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In school, I learnt that Potassium is the most reactive element in the reactivity series. I thought it was Francium?
In school, I learnt that Potassium is the most reactive element in the reactivity series. In the periodic table, the further you go down the more reactive the elemnt is. If this is true, then shouldn't francium be the most reactive metal? Or rubidium, or caesium? I am very confused.

https://chemistry.stackexchange.com/questions/9051/why-do-the-trends-in-reactivity-not-apply-for-francium

This is not so much a question about what's more reactive; the more important part is that this reactivity series which you learn is not meant to be exhaustive, unless it literally contains every (metallic) element in the Periodic Table. Why should a reactivity series, something you learn for everyday practical use, go out of its way to include francium, an element which we can't even get significant quantities of?

Additionally, reactivity order can be different thermodynamically-wise and kinetically-wise, and is also scenario dependent.

Francium is the most reactive. It cannot be pure because it can react with almost everything. I think you are using Potassium because it is the most reactive element used in most schools. Caesium and rubidium are more reactive than Potassium, but francium is even more reactive. You cannot legally use it withoutr approval from the government.
Potassium is not as dangerous as Caesium and rubidium, but you are allowed to witness it in schools with strict scientific control

Out of curiosity, why do you capitalize potassium and caesium, but not rubidium and francium?

Reactivity as a metal(i.e, electropositivity) DOES increase down a group and rubidium, francium etc are more reactive.
However,the reactivity series taught in school is meant to be a stepping stone towards the rest of the reactivity series, and understanding the trends that control them. Pottassium is one of the more common metals used in chemistry and is considerably reactive, comparing to iron or copper. In many cases, the reactivity is series is not meant to be all encompassing, but a quick trick to the most common metals used.

According to most schooling rules, the study of only the first 20 elements (i.e., from H to Ca) is included in the syllabus until the 10th standard, or higher; because after Calcium, the transition elements start, that have variable valency as a result of loss of electrons from the penultimate shell. Studying these elements thus becomes quite complicated. Now considering the first 20 elements, in the alkali metals, Potassium is present at the most bottom part, and is surely more reactive than either Lithium or Sodium. Hence, in our schools, it was taught that Potassium is the most reactive metal. But, considering the whole periodic table, we conclude that Francium is the most reactive. (Sometimes, it is also stated that Caesium is more reactive, because Francium is radioactive).


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Why don't we multiply Ksp and Kf?
My understanding is that for a process with two reactions having the equilibrium constants K1 and K2, K for the full process is K = K1 x K2. However, in this problem in my book where complexation increases solubility of CuS in water, they just argue that Kf >> Ksp and drives the reaction forward. Why is it not K = Kf x Ksp?
Thank you.
The problem:
A 0.1 mol sample of CuS is added to 1.00 L of 1.00 M NH3. What is the final concentration of the complex ion, tetraaminecopper(II)?
$$\ce{CuS(s) <=> Cu^2+ (aq) + S^2- (aq)}\ \ \  K_\mathrm{sp} = \pu{8E-37}$$
$$\ce{Cu^2+ (aq) + 4NH3 (aq) <=> [Cu(NH3)4]^2+ (aq)}\ \ \   K_f = \pu{1.1E13}$$
The answer key: $K_f$ will drive the dissociation of $\ce{CuS}$ forward such that the amount of $\ce{Cu^2+}$ available to react is [Cu2+] = .1 mol/1.00 L = 0.1 M, and due to the 1:1 relationship between $\ce{Cu^2+}$ and the complex ion, the concentration of the complex ion will also be 0.1 M. I believe this implies the overall K >> 1, rather than ~10^-24, which it would be if we multiplied Ksp and Kf.

Do not try to mechanically apply multiplication, think rather when we do it, when we do not and why for both. Analyze the task from the scratch, using both constants and forget the a priori multiplication for a while.

No one is stopping you from calculating the product between the two constants. However, whether this is relevant or not in the context of your problem is highly debatable.

source: Stack Exchange

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In a electrochemical cell, why does the anode (where oxidation occurs) have a lower concentration than the cathode?
I know that under standard conditions, the concentration in each half-cell is 1.0 M. However when the cell is not under standard conditions, let's say solution A is 0.1 M and solution B is 1 M. Solution A (0.1 M) will be the anode. Why is this?

Add more details. There is no such condition. Can you provide a specific example, 0.1 M of which ion?

Such a rule applies in concentration cell scenario, where the half-cells are the same but concentration.
The half cell (the first kind like M/M+, not the second kind like the reference silver chloride electrode)  with lower metal ion concentration has lower potential(approximating activity by concentration):
$$E = E^{\circ} + \frac{RT}{nF} \ln{c}$$
therefore is anode, as oxidation occurs there.
For different electrodes, the one with lower concentration can be an anode or a cathode - case specific, depending on their potentials.

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Relative stability of $\ce{V^2+}$ and $\ce{V^3+}$ and their Standard Electrode Potential
My book - NCERT Chemistry for Class XII -Page 218 Contains the following paragraph:

Initially they mention that the Low value for Standard Electrode Potential of $\ce{Sc^3+/Sc^2+}$ is due to the higher stability of $\ce{Sc^3+}$ which makes sense as it would have a lesser tendency to move to the $+2$ oxidation state.
But later on, they mention that the Lower value of SEP of $\ce{V^3+/V^2+}$ is due to the stability of $\ce{V^2+}$.
I get that the half filled $\ce{t_{2g}}$ configuration would make $\ce{V^2+}$ more stable but then, shouldn't the SEP value become higher rather than being low?  As $\ce{V^3+}$ would have more tendency to move to $\ce{V^2+}$

Do you see the fallacy of the word stability, so think a little deeper. When you say V2+ is more stable, but what is your reference point? Say, there is a lake X, which is deep but lake Y is deeper and I don't mention my reference starting point for measuring the depth. This textbook is exactly doing the same: repeating the word stability again and again but without telling you their reference point.

There are several problems in the textbook. It talks about the Sc half-cell for M3+/M2+ but it is not shown for Sc! "The low value for Sc reflects" ... there is no value there in Table 8.2. Did you see this value? The second biggest problem is the word stability. Stability with respect to what? Stability in chemistry can have different context and it can have so many meanings.

Yes I agree and did refer to the table for the values. Although there was no value given for Sc, if we still look at what they're trying to prove, the reason that $\ce{Sc^3+}$ is more stable justifies the statement made. But on the other hand, they've still given the value for V right? And the value given is negative... meaning $\ce{V^3+}$ has no tendency to reduce to $\ce{V^2+}$ but their relative stabilities(That $\ce{V^2+}$ is more stable than $\ce{V^3+}$) indicate the opposite tendency. That's where I'm stuck at!

Again, you will have to think about the meaning of stability for Sc. Without any electrode potential value, they should not be talking about it. Perhaps this is causing confusion. We do not know the value of Sc half cells as per your Table. In electrochemical language, something can be stable with respect to oxidation, reduction or disproportionation. Just saying V2+ is more stable V3+ does not mean anything in an absolute sense.

When we say, V3+ has no tendency to reduce to V2+, the reference point is the hydrogen cell. Of course, one can reduce V3+ to V2+ experimentally. So take your text with a grain of salt.

I remember as a young-chemist to perform experiment "color chameleon" with  NH4VO3 in diluted acid, reduced by zinc, causing colour transition for each step of $\ce{VO2^+  -> VO^2+ -> V^3+ -> V^2+}$.

The $\ce{E^°(M^3+/M^2+)}$ standard electrode potential for $\ce{V}$ has a low negative (that is, a relatively more positive) value. This is attributed to the greater stability of $\ce{V^2+}$ as given in the text.
This can be seen in the $\ce{E^°(M^3+/M^2+)}$ values for  $\ce{Ti,V,Cr}$ being $\pu{-0.37, -0.26, -0.41}$ respectively.
The book is not very clear in its meaning of low in this case. Looking at the table of values and comparing the $\ce{E^°(M^3+/M^2+)}$ of nearby elements can help you understand the text better.

An independent source gives $-0.37$ for titanium, $-0.26$ for vanadium and $-0.42$ for chromium, showing that vanadium is relatively high compared with its immediate neighbors. The book should have said "less negative" rather than "comparatively low".
In any event the effect of half-filling the $t_{2g}$ orbitals is weak compared with half-filling all of the $3d$ subshell. The above source gives $+1.5$ for manganese versus the aforementioned $-0.42$ for chromium and $+0.77$ for iron. Half-filling the $t_{2g}$ orbitals alone would have more impact with ligands that have a stronger splitting than water.


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Why should the cubic equation in volume, obtained from Van Der Waals equation of state, have equal roots at the critical point?
The video Edmerls —
Derive Equation for Critical Constants in terms of Vander Waal's Constants. | Physical Chemistry (YouTube) provides a derivation of the relationship between the van der Waals constants and the critical pressure, temperature, and volume.
It involves rearranging the van der Waals equation to form a cubic equation in the volume $V$ and comparing it to $(V - V_\mathrm{crit})^3 = 0$, where $V_\mathrm{crit}$ is the critical volume.
Why does the equation need to have equal roots? And can it also have equal roots for other fixed values of pressure and temperature (apart from the critical pressure and temperature)?

Do you know why the original  van der Waals equation has 2 equal real roots at and only at the critical point?

I don't, could you please elaborate on that?

1/ Converging 2 V values ( V_gas and V_liquid at given T,p) to the single V value 2/ Zero 1st and 2nd derivative dp/dV and d2p/dV^2 at T_c,p_c  3/ For T > T_c, function p=f(V) is monotonous = 1 root.   For T < T_c,  p=f(V) has the local maximu and minimum with 3 roots (the particular shape of the functions has just limited, physical interpretation here )


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Why my Infrared data acquired in transmission mode looks like it was acquired in absorbance mode?
I'm analyzing some carbon nanotubes with DRIFTS. I mix the CNT with KBr powder and proceed to measure the spectrum of the powder. After acquiring the KBr background (in %T, peaks facing down) I proceed and acquire the sample spectrum and it comes back with the peaks in the opposite direction . Once you apply  Kubelka-Munk correction you get the spectrum as in transmission again but sometimes the KBr background is not completely removed. Why is this?.

Please clarify your question and show the spectra. The question heading is different but you are also talking about background correction. Whenever you do something new, always run a simple sample (just a simple polystyrene powder), whose results are known. This gives confidence in the results. Carbon FTIRs are horrible anyway.

Eons ago, I used direct transmission mode by employing a small amount of carbon in KBr rather this complicated Kubelka-Munk correction. Use a large number of scans, say 64.

KM theory, is more valid for weakly absorbing materials. Carbon absorbs everything...recall a black body is a good absorber.

@AChem I mixed the CNT with KBr and ran 64 scans. I also acquired a spectrum of Polystyrene and it was fine. The issue I describe seems to be sample related; also the KM correction is done by the software in like a second, I don't understand your comment.

Is the sample and KBr bone-dry?

@AChem yes. The issue is if I acquire transmission spectra of any other sample, the peaks come facing down as expected. Once I analyze the CNT samples, the opposite happens as you can see from the image I uploaded

How do you account for negative KM units?

May I suggest getting a 1bounce or 3 bounce diamond atr cell

I will take your guys' suggestions, any insights into the actual question?

First of all I would suggest you to read the reference Validity limits of Kubelka–Munk theory for DRIFT spectra of photodegraded solid wood. Wood Sci Technol 45, 135–146 (2011). https://doi.org/10.1007/s00226-010-0314-x. The introduction highlights the how many spurious (or more crudely bogus) DRIFT results are out there just because researchers did not carefully think about the validity limits of KM.
The authors highlight the facts that KM is useful for weak absorbers. Certainly carbon nanotubes are not weak absorbers and one would expect problems (non-ideal behavior).
Although your question has been appreciated but it is missing a lot of information about the sample, its pre-treatment and many other critical instrumental settings. FTIR is a Fourier transform technique, and you have to choose (perhaps in advanced settings), the correct apodization function. Using an improper setting may cause final spectrum's artifacts. Check the manual which one is the right apodization for DRIFT. Make sure all the settings are correct from the manual.
Alternatively, a short-cut is that you repeat a well-behaving organic compound and reproduce a correct absorption spectrum which matches very well with a standard database and does not generate negative scales on the y-axis. You are also using 64 scans, which is fine.
The red-flag in that carbon spectrum is the negative y-axis. When the instrument generates an output which has no physical meaning...there is a problem somewhere. Even with simple double-beam UV-Vis spectrophotometers, some time the sample shows negative absorbance. It does not physically mean anything.
To troubleshoot one can re-visit the problem from the beginning.
(i) KBr is bone dry, FTIR grade and freshly dried.
(ii) The carbon nanotubes have been thoroughly washed with suitable high quality solvents and thoroughly dried > 24 hours. This will ensure the sample is clean and the background is not coming from random junk sticking to carbons. Carbon is an excellent non-selective adsorbent.
(iii) Mix the KBr and carbon nanotubes throughly (all IRs used to have mortar and pestle) so that the mixture is as uniform as possible.
(iv) After collecting background, use KBr as a sample and see how the baseline looks, and if this is reproducible or not. If it is reproducible, one could subtract this baseline from the final spectrum.
(v) Do a sample concentration study: Use different amounts of carbons in KBr (do it semi-quantitatively) and see how the spectrum changes. Overlay the spectra and see which regions of the spectrum change and when do they start to behave well. Carefully, use the same amount of the sample for DRIFT measurements.
(vi) In the worst case, if nothing works in DRIFT mode, use a simple KBr pellet method and use the transmission mode. A low sample concentration is required. Making a good pellet under high-pressure was an art in itself.
This is a little bit of work, but the end-results will be satisfying and you will be confident about the results.


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PVP-iodine and octenidine can stain skin purple, does the reaction create iodine radicals?
In this paper it is mentioned that

... octenidine can release iodine radicals from the
PVP complex, resulting in a tissue irritation as well as
strong brown to violet discolouration in the border areas.

but elsewhere I read that

... it turns out that ΔG > 0 for radical iodination.

Is it really plausible, that octenidine releases iodine radicals from povidone-iodine?
I have seen and experienced the purple discoloration of combining octenidine with pvp-iodine. The stain parts after a few days. However, I have not found any reports of skin irritation so far, except for the quote above. Iodine radicals could cause skin irritation, so if the first statement ist true, that would at least in theory support the hypothesis that skin irritation could occur.
Bonus question:
How would you explain the purple colour of the reaction?
Disclaimer: I am a medical doctor interested in understanding the chemical background of the drug interaction of these two bactericidal skin disinfectants, which is often warned about on the ward.
-- EDIT:
Additional background info, helped me to understand the problem:
This journal article mentions in its introduction:

Use of iodine as an antiseptic has gone into disrepute as a result of its short lasting action and irritant properties. Strong solution of iodine is corrosive and can cause blistering and necrosis of skin, commonly referred to as chemical burns or irritant contact dermatitis. Iodine has thus been replaced by substances known as iodophores that contain an iodine molecule linked to a large molecular-weight organic compound. A slow release of iodine from the polymer contributes to a long lasting action and a low irritant profile of such compounds. Povidone iodine is one such iodophore that has poor irritant property despite preserved antiseptic efficacy, and hence is advocated as a non-irritant, non-toxic compound for surgical scrubbing and disinfection of endoscopes.

https://en.wikipedia.org/wiki/Octenidine_dihydrochloride

An antiseptic that is only safe to apply to the intact skin, do I get that right? Why would I ever mix it with PVP-iodine? (I'm a chemist, and just asking. Planned surgery?)

Hi @Karl, yes that's correct. Octenidine is commonly used as an alternative to other skin desinfectants, like PVP-iodine and chlorhexidine. It is only safe to apply to the intact skin, because in deeper tissue it can cause necrosis of the tissue, which would then have to be resected surgically. PVP-iodine is contraindicated for neonates and rarely used in the oral cavity. Octenidine and PVP-iodine shouldn't ever be mixed, but it does happen by accident every now and again.

Thnx. I have no clue whatsoever why the described should happen. "Radicals" is most certainly nonsense. If copious amounts of hydrochloride strongly hydrolyse the lactame group in PVP, elemenal purple iodine will be freed.

What do they mean by "brown" discolouration? PVP-Iodine is very brown to begin with! (I get it, dicolouration from brown to violett.)

By discoloration they mean that the skin is stained. After normal application of pvp-iodine, the brown colour parts rather quickly from the skin. The "discoloration" stays much longer, like days up to weeks, depending on which skin area and how quick the skin is to renew there. If the stain is very deep, like where tattoo pigment is injected, the stain could also be permanent. This is only of concern for open wounds.

Elemental iodine (not iodine radical) is purple, and some iodine complexes are similarly colored: with starch, it's dark blue, almost black; with chitosan (a component arthropod exoskeletons) purple. Iodine's color also appears different in differing solvents.
You might experiment with a dilute solution of PVP-iodine and starch as a control, and with dilute PVP-iodine mixed with octenidine dihydrochloride (a surfactant, like soap) and starch, and see if the added surfactant makes for a much darker blue-black color. If so, that would indicate elemental (molecular) iodine is being freed from the PVP complex by octenidine dihydrochloride.

Thank you for the answer, would you agree with my interpretation of the second source, that it's more likely elemental iodine and not iodine radicals that are released from the complex?

Iodine radicals do not hang around long enough to see a color in solution. There are some stable radicals, such as nitric oxide (which is a vasodilator, https://en.wikipedia.org/wiki/Nitric_oxide), but not iodine, https://www.wiley.com/en-us/Stable+Radicals:+Fundamentals+and+Applied+Aspects+of+Odd+Electron+Compounds-p-9780470770832

Reactions with $\Delta G>0$ most certainly do occur; in fact green plants literally run such a reaction, conversion of carbon dioxide and water to sugar and oxygen, all day.
The catch, of course, is you must put in work energy (not just heat energy) to run the reaction, which is of course the definition of "nonspontaneous". In the case of green-plant photosynthesis, the work energy is provided by sunlight, whose photons carry much higher energy than would be the case for thermal radiation at equilibrium.
Work energy can thus be provided by light. Or it can be provided by an auxiliary chemical reaction, which is certainly possible on skin. Identifying this energy source precisely in the case of this question would require more background information from the OP.


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Is the method of Benji detecting plutonium realistic in Mission Impossible: Fallout?
In Mission Impossible: Fallout, there is a scene where Benjamin Dunn was examining a plutonium core brought by Eastern European gangsters. He touched the core with some sort of a rod and when asked what is that thing, he replies:

This is a Beryllium rod, which is causing a reaction with the plutonium inside the core.

There is a concept called startup neutron source which is a source for initiating a stable nuclear reaction inside nuclear reactors. There is an example of plutonium/beryllium reaction. Is this the concept used in the movie or is it a fictional method of detection? Maybe they could have a used a simpler device like Geiger counter to measure radioactivity but they opted for a fancy looking device. Has such device ever invented?

Beryllium is indeed used as main component of radiation detectors https://www.osti.gov/biblio/244639

It seems like a few related but different concepts were mixed in the movie as well as in your question.
1) Alpha/Be sources
Yes, it is possible to make a neutron source from plutonium and beryllium. Usually, plutonium oxide and beryllium powder are milled and mixed and sealed in a leak-tight stainless steel capsule. The involved reaction is:
$$\ce{^9Be(\alpha,n)^12C}$$
So actually, plutonium is not needed for this; any alpha source with sufficient energy would do. Typical laboratory sources (e.g. for the online boric acid measurement used for a pressurized water reactor) contain $\ce{^241Am/Be}$. The urchin in the first nuclear weapons was a $\ce{^210Po/Be}$ source. $\ce{^226Ra/Be}$ sources are easy to obtain but need more shielding of the gamma radiation.
Since most alpha sources would work, this method would not be suitable to identify weapons-grade plutonium as shown in the movie. Furthermore, because of the short range of alpha radiation, the materials need to be carefully mixed; placing a beryllium rod next to a plutonium sample would not work so well.
2) Primary and secondary neutron sources in a reactor
Yes, nuclear reactors have additional neutron sources. The primary neutron sources are usually $\ce{^252Cf}$ (the most expensive element I have ever handled in visible amounts). They have a half-life of only 2.6 a, but they are only needed for the first cycle. $\ce{^252Cf}$ is a spontaneous fission source. So this topic is not related to plutonium and beryllium as shown in the movie.

Secondary neutron sources are made from antimony and beryllium. During the operation of the reactor, antimony is activated by $\ce{^123Sb(n,\gamma)^124Sb}$. The high-energy gamma radiation of $\ce{^124Sb}$ is used for a different reaction with beryllium:
$$\ce{^9Be(\gamma,n){}2\alpha}$$
Since the antimony needs to be activated first, the secondary neutron sources are not available during the first startup of the reactor.
Although secondary neutron sources contain beryllium, this has nothing to do with the movie.
Note that there is a widespread misconception about such neutron sources. The primary and secondary neutron sources are actually not needed to make the reactor critical. The critical configuration depends on the fuel, moderator, absorbers, and reflectors; but it is independent of the presence of any additional neutron source. And there are always enough neutrons to start a chain reaction. However, the initial neutron count rates are very small; that's why additional neutron sources are needed so that the instrumentation can measure the neutron count rate during the startup process of a reactor with a fresh core. Otherwise, the instrumentation would be blind during first stage of the startup process.
Also note that no additional neutron sources are needed when the reactor contains a sufficient amount of irradiated fuel (i.e. after a partial reload) since irradiated fuel already contains various neutron sources (e.g. $\ce{^242Cm}$ and $\ce{^244Cm}$).
3) Neutron coincidence counting
Yes, it is possible to measure the amount and quality of plutonium (e.g. in weapons-grade plutonium or reactor fuel). In particular, this is done when you get a visit from an IAEA safeguards team. However, this is usually not a single measurement.
Coincident neutron detectors can be used to measure the amount of plutonium in non-irradiated fissile material (like MOX fuel). Neutron coincidence counting is used to
distinguish the neutrons emitted from a single fission event from neutrons created
from other processes. It seems like the measurement shown in the movie was inspired by a passive coincidence detector system.
However, spontaneous fission in plutonium is mainly caused by $\ce{^240Pu}$ and other even-numbered nuclides. Whereas for weapons-grade plutonium, you would be mainly interested in the fissile nuclide $\ce{^239Pu}$. Therefore, the isotopic
abundance must be known, typically measured by means of a high-resolution
γ spectroscopy. Using this isotopic abundance, the total plutonium amount in the sample can be determined from the coincident neutron count rates of $\ce{^240Pu}$.


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Buchwald-Hartwig coupling troubleshooting
I am conducting a synthesis using below scheme.

And Below is some information about my reaction history.

The basic condition is 1 eq of 1-aza-18-crown-6, 1.4 eq of 3-bromothiophene, 1 mol% of DavePhos Pd G4, 1 mol% of DavePhos, 2 eq of NaOtBu, toluene as solvent, and conducted at 100 ℃ for 24 h.
I chose the P ligand as DavePhos referring to this paper.
After 24 h, the reaction mixture is cooled to room temperature, taken up into 100 mL of dichloromethane, filtered through Celite, concentrated using rotovap, and dried in vacuum.
As I checked the GC/MS spectrum of the sample, the peak of 1-aza-18-crown-6 (SM2) is relatively much larger than that of 3-bromothiophene (SM1), and the product peak is also very small. That is, conversion is too low.
Below is the GC/MS spectrum. The unassigned peaks are unknown compounds or maybe impurities in GC column. The mass spectrum of each peak was omitted.

As I decreased the amount of catalyst and temperature, no product peak was observed.
As I increased the amount of catalyst to 3 mol% and temperature to 120 ℃, the peak of SM1 was largely decreased, but also the peak of product was.
Below is the GC/MS spectrum in the case of 3 mol% of catalyst loading. The unassigned peaks are unknown compounds or maybe impurities in GC column. The mass spectrum of each peak was omitted.

My doubtful point is that there are no remarkable byproduct peaks although the peak of SM1 is very small in contrast to the fact that the peak of SM2 is very large. In the case of 3 mol% catalyst loading the peak of bithiophene, which is the representative byproduct of Buchwald-Hartwig reaction, was observed, but it is also very small. If there were any large byprodut peaks, I would try to find a solution, however this situation is uninterpretable. What was going on in my reaction mixture?

In normal conditions there are about 20 different reasons why a Buchwald-Hartwig amination didn't go well, but your unique substrate makes me think of a 21st reason - could the coordination of $\ce{Na+}$ by the azacrown somehow be influencing its reactivity? It might make the N-H substantially more acidic, and my vague understanding is that acidic N-H makes for poor BH coupling partners. I wonder what would happen if you had a strong non-alkali metal base (assuming such bases have been successfully used in BH aminations). That said, when hearing hooves, think horses, not zebras I guess.

@NicolauSakerNeto Makes sense, in fact there were observed two spots from pure aza crown solution in TLC using Dragendorff reagent. And as far as I know the crown spot and crown+cation spot have different Rf values in TLC. But then I can't understand that where all the bromothiophene were consumed and where they went. Also, in the paper I linked, the condition using toluene and NaOtBu was successful in synthesizing N-aryl aza crown ether.

3-Bromothiophene is reasonably volatile. If it doesn't react, you may easily have vacuumed it out.

@NicolauSakerNeto Ohh...... That can be a reason...

@KrangLee  Have you run a sample of the 3-Br-thiophene starting material on the GC-MS? Does it show up well?

@Waylander I have tried only DIP-MS, but not GC/MS using pure 3-bromothiophene.

@KrangLee I think it is worth doing to get a handle on how well it shows up on the GC-MS. It is a technique that can sometimes give confusing results because conpounds do not show up as well as you expect.

@Waylander I would check it, Thank you. Also I think that I should check vacuum dried 3-bromothiophene sample too.


text:
What are the compounds that this article refers to?
The linked article studies the effects of a set of organophosphorus compounds on the bioelectrical activity in the digestive system. The OP compounds studied are referred to in the abstract as IMFF, PMFF, and MEDETFK. Some OP antidote referred to as N-5 (T) is also studied. Now, the original article is in Bulgarian, and only the abstract is available on the internet in English. Based on the fact that the English translation of the abstract refers to organophosphorus compounds with the abbreviation FOV, I would guess that the abbreviations IMFF, PMFF, and MEDETFK are transliterated from Bulgarian as-is. I would also guess that IMFF and PMFF refer to sarin and soman respectively. My question is: What compound does the abbreviation MEDETFK refer to, and what is the OP antidote N-5 (T)?
The article in question: https://pubmed.ncbi.nlm.nih.gov/4806515/
Update: I have managed to procure the full article. It does not elaborate on what compounds it refers to. Here is the article: https://i.stack.imgur.com/oBvVX.jpg

Have you tried to find the papers referenced in this one?

I think MEDETFK is the compound named as Me-DETPK on p 206 (fig 1) of this book: https://books.google.com/books/about/Selected_Technical_Publications.html?id=FjKY7pJeC0cC&printsec=frontcover&newbks=1&newbks_redir=0&source=gb_mobile_entity&ovdme=1#v=onepage&q=Organophosphorus&f=false

Their reaction produced primarily O,O diethyl S methylphosphorothiolate, but a minor product was O,O diethyl O methylphosphorothionate

@Andrew I don't know what journals References 3 and 5 are supposed to be for, and references 1 and 2 can't be found online. I'll try to procure the article in Reference 4.  However, I doubt MEDETFK would be the compound that's referenced in the book which you linked.


text:
Calculate rate constant of second order reaction and pseudo first order reaction

The reaction A + B -> Prod follows the second order kinetics. In an experiment, you initially have ten times as much B
as A. The concentration of A varies with time according to the following data:

How big will the error in the determined rate constant be if you choose to analyze the reaction as a pseudo-1st?

I began by plotting the first order reaction for this problem as ln[A] against t. From this I got that the slope = -$9.06 * 10^{-4}$ s$^{-1}$ and so the rate constant is k$_1$ =  $9.06 * 10^{-4}$ s$^{-1}$.
However, I am having trouble understanding how to plot for the second order reaction. I know that I should plot ln(a-x)/(b-x) but I don't understand how this is gotten from the given data.
How can I determine (a-x) and (b-x)?
I thought that x = [A]$_0$ - [A]$_t$ = $0.01$ - $1.11*10^{-6}$ = $0.0099$. But then do I subtract that from all the given data points? I did this and it gives me negative values and then I can't do the natural log in order to plot the points. Also if x = $0.0099$ and b = $10$*[A)$_0$ = $0.1$ then I get that (b-x) = $0.0901$ for all the points?

Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. Advantage is: Focus on principles, Easier spotting of mistakes, Improved orientation, Q/A reusability and the permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

The general rate law for a second order kinetics if $\pu{A_o < B_o}$, is $$\pu{\frac{1}{B_o - A_o} \ln\frac{A_o B}{A B_o} = kt}$$ If you only want to check that it is second oder, you can forget about the constants $\pu{A_o}$ and $\pu{B_o}$, and plot the logarithm of $\pu{\frac{B}{A}}$ versus the time $t$. The points should be well alined if it is second order.
Ref. Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, J. Wiley, NY, 1965, p. 16.


text:
Increasing the weight in metal with salt?
Is it possible to increase the weight in metal (such as bronze) by exposing it, or soaking it in salt?

The answer is no. The only possible effect is the oxidation of the metal due to the air. But the increase of mass is so small that it would even be difficult to measure. Soaking in salt has no effect on the weight. Anyway the mass of the metal will never increase, whatever the treatment, with or without salt. The increase in mass corresponds to the oxygen being fixed as oxide on the surface of the metallic piece. The mass of the metal itself cannot and will not change.


text:
Rearrangement vs. Ring Strain in Cyclobutanecarboxaldehyde Nucleophilic Reaction

Predict the major product of the following reaction:

Note: Product A is incorrectly marked as secondary halide, it should have been primary halide

This question appeared in an national level aptitude test.
My reasoning was that for the SN1 rearrangement carbon in the cyclobutane ring (which already has quite a bit of ring strain) undergoes the hybridization change from sp3 to sp2, which puts it at a much higher strain, hence this mechanism should theoretically be disfavoured. On the flip side, the SN2 mechanism in a primary alcohol does not have any such hinderance.
Which effect trumps which? Is the SN1 tertiary stability worth the ring strain?

Consider the possibility that the ring strain is sufficient that tertiary cation rearranges to relieve that strain

@Waylander Yes that is a possibility, I believe. However the alcohol is primary and hence Sn2 is highly favoured. I am looking for the proportion of products of A is to B. Like how much of A is formed and how much else is B.

see these questions  https://chemistry.stackexchange.com/questions/128941/which-mechanism-should-a-reaction-between-the-cyclopentylmethanol-and-the-hbr-sh   and https://chemistry.stackexchange.com/questions/112785/why-does-the-reaction-of-the-primary-alcohol-cyclopentylmethanol-with-hydrogen-b

@Waylander Yes, exactly. Kinetics seems to favour a ring expansion product to compensate increased ring strain. This appears possible in the above reaction. I do understand that Sn1 will be favoured over Sn2 if rearrangement is possible (since it is a polar solvent, MeOH/HBr system). If one takes into account the energy barrier for 1,2-hydride shift, and increased ring strain verses the increased stability of carbocation in polar solvent then how does the Sn1/Sn2 equilibrium look like. It is of my opinion that one will end up with a mixture of products A and B. I might be entirely wrong.

Also, just an additional curiosity: Will this reasoning hold if the -CHO group is replaced by a ketone ( - CO - CH3 ). What additional factors do I need to consider apart from increased steric hinderance.

The ring strain on a cyclobutane is considerable. I think relief of that strain will be the dominant factor, but until someone runs the experiment (I have not got database acces to search for it) it is all supposition.

@Waylander Is it of your opinion that a carbon in a ring will be relived after it changes its hybridization from sp3 to sp2? Also, I do not have access to any of the databases to search for reactions myself, which is why I posted this here.

My opinion is that you will get mainly cyclopentyl bromide and a small amount of the bromomethylcyclobutane


text:
Calculate the concentration of the reactant from the absorbance of the product without molar absorptivity

Given the reaction $\ce{A->B}$, where the absorbance of the product $\ce{B}$ has been measured (see the table below), and the initial concentration of $\ce{A}$ is $\pu{10^{-4} M},$ calculate the concentration of the reactant at the time $t.$
$$
\begin{array}{rr}
\hline
t/\pu{min} & A \\
\hline
0 & 0.000 \\
1 & 0.115 \\
2 & 0.188 \\
3 & 0.237 \\
4 & 0.273 \\
5 & 0.301 \\
6 & 0.321 \\
7 & 0.339 \\
8 & 0.353 \\
9 & 0.365 \\
10 & 0.375 \\
\infty & 0.500 \\
\hline
\end{array}$$

From the answer it is stated that one should use the following equation to calculate the concentration of $\ce{A}$:
$$[\ce{A}]_t = [\ce{A}]_0\left(1 – \frac{A_t}{A_\infty}\right)\label{eqn:1q}\tag{1}$$
I don't understand where this comes from since I have only ever converted absorbance to concentration using the equation:
$$A = \varepsilon lc\tag{2}$$
Could someone please explain the reasoning behind the equation \eqref{eqn:1q}?

Stoichiometry dictates that the amount of A and B are related. You're expressing the concentration of A by taking the absorbance of B and relating that to the concentration of B.

The equation can be written also as [A] = [A_0]-[B] = [A_0] - abs/abs_fin x [B_fin]

Because $A\to B$ final amount of B or $[B]_\infty=[A]_0=10^{-4}$, and as $Abs=\epsilon [C]l$ then $\epsilon l=0.5/10^{-4}$. At time $t,; [A]_t=[A]_0-Abs_t/(\epsilon l)$.

I bet you must be aware of the general concept of proportionality ($y = a.x \implies y_2 = x_2 \cdot \frac{y_1}{x_1}$ without need to know $a$) and of constant sum of 2 variables ( $a = x + y \implies y = a - x$ ). Why have not you tried to apply it?

Reactant $\ce{A}$ does not absorb the light, so the measured absorbance is only due to $\ce{B}$. Let $[\ce{B}]_t$ and $A_t$ be the measured values of $[\ce{B}]$ and $A$ at any time during the experiment:
$$A_t = \varepsilon_\ce{B} \cdot l \cdot [\ce{B}]_t \label{eqn:1a} \tag{1}$$
At the end of the experiment $t = \infty$, $A_\infty = 0.500$ and $[\ce{B}]_\infty = [\ce{A}]_0 = \pu{E-4 M}$. As a consequence
$$\varepsilon_\ce{B}\cdot l = \frac{A_\infty}{[\ce{B}]_\infty} = \frac{A_\infty}{[\ce{A}]_0} \label{eqn:2a}\tag{2}$$
Introducing \eqref{eqn:2a} in \eqref{eqn:1a} yields
$$[\ce{B}]_t = \frac{A_t}{\varepsilon_\ce{B}\cdot l} = [\ce{A}]_0 \frac{A_t}{A_\infty} \tag{3}$$
$$
\begin{align}
[\ce{A}]_t &= [\ce{A}]_0 - [\ce{B}]_t \\
  &= [\ce{A}]_0 - [\ce{A}]_0 \frac{A_t}{A_\infty} \\
  &= [\ce{A}]_0\left(1 – \frac{A_t}{A_\infty}\right) \tag{4}
\end{align}
$$
And this is the wanted formula.

I apologize for the late reply, but how did you derive [A]$_t$ = [A]$_0$ - [B]$_t$ ? I don't see how you could get a subtraction between the two from eq. ($3$)

@Katara. Suppose [$\ce{A]_0] = 1.00} M$. If $30$% of this amount has been transformed into $\ce{B}$ after a time $t$, [$\ce{B]_t = 0.300} M $. Simultaneously [$\ce{A]_t = 1.00 - 0.30 = 0.70} M$


text:
Methods to detect iron in highly concentrated ferric sulphate solutions
I’m writing with a query related to reagents used to detect iron in highly concentrated ferric sulphate solutions.
I am currently exploring iron detection in both solutions of ferrous $\ce{(Fe^2+)}$ and ferric $\ce{(Fe^3+)}$ sulphate. I have had success with the detection of $\ce{Fe^2+}$ by performing potassium permanganate titrations which lead to accurate results. However, I am struggling to identify an appropriate reducing agent so that the method can be applied to solutions of ferric $\ce{(Fe^3+)}$ sulphate.
A potential option that I was considering was hydroxylamine hydrochloride. However, I am concerned about the potential quantities of $\ce{N2O}$ which may be produced at such high iron concentrations in the ferric sulphate I am using $(\approx \pu{18 gl^-1})$. Is anyone aware of this area of experimentation or aware of any such standard method for the detection of $\ce{Fe^3+}$ in this way?

Hydroxylamine is not a good choice. Because if you add an excess in the ferric solution, the excess will react with permanganate like the ferrous ion. Better use metallic iron (iron wire). In hot solution, $\ce{Fe}$  reacts with ferric ions according to $\ce{Fe + 2 Fe^{3+} -> 3 Fe^{2+}}$. Here an excess of the reducing reagent $\ce{Fe}$ will not react with permanganate. You also may use zinc $\ce{Zn}$ instead of  iron, according to $\ce{Zn + 2 Fe^{3+} -> Zn^{2+} + 2 Fe^{2+}}$, although zinc may also react with $\ce{H+}$ ions, creating .. too many $\ce{Zn^{2+}}$ ions. Who knows ?

A classical titration of ferric ions is direct analysis with stannous chloride. Do you have analytical grade stannous chloride?


text:
Advice on reactions with copper sulphate and aluminium in swimming pool
I have a swimming pool in Portugal with a heating system of aluminium (good heat conductor) tubes, painted black and a small circulation pump. The sun shines, the pool gets warm, cost is minimal and it works well.
As the pool is private, I use a minimal level of chlorination and therefore I use a small amount of copper sulphate (<1 ppm) to control algae growth.
I notice that the heating circuit inlet pipe to the pool leaves white crystals on the floor of the pool.
Question - am I correct in thinking that the copper sulphate is reacting with the aluminium tubes, resulting in a copper coating on the inside of the tubes (not that I can see inside) and aluminium sulphate crystals appearing in the pool?
I understand that aluminium sulphate is essentially 'flocculant' and is not particularly hazardous: certainly the pool stays very clear.
Advice would be much appreciated, many thanks

There is the standard redox reaction, as I show here, but you have a very dilute solution of copper sulfate, and dilute  aluminum sulfate is harmless. I would not worry about it unless the situation worsens. In that case, use a different, non-corrosive, algaecide.

Your white crystals are not aluminum sulfate, because aluminum sulfate is soluble in water. It may be calcium sulfate produced by the calcium ions dissolved in the tap water after reaction with the sulfate ions from copper sulfate. Or it is calcium carbonate produced by heating the tap water. Because soluble calcium bicarbonate from tap water is decomposed in hot water into insoluble calcium carbonate, which produces nice white crystals. This deposit is often a problem in washing machines.

Thank you Gentlemen,
The tap water is soft here - negligible scale.
I doubt there is any danger from whatever it is, no skin irritation or anything like that, however, out of curiosity I think I shall take samples of water and crystals to the local lab. for analysis.
I'll post the results here for your interest.


text:
How to calculate concentration of a protein based on known concentration and mw of other protein
I have a Peptide A (mw = 5074Da), the table below showed
measured concentration by Qubit from 5uM up to 80uM solution.
All under 20uL solution.

Below is the standard plot for Peptide A (not sure if it's relevant or not):

Let's say I have another Peptide B (mw = 3102Da),
how can we calculate the expected Qubit read of Peptide B  based on Peptide A (just by calculation)?
For example just for 1 node: 20uM.

It does not seem to me to be linear.

@Poutnik Sorry do you mean it can't be calculated?

I mean there is consistent bending down and that questions justification of linear approximation.

@Poutnik If we ignore the linearity issue you brought up, how can my question be answered?

I was not going to address the question itself and I would let it on users involved in this area. But, I have noticed the question is very laconic in description and elaboration and has lack of signs of explicit own solving effort. It would not please users much.

What is "nodes"? Please add some clarification on the concentration units, as these are orders of magnitude different (measured vs nominal). If you read the manual you'll see that the measured concentration can vary ~12% between proteins, but this will apply to peptides only if the functional principle of the assay still applies. I would think it's ok.

https://www.science.smith.edu/cmbs/wp-content/uploads/sites/36/2015/09/Qubit-Protein-Assay.pdf

@BuckThorn Thanks so much for your reply. Nodes is the concentration in the peptide solution I put in the tube (before adding the working solution) to be measured with Qubit.

It would appear that the signal from the instrument is roughly proportional to the mass of peptide, although your other question shows this is not a simple linear relation, which explains your question. However, you may want to edit this post to make such details more clear. Also, is "nodes" common nomenclature in your lab? A node is a connection in my understanding.

If knowing the concentration of peptide B is important and you have pure peptide B available, you should measure standards of known concentrations of peptide B in the assay.
The assay works through a dye that has little fluorescence in solution and a lot when bound to proteins. The mechanism of binding is not known or not divulged. While there is a claim that there is little variation of mass of protein sample vs. fluorescence signal, the data presented by the company uses larger globular proteins.
We don't know what the amino acid composition or the structure of your peptides are. We also don't know whether the assay works under native or denaturing conditions (the assay buffer has a pH between 6 and 8, but there is no information about its composition).
If knowing the concentration of peptide B is not so important, you could assume that the same mass of peptide will have the same fluorescent signal. Or you could use the assay to get a relative concentration if the absolute concentration is unimportant.


text:
Comparison of the acidic nature between substituted benzoic acids and acetic acid
p-Methoxybenzoic acid $(\mathrm{p}K_\mathrm{a} = 4.47)$ is a stronger acid than acetic acid $(\mathrm{p}K_\mathrm{a} = 4.76)$ despite the presence of an electron donating group. I understand this is due to the greater electronegativity of $\mathrm{sp^2}$ carbon to which the carboxyl carbon is attached.
But is there a substituent with strong enough donating nature to make benzoic acid $(\mathrm{p}K_\mathrm{a} = 4.20)$ less acidic than acetic acid? If there are, which one might be just strong enough?

But is there a substituent with strong enough donating nature to make benzoic acid less acidic than acetic acid?

Yes, we can make benzoic acid less acidic than acetic acid by simply substituting strong electron donating groups like $\ce{-N(CH3)2}$ to the para position of benzoic acid. By substituting $\ce{-N(CH3)2}$ to the para position of benzoic acid we can reduce it's $\mathrm{p}K_\mathrm{a}$ to $6.03$! Which is much less acidic than acetic acid.

That's good to know, but considering -N(CH3)2 gets the pKa all the way to 6.03, there's probably a substituent that would make it just a little less acidic than acetic acid, like I mentioned in the 2nd part of the question. Do you have any idea about such a substituent? Or is there none, at least within the scope of a particularly hard college entrance exam?

@Skahoot p-aminobenzoic acid aka PABA has pKa for carboxyl 4.88, what is slighly less acidic than 4.75(4.76) for acetic acid.

This isn't a best example - amino groups complicate things - this 6 is pKa of zwitterion - N-acid.

@Mithoron, if so, could you mention a more suitable alternative? Although, TBH I think what has been mentioned would be adequate for my level.

@Skahoot Maybe trimethylammonium, or -BH2 group...


text:
Reaction of chloroform with DBU
There's a certain reaction I've been doing that involves DBU (I've drawn the structure out below). The reaction is done in a flow machine (ie. high temperature and high pressure).

I've read that chloroform can't be used as a solvent for this reaction, because under the flow machine conditions DBU reacts with chloroform to form a side product.
I was interested in what this reaction is, so I've had a bit of a look but can't work it out. Here's what I'm thinking so far:

I've read that chloroform can react with base to form dichlorocarbene. There's no strong base here in this reaction, but could the high temperature/pressure conditions be causing this species to form? Dichlorocarbene could then react with DBU.
The only alternative I can think of is that chloroform does a nucleophilic substitution (with the DBU acting as electrophile)? Under normal conditions chloroform doesn't do nucleophilic substitution, but maybe under high temperature/pressure conditions it can. Also, DBU could potentially act as an electrophile in different ways:

Or alternatively, the enamine tautomer could potentially be the nucleophile:

Thanks for any help!

DBU is stronger base than you think. Strong enough to generate dichlorocarbene from chloroform under flow conditions as this reference shows here.


text:
Solubility of water in toluene
I'm interested in studying the solubility of water molecules in toluene.
I already know that some water can dissolve in toluene. I've also done some calculations at DFT level using the C-PCM solvation model, but unfortunately, the DFT theory does not correctly describe all water's properties.
Is there somebody that has studied this phenomenon? I was able to find only toluene-in-water system.

Mutual Solubility Measurements of Hydrocarbon–Water Systems Containing Benzene, Toluene, and 3-Methylpentane is available from ACS at  J. Chem. Eng. Data 2008, 53, 1, 165–174
Publication Date:November 21, 2007
https://doi.org/10.1021/je700449z  but apparently it will cost you.

@JamesGaidis - depends if their institution has access. I get the full article with no issues. I'm sure a librarian can get it for them if the library doesn't have it.


text:
Chemical name of a solution of ions such as salt dissolved in water
Is there any chemical name for the solutions of ions in water? I want to find out the name given to an aqueous solution of ions such as $\ce{Na+}$ and $\ce{Cl-}$ ions in water. Would it be something like aqueous solution of ions?

For your specific case, how about "brine".

There is no specific name but some authors may use the word aquated when they wish emphasize that the ion is surrounded by /weakly bound/coordinated to water. Google Scholar shows that about 10,000 documents use this term.
Examples:
Extreme case (i) aquated electron (ii) aquated $\ce{Na+}$ etc.

Aqueous electrolyte solution.


text:
What is the net gain of charge carriers in a semiconductor if an atom is removed?
In pure germanium, all Ge atoms are bonded with a tetrahedral geometry. It has a band gap and is a semiconductor. If we remove one Ge atom, what type(s) and how many charge carriers will be generated? So far the explanation I have found is, by removing one Ge atom, 4 electrons from the valence band is removed, which is equivalent to say 4 holes have been added to the valence band. Hence, the answer is a net gain of 4 holes as charge carriers.
But we still have four non-bonding electrons left (from the 4 neighbouring Ge atoms): where do they go? Are they simply equivalent to the newly generated 4 holes or they should be counted separately? And by removing that Ge, four orbitals are removed as well, which I think would change both valence and conduction bands.
However, another contradictory scenario is for graphite: if we remove one carbon, what type(s) and how many charge carriers will be generated? If using the same rationale, 4 electrons from the filled band is removed, so we should expect a net gain of 4 holes as well? But considering the special band structure of graphite, perhaps only 1 hole is generated in the valence band, as $p_z$ is higher than $sp^2$ orbitals. But the suggested explanation is: we are removing 1 electron from the valence band and there are 3 more non-bonding ones generated, so there is a net gain of 2 electrons into the conduction band.
Or, trying to make the two situations agree, could it be that these net gain of 2 electrons equivalent to 2 holes? But it seems not very convincing.
Thanks in advance for any thoughts!

From https://journals.aps.org/prb/abstract/10.1103/PhysRevB.87.035203 one finds that there are various possible charge states for the germanium vacancy, with different levels ranging from inside the valance band to just below the conduction band. The occupancy of such states depends on where the Fermi energy is. Now, your thought process is very molecular-bonding like, not solid state physics like, so you are being led astray by it.

@JonCuster Thanks for your reply and really sorry that it is from a second year undergraduate chemistry course, so my reasoning is confined to the tight binding model, band structures, etc.... But many thanks for sharing the paper.


text:
Cl-P-Cl and Cl-S-Cl Bond angles in respective PCl3 and SCl2 compounds
Both $\ce{PCl3}$ and $\ce{SCl2}$ are $\mathrm{sp^3}$ hybrid. $\ce{PCl3}$ has 1 lone pair whereas $\ce{SCl2}$ has 2 lone pairs, so due to more lp-lp repulsion, $\ce{SCl2}$ should have smaller bond angle than $\ce{PCl3}$. But practically, $\ce{SCl2}$ has got larger bond angle than $\ce{PCl3}$. Why is it so?

Not so sure about $\ce{SCl2}$ to begin with.

You are worried about 3pm here. Possibly the slightly shorter bond causes slightly more repulsion between the electrons on the chlorines. There doesn't seem to be enough difference to be concerned about changes in hybridization. Personally I think simple use of hybrid orbitals and realization that unpaired electrons occupy such orbitals and take up space goes a long way towards understanding simple molecular structure.


text:
Internal Electrophilic Aromatic Reaction in 3-chlorobutylbenzene
I recently came across the reaction scheme: Aluminum chloride to 3-chlorobutylbenzene.
In Jerry March, reaction scheme is provided for 4-chlorobutylbenzene with $\ce{AlCl3}$ where ring closure occurs resulting in formation of a six membered ring. However, since $\ce{AlCl3}$ is being used:

Rearrangement of carbocation is possible as $\ce{AlCl3}$ forms carbocation unlike $\ce{FeCl3}$
A secondary carbocation is formed initially

Question: If the carbocation rearranges to a primary carbocation (formation of six membered ring over a five membered ring), then the reaction would proceed to give a 6 membered ring. However, if no rearrangement were to take place (as secondary carbocation is more stable in comparison to primary), the product formed would be a methyl substituted five membered ring.
So, would the reaction path be decided by the intermediate stability or by the product stability?

The carbocation rearrasngement (depending on conditions) occurs primary to secondery, not vise versa.

@MathewMahindaratne while i understand that concept that you're trying to say, my question also lies in the fact that in Jerry March, the primary carbocation wasn't rearranged to a secondary cation

@shukraditya bose , reaction scheme and page number with edition eoukd help in answering. And circumastance of reaction can be understood.

Apologies. It's the 4th Edition of Jerry March's Advanced Organic Chemistry, Page Number 547

Intramolecular Friedel-Crafts cyclization (either by acylation or alkylation) is an useful methord for the synthesis of bicyclic or polycyclic aromatic compounds. The cyclization of (4-chlorobutyl)benzene in the question is an exsample of such reaction. This perticular case has been studied (among other cases), and the study of rates of rearrangement versus ring closure is reported in Ref.1. The study indicate that in deed, the rearrangement to 5-membered ring has been occured via secondary carbocation, yet the closure to the 6-membered ring must be faster than closure to the 5 membered ring in the given case (as OP found in Jerry March book):

The abstract states that:

...Cyclialkylation rearrangements of both primery and secondery phenylalkyl chlorides were observed. The extent of rearrangement was less than in comparable intramolecular alkylations only in cases where a six-membered (tetralin) ring was formed directly. Five-membered (indan) rings were not formed directly from appropriate primery and secondery phenylalkyl chlorides; extensive hydride exchange and intermolecular polyalkylation interverned in reactions of these compounds. Direct cyclization of primary phenyl alkyl chloride (VIII) to tetralin was shown to take presidence over cyclization to an indene, either directly or over secondery intermediate...

So, what would be the answer to OP's question, "would the reaction path be decided by the intermediate stability or by the product stability?"?
The answer is not that simple. It is clear that the rate of cyclization to six-membered ring is much faster than the rare of cyclization to five-membered ring (the product stability) as well as the rate of cyclization to six-membered ring is faster than the rare of hydride transfer. The catalyst to substrate ratio is also playing a role (Ref.1).
For instance, the cyclization of (3-chloropropyl)benzene to indan has been occured in trace amount, but the yield can be increased to 10% with incresing amount of catalyst (Ref.1). Yet, the cyclization of (3-chloropentyl)benzene has been occured to give 1-methyltetralin as major product (not 1-ethylindan) in good yield. However, the appropriate tertialy carbocations (highly stable compared to secondery version) readily cyclized to give corresponding indans or tetralins (Ref.2). Some of these cases, rearrangements (even Wagner–Meerwein rearrangement) occurs to get the appropriate carbocation intermediate to facilitate the cyclization, but never a secondery to primwary carbocation rearrangement. Thus, it can be conclude that you cannot expect tetralin from (3-chlorobutyl)benzene, but 1-methylindan.
I'd recommend read some other references as well to get the mechansm point of view (e.g., Ref.3 & 4).
References:

Ali Ali Khalaf and Royston M. Roberts, "New Friedel—Crafts Chemistry. XVI. A Reconsideration of Cyclialkylation and Competing Reactions of Certain Phenylalkyl, Benzoylalkyl, and Acetylphenylalkyl Chlorides," J. Org. Chem. 1966, 31(1), 89-95 (DOI: https://doi.org/10.1021/jo01339a018).
Ali Ali Khalaf and Royston M. Roberts, "New Friedel-Crafts chemistry. XIX. Cyclialkylations of some phenylalkanols," J. Org. Chem. 1969, 34(11), 3571–3574 (DOI: https://doi.org/10.1021/jo01263a075).
L. R. C. Barclay, B. A. Ginn, and C. E. Milligan, "Friedel-Crafts Rearrangements. II. Rearrangements during cyclialkylation of ε-aryl sunstituted compounds," Can. J. Chem. 1964, 42(3), 579–590 (DOI: https://doi.org/10.1139/v64-08).
L. R. C. Barclay and E. C. Sanford, "Friedel–Crafts rearrangements. III. Deuterium tracer studies of the mechanism of cyclialkylation of phenylalkyl chlorides," Can. J. Chem. 1968, 46(21), 3315–3323 (DOI: https://doi.org/10.1139/v68-548).


text:
Can we compare absorbance of two peptides when the read concentration from Qubit Protein Assay are different?
I have two peptide A and B. And I would like to know the solubility level of these peptide in certain solute. For that purpose I use Qubit Protein Assay.
The Qubit readout I get is this:

Note that the Qubit Read Concentration and Fluorescence of Peptide B is lower than Peptide A.
As far as I understand, we use fluorescence to quantify the peptide (so it's equivalent to absorbance). High fluorescence means lower solubility.
Just by looking at fluorescence value I'm inclined to say that Peptide B has lower solubility than Peptide A.
But my worry is that the Qubit Read Concentration are different for both.
I wonder if this case will make the two peptide fluorescence incomparable?
If not, what is the better strategy to deal with it?

Update (based on @Karsten comment)
Here is the standard curve for Peptide A and B.
Can we still say that Peptide B is less soluble then Peptide A?

Please note that standard way of emphasizing in Markdown is using asterisks: *<emphasized text>* (). Backquotes are only used for inline code: \``.

Try plotting an overlay of concentration of the peptides in g/L vs determined (Qubit) concentration. In other words, multiply  conc(uM) by the molecular weight. I suspect the curves will overlap pretty well.

This data might tell you if your peptides self-associate, but not the solubility.

By the way please edit the title of your post to better reflect its contents.

If these two peptides are the same as those in this question, you would expect a higher reading and higher estimate of mass for solutions of A compared to solutions of B of the same concentration. This is because for every molecule of peptide A, you have a larger mass and a larger chance of fluorescent dye attachment compared to B (the molar mass ratio is about 5:3).
The question mentions solubility limit. While the concentrations are not linear with the signal, it does not seem that there is a ceiling to the signal at higher concentrations, so maybe the solubility limit is not reached. We would need a bit more context to understand what the experimental setup is to fully address that aspect.
Here is a sketch of what you would expect if you dissolve different masses of peptide in a given volume, spin down and measure the supernatant:

The solutions that are not saturated should show a signal roughly linear to the mass added. At the same amount (nanomoles), you would expect peptide A to have a higher signal because there are more amino acids to interact with the fluorescent dye. Once you reach the solubility limit (highlighted in the spreadsheet), some of the material precipitates, so there is a cap on the fluorescent signal. To figure out the solubility, extrapolate the signal until you reach the signal of the saturated solutions, and read off the amount and calculate the concentration. (I could have used concentrations and mass concentrations in the spreadsheet, with the same conclusion).

[OP] High fluorescence means lower solubility.

That does not make any sense to me. If the solubility is higher, there will be a higher concentration in a saturated solution, and the fluorescence signal should be higher.

thanks for your answer. So you meant the Qubit read cannot be used to determine which peptide has better solubility. Am I right?
How about calculating the fold change of Qubit read with initial concentration/mass before creating the solution? What other context we can give you so that you can suggest us the experimental setup?

The easiest experiment is to make a saturated solution, spin it down and measure the concentration of the supernatant. This gives you a rough idea of the solubility (mass per volume). In you experiments, did you see a pellet after centrifugation?

how do you measure the concentration of supernatant? Yes I can see the pellet after centrifugation.

In the Qubit. But maybe I just don't understand the data you are presenting. If you have known amounts of peptide, you can get a calibration curve for solutions that are not saturated. Then, you measure the supernatant of the saturated solution (diluting it if it is outside the range of the standard curve). This will give you a good estimate of the solubility.

That does not make any sense to me. My understanding is this: if a peptide has low solubility, it will create more pellettes (hence fewer mass being turned into saturated solution). This will cause lower fluoresence. Please correct me if I'm wrong.

If I understand you correctly, the table that I put in my OP, allows me to make the calibration curve for solutions that are saturated (i.e. supernatant). Right? How can you make curve for those not saturated?

The supernatant of a saturated solution has an unknown concentration (because some material is in the pellet). Not saturated solutions have a known concentration (everything is in solution), so you can use those to make the standard curve.

Let us continue this discussion in chat.


text:
In qualitative analysis for basic radicals, why are group 2 cations not precipitated along with group 4?
To ensure that group 4 cations are not precipitated with group 2, we add dil. $\ce{HCl}$ to ensure only group 2 sulfides of lower $K_\mathrm{sp}$ are precipitated.
For group 4, we add $\ce{NH4OH}$ along with $\ce{H2S}$ to shift the reaction in the forward direction since group 4 sulfides have higher $K_\mathrm{sp}$ which makes sense but why aren't group 2 sulfides of lower  $K_\mathrm{sp}$ precipitated along with group 4?

Pleyse explain what you call group 2, group 4. To my knowledge, based on Treadwell's classification, group 2 includes elements which are precipitated by $\ce{H2S}$ in acidic solution like copper, cadmium, etc.. Group 3 includes elements which are precipitated by $\ce{(NH4)2S}$ like iron,  zinc, etc. But those elements are in your group 4. You must have another classification. Explain.

In more modern nomenclature, Group 3 is used for metals that have low solubility hydrixides such as aluminum and chromium.

@Oscar Lanzi. With this definition, practically all metals would have been in Group 3, as the only metals that makes soluble hydroxides are from the first and partly from the second column of the periodic table.

Nomenclature aside, the full procedure calls for first making the acidic sulfide solution, precipitating the metals associated with that group, and then decanting the liquid and adding ammonia solution* to capture the ammonia-sulfide group. Thus the acid-sulfide precipitates don't stay in solution, rather they are already gone.
*Ammonia solution, not ammonium hydroxide. Ammonia dissolved in water is only a weak base, so the major species are $\ce{H2O}$ and $\ce{NH3}$, not $\ce{NH4^+}$ and $\ce{OH^-}$.

I get the confusion of grouping the cations. Some say there are 6 groups, other say 4 group but at the end of the day, it's the matter of convention.
Let's start with bunch of salts of unknown cations. You add dil. $\ce{HCl}$. If it forms insoluble chloride, then the cations present are $\ce{Ag+, Hg2^2+ and Pb^2+}$. If no precipitate is formed, add $\ce{H2S}$ (note the pH is already acidic). Sulfide having low  $K_\mathrm{sp}$ will form precipitate out in acidic solution (hence the group reagent you are observing is dil. $\ce{\bf{HCl}}$ + $\ce{H2S}$). If not, then make the solution basic with ammonia solution + $\ce{NH4Cl}$ (this is your group reagent). Sulfides having larger $K_\mathrm{sp}$ values such as ($\ce{ZnS, NiS, CoS}$ and $\ce{MnS}$ precipitate. Also, at basic medium, $\ce{Al^3+, Fe^3+}$ and $\ce{Cr^3+}$ form insoluble hydroxides and are separated from the solution. now comes cations which formed soluble sulfides and chlorides. They can be separated by forming insoluble carbonates (note the solution is basic so the group reagent is $\ce{\bf{NH_3(aq.)+ NH_4Cl}}$ + $\ce{(NH4)2CO3}$). The cations that are left are determined through flame test.
So, in my opinion, the group can be done in following way:
\begin{array}{c|c}
\mathbf{Group} & \mathbf{Cation} & \mathbf{Group~reagent} & \mathbf{Inference}\\\hline
\text{Group~I} & \text{$\ce{Ag+, Hg2^2+, Pb^2+}$} & \text{dil. HCl} & \text{insoluble chlorides}\\
\text{Group~II} & \text{$\ce{Hg^2+, Pb^2+, Cu^2+, Bi^3+, Cd^2+, As^3+, Sb^3+ , Sn^4+}$} & \text{$\ce{dil. HCl + H2S}$} & \text{insoluble sulfide}\\
\text{Group~IIIA} & \text{$\ce{Ni^2+, Zn^2+, Mn^2+}$} & \text{ammonia soln. + $\ce{NH4Cl}$} & \text{insoluble sulfide}\\
\text{Group~IIIB} & \text{$\ce{Al^3+, Fe^3+, Co^2+, Cr^3+}$} & \text{ammonia soln. + $\ce{NH4Cl}$} & \text{insoluble hydroxides}\\
\text{Group~IV} & \text{$\ce{Ca^2+, Sr^2+, Ba^2+}$} & \text{$\ce{NH3(aq.) + NH4Cl + (NH4)2CO3}$} & \text{insoluble carbonates}\\
\text{Group~V} & \text{$\ce{Mg^2+, Na+, K+}$}& \text{flame test} & \text{color}\\
\text{Group~0} & \text{$\ce{NH4+}$}& \text{-} & \text{since it doesn't have any group reagent, I don't think it needs a separate group}\end{array}
This was my previous grouping, but I suppose I will add a comment on the updated table.


text:
Is there a need to change sign for enthalpy value for a reaction that appears in reverse when using Hess's law?
Problem

Given the enthalpies of the following reactions
\begin{align}
\ce{2H(g) &-> H2(g)} &\quad \Delta_\mathrm{r}H^\circ &= \pu{-437.6 kJ mol^-1} \tag{R1}\\
\ce{C(s) + 2 H2(g) &-> CH4(g)} &\quad \Delta_\mathrm{r}H^\circ &= \pu{-75.2 kJ mol^-1} \tag{R2}
\end{align}
find $\Delta_\mathrm{r}H^\circ$ of
$$\ce{C(s) + 4 H(g) -> CH4(g)}\tag{R3}$$

Answer

 $\Delta_\mathrm{r}H^\circ = \pu{-950.4 kJ mol^-1}$.

Question
Why don't we change the sign of $\pu{-437.6 kJ mol^-1}$ (times two) if we go from $\ce{2 H2}$ to $\ce{4H}?$ Isn't it endothermic when we reverse the reaction $\ce{2H -> H2}$?

We do, but not in your case, as we are not interested in the reverse reaction..

There is no point in doing so here, just make sure the expression for the resulting enthalpy holds true algebraically for reactants and products. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Don't use ALL CAPS, it's rude and hearts readability.


text:
Why is effervescence endothermic?
The reaction of sodium bicarbonate and sulfuric acid is quite endothermic... it seems.

Of course, the chemical change is actually exothermic, but from what I've read, the production of carbon dioxide gas is what makes it appear endothermic. I'm assuming that as the gas is produced, it expands and this absorbs heat from the surrounding water. Does the gas expand because as the gas rises, the pressure drops? And does this expansion mean the gas is doing work against the surroundings, making the enthalpy change negative?

As a future scientist, always include a control experiment. Add "a lots" of solid sodium bicarbonate to water and see what is the temperature change. Second control is, mixing molar equivalent solutions and see what happens.

You might run it in a calorimetric bomb or other exceedingly strong container, preventing CO2 escape. Use caution, of course. But then the CO2 buildup will change equilibrium...


text:
Amino Acid Derivative Naming Conventions
Is there a trivial name for compounds based on glutamic acid, or other amino acids, where it is not the α-amino acid group that is functionalised? (I.e. that shown on the right)
Glutamylsulfonic acid, for instance, would seem to have a structure similar to glutamyl aldehyde, but with a sulfonic acid group appended to the aldehyde. Following on from the centre-right example, dimethyl glutamate could also be named: glutamic acid 1,5-dimethyl ester.
The issue seems to be that the glutamyl- group is no longer technically present, but the compound still strongly resembles glutamic acid. I am aware that amino acids do not normally exist in these neutral forms, and am only interested in the nomenclature.

I don't think even the names you came up with are OK. All this should be named systematically.

I would name the sulfonic acid as a homocysteine derivative rather than glutamate

Also the aldehyde is often called glutamate semi-aldehyde to indicate that only one acid group had been reduced

This one is glutamate-1-semialdehyde as opposed to glutamate-5-semialdehyde.

This is not an answer to the question but the systematic names of the shown structures are (2S)-2-aminopentanedioic acid, (4S)-4-amino-5-oxopentanoic acid, (4S)-4-amino-5-methoxy-5-oxopentanoic acid, and (2S)-2-amino-4-sulfobutanoic acid. Note that the last compound does not resemble glutamic acid anymore since it has lost a C atom so it's a butanoic acid and not a pentanoic acid.

As you're aware, biochemists typically do not use systematic naming for most compounds,  preferring instead to use common names and modifying them as needed for related compounds. In the case of amino acids, the names typically are created based on the names of the proteinogenic amino acid names. I do not know of any codified formal rules for this process, but a quick look at the names in use suggests a few guidelines that seem to be followed most of the time. Here's an imcomplete summary.
1: Priority seems to be given to the "R" functional group rather than, for example, chain length. Thus, $\ce{HS-CH2-CH2-CH(-NH2)-CO2}$ is named as "homocysteine" based on the presence of the -SH group rather than as a variant of methionine, which has the same number of carbons and a sulfur atom, but not a sulfhydryl group. Similarly, $\ce{HO-CH2-CH2-CH(-NH2)-CO2}$ is named homoserine rather than considered a reduced form of aspartate, which has the same number of carbons.
2: The prefix "homo" indicates an extra methylene group in the side chain (eg homoserine, homocysteine, homolysine)
3: The prefix "nor" indicates a structural isomer of an alkyl side chain (eg norvaline, norleucine)
4: Aldehydes tend to be named as carboxylic acid derivatives rather than alcohols, so $\ce{OCH-CH2-CH(-NH2)-CO2}$ is aspartate semialdehyde rather than an oxidized homoserine.
5: The suffix -ol designates reduction of the 1-carboxylic acid group to an alcohol.
The sulfonic acid compound you asked about poses an interesting challenge, because it does not have a methyl group on the S, but also is not an -SH group. One could make an argument for naming this compound as either a methionine or homocysteine derivative, but convention seems to favor homocysteine, based on similar compounds such as S-adenosyl homocysteine. Thus it would be "homocysteine sulfonic acid".
As noted in the comments, your aldehyde compound is typically named glutamate-1-semialdehdye to indicate that only one acid group (at the 1 position) is reduced.

Contradiction to "rule 1": Why glutamic acid is not named homoaspartic acid? :-)

@MathewMahindaratne It's a proteinogenic amino acid, so it gets its own name. The prefixes are for variants of the proteinogenic amino acids

I understood but wonder why not. I gave my upvote nonetheless.


text:
Do azaindoles primarily exist in anionic form at physiological pH and act as bases?
I am reading an article where they describe azaindoles (bicyclic pyridine/pyrrole rings) as being basic with $\mathrm{p}K_\mathrm{a} \approx 4.6.$
I suppose at physiological $\mathrm{pH}$ and after rearranging the Henderson–Hasselbalch equation
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\tag{1}$$
$$\mathrm{pH} - \mathrm{p}K_\mathrm{a} = \log\frac{[\ce{A-}]}{[\ce{HA}]}\tag{2}$$
$$10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}} = \frac{[\ce{A-}]}{[\ce{HA}]}\tag{3}$$
$$10^{7.4 - 4.6} > 1,\tag{4}$$
the numerator is greater than denominator, i.e. more product than reactant and it is a strong acid. Does this mean the compound would be negatively charged at physiological $\mathrm{pH}$?
What exactly makes this a base? Or are they referring to the lone pair on nitrogen acting as a Lewis base?

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

Which azaindole you are talking about. The $\mathrm{p}K_\mathrm{a}$ values may be different from different azaindoles.

The formula you have used implies the acid is neutral, like e.g. acetate/acetic acid conjugate pair. But it is not always the case, like for ionic acids like $\ce{NH4+}$ or $\ce{H2PO4−}$.
By other words the symbolic $\ce{HA}$/$\ce{A-}$ pair suggests $\ce{A-}$ has 1 negative charge, while more generally it can have any charge.
In the question context, we can apply the concept of Broensted-Lawry (B-L) acids/bases:

An acid is a substance able to release a proton ( in water a hydrated H+ ion ).
A base is a substance able to capture a proton.

A generalized B-L neutralization reaction is :
$$\ce{HA^{n} + B^{m} <=> A^{n-1} + BH^{m+1}}$$
where $m$ and $n$ are integer numbers indicating the charge. They can be positive, zero or negative.
Typically for $\ce{N}$-based bases, acids are positive ions and bases neutral. Like
$$\ce{H3O+(aq) + NH3(aq) <=> H2O(l) + NH4+(aq) }$$
or
$$\ce{H3O+(aq) + Py(aq) <=> H2O(l)  + PyH+(aq) }$$ where Py = pyridine


text:
Charge Transfer and Multireference Methods
I have read in literature that systems that exhibit "high charge-transfer character" are best treated using multireference/multiconfigurational methods. Why is that? What makes such a system not be adequately described by a single determinant? I understand why singlet biradicals, low-lying HOMO-LUMO gaps, etc, would fall into this category, but why would charge-transfer complexes fall into this category as well?


text:
How do other elements fit into a diamond's carbon lattice (e.g. nitrogen in yellow diamonds)
Traditional pure-carbon diamonds have a simple structure (tetrahedral, if I'm not mistaken). But common types of diamonds have other elements within them. For example, yellow diamond has nitrogen in it (at a rate of around 1 per million carbon atoms). My question is: what does the lattice structure look like with the occasional nitrogen atom thrown in the mix? The other one I'm curious about is blue diamonds, which have trace amounts of boron.
Ideally, I'd like to know what the structural/chemical formula might look like.

I assume it is very similar as for the group 13 and 15 dotting elements for P (holes) and N ( electrons) Si semiconductors, as Si has  AFAIK the same structure.

It looks the same, only with N or B atoms instead of some C.

The main difference is the extra valence electron of nitrogen. In a single site replacement, nitrogen will only form 3 bonds plus a lone pair, so one of the C-N bonds in broken, leaving an unpaired electron on a nearby carbon.

Converting my comment into answer
Your query has been answered in this article. So I am going to write the relevant parts.

Nitrogen atoms are very small and have the ability to substitute for the carbon atoms in diamond's crystal structure. Trace amounts of nitrogen substituting for carbon in the diamond crystal lattice will cause the diamond crystal to selectively absorb blue light and selectively transmit yellow. This will cause those nitrogen-bearing diamonds to have a yellow color. Nitrogen is the most common impurity that substitutes for carbon and can comprise up to 1% of a diamond on the basis of mass.
Nitrogen can exist in the diamond crystal lattice in a number of ways. One way that influences color is when a single nitrogen atom is shared by four carbon tetrahedrons. This defect is known as the "C center" and is shown in the accompanying illustration. In this configuration, just one nitrogen atom per 100,000 carbon atoms can produce a noticeable yellow color in the crystal.

When nitrogen atoms substitute for carbon in the configuration described above, it produces a defect in the diamond crystal which alters how light passes through. The defect causes a selective absorption of blue light. The remainder of the spectrum is transmitted, and that results in the perception of yellow color in the eye of the observer.
Yellow color in mined diamonds can be also caused by the N3 defect. It consists of three nitrogen atoms clustered around a vacant carbon position in the diamond crystal lattice. When an N3 defect is accompanied by an N2 defect, certain wavelengths of blue and violet light are selectively absorbed by the diamond, and yellow light is selectively transmitted. This gives the diamond an apparent yellow color in the eye of the observer.


text:
What is the number of acidic hydrogens and the n-factor of hypophosphorous acid?
My teacher said that the hydrogen which is directly linked with oxygen or more electronegative atom is an acidic hydrogen, and that the $n$-factor for an acid is the number of acidic hydrogens.
Then why is the $n$-factor of phosphinic acid $\ce{H3PO2}$ is 1? I saw its structure, all $\ce{H}$ atoms are linked to a more electronegative atom:

Please note that the concept of "n-factors" is only used in Indian chemistry books but not in real chemistry.

Related for phosphorous acid: https://chemistry.stackexchange.com/questions/122736/n-factor-of-phosphorous-acid

The structure of hypophosphorous acid is $\ce{HO-PH2=O}$. The electronegativity of phosphorus is near the same as for  hydrogen. The oxygen causes slight polarization of $\ce{P-H}$ bond, making $\ce{H}$ slighly acidic, but  acidity of the $\ce{H-P}$ is negligible in water environment.
Note that if putting rare exceptions and circumstances aside, we do not consider hydrocarbons as acids either.
Even $\ce{NH3}$ is very weakly acidic, reacting very slowly with solvated electrons in liquid ammonia.
Note also that electronegativity is a synthetic parameter. Relative element electronegativities are not the decisive and only factor, but just a rough indicator of possible acidity. Compare e.g. acidity of hypochlorous acid $\ce{H-O-Cl}$ and perchloric acid $\ce{H-O-ClO3}$. It is clear that the effective shift of electron density (leading to  bond polarization of acids) and charge delocalization of anionic conjugate bases have much bigger effect than the general element electronegativity.

Element
Pauling electronegativity

$\ce{H}$
2.20

$\ce{P}$
2.19

$\ce{C}$
2.55

$\ce{N}$
3.04

$\ce{O}$
3.44

As Poutnik explains, phosphorus is not more electronegative than hydrogen, but "electronegativity" is not the real criterion for rendering an attached hydrogen atom acidic.
Electronegativity is essentially a combination of electron affinity with ionization energy, whereas making an attached hydrogen acidic is instead a combination of electron affinity minus bond strength between the element and hydrogen (if the bond with hydrogen is weak, the hydrogen may be dissociated as a proton even if the attached atom has low electron affinity, as is the case with iodine).
The net result is that the acidity of an attached hydrogen atom correlates with the group the atom attached to the hydrogen is in: Group 16 and 17 elements, whose electron configurations are just short of a noble gas, have relatively high electron affinities compared with their bond strengths to hydrogen, and it is these elements that then tend to acidic hydrogen atoms in water solvent.
So hydrogen attached to sulfur, let us say, is likely to be acidic in water solution, whereas hydrogen attached to nitrogen is not given the different group/electron configuration associated with nitrogen.


text:
Transforming hydroxyl group to carboxylic acid in an aqueous environment
I am looking for a way of converting a hydroxyl group into a carboxyl group in an aqueous environment for eventual coupling to an amine. There seem to be plenty of ways of achieving this in various solvent e.g. using succinic anhydride (see below scheme for mPEG)

Would the above reaction using succinic anhydride form a stable carboxyl group if the reaction was done in water?
Or should I take a different approach, oxidising the -OH group using Potassium permanganate in a basic aqueous environment, as described here:
https://en.wikipedia.org/wiki/Oxidation_of_primary_alcohols_to_carboxylic_acids

Water is about as good a nucleophile as the -OH group of mPEG so I think attempting the reaction with succinic anhydride in an aqueous environment is unlikely to work. Do you have to do this in aqueous?

There is a good procedure in water/acetonitrile here http://orgsyn.org/demo.aspx?prep=v81p0195


text:
Does sodium percarbonate break down into 'safe' compounds in water?
I'm writing this question as a wine brewer (not a chemist), so apologies if I use inaccurate terminology.
I'm interested in using sodium percarbonate as a cleaning/sanitising solution for brewing equipment. It is my understanding that, when mixed with water, sodium percarbonate produces hydrogen peroxide for a short time, and then eventually breaks down into 'safe' compounds. The logic is that you can use the solution to clean brewing equipment, but after the reaction has completed then the resulting solution is 'safe' to pour down the drain and will not affect aquatic life.
We are setting up an off-grid winery and will be treating waste water using a reed bed system, so what may be deemed 'safe' to pour down a municipal drain may not necessarily be safe to put into a reed bed system. Therefore, I'd like to be entirely clear exactly what chemicals are left in solution after sodium percarbonate has finished reacting with water?

From Wikipedia:

Dissolved in water, sodium percarbonate yields a mixture of hydrogen peroxide (which eventually decomposes to water and oxygen), sodium cations $\ce{Na+}$
, and carbonate $\ce{CO3^2-}$
$$\ce{2Na2CO3*3H2O2 -> 3H2O2 + 4Na+ + 2CO3^2-}$$
$$\ce{2H2O2 -> 2H2O + O2}$$

…and to promote hydrogen peroxide decomposition one might want use a waste sump made of steel or copper.

So the solution will be quite alkaline, depending on the concentration of sodium carbonate.


text:
Can a spontaneous reaction have zero or negative activation energy?
Is it possible for a certain spontaneous reaction to have zero or negative activation energy?
My high school teacher explains that it shouldn't be possible since it breaks the Arrhenius equation, namely the Boltzmann factor
$$\exp\left(\frac{-E_\mathrm{a}}{RT}\right),$$
which gives the fraction of molecules having $E_\mathrm{kin} > E_\mathrm{a}.$
So, instead of putting in $E_\mathrm{a} < 0$, we will just say the Boltzmann factor is equal to 1.
The Boltzmann factor just doesn't make sense for a negative $E_\mathrm{a}$ as $E_\mathrm{kin}$ cannot be negative.
Instead of saying it breaks the factor, should we not just say that the factor is undefined for negative energy values and that $E_\mathrm{a}$ can be zero or negative?
Edit- so people are commenting on the question and I am very thankful. But I also want to know that if this was a test question with a simple true or false answer, would you say that $E_a$ can be $<0$ or not?

I would rather say that for low enough $E_a$ the Boltzmann energy factor is negligibly different to 1 and the reaction is driven by diffusion. I.e. whenever reactants meet and are properly oriented geometrically, then they react. Typical is neutralization reaction $\ce{H+(aq) + OH-(aq) -> H2O(l)}$.

Therefore, temperature dependency of reaction kinetics is not driven by the Boltzmann factor, but by thermal molecular speeds, medium viscosity and diffusion coefficients.

Bond formation between atoms (e.g. $\ce{H + H -> H2}$) is most often exothermic/exergonic and has zero activation energy.

IMHO, an eventual formal reaction with negative activation energy are 2 reactions closely linked to each other via a common major component, that is thermodynamically preferred to both reagents and products of the former single reaction.

Marcus Theory points to some cases of electron transfer that are in principle barrierless (or very close to barrierless).

Negative activation energy is not possible in a simple reaction, the smallest value is zero. Experimentally as @Zhe points out, activationless reactions are seen in solution for electron transfer, but only when the two species are held together, such as intercalation of a dye in DNA, otherwise if free to diffuse the rate constant is limited by how quickly they can diffuse together.

Possible duplicate https://chemistry.stackexchange.com/questions/95343/is-negative-activation-energy-possible

The short answer: yes, but it is not the most common case.
How it can be? You assume that you reaction has a simple reaction profile with a barrier and a single transition state between reactions and products, plus other approximations. When those approximations fail, the model is not helpful and you have to use a deeper model.
A longer, and hopefully, more nuanced explanation.
For many reactions the dependence of the reaction rate with temperature follows the Arrhnius equation
$$k(T) = A\exp \left(\frac{-E_a}{R\,T}\right)$$
where A, the preexponential factor, and $E_a$ are independent of temperature.
However, you may have that over an extended temperature range many reactions deviate from this behaviour and their behaviour are described better using more complex expressions that behaves as if $A$ and $E_a$ are functions of temperatura. For example, the Kooij equation that has the form
$$k(T) = A T^m \exp\left(\frac{-E_a}{R\,T}\right) $$
As an actual example you can consider the reactión OH + CH$_4$ in the 220-500 K which behaves as Ahhrenius but in the extended interval 220-2000 deviates from that behaviour as the following image shows.

Thus, at present the Arrhenius equation is understood as very useful empirical fitting equation, when it can be applied. But you have to interpret it carefully.
There are some important reactions that have no activation barrier, for example, radical recombinations CH$_{3}\cdot$ + CH$_{3}\cdot$ $\leftarrow$  CH$_3$CH$_3$, or the atmospheric reactions of O($^1$D) -e.g. O($^1$D) + CH$4$-.
Finally, there are some elementary reactions which rate decreases with temperature and fit Arrhenius, thus, its activation energy is negative ($E_a < 0$). For example, the gas phase reaction OH + HNO$_3$ $\leftarrow$  H$_2$O + NO$_3$. Its energy profile does not have a negative barrier, is exhothermic but the reaction profile is complex.

The reaction profile is adapted from the article Brown S.S. et al., J. Phys. Chem. A, 1999 103, 3031-3037


text:
Monochlorination of (R)-2-chlorohexane: Number of possible chiral dichloro products

How many possible chiral dichloro products can be formed when (R)-2-chlorohexane reacted with one equivelent of $\ce{Cl2}$ in the presence of UV radiation (monochlorination only)?
$$\ce{C6H11Cl + Cl2 ->[$h\nu$] C6H10Cl2 + HCl}$$

I am getting 8 as the answer but given answer is 9:

(2R)-1,2-dichlorohexane
(2R,3R)-2,3-dichlorohexane
(2R,3S)-2,3-dichlorohexane
(2R,4R)-2,4-dichlorohexane
(2R,4S)-2,4-dichlorohexane
(2R,5R)-2,5-dichlorohexane
(2R,5S)-2,5-dichlorohexane
(5R)-1,5-dichlorohexane

Am I missing something? Will the stereochemistry of the already chlorinated C change during the reaction?

Don't put your whole question in the title.

There are nine total dichloro products. Seven are chiral. the eighth does not have a chirality center. The ninth is a meso compound.

Related: https://chemistry.stackexchange.com/questions/93766/how-many-products-are-possible-when-chlorination-of-cyclohexane-takes-place ... https://chemistry.stackexchange.com/questions/153943/possibility-of-further-chlorination-of-hexachlorocyclohexane


text:
Gouy-Chapman model
Recently I have been investigating the problem of a neutral solutions (with equal number of anions and cations) interacting with an infinite surface $z=0$ with electrostatic potential $\phi(z=0)=\phi_0>0$.
If we assume only one type of cation and anion, with opposite charges, we obtain
\begin{equation}
\nabla^2\phi\left(\textbf{r}\right)=-\frac{1}{\varepsilon}\sum_{i=1}^{2}q_{i}n_{i}\left(\textbf{r}\right)
\end{equation}
with $q_1=-q_2=q$. If we assume that the density of charges is in thermal equlibrium, we have through Maxwell-Boltzmann distribution
\begin{equation}
n_{i}\left(\textbf{r}\right)=n_{i}^{0}e^{-\frac{q_{i}\phi\left(\textbf{r}\right)}{k_{B}T}}.
\end{equation}
After manipulating a bit, we obtain
\begin{equation}
\nabla^2\phi\left(\textbf{r}\right)=2qn_{0}\sinh\left[\frac{q\phi\left(\textbf{r}\right)}{k_{B}T}\right],
\end{equation}
which describes the very famous Gouy-Chapman model
with analytical solution given by
\begin{equation}
\phi\left(z\right)=2\frac{k_{B}T}{q}\ln\left[\frac{1+\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}{1-\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}\right]
\end{equation}
and anion (-), cation (+) densities given by
\begin{equation}
n_{\pm}\left(z\right)=n_{0}\left[\frac{1+\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}{1-\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}\right]^{\mp2}
\end{equation}, where I plot for different positive values of $\phi_0$

Contradiction: As we can clearly see from figure above, the neutrality condition does not hold anymore -- see imbalance between positive and negative charges. People from this field argue that there is no problem since a surface charged density was absorbed by the plate (positive surface charged if $_0>0$). However, how do we see this physically? I do not understand why positive charges were absorbed by the plate. Can we somehow show this mathematically? If we integrate the total density $\rho(z)$, we will not obtain zero, implying there is no charge neutrality within the problem.
Any thoughts on that?

The plate was charged in the first place. The imbalance of cations vs anions in the solution neutralizes that charge.

The surface charge is the origin of the non-zero surface potential $\phi_0$. Since it is positive anions are more attracted to the surface than cations.

Look for a relation between the surface charge density and the z potential at the surface, and evaluate what the surface charge density corresponding to the potential you used in your calc might be. The net countercharge density in solution should be opposite this.

@IvanNeretin it seems this is not quite right. If we have $\phi>0$, this implies in a positively charged plate. However, the bulk solution has an excess of negatively charged atoms rather than positive ones. So it seems the consistence remains.

@BuckThorn As I wrote previously, if we have a positively charged plate with $\phi>0$, we would have an excess of negative charges within the solution. This implies that more positive charges 'moved out of the solution'. So it seems the consistence is still present.

That's right: a positively charged plate is surrounded by the negatively charged solution. That's what the word "neutralize" means.

@IvanNeretin Right! But this seems to be leading to an inconsistency.
If the plate is positively charged, and assuming we started with a neutral solution, how can the solution have more negative charges after having them surrounding the positively charged surface?

The solution has as many negative charges as it had before, that is, exactly as many as positive ones. The layer around the plate has an excess of negative charges. See where this is going?

@IvanNeretin I am not sure I agree with this statement. The concentration of negative charges is larger than the positive one throughout the whole solution. So where is this imbalance coming from? Another way to put it: if we integrate the charge density over the volume of the solution, we will obtain a non-zero value, indicating the presence of a net charge different from zero.

There are models, and then there is reality. The picture is of the model that was deduced in assumption of infinite vessel with solution, bounded on one side with infinite charged plate. Is the model any good for a real, finite vessel? Yes and no. Yes for the layer near the plate, and no otherwise.

@IvanNeretin I understand what you mean, but I am not talking about reality here.
We first started with Boltzmann and Poisson's equation for a neutral solution. After solving the corresponding equations, we obtain a charge density profile with net charge different from zero. The obvious questions is: where does this charge come from?

This charge comes from far away. In an infinite medium filled with charges, the law of charge conservation does not hold.


text:
Understanding Cahn-Hilliard Equation in terms of Units
The Cahn-Hilliard equation may be formulated as
$$
\frac{\partial c}{\partial t} = M \nabla^2 \left(\frac{\partial \hat f}{\partial c}\right),
$$
where $c : \Omega \to [0,100]$ describes the concentration (mol-%) of an interesting substance, $M$ is the mobility coefficient (for simplicity $M$ is assumed to be constant) and $f$ is the generalised free energy per unit volume, i.e., $\hat f$ depends on the concentration $c$ and higher derivatives of $c$ (see e.g. Novick-Cohen & Segel (1984), p. 278 -- 282).
Problem: If I take a look at the units of this equation, I am confused. According to the equation, we have on the LHS
$
\frac{\text{mol-%}}{\text{s}}.
$
For the unit of $\nabla^2 \left(\frac{\partial \hat f}{\partial c}\right)$ on the RHS, I get
$$
\frac{\text{J}}{\text{mol-%}\,\text{m}^2}.
$$
Hence, the mobility constant $M$ should be given in
$$
\frac{\text{m}^2 \, \text{mol-%}^2}{\text{J} \, \text{s}}
$$
to end up with $\frac{\text{mol-%}}{\text{s}}$ on the RHS. However, I often noticed that the mobility is given in $\frac{\text{m}^2}{\text{V} \, \text{s}}$ which I cannot reformulate in the required unit.
In addition, I noticed that this Wikipedia entry considers the Cahn-Hilliard equation above (with a specific $\hat f$ and) with a diffusion coefficient which is given in $\text{length}^2 / \text{time}.$ Do I misunderstand something?

I think your mol% should be mole fraction and thus dimensionless?

Actually, that does not change anything (of the problem), does it?

This question is addressed here. For the units, $e$=energy (eg J), $t$=time, $l$=length.

The Cahn-Hilliard equation after the variational derivative takes the form$$\dfrac{\partial c}{\partial t}=V_{m}\nabla\cdot M_{i} \nabla\left(\dfrac{\partial f_{loc}}{\partial c_i}+\dfrac{\partial E_d}{\partial c_i}-\kappa_i \nabla^2c_i\right)$$where $V_m$ is the molar volume of the reference state of the material with units of $l^3/mol$.

The units of this equation are$$\dfrac{1}{t}=\dfrac{l^3}{mol}\dfrac{1}{l}\dfrac{l^2\cdot mol}{te}\dfrac{1}{l}\left(\dfrac{e}{l^3}+\dfrac{e}{l^3}-\dfrac{e}{l}\dfrac{1}{l^2}\right)$$ where the units of $M_i$ are $(l^2 mol)/(et)$. Note that some models include the $V_m$ in the mobility term, such that it has units of $l^5/(te)$.

The key bit is that last part about combining $V_m$ into $M_i$, which changes the units from what you'd expect for $M_i$. In your version of the equation, there is no separate $V_m$ term, so you need to account for it in your units for $M_i$.