text:
Gaussian calculation - PCM model for IRC compared to gas phase?
Finding a reaction pathway thanks to Gaussian calculations is tricky exercise requiring a lot of resources for each step of this time-consuming process. Thus, man needs to wonder if each calculation has to be performed or not in order to gain as much time and resource as possible.
After achieving a 3-step mechanistic path for my reaction in gas, meaning finding all optimized structures, TS, verifications by frequencies of the stationary points and IRC, I now need to use the PCM model in my calculations to provide a solvent correction to my mechanism. However, I was wondering which calculations were required for this purpose.
From the 2 options I have in front of me, which is more consistent to avoid wasting time and resources ?

Perform new optimizations for the minima and TS in my pathway using the PCM model. Then calculate frequencies to be sure of my stationary points.
Perform new optimizations of the minima and TS AND perform new IRC calculations from the fresh new TS (after solvent correction) with PCM model.

The first option means that I perform an optimization from the reverse and forward structures (already further optimized) obtained from my IRC calculation in gas.
The second option means that I calculate a new IRC to obtain new forward and reverse structures that I have to optimize again.

Option 2 is the most scientific and rigorous. You should re-run the IRC in solvent again because the reaction path can be different from what you got in the gas phase.
Now, will it be different in solvent? Unless you are dealing with some exotic system, different reaction path with implicit solvation is unlikely (provided you have the same TS and minima with PCM). However, the reaction barriers may significantly change, which can cause two subsequent steps to become one single step (the TS between them forming a shoulder in the IRC). I have seen this happen.
What matters though is that different reaction paths can happen, so you should run the IRCs in solvent if you want to be sure of your result (also because you want to publish it).


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How can over grinding lead to reduced API extraction in HPLC analysis?
I am looking for some advice regarding particle size distribution (PSD) & how this can be affected by grinding a tablet.
During an assay by HPLC, the method requires grinding of a tablet before dissolving in order to extract the active pharmaceutical ingredient (API) into solution. Of the tablet, the API is only 0.5% m/m of the tablet. I have been told that over grinding the tablets can affect extraction of the API and this is supposedly related to PSD as the API is the smallest particle found in the tablet amongst other excipients.
Could someone explain how "over grinding" would cause a detremential effect? Thanks.

Contrary to HPLC, readers of ChemSE may be not familiar with API and PSD; some are just starting to study chemistry.  Thus, if you use less common abbreviations, define them once.  If API or PSD are not correctly deciphered, revert the edit.

If the API is only 0.5% m/m of the total tablet, the active ingredient is not buried as a single grain into the matrix of e.g., potato starch, but finely dispersed.  Thus, there is no danger that if you miss the special grain your HPLC would yield a reading of zero.  But of course, the finer the grinding, the greater the surface area, which increases the rate the material may dissolve. Add information to add clarity to your question.

Can you give a reference which suggests that overgrinding reduces extraction? This must be very specific to the tablet matrix (polymeric coating?) etc.

If there is such an effect, could it be due to the localized heating during grinding? Though the overall sample may not appear to get much warmer, as particles are crushed, they can be heated intensely, leading to decomposition or hastening attack by air.

I agree with DrMoishePippik that localized heating may be part of the answer, but I'd suggest a different reason. It seems that only part of the tablet's contents are dissolved. I'd suspect that the grinding coats the active ingredient with an inactive one that doesn't dissolve.

I sort of feel this is actually more of a mathematical/statistical/modeling related question than traditional chemistry.
In essence, one argument is that the sample derived effective API concentration is impacted by possible changes in the particle size distribution due to grinding. Now, there are, in effect, known statistical distributions that address particle size distributions (PSDs), and associated methods to assess parameters of said distributions relating to sample means and variability, and changes thereto.  This would seem to be a possible component of the analysis.
As background, I reference Wikipedia on the topic, which importantly notes, to quote:

The PSD of a material can be important in understanding its physical and chemical properties...It affects the reactivity of solids participating in chemical reactions, and needs to be tightly controlled in many industrial products such as the manufacture of printer toner, cosmetics, and pharmaceutical products.

as both surface area (a function of particle size) and concentration influence a reaction.
Also, some detail on PSD distributions:

The log-normal distribution is often used to approximate the particle size distribution of aerosols, aquatic particles and pulverized material.
The Weibull distribution or Rosin–Rammler distribution is a useful distribution for representing particle size distributions generated by grinding, milling and crushing operations.
The log-hyperbolic distribution was proposed by Bagnold and Barndorff-Nielsen[9] to model the particle-size distribution of naturally occurring sediments. This model suffers from having non-unique solutions for a range of probability coefficients.
The skew log-Laplace model was proposed by Fieller, Gilbertson and Olbricht[10] as a simpler alternative to the log-hyperbolic distribution.

Now, for example, in the case of the Log-normal distribution, the best estimate of the parent population mean particle size (which here can be used to estimate total concentration) is obtained by working with the log values of the sample particle sizes. Wikipedia on the Log-normal distribution notes that the sample mean (say 'm') and variance (${s^2}$) of these log-transformed values can be employed to more accurately specify the parent population (aka, the Log-normal) mean, which turns out to be ${Exp(m + 0.5*s^2)}$. Note, simply taking the straight average of the observed Log-normals size values is not recommended as a Log-normal random deviate can be very noisy being the ${Exp()}$ of a Normal random deviate, hence the use of a log-transform. The Weibull distribution is another example of where the parent population mean is a function of two of the distribution's to-be-estimated parameters.
Further, it may indeed be the case that excessive grinding affects particle size distribution and reaction rate. However, this should be verified experimentally to determine the potential impact. Perhaps using a proxy, that produces a more readily measurable product (as in gas evolution, or strongly colored compound) may be of assistance.
Finally, this educational source comments on the inter-connection between reaction rate, mass and concentration, noting an exception, to quote:

The concentration of the reactants determines the speed of the reaction. In simple reactions, an increase in the concentration of reactants accelerates the reaction. The more collisions over time, the faster the reaction can advance. The small particles have less mass and more surface area available for the collisions of other particles. However, in other more complex reaction mechanisms, this may not always hold true. This is often observed in reactions involving huge protein molecules with large masses and convoluted structures with reaction sites buried deep within them that are not easily approached by collision particles.

So, still some experimenting required here in an obviously specialty discipline with a possibly large molecule.

source: Stack Exchange

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

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Which one out of a keto group or a hydroxyl group would get protonated first if both are present in a compound (attachment)

In the solution provided by my teacher he protonated the ketone, moved the +ve above by resonance did methyl shift and then did dehydration to form a benzene type system. Is this only because for this particular care aromaticity was being obtained or is the keto group always first protonated

It's generally a valid question to ask which gets preferentially protonated, but in this particular case, it actually doesn't matter at all. Protonation and deprotonation at oxygen is fast and happens all the time; if the proton gets onto the "wrong" oxygen (i.e. the hydroxyl group), it can't do anything apart from fall off again, so it just falls off and tries again.

@orthocresol So is there no standardized way to predict it theoretically rather than just hit and trial ?

It’s “trial and error”; and there are pKa values, you can look them up, so there is a “theoretical” way, but it’s simply not relevant for your question.

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


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Succinic anhydride esterification won't go forward
I'm trying to react succinic anhydride with a decyl alcohol to produce mono-decyl succinate. Literature and Google searches suggest that the monoesterification should proceed without catalyst. However I'm failing to get the reactions going. Reactions were in reflux, $\pu{80-140 ^\circ C}$ at $\pu{1 atm}$.
Any suggestions if anhydride esterifications even needs a catalyst? Would a solvent help?

Addition of catalytic pyridine should assist it.

Do you use solvent? if so, what is it? Would you please include those in your question?

The monoesters of succinic acid can be prepared using Steglich esterification method (Ref.1,2), which is specially suitable for alcohols with the $\ce{OH}$ functional group is in steric hindrance. In this method, dicyclohexylcarbodiimide (DCC) as the coupling reagent and 4-dimethylaminopyridine (DMAP) as a catalyst are used. The reference 1 gave a typical procedure as follows:

To a stirred solution of 10mmol carboxylic acid in $\pu{lO ml}$ anhydrous $\ce{CH2Cl2}$ (in the case of sparingly soluble acids, in DMF) is added $\pu{30-110 mg}$ DMAP and $\pu{20-40 mmol}$ alcohol or thiol ($\pu{10 mmol}$ with alcohols or thiols which are not easily removable without significant loss; $\pu{3.4 mmol}$ with glycerol). DCC is added to the reaction mixture at OT, which is then stirred for $\pu{5 min}$ at $\pu{0 ^\circ C}$ and $\pu{3 h}$ at $\pu{20 ^\circ C}$. Precipitated urea is then filtered off and the filtrate evaporated down in vacuo. The residue is taken up in $\ce{CH2Cl2}$ and, if necessary, filtered free of any further precipitated urea. The $\ce{CH2Cl2}$ solution is washed twice with $\pu{0.5 N}$ $\ce{HCl}$ and with saturated $\ce{NaHCO3}$ solution, and then dried over $\ce{MgSO4}$. The solvent is removed by evaporation and the ester isolated by distillation or recrystallization. Crystalline products can be obtained in pure form by filtration on a short silica gel column (eluent: $\ce{CH2Cl2}$ or $\ce{CHCl3}$).

For example, recently, this method has successfully used to couple Vitamin $\mathrm{D_2}$ to succinic anhydride in order to achieve succinic monoester of Vitamin $\mathrm{D_2}$ (Ref.3). However, there is an US patent publication revealing cheap solvent free method using alkali salt bases such as $\ce{NaOAc, KOAc, Na2CO3}$, and $\ce{K2CO3}$ as catalysts to efficiently prepare mono alkyl esters from cyclic anhydrides (Ref.4).
References:

Bernhard Neises, Wolfgang Steglich, "Simple Method for the Esterification of Carboxylic Acids," Angew. Chem. Int. Ed. 1978, 17(7), 522–524 (https://doi.org/10.1002/anie.197805221)(PDF).
Gerhard Höfle,  Wolfgang Steglich,  Helmut Vorbrüggen, "4‐Dialkylaminopyridines as Highly Active Acylation Catalysts. [New synthetic method (25)]," Angew. Chem. Int. Ed. 1978, 17(8), 569-583 (https://doi.org/10.1002/anie.197805691).
Victar Nyaarugwe, "Synthesis and Characterization of Hybrid Compounds for the Treatment of Bone Related Diseases," Journal of Trauma and Orthopedic Nursing 2018, 2(1:4), 16 pages (PDF).
Mark, B. Erman, Joe W. Snow, Mikhail Y. Lebedev, "Process for making monomenthyl esters," US Patent 2007, US 7,247,743 B1 (PDF).


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Why does delocalization (only) occur in molecules represented by resonance hybrids?
Before I start this question, I am aware tha

Electrons in all molecules are delocalized to some extent
Delocalization is a the phenomenon and resonance is an attempt to explain it.

When I asked this question from my professor, this is the explanation he gave me:

In VBT, it is difficult to approximate the molecular orbital with
localized bonding orbitals, so the electronic configuration is a
weighted average of the infinite possible electronic configurations
i.e Ψ= Aψ1 +Bψ2+Cψ3......
For most molecules, the most stable lewis structure has the majority
of the contribution and can sufficiently approximate the actual
electronic by itself. However, for some molecules, there are multiple
equally-stable lewis structures with an equal and significant
contribution and each of them has to be taken into account. For e.g,
for nitrate ion, Ψ=1/3 (ψ1 +ψ2 +ψ3)  where ψ1,2,3  represent the 3
equal lewis structures. As a result, the electrons MUST be delocalised
between the appropriate bonds

This explanation, in simple words, argues that  since the 3 lewis strucutes are identical/indistinguishable, the electron density must be equal and thus delocalised between the bonds by symmetry.
However, this doesn't really explain the physical reasoning behind delocalization. Why is (significant) delocalization more energetically stable for some molecules but not for others? And why dpes delocalization always occur in molecules with some kind of symmetry between the possible atoms for a double/triple bond?

From an MO point of view: Electrons in all molecules are delocalized to some extent is not correct- All electrons are delocalised in a molecule. Resonance is a consequence of the inability of Lewis-like structures (localised bonding) to account for that.

A short answer is that we mostly use resonance structures to represent delocalization when the bond orders or formal charges predicted by VBT are not approximately integers, simply because non-integer bond orders and charges are more cumbersome to draw. A typical example is a carboxylate, which we could draw with 1.5 bonds between both C-O pairs and -0.5 charge on each O, but it is simpler to just draw a resonance structure with one C=O double bond and one C-O single bond and a full -1 charge on one oxygen.
As others have noted, electrons are still delocalized across the entire molecule in molecules for which we do not use resonance structures. The difference is that we can approximate the bond orders in those cases as integers, so only one structure is needed to represent that.

I'll provide a partial answer: resonance and delocalization are invoked to explain the fact that certain molecules do not have specific bonding patterns expected based on Lewis bond theory. In particular, that theory would predict that some molecules exist as mixtures of symmetry-related  isoenergetic structures with some symmetry-related bonds differing in bond order, while experiment (and more advanced theories) instead show that there is only one structure and bond order rather than many. One explanation is that a molecule somehow exists as a "mixture" of the various possible structures suggested by Lewis bond theory, exchanging ("resonating") between these forms.
Resonance is a phenomenon in which energy is exchanged between tuned systems (such as radio antennae). In the VB theory case it is simply implied that the energy of the electron(s) involved in resonance is similar in the postulated geometries and that there exists a path allowing the electrons to shift between bonds involved in resonance. This shifting does not happen in practice, since there is really only one structure, but it clarifies what structures contribute to resonance.
Resonance hybridization can be used to explain why molecules do not behave as simple Lewis bond theory predicts. The answer to the question "when can I invoke resonance hybridization" is to generate all possible Lewis structures and then see if any of them are related by symmetry or whether there is a method to interconvert them by shifting bonds. If the answer is "yes" then it makes sense to assume that the real structure is a "resonance hybrid" of the Lewis structures (as opposed to each Lewis structure being an isomer).
Consider for instance the carboxylate ion, which can be represented as a resonance hybrid. It is straightforward to draw two possible Lewis structures for the ion. For the two possible structures not to be isomers a path must be available for exchange of valence electrons from one to the second CO bond center, that is, there has to be a common atom (in this case the carbon) serving as a "transit point" for the electrons.
Electrons are quantum mechanical particle-wave entities, and the resolution of the problem is that more advanced theory is required to properly describe the electron distribution in molecules. Lewis bond theory was developed before the full complexity of QM was understood. A simple QM model suffices however to explain why the energy drops when you increase delocalization of an electron: the particle in a box. If you double the size of the box you reduce the energy of the electrons by a factor of four, increasing their stability.

I understand the delocalization providing more stability due to particle in a box. What confuses me is why doesn't delocalization stabilize all molecules instead of a select few?

for example, the nitrite ion has delocalization, but nitrous acid molecule does not.

@OVERWOOTCH because you will run into the opposite problem. The electrons of molecules with no delocalisation (other than the fact that el are always spread as for their orbital representation) are just fine to accommodate themselves according to just one Lewis structure, or that Lewis structure has such a big dominance to appear as a good approximation itself. Symmetry is not a requisite, in your example it just explains why a particular Lewis structure has even more weight. It can be also a kind of driving force, too. Read my answer to this now closed thread, it might help:

@OP https://chemistry.stackexchange.com/questions/145993/what-is-resonance-exactly

@OVERWOOTCH and, as you mentioned the particle in a box picture, that is valid. But are the nuclei (and the other particles) which define what the potential, ie the box, is.

Probably one of the best posts on the topic: https://chemistry.stackexchange.com/q/51632/4945 // The 'real' electronic structure of nitrous acid certainly needs multiple resonance structures, but there is one Lewis-like structure that dominates the overall wave function. For almost all purposes that's enough. @OVERWOOTCH


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How to interpret this formula about hydrogen bond energy
I want to evaluate some chemistry related formulas, which I don't understand.
In proteins, hydrogen bonding often occurs between the oxo group = O oxygen of one amino acid and the α-amino group (N − H) of another amino acid.
The bond energy is determined by the distance between oxygen and hydrogen and by the angle enclosed by the N-H and H-Bonds.
Let's assume the $E = (\frac{1}{d_{HO}^{10}} - \frac{1}{d_{HO}^{12}}) * cos(\alpha)$ model describes the energy of the hydrogen bond.
What is the meaning of the number 10 and 12 in the formula?

I know the angle and the distances between one O, one H and one N atom (that's all the input we got).
maybe related:
https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/hydrogen-bond

Those are exponents. You are missing two parameters to replace the ones in the numerator, see e.g. slide 8 of this document: http://ncbr.muni.cz/~martinp/C3210/StructBioinf7.pdf

Thank you, it worked! (the numbers are powers)

These exponents $10$ and $12$ are a little bit arbitrary. They have been chosen so as to obtain an energy curve corresponding to the measured H-bond energies. Other attempts have been published, like the famous Morse curve, which is, for covalent bonds versus HO distance $d$ :  $$E(d) = D[1 - e^{-k(d - d_o)}]^2$$ where $D$ is the dissociation energy of the covalent H-O bond. Apparently the curve with the two exponents $10$ and $12$ gives a better fit.


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Can someone clear this confusion I have about the first law of thermodynamics
In our school we were taught that the first law of thermodynamics represented mathematically is $\Delta U=Q+W$. Also we were taught that for an isothermal process $\Delta U=0$ and therefore $Q=-W$, but we were taught that for an isothermal free expansion of a gas against vacuum $W=0$ and therefore $\Delta U=Q$ but shouldn't $\Delta U$ be zero here as well, I mean after all the process is still isothermal. Can somebody pls help with regards to this, also pls try to explain in a simple manner (High School Level). Thanks for the help

Everything you've written is correct (for a free expansion, $\Delta U = q = w = 0$), but it applies only if the system is an ideal gas, or has ideal-gas-like properties, namely that there are no attractive or repulsive interactions between the particles.
It also requires that the system is closed (no matter can flow into or out of the system). Internal energy is an extensive property, so its value is proportional to the amount of substance in the system. If matter is added to the system, internal energy can increase even if no there is no flow of work or heat.
Your teacher probably had both of these restrictions in mind.
Thus, if you expand an ideal gas isothermally into a vacuum, the system doesn't do any work, so $w=0$. And there is thus no need to flow heat into the system to compensate for the work to maintain the temperature, nor is there any need to flow heat into the system to provide energy to pull the particles apart (no interaction between the particles), so $q=0$.
If you want a molecular explanation, imagine an expansion against a constant external pressure.  As the piston retreats, the particles hitting the piston move more slowly after they reflect off of it (this is because they are colliding with a retreating object rather than a stationary object), which has the effect of cooling the gas (the temperature is a measure of how fast the particles are moving).  Thus, for an isothermal expansion in which work is done against some external pressure, to maintain the temperature, heat must flow into the system to compensate for this.
But for a free expansion, the gas isn't expanding against a retreating piston. So particles are not robbed of their kinetic energy and thus, in a free isothermal expansion (of an ideal gas), no compensating heat flow is required to maintain the temperature.


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How does Combinatorial Chemistry differ from Rational Drug Design?
I stumbled on Vertex Pharmaceuticals - Wikipedia.

It was one of the first biotech firms to use an explicit strategy of rational drug design rather than combinatorial chemistry.

Then I looked up Combinatorial Chemistry. What is Combinatorial Chemistry

     Combinatorial chemistry is a laboratory technique in which millions of molecular constructions can be synthesized and tested for biological activity.  It has generated massive numbers of targeted molecules for testing and the developing techniques of high throughput screening has automated the screening process so larger numbers of biological assays can be done.  All this together has reduced the discovery-to-market time from what used to be 10-14 years to 5-8 years.

But this doesn't explain how Combinatorial Chemistry differs from Rational Drug Design? Isn't Combinatorial Chemistry rational too? Mustn't chemists pick reactants rationally? Even if she can't predict the products, she won't picked her reactants randomly.

From a medical point of view the problem in drug discovery is to find a compound that causes a desired outcome, not one that interacts with a specific target. For instance, if you want to reduce mortality due to infection with Covid, you have many strategies that may be pursued at the molecular level, from vaccination to interference with viral propagation to reduction of symptoms or boosting of the immune response.
The difference between a combinatorial strategy and a rational strategy is to a first approximation simple: in a combinatorial strategy compounds are tested for activity without immediate concern for how they bring about an outcome. There is no "a priori" information to guide compound selection other than the common properties of known drugs (as encoded eg by Lipinski's rules) and the ability to procure or synthesize compounds. Therefore the "rational" approach in this case is to create as large a library and to screen as many compounds as is reasonable given cost and time.
Rational drug design takes into consideration a molecular target (such as a protein) and perhaps the mechanism of action of a known lead compound and designs future potential drug compounds around that target and/or lead. That narrows the set of compounds under consideration reducing time and cost.
In general, the strategies of combinatorial and rational design are not mutually exclusive. It is typical to start a rational strategy after early leads have been identified some other way, say through combinatorial chemistry. It is also common to design a set of compounds during a stage in rational drug design using a combinatorial strategy applied to develop next generation leads from a starting compound.


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Is Combinatorial Chemistry related to combinatorics in math?
If the answer is no, then I don't understand the adjective "Combinatorial". Barring decompositions, aren't the five main types of chemical reactions combinatorial? You must combine at least two reactants for a reaction!

Combinatorial chemistry usually refers to changing the choice of A and B, and when the reaction products differ according to the type of reaction, also the type of reaction.

Yes, the term "combinatorial" chemistry is related to mathematical combinatorics. Imagine that we have three positions (numbered 1, 2, 3) on a molecule, and at each position, we can place any one of 5 functional groups (A, B, C, D, E). We can synthesize all possible molecules in parallel by making all possible derivatives at position 1 (5 total), then reacting all of those molecules to make all possible derivatives at position 2 (5 for each of the 5 intermediates) and then again at position 3 (again, 5 new molecules for each product of step 2). In this example, we would produce 5x5x5=125 combinations as our final output.

In addition to the examples given, e.g., on Wikipedia's entry, combinatorial chemistry is not constrained to small organic and larger biomolecules.  Folks in catalysis screen by means of combinatorial chemistry libraries of e.g., mixed metal oxides, too.


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Mechanism of retro aldol condensation of hydroxycyclopropylketone
The reactants were treated with $\ce{NaOH} $. How the product here obtained. Can you show me the reaction mechanism for this. I found this is somewhat  Retro Aldol Condensation. But how?

You deprotonate the alcohol, and then flip bonds to create the enolate. There are two enolates possible, so swapping bonds gives you the terminal one, which can attack the aldehyde to form the 5-ring. Elimination of water gives you the product.


text:
Hindered carbon atom
In the hydration reaction that is addition of H2O , in the presence of (H.B.O/T.H.F/H2O2), where no rearrangement takes place,
Why does H is placed on more hindered C atom and OH on less hindered C atom ?

H is smaller and less cumbersome than OH.

And what chemical exactly is "HBO"?

The reaction Hydroboration-Oxidation follows anti-Markovnikov rule because of 2 factors:
1)Electronic consideration: When the borane ($\ce{BH_3}$) approaches the $\pi$ bond of the alkene, the attack of the $\pi$ bond to the empty p orbital of the boron triggers a hyride shift (the mechanism of this is similar to that of pericyclic reactions). There is a partial positive charge developed on the vinylic position of the alkene that triggers the hydride shift.

Steric considerations: In the first step of the mechanism of hydroboration, we add $\ce{BH_2}$ and $\ce{H}$ simultaneously. As the $\ce{BH_2}$ is bigger in size than $\ce{H}$, the transition state of this step would be less in energy and also less crowded if the $\ce{BH_2}$ is positioned at the less steric position.

Both factors suggest that the mechanism must follow anti-markovnikov addition.
After the attach of $\ce{BH_2}$, the further mechanism will replace the $\ce{BH_2}$ with $\ce{OH}$. Here I showed why the $\ce{H}$ attaches to the more crowded site and follows anti-Markovnikov addition.
For the further part of the mechanism, you can view the page 414 of the David Klein 2012 edition.
All the reference pictures are taken from the David Klein Organic Chemistry 2012 edition book.


text:
Why does the pKa - pKb relationship hold when Kw holds for the autoionization of water?
The relationship is commonly expressed as:
$$pK_a + pK_b = pK_w = 14$$
or
$$K_a * K_b = 10^{-14}$$
This holds because
$$K_a = \frac{[H^+][A^-]}{[HA]}$$
$$K_b = \frac{[OH^-][HA]}{[A^-]}$$
$$K_aK_b = \frac{[H^+][A^-]}{[HA]}\frac{[OH^-][HA]}{[A^-]}=[H^+][OH^-]=K_w=10^{-14}$$
$K_w$ is the autoionization constant of water, or the equilibrium constant of water as it reacts with itself in equilibrium. This has the measured value of $10^{-14}$.
But isn't this value measured when water is all by itself, reacting with itself in autoionization? Why does the value hold when, say, acid is added into the water? Why are the concentrations of protons and hydronium ions ($OH^-$) still both equal to $10^-7$?

"Why are the concentrations of protons and hydronium ions still both equal to 10^-7?" Nobody said they're both $10^{-7}$, only that its product is $10^{-14}$. If $ab = 100$, it doesn't imply that $a = b = 10$.

@orthocresol Okay, but still the product is measured for autoionization if water. Why does it hold for a solution where acid or base is added?

Because it's an equilibrium constant, which is always a constant at equilibrium (hence the name), regardless of the exact concentrations of the reacting species, and regardless of how that equilibrium was reached.

@orthocresol But arent equilibrium constants reaction specific? A different one for water reacting with water, and another for acid reacting with water?

Yes, but the first is $K_\mathrm{w}$, and the second is $K_\mathrm{a}$, which are different. The point here is not about whether $K_\mathrm{w} = K_\mathrm{a}$ (it obviously isn't), but rather about $K_\mathrm{w}$ being the same regardless of whether you're applying it to a neutral solution, acidic solution, or basic solution. Also, when I say $K_\mathrm{w}$ is the same, this means that $[\ce{H+}][\ce{OH-}] = 10^{-14}$, it doesn't mean anything about either $[\ce{H+}]$ or $[\ce{OH-}]$.

@orthocresol But why is $K_w$ equal to 14? It is so because for water in equilibrium in autoionization at room temperature the concentrations of both protons and hydronium are $10^{-7}$. But this is for one reaction only, the autoionization of water. Why is the product of concetrations the same in other reactions like acid in equilibrium with water?

The OP may mean Kw conventionally implicitly involves implied constant concentration(activity) of water, that decreases when a solute is added. But most of equlibriums are considered for rather diluted solutions where c << 55.x mol H2O/L, so the difference is neglected.

No, you have it the wrong way round. $K_\mathrm{w}$ is equal to $10^{-14}$ (not $14$), but this is not because $[\ce{H+}] = [\ce{OH-}] = 10^{-7}$! It is the exact opposite: in pure water, we have  $[\ce{H+}] = [\ce{OH-}] = 10^{-7}$ because $K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 10^{-14}$ and because in pure water, the stoichiometry of dissociation dictates that $[\ce{H+}] = [\ce{OH-}]$, hence both of them are $10^{-7}$.

Also, $\ce{OH-}$ is hydroxide, not hydronium. Hydronium is quite the opposite.


text:
Is standard change in activation Gibbs free energy indepedent of pressure?
For a chemical reaction the activation volume is defined as:
$$Δ^{\ddagger}V={V^\ddagger}-V$$
where both $V^\ddagger$ and $V$ refer to standard states (I have omitted the symbol for simplicity). One can calculate activation volume by taking the partial derivative with respect to pressure of $$Δ^{\ddagger}G={G^\ddagger}-G$$
(again for simplicity the standard state symbol is omitted). I can't understand why $Δ^{\ddagger}G$ must be a function of pressure if it happens at a specific pressure (standard state). I was expecting that it would be only a function of temperature like ${Δ_rG}^\circ$ is. The motivation to ask this question was how rate constant varies with pressure:
$$\left(\frac{\partial{\ln k}}{\partial{P}}\right)_T=-Δ^{\ddagger}V$$
The only thing that comes to my mind is if the standard state that is used to derive the above equation it means a state with fixed concentrations for example and not pressure.

In the last equation I don't think it can be true that the V daggers refer to standard states.  Thus the meaning of V dagger in your last equation is different than in your first one.


text:
How can chlorine be 'only' the third-most electronegative element yet have the highest electron affinity?
From Wikipedia:

It is an extremely reactive element and a strong oxidising agent: among the elements, it has the highest electron affinity and the third-highest electronegativity on the Pauling scale, behind only oxygen and fluorine.

I am confused with this quoting. Apparently, the somewhat vague concept of Linus Pauling's 'electronegativity' includes two concepts:  Both the relative, quantified tendency to grab an electron and the relative, quantified tendency not to lose one. Combined into the somewhat-quantified, relative concept of electronegativity.

https://chemistry.stackexchange.com/questions/6901/why-does-chlorine-have-a-higher-electron-affinity-than-fluorine?rq=1

Does this answer your question? Why does chlorine have a higher electron affinity than fluorine?

Electronegativity is a rather slippery thing, really.  Roughly speaking, it depends on both electron affinity (the tendency to take on a negative charge) and ionization enthalpy (the tendency not to assume a positive charge).  Chlorine has the highest electron affinity among halogens, true; but fluorine has a much higher ionization energy, a difference of about four electron volts according to typical ionization-energy tables.  By comparison the electron affinity of chlorine exceeds that of fluorine by less than one electron volt.  The way these factors balance out in the combined parameter called "electronegativity", the great difference in ionization energy dominates and puts fluorine at the top of the list.

Fluorine is a rather small atom. The incoming electron is put into a region of space already crowded with electrons. So there is a significant amount of repulsion, which lessens the attraction the incoming electron feels, and so lessens the electron affinity. A similar reversal happens between oxygen and sulfur. The first electron affinity of oxygen (−142 kJ/mol) is smaller that that of sulfur (−200 kJ/mol), for the same reason.

I will use electron gain enthalpy(ega) to explain this.
ega becomes less negative down a group in general. However, adding an electron to 2p orbital leads to greater repulsion than adding an electron to 3 p orbital and this, chlorine has the highest ega(the magnitude) and hence, the highest electron affinity.


text:
Acidity of Carbonyl group
I m a high school chemistry learner. I have learnt that acids are substance that can liberate H+ on hydrolysis. For that they must be connected to an electronegative atom.
In carbonyl ( mainly aldehyde and ketone) the H atom is not directly attached to Oxygen. So I may not consider it as an acid.
But today while performing a chemistry practical when I added Blue litmus solution to aldehyde it turned red.
This leads me to a confusion about whether they are acidic or neutral or What can I infer from the colour change of blue litmus.

How pure was the aldehyde? Aerial oxidation can easily produce carboxylic acid from aldehydes

It gave Silver mirror test. The Formaldehyde may have been converted to Formic acid. Am i right?

It would still give silver mirror if there was some acid present. Formaldehyde oxidises easily if not stored properly. It is good practice to purify your aldehyde before use in synthetic chemistry

But then why didn't it showed reaction with Na2CO3 after that??

How did you react it with Na2CO3 and what did you observe? If it was only midly acidic there would not be enough CO2 generated for you to notice bubbles.

Might as well make an answer of the comments.

Aldehydes are not only able to be oxidized to acids, they are quite reactive towards this oxidation.  Exposure to air would be enough to trigger enough oxidation and acid formation to get litmus to react, especially if the litmus is moist.  Aldehydes are not alone in this respect; for example iron(ii) salts can be oxidized by air too, especially if you try to store them as a solution or after breaking the original seal on the bottle.

Moist litmus paper is a good qualitative test for acidity or basicity because it is highly sensitive to small amounts of acid or base.  Reaction with sodium carbonate might not be as sensitive if you are looking for enough carbon dioxide to visibly form bubbles, especially as carbon dioxide has some solubility in most solvents.

Yes, you need to follow good practices inspecting, testing and if necessary, purifying sensitive chemicals if you want high precision or accuracy.


text:
is there a stead fast rule to predict hybridization or is it just an ad hoc concept
Before I start of this question, Ι want to make it clear that I know that hybridization, like pretty much all of chemical bonding, are just made up to qualitatively rationalize observations. I know that theres no actual orbital overlapping, resonance, excitation.... and what not, and the only real things governing "bonding" are electrodynamics and quantum mechanics
When I first came across hybridization, I learnt that it is just used to obtain the correct electron density distribution from atomic orbitals as the experimentally observe corresponding to least repulsion/energy. The go-to way to determine hybridization would then be to use VSEPR to find the most stable electron domain geometry. However, I've come across several molecules such as H2S, PH3 and Halogen molecules where there is no hybridization. How does VBT /hybridization theory rationalize this? And is there any way that I could predict it? (I obviously can't know the experimental observations for all possible molecules before hand)
Secondly, is rationalizing geometry the only reason for invoking hybridization? I have come across these 2 other reasons in several reliable textbooks:

Hybridization occurs as hybrid orbitals are able to overlap more effectively and the bond strength is proportional to the extent of overlap. While this seems somewhat logical, I can't think of a single molecule where better overlap needs to be the reason for a specific hybridization
Hybridization occurs inorder to make all orbitals degenerate before overlapping as this corresponds to the lowest energy "quantum mechanically". This one seems way too handwavy but is mentioned on wikipedia https://en.wikipedia.org/wiki/Orbital_hybridisation

Bonus question:
I have another small question that I dont want to start another thread about given the multitude of questions on hybridization already present on the site but feel free to ignore this.
Why exactly do sp,sp2 & sp3 hybrid orbitals have the same exact geometries as the ones observed in electron domains? Is it just a coincidence or are they forcefully combined in such a way to match them>

Hybridisation is a purely mathematical way to describe the electron density.

I know. That's why I clearly mentioned that right at the start of the question.

I hit enter by accident. Hybridisation has no physical meaning and is just a toy to play around. As such it is a point of view, a descriptive model. Therefore hybridisation does not occur. Both those quotes (? - please add citations and sources) are therefore wrong from the start. As a strict consequence: Hybridisation always follows molecular structure, never the other way around. Why VSEPR is still quite well at predicting general shapes is a different question, which I believe has been answered here before.

https://en.wikipedia.org/wiki/Orbital_hybridisation The second quote can be found under the sp3 subheading

Related: Is the hybridization only related to atomic orbitals but not to the molecular orbitals? // Please [edit] your question to add the citations.

The bit about more effective overlapping can be found in the introductory chapters of Klein's organic chemistry.

Please [edit] your question instead of posting more comments.

Regarding hybrid orbitals: the geometric forms they take is "real" insofar as they are the actual mathematical result of combining the starting s and p orbitals (in math representation) in the desired proportions.

@BuckThorn Does so are the electron geometries limited to those expressible as a linear combination of hydrogen-like wavefunctions ? Or can they be anything because expression as a linear combination of orbital is only OUR approximation and not a fundamental rule nature has to abide by?

No, there is no such limit, it is just a convenient way to intro the subject. Any proportion is allowed provided that they are properly normalized to reflect the electron density in the product orbital. Any (positive) combination of fractions is in principle ok. The same principle works when we solve for molecular orbitals generally. Solutions (molecular orbitals) are usually constructed as combinations of linear combinations of atom-centered orbitals (eg H-like functions).


text:
Entropy of Dissolution of Hydrocarbons
Here is what I think I know: The entropy of dissolution reactions increases as methylene groups are added (i.e. butanol has higher entropy of dissolution than propanol). Also, acyclic saturated hydrocarbons become more hydrophobic as the number of methylene groups increases.
Here is where I get confused:
Larger hydrocarbons have a higher enthalpy of dissolution because they are composed of a greater number of bonds. BUT why does the CHANGE in entropy of dissolution become more negative as the number of carbons increases? Doesn't the dissolution of a larger hydrophobic compound yield a more positive entropy value/become less negative because the initial entropy of the solvent is very low (many large, ordered solvent cages) and then solvent entropy increases after the molecules aggregate by the hydrophobic effect?
Super confused...thank you.

It is important to know the nature of the solvant, because H-bonds have the main role in water.  Is it water or a non-polar solvant ? You also never mention the nature of the solute. In the beginning it is an  alcohol. But in the title it is a hydrocarbon. So what ? Please explain !

Maurice: sorry for not clarifying. The solvent is indeed water, and I think that I am more interested in the dissolution of hydrocarbons (I used the alcohols misleadingly to illustrate my point). Specifically, I am hoping to figure out the trend in entropy as hydrocarbons with differing number of methylene groups dissolve.

Is it possible to measure the enthalpy of dissolution of hydrocarbons in water ? Their solubility is so weak. Does it make sense ? How can this dissolution be measured ?

Yes, the solubility of hydrocarbons is low in polar solvents. I am trying to figure out the thermodynamic justification of the hydrophobic effect. I added a picture of the textbook for reference.

I think that you should focus on what you wrote first, longer the hydrocarbon chain higher their entropy in solution. When dissolved, they can access a multitude of states both energetically and in spaces (think of all possible conformers).
This said, it does not mean that they must dissolve in, eg, water. Right because the H term.


text:
Why is a triangular cavity created during wire-cutting of an organic gel?
I conduct wire-cutting on a gel made with organic oil. When I use high cutting rates, the wire cuts the gel in half leaving two smooth cutting surfaces. In contrast, when I use a very low cutting rate, the oil inside the gel starts coming out and a cavity is also formed around the wire (as if, instead of cutting the fibers, the wire compresses the gel network and pushes the oil out). I'm wondering why the cavity is triangular.

I really hope someone could come up with a simulated CAD model illustrating displacement map–elastic modulus dependency and not a generic hand-waving explanation. I think we have a couple more of similar questions, and it would be nice to have a canonical answer for this type of questions.


text:
Will a crystal of iodine lower the yield of an Grignard Reaction
I discussed the following today:

If adding an iodine crystal to a Grignard to make the reaction go. will this tiny crystal force some of the metal or reagent to be used
up. ergo, therefore, there is a lower yield (I know it is microscopic
change, this is not the question)

It is more about if it takes some of the metal and uses it up, maybe forming Mg-Iodide, so it cannot be used in the Grignard reaction, and I am not sure if Mg-Iodide cannot be used or if it actually catalyzes the reaction.
It is in the first step of forming phenyl-Magnesium bromide (or other reagent of the same type) reagent for use in step 2 of Grignard. I write bromide on purpose, so you can see that it is another halide than iodide.
Could it be the iodine catalyzes the reaction, or actually, some of the phenylmagnesium bromide is turned to phenyl magnesium iodide, which can also be used in the reaction? Ergo there is no loss in yield at all?
Imagine for step 1, forming only the Grignard reagent, I used iodine equivalents to 10% of, for example, of bromobenzene, and used Mg (not in excess), but the exact amount of Mg equal to bromobenzene, as 1:1 ratio, and the reaction normally went to 100% yield (it does not) but imagine it. Would I lose 10% from the iodine crystals?

can you tell me what the iodine crystals forms in the reaction and will it able to be used in the reaction or not?
will it catalyze the reaction instead?

Will you show me, with drawings, the intermediates or products of the iodine crystal?
Will it interfere with the yield, as it takes some of the magnesium and forms most likely $\ce{Mg-I}$ that is not used in the reaction, or is it I am asking you? It doesn't give any hints regarding lowering of yield. Will it help catalyze or form Grignard reagent with iodide? Why?
A pointer to what I am actually asking is. what happens to the iodine that is not used to remove hydroxide or etch the surface then what comes out of that will it continually as catalytic etch give $\ce{Mg-I}$ then Grignard reagent with bromide and back or what happens with it? will it use op the Mg turnings lowering yield.

Note: 1) Caps Lock = SHOUTING. Need to avoid that 2) Spelling Mistakes: "Reagent", "Grignard".

Since normal practice is to use excess magnesium, why would a small amount of iodine affect the yield?

because the question got hijacked and they added a thump upon Maurice's answer, and a thump down on my question, I did shout, it happens every time because someone it only answering for points, and this is a serious question for someone in Metal-coupling reactions. it still amazes me that the whole question got ruined by these since it was a good question I asked!
now nobody wants to answer because of the negative 1  one my question and I have to ask it again somewhere else

if there is a spelling mistake and that is all you care about then you need to refocus yourself! there are also comma and period mistakes.

matt_black. this I already know. try to look at the example in the text and then what can you tell me, and are you saying then that it will lower yield but it doesn't happen in practice because there is still excess Magnesium turnings in the bottle?
all I want to know is "does Iodine lower the yield if magnesium is in a 1:1 ratio with an organic halide and they normally react 100% with each other  (they don't I know but imagine it)"

No. Iodine is used to help magnesium react with the bromoderivate. If pure magnesium is added to a solution containing a bromoderivate, it should react quickly. But sometimes it does not react at all, because magnesium is covered by a thin and protective layer of magnesium oxide. Added iodine will etch this layer and the underlying metal. This reaction breaks the oxide layer, so that the metal can get in contact with the bromoderivate, and quickly react.

Iodine does not diffuse through the passive film, it reacts.
Magnesium oxide, or the hydroxide if it reacts with moisture, is basic and will thus [disproportionate the iodine to form relatively soluble iodide and iodate salts] (https://www.researchgate.net/figure/Pourbaix-diagram-also-known-as-EH-pH-diagram-for-inorganic-aqueous-iodine-species-at_fig3_318164242).  We can use iodine because with magnesia being a relatively strong base, we do not need a more aggressive halogen to react.

@Oscar Lanzi. Thank you for the precision.

I cannot approve that as an answer like I did not know why we use it and how on earth did you guys misunderstand something very specific and complex, hard to know unless you worked with metal-coupling reactions to something anyone can find on the internet as to why we use Iodine hmm... point-hunters

@user53617 Maybe some magnesium is lost due to iodine having reacted with it. But this amount is small. And usually in Grignard synthesis metallic magnesium is added in excess, to compensate for this effect.

If the magnesium it is reacting with is MgO, then can't that magnesium already not be used for the Grignard anyway?


text:
What is non toxic to fish, has a low freezing point, and floats on water?
I am looking for a liquid that floats on water that will prevent the water from freezing (or at least slow it down).  The application would be a small area (less than a square foot).  It has to be non-toxic as I do not want to harm any critters.  Thoughts???

What's in your water? Because a simple solution is to add salt!

Maybe the question already has a solution on either https://pets.stackexchange.com/ (because of this), or https://gardening.stackexchange.com/?

Gardeners often use a tennis ball for this purpose - it preserves a gap in the ice.

What is the freezing point of a tennis ball?

There are oils sold for swimming pools to slow or stop evaporative cooling . That would be a problem for air breathers like mosquito larva. I don't know anything about it because I went with 1" thick Styrofoam panels. That is what I currently have on my pond ( 10' X 5 '), works very well .I cheat a little and have 150 watt aquarium heater in the 700 gallons. My pond is currently 59 F while nearby open water is 49 F. For your tiny area the Styrofoam is fine leave about 1/2" on all sides for air . On the other hand why not do the obvious and put in a 25 watt ( if they make them that small) aquarium thermostat/heater ?

A possible more natural solution allowing continuing oxygen exposure, use a layer of leaves resting on a frame (say wood branches).
Supporting related comments can be found here, to quote:

Leaves can be used to insulate plants from cold weather. To provide protection for plants, including container plantings, circle a plant with wire fencing. Then, stuff leaves inside the fencing all the way around the plant...Moreover, if you have a root cellar or storage basement, you can use dry leaves to layer your vegetables in, rather than using sawdust or newspaper. This should allow them to be stored through the winter.

No guarantee of success, but perhaps better than being completely exposed.


text:
Purging submersible with nitrogen
I do a lot of short term ocean field deployments where system electronics are self contained in a submersible enclosure. A new project I'm working on requires a long term deployment in the deep ocean. Therefore, once the enclosure is deployed, I can not just go recover the instrument if something fails. I've spoken to a lot of colleagues about how they prepare electronic enclosures and one of the standard protocols is purging the enclosure of atmospheric air. For one, any moisture in the air can condense at cold, deep depths, which can cause electrical problems. I think another reason they do it is to reduce oxygen exposure. Typically the purging is done with argon from an air cylinder. However, this is something I don't have. Now I know I can go out and get one...that's not the point of this question. What I do have is a small nitrogen generator that was needed for other projects.
The question I would like to ask here is, can I use the 99.x% dry nitrogen air from the nitrogen generator as a substitute for argon? I know argon is denser than nitrogen, so clearly is a better choice, but I'm not concerned with the better choice. I want to know if I can use what I have at my disposal to achieve the same goal: replace atmospheric air with 99.x% dry nitrogen air?

It sounds like a costly project; seems strange they don't have a few dollars to rent a clean dry bottle of argon from  a local weld shop supply. and know exactly what you have.

What is the temperature in the ocean depths?  If it is too low and the x in 99.x% is too high, the water vapor can condense on the equipment.

Looks like the main issue is the presence of oxygen in the atmosphere of the enclosure. It looks like dense argon displaces more that just nitrogen. Nitrogen gas, even in the liquid form can hold a certain amount of oxygen. I recommend checking out dissolved oxygen topic at Science Direct.

A small bottle of highly pure argon will set you back perhaps 200 dollars, and that includes buying the gas bottle in the first place. Refills will be more like 60 thereafter. Just purge with argon like your colleagues - you want your experiments to be on your subject, not the best way to purge your enclosures...

In my judgment, the water vapor in the nitrogen is also a major issue.

@ChetMiller It sounds like you're suggesting there is water vapor in the nitrogen, even though it says "dry" (which I realize is a vague label and I might be misinterpreting that as "no water vapor").

@thatsrightjack it says 99.x% dry.

I am not going to write an answer for this. Argon is preferable to nitrogen just because being heavy it tends to fill the volume you want to be controlled and stay there. This is important for containers that are not perfectly sealed. It can be a chemical flask that must be periodically opened, or a glove-box, etc. In your case it won't make much difference if the degree or purity of the gases is comparable. Perhaps it takes longer for nitrogen to displace the air, but that is all.

Using argon is, in effect, a costly physical path to remove oxygen, hence your interest in using nitrogen.
However, I suspect, a more efficient, cost-effective, and likely safe chemical path is also possible. For example, assuming a control structure (so as to mitigate potential deaths from O2 depletion, as has been reported, for example, among sewer workers working in a recently drained iron pipe), oxygen scavenging can result from a limited exposure to iron surface previously scrubbed with an acid. Here, lemon juice appears to be a friendly, pleasant, and effective choice.
Note, iron powder and ferrous salts have been used in the food industry to remove problematic oxygen (see, for example, THE USE OF OXYGEN SCAVENGERS AND ACTIVE PACKAGING TO REDUCE OXYGEN WITHIN INTERNAL PACKAGE ENVIRONMENTS).
One should be able to stoichiometrically compute, and experimentally validate, a safe amount of Fe (or its surface area) to employ. This is to limit a continuing unsafe scavenging action if, for example, subsequent maintenance work is required.
You can also employ commercially available products containing Calcium Chloride, for example, to remove water vapor.


text:
Is selenite (hydrous calcium sulfate) dangerous when you grind it up?
I was wondering about the safety of a specific mineral, Selenite. From my research, I have read that selenite is the crystalline form of gypsum, which is hydrous calcium sulfate. I was wondering if I were to grind selenite and mix about a 1/2 tsp with 3 ounces of water and spray around my home, is that dangerous? Also will the grinding process be dangerous if any small amounts of dust are inhaled?

You should check its msds:https://beta-static.fishersci.com/content/dam/fishersci/en_US/documents/programs/education/regulatory-documents/sds/chemicals/chemicals-c/S25230.pdf

Gypsum is the major constituent of plaster. It is non-toxic.
Our lungs are capable of filtering out particles greater than 50 micrometers in diameter. Any dust finer than that is potentially dangerous.


text:
Are all amphoteric substances weak acids and weak bases?
I saw as an example that water is amphoteric (to be more precise amphiprotic), and I came up with the following questions:
Is there any substance that acts as a weak acid and a strong bases at the same time, or vice versa?
If so, or if not so, please give some explanations on the possible reasoning using their structures.
Also I have another question which is quite related but too short to be posted as an independent question, so I will ask here. The necessary condition for a substance to be amphoteric, as I heard, is that for it should have a hydrogen atom in it, and also the molecule should either have a lone pair of electron or be negatively charged. I understand the conditions about hydrogen atom and an electron pair, but why can the negatively charged property possibly be a necessary condition for a molecule to be amphoteric?
Thanks in advance.

OH- in NaOH is a very weak acid and O^2- extremely strong base, but NaOH melted with Na would produce Na2O and H2.

@Poutnik So is there any systematic rule for amphoteric substances or do properties differ for every substance?

Aside of amphoteric properties they share, their properties differ, but that is obvious. There must be 2 conjugate acid/base pairs that share 1 member.

@Poutnik Sorry I was being ambiguous, I was talking specifically about amphoteric properties

Ask yourself, what do acidities of 2 different hydrogens of the same molecule/ion structure have in common ? Nothing in general. They can reside on the same atom as one extreme, or can be separated by many different atoms as the other extreme.

@Poutnik Hmm could you please elaborate possibly with examples on what you said on your last comment? I'm afraid it's quite unclear to me.

@Poutnik And so can I understand your conclusion as saying that there is no general laws like amphiprotic substance should always have both weak acidity and basicity? Also the groups of amphiprotic substances in the Bronsted-Lowry context are limited but can I think that every molecule can always (theoretically) act as an amphoteric substance from Lewis acid/base point of view? Because it's merely about losing or gaining electrons.


text:
Autoionization of water
Consider as solution of pure water at $\pu{25 ^\circ{}C}$ with a $K_{\mathrm{w}} =  10^{-14}$.
If we are to add an acid to the mix we would observe an increase in hydronium and an equal decrease in hydroxide content.
Yet, I fail to see how the hydroxide ion concentration can decrease with an increase in hydronium concentration.
Lets add $\ce{HCl}$ to our solution: $$\ce{HCl + H2O -> Cl- + H3O+}$$
Concentration $[\ce{H3O+}]$ increases,  concentration $[\ce{OH-}]$ decreases in order to fulfill the $K_{\mathrm{w}}$ constant of $10^{-14}$.
Through what reaction exactly then will the hydroxide content decrease (by turning into water I assume)? Where does $\ce{OH-}$ gets an $\ce{H+}$ from? Not from $\ce{H3O+}$, I assume seeing that every molecule of a strong acid is said to turn into an hydronium ion 1:1.
Please enlighten me.

H+ is in acidobasic reaction context just convenience for H3O+ ( which is exactly convenience for H3O+ . n H2O ). As H+ is nothing else but a proton/deuteron, which creates locally so strong electrostatic field which does not allow its free existence in condensed phases.

@Lars To facilitate rendering chemical equations in the body of a question, answer, or comment, ChemSE offers you to use mathJax and mhchem syntax.  Which I applied to the current form of your question.  You may familiarize yourself with the underlying rules e.g., here.  Because it is not understood by all internet browsers, you should not use this peculiar syntax in the title of a question.

@Buttonwood Thanks, message deleted.

Note that the kinetic rate of $\ce{H3O+ + OH-  -> 2 H2O}$ is $\frac{\mathrm{d}[\ce{OH-}]}{\mathrm{d}t}=k \cdot [\ce{H3O+}][\ce{OH-}]$ and it is one of the fastest ever chemical reactions, controlled by diffusion.

To answer your question, yes the $\ce{H+}$ cations that react with the $\ce{OH-}$ anions come from the acid. Remember that in pure water the $\ce{OH-}$ anion concentration is only $1\times 10^{-7}$ so if you have a 0.1 molar acid solution then a minuscule fraction of the acid is consumed.

Maybe a numerical example will help you. Let's start from pure water, with the following concentrations :$\ce{[H+] = [OH-] = 10^{-7} M}$. Now we will suppose you add $\ce{10^{-7}} mol$ $\ce{HCl}$ in one liter of this water. Suddenly, the concentration of $\ce{H+}$ should double. This is not possible in the long run. A part $a$ of these supplementary $\ce{H+}$ ions must react with the same amount $a$ of the $\ce{OH-}$ ions to form water. The final molar concentrations of both ions are given by  : $$\ce{[H+] = 2.000·10^{-7}} - a$$ $$\ce{[OH-] = 1.000·10^{-7} } - a$$   This amount $a$ may be obtained by solving the equation : $$\ce{ (2.000·10^{-7}} - a)\ce{(1.000 10^{-7}} - a)= 1.000·10^{-14}$$ The solution is : $a = 0.382·10^{-7} mol/L$. So about $38.2$% of the initial $\ce{OH-}$ ions have been destroyed by the addition of the new acidic ions. And the numerical values of the final concentrations are $$\ce{[H+] = 2.00·10^{-7}}- a = 1.618·10^{-7} mol/L$$$$\ce{[OH-] = 1.00 10^{-7} } - a = 0.618· 10^{-7} mol/L$$ And of course : $\ce{[Cl-] = 1.000·10^{-7}$ $ mol/L}$. If you want you may check the product of the two final concentrations $\ce{[H+]}$ and $\ce{[OH-]}$. It is equal to ${1.000·10^{-14} mol^2L^{-2}}$.


text:
What are the benefits of using zero capillary voltage for discarding fractions in LC–ESI–MS/MS?
In the second episode of The Association for
Mass Spectrometry & Advances in the Clinical Lab (MSACL) podcast, namely Getting going with mass spectrometry : Josh learns chromatography (aired 2021-01-07), Dr. Russell Grant mentioned the following tip at 01:03:11 (I apologize in advance for any inaccuracies in the transcript):

The other thing that is often forgotten there is that you can drop the capillary voltage to zero. It's not as good as the diverter valve, but you are not going to be pulling the stuff towards the mass-spectrometer. If you only have a couple of windows and a bunch of junk, you can just keep capillary voltage on during these desired windows and save your diverter valves' maintenance time because they can become maintenance-limiting.

I would think it would lead to a more frequent cleaning of the ionization chamber. Is the gain in postponed maintenance of a diverter valve really worth it?
Are there any other use cases when disabling capillary voltage could be a superior alternative to using a diverter valve?

I think there is no perfect answer to this situation. In general when the flow is stopped to any detector in an HPLC experiment, there is a change in baseline, and when the flow is re-started, ghost peaks appear in chromatogram. I am talking about optical detectors such as UV, fluorescence, refractive index etc. Same thing happens in MS and perhaps LC ESI-MS is not an exception, in the sense that diverting the flow can lead to the appearance of so-called "ghost" peaks or raised baseline. I have seen that in GC-MS experiments where a change in flow causes weird baseline shifts. This is an interesting phenomenon worth investigating.
So the authors suggest that let us continue the flow, but let us stop the ionization process so that the junk does not go inside the quadrupoles and stay there forever. It is always easy the clean the external ESI system but the internal system needs professional service.
Someone talks about these problems in this forum Diverter valve issues

Great point about the ghost peaks. I suspected that alternating the flow could have an impact on registration, but I wasn't sure how to clearly articulate this effect.


text:
How to prepared saturated salt solution (NaCl)?
What do they mean by "add 321 g saturated NaCl solution"? Description does not contain volume.

Welcome to Chemistry SE! As NaCl solubility is temperature-dependent, more details are needed in the OP. I recommend posting the source of this quote to provide context.

If you know the density (i.e., mass divided by volume) it suffices to weigh the solution on a balance instead of using e.g., a graduated cylinder to record a volume.

Based on your description, I would suggest to add $\displaystyle\frac{\pu{321 g}}{\pu{1.1998 g cm^{-3}}} = \pu{267.5 mL}$ of saturated salt solution (assuming approx. $\pu{25 °C},$ ambient temperature).
A report The Densities of Saturated Solutions of
NaCl and KCl from 10° to 105 °C by U.S. Geological Survey (PDF) contains a collection of density values for saturated NaCl solutions at various temperatures, of which I used the value at $\pu{25.37 °C}.$


text:
Draw the enantiomer products of the reaction between 1,3‐butadiene and (3Z)‐pent‐3‐en‐2‐one

From this reactants I need to draw two products with cyclohexene rings that are enantiomers to each other with chemical formula $\ce{C9H14O}.$ And also draw the reaction mechanisms. Why does it form enantiomers? I'm not sure how I should approach this.

Welcome to Chemistry SE! It would be helpful to add more specifics. What are the conditions of temperature, solvent and pH? What are your attempts at solving this problem? Help us help you!

Do you know how a [4+2] cycloaddition works?


text:
Do I have to systematically show the protons in my reaction equation to find the Pourbaix diagram's line equations?
I was trying to get the line equation for the Pourbaix diagram for the couple $\ce{Ni(OH)2}/\ce{Ni}.$ I can write two chemical equations for basic and acidic conditions, respectively:
$$
\begin{align}
\ce{Ni(OH)2 + 2 e- &-> Ni + 2 OH-}\tag{R1}\\
\ce{Ni(OH)2 + 2 H+ + 2e- &-> Ni + 2 H2O}\tag{R2}
\end{align}
$$
Now I can determine the line equation for both chemical equations using Nernst equation:
$$
\begin{align}
E^\circ + \frac{0.06}{2}\log\left([\ce{H+}]^2\right) &= E^\circ - 0.06\,\mathrm{pH};\tag{1}\\
E^\circ - \frac{0.06}{2}\log\left([\ce{OH-}]^2\right) &= E^\circ + 0.06\,\mathrm{pOH}.\tag{2}
\end{align}
$$
I need to make the $\mathrm{pH}$ appear in the previous equation. To do so, I use the relation
$$\mathrm{pH} + \mathrm{pOH} = 14 \quad\Leftrightarrow\quad \mathrm{pOH} = 14 - \mathrm{pH},\tag{3}$$
which gives
$$E^\circ + 0.06(14 - \mathrm{pH}) = E^\circ + 0.84 - 0.06\,\mathrm{pH}.\tag{4}$$
I've found two slightly different expressions depending on whether I choose basic or acidic conditions. I've spend quite a few hours trying to figured out what is wrong here, and I've found a PDF Diagrammes E-pH = Diagrammes de Pourbaix (courtesy of Académie de Clermont-Ferrand) stating on page 4 that even if the species only exists in basic conditions I need to make the $\ce{H+}$ appear in the half-reaction.
Is there a convention when writing chemical equation prior to the determination of the line's equation for Pourbaix diagram? Or did I make a mistake somewhere? If yes, could help me please?

the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $


text:
What is the difference between resonance and π conjugation
I have seen electron delocalization mentioned frequently in 2 separate contexts: resonance (ozone, carbonate / nitrate...) and π-conjugation systems (aromatic compounds).
While the 2 initially sound the same, 2 different points lead me to believe otherwise:

π conjugation requires p- orbitals to be in the same plane, whereas many compounds said to have resonance ($\ce{ClO4-}$, $\ce{SO4^{2-}}$) are not planar so resonance can't be a subset of π conjugation
All molecules with (significant) resonance have multiple equivalent stable-lewis structures (nitrate has 3, sulfate has 4...) whereas many molecules with π-conjugation do not (haxa-1,3-diene has only one "most stable" lewis structure) so π-conjugation cant be a subset of resonance

NOTE ON POINT 2: I am aware that hexa-1,3-diene has minor contributing structures, but so do all possible molecules. However, the only molecules for which resonance hybrids are used are those with multiple major contributing lewis structures.
The only intersection between the 2 concepts I find is benzene; it has 3 equivalent major contributing structures implying resonance and 6 overlapping p-orbitals lying in the same plane, implying π conjugation
I feel like I'm missing some part of the bigger picture. What's actually happening?

Resonance is one mechanism for checking if a system has π conjugation .

Resonance and conjugation are defined in the IUPAC gold book.

You might be confused but not missing anything big or fundamental.
A pi-conjugated system is one in which delocalisation of pi-electrons can take place. It is all.
When effective, pi-conjugation is a subset, or a type, of delocalisation than can be described by resonance.
However:
to be strict, conjugation refers to bonds, and specifically to an alternating single and double bonds structure. It is just a positional nomenclature.
It turns out that in cases and in different extent the structure with such conjugated alternating single and double bonds does not represent the real molecule right because delocalisation takes place.
While the terms are linked, a not very accurate use of them leads to their treatment as if they are synonyms. That is what I have more or less done above, but just using a more nuanced term as pi-conjugation, thereby referring to electrons rather than bonds.
This is as to say that a sequence of alternating single and double bonds (as in at least one limiting form) is conjugated by definition. The conjugation (would be better saying delocalisation, here) can eventually/effectively be so true to the point of erasing the original meaning, i.e. in fact there are even not localized conjugated bonds.
Whit all the above in mind, a perfect organic metal with equalised bonds would be seen as "the king" of conjugated polymers.
Finally, within the context and dictionary of  VB the above delocalisation is described by resonance, This again lead to the use of delocalisation and resonance as interchangeable terms.

Oh ok so alternating single and double bonds are conjugated by definition and have delocalization only to a small extent (due to several but minor resonance structures). And conjugated molecules like benzene withe multiple major contributing structures  have significant delocalization and are best described by a resonance hybrid. Is this understanding correct?

@OVERWOOTCH yes. Aromatic indeed are not really called or seen as conjugated moleculed, but you can somewhat say there is pi-conjugation to intend resonance /delocalisation bla bla bla. Case in between are conjugated polymers, they do have delocalisation but not the point of removing single and double bonds alternation. i.e. the actual hybrid has localized bonds, just for various reasons the weight of the opposite facing limiting forms isn't equal. I would say you got it.

I would simply say that conjugation (alternate single and double bond in a molecule) results to resonance (delocalization of pi electrons) in that molecule.
Fair and simple, conjugation is the cause and resonance is its effect meaning that conjugation in a structure causes resonance...

One should be careful while dealing with these concepts. On first look it might appear that simple, but it usually is not. This answer also reads more like an opinion than actually something fact based. The other answer is at least somewhat more detailed.

Resonance and conjugation are defined in the IUPAC gold book.


text:
Is it valid to calculate concentrations given equilibrium constant and initial concentration?
Suppose we have the following reaction:
$$\ce{2A <=> B + C}\tag{1}$$
which can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$
with equilibrium constants $K_1$ and $K_2$ respectively. The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:
$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$
If we solve (4) for a given $x$ (initial amount) then concentrations of $\ce{B}$ and $\ce{C}$ must be equal. But how can this be possible if $K_1$ and $K_2$ have different values? I mean the following equality must holds:
$$[\ce{C}]= [\ce{B}] \Longleftrightarrow K_1{[\ce{A}]} = K_2\ce[{A}] \tag{5}$$
Edit I am asking if it is valid to use the equilibrium constant $K$ to find the concentrations of $\ce{B}$ and $\ce{C}$ because every reaction can be thought of as sum of other reactions. It is a common exercise in many general Chemistry textbooks where you are given a reaction with an equilibrium constant and you must find the concentrations at equilibrium. Well the substitutions into equilibrium constant that they make is the one that I have also did (assuming 0 initial concentration for both $\ce{B}$ and $\ce{C}$. So what is going wrong?

It's wrong to set both [B] = y and [C] = y.

The first equation means that two A molecules are needed to produce one B and one C, the second two equations mean that one A splits to produce one B and one C because you cannot distinguish one A from another.

Remember that $\frac{[B]}{[C]}=\frac{K_1}{K_2}$. Summation of the concurrent reactions  is wrong, as it creates false impression that it is a bimolecular reaction producing equal molar amounts of products.

I disagree that (1) = (2) + (3). What if A is a dimer (e.g., $\ce{D2}$) with $\ce{B}$ = $\ce{D}$ and $\ce{C} = \ce{D3}$? Then (2) and (3) would not even be valid equations.

The idea of reaction chains is correct, but your example is wrong.

@Poutnik But if I am given a reaction how could I know if it is a bimolecular reaction or just two concurrent reactions?

@Anton For the former, $\Delta [\ce{B}]=\Delta [\ce{C}]$, for the latter, $\frac { [\ce{B}]}{ [\ce{C}]}=\frac{K_1}{K_2}$

@anton P.S.: if the letters are substituted by particular  compounds, you would see it at the first glance. If B and C do not have the same summary formula, parallel reactions are excluded.

The equation
$$\ce{2A <=> B + C}$$
implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be
$$\ce{(x + y) A <=> x B + y C}$$
Here is an example that would be well-described by reaction (1):
$$\ce{2 H2O <=> H3O+ + OH-}$$
One water molecule cannot make a hydronium ion (goes against the conservation of mass and charge). The same goes for a water reacting to form a hydroxide. Instead, for every hydronium that forms, exactly one hydroxide is formed. So the separate reactions do not exist, and the separate equilibrium constants do not either.
Of course, there are other reactions where two independent products exist, such as linear glucose forming the alpha or beta anomer of the pyranose ring. In this case, there are two equilibrium constants, but there is no combined equilibrium constant because the two reactions are not the steps of an overall reaction, but rather two distinct reaction paths.

Your last paragraph shed light on my confusion. But how it is possible to know if a chemical reaction (whose formula is consistent with the laws of mass and charge conservation) isn't one of the examples you described in your last paragraph where two inepedent products exist?

A would have to be an isomer of B and of C. Also, you would be able to measure the two individual equilibrium constants. In general, if balancing an equation is ambiguous (has multiple solutions apart from a common factor), it is made up of stoichiometrically independent reactions. @Anton

You have stated that the reaction:

$$\ce{2A <=> B + C}\tag{1}$$
[...] can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$

However, this is not always the case. In particular, your assertion that the overall reaction (1) is a linear combination of reactions (2) and (3), is based upon the assumption that the rate of forward reaction ($r_f$) and rate of reverse reaction ($r_r$) are equivalent.
I have re-written your reactions (2) and (3) in more detail below (including the forward and reverse reaction rates):
$$\ce{A <=>[\ce{r_1}][\ce{r_2}]}B\tag{2*}$$
$$\ce{A <=>[\ce{r_3}][\ce{r_4}]}C\tag{3*}$$
Chemical systems are dynamic. At equilibrium, even though the concentrations specified by the reaction equation may not be changing, it does not imply that reaction is NOT occurring. In actuality, at equilibrium, all reactions are occurring, just at equal and opposite rates. If one looks at the equation for $K$, the equilibrium constant, one would see that this can also be expressed as the ratio of the forward and reverse reactions for the reaction system, such that:
$$K = \frac{r_f}{r_r} $$
In your case, you are assuming that in equations (2*) and (3*), that $r_1=r_3$ and $r_2=r_4$. Assuming this allows equations (2*) and (3*) to be "added" together. As such, these equations revert to the forms you've outlined in (2) and (3) and when combined they constitute equation (1).

Now, regarding the portion of your question:

The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:
$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$

The equilibrium constant, $K$, is specified in terms of the equilibrium concentrations of each reaction constituent. For the reaction system below, we wouls specify $K$ as being,
$$\ce{2A <=> B + C}\tag{4*}$$
$$K = \frac{[B][C]}{[A]^2}$$
In particular, $[A]$, is the equilibrium concentration of A. Modifying $[A]$ to be dependent on the initial amount of A is contradictory, as $[A]$ should not be dependent on the initial amount of itself.


text:
How to assign overlapping multiplets in 1H NMR spectra?

In the above $\ce{^1H}$ NMR spectrum ($\pu{400 MHz}$, $\ce{CDCl3}$) the methyl, alkene, and $\ce{CCl3CH2}$ protons are fairly easy to assign however, the protons in the cyclohexene ring show overlap and complex splitting (which I assume is due to the axial and equatorial arrangements). How would one go about differentiating between and assigning the remaining protons to the overlapping signals (if possible)?
I've actually been trying to find more generalised information to help with the assignment of a different (but similar) cyclohexene compound but couldn't really find anything, so took a suitable example from a paper I found.

Referring to your comment on Buttonwood's answer:

Unfortunately, in this case I only have 1D HNMR and IR available to analyse the product

If this is really the case, then you are essentially out of luck. It is not possible to extract much information from a series of overlapping multiplets, which may well be coupled to each other (these strong coupling effects are likely the cause of the "complex splitting" you mention). It may be the case that by going to a higher field you can alleviate some of this, as multiplets will be more dispersed along the frequency axis. However, you will quickly run into a hard limit this way, since our maximum field (as of the time of writing) is ~1.2 GHz and even that is likely not enough to fully resolve your overlapped multiplets.

With that out of the way, let's talk about some other NMR techniques which will be useful in this situation. I want to preface this with a few points:

These are general suggestions which are broadly applicable to a variety of different problems; they are not tailored to the specific molecule which you asked about, and indeed you may find that some of them are less suitable for your specific compound. For the specific case at hand I would suggest talking to a NMR specialist at your institution, if this is at all possible.

There is a caveat in that many of these techniques have slightly poorer performance in situations where there is strong coupling, although it by no means makes them useless.

You may find that you need more than one of these to get a complete understanding of the $\ce{^1H}$ spectrum (especially in light of the previous caveat).

I assume here that the overlapping multiplets belong to the same species, as is the case in the question. In the situation where you are studying a mixture of different species which may have peaks that overlap with each other, the following techniques will still remain very useful, but there are more options which will help to separate the different subspectra based on molecular properties (e.g. relaxation rates or diffusion coefficients). These are outside the scope of the current answer.

Finally, or further reading about all of these techniques I recommend Claridge, T. D. W. High-Resolution NMR Techniques in Organic Chemistry, 3rd ed.; Elsevier: Amsterdam, 2016..

(1) 2D correlation NMR
This has already been mentioned by Buttonwood, and is the standard answer to issues of resolution in 1D spectra. The downside is that 2D spectra are very, very rarely run with high resolution, partly because of time constraints, but especially so if heteronuclear decoupling is used (e.g. in a HSQC). Thus, instead of measuring well-resolved multiplet structure (which would be transferable to the 1D spectrum, thus allowing you to assign the 1D), you are more likely to get broad blobs that are centred on a certain frequency. This tells you nothing about the multiplets actual proton spectrum, it only tells you that at (say) 1.46 ppm you have some kind of multiplet that corresponds to a certain proton in your structure.
Despite this, one should not neglect the amount of information that is provided by standard 2D spectra (COSY, HSQC, HMBC, NOESY). Often even having a blob is enough to figure out which proton is which, even if it may not tell you the exact shape of the underlying multiplet.
(2) Pure shift NMR
This is a series of techniques which remove all coupling information, thereby collapsing multiplets into singlets and giving a 1D spectrum with exactly one singlet per (chemically inequivalent) proton. I would consider the TSE-PSYCHE method[1,2] to be the "state of the art" in this regard, especially with regard to strong coupling. This is very powerful but suffers from the same drawback as 2D NMR does: it removes all multiplet structure, so on its own does not lend any insight into the original $\ce{^1H}$ spectrum. Nevertheless, I would say that it is really worth having a pure shift spectrum in addition to some of the other methods, as it can greatly aid in assigning peaks.
Here's an example of a pure shift spectrum, versus the original proton spectrum of the same compound (personal data).

(3) 2D J NMR
A 2D J spectrum shows proton chemical shifts along one axis and proton multiplet structure along another axis. What this means is that every multiplet in a $\ce{^1H}$ spectrum is separated along a secondary axis according to its chemical shift. This allows you to not only figure out the position of each multiplet (which both of the above already provide you) but also the shape of each individual multiplet.
In particular, pure shift J spectra[3] are really the best versions in this regard as these provide much higher resolution and do not take a lot of time. To be technical, these don't provide "pure shift" characteristics because the 2D J $f_2$ axis (the horizontal axis) is already "pure shift", with the coupling information already separated out onto the $f_1$ axis. What they do is to convert the lineshapes in the spectrum from a "phase twist", which has long "tails" in both dimensions, to a pure absorption Lorentzian which is a single sharp peak. This is an example of how it looks (personal data again):

You can extract vertical traces from this spectrum to view each individual multiplet, with its chemical shift given simply by the value along the $f_2$ axis (horizontal axis).
(4) Selective excitation experiments
This refers to a family of 1D experiments where you selectively excite one multiplet and then perform some kind of mixing in order to look at spins that are coupled / near to the spin that you excited. A selective 1D NOE experiment (for example) would tell you which spins are in close proximity to a given spin that you chose to excite; it is analogous to acquiring only one row of a 2D NOESY. You could also run selective 1D TOCSY experiments, which would tell you which spins belong to the same spin system as the one you excited. By choosing an appropriate TOCSY mixing time you can limit the "spin diffusion" to one or two coupled spins away.
The benefit of this over the original 2D experiment is twofold: first, it takes much less time; and secondly, you can acquire much higher-resolution data. This allows you to disentangle different overlapping multiplets based on their coupling partners or spatial proximity to other spins. For example let's say you have two multiplets, corresponding to protons A and B, that are overlapping, but only proton A is in close proximity to proton C (whose multiplet is well-resolved). Then you can run a selective 1D NOE experiment where you excite proton C, and the resulting spectrum should include only multiplet A and not B. The same can be applied for scalar couplings via 1D TOCSY.
(5) Chemical shift-selective filters
If you find that you need to excite a peak inside an overlapped region, then the idea of a chemical shift selective filter may be useful. This seeks to excite one multiplet out of a series of overlapping ones, and traditionally involves recording multiple spectra and adding them up, allowing the multiplet of interest to add up while averaging all other multiplets to zero by virtue of them having a different phase. However, there is a recent implementation called GEMSTONE[4] which does this very cleanly in just a single experiment without needing to sum anything up. In my limited experience it works quite well although some of the experimental parameters need to be optimised for each peak to obtain best performance. This particular demonstration that I have (personal data, again) isn't so impressive because the overlap isn't as severe, but the paper has slightly more interesting examples.

References

Foroozandeh, M.; Adams, R. W.; Meharry, N. J.; Jeannerat, D.; Nilsson, M.; Morris, G. A. Ultrahigh-Resolution NMR Spectroscopy. Angew. Chem. Int. Ed. 2014, 53 (27), 6990–6992. DOI: 10.1002/anie.201404111.
Foroozandeh, M.; Morris, G. A.; Nilsson, M. PSYCHE Pure Shift NMR Spectroscopy. Chem. Eur. J. 2018, 24 (53), 13988–14000. DOI: 10.1002/chem.201800524
Foroozandeh, M.; Adams, R. W.; Kiraly, P.; Nilsson, M.; Morris, G. A. Measuring couplings in crowded NMR spectra: pure shift NMR with multiplet analysis. Chem. Commun. 2015, 51 (84), 15410–15413. DOI: 10.1039/c5cc06293d.
Kiraly, P.; Kern, N.; Plesniak, M. P.; Nilsson, M.; Procter, D. J.; Morris, G. A.; Adams, R. W. Single‐Scan Selective Excitation of Individual NMR Signals in Overlapping Multiplets. Angew. Chem. Int. Ed. 2021, 60 (2), 666–669. DOI: 10.1002/anie.202011642

Great answer. Cool techniques. I think I'm out of touch for so long now do not know these exists!

@MathewMahindaratne It so happens that I work on some of these things, which also explains why I have data to hand ;-)

In short:

Recording NMR is not limited to $\mathrm{^1H}$, but for isotopes of other nuclei, too if their magnetic spin is not zero.  $\mathrm{^{13}C}$, for example is an other routinely recorded sample.

Literally, you don't have to stick to 1D, but may record correlation spectra,  (not exhaustive list:) either between nuclei of same type (e.g., $\{\mathrm{^1H,^1\! H}\}$-COSY), or different type like $\{\mathrm{^1H,^{13}\! C}\}$-HSQC in 2D.  Depending on sample, spectrometer, and NMR experiment, these maps may resolve better / completely the overlap of the superimposed signals, than your 1D experiment if you select a good trace along one of the two dimensions and direction of observation (projection).  For example in the $\{\mathrm{^1H,^{13}\! C}\}$-HSQC below:

(credit)
you not only see the sum of all signals in the $\mathrm{^1H}$ domain on top, but you may see this map «in slices» for different resonance frequencies / ppm for the $\mathrm{^{13}C}$ domain (here: vertical direction).  Thus the signal (6,6') has a correlation around $\pu{3.8 ppm}$ ($\mathrm{^1H}$) and $\pu{72 ppm}$ ($\mathrm{^{13}C}$); while (5,5') displays a correlation around $\pu{3.9 ppm}$ ($\mathrm{^1H}$) and $\pu{82 ppm}$ ($\mathrm{^{13}C}$).
Rules you have learned, e.g. about coupling constants of vicinal protons and coupling patterns, apply again.  But the topic is too large for a one-fits-all answer on ChemSE.  For example for structure elucidation of e.g., proteins, correlation spectra routinely observe three domains simultaneously (H,C,N) in 3D NMR, too:
[
(credit)
It is the careful combination of multiple NMR experiments which eventually leads to the assignment of a structure; typically in conjunction with other techniques (MS, IR; X-ray) and knowledge of the sample's history (previous steps of isolation / synthesis).

This is a great answer, and I am fully aware that the assignments would be a lot more clear using higher dimension NMR. Unfortunately, in this case I only have 1D HNMR and IR available to analyse the product that I've obtained (which is a similar but different compound as to the one I posted).

In a simplified 1H NMR spectrum (no multiplets at all), a prediction e.g., on nmrshiftdb already may be read like «likely some of the experimentally recorded signals will overlap a lot» (see). Check if you really need to attribute all signals in the 1H NMR to a specific atom. Often, multiple 1H signals are reported like 1.5--1.7 (br., 5H), with more detail in 13C about individual C atoms (frame).

(continued)  This abbreviated report is good enough if there are sufficient complementary data available, either a) for this very molecule (e.g. a set of mp, MS, X-ray matching a reference). Or b) the description may contain more details for a step later in the course of the synthesis (e.g., a plane of symmetry in the molecular structure simplifies the shape of NMR spectrum).

#3: If still not satisfied, see if an other institution may record the spectra missing for you; some offer academic discounts (if applicable).  Double resonance experiments equally may be used to simplify regions of 1H NMR spectra, too (10.1063/1.1717118, https://www.degruyter.com/document/doi/10.1351/pac197440010159/html or old.iupac.org/publications/pac/1974/pdf/4001x0159.pdf).

I'll try to give some information that would help you according to your statement in the question:

I've actually been trying to find more generalised information to help with the assignment of a different (but similar) cyclohexene compound.

First of all, you may have found now from the two answers given elsewhere that there are no way you can assign all peaks without additional information such as 2D-COSY or other experiments. However, in general, for cyclohexane (or cyclohexene) ring protons, equatorial protons give rise to resonances downfield from their axial counterparts. For instance, the axial and equatorial homoallylic protons in cyclohexene (e.g., two protons at $\ce{C}$-5 in your structure), which give rise to a resonance at $\pu{1.65 ppm}$ at room temperature (the sample is a rapidly inverting mixture of conformers at room temperature). If you are able to gradually reduce the working temperature, the corresponding axial and equatorial resonances would appear to be separated, at very low temperatures, by approximately $\pu{0.4 ppm}$ (this value is an equivalence of $\pu{160 Hz}$ in $\pu{400 MHz}$ spectrum). It is interesting that the value is not greatly different to that in cyclohexane itself (Ref.1). Of course, this is the case only in the absence of complicating factors. Following cases are few exceptions to the general cases:

The axial-2-proton in 2-bromo-4-tert-butylcyclohexanone (where 2-bromo and 4-tert-butyl-groups are in cis-orientation), displays a resonance at around $\pu{4.50 ppm}$ while the equotorial-2-proton in its stereoisomer (where 2-bromo and 4-tert-butyl-groups are in trans-orientation) displays the corresponding resonance at around $\pu{4.26 ppm}$ (Ref.1).
Similarly, the axial-2-proton in 2-bromo-4-phenylcyclohexanone (where 2-bromo and 4-phenyl-groups are in cis-orientation), displays a resonance at around $\pu{4.87 ppm}$ while the equotorial-2-proton in its stereoisomer (where 2-bromo and 4-phenyl-groups are in trans-orientation) displays the corresponding resonance at around $\pu{4.38 ppm}$ (Ref.1).
Also, if you consider 2-methoxy-trans-decalin-1-one, there are two stereoisomers where 2-methoxy group is either axial or equatorial. Since trans-decalin confirmation is relatively "fixed," its corresponding 2-peoton is also either equatorial or axial, respectively. Accordingly, axial-2-proton gives rise to a resonance at around $\pu{3.58 ppm}$ while its equatorial-counterpart gives rise to a resonance at around $\pu{3.36 ppm}$ (Ref.1).

Now, if we concentrated on your spectrum, as you stated that one can easily assign resonances at around $\pu{5.80 ppm}$ (1H) and $\pu{3.34 ppm}$ (2H) for 2-H olefinic proton and methelene group on $\ce{CH2-CCl3}$ function, respectively. The two singlets at around $\pu{1.80 ppm}$ (3H) and $\pu{1.75 ppm}$ (3H) are also assigned as two methyl groups on $\ce{(CH3)2CBr}$-function. Keep in mind that these two methyl groups are diastereomeric since the function is attached to a chiral carbon.
Then, the rest of the protons are in the ring. A three proton multiplet at around $2.48$-$\pu{2.23 ppm}$ and two proton multiplet at around $2.15$-$\pu{2.00 ppm}$ can be tentatively assigned as allylic protons due to their chemical shift values. However, the structure has only four remaining allylic protons (two proton each on $\ce{C}$3 and $\ce{C}$6). Thus, the extra proton should be one of remaining three homoallylic protons (one on the chiral $\ce{C}$4 and two on $\ce{C}$5). A one proton multiplet at around $1.48$-$\pu{1.35 ppm}$ is strongly suggested that it could be the homoallylic proton at $\ce{C}$4. However, I hesitate to assign it is as so because we don't know the stereochemistry of $\ce{C}$4. Finally, another one proton multiplet at around $1.68$-$\pu{1.58 ppm}$ can be also tentatively assigned as the last homoallylic proton. However, it is worth noting that the correct assignments of those ring protons are highly depend on advance 2D-analysis.
References:

L. M. Jackman, S. Sternhell, In International Series in Organic Chemistry, Volume 10: Applications of Nuclear Magnetic Resonance Spectroscopy in Organic Chemistry, 2nd Edition; Pergamon Press, Ltd: Oxford, United Kingdom, 1969 (ISBN: 0 08 012542 5).


text:
What is the carbon content, by weight, of vegetable oil?
I have only a high school understanding of chemistry, so please bear with me and my ham-fisted question.  I’m trying to understand the carbon content in vegetable oil.  I get that various oils have differing densities, so the measure needs to be “by weight”.  Molecular weight doesn’t help, I need the measure in useful measures like “grams per liter” (preferred).
(Also of note, I’m not certain I got the right tags)
UPDATE: Based on feedback, I looked up oil production per hectare to determine an oil type.  It appears that the highest production per hectare (which feeds into the analysis I'm working on) for grain-based oils is rapeseed.  For tree-based fruits, it's oil palm, and then there's algae oil.
I'm trying to establish how much carbon is embodied in the oil production of one acre of crop (a bit of information that probably would have been helpful to say first).

It's different depending on particular oil, also what you prefer isn't by mass.

Which vegetable oil ? There exist so many vegetable oils ...

Two key problems: there are lots of different oils so no general answer is possible; most "oils" contain more than one chemical component (even ignoring impurities). So you need to be specific in asking the quetion.

However, all the main oil molecules are long-chain (say like https://en.wikipedia.org/wiki/Omega-3_fatty_acid with 18 carbons on the backbone), with mostly hydrogen filling it out, and an occasional OH or O group added on. So roughly 2-plus a bit hydrogen per carbon, ignore the OH or O, and you get a carbon mass ration of a bit less than 12/14, or 80 wt.% carbon (rounding down).

@JonCuster, this is a sufficient answer for me (presuming it's reasonably accurate across the oils I identified in the update).  Turn it into an answer and I'll accept it.

Vegetable oils are mainly made of glycerol esters of oleic acid, linoleic acid and palmitic acid. For example, the fatty acids extracted from olive oil are a mixture of $74$% oleic acid, $11$% palmitic acid, and $10$% linoleum acid, plus 5% other acids. These proportions are average values that may change from one sample to the next one.
The mass proportion of carbon atom is $77.2$ % in glycerol trioleate, $78.3$ % in glycerol trilinoleate, and $75.8$ % in glycerol tripalmitate