text:
How stable is Remdesivir when vial vacuum is broken and/or it is reconstituted with sterile water?
I was recently looking at how long can Remdesivir be stored in its different forms.
I came up across several articles provided by the FDA and Gilead Sciences and others.
What caught my eye was the fact that after you reconstitute Remdesivir with sterile water you cannot store it at all! The manual and articles provided by numerous sources all say the same thing. they clearly state that you should use the reconstituted solution right away, meaning combine it with a NaCl serum. The serum then can be stored for up to 48 hours inside the fridge (no more than that). As stated by the FDA article and also another article by the EU medicines agency this is mainly because of bacterial and parasitical contamination, which is irrelevant to this question.

Care should be taken during admixture to prevent inadvertent microbial contamination. As there is no preservative or bacteriostatic agent present in this product, an aseptic technique must be used in the preparation of the final parenteral solution. It is always recommended to administer intravenous medication immediately after preparation when possible.
-- FDA

Chemical and physical in-use stability has been demonstrated for 24 hours at 25°C. From a
microbiological point of view, unless the method of opening/reconstitution/dilution precludes the risk of
microbial contamination, the product should be used immediately. If not used immediately, in-use
storage tomes and conditions are the responsibility of the user.
-- European Medicines Agency

So my question is putting aside the problem of keeping the product aseptic, How long does the Remdesivir molecule itself manage to be stable inside a vial reconstituted with sterile water. (Note that the vial vacuum has also been broken)
It would also be helpful to describe the chemical processes/reactions that happen, causing the product to lose its efficacy.

Without speaking to a specific drug, the storage times may be "generic", with a large safety margin. I.E.,  A particular mRNA vaccine might not have safe storage checked individually, but  be given a storage plan found suitable for similar mRNA vaccines. Note that after release, some of these vaccines have had storage options change!

I can see an apple going "bad" after several hours, however, I imagine an apple being a living thing that dies while outside. but, I don't know how a chemical compound which is a dead thing such as Remdesivir can go bad.

From the FDA document you link to : "Care should be taken during admixture to prevent inadvertent microbial contamination. As there is no preservative or bacteriostatic agent present in this product, aseptic technique must be used in preparation of the final parenteral solution. It is always recommended to administer intravenous medication immediately after preparation when possible." Basically you want to avoid contamination because it is administered intravenously - see the highlighted part above.

I admit it sounds mysterious. Presumably the vial is not an aseptic container once vacuum has been breached, whereas the saline bags are considered somewhat safer. But that is an opinion. You may want to ask this over at medical sciences SE: https://medicalsciences.stackexchange.com/

@BuckThorn Could one, smear some cotton with alcohol and tape it on top of the vial in order to avoid contamination?

I actually asked this from several doctors, however, all of them were clueless as to why it is so, most of them were like I could try to store it inside the fridge and see what happens. Some drugs such as Ceftriaxone actually do go bad, turning from white to yellow and then orange to red after reconstitution and storage inside the fridge, however this did not happen with Remdesivir and it managed to keep its clear color through 5 days. However, I imagine color is not a reliable indication of whether a drug has gone bad or not.

Treating it this way might be standard best practice ("better safe than sorry"). "Go bad" can mean many things. Remdesivir is a prodrug and is supposed to undergo enzyme-catalyzed conversion into an active drug.  Going bad might mean a reaction like oxidation or hydrolysis or some unwanted phase transition (precipitation) coupled to a chemical change. The fact that saline is ok makes it mysterious. I suspect it minimizes bacterial contamination. I cannot advise handling it in any way other than what is suggested by the manufacturer.

If you put aside the issue with keeping the product aseptic, there is a potentially interesting chemistry question here: how stable is remdesivir the molecule if the vacuum seal is broken, or if sterile water is added. I can't say what kind of reception such a question will get, but I think people here can at least guess that just by looking at the molecular structure.

@BuckThorn If you think I forgot to add anything please feel free to edit.

@ATheCoder Well at least the focus is on chemistry now and I (personally) find it useful. Note that I found an answer by searching for the keywords "remdesivir stability testing", see eg https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjAt_3bpb7wAhVXAhAIHdm_DbsQFjAEegQIBBAD&url=https%3A%2F%2Fwww.mdpi.com%2F1999-4923%2F12%2F11%2F1002%2Fpdf&usg=AOvVaw1Zc2FAtVfn_-En6qrJsITp Look into the introduction of that article and you will find your answer...

Let us continue this discussion in chat.


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Confusion regarding mechanism of dehydration of alcohols
1)Today I came across a problem which asks to find the order of dehydration of 2-hydroxycyclohexanone(1), 3-hydroxycyclohexanone(2),4-hydroxycyclohexanone(3). The answer is 2>3>1 But I thought answer should be 2>1>3 cause in 2 and 3 the product would be conjugated and hence more stable.
2)My second confusion is regarding the pathway taken by such hydroxyketones. Do they follow E1 pathway and form carbocation or do they take some other pathway?

I have provided what I think but I am not sure whether the same mechanism will be followed or not.

Please substantiate / illustrate your lines of thought by a mechanism drawn.  It needn't be complete, but strengthens the ground to engage in a discussion here.

I have added a picture of what I think

I think your mechanism is reasonable.
Coming to rate of dehydration,
In first case Alpha hydroxy ketone will have the least rate of dehydration because a carbocation is formed which is destabilized by Ketone group and even notice there is hydrogen bonding in reactant making reactant more stable and thus rate of dehydration is decreased in first case.
In second case, though carbonation is not stabilized,  here breaking of H will be easier because the H are more acidic being next to electron withdrawing group , so 2 has highest rate .
In third case, since carbocation is not so unstable and even the H is not much acidic. So  3 has moderate rate of dehydration.
Therefore,  rate of dehydration:2>3>1

Product of all the three compounds would be same. You have missed space tautomerism in the third one.
Besides,charge beside a carbonyl is considered very unstable. Thus, first one is lowest in rate. Third one is placed second because there is an extra step of tautomerism.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Electron Affinity of Lead as compared to Bismuth
For "nitrogen family" and "carbon family" the trend goes that in a period, the electron affinity in case of group 15 is less than that of group 14. This is attributed to the half-filled p-subshell which tends not to gain an electron. But this true till antimony only. Bismuth has a greater electron affinity as compared to lead.
My question is, why so? What is there that brings about this particular exception?

I think you got it backwards - it's the other ones that are exception. BTW I still don't get why people ask about electron affinity so often.

In a sense this is an exception to an "exceptional trend". Looking at whole of the periodic table trends, the case of Pnictogen's electron affinity is an exception. But within this exceptional trend too, we find this case that contradicts what we would expect based on the ongoing trend.                                        But the question remains that why such a contradiction is seen one would expect that owing to the half-filled p-subshell, bismuth would have less electron affinity. But it doesn't, why so?

Because those electrons are very weakly shielded by by the d and f electrons. They are just too tightly bound that the half filled logic just doesn't account for much.

Relativistic effects play a role here. Electronic configurations ending with np2 are relativistically stabilised for large values of n (like [Xe]4f14 5d10 6s2 6p2 for lead). This is often called the "inert pair effect". The comments to this answer are related.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is Maleic Acid more soluble in water than its trans isomer Fumaric Acid?
My teacher taught me that compounds having intramolecular H-Bonding are relatively less soluble in water as they will not involve in H-Bonding with water appreciably while those able to do intermolecular H-Bonding will appreciably form H-Bonding with water, and thus the latter ones will be relatively more soluble in water.
Applying this logic will make maleic acid less soluble than fumaric, but that's not the case. Wikipedia shows 478.8g/L for maleic, while just 4.9g/L for fumaric. Can someone help me with where I am going wrong?

You need to consider also the crystal lattice energy of each

https://chemistry.stackexchange.com/questions/66771/does-fumaric-acid-form-stronger-hydrogen-bonds-than-maleic-acid

https://chemistry.stackexchange.com/questions/51503/which-isomers-cis-or-trans-are-more-water-soluble

So the teacher should have been more cautious? @Andrew

That means the crystal structure factor is outweighing the H-bond factor?? @Andrew

Several factors could be important, but the first that comes to mind is polarity. The dipole moment of maleic acid is much larger than fro fumaric acid.

Determining the crystal lattice energy would be helpful in comparing the starting points for a solubility investigation. Perhaps an approximation to the crystal stability is provided by the melting points: for fumaric acid, the mp is 287$^o$C; for maleic acid, 135$^o$C. 135$^o$C is a sort of normal mp for organic compounds. 287$^o$ seems quite high, suggesting a high lattice energy, and therefore, lower solubility for this isomer.

Even if molecules have the possibility of intramolecular bonding, as indicated by molecular models or drawings on paper, the existence of intermolecular bonding is not precluded.
Sometimes explanations of an observation are used as predictions which then turn out to be incorrect. The explanations may still be useful, but as explanations of actual data, rather than as reliable predictions.


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Stability of complexes based on number of rings
$\ce{[Co(edta)]^-}$ has 5 rings in total and in $\ce{[Co(en)_3]^{3+}}$, there are 3 rings. The stability of the former is more than that of the latter. The reason I was told was that a complex with more number of rings is more stable.
Why is this so?

https://chemistry.stackexchange.com/questions/31005/whats-so-special-about-chelation https://chemistry.stackexchange.com/questions/78895/whats-the-explanation-for-the-chelate-effect

When polydentate ligands binds to a metal in a ring form, it is called chelate ring and the corresponding complex is called chelate complex.
Experimentally, it is observed that metal complexes of polydentate ligands with more rings are significantly more stable than the corresponding complexes of similar ligands but with less rings. Because, the larger the number of ring closures to a metal atom, the more stable the compound. This phenomenon is called the chelate effect.
But, do note that ring number is not the only factor to determine the stability of a complex. There are several other factors like nature of central metal ion or the ligands, ring size, ionic size. For this question, let's consider ring size.
The more the ring size, the less stable is the complex. The entropy of complex is changed if the size of chelating ring is increased or decreased. Four membered ring compounds are unstable, whereas five-membered are more stable because of negligible ring strain. Likewise, ligands with a flexible organic backbone like ethylenediamine are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings.
Refer to the following articles for more details:

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Stability_of_Metal_Complexes_and_Chelation
https://old.amu.ac.in/emp/studym/99996027.pdf


text:
Dead trees or charcoal is generally used as a reductant in the production of pure silicon for use in industry what other reductants could be used?
What other known reductants for SiO₂ could potentially be used for producing silicon from silica in an arc furnace?

Read more: http://www.madehow.com/Volume-6/Silicon.html#ixzz6uKynirns

Industrial or laboratory scale  ?

The wikipedia article on “silicon” lists several methods used to produce silicon for semiconductor production. No mention of using hydrogen gas as reductant, possibly because combining arc furnaces with high pressure hydrogen gas is a glaringly obvious disaster waiting to happen.

@EdV More likely because H₂ is reasonably expensive (ATM) and there's no price on polluting the atmosphere with CO₂ emissions. Several steel production facilities are now trialling injection of Hydrogen to eventually replace coking coal entirely. e.g. in Sweden https://www.h2greensteel.com/newsroom/h2greensteel and Germany https://www.steelguru.com/steel/arcelormittal-advancing-on-green-steel-production-in-germany
If it can be safely done in iron ore and steel smelting  with arc furnaces can it be done in silica processing to produce pure Is I wonder?

I did find one patent application for use of H₂ as a reductant in silicon production but it's only in the final stage of reducing a silicon halide that is to made by furnacing SiO2 + C in a chlorination process, so it's not the full process just the end point using H₂ in an arc furness where the gases are separated using a vortex apparatus.

Glad to hear hydrogen can be safely used in steel smelting! I had a colleague who worked in a steel mill before becoming a professor. The stories he told made me happy to never visit a working steel mill.

They already have one H₂ injection jet operational as a retro fit at a steel smelting furnace in Germany and will scale up bit by bit, it's a pilot operation to prove and improve the technology for roll out at their plants elsewhere.

On a small scale, hydrogen gas is certainly an option. Here is a paper: Reduction Kinetics of Metal Oxides by Hydrogen. High temperatures induce the formation of monoatomic hydrogen as the active agent.
However, on a practical note, explosive mixtures of hydrogen with air exist over a wide range (Wikipedia has a good discussion on this topic), so this path is clearly not without danger.
[EDIT] To address a question on trials, not that am aware of, however, here is an interesting experiment one may conduct at a seemingly much lower temperature that is likely safe. For reducing, it is known that in situ creation of the hydrogen atom radical (cited as the major reducing species by Buxton, see Eq 3 on page 2 here). Suggestively, the chemisorption of the hydrogen radical on the surface of metals like Zn, Al and Mg (albeit, effective for a limited lifespan) may be an interesting trial path.
So, as an experiment, proceed to imbue the surface of Mg or Al metal with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH). Place the SiO2 on the imbued metal surface and heat.
As to the reaction mechanics, the •H radical functional behaves based on the equation:
$\ce{ e- + H+ <=> •H }$
where this seemingly reversible formation reaction apparently acts as (e-,H+) pair on ions. For an example, citing a work in Hydrometallurgy 2008: Proceedings of the Sixth International Symposium on p. 818, a suggested commercial reductive leaching equation, to quote:
$\ce{PbS + 2 •H = Pb + H2S (5) }$
Now, could a related reaction path exist with SiO2 at a much lower temperature? To answer, this suggested trial could be quite interesting.

thanks @AJKOER. I'm aware of several pilots for injecting hydrogen into steel furnesses  and intentions to transfer the production process to entirely avoid coal eventually. Are there any such trials you're aware of for the production of silicon? And disadvantages other than the cost and flammability of hydrogen?

Thanks again for the additional information @AJKOER. I did locate one patient claim using hydrogen as the reductant, but it only uses it on a silicon halide and carbon is used in the chlorinator and results in CO₂ production, not sure if its the typical ratio of  C to SiO2  but would seem so if the oxygen is all captured in a resulting CO₂ stream. https://www.freepatentsonline.com/4102985.pdf


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How can a compound having a plane of symmetry be Chiral?
The compound( Given in the image below) is treated with $\ce{excess H2 in Pt}$.
Here, (a) - The given compound. (b) - What the answer should be according to me.
(c) - What the actual answer is:

Q1 - How is the chirality center at the top right even a chiral center, as the carbon has 2 identical groups attached to it?

Q2 - How is the compound as a whole a chiral compound? As far as I'm aware, compounds having a plane of symmetry (passing through the two 'chiral' centers in this compound) are supposed to be achiral?

Who wrote diastereomers under c & d. The book? Because if it was you is surprising that you ask this question. See the detailed answer by Lycoris, anyway.

If you draw the methyl group attached to the cyclohexane with a dash, then either a) you do not know (e.g., lack of experimental evidence), or b) you do not want to assign the configuration of the carbon atom of the cyclohexane binding to the methyl group (e.g., because it is irrelevant for the discussion, or you know it is racemic mixture about this carbon atom).  Wedges as in c) or d) however specify the absolute configuration to be either (R), or (S).

I think you are getting confused between chirality and E-Z isomerism in cyclic structures. So let's clear that up first.
According to IUPAC:

chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

Chiral molecules are always dissymmetric (lacking $S_n$) but not always asymmetric (lacking all symmetry elements except the trivial identity). However, asymmetric molecules are necessarily always chiral.
For a basic E-Z isomerism definition:

A cycloalkane has two distinct faces, and any substituent on a ring lies toward one of two faces. When two substituents on a ring point to the same face, they are cis. When the two substituents point to opposite faces, they are trans.

Now, I'll address your second question first:

[OP]How is the compound as a whole a Chiral compound because afaik Compounds having a plane of symmetry(Which passes through the two chiral centers in this compound) are supposed to be achiral.

You're right. This compound does possess a plane of symmetry. However, that only means it can't form enantiomers. It can very well form diastereomers (in this case a pair of E-Z isomerism).
Some further examples to illustrate the point:

The trans isomer is achiral, meaning this structure doesn't have enantiomers. However, it still has a cis diastereomer which can show chirality.
An example closer home:
Image courtesy
I'm sure you must've figured out by now, that the carbon on top isn't a capable of showing optical isomerism. However, due to the dissimilar substituents present ($\ce{H\text{ and } CH3}$), the compound can show cis-trans isomerism.
Further reading:

I believe in the 1,4-dimethyl example, the carbons with the methyl groups are considered pseudo-asymmetric centers, and the isomers can respectively be denoted (1 s,4 s) and (1 r,4 r) in the CIP system.


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What is an example of chemical reaction that can be assisted by both an inorganic catalyst and an enzyme?
I have been researching chemical reactions of inorganic catalysts and enzymes and cannot find a chemical process where an inorganic compound can be replaced by an enzyme (or vice versa) and have the same products. I have looked at article after article with no luck.
What is an example of chemical reaction that involves an enzyme that can be replaced by an inorganic catalyst?

Can't find an example? I'd be surprised if there's any reaction that wouldn't qualify here, at least potentially.

@Mithoron the difficulty is finding a research paper comparing both in a single reaction

Well, you should maybe mention it explicitly in question? But why seek a paper on that? I don't see why would write one, difference is so big, I doubt they would replace one with another.

@Puzzler, you can look for papers on mimics of enzymes for phosphate ester hydrolysis. Most of those try to prepare a small molecule (which you could call inorganic I guess) mimic of the enzyme and then compare it's reactivity to the original enzyme

A good example is the conversion of hydrogen peroxide to oxygen and water
$$\ce{2H2O2 ->2H2O + O2}$$
It can be catalyzed by catalase, which has one of the highest known turnover numbers, or by an inorganic catalyst such as manganese(IV) oxide or lead(IV) oxide.
This has been suggested for instance as an educational project by the University of York (see references).
References

Decomposing hydrogen peroxide. Chemistry Review (1995), Volume 5, Number 1, p. 30. https://www.york.ac.uk/chemistry/schools/chemrev/

There are many many examples of reactions that are usually done in biological systems with an enzyme, but can be done outside in the lab with general acid or base catalysis.
Take for example the hydrolysis of phosphate esters, which form the basis of the links between successive bases on the same strand in DNA or RNA.
 [1]
The general phosphate hydrolysis reaction is:
 [2]
Usually $\mathrm{Nu\!:}\,$ is $\ce{H2O}$ for hydrolysis. Inside the body, the reaction is done by alkaline phosphatase enzymes. You can also carry out the reaction with strong acid or strong base in aqueous solution.
However, the enzyme is extraordinarily fast with a 2nd order rate constant of $\mathrm{1.2\times 10^6\, M^{-1}s^{-1}}$ [2]. A lot of current research is focused on trying to make mimics of the phosphatase enzyme that would be able to cleave DNA outside of biological conditions.
Apart from this, there are other reactions such as carbon dioxide hydrolysis (carbonic anhydrase), starch digestion (amylase) etc. which can be done with acid/base catalysis without the enzyme.
References:

https://en.wikipedia.org/wiki/DNA
J.K. Lassila, J.G. Zalatan, D. Herschlag, Annu. Rev. Biochem, 2011, 80, 669-702

Sucrose inversion (hydrolysis of sucrose into fructose and glucose).
Done with invertase — that's how human body breaks it down and that's how bees make honey.
Done with any acid at elevated temperature (like 100 to 150 °C) — that's how one can make a sugar syrup in the kitchen.

Fixing Nitrogen
What "fixing" nitrogen means it taking the–very unreactive–dinitrogen molecule which is the major component of air and turning it into something that is more reactive and can be used by plants to make, well, more plants (since nitrogen is an essential part of most important biological molecules).
Not every plant can do this (which leads to important discoveries in agriculture like crop rotation). The plants that can fix nitrogen don't actually do it themselves but use symbiotic microorganisms (which, in the case of legumes, live in the plants' roots). These microorganisms use a series of nitrogenase enzymes to do the conversion from the atmospheric nitrogen. Until the 19th century this was the major source of nitrogen for biological systems and it severely limited agricultural yields (there just isn't enough animal poop on the planet and even that, ultimately, is derived from nitrogen fixed by plants).
In the early 20th century a german chemist, Haber, worked out how to do it with a chemical catalyst based on iron and some expensive industrial plant (the other sort of plant, pipes and machinery, not the green photosynthesising kind). The process was improved and commercialised by BASF where Haber worked with Bosch (both got Nobel prizes for their work).
The process turns nitrogen into ammonia which is still one of the major bulk products of the chemical industry and can be turned into many other useful things. Initially the goal was fertiliser. For the first time agricultural yields could be greatly improved without relying on crop rotation or a supply of other existing nitrogen compounds. Of course ammonia is also useful for making nitric acid, essential in explosives, so the world immediately used it to kill many of the extra people it could now feed.
Even today the industrial process is important (we still need fertiliser and we still like to kill each other with bombs). But many useful chemical feedstocks for more benign compounds (plastics, drugs, cleaning agents...) are, ultimately, derived from the process.
So, for the production of ammonia, plants can do it with enzymes and we can do it with industrial catalysts. This is probably the biggest volume chemical where a natural enzyme can be replaced by an industrial catalyst and vice versa.


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Does mass affect rate of reaction?
Suppose I am reacting HCl with CuCO3.
Experiment 1: I use 2g powdered CuCO3
Experiment 2: I use 4g powdered CuCO3
All other variables are kept constant.
Will this have any affect on rate of reaction? Will the 4g react faster because of more particles available for collision?

More powder doesn't mean higher rate but a more saturated solution does.

Higher mass means high surface contact area and therefore higher rate, especially if HCl is in sufficient excess.

$1$)
Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever the order.
For $1$st order, the rate is : ${r = - dn/dt = k_1 n}$. So the rate ${r}$ increases by increasing the amount of reactant ${n}$.
For $2$nd order, the rate of rection is : ${r = - dn/dt = k_2 n^2}$. Here too, the rate of reaction increases when the amount of reactant $n$ increases.
$2)$ As the reactant is a solid, the reaction rate may not be proportional to the amount of reactant (in mole). Let's suppose that it is proportional to the surface of the grains. Even so, the surface increases by increasing the amount of powder. As a consequence, the reaction rate does also increase.


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Does sodium hydroxide react with ozone to produce sodium ozonide?
I read two reactions which I want to specify first:
\begin{align}
\ce{Na2O2 + O3 + H2O &-> 2NaOH + 2O2}\\
\ce{2KOH +  5O3 &-> 2KO3 + 5O2 + H2O}
\end{align}
Does $\ce{NaOH}$ also undergo a further reaction to produce $\ce{NaO3}$ or not? Since both are alkali metals and with relatively similar sizes, I thought it could, but Google doesn't show me anything apart from a reaction balancer site which just shows me the reverse equilibrium of the above reaction.

See: https://chemistry.stackexchange.com/questions/98179/reacting-naoh-in-solution-with-o3

My own Google search does show that sodium ozonide can form [1].  But it requires low temperature:

Sodium ozonide was produced by the ozonization of sodium hydroxide at -80 to -100°[C] followed by extraction with liquid ammonia and removal of the latter under vacuum at-50° [2]. Composition of the product obtained: $\ce{NaO3}=80.1, \ce{NaOH•H2O}=11.2, \ce{H2O}=8.6$ [%].

Upon warming it decomposes to give the superoxide, and then in the presence of water back to sodium hydroxide:

The process of thermal decomposition of sodium ozonide to $\ce{NaO2}$ occurs at −20 and −10°.

At temperatures above 0°, the process of thermal decomposition of sodium ozonide is accompanied by the process of interaction of the $\ce{NaO2}$ formed with $\ce{H2O}$, and decomposition proceeds to the formation of $\ce{NaOH}$.

References
1.
Tokareva, S.A., Pilipenko, G.P. Thermal decomposition of sodium ozonide. Russ Chem Bull 13, 686–688 (1964). https://doi.org/10.1007/BF00845322.
2.
S. A. Tokareva and M. S. Dobrolyubova, Izv. AN SSSR, Ser. Khim. 1964, No. 4.

Yes, indeed. Sodium ozonide is formed in high percentage by the reaction of sodium hydroxide and ozone.
The reaction of ozone and anhydrous alkali metal hydroxides were studied by I.A. Kazarnovskii which confirmed the formation of red colored products known as ozonides. He proposed the following reaction mechanism:

\begin{align}
\ce{2MOH + O3 &-> 2MOH.O3} \tag{a} \\
\ce{2MOH.O3 +2O3 &-> 2MO3.2HO2 + 2O2} \tag{b} \\
\ce{2HO2 &-> H2O + 1.5O2} \tag{c} \\
\ce{MOH + H2O &-> MOH.H2O} \tag{d} \\ \hline
\ce{3MOH + 4O3 &-> 2MO3 + MOH.H2O +3.5O2} \tag{e} \\
\end{align}

So, a mixture of both ozonide and alkali metal hydroxide hydrate is formed as products with the simultaneous transformation of a large amount of ozone to oxygen.
Some information is also there about the structure and chemical properties of sodium ozonide:

Kazarnovskii et al. investigating the reaction of anhydrous sodium hydroxide powder and ozone-oxygen mixtures at -50 to -60°C, observed that, indeed, the hydroxide acquired an intense yellow color which, when extracted with liquid ammonia, resulted in the formation of a dark red solution. The color fades at r.t. Upon evaporation of the ammonia, crystals were recovered. Analysis of the recovered product showed $\ce{90 wt.\% NaO3}$.
Kazarnovskii  further reported that $\ce{NaO3}$ is unstable, and that at room temperature it decomposes.
$$\ce{2NaO3 -> 2NaO2 + O2}$$
They also noted, however, that the ozonization of $\ce{NaOH}$ can be done in room temperature but the $\ce{NaO3}$ obtained from the room temperature reaction is not soluble in liquid ammonia, and the ozonide-hydroxide mixture is quite stable. Subsequent studies have explained the difference in the solubility and stability
characteristics of $\ce{NaO3}$ prepared at room temperature and low temperatures. The reason proposed that $\ce{NaO3}$ can exist in two crystalline forms, one being soluble in liquid ammonia but unstable at room temperature, and the other being insoluble and stable although not confirmed experimentally.
The $\ce{NaO3}$ lattice is tetragonal, with $a = \pu{11.61 Å}$ and $c = \pu{7.66 Å}$. The space group is 14/mmm  Its density varies from $\pu{1.56-1.60 g cm-3}$ . The standard heat of formation of $\ce{NaO3}$ is estimated as $\pu{45 kcal/mole}$.

Lenntech suggests that when ozone is exposed to aqueous sodium hydroxide it forms dioxygen species (peroxide, superoxide).  Might that be occurring in the room temperature reaction you quote here?

@OscarLanzi Essentially yes. If the stoichiometry of the reaction is altered or the reaction is done in r.t., various other species is noted to form, especially the alkali hydroxide hydrate $\ce{MOH.H2O}$ which being unstable at r.t. form $\ce{HO2.}$ radical along with side products like superoxide and peroxides. So, for optimal results, the reaction is done in sub-zero temperatures where the majority of product is the ozonide and oxygen.


text:
What are counterion effects?
I was reading the following article on wikipedia about Hydroponics

Unlike most nutrients, plants can tolerate a high concentration of the
$\ce{SO4^2−}$, selectively absorbing the nutrient as needed.
Undesirable counterion effects still apply however.

Can somebody explain me what is meant by counterion effects? Is it just the presence of the counterion, i.e. the concentration of the counterion in solution?

Consider general effects as osmotic pressure and tonicity, and ion specific effects, depending on biochemical ion influence on plant cells.

Here is a passage from Marschner P. (2012). Marschner’s mineral nutrition of higher plants. 3rd Edn. London: Academic Press:

In the context of the wikipedia table, it means that while you can have a high concentration of sulfate, you should consider the cation of the substance (e.g. calcium sulfate, potassium sulfate, ammonium sulfate) as well when trying to figure out the effects on a plant.


text:
Why is the aqueous solution of KMnO4 slightly basic?
This is in the context of the various changes in oxidation state the permangananate ion can undergo with the variation of the pH of the medium. Besides the cases of strongly basic and strongly acidic media, the case for neutral media (with reduction to $\ce{MnO2}$) in my notes has a footnote; that this medium is in fact slightly alkaline, due to it being a solution of $\ce{KMnO4}$.
I tried to reason this out. It looks like $$\ce{K+ + MnO4- + H2O -> K+ + OH- + HMnO4}$$ is how it works out.
A quick google search establishes the pKa of HMnO4 to be -2.25. So it makes a pretty good acid in comparison to water.
Thing is, in that case, the product side should have an $\ce{H2O}$ instead of the constituent ions, with the potassium permanganate dissociated. That means the solution's neutral. Which apparently it isn't, going with my notes.
Where did I mess up here?

The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does not work with the latter.
$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$ (in molten KOH solvent; reactants are fused together)
$\ce{2 K2MnO4 + 2 H2O → 2 KMnO4 + 2 KOH + H2}$ (electrolysis in alkaline media; a direct chemical reaction would convert some material back to $\ce{MnO2}$)
Samples are of course purified but must be assumed to contain some leftover alkali from this preparation. Some alkali may also be left in solution on purpose because potassium permanganate is most stable in alkaline solution and shelf life would be improved by providing a buffer against absorption of carbon dioxide from the air.

Excess alkali is probably a good thing, as acidic permanganate would degrade to oxygen and MnO2, making the KMnO4 a little dirty-looking in solution.

I assume alkalization is due the decomposition. $\ce{4 KMnO4(aq) + 2 H2O -> 4 KOH(aq) + 4 MnO2(s) + 3 O2(g)}$


text:
Why in positive deviation from Raoult's Law partial vapour pressure of a liquid component is greater than what is predicted by Raoult's Law?
When my teacher was teaching me positive deviation from Raoult's Law, he told me that in this case
$P_{solution}>P_A^°χ_{A} + P_B^°χ_{B}$
He told that here the solute-solvent interactions are weaker than the solute-solute interactions and solvent-solvent interactions due to which more vapours can be formed from the solution and hence the observed vapour pressure is greater than the theoretical vapour pressure. I was pretty satisfied with the answer and was happy. But after that he told me that in this case, $P_{A} > P_A^°χ_{A}$ and $P_{B} > P_B^°χ_{B}$. He did not give any explanation about it. Okay, the previous explanation makes sense, but how are the above two equations possible? I mean in positive deviation more vapours are formed because solute-solvent interactions are weaker. But there is no change in the interactions between solute-solute and solvent-solvent. So, why are $P_{A} > P_A^°χ_{A}$ and $P_{B} > P_B^°χ_{B}$ true? Shouldn't these equations actually follow Raoult's Law and follow $P_{A} =P_A^°χ_{A}$ and $P_{B} = P_B^°χ_{B}$? Please someone explain as to why $P_{A} > P_A^°χ_{A}$ and $P_{B} > P_B^°χ_{B}$? I am so confused. Please help.

Baruah I don’t think (Maybe) you actually understood what a solution is in first place, take a solution of A and B, here molecules of B are surrounded by not only B but A as well. Now consider an ideal solution of A and B components$\ce{Interaction_{B-B}=Interaction_{A-B}}$. So here vapour pressure of B is as expected according to its mole fraction in the solution. But for a positive deviation we know that interaction of A-B will be less, thus Since B is also surrounded by A molecules as well. Thus B will have more vapour pressure than expected.

Also the inequality you seem to accept implies the second one, at least for one term.


text:
Why is a C-N bond weaker than a C-C?
As far as I could remember, in the same period, a bond to a more electronegative atom implies in a more polar and stronger bond, due to the greater ionic character. However the bond dissociation energy for the C-X series follows the order: C-F > C-O > C-C > C-N. The same for H-X: H-F > H-O > H-C > H-N.
Why are the bonds to nitrogen weaker than expected? (Or, more speficically, weaker than bonds to carbon, where no dipole is formed)
Values of average bond enthalpies I based my question off:

Bond
BDE (kJ/mol)

H-F
565

C-F
484

H-O
463

C-O
360

H-N
388

C-N
305

H-C
412

C-C
348

Sources: Atkins' Chemical Principles, Housecroft's Inorganic Chemistry, http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html

By the way, I found https://chemistry.stackexchange.com/questions/85767/why-nitrogen-bond-is-stronger-than-carbon-bond but I don't think the answer is satisfactory

I think it has to do with Nitrogen being half-filled which is a rather stable state.

Where did you get your BDE data from? The series you've presented have little in common with the values reported in 97th edition of CRC handbook of chemistry and physics (p. 9-73):
$$
\begin{array}{lr}
\hline
\ce{A-B} & D^\circ_{298}/\pu{kJ mol^-1} \
\hline
\ce{C-F} & 513.8 ± 10.0 \
\ce{C-O} & 1076.63 ± 0.06 \
\ce{C-C} & 605.03 ± 0.28 \
\ce{C-N} & 749.31 ± 0.14 \
\hline
\ce{H-F} & 569.680 ± 0.011 \
\ce{H-O} & 429.74 ± 0.03 \
\ce{H-C} & 338.72 ± 0.11 \
\ce{H-N} & 358.8 ± 0.2 \
\hline
\end{array}
$$

I'm sorry, but 1000 kJ/mol for a C-O bond is absurd, that may be the value for carbon monoxide. I took a look at Housecroft and Atkins Chemical Principles, and I will update the question with values

Of course the table is for diatomic molecules. I'm not sure it makes sense to compare average BDEs for random polyatomic species. Not only the values are less accurate, but also the difference in bond energies for the same bond for for various compounds may vary in hundreds $\pu{kJ mol^-1}.$ I suggest you either compare series of equivalent polyatomic molecules, or stick with diatomic ones from the table I provided. Otherwise it's no different than comparing average patients' temperatures at the hospitals (persons in the morgue and the ones with fever contributing to theaverage 36ish °C).


text:
Assigning E–Z configuration for alkenes with cyclic substituent groups
I was taught that according to CIP rules, we must check the priority order for the substituents. But how can we do the same for molecules with cyclic groups as substituents?
How do we compare the substituents for the following molecule?

How to determine E–Z configuration for such molecules?

Follow the same CIP as straight chains.

How do I break the cyclic compound though?

I feel more and more that many teachers don't really know what to teach. This exercises with no didacticak value resemble sudoku and staff like that. They should be reserved, quite paradoxically, to future chemists and later in the course of their studies and not part of a general teaching chemistry. Even a professional organic chemist dig into this just because a IUPAC acceptable name has to be given after isolation and in publication. But that is.

The compound is a tetra-substituted alkene with four different groups. Thus, it could be a (E)- or (Z)-ethene. To find the priority of the groups, you must follow CIP rules. The four groups are cyclobut-1-enyl, cyclobut-2-enyl, 2-methylcyclobutyl, and 1-methylcyclobutyl. Since it is an alkene (planner sp2–sp2 bond), we don't have to consider all four groups at once. What you have to do is consider two groups at once on each side of the double bond (say they are a and b):

On side a the first carbon of each cyclobutyl group (cyclobut-1-enyl and 1-methylcyclobutyl) has equal priority ({C,C,C} and {C,C,C}) as indicated in the image. However, the second carbon of cyclobut-1-enyl group has the priority {C,C,H}, while the second carbon of 1-methylcyclobutyl group has the lower priority {C,H,H} compared to that in cyclobut-1-enyl group. Therefore, between these two groups, cyclobut-1-enyl group has the higher priority.
On side b the first two carbons of each cyclobutyl group (cyclobut-2-enyl and 2-methylcyclobutyl) have equal priority: First carbons of both rings have {C,C,H} and the second carbons of both rings have {C,C,H} as well (see the image). Yet, the third carbon of cyclobut-2-enyl group has the priority {C,C,H} while the that of 2-methylcyclobutyl group has the lower priority {C,H,H} compared to cyclobut-1-enyl group. Therefore, between these two groups, the cyclobut-2-enyl group has a higher priority.
Now, you see higher priority groups of a and b sides are oriented on the same side of the double bond. Therefore, the compound is (Z)-isomer. The name is (Z)-1-(cyclobut-1-enyl)-2-(cyclobut-2-enyl)-1-(1-methylcyclobutyl)-2-(2-methylcyclobutyl)ethene.
Also, note that this molecule also has three chiral centers as indicated in the image. However, their stereochemistry is not given, so I neglect them.

Complementary to this answer, see e.g., this and this example on chemistry.se.

"the second carbon of cyclobut-1-enyl group has the priority {C,C,H} while the second carbon of 1-methylcyclobutyl group has the lower priority ({C,H,H}) compared to that in cyclobut-1-enyl group"
How did you find which side to go to for the "second" molecule, when the first molecules had the same priority. You went clockwise for the top case and to another substituent for the bottom one while comparing priorities for the second molecule.
I don't understand on what basis you did those.

@Natru You go both ways around, treating each direction as a separate branch, then make the comparison based on the higher-priority branch. I've tried to illustrate it with this image: the red atoms are the ones which distinguish the ligands, and the blue atoms are the ones which must be compared to find the difference.

@LegionMammal978 wow I found the image really helpful, great visualisation. Thanks!

Can you send the link to any famous book or wiki page or article on the splitting into tree diagram?

Natru: They  are called digraphs. Look here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/pinene.html


text:
Ambiguity in the direction of dissolution equilibrium
The following reaction is usually carried out to test for the carbonate anion in an inorganic salt;
$$\ce{CO3^2- +BaCl2/CaCl2 -> BaCO3/CaCO3(s) + 2 Cl-}$$
The resulting carbonate is stated to be an insoluble precipitate, and also to be dissolvable in mineral acids like $\ce{HCl}$.
How does that happen? If the initial test is to actually work, the dissolution equilibrium should lie to the right; in the favour of the carbonate being formed. The dissolution requires the metal ions in the carbonate's solid lattice to break up and get solvated by the water molecules, ending up just like the strong electrolyte that we started with($\ce{BaCl2}$ or $\ce{CaCl2}$).
If you're going to just take some hydrochloric acid and treat the precipitate with it, and the reaction we just saw works out in the reverse, what does that mean about the equilibrium constant? It's supposed to lie to one side no matter what you start with, right?

The ion $\ce{CO3^{2-}}$ introduced in the first reaction must be produced by dissolving a soluble carbonate like $\ce{Na2CO3}$. On the other hand $\ce{BaCl2}$ or $\ce{CaCl2}$ have to be dissolved in water to react. If they are introduced as solid these substances will not react, unless they slowly get dissolved. Nobody really knows why these chlorides are soluble in water, and the corresponding carbonates are insoluble. So the precipitation reaction should be rewritten as $$\ce{Ba^{2+} + CO3^{2-} <=> BaCO3(s) \ (1)}$$ $$\ce{Ca^{2+} + CO3^{2-} <=> CaCO3(s) \ (2)}$$ Usually the equation is displaced to the right side, and a mixture of some carbonate plus a solution containing Ba or Ca produces a dense precipitate  of carbonate. But by adding an acidic solution, containing the ions $\ce{H+}$ (or $\ce{H3O^+}$), the following reaction happens that consumes the acidic ions : $$\ce{2 H+ + CO3^{2-} -> CO2(g) + H2O}$$ Here the ions $\ce{CO3^{2-}}$ are all destroyed and produce a gas which leaves the solution. This displaces the first equilibriums ($1$) and ($2$) to the left-hand side, and the precipitates disappear.

Oh, so this is how dissolution actually works; it's not the dissolution of the solid substance, but the reaction with its solvated ions that causes the precipitates to disappear using Le Chatelier's principle. Can this be generally stated for all dissolutions?

That said, I guess what I did wrong in my answer was taking the dissolution to be just "precipitate+acid". Is that correct?

@Harry Holmes. Honestly my explanation explains the observed phenomena. But this does not prevent your explanation to be as valid as mine. Who knows ?

On writing the equations for the initial precipitation and the dissolution reactions, it can be seen that their equilibrium constants are not the same. Taking $K_1,K_2$, and $K_3$ to be the constants for each of these and the double dissociation of $\ce{H2CO3}$ respectively, you get $$K_2=\frac{1}{K_1K_3} \neq K_1$$ This is why you can't reason out that the equilibrium in both reactions should be to one side; they're not the same reaction.


text:
Comparing the degree of dehydration
I am confused regarding the following problem  which asks to arrange the given compounds in decreasing order of their degree of dehydration

My attempt at the solution:
I tried to find out the products and compare their stability:

Now in the first case product is conjugated. So I think it is the most stable.
The second one has 6 alpha hydrogen while the third one has only 3 alpha hydrogen.
Hence order should be 1>2>3
But the answer given is 1>3>2

My second confusion is whether such alpha-hydroxy ketones will follow the E1 or E2 dehydration process. I thought that E1 elimination won't be favourable cause the positive charge next to the keto group will be destabilised. Is it correct?

What is "decreasing degree of dehydration"? Do you mean rate?

Yes in other words "the rate or ease of dehydration"

@Arpan For considering rate of these reactions you have to compare the rate of intermediate steps, and not consider the thermodynamic stability of the final products. Here, rather than formation of carbocation you should consider, the relative rate of abstracting the acidic hydrogen. Since in 1 the hydrogens are acidic due to strong inductive effect(negative) thus the rate for $1st$ is highest, and $2nd$ has the least acidic hydrogens thus $1>3>2$

@Rishi Thanks. I got your logic.Are we doing so(not comparing the carbocation stability) cause this process does not follow E1 pathway?

As a general rule in organic chemistry problems, rate (a kinetic concept) should only be tied to thermodynamics by the appropriate application of the Hammond Postulate.


text:
Hydration vs Dissolution
While revising Thermodynamics and Thermochemistry from revision material published by Arihant publications I encountered the following mentioned statement.

During dissolution, the physical state of the compound changes while during hydration, there is no change in the physical state of the compound.

How does dissolution result in a change in physical state?

Should not hydration (or solvation in general) be the same as dissolution as in both the bonds of solid salt to be dissolved or solvated weaken and a new bond is formed between constituents of salt and the solvent? Could someone please explain what I'm missing?

Solvation starts from gas phase (separated) molecules or atoms. See e.g. https://opentextbc.ca/chemistry/chapter/11-1-the-dissolution-process/

Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed.
Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has not changed : it is still a solid –

Hydration has 2 meanings. One is the hydration provided in this answer, e.g. $\ce{CaCl2(s) + 2 H2O(g) -> CaCl2 . 2H2O(s)}$. The other one is a part of dissolution, mainly of ionic compounds, wrapping "naked" ions by oriented layers of water, being a special case of solvation.

I indeed agree with @Poutnik 's comment. For example, differential hydration energy indeed refers to the energy release in solvating 1 mole of salt in excess of water. So in such a case is there a way ionic salt can stay a solid, or be even capable of maintaining a strong bond between both the ions?


text:
Retention Time of Isotopologues
In several sources I have read that heavier isotopologues eluate earlier than lighter ones.
As explanation vapor isotope effects are given.
But this is something I do not understand. I tought in a GC column substances are separated by their different vapor pressures and the longer a substance stays in the gas phase the earlier it eluates. Heavy isotopologues have have a smaller vapour pressure and should therefore eluate later.
What I am missing?

You may miss the shift of the distribution constant between phases.

Could you please explain this a little bit more.

Imagine there is a polar and nonpolar compound with the same saturated vapour pressure, having nonpolar SF. Which would have the longer elution time ? Now generalize for compounds with different distribution constant between MF/SF, or the equivalent Henry constant.

Isn't the boiling point what matters.

Think longer.... Imagine you have initially 2 containers with the same initial analyte vapour pressure. Then you insert there the SF. Both analytes have different KD. Will they have the same final vapour pressure ?

Sorry what does KD mean?

$K_\mathrm{D}=\frac {c_\mathrm{SF}}{c_\mathrm{MF}}$ means distribution constant.


text:
Carbocation stability + determining position of attack on a carbocation in SN1 reaction
It's implied that the right structure is the more stable carbocation because it undergoes the second half of an SN1 reaction, but I'm not entirely sure why:

I'm also not sure why the $\ce{MeOH}$ attacks the carbon next to the carbocation and not the oxygen itself, where it has the positive charge (doesn't the nucleophile attack the carbocation, or whichever atom bears the positive charge?)

the carbocation is in reality the first structure, you have an empty orbital in (left) over the right structure. the nucleophile attacks this empty orbital

@SafdarFaisal That makes a little more sense, although my professor says that the nucleophile attacks "the more stable carbocation"
Also when you mean empty orbital, you mean 2p, correct?

It's a mistake to think of these two as different structures. They are representations of the same thing as indicated by the double-headed arrow. Note that where the positive charge is written is only the formal charge, not the real charge. The charge here happens to be more localized on the carbon.

@Zhe I understand that thank you, I just don't know why the methanol attacks the sp2 carbon in the right structure instead of the carbon with the +1 formal charge on the left structure. An SN1 reaction has the leaving group leave, forming a carbocation, and the nucleophile "filling" in that carbocation

Again, the confusion here is to think that the left structure and the right structure are different. They are not different. It's the same structure. The nucleophile reacts with the carbocation on the left for the same reason that it reacts with the oxocarbenium ion (similar to a carbonyl) on the right. They are two views of the same thing.

It is important to note that the two structures shown aren't actually different. That's what resonance implies that in reality, it exists in between both of the as a 'hybrid'.
So it is not right to state that the nucleophile attacks on the left structure 'or' the right.

Coming to the carbon, (again, it's not right to take the structures to be different)

"I just don't know why the methanol attacks the $\ce{sp2}$ carbon in the right structure instead of the carbon with the $+1$ formal charge on the left structure"

It's actually the same carbon! and the hybridisation is still $\ce{sp2}$ no matter what, as the carbocation is also a planar species($\ce{sp2}$). And this is where the attack occurs.

"I'm also not sure why the MeOH attacks the carbon next to the carbocation and not the oxygen itself, where it has the positive charge"

'Carbon next to the carbocation' ? I believe you are confusing the +1 formal charge on oxygen to be a carbocation which is not true( Carbocation has the positive charge exclusively on Carbon) .
The attack is ocuring at the carbocation.

Our nucleophile here, is $\ce{OH-}$ while oxygen only has a +1 formal charge and also, the positive charge is more localised on the carbon as Zhe mentioned.
This is because oxygen being a more electronegative species, doesn't really 'like' the positive formal charge on it.
But still,

"doesn't the nucleophile attack the carbocation, or whichever atom bears the positive charge?"

Well, a nucleophile is a species that donates electrons at electron deficient sites. But oxygen still has it's lone pair! and the $\ce{OH-}$ will face repulsions from them if it tried to attack here.
The real electron deficient species here is only the carbocation and thus it is the preferred site for attack by the nucleophile.


text:
Determining the most stable enolic form of 1-phenylbutane-1,3-dione?
Question

Which form of 1-phenylbutane-1,3-dione is more stable?

Answer

3-hydroxy-1-phenylbut-2-en-1-one

My question
I've been told that between the above two, 3-hydroxy-1-phenylbut-2-en-1-one is more stable because the double bond is in the middle which is unclear to me.
Could someone explain the above statement?

In opposition to the line of thought presented, I argue that the 4-hydroxy isomer is more stable than the 3-hydroxyl form for that the conjugated $\pi$-electron system is larger, than in the former.
This is based on multiple assumptions.

For one, given a molecular framework which may have multiple $\ce{C=C}$ bonds, if other parameters are not altered, the system with double bonds in conjugation is more stable than the one with isolated double bonds. As an example, the computed free enthalpy of formation, $\Delta{}_RG$ about 1,5-hexadiene (two isolated $\ce{C=C}$ bonds) is $\pu{92.70 kJ/mol}$ (link to computation on MolCalc.org), the about 1,3-pentadiene (two $\ce{C=C}$ bonds in conjugation) $\pu{86.57 kJ/mol}$ (link to computation).  And the less endothermic a compound, the more thermodynamically stable.

For the other, I said the 3-hydroxyl isomer would contain a less extended conjugated $\pi$-system, than the 4HP isomer.  Elements to participate in this system are the benzene ring, the carbonyl group, and the simple $\ce{C=C}$ bond, present in both.  If you imagine a water wave running from one to the other other end of the $\pi$-system, $\ce{C=O}$ only is tangent in 3HP, while in 4HP the wave runs all across these bonds.  Thus, equally a reason why the 4HP isomer is more favourable, than 3HP.
For a raw simplification, you may consider the benzene ring as "super atom" and count the longest pathway accessible. 3HP: 4 atoms, 4HP: 5 atoms participate in the $\pi$ system; again energetically more favourable for 4HP, than for 3HP.

Sadly, the engines to run MolCalc.org's web site constrain the computations to molecules with 10 non-H atoms maximum.  Thus, contrasting to earlier posts by others (e.g., this by Ben Norris), I'm not able to support my arguments by a similar computation for your substrates (12 non-H atoms).

I understand that tangent concept you presented, but due to the presence of the carbonyl group the C attached to the group in 3HP is polarized and has a partial positive charge on it, which I think will let it participate in resonance with the benzene ring.

You can find a similar resonating structure in Benzoic acid too

Kindly elaborate :(

The carbonyl group will polarize the $\pi$ system (EWG), as does the phenyl ring (EDG). In 3HP, these two effects partially annihilate each other with lesser interaction of the C=C bond than in 4HP; in the later, this push and pull is right across the C=C bond.  The resonance the mesomeric formulae conceptualize for the benzoate anion do not extend the dimension of the $\pi$ system; here, isomerism however affects this dimension.

Buttonwood, your argument makes sense.  This is a case of cross-conjugation (sometimes called Y-conjugation) vs. linear extended conjugation, the latter is generally energetically preferred.  Perhaps you could run MolCalc with a double bond replacing the aromatic ring?

@ron The idea to simplify Ph to C=C is appealing. However with the polar groups present (intramolecular binding / orientation of OH to C=O shown in the textbooks seems missing in the 2nd form), I'm not sure if the minima found are the global ones, or close to GIGO. 3HP_B, $\pu{-260.26 kJ/mol}$; 4HP_B, $\pu{-232.03 kJ/mol}$.  It is published in 2013JChemEduc1093 with GAMESS' PM3 implementation.


text:
Equilibrium constant expression for an acid base reaction
While reading an organic chemistry textbook I saw the following Formula for calculating equilibrium constant of an acid-base reaction
Consider following acid base reaction  in which $\ce{HA}$(reactant acid) and $\ce{HB+}$(product acid) are acids in forward and backward reaction respectively.
$$\ce{HA + B <=> A- + HB+}$$
$$\mathrm pK_\mathrm{eq}= \mathrm pK_\mathrm {a_\mathrm{reactant}}-\mathrm pK_\mathrm {a_\mathrm{product}}$$
$$\implies K_\mathrm{eq}=\frac{K_\mathrm{a_\mathrm{reactant}}}{K_\mathrm{a_\mathrm{product}}}$$
Now I want to show  above  expression is equivalent to $$K_\mathrm{eq}=\frac{\ce{[A-][HB+]}}{\ce{[HA][B]}}$$
So I have to write expressions of $K_\mathrm{a}$ of both acids to proceed further.
The way $K_\mathrm {a}$ is defined in my textbook is like this

$$\ce {HA + H2O <=> A- + H3O+}$$
$$K_\mathrm {a}= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

So In our acid base reaction we should write expressions of $K_\mathrm {a}$ of acids in a similar way as we have no other option left.
$$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$
$$\ce{HB+ + H2O <=> B  + H3O+}\tag{2}$$
The above 2 equations I have written are for defining $K_\mathrm{a}$ of acids ($\ce{HA}$ and $\ce{HB+}$).
They are not related to our main acid-base reaction in any way.
As people are pointing out in the comments, concentration of $\ce{H3O+}$ should be same in expressions of $K_\mathrm a$ as they are in the same solution but I have already pointed out that equation 1 and 2 are written only for the sake of defining $K_\mathrm a$ of acids.
They are altogether different from our main acid base reaction. Then the concentration of $\ce{H3O+}$ can be different in equations 1 and 2 as they are two different solutions for determining $K_\mathrm a$
$K_\mathrm {a}$ of $\ce{HA}$ which is the reactant is $$K_\mathrm a = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}}$$
$K_\mathrm {a}$ of $\ce{HB+}$ which is the acid being produced is $$K_\mathrm b = \frac{\ce{[B][H3O+]}}{[\ce{HB+}]}$$
On substituting  respective $K_{a}$ into the obtained expression we get:
$$K_\mathrm {eq}=\frac{\ce{[A-][HB+]}}{\ce{[HA][B]}}$$
But during this process of substitution we are required to cancel $\ce{[H3O+}$ term in numerator and denominator
My question is:
How can we cancel the term of concentration of $\ce{H3O+}$ as concentration of protonated water in expression of $K_\mathrm {a}$ of reactant and product can be different?

The issue as I can see it here is a simple misunderstanding that can rectified:
$K_\mathrm a$ is a constant even if $\ce{[A]}$ is not. (which is how it was defined)
The definition of $k_\mathrm a$ is as follows:

$K_\mathrm a$ , the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution.

Nowhere does this definition state that the concentration taken for water has to be constant, in fact, only $K_\mathrm a$ is constant.
Once we find the value of $K_\mathrm a$, the value cannot change for a given temperature. This means that no matter what solution you add the acid to - in the end, the value of $K_\mathrm a$ remains constant. i.e:
$$K_\mathrm a = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}} \tag{constant}$$
Now, in your solution, the value of $\ce{H3O+}$ may not be the same as in the place where we determined the $K_\mathrm a$ in the beginning but that only means that are $K_\mathrm a$ (which is still constant) is the same ratio but with different numbers.
Now, as to the reason why $\ce{[H3O+]}$ can be cancelled. Imagine you take a glass of water and add two acids $\ce{HA}$ and $\ce{HB}$.
What is the concentration of protonated water when we try to find the $K_\mathrm a$ of these acids in this solution, we get:
$$K_{\mathrm a_\mathrm A} = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}} \tag{1}\label{1}$$
$$K_{\mathrm a_\mathrm B} = \frac{\ce{[B-][H3O+]}}{\ce{[HB]}} \tag{2}\label{2}$$
Now what is $\ce{[H3O+]}$ in ($\ref{1}$) and ($\ref{2}$)?
They will be the same since you are measuring the $K_\mathrm a$s in the same glass of water.
Therefore you can cancel $\ce{[H3O+]}$ out since the value is same for the one glass of water that you have.


text:
A possible reaction of heating Terephthalic acid

There is no reaction of terephthalic acid heating effect on the internet (at least Wikipedia). I have written the most probable reaction possible as per my knowledge. Please tell if it is spontaneous and possible or not. Also, give real conditions of doing the reaction if possible.

It looks rather challenging to get those oxygens and hydroxies to face each other long enough for any dehydroxylation to happen. Like herding cats

Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861:

On heating, terephthalic acid sublimes without previously
fusing(melting). The sublimate, which is indistinctly crystalline, has
the same composition and properties as the original acid, and
therefore, unlike other bibasic acids, terephthalic acid cannot be
converted into an anhydrite by merely heating it

Terephthalic acid when pure is white opaque powder, insoluble in water and it sublimes at $\pu{300^{\circ}C}$ without melting.
When heated to decomposition it emits acrid smoke and irritating fumes. $\ce{FeOOH}$, nickel hydroxide and nickel oxide are suitable catalyst for thermal decomposition of terephthalic acid (at low temperature) but since nickel compounds are expensive so $\ce{FeOOH}$ is generally used. At higher temperature (in absence of catalyst) Terephthalic acid$\ce{_{gas}}$ is decomposed to carbon dioxide,benzene and benzoic acid.
For further reading refer to this more in depth paper
"Determination of the Thermal Decomposition Products of Terephthalic Acid by Using Curie-Point Pyrolyzer" by Mehmet uluturk

+1 Thanks for the answer and welcome to Chem.SE. Please note that $\ce{...}$ is used to render chemical formulae and equations, as they are not supposed to be in italic.

@Jay Heating without catalyst, catalyst are only required at low temperature for this reaction.

@user1055 answered the question pretty much. I am just going to add a complementary answer which has the abstract to the paper which @user1055 mentioned. The abstract of the paper mentions the various products of the decomposition reaction:

The thermal decomposition behavior of terephthalic acid (TA) was
investigated by thermogravimetry/differential thermal analysis
(TG/DTA) and Curie-point pyrolysis. TG/DTA analysis showed that TA is
sublimed at 276°C prior to decomposition. Pyrolysis studies were
carried out at various temperatures ranging from 160 to 764°C.
Decomposition products were analyzed and their structures were
determined by gas chromatography–mass spectrometry (GC-MS). A total of
11 degradation products were identified at 764°C, whereas no peak was
observed below 445°C. Benzene, benzoic acid, and 1,1′-biphenyl were
identified as the major decomposition products, and other degradation
products such as toluene, benzophenone, diphenylmethane, styrene,
benzaldehyde, phenol, 9H-fluorene, and 9-phenyl 9H-fluorene were also
detected. A pyrolysis mechanism was proposed based on the findings.

Reference

A. Begüm Elmas Kimyonok & Mehmet Ulutürk (2016) Determination of
the Thermal Decomposition Products of Terephthalic Acid by Using
Curie-Point Pyrolyzer, Journal of Energetic Materials, 34:2,
113-122, DOI: 10.1080/07370652.2015.1005773


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Equilibrium in Gas Phase and Pressure
In one task/example there is gas phase reaction and it is said that reaction takes place at pressure of 1 bar and standard Gibbs energy of reaction is needed to be calculated. In this task 1 bar is meant to be equilibrium pressure of a mixture. Problem I have is that total pressure initially is less than 1 bar since number of moles is increasing in reaction, but if so why do we calculate standard reaction Gibbs energy since such quantity is applicable for predicting equilibrium conditions at constant pressure? How can standard reaction Gibbs energy predict equilibrium state if pressure during reaction is changed?

Didn't you just say that reaction takes place at pressure of 1 bar? This typically means that the pressure was 1 bar both before and after the reaction. The changing number of moles is irrelevant.

Number of moles of gas being changed doesn't necessarily mean pressure is being changed.  If the container were at constant V, then, yes, p would have to change. But if you are conducting the experiment in an open container (a typical scenario), the pressure is always going to be determined by the ambient pressure (i.e., the pressure in your lab), which will remain constant at about 1 bar (assuming your lab is at about sea level).

Actually, this is not the case. While at equilibrium pressure is 1 bar, I needed to calculate what was initial pressure of reactant because I need to calculate equilibrium constant (Kp). After doing famous analysis in equilibrium tasks: initial, change and equilibrium, what I've gotten is that initial pressure of a reaction mixture is lower than equilibrium one.

@theorist, yes I agree with you. But, in this task after doing famous analysis in equilibrium tasks: initial, change and equilibrium, what I've gotten is that initial pressure of a reaction mixture is lower than equilibrium one.

If that happened, then the reaction by definition wasn't being conducted at constant pressure. (Of course real reactions are not conducted at exactly constant pressure, but a typical open container in a lab that isn't, say, exploding will keep its pressure close to the ambient pressure.)

@Ian Well, if this is so (and I agree with you), how can standard reaction Gibbs energy predict equilibrium condition since in the task I am asked to calculate it?

You're asked to calculate what? The Gibbs energy change? Or the equilibrium pressure? The equilibrium pressure is just the pressure that the reaction is held at from outside (to within this approximation). Under that condition (and constant temperature also), you can define and calculate the Gibbs energy change.

Standard reaction Gibbs energy from equilibrium constant (Kp) through famous expression, deltaG(standard) = -RTln(K). Equilibrium pressure of a mixture is needed along with dissociation degree to calculate partial pressures initially and in equilibrium since I need these to calculate Kp.

I may be misunderstanding the experimental design that this computation is meant to serve. Is it a gas phase reaction in a sealed container whose pressure is held fixed by a mobile piston?

@ Ian, you didn't, I didn't explain the task in detail. So, there is a dissociaton of N2O4 to 2NO2. What you asked is  reaction happening in a sealed container with mobile piston isn't specified. Equilibrium pressure (1 bar) and degree of dissociation of N2O4 are given. Standard reaction Gibbs energy is needed to be calculated. To do so we need Kp, for which we need partial pressures at equilibrium. To find equilibrium pressures we need to do standard initial, change and equilibrium analysis (Do you know what I mean?).

After you do this analysis, it turns out that pressure of reaction mixture in beggining (when you have only N2O4)  is not the same as equilibrium. If so, why do we calculate standard reaction Gibbs energy since it isn't applicable to predict equilibrium in these conditions? We can use this quantity to predict equilibrium if this reaction is carried at 1 bar from start to equilibrium which isn't the case in our task. So, as you said, we need to increase the volume of our system to keep the system at constant pressure since number of moles is increasing.

If the system is actually held at constant pressure, such as by conducting the reaction in a sealed container with a mobile piston with a fixed external pressure on the piston, then Gibbs energy calculations can be used to predict equilibrium conditions. This also approximately works if, say, the reaction is in a sealed container and the total pressure of the reactants/products is much less than the fixed pressure from the other gas in the container.

@ Ian, yes. In this example, volume was probably held constant since pressure is changing, but I was still asked to find Gibbs energy since I can use it to predict equilibrium, but in conditions of constant pressure at equilibrium, not in conditions of a task.

You can compare the actual values of the Gibbs energy as a state function regardless of anything, though again in order to use this to predict equilibrium conditions, you need to impose external pressure and external temperature control on the system. I still think you are most likely misunderstanding something about the situation being modeled.

As I said, task doesn't specify any kind of conditions that you asked me like is it sealed container or does it have mobile piston. Given the equilibrium pressure of a mixture and dissociation degree for specific reaction, calculate Kp and standard deltaG at 298 K. That is all.

OK, then if you want to justify why you would do such a calculation you can imagine a mobile piston scenario where the volume of the system increases as you run the reaction so that constant pressure is maintained. As for actually doing the calculation, the partial pressures are just modeled as the associated mole fractions of each gas in the system times the total pressure.

You wrote: "While at equilibrium pressure is 1 bar, I needed to calculate what was initial pressure of reactant because I need to calculate equilibrium constant (Kp)."  Based on this, I think you are confusing the total pressure (1 bar) with the partial pressures of the individual chemicals involved the reaction.  The partial pressures of the individual chemicals will change, of course, as the reaction proceeds.  But if the reaction is in an open container, the total pressure at the end will be the same as the total pressure at the start, since that's just the ambient pressure.

Yes, I figured everything out as I was talking with others here. Thank you.


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Meta-directing nature of anilinium cations
The comments on this answer seem to detail a trend in which $\ce{NR3+, NHR2+, NH2R+}$ all show meta-directing nature towards electrophilic substitution, with the slightly stronger para-direction of $\ce{NH3+}$ being the odd one out, as cited from March's Advanced Organic Chemistry.
The 2020 edition has this to say, on pages 617 and 618 under "Aromatic Substitution, Electrophilic" (emphasis mine);

Groups that lack an unshared pair on the atom connected to the ring are electron
-withdrawing with a (+) charge or (δ+) dipole. In this category are, in approximate order of decreasing deactivating ability, $\ce{NR3+}$, .... and $\ce{NH3+}$. Also in this category are all other groups with a positive charge on the atom directly connected to the ring... and many groups with positive charges on atoms farther away, since these are often still powerful −I groups. The field-effect explanation predicts that these should
all be meta directing and deactivating, and (except for $\ce{NH3}$
+) this is the case. The $\ce{NH3+}$ group is an anomaly, since this group directs para about as much as or a little more than it directs meta, often because the ammonium ion is in equilibrium with the amine. The $\ce{NH2Me+, NHMe2
+}$, and $\ce{NMe3+}$ groups all give more meta than para substitution, with the percentage of para product decreasing with the increasing number of methyl groups.

@ron's explanation coupled with the equilibrium argument makes a pretty good reason for why aniline behaves this way.
But why do the other groups meta-direct? The field effect mentioned above seems to refer to inductive effects, which should make them para-directing.

This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations.
First of all, it is not true that the para-product is always favoured with the $\ce{NH3+}$ ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining step of the reaction. If the reaction is under thermodynamic constrol, the most stable isomer will form regardless of the directing property of the substituent. If it is under kinetic control, then you have to look at the rate determining step of the reaction. The electrophilic aromatic substitution is a two step process in general, and if the first step is rate determining, then the nucleophilicity of each postion of the ring will be the main factor. If the second step is rate determining then the energy of the intermediate will also matter. Obviously the ortho and para intermediates will be unstable as the positive charge is delocalized next to the electron withdrawing group. So, it's not always true that inductive effect  by itself cannot be meta-directing.
Anyway, back to your original question. Let's consider the nitration of aniline in presence of conc. $\ce{H2SO4}$.

Brickman and Ridd[1] has noted the relative ratios of products for different sulfuric acid concentrations.
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 85 & 92.4 & 96.4 & 100 \\ \hline ortho(\%) & 6 & - & - & - \\ \hline meta(\%) & 34 & 53 & 58 & 64 \\ \hline para(\%) & 59 & 47 & 42 & 36 \\ \hline  \end{array} $$
As you can see, the meta product becomes more favoured as the concentration of $\ce{H2SO4}$ increases.
What about the N-methylated anilines? In another paper, Brickman, Utley and Ridd[2] have mentioned the percentages of meta and para products (ortho does not form at high acid concentrations, probably due to the steric effect).
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 90.9 & 96.2 & 99.8 \\ \hline ortho(\%) & - & - & -  \\ \hline meta(\%) & 61 & 67.5 & 70.3 \\ \hline para(\%) & 39 & 32.5 & 29.7 \\ \hline  \end{array}$$
$$\text{For N-methylanilinium ion }\ce{Ph-NH2Me+}$$
Here, we again see the same trend that the increasing sulfuric acid concentration increases the meta-product percentage. However, the meta-product is now the major product.
$\ce{Ph-NHMe2+}$ also shows similar results. The N,N,N-triemthylanilinium ion $\ce{Ph-NMe3+}$ shows $89\%$ m-product and $11\%$ p-product at $98\%$ conc. $\ce{H2SO4}$.
In the two papers, the authors mention that the analysis of acid concentration vs. rate plot seems to indicate that the substitution predominantly happens on the anilinium cation i.e. the protonated state, so there is no possibility of the para-product arising from the reaction of the deprotonated compound. Sulfuric acid is a pretty strong acid, so it's unlikely that there is much deprotonated aniline left at such high concentration. The p/m ratio of aniline nitration changes by a factor of less than $2$, when the acid concentration is changed from $90$ to $98\%$, which reduces the proportion of deprotonated form by a factor of $31$. So, it's not possible that the para-product is coming from aniline, because then the rate profile would have been much more sensitive to the concentration of acid.
They have also ruled out significant hyperconjugative effects, by comparing the rates of the benzylammonium variants of all of these anilines. So, the effect must be mostly from inductive effects.
After a lengthy discussion, it is noted that the $\ce{-NH3+}$ group is much more solvated than the $\ce{-NHMe2+}$, $\ce{-NMe3+}$ etc. groups. So, it is possible that the solvent forms hydrogen bonds to the $\ce{-NH3+}$ group and allows the positive charge on it to spread away. This means that the inductive effect from $\ce{-NH3+}$ is actually weaker than the methylated groups. So, the methylated anilinium ions are simply more strongly averse to reaction on ortho/para positions than anilinium.
However, the authors note after that[2]:

The above explanation  accounts for the major differences in the substituent effects recorded in Table 3 but it is not completely satisfactory, for the extent of  para-substitution in the anilinium ion is  more than would  be  expected if the effect of replacing $\ce{-N+Me}$, by $\ce{-N+H}$, were simply to "scale down" the effect of the positive pole;  such an effect would  never  cause the para-position to be more reactive than one meta-position. This anomaly is also illustrated by the comparison of  the anilinium ion with the benzyltrimethyl-ammonium ion: the latter ion is more reactive by a factor of ca. 25 but gives less para- substitution. It appears that the protonated poles give significantly higher p/m ratios than expected  from their reactivity relative to the methylated poles.
The reason for this is not entirely clear ...

So, it's not completely known why aniline gives more para-products than its N-methylated derivatives.
This is only considering the nitration reaction. If you consider other reactions, there are varied directing nature. For example the sulfonation reaction (done with $\ce{H2SO4}$ and $\ce{SO3}$) produces almost entirely the para-product[3] (sulfanilic acid), and this has been attributed to the fact that the sulfonation is reversible (thermodynamic control). I haven't been able to find any data on halogenation or other substitutions on the anilinium ion.

M. Brickman, J.H. Ridd, J. Chem. Soc., 1965, 6845-6851
M. Brickman, J.H.P. Utley, J.H. Ridd, J. Chem. Soc., 1965, 6851-6857
J.J. Jacobs Jr., D.F. Othmer, A. Hokanson, Ind. Eng. Chem., 1943, 35, 3, 321–323


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Trying to understand phrase in paper: preventing nonspecific binding of AuNP antibody conjugates using a blocking solution
I am working through a paper, Lateral flow immunochromatographic assay on a single piece of paper. I'm having trouble understanding one of the points!
Under the Device fabrication question, it says:

The sample and conjugate release regions were treated with a blocking solution to prevent nonspecific binding of AuNP-antibody conjugates.

I didn't understand this phrase, so I did a bit more research and found a question by the author on ResearchGate asking how to prevent nonspecific binding.
However I still don't quite understand. What exactly is nonspecific binding of AuNP-antibody conjugates, and how does a blocking solution prevent this?
Any overview/pointers about where to research further would be very much appreciated.

Isn't Biology SE more suitable? But it is hard to understand your question. Isn't clear if you don't understand the sentence (which I do understand) or you want to understand details (which I don't know). You could have a title like "How to prevent.....

@Alchimista the former: I don't understand the sentence. Probably the basics: what is a blocking solution? what is nonspecific binding? how does a blocking solution prevent this?
This is probably pretty basic stuff, I would love to know how to learn more about these points.

Biology SE more chances to get an answer.

@Alchimista The answer on Chemistry SE posted today is pretty good. And it is pretty chemical (conjugated proteins, surfactant, buffer).

Let's start with defining the terms.

The sample region is the portion on a lateral flow immunoassay that receives a sample (of blood, urine, oral fluid, ...). This has often been treated to adjust the sample's pH, viscosity, ionic strength, or protein content.
The conjugate release region is the portion on a lateral flow immunoassay that holds immobilised, conjugated antigens. It provides a site for the conjugated antibodies to bind to the target before migrating further.
A lateral flow membrane is commonly made of nitrocellulose. In this specific paper, the authors evaluate nitrocellulose membrane and cellulose chromatography paper. They mention, "Nitrocellulose membrane and cellulose chromatography paper both offer excellent protein binding capabilities".
Protein binding is where the protein 'sticks' to a surface. This is great for binding to the test and control lines (specific binding), but no good when binding to the membrane (non-specific binding).

The substance tested for is usually a protein. When the sample travels up an untreated nitrocellulose strip, the protein will bind to the membrane instead of the test line.
To prevent this, a nitrocellulose strip is treated with a blocking buffer. This is a mixture of proteins that binds to the membrane, preventing binding from the protein being tested for.
In this paper, the protein tested for was PfHRP2 (Plasmodium falciparum histidine-rich protein 2), a biomarker for malaria detection. Unfortunately, the authors were getting false-positive results (no test line showing up) because the PfHRP2 was binding to the nitrocellulose membrane and not to the test line.
This was prevented by treating the membrane with a blocking buffer of 20% sucrose (stabilises proteins), 0.25% Tween-20 (a surfactant, enhances the flow speed of the sample), and 2% BSA (bovine serum albumin, the protein which binds to the membrane) in PBS (phosphate buffer saline, a buffer solution).


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Evaluating inductive effects in cyclic compounds
I was required to find the products on heating the hydroxide of;
Now it's easy enough to find the answer; Hoffman elimination isn't going to happen from the methyl, and the n-propyl doesn't have $\beta$-hydrogens as acidic as that of the ethyl. It doesn't feel like the hydrogens in the ring are more acidic, either, so the product involves ethene. Simple enough.
But this lacks rigor. How do I actually evaluate the inductive effect on the $\beta$-carbons on the ring? If I have to compare it to some other fragment less obvious to win than ethyl, where do I start counting the carbons that inductively push electron density onto the $\beta$-carbons on the ring?

Is that a boat conformation? In any case I interpret the drawing as a trans methyl amino cyclohexane. That means the methyl and amino groups are both equatorial and not optimally arranged for a Hoffman elimination in solution. I have no idea of the physical arrangement with the hydroxide ion in a melt. If a boat configuration is intended the ring is set up for a syn elimination. This is best settled by finding the publication if this has been run or by synthesizing the compound if it hasn't.

@jimchmst: isn't cis-trans nomenclature for unsaturated systems?

It is also pertinent to saturated simple ring  compounds


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Extract seaweed iodine from petroleum ether and cyclohexane
I'm working on iodine extraction from seaweed by using petroleum ether and cyclohexane. I have already observe the presence of iodine in the solution but I don't know how to extract the iodine in a solid form. I try to boil it but it does not work because there is no iodine left after the process. Hope to get the answer ASAP since I'm at nick of time

What is your goal. Do you want to quantitivly measure how much iodine is extracted? Or you just want to qualitivly show, that it is iodine?
One idea would be, tro transfer the iodide into a salt which can not evaporate.

Be as descriptive in your question as possible, considering the question background. Every omitted word is usually figurally expensively paid by extra clarification chain.

Perhaps use the highest volatile petroleum ether fraction and avoid heating. For small vials it should go away easily. Consider that the amount giving deep color to your extracts might be minuscule, so no reason to expect iodine grains as residue.

Indeed in this protocol the extract is just let aside. https://edu.rsc.org/experiments/extracting-iodine-from-seaweed/1915.article. You should have find it, btw.


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Decide major product for first order elimination reaction
This is my own created question.

The favourable thing for (c) to be a major product is  conjugation of double bonds and highly acidic nature of hydrogen (resulting in elimination of hydrogen at a higher rate) adjacent to carbonyl group (credit $-$ Rover). According to information I have the above reaction is most probably reversible in nature.
(b) should be major product as per mechanism I have given. So which will be the major (or only) product. You may take high temperature conditions if temperature is required as we are doing elimination reaction.

I think your mechanism is correct.

Have you considered acidic nature of H ?

The H attached to carbonyl is more acidic.

The product that you have drawn is very unlikely to be the major product of a mixed aldol condensation. The actual major product in said reaction would be (assuming I get the substrates right) would be as follows:

However, your question is a valid one if we take this out of consideration. Now, If we consider just the elimination reaction and your reaction conditions, your mechanism is correct. However, if this reaction were to take place via E2, the rate determining step would be that of the removal of the hydrogen and this would mean that the product formed would be (c). In case of E1, (b) would be the major product since the rate determining step is that of the formation of the carbocation.


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Why lactols being tetrahedral complexes do not break?

The answer given is (D) but I think (B) should be the answer as the concerned benzylic carbon in option (D) is an unstable tetrahedral complex and it should break. Besides, the medium is acidic so it should catalyse this process.
In generalising, my question is: are lactols stable?
Edit:
I found online that there is an equilibrium between lactols and corresponding hydroxyaldehyde. So, why lactol is favoured more in this case?

If there's something wrong with the question then please suggest edits.

What are tetrahedral complexes and why must they be unstable? I heard of no such law. Then again, you might point out that hemiacetals should be prone to hydrolysis, which is generally true, but not when the alcohol and aldehyde are both contained in one molecule.

That is a term one of my teachers uses. I checked the meaning online and it does not match with the meaning implied. By unstable tetrahedral complex I meant that a carbon on which there's a group which has leaving tendency and -OH which can kick it out. I do not know another word to edit it to.

Sir, the other part that you talked about "...but not when the alcohol and aldehyde are both contained in one molecule." I did not understand that. Why does that happen.

This happens just because the molecule (B) happens to contain an alcohol group and an aldehyde group, as simple as that.

You may notice that B is an intermediate to eventually reach D.  Indeed, it is quite possible, performing the reaction in the lab, B would spontaneously yield D with only little chance to isolate this intermediate as such.
$\ce{NaBH4}$ reduces the aldehyde to an alcohol.  Then, mediated by the acid, the actetal is converted into an aldehyde to yield B.

The aldehyde in B already is more reactive than a ketone would be, but the presence of acid (still present from converting the acetal) may increase the reactivity of the carbonyl group even further.  The subsequent cyclization is favored

because the chance that alcohol and aldehyde are close enough to form a new bond is much higher for an intramolecular reaction, than for an intermolecular one.  This is a general rule worth to retain. In addition, the formation of five- and six-membered rings equally is energetically favorable.

It is also worth noting that cyclic hemiacetals (lactols) are often stable. See the cyclic forms of most simple sugars for examples.

@BenNorris Based on your answer given here, supporting a postulate by computations, I notice a little bit of envy with me for those capable to perform such a reaction in the lab .and. to use such an in silico yardstick to estimate quantitative likelihoods in advance.


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Determining the mechanism followed when alcohols undergo a reaction with sodium bromide and sulphuric acid
$$\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}$$
Due to sulphuric acid, the $\ce{-OH}$ group will get protonated and convert into $\ce{-H2O+}$ which is a great leaving group in comparison to the hydroxyl group. Therefore I assumed it would readily form a carbocation and then $\ce{Br-}$ could attack to form the alkyl bromide.
That is, it would follow the SN1 pathway (since the reaction is happening in 2 steps, first the formation of carbocation and then the attack of $\ce{Br-}$) but I also read somewhere that we might need to have a look on the substrate, if it is primary then it would follow SN2.
I am a bit confused after so many conditions and situations being introduced so I was hoping if someone could give me a brief and crisp explanation about this, and is there any particular factor that needs to be focused on before judging this mechanism?

Both reaction happen but the extent of each reaction depends on what the R is and to some extent on the solvent.

Alright, so is it safe to assume that it would majorly follow SN2 if it is a primary alcohol and SN1 for a tertiary alcohol? Also, what should be our approach for a secondary one? Do we simply conclude that SN2 and SN1 both happen in a considerable amount? I'm not sure about the solvent part because my book didn't mention anything else but the reaction that I mentioned.

What you are asking is too broad and would be difficult to cover in one answer. Read this (and the linked pages) and see if it answers your question.

yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall reaction follows SN1 (as it is acid-catalyzed) hence we would compare the rate of reaction in terms of RDS (FORMATION OF CARBOCATION) and hence the more stable the carbonation is the faster the reaction proceeds.


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Will potassium permanganate react with ferric chloride?
I came across a question :

Which of the following compound does not decolourise an acidified solution of $\ce{KMnO4}$?

Answer was:

 $\ce{FeCl3}$

After doing a bit of research I found that $\ce{KMnO4}$ oxidises $\ce{HCl}$ to $\ce{Cl2}$.
I doubt whether $\ce{KMnO4}$ would also oxidise chloride ions in case of acidified solution of $\ce{FeCl3}$.
I know that the ferric ion cannot be oxidised further.

Are you sure that "[you] know iron cannot be oxidised further"? Ferrate (IV), Ferrate (V) and Ferrate (VI) oxidation states exist even though the (VI) is the most encountered among the three. $FeCl_3$ has iron (III).

Have you considered ferric chlorocomplexes ?

@Kav how likely is it that $\ce{Fe(III)}$ can be oxidized further ? Will it do under normal conditions with $\ce{KMnO4}$ or do you require drastic conditions ?

Iron can be oxidized further, but not by KMnO4.

@Kav ...Actually the conditions are normal as the book has not dealt with complexes yet.

If it reacts with chloride ions of HCl then why not chloride ions of FeCl3?

@Curious.SWARNIM complexes were mentioned by Poutnik.....

@Ashish your content of answer suggests that I meant ferrates will be formed when I was suggest show their existences

@IvanNeretin how do you know in this reaction condition. The OP was thinking," if MnO4- can oxidize the chloride ion in HCl, can it also oxidize Cl- of FeCl3, can't it?" The OP's sentences of reasoning was promoting thought by analogy while yours is stating the chloride cannot be oxidized. So yours need a support.

@Kav I never said a thing about chloride.

@IvanNeretin my sorry about the chloride (better quote next time) but yours is stating KMnO4 cannot oxidize Fe. I guess my comparison still holds. To your benefit, the OP narrowed the scope of the question in stating the OP knows the $\ce{Fe^3+}$ cannot be oxidized further.

@Kav Well, ferrates are stronger oxidants than permanganates, and other higher oxidation states of Fe don't exist in water.

@IvanNeretin Thank you for helping out.

Bleach can oxidize a Fe(III) salt to Fe(VI)

@Nilay Ghosh ..I am talking about KMnO4 being the oxidising agent not bleach.

Did you not read Ivan's comment: "Iron can be oxidized further, but not by KMnO4". KMnO4 can't but bleach can.

@Curious.SWARNIM What were the other compounds given?

@Rishi ..the other options are:SO2,H2O2,FeSO4


text:
question about photodamage on plastic
I know that plastic are susceptible to UV attack. They create free radicals when put under sunlight even when kept for a short time.
I'd like to know if the photo-damage will stop once it is taken away from the sunlight or will it continue the photo-damage to the plastic. Does the rate becomes very slow because of the absence of light even when there is presence of free radicals?

What do you mean by "plastic".

Although radicals are involved this does not necessarily involve a chain propagation. The aging by UV should stop or slow down in absence of irradiation. Also, at least in term of viscosity, ,the eventually initiated process should be rather confined. In plastic items there will be the appropriate amount of scavengers so they should be more than effective in dark. Different is the case of an oil. Note that I am not a specialist by this is what I can think of.


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Major product in reaction of butan-2-one + selenium dioxide
When reading about $\ce{SeO_2}$ I read this following reaction:
$\ce{CH3CH2COCH3 ->[SeO2] CH3CH2COCHO + CH3COCOCH3}$
In the following reaction it was stated that $\ce{CH3CH2COCHO}$ was the major product. But I fail to understand why ?
The two possible reasonings I can think of are that the terminal hydrogen is either less sterically hindered or that it is more acidic. But I am not quite sure of either.
What is the reason for the same ?

To understand the major factor we have to consider the mechanism of the reaction first. The reaction you mentioned is popularly know as Riley oxidation. The mechanism is as follows:

As you can observe the RDS of the reaction is the tautomerization of the keto form into an enol. Thus the extent of enolization determines the major product in this reaction.
As the reaction takes place in an neutral environment the acidity of the alpha-hydrogen plays the major role in deciding the major enol formed and so the corresponding oxidized product is major.
In short, the major product in Riley oxidation is the product associated with the major enol content which in the given conditions is determined by the acidity of respective alpha-hydrogens.

This is known as the Riley Oxidation.
The Wikipedia entry states that ketones with two alpha methylene positions react more quickly at the least hindered site which references Riley's original paper1.
Reference

Selenium dioxide, a new oxidising agent. Part I. Its reaction with
aldehydes and ketones by Riley et.al. J. Chem. Soc., 1932,
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