect will operate in this case? The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$. In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ ...


2

Chlorine is more stable in ion state than in the neutral atom. But it is only more stable with respect to its own electrons. For the rest of the world it has one negative charge in excess. So, stable or not, it attracts positive charges like Na+ or any other positive ion. You cannot avoid that it will produce neutral assemblies of atoms, like NaCl. It is not ...


2

(a) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Even though $\ce{O}$ is more electronegative than $\ce{C}$, its octet is fulfilled while $\ce{C}$'s is not (only 6 $e^-$s on positively charged $\ce{C}$). Therefore, second structure is more favored over the first one. (b) Negative charge on $\ce{O}$ versus negative charge on $\ce{C}$: Here, ...


2

What's meant by stability is whether the compound "wants" to be in current state. If it wants to change the state (separated ions want to bond) then the system is less stable. In your example you can simply imagine 2 spherical magnets placed close to each other - they would rush towards one another. They had potential energy that turned into ...


2

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium $$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$ that can be expressed also as: $$\ce{acid1 + base2 <=> base1 + acid2}$$ The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous ...


2

Your basic assumptions are correct. It can be observed as such: In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest. Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also ...


2

@ Martin has addressed the shortcomings of assessing “stability” by counting hyperconjugating C-H bonds. If by “more stable” one means which one of the two isomers, 1,2,3,4,5,8-hexahydronaphthalene 1 or cis-1,4,4a,5,8,8a-hexahydronaphthalene 2, has the lower heat of combustion, formation or hydrogenation, then a decision can be made. Unfortunately, there ...


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