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Dehydration of primary alcohol with a cyclobutadiene substituent

Note: Primary alcohol will follow E2 mechanism, therefore, removal of protonated water and proton (RDS) shall proceed in a single step in the above mechanism.
My questions are:
1- According to wikipedia "Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable."
As cyclobutadiene is not a stable compound, Is the primary alcohol (reactant) a stable compound?
2- Is the ring expansion step and the final product (benzene) correct?

Your note doesn't make sense. An E2 elimination would immediately lead to an alkene with no intermediates that could react via side pathways. The ring expansion (aside from the issues of arrow pushing described in one of the answers) is extremely unusual. Do you know of a literature precedent for such a mechanism?

Some context would help. Are you proposing a mechanism for a known reaction? If so, do you have a reference or at least conditions? Or you are suggesting a transformation?

@Zhe (1) So, is ring expansion not possible in E2 mechanism because elimination of protonated water and β H occur in a single step? (2) I have proposed this reaction for this unstable alcohol; it's not a known mechanism.

(1) Yes. (2) In that case, my opinion of the ring expansion being unusual has changed to it being highly unlikely. Note that that compound is unstable merely because it has a cyclobutadiene, which will dimerize via Diels-Alder.

@zhe So, the product will be 1-ethylidene cyclobut-1,3-diene which may be stable due to conjugation?

Whether your mechanism is right or wrong, the last step is not the RDS. It is the most obvious and fastest reaction among all other steps.

o_O This is all absurd! Doesn't make sense, no, not even hypothetically. Even if one used some powerful magic to conjure starting molecule, it would instantly react intramolecularly.

You have misinterpreted what Wikipedia has said. I'd put whole quote here:

Cyclobutadiene is an organic compound with the formula $\ce{C4H4}$. It is very reactive owing to its tendency to dimerize. Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable. Since the compound degrades by a bimolecular process, the species can be observed by matrix isolation techniques at temperatures below $\pu{35 K}$. It is thought to adopt a rectangular structure.

These somewhat stable cyclobutadiene derivatives are not just derivatives with a simple alkyl substituent. They are with bulky substituents, e.g., tetrakis(tert-butyl)cyclobutadiene. Therefore it is safe to assume that given compound (2-cyclobutadienylethanol) does not exist. However, as Aniruddha Deb has said, we can treat it like it is existing only for this theoretical question (we don't have other choice):
Since it is a primary alcohol it can't undergo usual $\mathrm{E1}$ elimination with an acid. The $\mathrm{E1}$ pathway through formation of a primary carbocation is not the most likely pathway here, because primary carbocations are extremely unstable. Therefore, more likely, this reaction undergoes through an $\mathrm{E2}$ mechanism where the transition state should be lower in energy. Accordingly the protonation of the alcohol group to give the good leaving group $\ce{-OH2+}$:

And then, a weak base (which could be either $\ce{H2O}$ or $\ce{H2PO4-}$, or even another molecule of the alcohol) could abstract $\beta-\ce{H}$ from nearby $\beta$-$\ce{CH2}$ group, leading to give the conjugated alkene with cyclobutadienyl group (vinylcyclobutadiene):

We really don't know how stable vinylcyclobutadiene is, but it should be more stable than original alcohol since it has additional resonance contribution with external vinyl group. This compound can undergoes either Diels-Alder or Cope rearrangement to give a bicyclo[2.2.0]hex-1,5-diene derivative. This derivative should be unstable due to severe strain, and expected to rearrange to Dewar benzene derivative, which then rearrange to benzene upon heating (Ref.1).
Reference:

L. Watts, J. D. Fitzpatrick, R. Pettit, "Cyclobutadiene," J. Am. Chem. Soc. 1965, 87(14), 3253–3254 (https://doi.org/10.1021/ja01092a049).

In real-life, Cyclobutadiene dimerizes at temperatures above 35 K by a Diels-Alder reaction (source: Wikipedia). However, this seems more like a theoretical reaction rather than a practical one, so treat it as such and stick to the task of finding the dehydration product of this cyclobutadiene derivative.
Yes, this step is correct (although the arrow-pushing mechanism seems a bit wrong). Benzene has a large resonance energy and the dehydration would favour the formation of Benzene. This is another question on similar lines, where toluene is formed by the dehydration of (cyclobutane-1,2,3-triyl)trimethanol.

+1, Can you please point out where the arrow-pushing went wrong?

arrow pushing doesn't show thr bond being formed between (1) and (6)

The linked mechanism is very different from what is being posited here.

@Zhe linked mechanism explains how benzene is formed from a smaller ring by successive ring expansion. Its not a point-to-point answer for this question, just an example that may be helpful.

But the linked example is a 1,2-shift, which is well-known in the literature. The example here is a 1,3-shift (?) coupled with a hydride migration that is highly unusual.

@Zhe This mechanism involves: (i) Hydride shifting between (2) and (3), (ii) Breaking of C3-C6 bond and (iii) Formation of C1-C6 bond. Which 1,3-shift you are referring to?

@Apurvium That's like 3 steps shoved into one. That's not the usual pattern for a hydride shift, which normally occurs onto a cation, which you don't have. What's the driving force for bond breaking/bond forming? Are you going through an intermediate? That seems unlikely. And if it's single step, that is a 1,3-shift. My advice to you is to not write mechanisms for reactions that you don't know take place...

@Aniruddha Deb: Your answer for #2 question is incorrect. Whether OP's mechanism is right or wrong, the last step is not the RDS (OP's assignment). It is the most obvious and fastest reaction among all other steps.

@Zhe You were correct, which is now clear from the accepted answer. I had doubts and that's why I asked question by writing the mechanism. If I didn't mention the mechanism then it could fall under Homework question. I believe that is the purpose of this forum and that's how we learn. Thanks for your time!


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Is it possible to manipulate some specific atoms of a liquid metal by dint of the magnetic field, electric field, etc?
Let's deem, we have a liquid metal that is multicomponent, for example, type A, B, and C. Is it possible that we single out the type A and put them in a specific shape, for example, a circle?
Let me clarify it, I have a liquid metal, having three different elements, and I poured it in a square framework. I want to separate element C and then set them in several lines that are parallel with each other; so that, they play the role of reinforcement and the alloy of A and B is my matrix.

While Field Structured Composites are a thing, they use nano- to micro-scale particles in a polymer matrix, with particle alignment controlled/influenced by applying electric or magnetic fields during curing.
Trying to do this at the atomic level, as you ask about above, runs into several problems. These include:

Overcoming the desire of the liquid to mix, so you need your desired element to have a fairly well tuned positive heat of mixing to just balance the entropic contributions to the free energy. Or, in other words, just barely soluble.

Overcoming diffusion in the liquid would require strong interactions of your field with the specific element (and only that element). This runs up against the limits below.

In a metallic liquid, applying an electric field isn't going to specifically target any one atom type since the free electrons will shield the bulk from the field. Ions in a dielectric liquid would be a better choice (much like the field structure composites above), but that is just the reinvention of ion chromatography.

The magnetic moments of atoms are pretty darned small, so you won't get the ferromagnetic boost that can be achieved with particles. Applied magnetic fields will have to be enormous if they would even work at all.

source: Stack Exchange

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Detectable elements using flame AES
I've been looking for extra information concerning flame atomic emission spectroscopy. I need a clear overview of which elements are detectable with this technique, but I can't find anything because flame AES seems to be quite outdated.
If there is no such overview, could there be a list of elements not detectable by flame AES, but detectable by ICP-AES? That would also work for me.

It is not outdated. The vendors of ICP-AES are happy to share the information about limits of detection of the corresponding elements in surveys like this one https://www.ssi.shimadzu.com/industry/environmental/icp-aes.html

but I can't find anything because flame AES seems to be quite outdated.

Please don't say that you cannot find anything, this is the reason (i) why physical libraries still exist and (ii) right key words are needed for proper literature search and (iii) not everything is available on the web free of cost.
Most of the metals are detectable by flame atomic emission spectroscopy. All you need is the right flame. The instrument, called the atomic absorption spectrophotometer, works perfectly in atomic emission mode. There is no separate instrument for Flame-AES.
The flames are quite "cold" for non-metals like sulfur, carbon, nitrogen, and noble gases. ICP also works poorly for non-metals.

So, if I get it right, you'd say most elements are detectable by both flame and plasma? About the searching: sadly enough, in corona times and being a student without research budget, physical libraries and paying content are not really an option. Thanks for the help!

Yes, most of the metals and some non-metals can be detected. Search the term Atomic Absorption Spectrophotometer Cookbook on the web. It should be the first result. This is a thick book available on the web for free from companies. It lists the entire set of elements and the conditions under which they can be analyzed by flame atomic absorption AND emission.

@WilGos Your capital is diversified, and money is only one form.  An additional your experience.  If school/library/institutes in your area are not an option now, the combination of "atomic emission spectroscopy" "detection limit" in google still yields, for example, the Chemistry Libre Texts (https://chem.libretexts.org/) including the AnaChem by Harvey with chapter 10.7 about AES (https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Book%3A_Analytical_Chemistry_2.1_(Harvey)/10%3A_Spectroscopic_Methods/10.07%3A_Atomic_Emission_Spectroscopy).  By the way, last update: Jan 7, 2020.

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Approximating entropy change in a reaction using Arrhenius equation
Recently my teacher finished teaching the chapter Chemical kinetics in which I have gone through Arrhenius equation.Can we estimate the change in entropy using it? I have done my calculation, considering the reversible reaction, $\ce{A <=> B}$ which has forward and backward rate constants $\mathrm{k_f}$ and $\mathrm{k_b}$ with pre-exponential factors $\mathrm{A_f}$ , $\mathrm{A_b}$ with activation energies of $\mathrm{E_f}$ and $\mathrm{E_b}$ respectively.
According to Arrhenius Equation,
$\mathrm{k_f= A_f.e^{-E_f/RT}}$ and
$\mathrm{k_b=A_b.e^{-E_b/RT}}$
$\mathrm{ k_f/k_b = {A_f/A_b}.e^{{-E_f+E_b}/RT}}$
and $\mathrm{k_f/k_b}$ = $\mathrm{{A_f/A_b}.e^{-\Delta H^0/RT}} $
$\mathrm{k_f/k_b}$=$\mathrm{k_{eq}}$=$\mathrm{e^{-\Delta G^0/RT}}$
but, $\mathrm{\Delta G^0 = \Delta H^0 - T \Delta S^0}$
which gives $\mathrm{\Delta S^0 = Rln(A_f/A_b)}$
Will the final expression give an approximate value of change in entropy of reaction?

The thermodynamic quantities are the enthalpy and entropy of activation and are different to 'normal' $\Delta H^\text{o}$ etc. but  come from the transition state approach to reactions where reactants and products are assumed to be in equilibrium. We write down directly $K=(k_BT/h)\exp(-\Delta G^{o*}/RT)$


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activating and deactivating groups and directing effect
In electrophilic aromatic substitution:
I have just learnt a 'rule' about benzene with more than 1 substituents attached (excluding the effect on steric hindrance):

if a strong activating group (EDG) and a strong deactivating group (EWG) are attached, activating group wins hence the electrophile adds on ortho & para position relative to the activating group.

if a weak activating group (EDG) and a strong deactivating group (EWG) are attached, deactivating group wins hence the electrophile adds on meta position relative to the deactivating group.

I understand rule 2 (strong must dominate the weak), but I do not understand what is the principle behind rule 1:

e.g. methoxide group wins over nitro group, although they are both strong EDG & EWG. I understand if the methoxide group is replaced by a methyl group, nitro group wins as $\ce{R-CH3}$ only has a much weaker inductive effect (compared to strong resonance effect). However, both methoxide and nitro groups use resonance to exert their respective directing effect, why activating groups always win?

Is there any information on the positions of EDG and EWG in the first case? As in, was it referenced using an example?

@Safdar Do you mean the rule 1? The molecule shown in the photo (with methoxide & nitro) is the example I just came across

Came across this on the related sidebar. Does this answer your question? Electrophilic substitution on disubstituted benzene rings

@Safdar Thanks, I have read this post before posting this question. The solver tells us that a strong activating group has lone pair electrons for charge stabilization, hence it trumps over the deactivating group (the most-stable extra resonance structure). However, I don't think this is the sole factor, as the solver also says it maybe overall deactivating.

@Safdar I began writing the answer but then got mixed. The neutralization of charge accumulation is observed at the same location when the substituents are meta. The only reason I can think of is simultaneous resonace structures cannot be constructed. If electron density increases at a point ortho or para then that same charge accumulation cannot be again decreased by the EWG at the meta position, by drawing resonace structures. Thus the M- effect is lost. And only I- effect occurs wrt the EWG. Does this seem somewhat correct?

@ADITYAROYCHOWDHURY But that could also be true in the opposite direction, so it isn't the best argument.

@Safdar Ok Lets take it to chat.

@Safdar In most cases but not methoxy, I+ effect accompanies M+ effect. By drawing resonace structures, it appears that the two effects of M+ and M-  cannot occur simultaneously, and only one prevails. When M- occurs, the I+  of the other group may cause increase in e density at the ortho and para. This perhaps may explain the dominace of the meta directing effect in the second case as I + of EDG may be weaker then.

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present at exactly an ortho or para position where the $\ce{+M}$ effect operates, and the negative charge lies on the carbon alpha to the electron-withdrawing group.
It can be seen here that the negative charge in the resonance structures of $2,4$-dinitroaniline can be taken up by the nitro group only when the negative charge becomes alpha to it:

So the only way an electron-withdrawing group can "win" is if it placed against a very poor electron-donating group.
First, in the case of $\ce{-CH3}$ or halogens (who neither use $\pi$ conjugation for electron donation nor have an electron-donating nature), the electron density donated to the ring is unable to emerge in the intermediates without being delocalised by the EWG, or second when the EWG is present exactly at positions where the negative charge reaches by $\pi$ conjugation(in case of "strong" electron donators).
If the group donating via the $\pi$ bonds is even a relatively better donating group, then unless all the ortho and para positions are having electrons being withdrawn by the presence of electron withdrawing groups, there will always be resonance structures donated having some partial activation at ortho or para positions.
Hence, the electron-donating group "wins" in such situations unless there is no steric hindrance during the substitution taking place.

Thanks. In the presence of strong EDG & EWG, from resonance structures, the EDG donates electron density to ortho & para positions (those positions become more nucleophilic = activation) unless EWG is bonded to ortho/para, which withdraws the delocalized electrons. Therefore, in my case of meta-position nitro group, EDG trumps?

And in the case of weak EDG & strong EWG,  since the EDG only has an inductive effect only (no donation of electrons by resonance), EWG meta-directing wins due to the partial positive charge (less nucleophilic) from the resonance structures (resonance effect > inductive effect)?

Yep, pretty much what you have said sums it up @234ff


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I calculated internal energy at various temperatures and pressures. My results differ from someone else's. Is this OK?
I would like to make sure that my understanding of internal energy is correct.  I'm not a thermodynamicist, so apologies in advance for what is probably a basic question.
I previously calculated density for a liquid system via molecular dynamics (NPT ensemble).  I needed the results for use in fluid flow simulations.  The calculations also generated internal energy, which I have so far ignored.  However, someone has expressed interest in comparing my internal energy results with his.  He looked at the same system, but used a different calculation method.
His internal energies, E, for all temperatures and pressures are between -11000 to -12000 kJ/mol.  Mine are between -38000 and -40000 kJ/mol.
Here's my question.  Is it correct that I don't need to worry that one of us has made a mistake--that it's OK for the absolute values to differ as long as the differences are consistent?  For example, his highest-temperature, highest-pressure data point differs from his lowest-temperature, lowest-pressure data point by about the same amount as mine does.

Generally, absolute energies aren't meaningful and we are mainly interested in energy differences. The concern I would have in comparing your results is that yours seem to have a much larger range (2000 vs 1000).

@Tyberius, thanks very much for helping with such a basic question.  If you want to put that in an answer, I will mark it correct.

You may want to post this question at the material modeling SE site instead (I think Tyberius for one is a regular there).

@BuckThorn, thanks.  I'll do that if I haven't heard back from him by tomorrow--don't want him to think I deleted his current effort in case he checks back today!


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Difference in peaks in EDS and XPS spectra
I have been reading some papers on XPS and EDS and I saw that there's a difference between the width of the peaks (of the same material) in XPS and EDS. Is the reason for this difference because of the better energy resolution of XPS (spectral resolution)? EDS uses the signal from emitted X-rays, whereas XPS uses the emitted electron. There's a limit to what how much energy of the X-rays can be detected (about 2 eV) and electrons are easier in that respect and allow for measurement of lower eV.
Or is there another reason for the difference in the peaks?

Bluntly EDS measures energy of core electrons, and XPS measures the energy of valence electrons.

Recommend you start reading in a textbook, and perhaps before also here https://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy

I am reading a textbook and I have also searched for the answer on the wikipedia page of course, but I can't find the answer to my question. Perhaps I am missing a point, that is why I am asking it here.

The two techniques are used for different purposes.
EDS - Is a "bulk" analysis technique.

Since the exciting electron is of high energy it penetrates relatively deeply into the sample.

The emitted x-rays have a characteristic energy which is unaffected by the chemical bonding. (There is some effect due to bonding of course, but the difference in energy is infinitesimally small and hence insignificant.)

High energy x-rays can escape through the sample to a "great" depth.

XPS is a surface analysis technique.

Exciting electrons of low energy and only penetrate the surface layers of the sample.

The bonding energies of the valence electrons are low in energy.

The emitted electrons have a low energy (few eV) and hence don't penetrate out of the sample except for the surface layers.


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Resonance hybrid and octet
I am a student of grade 11. I am learning about resonance phenomena. I am a beginner with this and I am a bit confused about this. I wished to know whether atoms in resonance hybrid (not resonating structure) have complete octet or not. For eg in the resonance hybrid of the following resonating structure:-

If not, then does it gain stability by resonance or something else?
I request you to please help me.

Resonance with lone pair of nitrogen. Same as this

"Octet" starts to become a rather more fuzzy term when you start talking about delocalisation. Think about it: when you say "octet", you mean that each atom has eight valence electrons. Now if you can't assign every electron to a specific atom (or pair of atoms), then that starts to break down. You don't even know which atom has which electrons, let alone how many of them it has. But, if you really wanted to, one could very well argue that because the major resonance contributor(s) are formally octet complete, the molecule as a whole does satisfy the octet rule.

In that question, I wished to kow why oxygen, being electronegative and an element which completes it's octet by attracting electron disperses its electron density.

So, this means that a molecule would have many as electron as is required to fill up octet of every atom. But there is no guarantee of each atom having filled octet as it is senseless to talk about octet in resonance. Please confirm if this is the answer.

That's one way of putting it. I think the molecule you chose is not really the best example. The resonance form which you drew is probably the greatest contributor by quite a bit, so there isn't all that much resonance going on. Benzene is a more challenging example, and in that case, I think your summary would be quite apt.

Okay. Got it. Thank you so much for helping me.

@orthocresol do they not already all have a complete "octet" here? There seems to be a minor flaw in the understanding of OP.

@Safdar That’s kinda why I said this isn’t the best example.


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Thermochemistry questions regarding enthalpy
I have a few questions about heat of combustion. I have two different scenarios that I was to present:
Scenario 1: Given heat of combustion in the equation
$\ce{CH4 + 2O2 -> CO2 + H2O}$ and $\Delta H =\pu{-889kJ mol-1}$
Q1: Would $1$ mol methane produce $\pu{889 kJ}$; whereas, $1$ mol of oxygen would produce $\pu{444.5 kJ}$?
Scenario 2: Given the energy produced per mol of a molecular compound
$\ce{C2H6O + 3O2 -> 2CO2 + 3H2O}$ and $\Delta H=?$
However, we are given that the oxygen produces $\pu{400 kJ}$ of energy.
Q2: Would that mean that the $\Delta H=3 \times400=\pu{1200 kJ mol-1}$?
It would be really great if someone could explain the theories behind thermochemistry.

What do you mean by theories behind thermochemistry?

You should not separate the species on the left and assign enthalpy variations to each of them. On Scenario 1, you say that 1 mol of methane reacts with 2 mols of oxygen, and the variation of enthalpy for this is −889 kJ (since it is 1 mol of methane). If you had 2 mols of methane, and 4 mols of oxygen, $\Delta$H would be 2*(-889) kJ, and so on. As you mentioned, if only 1 mol of oxygen were available, only 0.5 mol of methane would react and $\Delta$H would be 0.5*(-889) kJ
So, for Scenario 2, you said it was given that

the oxygen produces 400kJ of energy

This probably means that, for each mol of oxygen that reacts with ethanol, 400 kJ of energy is released. So yes, if you wanted the $\Delta$H for the reaction, you should multiply by 3 and get that $\Delta$H = -1200 kJ mol-1.
This is in agreement with the measured molar enthalpy of combustion of ethanol (-1211 kJ mol-1)

Thank you @Rafael L ! Appreciate the explanation!


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Phenol vs 4-hydroxypyridine - Enol Content

Which has a higher enol content; phenol or 4-hydroxypyridine?

The solution to this question in my book says that phenol has a higher enol content due to aromaticity. But 4-hydroxypyridine is also stabilized by resonance. So how does that affect enol content?

Aromaticity in the keto form, if present, does matters too.

Does this answer your question? Stability of 4-pyridone vs 4-pyridinol

Your solution manual correctly predict that phenol has a higher enol content due to aromaticity. For example, you won't see strong carbonyl peak around $\pu{1700 cm-1}$ in FT-IR, which proves keto-form does not exists. It is 100% enol-form as phenol.
On the other hand, 4-pyridone is a different story. The dominance form, whether it is 4-pyridone or 4-hydroxypyridine, would be depend on which phase the compound is in. For example, the Ref.1 states that:

In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to the right in favor of the oxo-form, 2- and 4-pyridone, respectively in polar solvents as well as in the crystalline state.

Thus, it is safe to say, it is only in the gas phase, 4-hydroxypyridine dominate. However, in liquid phases, both form exist where predominance is depending on the solvent. In solid phase, 4-pyridone dominate. Since, both phenol and 4-pyridone are solid at ambient conditions, phenol has more enol form than 4-pyridone, which exist in keto-form as a solid.
Note that a public database set like Tautobase, implemented in DataWarrior, may be used to train yourself, too.
Reference:

Jiali Gao, Lei Shao, "Polarization Effects on the Tautomeric Equilibria of 2- and 4-Hydroxypyridine in Aqueous and Organic Solution," J. Phys. Chem. 1994, 98(51), 13772–13779 (https://doi.org/10.1021/j100102a051).
Oya Wahl, Thomas Sander, "Tautobase: An Open Tautomer Database", Chem. Inf. Model. 2020, 60, 1085–1089 (doi 10.1021/acs.jcim.0c00035).


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Why do atoms on the periodic table have the number of valence electrons that they do?
I'm studying year 11 and am learning about valence electrons and how these electrons drive intermolecular bonds.
What I have not been able to find an explanation to is why all the elements on the periodic table have the number of valence electrons that they do.
To me, it seems intuitive that if the halides are so eager to gain an electron, why would their natural state not be to have the 8th electron in their shell?

Well, but their natural state is to have the 8th electron (mostly). You might be talking about the elemental state, but elemental halogens are not commonly found in nature.

I see, I was confusing the elemental and natural states. If their natural state is to mostly have the 8 valence electrons, why do we mention the elemental state at all? Why is this what is displayed on the periodic table?
If the elements natural state is to be stable as such, how do chemical equations work? Should we not assume that the atoms of these elements in the world would then be stable by default and therefor less reactive?

We mention elemental state because it does exist, albeit sometimes only in the lab. This is what characterizes an element by itself, when not in a compound with other elements; that's why it is displayed on the periodic table for that element. The chemical equations work because they are true. The chemical reactions occur because... well, look around you, and you'll see a plenty of examples. There are no such things as "more reactive" and "less reactive". Reactive towards what, that is the question?

Atoms have protons (and neutrons) in the nucleus that define the element. For example, Fluorine always nine protons. The number of electrons must balance the number of protons in the atomic state. If fluorine had ten electrons, i.e. a full inner and outer shell, it would be a negative ion with net charge -1, so the electrostatic charge would repel the last electron.
It is only with the help of other atoms nearby that charges can be (almost) balanced. In $\ce{LiF}$, the fairly positively charged $\ce{Li+}$ ion attracts the comparatively negatively charged $\ce{F-}$ ion, so there is no net unbalance in a crystal of $\ce{LiF}$, though in each pair, the valence electron spends most time in the outer shell of fluorine.


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What exactly is peptone (tryptone?) and why does it not get precipitated?
According to Wikipedia Peptone (also called Tryptone) is a form of amino acid that has been treated by stomach acids. According to Google it also stays in solution when treated with acids or heat. Why is that so?

Sigma-Aldrich has listed long list of their Basic Ingredients of Microbiology, which are called Protein Sources (Peptones):

Peptones and extracts are excellent natural sources of amino acids, peptides, and proteins in growth media. They are most often obtained by enzymatic digestion or acid hydrolysis of natural products such as animal tissues, milk, and plants or microbial cultures. The number of available peptones and extract is enormous, and can promote and sustain the growth of most organisms.

Some protones are animal-based and some are plant-based. For instance, peptone water is a microbial growth medium composed of peptic digest of animal tissue and sodium chloride. The $\mathrm{pH}$ of the medium is $7.2\pm 0.2$ at $\pu{25 ^\circ C}$ and is rich in tryptophan. Peptone water is also a non-selective broth medium which can be used as a primary enrichment medium for the growth of bacteria (Ref.1).
References:

J.E.L. Corry, G.D.W. Curtis, R.M. Baird, Eds, Culture Media for Food Microbiology, Volume 34; 1st Edition, Elsevier Science: London, UK, 1995 (ISBN: 9780444814982).


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Is there a way of finding solubility/solvent from basic information?
Let's say I have a compound and I want to find a solvent for it.
Looking on some where like Wikipedia, it is quite easy to find solubility, in various solvents for common compounds. However it can be quite lacking for less well know compounds (For example Glycin).
Although one can quite often find a list of some basic metrics (Polarizability,  log D, etc) for the compound. Say on:
Chemspider
Given those basic metrics is it possible to infer what some likely solvents would be? Or What would be the name of those techniques?
*Assuming normal temperature and pressure etc

The CRC Handbook of Chemistry and Physics is a good source of info like this.  It lists solubility info for many compounds.

This is a broad question as solubility equally depends e.g., on the temperature of the solvent(s), presence of other (already dissolved) compounds. A database like Reaxys used by research institutes may indicate the publications with these (measured) values, or contain these values itself. Based on known data, programs may be trained to anticipate solubility of unknowns based on a structure drawn. E.g., for X in water background of a relevant program, Marvin, in function of the pH value, or an implementation of ESOL.


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Acids used in Friedel-Crafts alkylation
I am learning adding alkyl groups to benzene using Fridel-Crafts alkylation:

I know that it is very common to use $\ce{AlCl3/AlBr3}$ as Lewis acid catalyst in this reaction. But I am just wondering whether I can use a Bronsted acid, e.g., $\ce{H2SO4}$, as the catalyst or not. For example, in a $\mathrm{E1}$ or hydration reaction, $\ce{H2SO4}$ is added to alkene or alcohol, respectively, forming a carbocation, which is the same ultimate result when $\ce{AlCl3}$ is used (make the alkyl group more electrophilic).
I am thinking $\ce{AlCl3}$ is the only way if I am adding primary alkyl halide since in  $\mathrm{E1}$/hydration reactions, an actual primary carbocation is impossible to be formed (too unstable), while it exists in a complex with $\ce{AlCl3}$ (not entirely a carbocation, so can be formed).
SO, what is the problem with secondary/tertiary alcohols if I use the Bronstad acid? If there is a problem, way? Both Lewis and Bronsted acid ways will lead to carbocation rearrangements (I have not learnt the acylation).

Brønsted Acid has been used in Friedel-Crafts alkylation before (for example, Ref.1). According to this reference, when toluene use as the solvent, the F-C alkylation reaction of toluene is fruitful for tertiary and secondary alkyl bromides, tosylates, and alkenes (which give both carbocations upon protonation). However, the only primary bromide, which worked in this alkylation are benzyl and cinnamyl bromides. Other active primary bromides such as allyl and crotonyl bromides gave ditolyl products (e.g., crotyl group substituted first, then underwent double bond protonation followed by hydride shift to give secondary benzyl carbocation, which then reacted with the second toluene nucleus to give 1,1-ditolylbutane). Keep in mind that these reactions worked only with activated phenyl nucleus. For example, chlorobenzene gave only trace of products even with higher temperatures.
Later, this reaction is successfully used in Tandem-Friedel-Crafts alkylation reactions to synthesize hitherto unknown perylene derivatives (Ref.2):

References:

Mathew P. D. Mahindaratne, Kandatege Wimalasena, "Detailed Characterization of p-Toluenesulfonic Acid Monohydrate as a Convenient, Recoverable, Safe, and Selective Catalyst for Alkylation of the Aromatic Nucleus," J. Org. Chem. 1998, 63(9), 2858–2866 (https://doi.org/10.1021/jo971832r).
Mark A. Penick, Mathew P. D. Mahindaratne, Robert D. Gutierrez, Terrill D. Smith, Edward R. T. Tiekink, George R. Negrete, "Tandem Friedel–Crafts Annulation to Novel Perylene Analogues," J. Org. Chem. 2008, 73(16), 6378–6381 (https://doi.org/10.1021/jo800558c)(PDF).

Thank you. This means in general, the Bronsted acid pathway (i.e. protonation to give carbocation as you said) mentioned in my question is actually valid only if the benzene is activated by activating groups (EDG)? If it is not activated (just benzene), Lewis acid is the only option? Why is that?

@Rahul Verma: Yes, that's me. Thr ref.1 is part of my PhD work.:-) The mechanism of Lewis acid catalysis is different from Bronsted acid. It is always superior on both F-C alkylation and acylation (I attached PDF of Ref.2 to read; It is also Mark Penick's PhD work).


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What are the least dense fluids?
This question left me wondering what other very light (condensed) fluids could exist, at any environmental condition.
Obvious candidates for the list are (at their Bp./1 bar)

LH2 (0.07 g/ml)
LHe (0.125)
LNe (nope, 1.21)
methane (0.42)
ethane (0.54)
propane (0.58)

plus those mentioned in the other question

$\ce{4NH3 \cdot Li}$ (0.48)
isopentane (lightest "normal" liquid at 25°C, 0.62 g/ml)

To give a definit boundary on possible answers, let´s say the density must be below that of isopentane.
The question is, is there anything else? Perhaps at very high temperatures?
(The term "fluids" of course also includes supercritical fluids and gases, those are obviously out. You can always reduce the pressure, and thereby density, on those, until you either have a regular liquid, or you hit the bottom line of your phase diagram at $p=0$. That makes no sense, so I want actual condensed matter.)

What are LNe, LH2,LNe? Searching on google yields the value of $ \ln e$. Do you mean liquid Ne, $\ce{H2}$, He?

I believe that's pretty much the size of it. Maybe you want to add liquid diborane (at low T) and liquid Li (at high T), but those won't break the record.

@IvanNeretin Diborane would be a nice answer, it seems. I found a hardly readable scan of a NASA report that would land it in third place, but can´t get at the references.

What about the supercritical regime? The question needs to be bounded. Define "fluid".

@BuckThorn Supercritical makes no sense, obviously. Perhaps I should have said "liquid", but I didn´t want to exclude anything that´s a bit out of the box.

Well, the boundary between liquid and supercritical fluid is not well defined as far as I am aware, so it would appear to remain a question of semantics :-) In fact, I would be surprised if the theorem of corresponding states didn't have something universal to suggest about this.

According to Physics Reviews (2013), the lowest density liquid in nature is He-3 adsorbed on alkali metals as a liquid layer (at least in theory) https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.111.045303

@BuckThorn If you are somewhere within the supercritical regime, you can always reduce the pressure, which will inevitably reduce the density, until you are definitely in the liquid, or the density approaches zero. The former case then is relevant for this question, the latter sort of doesn´t matter. Sorry for the pun. ;-)

Well, these limitations were in original question to make it not overly broad. I think there may be need to make some other besides not gas or supercritical...

Fluid is anything that flows, i.e. liquids, supercritical fluids, gases and plasma. The least dense fluid, then would be lone atoms (or ions), in intergalactic space.

Some alkylboranes are light, around ~0.7 . But some are not, closer to 1.4-1.6

@permeakra Trimethylborane would be a good candidate, en:wp gives it 0.63 at -100°C, and it only boils at -20. The adduct with trimethylamine might be another idea. I was thinking about $\ce{BeH2}$, but wasn´t able to find if it would melt without decomposing at higher pressure.

@Karl Triethylborane is 0.68 with bp +95 mp -93. Not isopentane, but close enough in my opinion and a nicer liquid phase range

Okay, I need some clarification on this.

Do you mean liquid (like water or oil) or fluid (basically anything that flows, including gas and plasma)?
Do you mean density in the term of mass of mole?
Because @DrMoishePippik is correct, if you mean fluid, then intergalactic medium would be the least dense.

The least dense fluid is liquid Hydrogen $\ce{H2}$, at $-255$°C and $1$ bar. Its density is $0.071$  g/mL.

Actually, helium-3 has a density of $\mathrm{0.059\ g/mL}$ at its normal boiling point of $\mathrm{3.19\ K}$. For comparison, in helium-4 the respective values are  $\mathrm{0.1247\ g/mL}$ at $\mathrm{4.23\ K}$.

@NicolauSakerNeto No idea where I got the 0.18 for helium in the question. Tnx!


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Why don't phenol behave like aniline in electrophilic substitutions?
As far as I know, many electrophilic aromatic substitution reactions fail to undergo the usual way in aniline. This is due to $\ce{-H2N}$ group being basic (pKa of $\ce{NH3}$ = 38), even it is stabilized by the aromatic ring (resonance). Therefore, it often acts as a Lewis / Bronsted base (becoming a deactivating group), rather than being an activating group. Therefore, reactions like Friedel-Crafts reactions, nitration, sulphonation etc. will not undergo, unless the $\ce{-H2N}$ undergoes acylation.
e.g. Failure of Friedel-Crafts alkylation:

I am pondering why phenol does not behave the same way as aniline does. In my mind, fast proton transfer can happen during nitration and sulphonation (pKa of phenol = 10). It can also behave as Lewis Base and attack AlCl3 (presence of lone pair electrons). (or any other activating groups that the aromatic carbon is directly bonded to an atom that has lone pair electron)

But it seems the truth is that phenol just behaves as a strong activating group. What is wrong with my deductions?
edit:
This post says phenol actually behaves the same as aniline (according to the first sentence from OP), echoed by this website.
This website and Organic Chemistry 9th edition by Wade Jr. only says aniline fails, but not mentioning phenol.

Phenol can protonate on oxygen but the protonation is reversible.

@user55119 I know the -H2N in aniline is much more basic. But e.g. nitration, even with leveling effect, pKa of H3O+ is 1.7 (highest among all acids in solution), compared to 10 in phenol, so I think it is still 10^8 times more favorable for protonation of phenol, which is huge.  Please correct me if I am incorrect

Are you saying that the oxygen of phenol is 10^8 times more basic than the nitrogen in aniline? if so, I think not. Is ammonia more basic than water?

I know aniline is much more basic, so the protonation is irreversible. I mean in the solution of nitration (conc. HNO3 + conc. H2SO4), H3O+ is present, which has a pKa of 1.7. When pKa = 1.7 compared to pKa of phenol = 10.0, the equilibrium favors the phenol being protonated by around 10^8 times, which I think can be said irreversible too. This causes me to wonder why phenol does not behave as aniline in this case. Sorry for causing misunderstanding

Compare pKa of H2O (15.7) with phenol (10). H3O+ (-1.7 !!!!) with PhOH2+, whatever value it has. You are comparing apples and oranges.

Thanks! Now I see the mistakes I have made.


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What is the pka of dibromoacetic acid?
There is a reference of 1.48 value at room temperature which is a different value from an experimental study.

Could you tell us what reference this is? and what experimental study?

That's a reasonable value for water pKa of such acid. BTW if the difference you found isn't more then 0.1-0.2 it's not like either of them is necessarily "wrong".

An IUPAC reference of ionization constants. An experimental study that shows the breakdown pathway of another chemical.

It's very different than the reference.


text:
What is exposed in the surface tension of water?
(I guess) I understand the covalent bonding of water and the hydrogen bonding of water between two different molecules, but I would like to know which part is the part that that exposes itself to the air in the surface tension of water. There should be something different to the rest of the liquid, otherwise there would be no frontier, right?
Is it the electropositive part of hydrogen or is it the electronegative part of oxygen that faces the air? (or may be none of them).
Surface tension in water owes to the fact that water molecules attract one another, as each molecule forms a bond with the ones in its vicinity. At the surface, though, the outmost layer of molecules, has fewer molecules to cling to, therefore compensates by establishing stronger bonds with its neighbors, this leading to the formation of the surface tension.... more here:

https://www.usgs.gov/special-topic/water-science-school/science/surface-tension-and-water?qt-science_center_objects=0#qt-science_center_objects
Thanks to all for your help (I'm in chemistry '101' for dummies so please try to be as basic as possible.
Video: https://www.youtube.com/watch?v=moITG5Q7zzI
(Hydrogen bonds https://www.khanacademy.org/science/high-school-biology/hs-biology-foundations/hs-water-and-life/v/hydrogen-bonding-in-water)

Hmm, I don't know if one orientation is more prominent on average. But I wanted to say that water molecules are moving all the time, faster than we might think. So they can't, for example, sit in one particular orientation with the oxygen always facing the air, and then stay there for good. But over time, on average, it seems plausible that you may have some imbalance in the orientation ("anisotropy"...?), or indeed you may have none.

It may be either part; it does not matter. The molecules on the surfaces are no more ordered than those in the bulk.

They could be sideways. The surface is different from the bulk. You can see that it has a skin-like nature by overfilling a glass with water. This effect is more dramatic if you put some water in a dish, sprinkle some black pepper grains on it (no effect), then touch the surface of the liquid with a bar of soap. ZAP! The skin tears and is pulled to the edges of the water layer - pulling the pepper grains with it.

Surface tension may be described regardless if all / some / none of the water molecules expose one or both hydrogen atoms or the oxygen atom toward air, even if there are studies in this field (e.g., example, example).
Key points however are

the movement (translation, vibration) of the water molecules in the bulk of water droplet in restained, because of intermolecular attractive forces (e.g., hydrogen bonds), as well as intermolecular repulsion (molecules normally do not come closer to each other than the sum of the corresponding van der Waals radii).  Like with humans, there is an optimal intermolecular distance where overall the energy passes a minimum.
In extension of your drawing of the situation in the water droplet, I add the following figure where molecules are reduced to a black sphere; these are either in the inner of the droplet (light blue background), or at the interface (dark blue background) exposed to air.  The black arrows indicate intermolecular attractive forces (responsible for cohesion).

(derived from wikipedia)

For a molecule in the inner of the droplet, surrounded completely by molecules of the same type, the net force on this molecule is zero.

For a molecule at the interface water / air, the attractive forces to molecules of water in the inner of the water droplet exceeds the attractive forces between this water molecule and air molecules.  Thus, there is a net force pointing to the inner of the water droplet; this is symbolized by the red arrows.
Because of this inward directed force, the droplet thus tends to contract, building an inner pressure until compensated by the earlier mentioned intermolecular repulsion.

Overall, the lowest energy of the water droplet would be achieved as a sphere.  Sometimes, this is shown in microgravity (Scott Kelly at 8:28min).  On earth, however, things like gravity, contact with a surface (adhesion), relative air speed (e.g., if a drop falls down the rain gutter) etc. lead to deviations from this ideal.  However no, because the molecules in the inner of the water droplet are the same as the ones at the interface exposed to the air, there is no difference in the intermolecular, pairwise attractive forces between molecules of same type.


text:
Acid catalysed dehydration of 5-methylidenecyclopent-2-en-1-ol

I think the following reaction is NOT possible.

I could not find a mechanism for ring expansion. Is ring expansion possible in any way?
Is the following mechanism correct?

Think over resonance b/w charge and pi-bond.

@RahulVerma I have edited the question based on your comment.

Ring strain is not very significant here. For ring expansion to occur methanide shift should take place instead of hydride shift which has higher migration amplitude and that too if a primary carbocation forms violating markovnikov rule.

I'd say your mechanism is acceptable. However, the final product, pentafulvene (PIN: 5-Methylidenecyclopenta-1,3-diene) is said to be not stable at room temperature. Generally, fulvenes are thermally unstable, sensitive to oxygen and photosensitive (Ref.1). They are also prone to acid- and cation-catalysed polymerisations. For example, there have been numerous reports of pentafulvenes undergoing dimerization via Diels–Alder cycloadditions at room temperature similar to that of cyclopentadiene (e.g., Ref.2):

Based on these facts, I'd like to suggest an alternative path for the reaction giving much stable cyclopentaenone derivative:

References:

Ellen Swan, Kirsten Platts, Anton Blencowe, "An overview of the cycloaddition chemistry of fulvenes and emerging applications," Beilstein J. Org. Chem. 2019, 15, 2113–2132 (doi:10.3762/bjoc.15.209).
Beat Uebersax, Markus Neuenschwander, Hans‐Peter Kellerhals, "Fulven‐Dimere: Synthese, Strukturbeweis and thermisches Verhalten," Helvetica Chimica Acta 1982, 65(1), 74-88 (https://doi.org/10.1002/hlca.19820650109).

It is also likely that the 5-methylcyclopentenone might isomerize via the deconjugated isomer to 2-methylcyclopentenone and then undergo the Diels-Alder.

@user55119: Yes, it is possible based on the conditions used.

Is there some aromaticity angle to the instability of fulvene that is, the external pi bond can enter into some kind of conjugation with the internal pi bonds and make an antiaromatic system with 4 pi electrons?

@Yusuf Hasan: Pentafulvene ($\ce{C6H6}$) is a polarmolecule with carbon-6 is having partial positive charge and carbon-6 is having partial negative charge.

@MathewMahindaratne: I believe you mean carbon-5, the ring for that matter, has a partial negative charge.;)

@user55119: Yes, that's what I meant. Carbon-6 is exocyclic carbon.

@MathewMahindaratne Does this factor of partial charge so great that in the first step $\ce{H+}$ would prefer to attack the exocyclic $\ce{C}$ instead of electron rich $\ce{O}$ atom?

Also, as the required temperature is high for dehydration; does this make 5-Methylidenecyclopenta-1,3-diene a stable compound at that temperature? Cyclopentadiene is stable at high temperature.

Pentafulvene is produced at last step, so intial protonation has nothing to do with the partial charge; 2) Sure, cyclopentadiene is stable at high temperature, but here, pentafulvene reacts with oxygen as low as $\pu{-35 ^\circ C}$ (Ref. 1). If interested, you should read this paper (PDF available).

@MathewMahindaratne For 1) I was referring to the mechanism proposed by you where initial protonation occurs at C atom.

Protonation is in equilibrium including protonation of hydroxy group and double bond. Then, the lowest energy path in next steps wins. If reaction is in thermodynamic control (this reaction is as it is heating), most stable product is the result.

@MathewMahindaratne Is initial protonation permitted in cyclic double bond?


text:
Molar conductivity of coordination compound
The compound is $\ce{[Cr(H2O)6]Cl3}$. Which isomer has least molar conductivity ?
Correct answer : $\ce{[Cr(H2O)4Cl2]Cl.2H2O}$
My answer : $\ce{[Cr(H2O)3Cl3].3H2O}$
Molar conductivity is directly proportional to the number of ions on dissociation. If number of ions is same, the one with more net charge  has more molar conductivity. Therefore the compound I stated should have least molar conductivity as the only dissociation it undergoes is separation of coordination entity from water molecules. As none of them are charged ions, they wont increase molar conductivity. What is wrong with my concept ?

@MathewMahindaratne yes it does

You are right. But for some reason, your compound has never been synthesized, although theoretically ti should exist. It may be a problem of entropy.

https://chemistry.stackexchange.com/questions/66932/number-of-hydrate-isomers-of-crcl3-6h2o?noredirect=1&lq=1 This would be a plausible reason.


text:
Rules to follow while writing a chemical formula for polyatomic ions
In Science, textbook for class IX, chapter $3$ [1,p.38], it states:

The rules that you have to follow while writing a chemical formula are as follows:

in compounds formed with polyatomic ions, the number of ions present in the compound is indicated by enclosing the formula of ion in a bracket and writing the number of ions outside the bracket. For example, $\ce{Mg(OH)2}$. [...]

Now my book further states that hydroxide ($\ce{OH–}$) and nitrate ($\ce{NO3-}$) are polyatomic ions. And then it says:

Formula of sodium nitrate:  $\ce{NaNO3}$

But the formula of calcium hydroxide:  $\ce{Ca(OH)2}$
My question is that even though both nitrate and hydroxide are polyatomic, why are we using ( ) in notation for latter in their symbolic representations?

Which book is this from?

https://ncert.nic.in/textbook/pdf/iesc103.pdf - please look at pg 29 and also google search says so - but I could not get the reason why () are used for later and not for former?

Well, simply put if you wish to write it as so, you would have the compounds to be $\ce{NaNO3}$(sodium nitrate) and Calcium hydroxide as $\ce{CaOH2}$ instead of $\ce{Ca(OH)2}$. The former having 2 hydrogens and one oxygen, wherein reality, it has two oxygens and two hydrogens.

So CaOH2  instead of Ca(OH)2 are both same? So it is not important to have the brackets - google says Ca(OH)2 or CaH2O2 are same?

No.. $\ce{CaOH2}$ means that the compound has two hydrogens, one oxygen and one calcium. However we know that Calcium hydroxide should have two oxygens, two hydrogens and one calcium.. The entire composition changes. $\ce{OH-}$ is a hydroxide group. and we need two of them to balance charges.

So NaNO3 means one atom of Na and N but 3 atoms of O so thats why we do not use the bracket in it? Is there a way that we can understand when for polyatomic we need to make use of brackets and not - if we take the one in the question as example then one should be knowing that one atom of N is used instead of 3 for O but not in case for Ca(OH)2 - I mean is () means like group in mathematics?

Let us continue this discussion in chat.

I'll add a odd clarification to the question. The book chapter linked by the OP does indeed refer to brackets, [ ], but I've never seen anything but parenthesis, ( ), used to defined a polyatomic functional group. // Maybe this is a US vs British English thing??

@MaxW They are ( ), it is just that the font makes it look different. [ ] is used for co-ordination complexes.

@MaxW, comment in comment, lol

What they are trying to say is that hydroxide $\ce{OH^−}$ and nitrate $\ce{NO3^−}$ ions, which are polyatomic anions react bond with cations as a group.
When $\ce{NO3^−}$ bonds to $\ce{Na^+}$ only a single nitrate anion bonds to a single sodium cation, to form $\ce{NaNO_3}$ so brackets are not needed. You can still use them, in which case it would be $\ce{NaNO_3}$ but it is redundant and current convention doesn't use it.
However, when $\ce{OH^−}$ bonds to $\ce{Ca^{2+}}$ two hydroxide anions bond to a single calcium cation, to form $\ce{Ca(OH)_2}$ so brackets are needed to avoid confusion and show the structure of the compound.
Old ways of writing it (before my time) before the conventions were standardised, such as $\ce{CaO2H2}$ or $\ce{CaH2O2}$ don't immediately show that it is a hydroxide, so would have been confusing to non-experts.
To reiterate: the brackets are used when there is more than one polyatomic ion in a compound.  So if you write down sodium hydroxide it is $\ce{NaOH}$ but calcium hydroxide is $\ce{Ca(OH)2}$

See here on how to write professional chemical formulas, using the mhchem extension.

https://chemistry.meta.stackexchange.com/questions/443/which-symbols-are-written-in-roman-upright-font-and-which-are-italicized. According to IUPAC guidelines, they have to be upright.

There is a direct analogy in mathematics.  We write $2\cdot(a+b)$, but we do not write $1\cdot (a+b)$. We write just $a + b$.
So we do not write $\ce{Na(NO3)1}$, but just $\ce{NaNO3}$, as the meaning is obvious even without the explicit parentheses. Sometimes, the parentheses are used with implicit index 1, if it supports the structural clarity of complexes with mixed ligands, like e.g. $\ce{[Fe(H2O)5(OH)]^2+}$ or $\ce{[Fe(CN)5(NO)]^2-}$
OTOH, as there are 2 $\ce{OH}$ groups in calcium hydroxide, we writ0e $\ce{Ca(OH)2}$.


text:
Is it possible to achieve a level of "truly zero" concentration?
If we take some aqueous solution and dilute it further and further, will the concentration of the solution ever get to zero? I would say no, simply because total dilution implies that all the molecules of the solute have literally disappeared. But, the fact that I am unable to figure out where the molecules have gone doesn't make my argument compelling at all. I am led to believe that there is a far better answer and/or explanation to my questionUPDATE: While it is true that this question closely resembles the linked one, the answer provided in this question is a lot better as it gives a much deeper insight into the dilution process.

Homeopathy does it all the time :)

Sorry, Alchimista; that's simply not true. Homeopathy not only recognises but to a large extent depends on the idea that no such thing could ever be possible.

@RobbieGoodwin You're taking the the logic of homeopathy at face value and using it to justify homeopathy when you say that, so of course that would be the conclusion.

Of course you might find a bottle among others billions of billions containing 1 molecule of the diluted stuff. It does not classify as a solution of that particular stuff as for you can find whatever exist. I did not want to be too serious because the topic doesn't deserve. But there is a nice comment by @ Karl below. Also as a curiosity. Starting from a spoon of NaCl there is not enough water in the known universe to get a C 60 homeopathic solution in one step!!!

This is one of the questions where "In theory, there is no difference between theory and practice. In practice, there is."

@DKNguyen I wasn't taking anything at any value… simply repeating one of the basic tenets of homeopathy. That doesn't, and I didn't mean to, justify anything. When you say "of course that would be the conclusion" what conclusion were you thinking of?

It depends how you dilute it. If you take an aqueous solution of A and just add pure water (absolutely 100% water), the concentration of A will never quite be null. In this case however, you will reach a point where the concentration of A is so small that it can be considered null for your applications.
If, however, you dilute the solution, take a sample, then dilute that sample (and so on), you could reach a concentration of exactly 0M. Imagine you have diluted the solution enough so that it contains exactly 1 molecule of A. When you take your sample for the next dilution, if this molecule isn't in the sample, the concentration will be exactly null. If it does happen to be in the sample, it could be left behind when you draw the next sample, or the next, and so on.
In practice though, the water you use for the dilution will likely contain impurities. You will maybe not achieve exactly 0M, but the concentration could be so small that it is undetectable and have no measurable consequence.

"...dilute the solution, take a sample, then dilute that sample (and so on), you could reach a concentration of exactly 0M." That's amazing. That's the type of solution I was looking for!

@KaraKirkland It's worth mentioning, explicitly, that the reason this works is because a solution is not a continuous medium—if it were, you could never get to zero.  You can only get to zero because molecules are discrete.

I get that it wouldn't be in the spirit of the answer, but if you are doing serial dilution with the explicit aim of getting to zero concentration, you could just skip the part where you add any of the original solution to the solvent and you'll get there faster

@Paul But water have memory! Ask any homeopath!

@Raphaël I suggest it doesn't depend on "how you dilute it"; the difference between your examples is not final concentration but merely time taken; how pure the water solvent is has little to do with how much solute remains…
"… so small it can be considered null…" is a judgement, not a technicality.
From a dilute sample diluted further, you could no more reach a concentration of exactly 0M; just more quickly.
How can you imagine real dilution to 1 molecule? Guess that 12, or 397 or however many dilutions do the trick?

@RobbieGoodwin if I have a box with 99 red balls in it and one yellow ball, the concentration of yellow balls is 1%. If I add a million red balls, the concentration is about 1 ppm. If I dump half of the contents of that box into another box, I will have one box with about 2 ppm yellow balls and another with exactly 0 ppm yellow balls. The same is true of molecules in solutions, they ARE different processes.

For an analytical chemist, the concept of zero concentration does not exist. The concentration cannot be exactly 0! Only a limit of detection can be developed in terms of statistics. This is why a senior respectable user here has written an entire monograph on this topic. Suppose you have a NaCl solution, and so called "pure water"*, matter how much you dilutions you perform, you cannot say with 100% confidence that there is no single ion sodium left in the solution now.
*Pure water cannot exist in any ordinary laboratory.

There's plenty of room at the bottom, as they say, and I suppose this includes room for impurities. The lowest (indirectly) measured concentration I know in a condensed phase solution comes from a dark matter detection study, where the authors propose a 14 yM (that's yoctomolar) concentration of tritium in liquid xenon may have produced enough noise to mask signals from dark matter decay. Somehow it still remains after very stringent purification, and even though tritium itself is a trace radioisotope with modest half-life.

Very nice example. No reagent is 100% pure.

@NicolauSakerNeto unanswered in Physics SE: How will the possibility of tritium contamination decay be removed from XENONnT data?

@uhoh, an unrelated question. Do you anything about echelle grating and cross dispersion. I want to ask a question about it but what about a relevant site? I have a echelle grating (for fun) and I want to see a 2D spectrum in a simple set up which could a simple cardboard box.

Searching, there are 10 posts in Astronomy SE and 8 in Physics SE that contain "echelle" and many are interesting. It looks like the ones in Physics SE indicate there is more of a chance of a "DIY-based answer" there but there are some observational astronomers active in Astronomy SE who might really enjoy answering this as long as you mention it might use it to analyze sunlight or something astronomical.

Let us continue this discussion in chat.

You need to go for larger molecules, which are also not ubiquitous in the environment.

If you are doing serial dilution, still there is a possibility that at least one single molecule is there. Hence the point that in analytical chemistry, zero concentration does not exist. Nobody can "measure" zero concentration. It has no legal meaning either.

Can you provide a reference to back up a general statement such as that made in your first sentence? It sounds like an opinion. I would argue that if you make your samples by diluting with a known buffer not containing the solute (the solute stock solution is the only source of solute), then the buffer would have zero concentration of the solute. Circular, but not controversial.

BTW, it is commonly known as a "blank" measurement.

Buck Thorn, there is a lovely thin book, "Two papers on the limit of detection of a complete analytical procedure" H. Kaiser.

@M.Farooq It seems your «Kaiser» (with «ai») is not the same the kayser units prior $\mbox{cm}^{-1}$ / wavenumbers in spectroscopy is named.

No two different Kayser and Kaiser. H. Kaiser was at the Institute for Spectrochemistry Dortmund. That Kayser is quite old story! See Zum Problem der Nachweisegrenze in Z. Analyt. Chem. 1965, 209, 1. This book is the translation of his two famous papers.

@uhoh, I posted the query on echelle, if you want to check!

Thanks Farooq I'll have to take a look.

I'd have formulated a bit differently: it's not that it cannot exist - just that we cannot be sure that (where = in which bottle) we have a zero concentration. Then enter practical considerations like impurities and homogenization. The latter will make it particularly difficult to set a valid upper bound on the concentration one may actually have in one of the "zero conc" bottles.

@M.Farooq We can detect the spectral lines from a single isolated atom. Would it not be possible, with a sensitive enough spectrometer, to detect the spectral lines from a single molecule in solution?

Yes, may be in principle and on a piece of paper, but tell me how would you be "100%" sure that there is a single atom in that solution and how would we practically prepare such a solution?

Comments are not intended for extended discussions. There is already a chat room associated with this post: https://chat.stackexchange.com/rooms/111555/discussion-between-uhoh-and-m-farooq Please use this room.

Unequivocally yes! If your solute is something of which you can identify a single molecule in a macroscopic solvent sample (let´s say a fluorescent dye), then you can dilute down to zero, and be sure about it.
(For the nitpickers: This can be done in a finite number of steps, if you split the solution in two after each step, and keep diluting the part which shows a lower concentration. Divide et impera.)
If you´re unable to measure the concentration down to actual zero, then the statement "this has been diluted to zero" is not falsifiable, therefore unscientific (see Karl Popper), and as such not worth pondering. ;)

I'm also a fan of Popper!  (and Kuhn, and Lakatos).  Anyways: If you knew the initial concentration, and the number of splits, then even if you couldn't measure the final concentration, you could make a scientific statement about its probability of being zero :).

I would take a minor issue with your last paragraph: just because you (the hypothetical scientist doing this) are unable to measure the concentration down to zero doesn't mean that the hypothesis that it's zero is unscientific. Maybe someone else can measure it, or people in the future will be able to measure it, or it has some physical consequence that you're not aware of, or so on.

@DavidZ Can you explain how any of those would make the statement more "scientific"?

@Karl Because if someone in the future, with better technology, could falsify it, then the statement is falsifiable. "There are zero molecules of X in this solution" is a testable prediction when you have a device precise enough to detect whether there is a single molecule of something in a solution.

@probably_someone Balderdash. A statement that might become correct in five or fifteen years is wrong today, period.

@Karl The statement is not "becoming correct", it is becoming measurable. Whether it's correct or not doesn't change. You can make correct statements that aren't measurable at the time (for example, Einstein in 1916: "Gravitational waves exist", or Fermi in 1934: "Neutrinos exist"). The fact that they weren't measurable at the time of proposal doesn't mean they weren't correct.

@probably_someone Your point being? Esp. in relation to the original topic?

@Karl I answered your question: a statement is "scientific" if it is falsifiable. Falsifiable means that someone could conceivably measure it, not that someone can do that right now. And falsifiability doesn't really have anything to do with correctness, so talking about a statement "becoming correct" isn't really relevant here.

@Karl For example, the falsifiable statement "sterile neutrinos with certain parameters exist" is just like the statement "there are zero molecules of X in this solution". If sterile neutrinos with those parameters exist, there's an identifiable experimental signature, and if they don't exist, we wouldn't see that signature. The fact that our detectors aren't precise enough to measure that signature yet is immaterial. Similarly, there are spectrographic consequences for having a single molecule of X in solution, and an experiment with a precise enough spectrometer would be able to detect that.

Let us continue this discussion in chat.

Something like a zero concentration is achieved in the vapor phase in some materials.  In this answer the triple-point vapor pressure of gallium is identified as so low that a measurement thereof defaults to zero with a high probability.

Hypothetically, I think limitation for zero-dilution would be when you take a sample of diluted sample the possibility of having a solute molecule in there should be zero. To achieve this goal, the concentration of your solution should be less than $10^{-24}$ in power $\left(\frac{1}{6.022 \times 10^{23}}=1.66 \times 10^{-24}\right)$. To do so, if you have $\pu{1 M}$ of solution, you should add another $\pu{1.66 \times 10^{24} L}$ of water to get that dilution. Considering the whole water body of the Earth is about $\pu{1.3859 \times 10^{21} L}$ ($\pu{2.5511 \times 10^6 mile^3}$ according to USGS) you may run out of water before you get right concentration.:-)
That's why Raphaël has said its depends on how you want to do it. So, you may use the procedure called serieal dilution.

If you sample substantially less than a litre (say, 1 µL), then the concentration can be proportionately higher and you can still statistically expect to find zero molecules of the solute in some samplings.


text:
Entropy paradox and Confusion with Gibbs energy
Define $G=H-TS_\text{sys}$. Then,
\begin{align}
\mathrm dG &=\mathrm dH-S_\text{sys}\,\mathrm dT-T\,\mathrm dS_\text{sys}\\[3pt]
&=-T\left(\frac{-\mathrm dH}T+\mathrm dS_\text{sys}+S_\text{sys}\,\frac{\mathrm dT}T\right)
\end{align}
At conditions of constant temperature and pressure, $\frac{-\mathrm dH}T=\mathrm dS_\text{surr}$ and $\mathrm dT=0$. Thus, we end up with:
$$\mathrm dG=-T(\mathrm dS_\text{sys}+\mathrm dS_\text{surr})=-T(\mathrm dS_\text{universe})\tag{E-01}$$
Using $H=U+pV$, and by using the first law of thermodynamics, it can be shown that:
$$\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$$
At constant $p,T$, this equation reduces to $\mathrm dG=0$. Using this in $\text{E-01}$ yields $\mathrm dG=\mathrm dS_\text{universe}=0$, at constant $T$ and $p$.
So how exactly do we obtain the so-called "spontaneous processes", which have a negative value of $\mathrm dG$ at constant $T$ and $p$?
I can somewhat see why the result makes some sense: The conditions assumed in the derivation resemble a reversible thermodynamic process, for which it's known that $\mathrm dS_\text{universe}$ is zero. The question in bold though is still unanswered. Maybe I am confused regarding what is actually meant by $\Delta G$.

You said it already with $dG = -TdS_{\mathrm{univ}}$. The case of spontaneity is $dS_{\mathrm{univ}}>0$ implies $dG < 0$.

Also, I am a bit surprised by this question. I would think that satan would be the master of entropy...

@Zhe But my derivation suggests we always get $ds_{univ}=0$. Thats the entire problem!, As in, My derivation suggests its not possible to attain $ds_{universe}>0$, at conditions of constant $p$ and $T$.

Oops. I read this too fast. I think what you're saying is that an isothermal, constant pressure process is reversible?

I think I see your problem. Are you claiming that $dS = \frac{dH}{T}$?

yes. Isnt that exactly how the first equation is supposed to be derived?

But $dS = \frac{\delta q}{T}$.

isnt dq=dH at constant pressure?

$H=U+PV$, so no.

During the derivation of the stated equation, you need to make two assumptions : 1. Non PV work is zero.
2. The process is reversible as then only you would be able to write PV work as -P.dv That's why it yields zero at constant pressure and temperature , which indeed it should for a reversible process.

The essence of what's going on is this:
Your equation, $dG = Vdp -SdT$, is correct, but has restrictions you don't realize. Specifically, it applies only to systems that are (a) closed (so no adding or subtracting material), (b) have only a single component* (so no mixing, phase changes, or chemical reactions) and (c) that can do $pV\ \hspace{-.45em}$-work only.
[*It also applies to multi-component systems where the composition is fixed.]
To understand why, under such conditions, $dT = 0 \text{ and } dp = 0 \Rightarrow dG = 0$, let's apply the Gibbs phase rule to such a system.  The phase rule says:
$$F = C − P + 2,$$
where C is the number of components, P is the number of phases at equilibrium (not to be confused with $p$, pressure), and F is the number of degrees of freedom.
Since $C = 1 \text{ and } P = 1$, we obtain $F = 2$.  That means we only have two degrees of freedom, i.e., two independent ways in which we can vary the intensive properites of the system. If the only type of work we can do is $pV\ \hspace{-.45em}$-work, the only ways we can adjust the system's intensive properties are by changing its temperature or its pressure.
Hence, if we apply the restrictions $\boldsymbol{dT=0 \textbf{ and } dp =0}$ to a closed single-component system where we only allow $\boldsymbol{pV}$-work, the system can't change!  And if the system can't change, then of course $\boldsymbol{dG = 0}$!
But, you might protest, $dG$ is not generally zero, even at fixed  $T\text{ and } p$.  So how do we reconcile this with what we wrote above?  Well, we need a more general expression for $dG$ that allows for non-$pV\ \hspace{-.45em}$-work, addition and subtraction of material, and changes in composition:
$$dU = \text{đ}q + \text{đ}w +\sum_i \mu_i dn_i
= \text{đ}q + \text{đ}w(pV) + \text{đ}w (non\text{-}pV) + \sum_i \mu_i dn_i$$
Since we can calculate $dU$ using any path, let's use a reversible path:
$$dU = TdS - pdV + \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
And since:
$$G = U+pV-TS \Rightarrow dG = dU + pdV +Vdp - TdS - SdT$$
$$\Rightarrow  dG =  VdP -SdT+ \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
Here, $\sum_i \mu_i dn_i$ is the sum of the chemical potential of each species $i$ $(\mu_i)$ times the change in the amount of species $i$ $(dn_i)$. This accounts for the change in $U$, and thus in $G$, as we change the composition
Hence, even if $dT = 0 \text{ and } dp = 0$, if we have non-$pV\ \hspace{-.45em}$-work and/or a change in composition (e.g., mixing, a change in phase, gain or loss of material, or a chemical reaction), it will not be the case that $dG$ is restricted to be zero.

excelent answer, +1. Just one clarification: https://chemistry.stackexchange.com/questions/109951/pressure-at-which-graphite-and-diamond-are-in-equilibrium. The answer here seems to have only considered the equation dG=VdP-SdT, where dT=0. How does he then end up with the correct answer?

@satan29 He's calcuating the molar free energy (which is the chemical potential) for each phase separately, and then equating them.  $d\frac{G}{n}=\frac{V}{n}dp-\frac{S}{n}dT= d\mu=vdp-sdT$ is applicable to each phase, because each phase meets requirements (b) & (c) indicated above; requirement (a) is not met, because material can move from one phase to another; however, since we are using molar quantities, and the composition of each phase is fixed, adding or removing material does not change the free energy per mole, and thus we don't need to meet requirement (a).

For systems of variable composition (mixtures)
$$\mathrm dU=-p\mathrm dV + dw_{other}+ T\mathrm dS + \sum_i\mu_i\mathrm dn_i $$
This leads to
$$\mathrm dG = V\mathrm dP -S\mathrm dT + dw_{other}+  \sum_i\mu_i\mathrm dn_i$$
or, at constant temperature and pressure,
$$\mathrm dG = dw_{other}+ \sum_i\mu_i\mathrm dn_i$$
With regard to the question of spontaneity, the explanation lies in a comment by Zhe. For a process involving a pure substance in a closed system that undergoes only expansion work at constant T and p,
$$dH=dq$$
but
$$dS=\frac{dq_{rev}}{T}$$
The "rev" label is important, the equation $dS=\frac{dq}{T}$ does not hold for just any process.
At constant T and p but without any constraint on reversibility,
$$dG=dH-TdS=dq-dq_{rev}$$
The second law $dS_{universe} \ge 0$ implies that
$$\begin{align} 0 &\ge -dS_{surroundings}-dS_{system}\\  0 &\ge \frac{dq}{T} -\frac{dq_{rev}}{T}\\ 0 &\ge dq - dq_{rev} \end{align}$$
It follows that
$$dG\le 0$$
and only for a reversible process is $dG=0$.

This is what I think to be correct. Probably need confirmation from Chet Miller.
The fallacy in the proof (for dG always = 0 at constant p and constant T) is at the line
dG = Vdp - SdT......(E-02)
This is not always true. You understand that it is from First Law and the definition of enthalpy, but you have forgotten the assumption that you have made to reach E-02.
E-02 is from the fundamental equation
dU = TdS - pdV
If you substitute enthalpy definition and Gibbs energy definition into this, you will get E-02. But the problem is, when deriving the fundamental equation, you have assumed non-expansion work to be 0:
First Law: dU = dqrev + dwrev = TdS + dwrev;
dwrev = dwexpansion, rev + dwnon-expansion, rev = -pdV if dwnon-expansion, rev = 0
So dU = TdS - pdV is true only if dwnon-expansion, rev = 0
So  dG = Vdp - SdT is true only if dwnon-expansion, rev = 0
So at constant p and constant T,
dG = dwnon-expansion, rev = dwnon-expansion, max, because the complete form of fundamental equation is actually
dU = TdS - pdV + dwnon-expansion, max
If at constant pressure and temperature and the process won't do non-expansion work, then dG = 0. (You proved this)
Else if the process can do non-expansion work at constant temperature and pressure, dG will smaller than 0 then the process will be spontaneous (itself does work, no need work input), and hence I proved that dG can actually < 0.

So consider a chemical reaction, (this is a typical example of a system with constant $T$ and $P$). Whats the non-expansion work here?

I am not sure what they call it, it is the work due to the difference in chemical potential, μ, the work due to change in composition and mixing.

You can find out more in Atkins text, there's a chapter called chemical equilibrium.


text:
When is the process exothermic and when is it endothermic?
When electrons are being added/removed from atoms or ions how can I tell if the process is exothermic or endothermic? For example in the question below, how could I tell for sure which of the processes are exothermic and which are endothermic?
$\ce{Ca -> Ca+ + e–}$
$\ce{I + e– -> I–}$
$\ce{O– + e– -> O^{2–}}$

How much you already know about this topic?

Your title needs work!

You need to know that thermodynamics are greatly affected by state.
Hence you need to give precise states of all the reactants and compounds.
Ca(g)⟶Ca+(g)+e− and Ca(s)⟶Ca+(s)+e− has quite a difference.
Secondly for any element, ionization energy is always endothermic. Let me define ionization energy.
Ionization Energy is the energy required to remove one mole of electrons from one mole of gaseous atoms or ions.
X⟶X+(g)+e−
Enthalpy Change = positive Hence process is endothermic
Electron affinity is the enthalpy change which occurs when a mole of electrons combine with a mole of gaseous ions or atoms.
X(g)+e−⟶X-(g)
Now, electron affinity depends on the element under study. This webpage mentioned below will give detailed insight about electron affinity and its trends.
https://byjus.com/chemistry/electron-affinity/


text:
Creating 100mg/L aspirin solution
I plan on creating 100 mL of 100mg/L aspirin solution by dissolving 10 mg of pure aspirin powder in 100 mL of buffer solution (of different pHs). Problematically, aspirin doesn’t tend towards dissolving in polar solution easily, so I thought about dissolving the 10 mg of aspirin in 1 mL of ethanol first, and then taking that ethanol-aspirin solution and adding it to 99 mL of the buffer solution.
I was wondering if this method seems reasonable/feasible. If not, are there any better suggestions?

The salts of aspirin are much more water soluble

According to Wikipedia, solubility of aspirin in water is $\pu{3 g/L}$ or $\pu{3 mg/mL}$ at $\pu{25 ^\circ C}$. Thus, it is not hard to make a $0.3\% (w/v)$ aquious aspirin solution (maximum concentration, which equals to $\pu{300 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$). Therefore, your target, $\pu{10 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$ can be easily achieved without any modifications. You have not mentioned the what different $\mathrm{pH}$ values of buffers you are planning to use. Since $\mathrm{p}K_\mathrm{a}$ of aspirin is about $3.5$ and the amount of it above your target value would dissolve in water (so final $\mathrm{pH}$ would be around $2.12$ for $\pu{300 mg}/\pu{100 mL}$ soution), it is guaranteed to dissolve your desired amount in any buffer with  $\mathrm{pH}$ above $2.12$.
None the less, dissolve aspirin first in small amount of ethanol and diluted it with desired buffer is not bad idea. As a matter of fact, in aspirin preparation lab, we use ethanol to recrystallize the final product, therefore, would not arise any problem on dissolution. Thus, this would be an ideal alternative procedure.


text:
Adiabatic expansion of gas
If a real gas is adiabatically expanded against constant pressure, then which of the following will definitely increase?
Compressibility factor (Z) or Entropy (S)
Correct answer: Entropy
My answer: Compressibility
My Reasoning :
$Z=\frac{PV}{nRT}$
In adiabatic $TV^{γ-1}=\text{constant}$
Therefore, $T$ ∝ $\frac{1}{V^{γ-1}}$
$γ-1$ is greater than zero always. So we can say is volume increases, temp decreases. Therefore Z should increase as its numerator increases and denominator decreases. Entropy for isobaric adiabatic has $\ln\frac{T_2}{T_1}$. Since $T_2$ is less than $T_1$, entropy decreases. What am I doing wrong ?

First of all, that TV equation is for perfect gas only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change.

@The99sLearner so how do i proceed to solve it then ?

The compressibility vs pressure shows a decrease then increase so depending on what the initial pressure is the change could be positive or negative. The entropy change in a reversible adiabatic expansion is zero for an ideal gas and positive for a non-reversible adiabatic change (positive because less work is done so the temperature is lowered less than the reversible case and no longer balances the volume change). The entropy in the non-ideal gas case will increase as work always has to be done against intermolecular interactions.

First note that the process is an irreversible adiabatic expansion. If it were a reversible one, the gas has to expand against a varying pressure because pressure of gas decreases during expansion and this varying gas pressure must be matched with external pressure in all infinitesimal steps of expansion. "The gas expanded against constant pressure" ⇒ the gas must have a pressure which is always higher than the constant external pressure during the expansion ⇒ irreversible.
Will entropy definitely increase?

ΔSsur = 0 as q = 0. Irreversible ⇒ ΔStot > 0. ΔStot = ΔSsys + ΔSsur = 0 + ΔSsys = ΔSsys. So ΔSsys > 0, so entropy of system will definitely increase.

Will compressibility factor definitely increase?

In adiabatic irreversible expansion all p, V, T changes. If Z is to increase, the increment in pV must be greater than the increment in T, or the magnitude of decrement in T must be greater than that in pV, which cannot be sure will happen as all p, V, T change.

What did you do wrong?

First of all, TV equation is for perfect gas doing reversible adiabatic expansion only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change. All p, V, T will change. You said as volume increases, temperature decreases, but pressure of gas will decrease too.

You are correct in saying this process is irreversible.  But, I think you should add a sentence about why the process is irreversible.

Note that the TV equation for adiabatic processes has two restrictions.  First, as you mentioned, it applies only to ideal gases.  Second (and it's important to state this explicitly), it applies only to reversible processes.  Those are the two fundamental ways the OP went wrong in applying that equation.


text:
Can a chiral molecule be optically inactive?
Does there exist a chiral compound with zero optical activity? It seems possible that for some (chiral) arrangement of atoms, the optical activity of one section of the molecule could be exactly compensated by another section of the molecule, even if these two sections are not mirror images of each other. Although this is just a mathematical coincidence, does any chiral molecule have an optical activity of exactly zero?

Are you asking for compound with smallest specific rotation?

Edited it... I wanted to ask something else

It seems you're asking if "a molecule with chiral centres can have zero optical activity?". But, I'm not sure.

I very much doubt that there is a compound with exactly zero activity. Meso compounds are achiral (by definition) even though they contain stereo centers.

@RahulVerma yes, but not just limited to chiral molecules that have chiral centres alone. There exist molecules without chiral centres that are chiral.

Do you mean allenes? they do have a stereocenter.

@Safdar allenes, spiro compounds, and combinations of rings and double bonds are a few examples

The likelihood of exact cancellation of optical rotation in a chiral molecule is extremely low, so much so that the answer to this question is practically no.

With infinite measurement precision then no, but within the limit of experimental error then yes - see cryptochirality

The intent of the question appears to be to ask whether any chiral molecules could have exactly no optical activity. But this is problematic and the best we can do is test whether any compounds don't have observable optical activity.
As the Link by Nicolau Sater Neto to the concept of crypto chirality shows, on this criterion the answer is yes. The molecule 5-ethyl-5-propylundecane shown here has no observable optical rotation even when a pure enantiomer is isolated.

One can intuit that the nature of the compound is such that its rotation probably exists but is very very small.
it seems improbable from first principles that non trivial examples (eg not meso compounds) exist where the optical rotation is exactly zero.

If the reason for the chirality were far from the center of chirality I would think the effect would approach zero or become not measurable.

Yes, tertiary amines where all groups are different are optically inactive  (due to amine inversion) even though it's a chiral molecule. Eg. N(CH3)(C2H5)(H) or NHDT

While technically true this feels like a cheat as the reason why optical activity isn't observed is the lability of the molecules under normal conditions. At lower temperatures they might well show optical activity.


text:
Diastereomers or not
Why is the first one not a pair of diastereomers while the second is?

The only reason I could think of was in the first $\ce{Me}$ is the same as $\ce{CH3}$ that is same groups even though differently oriented are considered same. While in second $\ce{COOH}$ and $\ce{OH}$ are obviously different. Is my reasoning correct? If it is could someone please state it in better words so that I understand the minute details.
Also I was able to reason this ONLY AFTER the correct answer was told. How to identify if a certain pair of compounds is a diastereomeric pair?

The first pair is not stereoisomers. You have correctly explain why. When you are dealing with stereoisomers, that's what you have to do first: Identify stereo centers. Accordingly, each compound of second pair has two stereo centers. Then, mark (R,S) configurations of all stereo centers. As I marked, the first compound has (R,S) configuration, while second compound has (R,R) configuration.
Fact: The mirror image of any stereo center has the opposite configuration. For example, mirror image of (R)-glyceride is (S)-glyceride.
Now, let's look at second compound of this pair:

The configurations of this compound's two stereo centers (right structure) are (R,R). Thus, the configurations of two stereo centers of its mirror image should be (S,S). Thus, the stereoisomers with (R,R) and (S,S) configurations are enanthiomers. All other stereoisomers related to them are diastereomers. Therefore, the pair is diastereomers since they are (R,R) and (R,S).
I also want to emphasize one more point. When a compound has two or more stereocenters, it is beneficial to see any symmetry within them. Each stereo center of the pair contains identical set of groups attached to it ($\ce{H, OH, COOH}$). Thus, if these two stereo centers have opposite configurations, each stereo center is the mirror image of the other (see the left structure). Therefore, it has a plane of symmetry and hence, optically inactive. These compounds are called meso-compounds (They do not have their relevant enanthiomer).

Your first set of compounds are identical.
In the Fischer projection of a chiral centre, you can rotate three of the connected groups in a clockwise/anti-clockwise direction without changing the configuration of the chiral centre.
Upon performing this operation (keeping the top fixed, rotating bottom three clockwise) on the first compound in the first set, we see that the methyl group now occupies the position of the $\ce{-CH3}$ group in the second compound and the $\ce{-CH3}$ in the first occupies the position of the methyl group of the second.
Now, $\ce{-CH3} \equiv -\text{Me}$. Therefore, the first two compounds are now identical. As yukelid stated, identical compounds are not diastereomers.
However, repeating the same for the second case does yield two different compounds that are non-mirror images yet non-superimposable. Hence they are diastereomers.
Enantiomers and diastereomers are only seen for chiral compounds since achiral compounds would have symmetry that makes them identical. So, chirality is needed for enantiomers and diastereomers.

First pair consists of 2 identical compounds. Identical compounds are not diastereomers.
Second pair is diastereomeric because it consists of 2 stereoisomers that are not mirror images.
You could read on stereoisomerism: https://en.wikipedia.org/wiki/Stereoisomerism


text:
Acid catalysed dehydration of alcohol with a bicyclo group

Pathway 1: Rate determining step is the formation of carbocation. In this pathway, a conjugated tertiary carbocation is generated as an intermediate.

Pathway 2: One of the resonance form of intermediate carbocation proceeds to produce a more stable  product (due to benzene ring) through alkyl migration. Though, the resonating form of carbocation is conjugated secondary carbocation which is less stable than the carbocation formed through pathway 1.

My questions are:
1- Is tert-butyl (don't know the correct term) migration possible in pathway- 2?
2- What are the major and minor products? Which factor will decide the major product: stability of final compound or the most stable intermediate?

The reactant/intermediate is clearly not aromatic, but the product is. What about the $\ce{\Delta H, \Delta G}$ of reaction?

This reaction is under thermodynamic control. All of steps are reversible but the last step to form aromatic compound. Hence, aromatic compound is the major. The product from first path is still susceptible to protonate under acidic conditions.

" resonating form of carbocation is conjugated secondary carbocation which is less stable than the carbocation formed through pathway 1" - that's wrong on many levels.

I have never seen a migration of a t-butyl group before. That seems sterically problematic.

@Zhe Migratory aptitude might depend on the reaction. According to wikipedia for  pinacol rearrangement the order is Hydride > $\ce{(CH3)3}$> Phenide > $\ce{C2H5}$ > $\ce{CH3}$ and per this site for Baeyer-Villiger rearrangement the order is 3º-alkyl > 2º-alkyl ~ benzyl ~ phenyl > 1º-alkyl > methyl (depends on electron density).

Therefore based on the electron density, t-butyl group should have a high migratory aptitude. But what factors can decide the migratory order for dehydration of alcohols?

That's fair, but we're not talking about relative aptitude for migration. We're talking about whether elimination is faster than the shift. Also, thanks for the reference.

@MathewMahindaratne What would be the outcome if t-butyl substituent is replaced by a substituent which cannot migrate?

Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on labeling to elaborate the mechanism.

(a) I have gone through your linked answer (but could not access the references mentioned therein) and what I understood is that isotope labeling can decide the correct mechanism between the two discussed above. So, is it determined in the cited paper? (b) In your mechanism, is there any driving force between (1) and (2)? Or this is not important and final stable product decides it?

(a) In 1950 they didn't use isotopic labeling which is one way of testing the mechanism. They expected 4 but got 6 in the link. That result led to the spiro mechanism. (b) The cation 2 is resonance stabilized. The migratory aptitude of methyl is less than RCH2-.
https://chemistry.stackexchange.com/questions/116051/possible-nonclassical-ion-from-a-bicyclic-system/116066#116066

Sorry, I should have asked for the driving force between (2) and (3).

There is not much difference (strain) between a 5- ans 6-membered ring. The delocalized cation is the same.in 2, 3 and 4. Aromatization drives the reaction.


text:
Different solutions in cyclic voltammetry
CV is usually used to monitor corrosion (from what I have learned) because, if it is a reversible reaction, we can see the cathodic and anodic peak in the voltammogram. I know that the scan rate influences the intensity of the current peak and I also know why that happens. However, I just can't seem to find the reason why different solutions result in different current peaks in the voltammogram. Especially if the solutions get more acidic the current peak changes.
So my question is why do the peaks in the voltammogram change when we change the solution?

Recording a CV assess both so-called half cell reactions.  If balanced individually, some of them only need only the transfer of an electron, e.g. for
$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$
Other half-reactions include either $\ce{H+}$ or $\ce{OH-}$, and their consumption / generation alters the pH value of the solution, e.g.
$$\ce{MnO4^- + H2O + 3 e- -> MnO2 + 4 OH-}$$
or
$$\ce{MnO4^- + 4 H3O+ + 2 e- -> Mn^{2+} + 6 H2O}$$
Since each of them are characterized by a standard electrode potential, altering the pH value of the solution thus may alter the reaction pathway accessible; the reaction of permanganate to manganese dioxide is typical for a reaction in a basic aqueous reaction, the one to yield $\ce{Mn^{2+}}$ typical for an acidic one. Indeed pH electrode is an application of this pH dependence which may be quantified by the Nernst equation.
While the examples shown deal with small inorganic ions, the same principle is valid for recording a CV for organic molecules, too.  For the couple quinone / hydroquinone, the variation of the pH value thus may yield a plot like the following:

(reference)
The changes for the redox couple, recorded in different solutions of different pH value equally were plot with the cell potential in function of said variation of the pH value of the analyte:

Note: compare the half reactions with each other.  If both the number of electrons, as well as the number of protons exchanged are equal, the potential of the cell will not be affected by a change of the pH value.
reference: pH Dependent Redox Couple: An Illustration of the Nernst Equation, Walczak et al., J. Chem. Educ. 1997, 74, 1195-1197; doi 10.1021/ed074p1195

Thanks for the answer, but I still don't really understand it. In the figure the peaks barely change, it's just that the potential has to change to get the reaction going. I have found other figures in papers where the peaks change due to a different solution. I don't understand why that happens.

The peak position, e.g. about 0.38 V (pH = 1.6), 0.27 V (pH = 3.4), ..., 0.80 V (pH = 6.3) changes significantly because these are recordings in different, independently prepared samples.  It is not that there was one beaker with the hydroquinone / quinone couple at pH 1.6 you record one CV which after this cycle happens to have a pH = 3.4 because of the previous recording of the CV.  The buffer acetate/phosphate concentration was about 0.1 M, hydroquinone only 1 mM.  CV does not convert synthetically useful amounts of matter, as electrosynthesis (https://www.youtube.com/watch?v=8UPQLiR4Fsk)