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Detection of oxygen by Lassaigne test
I have read that compound with nitrogen and carbon atoms gives Prussian blue color when sodium extract of the compound is added with $\ce{FeSO4}$ . Will the test work even if the compound has oxygen atoms?

Yes ! The test works for any nitrogen compounds, with or with our any oxygen atoms attached in the analyzed substance.

The question is unclear to me. Are you asking if Lassaigne test will give positive result to organic compounds containing oxygen only or to organic compounds containing nitrogen with oxygen atoms included? Lassaigne test will give positive result if there is either nitrogen/sulfur/halogen atoms present. So, presence of oxygen doesn't matter if there is either of above mentioned atoms.

Sodium-Fusion test or Lassaigne test was originally introduced by Lassaigne in 1843 (Ref.1), and was subsequent modified by Jacobsen in 1879 (Ref.2). The major modification of Jacobson was the replacement of potassium by sodium metal as the active reagent. Since then, the test has altered very little, and basically is still in use today. The difficulties associated with the test have been reviewed in 1969 (Ref.3).
The sodium-Fusion test or Lassaigne test is a general test for the detection of halogens, nitrogen, and sulfur in an organic compound. These elements are usually covalently bonded to the organic compounds such as halo, nitro, amino, and thio compounds. In order to identify their presence, each covalently bonded element has to be converted into their ionic forms. This task is achieved by fusing the organic compound with melting sodium metal. The ionic compounds formed during the fusion are then extracted into aqueous solution and subsequently identified by simple chemical tests. The aqueous extract is called sodium fusion extract or Lassaigne's extract.
The reference 3 gives good procedural practice for preparation of Lassaigne's extract, thus I'm not going to elaborate too much unless it is necessary and going to address the question directly: Test for nitrogen usually involves boiling an alkaline sample of
the water-extract from the sodium fusion with ferrous sulfate and then cooling and acidifying with dilute sulfuric acid. The development of a blue coloration or a dark blue precipitate is confirmatory for nitrogen.
The theory: The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide ($\ce{NaCN}$), which is soluble in water. This is converted into sodium ferrocyanide ($\ce{Na4Fe(CN)6}$) by the addition of sufficient quantity of ferrous sulphate ($\ce{FeSO4}$). Ferric ions ($\ce{Fe^3+}$) generated during the process react with ferrocyanide to form $\color{blue}{\text{prussian blue}}$ precipitate of ferric ferrocyanide ($\ce{Fe4[Fe(CN)6]3}$):
$$\ce{Na + C + N -> NaCN }$$
$$\ce{6NaCN + FeSO4 -> Na4Fe(CN)6 + Na2SO4}$$
$$\ce{3Na4Fe(CN)6 + 4Fe^3+ -> Fe4[Fe(CN)6]3 + 12Na+}$$
There are two issues with this test, according to Ref.3:

It seems that the development of a blue coloration and/or precipitate in a positive test depends on pH and on temperature. With $\mathrm{pH}$ only just below 7, the development of coloration is inhibited. It is, thus, necessary to ensure the presence of an excess of acid. Similarly, cooling of the solution is advisable because development of the blue coloration appears to he inhibited in hot solutions.

During the Lassaigne's extract, if the compound has too much nitro groups and too little carbons, it is advisable that to add a little paraffin or a hydrocarbon before heating with sodium. Carbon in them provide enough carbon to make cyanide with nitrogen. In this praise, there is a answer for your question: Nitro group contains oxygen but did not interfere with the test.

References:

J. L. Lassaigne, "Mémoire sur un procédé simple pour constater la présence de l'azote dans des quantités minimes de matière organique (Memoir on a simple procedure for confirming the presence of nitrogen in minimal quantities of organic matter)," Comptes rendus 1843, 16, 387-391.
Oscar Jacobsen, "Ueber die Oxydation der Parasulfamintoluylsäure (On the oxidation of 3-methyl-4-sulfamoyl-benzoic acid)," Berichte der Deutschen Chemischen Gesellschaft 1879, 12, 2316-2320.
R. P. Gower, I. P. Rhodes, “A review of techniques in the Lassaigne sodium-fusion,” J. Chem. Educ. 1969, 46(9), 606-607 (https://doi.org/10.1021/ed046p606).

Whoever down voted this answer should give a reason, because the answer is to the point.

Hi Mathew. Actually it was me but it was a misclick. I realized after 5 minutes that I clicked the downvote insted of upvote. Thank you for editing. I was able to revert the vote.


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How does the charge imbalance affect the cell reaction when there is no salt bridge in a galvanic cell?
As mentioned in the answers to this question Why is it important to use a salt bridge in a voltaic cell? Can a wire be used? a salt bridge is used to keep the two half cells neutral. i understand how the imbalance occurs without a salt bridge, but i don't understand the way this imbalance affects the cell reaction. Can someone please explain? Much thanks.

The charge disbalance caused by the current 1 A per 1 min is big enough to create 10 km long GC+ lightning with potential diffference up to 1 gigavolt. I guess it is strong enough force for a cell going against it. Potential difference would stop cell net reactions very soon, with external cell voltage quickly converging to zero.

If there is no bridge in a voltaic cell, this will be no cell any more. The arrangement will just be a metallic connection between two different containers. It will not deliver any continuous current. And replacing the salt bridge by a wire will simply add a second metallic connection between the solutions. There will be two parallel but different metallic connections between the two solutions. It is not a cell.
To get a cell, a liquid connection must exist between the two solutions through which ions can move.
Let's consider the famous Daniell cell, without this bridge. In the anode solution, a piece of zinc metal is dipping in water or in a zinc sulfate solution. Zinc has a tendency to loose electrons and produce ions $\ce{Zn^{2+}}$ which pass in water. The electrons are following a long wire connected to the zinc plate up to the cathode (Cu in $\ce{CuSO_4}$ solution). But this process will be stopped after some microseconds, because the first $\ce{Zn^{2+}}$ ions appearing in the anodic solution are not counter balanced by any negative ions. The solution becomes positively charged and repels any other $\ce{Zn^{2+}}$ ions form being produced. $$\ce{Zn -> Zn^{2+} + 2 e^-}$$
The same thing happens in the cathodic solution ($\ce{CuSO_4 solution}$). The electrons produced at the zinc anode discharge $\ce{Cu^{2+}}$ ions : a deposit of metallic copper appears on the cathode. But this may last a couple of microseconds, and not more, because the solution around the copper electrode becomes negatively charged due to the remaining sulfate ions. These negative ions are not counter-balanced by positive ions : they will preventl new electrons from coming to the copper electrode. $$\ce{Cu^{2+} + 2 e^- -> Cu}$$
So the chemical reactions will be stopped in both electrodes after a couple of microseconds. To transform this arrangement in a cell, the remaining sulfate ions $\ce{SO_4^{2-}}$ must be allowed to move to the anodic cell where there is an excess of positive charges (the $\ce{Zn^{2+}}$ ions). The only way of allowing this migration is to build a wet bridge between the two solutions. This bridge should allow too numerous negative ions from the cathode to migrate to the anodic solution where $\ce{Zn^{2+}}$ are attracting them.
But the solutions should not be mixed. If mixed, the anodic and cathodic reactions occurs  in the solution on the zinc plate : no electrons will be emitted in the outer circuit. $$\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$$

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Stability of α-chlorocarbocations
I'm trying to compare the relative stability of two carbocations:
$$\ce{CH3-\overset{+}{C}H-Cl}\quad\text{vs}\quad\ce{CH3-\overset{+}{C}H2}$$
Does the $\ce{Cl}$ exert a +M effect to stabilise the carbocation, or does it exert a −I effect to destabilise it?

As Oscar Lanzi suggested, both $+M$ and $-I$ applies here, but $\ce{Cl}$ stabilizes carbocation, meaning $+M$ is more effective than $-I$. This fact was confirmed by this peer-reviewed paper (Ref.1):

The lowering of $\ce{C_\beta–H}$ stretching frequencies in carbocations 1a–d and 2a–c induced by hyperconjugation was tested as a possible probe for estimating the electron donating ability of $\alpha$-substituents. Conclusions are based on the results of high level quantum chemical calculations confirmed with experimental FT-IR spectra. Because the decrease in the $\ce{C_\beta–H}$ stretching frequency is comparable in 1b and in 1c, and in 2b and 2c respectively, it follows that $\alpha$-substitution $\color{red}{\text{by a methyl group}}$ or $\color{red}{\text{by chlorine}}$ stabilizes a carbocation with $\color{red}{\text{almost the same effectiveness}}$.

The effect of the chlorine $n$-electron back donation is evident by a partially double bond character of the $\ce{C+–Cl}$ bond in $\alpha$-chlorocarbocations observed experimentally in IR spectra (Ref.2). In this experiment, FT-IR spectrum of $\ce{Cl3C+}$ cation has displayed the $\ce{C–Cl}$ stretching frequency at $\pu{1040 cm–1}$, which is $\pu{250 cm–1}$ higher than in neutral alkyl chlorides (e.g., characteristic $\ce{C–Cl}$ stretching frequency of $\ce{CCl4}$ is $\pu{785 cm-1}$). This is indicative of partial double bond character of $\ce{C+–Cl}$ bond suggested by the resonance structures:
$$\ce{Cl2C+-Cl <-> CH2C=Cl+}$$
The pioneering work of Olah and coworkers has also predicted this phenomena by $\ce{^{13}C}$-NMR studies of trihalocarbocations (Ref.3):
$$
\begin{array}{c|c|c|c}
\ce{HCX3} & \delta\ce{^{13}C}\text{ of }\ce{CHX3}  & \ce{^+CX3}  & \delta\ce{^{13}C}\text{ of }\ce{^+CX3} & \Delta \delta\ce{^{13}C} \\
\hline
\ce{HCCl3} & 77.7 & \ce{^+CCl3} & 236.3 & 158.6 \\
\ce{HCBr3} & 12.3 & \ce{^+CBr3} & 207 & 194.7 \\
\ce{HCI3} & -139.7 & \ce{^+CI3} & 95 & 234.7\\
\hline
\end{array}
$$
Olah has suggested that the decreasing trend of $\Delta \delta\ce{^{13}C}$ by going $\ce{I}$ to $\ce{Br}$ to $\ce{Cl}$ is in agreement with the positive charge stabilization by back-bonding in order of $\ce{Cl > Br > I}$.
References:

Milan Mesić, Igor Novak, Dionis E. Sunko, Hrvoj Vančik, "$\ce{C–H}$ Hyperconjugation in $\alpha$-chlorocarbocations," J. Chem. Soc., Perkin Trans. 2 1998, (11), 2371-2374 (https://doi.org/10.1039/A805772I).
Hrvoj Vančik, Ksenija Percač, Dionis E. Sunko, "Chloromethyl cations in cryogenic antimony pentafluoride matrixes and the generation of carbocations from hydrocarbons," J. Am. Chem. Soc. 1990, 112(20), 7418–7419 (https://doi.org/10.1021/ja00176a065).
George A. Olah, Ludger Heiliger, G. K. Surya Prakash, "Stable carbocations. Part 276. Trihalomethyl cations," J. Am. Chem. Soc. 1989, 111(20), 8020–8021 (https://doi.org/10.1021/ja00202a056).

but a later dated paper by the same author (https://hrcak.srce.hr/file/195027) says that chloro substituent destabilizes the carbocation intermediate (see the discussion under  Solvolysis of bridgehead substituted methyl- and chloroderivates of 7-nor
bornyl esters) !

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

I would say that many books suggest that +m effect overshines the -I effect but I feel it depend upon the reaction where thee intermediates are formed and type of reaction is happening and stability of intermediate involves the thermodynamics for a general exam like situation you can mark A is more stable than B.

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Why do we subtract the mass of one proton in mass spectrometry (negative mode)?
This is probably very basic, but I have been given the understanding that we subtract the mass of only one proton when looking for our compound in a mass spectrometry (negative mode). I dont really understand why we are only subtracting the mass of one proton when our compound could potentially loose more than one proton at higher pH?
In example, the amino acid glutamate, would it not loose or gain more than one proton is a negative (higher pH) or positive (lower pH) mode mass spectrometry (MS)? Perhaps I have misunderstood the concept of negative and positive mode for MS?

Higher pH is rarely used in LCMS due to possibility of damaging column packing usually made of $\ce{SiO2}.$

This question concerns the MS, not the column, but some columns do tolerate higher pH as far as I understand it?

I concluded from "mass spectrometry chromatogram" that you are asking about LCMS meaning the backend is a chromatographic column. And from from I heard, there is rarely a need to go with higher pH for amino acid assays and switching to other packing material like $\ce{ZrO2},$ which are usually considerably more expensive and have their own quirks.

ah okay, thanks for the input! I also assume the higher or lower pH is relative (i.e. not always extreme), but in general sticking to a neutral pH range is more common.  Im not all that familiar with this topic, so I will remove the 'chromatography' word to clarify my question.

Interesting question, which originates from a major misconception related to pH in LCMS. Nowadays, there are dozens of ionization techniques so in the question always mention the ionization method.
Anyway, LCMS is usually done with electrospray ionization (ESI) technique and so are amino acids. First of all nobody completely knows the exact process in  electrospray ionization as yet. This is not unusual in mass spectrometry where even a common ionization technique like ESI and electron impact, the picture is not 100% clear.
Second important point is the pH in ESI does not determine whether a negative ion or a positive analyte ion will be formed. I also used to have this misconception a decade ago about pH. The final ionization process in ESI takes place in a gas phase. However it is well known that certain ion can suppress ionization very badly. Don't try to connect mobile phase pH with the state of ionization of the molecule in the gas phase.
In negative ion, you are simply observing a net negatively charged species. Nothing
For details, I'd encourage you to read "pH Effects on Electrospray Ionization Efficiency" in the Journal of the American Society for Mass Spectrometry· Go to the journal website and locate the paper with this title. I am not providing a link so that students develop the habit of searching themselves.

Probably worth adding that in negative mode Mass Spectrometer doesn't change ionization - it changes whether the negative ions are captured.

Thanks! This was very clarifying on a few misconceptions that I had - particularly the whole charge of molecule. I also found the reference article!


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Gibbs function for pure substances
I know that the Gibbs function for pure substances is given by
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT.\tag{1}$$
However, I also saw the same equation in the form
$$ΔG = VΔp - SΔT,\tag{2}$$
however I am doubtful if the second equation is correct as now $Δ$ represents finite changes which is different from the infinitesimal changes in the first equation. Is the second equation correct?

Maybe.  It depends on whether you are OK approximating $V(P, T)$ as a constant $V$ and $S(P, T)$ as a constant $S$.

The 2nd equation is not correct.

Well,i saw in an pdf from Massachusetts Institute Of Technology where they have used d(∆G)/dP=ΔV (at constant temperature)..from where did they get this equation then?

What is the source of the second equation in your post? The equation from the MIT pdf is ok, it is taken from eq 1 at constant T, when looking at the difference in G for two states.

BTW make sure to use handles when targeting someone with messages ie @ChetMiller


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Why is biphenyl not planar?
I've seen in a lot of books that biphenyl is not planar. I don't know why that is because both the carbon are $\mathrm{sp^2}$ hybridized so it should be planar. Can someone explain it to me?

Did you google your question? this link explains it quite nicely

https://chemistry.stackexchange.com/questions/55270/why-are-substituted-biphenyls-optically-active https://chemistry.stackexchange.com/questions/79155/can-resonance-take-place-in-between-biphenyls

https://doi.org/10.1021/jp0122124 I´d like to know how large the actual rotational barrier is for biphenly. Anyone have access?

@Karl Depending on which model you're using, it's apparently somewhere around 2-3 kcal/mol.

@probably_someone 10kJ/mol, that´s ~ 4kT at RT. More than I expected. Benzene has a resonance energy of  150 kJ. The question is, would the molecule become planar if one replaced two CH groups by N? And what would the (negative) rotational barrier be then?

@Karl Well, based on the answer now below, I might have misread what the paper was saying.

Causally related, possible duplicate: Is biphenyl considered to be aromatic?

@probably_someone, as far as I can tell, you read it right. I'm mildly curious now as to why this value differs from the 6 kcal/mol experimental value given below, although that's really a separate question to be asked.

Biphynyls are a class of compounds with two phenyl rings attached by $\ce{C-C'}$ carbons in each phenyl ring (See Figure $\bf{\text{A}}$):

The phenyl rings can rotate around this $\ce{C-C'}$ bond (acts as an axis). Keep in mind that any molecule capable of rotating one part of it around fixed axis (a bond) is never planner. As a rule, if the rotational energy barrier is less than an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ at room temperature, the part of the molecule constantly rotate around the said axis. The experimental barriers for rotation about the $\ce{C-C'}$ bond of parent biphenyl (no substitution) are $\pu{6.0 \pm 2.1 kcal mol-1}$ at $\theta = 0^\circ$ and $\pu{6.5 \pm 2.1 kcal mol-1}$ at $\theta = 90^\circ$ with minimum energy barrier at $\theta \approx 44^\circ$ (Ref.1 & Ref.2; See Figure $\bf{\text{B}}$ for $\theta$). This values are less than required energy barrier to prevent rotation ($16$ to $\pu{19 kcal mol-1}$) at room temperature, thus, the biphenyl is rotate around $\ce{C-C'}$ axis (See Figure $\bf{\text{B}}$).
In addition, the rotation about $\ce{C-C’}$ can be restricted if each of the phenyl rings of a biphenyl has at least one same or different ortho substituent (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). However, if these substituents are not large enough, it won’t prevent the rotation about $\ce{C-C’}$ and relevant conformers would not be able to isolate at room temperature (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). If there are two different ortho substituents on each of two phenyl rings of a biphenyl (e.g., 6,6’-difluro-2,2'-dicarboxylic acid in $\bf{\text{C}}$), the rotation about $\ce{C-C’}$ would be not possible at lower temperatures since thermal (kinetic) energy is not sufficient to provide the necessary activation energy for racemization. As a result, the twisted $\theta = 90^\circ$ dihedral angle conformer becomes chiral. In order to inter-convert such conformers with their mirror image structures, a rotation through the higher energy coplanar form must be made. How easy this inter-conversion would be depends on the size of the ortho substituents. Unlike 2,2'-dicarboxylic acid, the two additional substituents in 6,6’-difluro-2,2'-dicarboxylic acid have a higher activation energy for racemization, and hence, can be resolved if care is taken to avoid heating them (Note: As a rule, an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ is required to prevent spontaneous room temperature racemization of substituted biphenyls). Since fluorine is smaller than a nitro group, 6,6’-difluro-2,2'-dicarboxylic acid racemizes more rapidly on heating than does the nitro compound (6,6’-dinitro-2,2'-dicarboxylic acid in $\bf{\text{C}}$). In fact, two conformers of 6,6’-dinitro-2,2'-dicarboxylic acid can be isolated at room temperature. Those conformational isomers that are isolable due to high energy barriers are called atropisomers.
For example, those conformational isomers have been studied comprehensively and computationally (Ref.3):

Using B3LYP/6-311+G* and other methods, twist angles and torsional energies were obtained for biphenyl (BP), 2-halogen BPs, 2,2'-dihalogen BPs, and 3,3'-dihalogen BPs, the halogens ranging from $\ce{F}$ to $\ce{I}$. The results were compared with available gas phase and X-ray data. 2,2'-difluoro BP has a rotational double minimum, at $57.9^\circ$ and $128.9^\circ$, whereas the other 2,2'-dihalogen BPs have a single minimum at dihedral angles ranging from $84.9^\circ$ to $94.8^\circ$. All 3,3'-dihalogen BPs have a double minimum at about $45^\circ$ and $135^\circ$. Optimized twist angles and energy barriers were also calculated for 2,2'-dimethyl BP and for perfluoro as well as perchloro BP. Most structures are accounted for by steric effects. For 2,2'-dihalogen BPs, however, attractive forces also appear to play a role, as evidenced by the dihedral angle of 2,2'-dichloro BP lying well below $90^\circ$.

References:

Mercedes Rubio, Manuela Merchán, Enrique Ortí, “The internal rotational barrier of biphenyl studied with multiconfigurational second-order perturbation theory (CASPT2),” Theoretica chimica acta 1995, 91(1), 17–29 (https://doi.org/10.1007/BF01113859).
Otto Bastiansen, Svein Samdal, "Structure and barrier of internal rotation of biphenyl derivatives in the gaseous state: Part 4. Barrier of internal rotation in biphenyl, perdeuterated biphenyl and seven non-ortho-substituted halogen derivatives," J. Mol. Struct. 1985, 128(1-3), 115-125 (https://doi.org/10.1016/0022-2860(85)85044-4).
Friedrich Grein, “Twist Angles and Rotational Energy Barriers of Biphenyl and Substituted Biphenyls,” J. Phys. Chem. A 2002, 106(15), 3823–3827 (https://doi.org/10.1021/jp0122124).

This is because the hydrogen atoms at the ortho- positions of both phenyl rings prevent rotation about the single bond connecting the two phenyl rings (due to steric hindrance).
Additionally, ortho-disubstituted/ tetrasubstituted derivatives of biphenyl (where all four substituents are distinct) are chiral (they do not possess a plane of symmetry) and hence are conformational isomers. However, any one isomer of a tetrasubstituted biphenyl system may racemize at higher temperatures (the passage of the compound through a planar configuration contributes to a large activation energy). Conformational isomers that can be isolated at higher temperatures due to these high energy barriers are called atropisomers.
Footnotes:
Conformation of biphenyls


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Scope of Haloform reaction
We know that methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and acetaldehyde, respectively. 1,3-Diketones such as acetylacetone also give the haloform reaction.
Do Terminal Alkynes give Haloform Reaction if Yes then how the mechanism will follow?

Maybe if you react with H2SO4/HgSO4 to make an aldehyde beforehand. I don't think a plain terminal alkyne will give the haloform test.

And if it is attached to some aromatic ring ?

You should read this: https://chemistry.stackexchange.com/questions/135626/do-cyclohexane-1-3-diones-give-positive-haloform-reactions

Terminal alkynes, including phenyl acetylene, react with iodine (in MeOH) to give 1,2-diiodoalkenes according to this JOC paper  here. Presence or absence of base will not alter the outcome. Presence of AgNO3 gives some di-iodo ketone which can then give rise to the iodoform reaction.


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Is it possible to graph the equivalence of p[Cu²⁺] with pH?
In no way I am a chemistry major, so please forgive me if my question seems stupid. I just need a way to find the "equivalent $\mathrm{pH}$" of $\mathrm p[\ce{Cu^2+}]$ if that makes any sense to you.
I'm making a paper about a transistor that can sense ions (ion-sensitive field-effect transistor); i.e. it's a potentiometric transistor. I was able to simulate voltage values with respect to $\mathrm{pH}$ activity in a simulation software named LTspice, but in order for this paper to be passable to whoever shall evaluate it, I must compare my data with a preexisting experimental data.
The only experimental data that I have is from my co-advisee. He measured the voltage values with respect to $\mathrm p[\ce{Cu^2+}]$ $(\ce{CuCl2 · 2 H2O}).$ I think it wouldn't be proper to compare the voltage vs $\mathrm{pH}$ graphs and voltage vs $\mathrm p[\ce{Cu^2+}].$
So, I figured that if I can express $\mathrm p[\ce{Cu^2+}]$ to its equivalent $\mathrm{pH}$ (I don't think this is possible and makes any sense), then I'd be able to compare the two graphs.

Would it make sense to compare simply on the basis of charge, ie 1 Cu2+ = 2 H+ ?

I think so? Then I'll just get the power of both Cu2+ and H+ to form a graph for their equivalence?

No it wouldn't. Sensitivity to different ions would not be equal.

TTbulaski, I gather that you are developing a sensor for copper ions whose output is proportional to -log [Cu(II)]? I believe this is what you are trying to say.
What you can test is whether the output of your sensor gives any linear relationship between actual Cu concentration and the antilog of your sensor signal or not. Plot the Cu ion concentration on the x-axis and the anti-log of your sensor signal on the y-axis? Do you see a linear relationship?
pH and -log [Cu(II)] connection is pretty tricky and it depends what level of depth you want to go to. There is no point in looking for a similarity between the two measurements.
People have done calculations to show what species of Cu is present as a function of pH. It is called speciation diagram of copper.
P.S. The most important thing to test any metal ion sensor is something called as "selectivity". No analytical chemistry reviewer will accept the paper until and unless you show how selective is your sensor. For example if you have Zinc or Cd or other transition metals, do you still get a signal for copper in a correct way?
Not long time ago, NASA did a mistake in ion-selective electrodes. They designed or used a nitrate selective electrode and it turned out that is more selective towards perchlorate ion. It was an interesting news.

Regarding the second paragraph, yes I have both the antilog of the sensor signal and the normal, unaltered graph. The relationship with the Cu2+ is linear at R^2 = .987.
Regarding the 3rd paragraph: so I assume you're saying it is not proper to use the experimental data which makes use of copper ions, to verify a simulated circuit which takes pH as "input"?

Yes, I think so. Do not try to connect Cu(II) and pH by any means.

Welp, I guess I'll have to go to the university laboratory despite the ongoing lockdown. Thank you for your answer!


text:
Need to specify cyclic alkenes as cis or trans
I'm studying organic chemistry from John McMurry's book. Double bond's substituents can be on cis or trans configuration. An example for cyclic molecules is trans-cyclooctene and cis-cyclooctene, where the cis compound is more stable. On the other hand, in a naming exercise this molecule is named simple as -cyclohexene. None of the other molecules  in the same exercise was named as cis or trans.

So, my question is: In small rings is not necessary to specify cis or trans orientation because the trans orientation is less stable (maybe impossible to obtain)?

Impossible to obtain trans in rings smaller than 8-membered

IUPAC nomenclature does not use E/Z (cis/trans) for rings smaller than 8-membered.

Can one of you post as response and I will accept it.

Both @Waylander's and my Comment are true as well as your conclusion. From the IUPAC Blue Book 2013:


text:
Proper notation of doubly charged ions
I have always notated doubly charged ions with a superscript $++,$ such as $\ce{He^{++}}.$  For more highly-charged ions I would denote them with the number of charges, such as $\ce{Pt^{4+}}.$
Recently I have been challenged that doubly charged ions should be similarly written as e.g. $\ce{He^{2+}}.$  My internal consideration of the matter is that the fully consistent application of such a convention would necessitate writing $\ce{He^{1+}},$ which wastes space.  I have a vague recollection from my student days of being told that we use a superscript $++$ because it is the same character count as $2+,$ while avoiding the logical conclusion that one should use superscript $1+,$ and thus the logic of the naming convention was to preclude excess typesetting.  However, my Google skills have failed to turn up documentation of such a rationale $−−$ or any rationale, for that matter!
Can anyone point me to some documentation that might describe/explain the notation convention?

Just as $x+x=2x$, nobody writes $1x + 1x=2x$, it is completely understood that with a single plus sign, 1 is written before it. If you read spectroscopic literature you might see He(I) written for neutral helium ion, He(II) for singly charged helium atom. You see there are different acceptable ways of writing the same thing. All you have to do is to accept them :-)

All your notations are acceptable. Only $\ce{He^1+}$ is odd.

M. Farooq, thanks for the comments.  I am aware the X$^{++}$ is used throughout the literature.  So is X$^{2+}$, which seems to be decidedly less common in usage for ions while being the standard notation for oxidation states.  I was hoping for some documentation (such as IUPAC guidance or even a random textbook) to explicitly describe the general convention on the notation.

Related: https://chemistry.stackexchange.com/questions/24769/why-ca-2-and-not-ca-2

TL;DR: $\ce{He^2+}$ is the only preferred notation.
Notations $\ce{He^{++}},$ $\ce{He^{+2}}$ or $\ce{He^{1+}}$ are obsolete and should be avoided.

From IUPAC “Green Book” [1, p. 49], section 2.10.1 Other symbols and conventions in chemistry, subsection (i) The symbols for the chemical elements:

The ionic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.
$$
\begin{array}{lll}
\textit{Examples} & \ce{Na+} & \text{a sodium positive ion (cation)} \\
  & \ce{^{79}Br−} & \text{a bromine-79 negative ion (anion, bromide ion)} \\
  & \ce{Al^3+}~\text{or}~\ce{Al^{+3}} & \text{aluminium triply positive ion} \\
  & \ce{3 S^2-}~\text{or}~\ce{3 S^{-2}} & \text{three sulfur doubly negative ions (sulﬁde ions)}
\end{array}
$$
$\ce{Al^3+}$ is commonly used in chemistry and recommended by [74]. The forms $\ce{Al^{+3}}$ and $\ce{S^{−2}},$ although widely used, are obsolete [74], as well as the old notation $\ce{Al^{+++}},$ $\ce{S^=},$ and $\ce{S^{−−}}.$

For the historical perspective, see the corresponding article by Jensen [2]:

In contrast, the German chemist, Walther Nernst, in his equally inﬂuential 1893 textbook of theoretical chemistry, chose to place an appropriate number of superscripted $+$ or $−$ signs directly above the ion’s atomic symbol (4), a practice which was soon modified by placing them instead to the immediate right of the symbol, as in $\ce{Ba^{++}}$ and $\ce{PO4^{---}}$(5).
The IUPAC guide to Quantities, Units and Symbols claims that yet a third “algebraic” method of indicating ionic charges was also used in the past in which the charge preceded the numerical value, as in $\ce{Ba^{+2}}$ and
  $\ce{PO4^{-3}},$ even though this particular sequence of symbols was originally intended to represent the inherent sign of a number or exponent and not the number of signs (6). However, inspection of nearly three dozen general, inorganic, and analytical textbooks, spanning the period 1909–1975, revealed that the vast majority employed the modified Nernst notation, with a smaller number — mostly of European or Russian origin — using the Ostwald notation instead. Rather surprisingly, very few examples of texts using the algebraic notation could be found, all of them post-1970 (7).
Since at least the 1950s IUPAC has ruled that ionic charges or “charge numbers,” as they are now officially called, should be written instead with the number preceding the charge sign, as in $\ce{Ba^2+}$ and $\ce{PO4^3-}$
  (6, 8, 9). There are several reasons for this decision. It is more concise than the typographically inelegant Nernst approach and more physically meaningful than the Ostwald notation. Unlike the algebraic notation, it
  avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Jensen, W. B. The Proper Writing of Ionic Charges. J. Chem. Educ. 2012, 89 (8), 1084–1085. DOI: 10.1021/ed2001335.

The best part of this post was learning that $\ce{S^=}$ was considered acceptable at some point. :)

@Zhe $\ce{S^=}$ is not that bad. Imagine future generations using something like $\ce{S^{}()}$ in place of $\ce{S^2-(aq)}.$ That's horror.

I can't unsee that.

Thank you for this.  While it is not the answer I was hoping for, it is good to know.


text:
Predict the products of the following reactions with Mechanism

1

My Attempt : The correct answer is C. I tried rearranging similar to claisen rearrangement but I have no clue how to go forward. I know that option A is wrong and that the product is pretty easily predictable.
But I cannot predict either the product or the mechanism.A detailed solution for each reaction  would be very helpful.
Thank you.

fyi (D) gives pyrazole. It is known as the Pechmann pyrazole synthesis  https://www.drugfuture.com/OrganicNameReactions/onr296.htm

It's hard for me to imagine that (C) does not take place at some level.

Can you imagine that C takes part to form something?

Unlikely that B is the correct answer because of the oxidation state. Loss of acetic acid gives a nitrile but there is no way to add acetate to the ring. The oxime of benzaldehyde is dehydrated with Ac2O to benzonitile and acetic acid, i. e., the intermediate is B's starting material. C, as @Zhe has suggested, is the likely answer.

Ok then Guys,I have edited my question to C being the answer,I'll accept it if you can give some mechanism predicting each product.

I think we have helped you enough. As to a mechanism for C, I suggest you Google thermolysis (pyrolysis) of esters.


text:
Does 1,3-dichloroallene possess a C2 symmetry axis?

Is there an $C_2$ axis of symmetry in $\ce{Cl-CH=C=CH-Cl}$?

A $C_2$ axis of symmetry means that I should be able to rotate the molecule about an axis by $180^\circ$. But I would need a composition of the following two rotations

$90^\circ$ rotation about $\ce{C=C=C}$ axis
$180^\circ$ about the perpendicular bisector of $\ce{C=C=C}$.

Would that count as a $C_2$ axis of rotation?

The rule is that the molecule must be indistinguishable before and after the symmetry operation. Therefore is not the C$_2$ axis  a line from the midpoint between the two Cl atoms  (or H atoms) to the central C atom ? The only other operations are E and S$_4$

Sometimes, the perfectly OK and preferred formula does not help. In this case it's useful to draw it with 4 stereochem. wedges insted of 2.

A $C_2$ axis is indeed here—although not the one that you are thinking of: your suggestion of a $C_4$ then $C_2$ rotation does not qualify as a $C_2$ symmetry operation. There is a true $C_2$ axis which bisects the allene and is tilted by $45^\circ$ from both the $\ce{Cl-C-Cl}$ and $\ce{H-C-H}$ planes.
One can visualize symmetry elements for small molecules using, for example, Jmol.
Launch the program GUI and do the following:

File → Get MOL
Type in an identifier, e.g. the name, 1,3-dichloropropa-1,2-diene, or CAS number, 83682-32-0.
Open console (File → Console) and enter show pointgroup to list symmetry elements and draw pointgroup to add them to the picture:


text:
How many nearest neighbours does a corner atom have?

Figure shows a unit cell of a cubic lattice. The atom in the middle has 8 nearest neighbours (the atoms in the corner). How many nearest neighbours does a corner atom has ?

I identified the given lattice is a BCC lattice. But I have trouble finding the number of nearest neighbour atoms of the corner atom. Is there a definition for nearest neighbour atoms ? (Please note I don't know much about crystal structures. I am a student in the first semester of an engineering degree. Please can someone explain how to solve this in simple terms.)

The nearest neighbor of corner atom is at a distance √3a/2 where a is the length of side of unit cell. The no. of nearest neighbor is 8. Nearest neighbor of an atom means those atoms which surround the given atom at the closest distance to that atom.

A corner atom has 6 neighbours at distance a, two per axis : one before, on behind, one left, one right, one above, one under. The central atoms are too far away from corner.

@Maurice $\frac{\sqrt{3}}{2} < 1$, therefore the central atoms are closer than the other corner atoms.

@Manu So the 8 neighbour atoms that a corner atom have are the atoms in center of each unit cell surrounding the corner atom. Am I correct ?

Let's just take a picture!
For that I took the crystal structure data (CIF) of CsI from literature and changed the $1b$ position from iodide to another cesium, so all atoms are the same, like in your picture above. I then created a $2 × 2 × 2$ cell and took the central cesium atom which used to be the corner of one unit cell and colored it yellow.
And then I searched for the closest neighbours, those on the vertices of the unit cells (in blue) forming an octahedron and those in the center of the former unit cells (in red) forming a cube.
If you check on the distances you will find that for this case of a $\ce{CsI}$ lattice, the distances are $d_\mathrm{yellow-blue} = \pu{4.56 Å},$ while $d_\mathrm{yellow-red} = \pu{3.95 Å}$. So the answer should be 8 neighbours here.

BCC is a body-centered lattice. The origin and the center of the cell are crystallographically equivalent, i.e. there is a translation operation relating them and the entire crystal. So if the atom in the center has 8 nearest neighbors, the atoms at the corners have 8 nearest neighbors as well.


text:
Synthesis of a lactone
I'm solving some synthesis problems for practice and I came across this one:

I thought of going for compound A to B(aldehyde) with this procedure:

$\ce{t-BuOK}$
$\ce{BH3,THF,NaOH,H2O2}$
$\ce{PCC,CH2Cl2}$

For the second path, in order to get to a lactone, I don't know what to do. I see that the carbonyl bond is moved to another carbon. I've learnt a way, in order to get an ester, is to add an alcohol $\ce{ROH}$ to an acyl chloride. But I'm so confused.

What do three arrows signify? Three synthetic steps, or just a very fast transition?:)

You have added a 4 carbon unit to your aldehyde. How many steps are you contemplating?

@andselisk I really don't know, the exercise just asks to write the reagents which will lead to the wanted compounds.

@Waylander Ι don't know where to start, im really confused. The part that confuses me the most, is that there are 4 carbon atoms added, but they are in a ring and i wonder if it is an intramolecular type of thing or not

@chemistryluvv Would it be possible  for you to attach the image of the original problem, or literally quote it? Does the currently shown reaction scheme also appear like this in the original?

@andselisk Im so sorry, i can't post it because it comes out very blurry. The compounds are shown exactly the same and the exercise says: ''Starting from compound A, predict the reagents required for the synthesis of the other two''.

Consider also https://old.reddit.com/r/chemhelp/comments/gvuz0l/synthesis_of_a_lactone/ and https://old.reddit.com/r/chemhelp/comments/h093mj/synthesis_of_this_ester/ for some other guidance

It's a bit hard to know what to suggest for your lactone without knowing how much you know. Personally, I would try 1-(trimethylsiloxy)-1,3-butadiene and attempt a [4+2] cycloaddition, but that's weird to suggest if you haven't learned about that class of reactions yet. Also, we generally try to avoid PCC unless other methods aren't working. Chromium waste is not good for the environment. These days one might attempt something like a Swern oxidation.

Add a 5-carbon unit via the Grignard reagent of 5-bromo-1-pentene to the aldehyde. Ozonolysis of the double bond to a hydroxy aldehyde that will likely form the hemiacetal. Oxidize with PCC

Here is a way to make the lactone:

To your aldehyde in dry THF at 0 °C, add the commercially available 3-butenylmagnesium bromide. Work up and collect the resulting secondary alcohol.
React the alcohol with dihydropyran in dry $\ce{CH2Cl2}$ to form the THP ether.
React this THP ether with $\ce{BH3.THF}$ as you did before, but do an oxidative work up with PDC to get the carboxylic acid according to Brown et al. [1], or do the basic peroxide as before to get the aldehyde, then oxidise that.
Treat the product acid with PTSA in MeOH to remove the pyran ether, the lactone will form spontaneously.

Your choice of alcohol protecting group will be driven by the oxidation conditions you use for the borane. THPO is known to be stable to PDC/$\ce{CrO3}$ in pyridine.
Reference

Brown, H. C.; Kulkarni, S. V.; Khanna, V. V.; Patil, V. D.; Racherla, U. S. Organoboranes for Synthesis. 14. Convenient Procedures for the Direct Oxidation of Organoboranes from Terminal Alkenes to Carboxylic Acids. J. Org. Chem. 1992, 57 (23), 6173–6177. DOI: 10.1021/jo00049a024.


text:
Do tert-butylcyclohexanes with a leaving group locked in the equatorial position undergo E1?
Background
Elimination of 1-bromo-tert-butylcyclohexane is a prototypical example of how important the anti periplanar relationship is for E2. From Organic Chemistry: Structure, Mechanism, Synthesis
By J. David Rawn, Robert J. Ouellette [1, p. 281],

The effect of conformation on E2 reactions is demonstrated by the difference in rates of the dehydrobromination of cis- and trans-1-bromo-4-tert-butylcyclohexane. The axial bromine atom of the cis isomer is eliminated about 500 times faster than the equatorial atom of the trans isomer. Figure 9.8 shows that the axial bromine atom of the cis isomer is anti periplanar with respect to the axial hydrogen atoms at C-2 and C-6. Therefore, the most favorable geometry for elimination is "built into" the molecule, and no conformational equilibria are required to achieve it. In the trans isomer, however, no hydrogen atoms are anti periplanar to the equatorial bromine atom. To eliminate HBr in the trans isomer, a hydrogen atom that lies at a 60° dihedral angle with respect to the bromine atom must be removed. This process or any other reaction of a distorted ring that moves the two atoms into a better geometry for elimination requires more energy. Hence, the reaction rate is slower.

The tert-butyl group is very bulky, resulting in strong 1,3-diaxial interactions when in the axial position. This makes the A value of the tert-butyl group exceptionally high—about 5 kcal/mol. Compare this with the A values of other groups, almost all of which lie under 2 kcal/mol. This effect is so prominent that it results in conformation "locking" in tert-butylcyclohexanes, where the ring is forced to almost entirely take on the conformation with the tert-butyl group in the equatorial position. Quantitatively, there is a ratio of about 10,000 to 1 equatorial conformers to axial conformers for tert-butylcyclohexane. In fact, sufficiently nucleophilic bases will prefer to react with trans-1-chloro-4-tert-butylcyclohexane via substitution rather than elimination even though nucleophiles prefer to substitute axial leaving groups.
The effect of the tert-butyl group on E2 is quite clear. Since it locks the bromine in the equatorial position in trans-1-bromo-4-tert-butylcyclohexane, E2 is significantly harder to carry out. However, this conformational locking shouldn't have much of an effect on E1 as in that case an anti periplanar relationship between the leaving group and the abstracted hydrogen is not required. Other factors seem to be in favor of E1 in the elimination of trans-1-bromo-4-tert-butylcyclohexane. Bromine is a good leaving group, and the resulting carbocation would be secondary. Of course, whether E1 or E2 occurs also depends on our solvent and base. I'm particularly interested in the reaction conditions from the excerpt above, KOtBu/HOtBu.
Strangely enough, I've found information on this quite scarce. Almost every instance I've seen the elimination of trans-1-bromo-4-tert-butylcyclohexane come up has been in the context of assessing which isomer of 1-bromo-4-tert-butylcyclohexane will undergo E2/elimination faster. There seems to be little information on the actual rate law of the elimination.
I've managed to find a few places where it's said that 1-bromo-4-tert-butylcyclohexane or similar molecules with different leaving groups will undergo E1 elimination. Slide 37 of this presentation says that trans-4-tert-butylcyclohexyl tosylate undergoes E1/SN1 with NaOEt/EtOH:

Unfortunately, there don't seem to be any references, but the exact same thing is said in Organic Chemistry, Part 2 of 3 by Richard Daley [p. 582]:

The difference in elimination pathways between cis- and trans-4-tert-butylcyclohexyl tosylate illustrates the importance of molecular conformation. Because of its size, the tert-butyl group actively locks the cyclohexane ring with the tert-butyl group in the equatorial position. In the cisisomer, the smaller tosylate group is in the axial position; and in the trans isomer, it is in the equatorial position.
Both compounds react with ethoxide ion in ethanol to eliminate the tosyl group and a β hydrogen to form 4-tert-butylcyclohexene. However, the reaction of the cis isomer follows E2 kinetics, whereas the reaction of the trans isomer follows E1 kinetics. The cis isomer has the tosyl group and the β hydrogen in an anti-coplanar conformation; thus, allowing the base to easily remove the β hydrogen.
Because of the size of the tert-butyl group, the trans isomer does not easily get into either the anti- or syn-coplanar conformation. Thus, the rate of the reaction is quite slow.

Again, there weren't any references to papers confirming this. Aside from these two places, I could not find anything else claiming that trans-4-tert-butylcyclohexyl tosylate undergoes E1 elimination. On the contrary, I did find a paper that suggests trans-4-tert-butylcyclohexyl tosylate undergoes E2 rather than E1 elimination in HFIP:

Reactions of the epimeric 4‐tert‐butylcyclohexyl tosylates in hexafluoroisopropyl alcohol (HFIP) (1e, 1a ) allow for the first time to observe SN1‐type non‐stereospecific substitution, whereas conventional solvents including even trifluoroethanol yield SN2‐type inversion by solvent assisted pathway (≙k s). Large differences are found between the epimers in the solvent participation, measured kinetically by the Schleyer‐Bentley equation. This, as well as the even enhanced elimination with the equatorial isomer 1e as compared to 1a (98% vs. 96% in HFIP) points to the occurrence of non‐chair intermediates from 1e ‐derivatives, and to more E2‐ than E1‐type reactions.

The two places I was able to find claiming that 1-bromo-4-tert-butylcyclohexane undergoes E1 were pdfs of tests/exercises, presumably from undergraduate organic chemistry classes. This one makes the claim in problem 4, and this one (original pdf link is dead so only cached view is available) makes the claim in problem 7. Searches for literature confirming or denying whether this claim is true proved mostly futile. I did find this paper, which suggests an E2 mechanism for the elimination of 1-bromo-4-tert-butylcyclohexane by NaDA (the sodium counterpart to LDA):

The cis- and trans-4-tert-butylcyclohexyl bromides show a preference for the elimination of the cis isomer (kcis/ax/ktrans/eq = 10). Rate and isotopic labeling studies are consistent with a trans-diaxial E2 elimination via trisolvated monomer for the cis isomer and a carbenoid mechanism via disolvated monomer for the trans isomer.

However, NaDA is significantly more basic than alkoxides (compare the pKa of diisoprpylamine, 36, to the typical pKa range of alcohols, 16-18), so the elimination may go via E2 in large part because the base is so strong.
Question
I am curious whether conformational locking by a tert-butyl group is strong enough to force molecules like 1-bromo-4-tert-butylcyclohexane to undergo E1 elimination in conditions that otherwise favor E2, particularly in the case where the base/solvent is an alkoxide/alcohol like tBuOK/tBuOH. If so, to what extent can we see E1 creep its way into E2 territory as a result of this type of conformational locking? And in particular, is there some literature on this subject, especially for the case of 1-bromo-4-tert-butylcyclohexane?

It's a bit difficult to understand what you are asking, mostly because of your choice of words. "E1 creep ints way into E2 territory" You're making the E2 reaction so slow that other mechanisms (E1, specifically) are now fast, by comparison. We're not making E1 faster by itself, just in relation.

@Zhe sorry about that. Obviously what’s happening E2 is becoming slower by comparison, but I’m talking about elimination reactions that would otherwise favor E2 if the cyclohexane wasn’t conformationally locked now favoring E1.

You should check Reaxys or Scifinder Scholar. I'm sure that those examples are based on literature precedence.


text:
Is HBrO2 a reductor?
I'm working on a chemistry project and that features a reaction where $\ce{Br-}$ participates in a redox reaction as a reductor. Eventually, $\ce{HBrO2}$ competes with $\ce{Br-}$ to react with the same chemical as $\ce{Br-}$ reacts with.
Is is therefore true that $\ce{HBrO2}$ is also a reductor? If yes, what is the redox potential of $\ce{HBrO2}$?

Well, it obviously can be further oxidised - for example to BrO3 and BrO4 anions - with different potentials.

Bromous acid ($\ce{HBrO2}$) is an inorganic compound, which is an unstable compound. However, the salts of its conjugate base, bromites (e.g., $\ce{NaBrO2.3H2O}$) have been isolated. Bromous acid is considered to be an oxidizer. However, I can't find its reduction potential, probably because of its unstable nature (Ref.1). Based on the studies of Faria, et al. (Ref.1), $\ce{HBrO2}$ decomposes according to following mechanism:
$$\ce{HBrO2 <=>[$K_\mathrm{a}$] H+ + BrO2-} \tag{E1}$$
$$\ce{HBrO2 + BrO2- ->[$k_1$] HOBr + BrO3-} \tag{R1}$$
$$\ce{HBrO2 + HBrO2 ->[$k_2$] HOBr + BrO3- + H+} \tag{R2}$$
It was found that $k_1 = \pu{39.1 ± 2.6 M-1 s-1}$ and $k_2 = \pu{800 ± 100 M-1 s-1}$ for reactions $(\mathrm{R1})$ and $(\mathrm{R2})$, respectively. Hence, authors have calculated and given equilibrium quotient for fast equilibrium $(\mathrm{E1})$ as $K_\mathrm{a} = \pu{(3.7 ± 0.9) \times 10^{-4} M}$ at ionic strength $\pu{0.06 M}$ and $\pu{25.0 ± 0.1 ^\circ C}$.
I think OP is working on Belousov-Zhabotinsky reaction (oscillator), which would involves both $\ce{HBrO2}$ and $\ce{Br-}$. For example, usual reaction conditions are for reaction volume of $\pu{28 mL}$ at $\pu{39.6 ^\circ C}$ are: $\ce{[NaBrO3]} = \pu{1.8 \times 10^{-3} M}$; $\ce{[CH2(CO2H)2]} = \pu{5.6 \times 10^{-3} M}$; $\ce{[CeSO4]} = \pu{5.8 \times 10^{-4} M}$; and $\ce{[H2SO4]} = \pu{1.5 M}$ (Ref.2; yet this reference did not mention adding $\ce{[NaBr]}$ or $\ce{[KBr]}$ solution to the mixture). Note that either $\ce{Ce(IV)}$, or $\ce{Ru(II)}$, or $\ce{Fe(II)}$ complex can be employed as the catalyst. The Ref.3 proposed three mechanisms for the reaction:
Process A (consumption of bromide ion):
$$\ce{Br- + BrO3- + 2H+ -> HBrO2 + HOBr} \tag{A1}$$
$$\ce{Br- + HBrO2 + H+ <=> 2HOBr} \tag{A2}$$
$$\ce{Br- + HOBr + H+ -> Br2 + H2O} \tag{A3}$$
Process B (Oxidation of catalyst(R), autocatalytic reaction):
$$\ce{2HBrO2 + 2H+ -> HOBr + BrO3- + H+} \tag{B1}$$
$$\ce{HBrO2 + BrO3- + H+ <=> 2BrO2 + H2O} \tag{B2}$$
$$\ce{BrO2 + cat.(R) + H+ <=> HBrO2 + cat.(O)} \tag{B3}$$
Process C (Reduction of catalyst(O), production of bromomalanoic acid):
$$\ce{BrO2 + cat.(O) + H2O -> BrO3- + cat.(R) + 2H+} \tag{C1}$$
$$\ce{Br2 + CH2(CO2H)2 -> BrCH(CO2H)2 + Br- + H+} \tag{C2}$$
$$\ce{6cat.(O) + CH2(CO2H)2 + 2H2O -> 6cat.(R) + HCO2H + 2CO2 + 6H+} \tag{C3}$$
$$\ce{4cat.(O) + BrCH(CO2H)2 + 2H2O -> 4cat.(R) + HCO2H + Br- + 2CO2 + 5H+} \tag{C4}$$
As a consequence of this chain of reactions, it is evident by the equation $(\mathrm{A2})$ that $\ce{Br-}$ is oxidized to $\ce{HOBr}$ by $\ce{HBrO2}$ (which has formed as an intermediate):
$$\ce{Br- + H2O <=> HOBr + H+ + 2e-} \tag{A2'}$$
$$\ce{HBrO2 + 2H+ + 2e- <=> HOBr + H2O} \tag{A2"}$$
The sum of the equations $(\mathrm{A2'})$ and $(\mathrm{A2''})$ gives:
$$\ce{Br- + HBrO2 + H+ -> 2HOBr} \tag{A2}$$
Since this reaction is spontaneous for oscillation to be continued, it is safe to say $\ce{HBrO2}$ is a strong oxidizer compared to $\ce{Br-}$ since $\ce{Br-}$ acts as a reducer in this particular reaction.
However, it is also evident by the equation $(\mathrm{B2})$ that $\ce{HBrO2}$ has reduced to $\ce{BrO3-}$ to $\ce{HBrO2}$ (all formed as intermediates except $\ce{BrO3-}$):
$$\ce{BrO3- + 2H+  + e- <=> BrO2 + H2O} \tag{B2'}$$
$$\ce{HBrO2 <=> BrO2 + H+ + e- } \tag{B2"}$$
The sum of the equations $(\mathrm{B2'})$ and $(\mathrm{B2''})$ gives:
$$\ce{HBrO2 + BrO3- + H+ -> 2BrO2 + H2O} \tag{B2}$$
Since this reaction is also spontaneous for oscillation to be continued, it is safe to say $\ce{HBrO2}$ can be a reducer compared to $\ce{BrO3-}$ since it acts as a reducer in this particular reaction.
Also note that according to the equation $(\mathrm{R2})$, $\ce{HBrO2}$ self propagates. Therefore, whenever $\ce{HBrO2}$ forms, it could be acting as oxidizer or reducer, based on the conditions.

References:

Robert de Barros Faria, Irving R. Epstein, Kenneth Kustin, “Kinetics of Disproportionation and pKa of Bromous Acid,” J. Phys. Chem. 1994, 98(4), 1363–1367 (https://doi.org/10.1021/j100055a051).
Y. Pomeau, J. C. Roux, A. Rossi, S. Bachelart, C. Vidal, “Intermittent behaviour in the Belousov-Zhabotinsky reaction,” J. Physique Lett. 1981, 42(13), 271 – 273 (https://doi.org/10.1051/jphyslet:019810042013027100).
Yu Chang, Nannan Zhang, Yuxin Yang,  Jun Du,  Xing Fan, Changyuan Tao, “Time-periodic oscillation reaction in an organic-solvent dominated electrolyte,” Physical Chemistry Chemical Physics 2017, 19(40), 27643-27650 (https://doi.org/10.1039/C7CP05414A).

Thank you Mathew for your very extensive answer. You indeed guessed correct that it was about the BZ (or you looked at my other questions ;-) ). My question rose when I was reading this paper: https://scholar.rose-hulman.edu/rhumj/vol3/iss1/1/, it says that ‘the concentration of bromide eventually falls below some critical level [Br−]cr. It is at this point that Process B begins to dominate Process A: the hypobromous acid (HBrO2) begins to compete with the bromide to reduce the bromate.’ (citation). So I thought, if Br- is a reductor, is HBrO2 that too?

@ralphsmit: So my guess is right! :-) By the way, the rose-hulman reference is very interesting. Yet, there are some errors I found such as not balancing chemical equations. And also, keep in mind that hypobromous acid is $\ce{HOBr}$, not $\ce{HBrO2}$.

Thanks for checking it out. I already knew about the wrong name for HBrO2, someone else pointed it out in an other question. It does, however, say sth about the quality of it. The paper was my starting point for it. It is for secondary school (last year of pre-university-education). In the Netherlands it is common for sec. school students to do some limited kind of research as a preparation for university. As you can probably imagine, I didn't have any experience with ODEs (we're doing it about the Oregonator), so it was a good starting point. I've based most of it on more trustworthy sources.

If you need a good answer, you should not put an information embargo on the context of your question.

This is to be taken rather as an expanded comment than a full answer:
$\ce{HBrO2}$ oxidizes $\ce{Br-}$ to $\ce{Br2}$.
$$\ce{HBrO2 + 3 Br- + 3 H+ -> 2 Br2 + 2 H2O}$$
To tell if it may compete with $\ce{Br-}$ to react with the "chemical", we would have to know that chemical.
Regarding standard redox potentials, an exhaustive list is on Wikipedia, but $\ce{Br^{III}}$ species are not there.
But be aware $\ce{HBrO2}$ is unstable, strong oxidizer, while $\ce{Br-}$ is a weak reducer. I personally think the info may have been misinterpreted, having meaning $\ce{HBrO2}$ may compete with the chemical to react with $\ce{Br-}$.


text:
Copper brightness change
In an electrolysis situation I have identified a slight brightness change on a copper electrode.
What might have happened?
I'm dealing with different experiments and this reaction occurs in some of them.
In this experiment I am doing an electrolysis using a water solution with $\ce{NaCl}$, a Lechanché and two copper bars. The brightness change happens on the anode.
I also get bubbles and a color change in the solution of the cathode, I'm guessing it is due to the reduction of water.

Related: Why does copper precipitate out as a pink solid?

If an aqueous solution of $\ce{NaCl}$ is electrolyzed between copper electrodes, the most important reaction taking place on the anode is the dissolution of copper, according to : $$\ce{Cu -> Cu^{2+} + 2e^-}$$ If the copper plate was well polished and shiny like a mirror, the corrosion reaction will first occur on the smallest crystals, which are inserted between bigger regular crystals. As a consequence, the surface will be irregularly attacked, and its aspect will loose its shiny appearance.
Of course, at the cathode, some Hydrogen gas is produced.

My take on the possible chemistry starting with the electrolysis of salt water, which can produce some chlorine. The latter can further react with water as follows:
$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$
The created Hypochlorous acid may react with cuprous (here present as a $\ce{Cu2O}$ on the copper metal electrode). A fenton-type reaction based on $\ce{HOCl}$ could then proceed as follows for $\mathrm{pH > 5}$:

$\ce{Cu(I) (s) + HOCl (aq) -> Cu(II) (aq) + ^.OH (aq) + Cl- (aq) }$

The dissolving of the $\ce{Cu2O}$ results in a de facto cleaning of the copper metal surface, which per a reference to quote:

Copper is a beautiful burnished-gold color when it is clean and well maintained, but like all metals, copper can become discolored when exposed to air and water.

For background on the fenton-type reaction, see "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, to quote:

For $\ce{Fe(II)}$ and $\ce{Cu(I)}$, this situation can be generally depicted as follows [20,39]
$\ce{ Fe(II)/Cu(I) + HOX -> Fe(III)/Cu(II) + HO^. + X- }$ (15)
where $\ce{X = Cl, ONO}$, and $\ce{SCN}$.


text:
Is entropy maximum in equilibrium?
Usually, in the context of non-equilibrium thermodynamics, it is said that entropy achieves a maximum in equilibrium, so the Taylor series expansion of entropy around the equilibrium state as a function of fluctuations $\xi_i$ of internal extensive variables does not have degree 1 (linear) terms, just the contributions of the Hessian matrix (which is symmetric negative definite).
\begin{align*}
S\left(\boldsymbol{\xi}\right) &\approx S\left(\mathbf{0}\right) + \boldsymbol{\xi}^{\top} \nabla S\left(\mathbf{0}\right) + \frac{1}{2 \, !} \boldsymbol{\xi}^{\top} \mathbf{H} \left(S\left(\mathbf{0}\right)\right) \boldsymbol{\xi} \\
&= S_0 + \frac{1}{2} \boldsymbol{\xi}^{\top} \mathbf{H} \left(S\left(\mathbf{0}\right)\right) \boldsymbol{\xi}
\end{align*}
This is necessary to derive Onsager relations. But isn't there a flaw in this reasoning applied to open and closed systems? The entropy of the system should achieve max value in equilibrium only in isolated systems, shouldn't it?

The main point is that, if you assume that the entropy belongs to the universe (isolated system), then $\nabla S(0)=0$ only at equilibrium is a valid statement. While if the entropy belongs to any other kind of system, then $\nabla S(0)=0$ is still valid, and just tells you that you have a set of local variables in thermodynamic space "$\mathbf{0}$" that form a local extremum point for the entropy, and nothing else


text:
How does the entropic force arise within the hydrophobic effect?
I know that the hydrophobic effect is usually explained by the entropic effect originating from the disruption of hydrogen bonds between molecules of water and the nonpolar substance. The hydrogen bonds are reoriented to minimize the disruption of the hydrogen bond network of water molecules.
What I don't understand here is how this explains the tendency of non-polar substances to aggregate in water? Supposedly they disrupt the hydrogen bond network, but what forces them to clump together?
Then how does the above explain the tendency of water to decrease its surface area toward the nonpolar substance? Yes, the molecules of water around the nonpolar surface have lower entropy, but how does a decrease in entropy (in sense of a higher structural order) force them to minimize their exposure towards the nonpolar substance?

Nothing "forces" them. If molecules didn't have enough energy of themselves, they wouldn't do anything. It's just that while they randomly bump around, they get to state of lower Gibbs' enthalpy.

https://chemistry.stackexchange.com/questions/10210/whats-the-relationship-between-van-der-waals-forces-and-hydrophobic-interaction

Does this answer your question? Basis for the hydrophobic effect?

I've seen the other question and the respective answers, but they don't really answer what I'm asking. What you're saying, does this imply that water molecules around non-polar substance have a higher Gibbs free energy? Is this increase in energy (compared to other water molecules) induced by the introduction of the nonpolar substance?

Whenever a non-polar substance is put into water, water molecules will organise themselves around it in a cage-like formation. The molecules exposed to the non-polar substance at any given time will orient themselves to form as many hydrogen bonds with the rest of the solution as possible.
(Image from here)

The way I understand it intuitively is that these water molecules are restricted in the orientations they can assume, becoming more ordered to maximise hydrogen bonding, so have an overall lower entropy than if they were free to tumble like in the bulk solution. Similarly, the aggregation of non-polar molecules also allows more free movement within the non-polar 'blob'. If they were distributed evenly, there would be a higher surface area and therefore amount of 'cage' water molecules and non-polar solutes would be more restricted in their movement (trapped in a polar environment). This is the entropic factor that drives the hydrophobic effect.
It is also favourable enthalpically; hydrogen-bonding interactions and non-polar interactions are maximised respectively.
These two approaches demonstrate why it is thermodynamically favourable for non-polar substances to aggregate, not disperse. To clarify, it is still thermodynamically unfavourable to mix the two in the first place, that's why they separate; this is just the better option if they were to mix.
It is also worth noting that oil 'blobs' in water tend to be very round, due to the fact that a sphere is the shape with the smallest surface area to volume ratio. This supports the above reasoning, minimising the number of 'cage' water molecules whilst maximising movement of both water and the hydrophobe.

Very nice qualitative explanation but it also opens a can of worms. Entropy and order :-(

@M.Farooq How do you mean? (Not in a stand-offish way I'm genuinely curious)

Frank Lambert has written a lot on entropy and the associated myths with disorder. He passed away some time ago. A lot of gen. chem. (which I consider as a useless course) corrected their notions including Atkins. Thinking deeply, how entropy and so-called disorder are connected? What do we mean by disorder? I am still searching  :-)

Does this from a statistical viewpoint of entropy then mean that when nonpolar molecules are added to water they momentarily decrease the total number of possible microscopic configurations of the system? And afterwards the molecules by randomly moving around at the same time also arrange themselves to a state where the entropy is the highest? And then going back to a state of lower entropy among all possible states is of such low probability that of all configurations we mostly observe the other states?

@Gljiva A system will always tend to a state where its Gibb's energy is lowest (or entropy in the universe is highest) by random movement, in accordance with the laws of thermodynamics. It is a case of probability, the more disordered state is much more likely than any other state, so none others are really observed.

I think there is a typo: "It is also favourable entropically". You probably mean enthalpically.

@BuckThorn You're right! I'll change that :)

@M.Farooq Another less ambiguous synonym for disorder might be "regularity". I think the answer nicely explains how this works. Waters have less choices wrt hydrogen bonding partners when involved in caging a hydrophobic molecule, so there is more regularity (less variability) in the local bonding patterns.

I should of course have written "antonym"


text:
Can analytical precision be finer than a lower detection limit?
Can analytical precision be finer than a lower detection limit?
For example, if the lower detection limit of an analytical method is reported as 0.010 ppm. Is it possible to have an analytical precision of 0.001 ppm, and therefore a measurement like 0.116 ppm?
Obviously this depends significantly on the method, but in theory is this possible?

There's no inherent contradiction, I think. Detection limit doesn't necessarily stem from lack of precision.

@JoshuaWright, The question is how did you determine the detection limit? The problem is multiple methods will yield different detection limits for the same analyte or the same analytical method. There is no true "definition" of detection limit nor there is any ideal way of determining it.

Yes this is obviously possible.  For many methods of interest, the method will be precisely tailored to a particular measurement range.  For example, a technique might have a precision of 0.001 ppm when analyzing solutions that are between 0.114 and 0.118 ppm in concentration, and have rapidly declining precision outside of this range.  Isotope ratio mass spectrometry comes to mind as a related example.

@CurtF. Thanks for your comment, I suspected this was the case. Any chance you know where I could find a isotope ratio mass spectrometry data set that shows this?

@CurtF. Wouldn't that allow you to spike a sample to increase the concentration of an analyte and then, by extrapolating from higher concentration, to determine an analyte concentration in the original sample that is below the detection limit? For instance, if I can determine the concentration of original sample + spike as $x+dx$ and that of spike solution as $x$, then by simple subtraction I know $dx$. If what you write is true $dx$ could be below the detection limit

@Buck Torn.  Good thought experiment.  I think a key consideration is how exactly you know $x$.  Is it really practicable to get infinite precision in your assumed value of $x$?

One situation that illustrates how infinite precision in $x$ is impossible is shot noise in counts-based data.  Suppose the application is DNA sequencing, and you are measuring counts of covid amplicon from a PCR and counts of a control amplicon from the same PCR.  If you get 100,004 counts for the viral genome-derived molecule, and 100,000 counts for the spike-in amplicon, you can be sure the ratio of concentrations in the original sample is 0.500 (maybe even better, I havne't done the math.  But you can't then dilute the sample before spike-in by 100,000-fold and hope to get 4 counts.

@CurtF. One does not need infinite precision to do what I suggest, just precision higher than the lower limit of detection. I'm sure there is a flaw in the argument, such as problems with extrapolation (uncertainty growing when you extrapolate). I'm also starting to suspect this question might be more appropriate at cross validated SE


text:
What reagents change the time of a chemical clock?
I am doing a research task involving an iodine chemical clock reaction. I need to figure out which reagents in the experiment are responsible for changing the time before the colour change in the clock. The current method (see image attached) shows the quantity of materials required for a ~2 minute iodine clock. For this research task, we are required to find the quantity of materials necessary for a ~10 minute iodine clock. Can you please clarify which reagents cause the increase/decrease in time of the chemical clock reaction?

No additional reagent is required.  Decreasing the pH or increasing the concentration of iodide or hydrogen peroxide will shorten the time.

This may help you. Also read https://doi.org/10.1021/ed079p41.

First, a review of the underlying chemistry per an educational source:

A clock can be formed from molecules that react at a rate from the time the chemicals are mixed to the time the reaction occurs.
Explanation: There are two forms of iodine:
Element form (Iodine): I2 + starch -------> blue
Ion form (Iodide): I- + starch --------> colorless
There are two reactions taking place simultaneously in the solution:
#1. 2I- + H2O2 ---(slow)-------> I2 + H2O
#2. I2 + Vitamin C ----(fast)----> I (colorless)
In Reaction # 1 iodide ions react with hydrogen peroxide to produce iodine element which is blue in the presence of starch. BUT, BUT......
In Reaction # 2 The Vitamin C is immediately reacting with any iodine formed in reaction # 1. The net result, at least for part of the time is that the solution remains colorless with excess of iodide ions being present.
Now after a short time as the reactions keep proceeding in this fashion, the Vitamin C gets gradually used up. The Vitamin C creates a clock reaction (1 - 2 minutes) and once it is used up, the solution turns blue, because now the iodine element and starch are present.

So, to answer the question: "which reagents in the experiment are responsible for changing the time before the colour change in the clock", the answer is H2O2 and Vitamin C. Now, while the rate-controlling step, associated with the slowest reaction is H2O2 (creating the colourful I2 highlighted by starch), some Ascorbic acid must be present to turn the solution clear again.
On the question:"Can you please clarify which reagents cause the increase/decrease in time of the chemical clock reaction?", the strength of the H2O2 (try changing from 3% to 10%) and the pH of the hydrogen peroxide (try adding alkaline Washing Soda Na2CO3 to the H2O2), both I would likely expect to have impacts on the Vitamin Clock timing.


text:
Reduction potential for reduction of O2 to H2O2 in alkaline medium
I've read that $\ce{H2O2}$ is a good reducing agent in alkaline medium and can reduce $\ce{Fe^3+}$ ions to $\ce{Fe^2+}$ in alkaline conditions. But I need to verify this using the standard reduction potential values (at 298 K) for $\ce{Fe^3+}$ to $\ce{Fe^2+}$ and $\ce{O2}$ to $\ce{H2O2}$(in basic conditions). But I haven't been able to find any reduction value for $\ce{O2}$ to $\ce{H2O2}$(in basic medium). Can someone provide me the data for this?

Search the book by the title "Oxidation Potentials" on "Internet Archive". The whole book is viewable after registration. It must have this type of information.

Can you state the name of the author, as there are multiple books with the same title.

I've tried to find it in the exhaustive list provided in Paula Atkins' Physical Chemistry, but I couldn't find it.

Same issue. No reduction potential for O2 to H2O2 reduction(in basic conditions) is given.

Here is it: $\ce{O2 + 2H2O + 2e- <=> H2O2 + 2OH-} \quad E^\circ = \pu{-0.146 V}$.

Thanks a lot! Can you mention the source please?

I believe the name of the author is Latimer's "Oxidation Potentials"

The main trouble about this problem is the fact that $\ce{Fe^{3+}}$ and $\ce{Fe^{2+}}$ are both insoluble in alkaline conditions. So that it would be difficult to check reduction potentials as the concentrations of these ions are extremely low. In NaOH 1 M, the solubility of $\ce{Fe^{3+}}$ is $\ce{2·10^{-39}}$ M, and the solubility of $\ce{Fe^{2+}}$ is $\ce{8·10^{-16}}$ M, according to J. Stark,  Chemistry Data Book, John Murray London, 1982.


text:
Calculating force and pressure of an ideal gas on a wall
The text below is from a discussion on ideal gases and the kinetic theory of gases:

After the collision the molecule must travel a distance $l$ to the
opposite wall, and then back across this same distance before
colliding again with the wall in question. This determines the time
between successive collisions with a given wall; the number of
collisions per second will be $v / 2 l$. The force $F$ exerted on
the wall is the rate of change of the momentum, given by the product
of the momentum change per collision and the collision frequency:
$$F=\frac{d\left(m v_{x}\right)}{d t}=\left(2 m v_{x}\right) \times\left(\frac{v_{x}}{2 l}\right)=\frac{mv_{x}^{2}}{l}$$
Pressure is force per unit area, so the pressure $P$ exerted by the
molecule on the wall of cross-section $l^{2}$ becomes
$$P=\frac{m v^{2}}{l^{3}}=\frac{m v^{2}}{V}$$
in which $V$ is the volume of the box.

To find pressure, the text considers the area of the wall instead of the area of contact between the molecule and the wall. Why is that the case?
According to the text, force on the wall exerted by the molecule is the change in momentum times frequency of collisions. Isn’t force supposed to be momentum over the time in which the molecule is in contact with the wall?

@Mathew Mahindaratne I’m just not getting one thing -why is the time that’s considered to find the rate of change of momentum not the time for which the gas particle is in contact with the plane?

Please don't use images instead of text: 1. Images are large and hard to read on mobile devices, and often cost mobile users valuable data that is limited. 2. Images are often blocked by corporate proxies. (This can also apply to some non-US countries that do not allow access to some sites, including image sites.) 3. Images can't be searched and therefore aren't useful to future readers. 4. Images are harder to read than text. 5. Images cannot be interpreted by screen readers for those with visual impairments.

@Richie Harvy: I thought I explained it clearly. Newton's secondlaw says $F = ma$ where $a$ is acceleration, which is rate of change of velocity, $v_x$. Each collision change velocity. Time take to change this velocity change is time between two collisions, meaning time take to travel $2\ell$. That equals to $2\ell v_x$.


text:
Nucleophilic substitution in polar protic/aprotic solvents
I am confused as to why nucleophilic substitution reactions occur. Does a stronger nucleophile substitute a weaker nucleophile in nucleophilic substitution reaction?
For example, I reasoned that reaction
$$\ce{R-Br + KF ->[DMF] R-F}$$
is correct because $\ce{F-}$ is a stronger nucleophile than $\ce{Br-}$ in polar aprotic solvent.
Therefore,
$$\ce{R-Br + KF ->[Polar protic solvent]}$$
should give no reaction because $\ce{F-}$ is a weaker nucleophile than $\ce{Br-}$ in a polar protic solvent. Is my reasoning correct?

It depends whether the reaction progresses through SN1 or SN2. In both, the departing of a leaving group is involved in the rate determining step, but only in SN2 is the nucleophile included in the rate determining step. In the following mechanisms, both of the first steps are rate determining.
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

It is important to remember that a good leaving group is defined by poor basicity, which can be quantified by a low pKa of the conjugate acid of the leaving group. Intuitively, this means the conjugate acid can lose a proton and the resulting base - our leaving group - will be reluctant to re-react with the proton. Or in our case, re-react with the electrophilic carbon. This generally means that good leaving groups are large, low charge density species; good examples are neutral molecules or even charged species like iodide ions.
This idea of strong and weak nucleophiles is situational. Hard-soft acid-base theory becomes a very useful reference in a lot of situations. It generally states that low charge density 'soft' nucleophiles and electrophiles have an affinity for eachother, and high charge density 'hard' nucleophiles also have an affinity for eachother. You can read more about exactly why this happens here:
https://en.wikipedia.org/wiki/HSAB_theory
Generally, we consider a saturated carbon a 'soft' electrophile, which will react very well with low charge density nucleophiles like iodide ions. What this means, is that in SN2 reactions, species that are considered good leaving groups are also good nucleophiles. For instance, you may predict - correctly so - that iodide ions are good leaving groups as well as good nucleophiles for SN2; this is why KI can be used to catalyse SN2 reactions by acting as both (usually better than your actual nucleophile and leaving group)!
(Image from Clayden Organic Chemistry)

These reactions can end up reasonably reversible, and can be driven to one side by modifying conditions. For instance, the reaction below (from Clayden organic chemistry) exploits the insolubility of a byproduct, effectively removing it from the equilibrium, shifting it to the right:
(Image from Clayden Organic Chemistry)

In SN1, nucleophilicity is irrelevant, since the nucleophile doesn't appear in the rate determining step, but I'd imagine the equilibrium position can be roughly estimated by relative leaving group abilities.
In answer to your specific example, fluoride would actually be a poorer SN2 nucleophile than you might think, worse than the bromide leaving group would be. The polar protic solvent will even further inhibit the nucleophilicity of the fluoride due to hydrogen bonding, slowing it down dramatically (so your assumption is accurate for kinetic reasons); and once it has been substituted, the fluoride ion is not a particularly good leaving group. In DMF, you don't have the hydrogen bonding issue, and due to poor solvation of anionic species in DMF, the fluoride is made slightly more reactive than it would be 'on its own'. And leaving group ability is also reduced due to poor solvation. This seems like it would be clear-cut; the fluorine will replace the bromine, right? However the bromide ion is a good nucleophile in this situation, and has the potential to kick off the fluoride in either solvent.
As you can see, there are a lot of factors to juggle here, and the actual degree to which the reaction progresses is specific to the exact conditions of the reaction. At a first glance, it is very difficult to predict, and is not necessarily as simple as the 'better' nucleophile replacing the 'worse'.


text:
Is alcohol based sanitizer flammable?
After applying sanitizers can they catch fire, if yes, then after what time is it safe to use the hands, say in the kitchen.
Also one more question is that if they are flammable then the flammable component is alcohol. However as soon as we apply the sanitizer the alcohol evaporates? Then how can they remain flammable on our hands?

As soon as alcohol evaporates, they cease to be flammable. Soon as it is, this does not happen instantly, though.

Hm, their medical application Is off-topic, but their physical and chemical properties are still on-topic.

@Poutnik It's still the same bag. Being too broad, this should be already closed.

Firstly, we should realise that sanitizers should not be considered as pure alcohols, alcohols are indeed a major component of it but many other important components are also present. A typical sanitizer contains these things -

Isopropyl alcohol/ Ethyl alcohol, Glycerin, Water, Carbomer(stabilizer) and some other components mostly related to cosmetics.

Now, i will answer your questions one by one

After applying sanitizers, can they catch fire?

Yes, there is 60%-70% alcohol content which is highly flammable. Organic compounds like alcohols, alkanes etc have very high enthalpy of combustion ($\Delta_\mathrm{c}H^\circ$ = -2000 kJ/mol for Isopropyl alcohol) and they can evaporate very easily due to the heat of our hands($\Delta_\mathrm{vap}H^\circ$= 40kJ/mol for isopropyl alcohol).

As soon as we apply alcohol, it evaporates so why is it still flammable?

This is actually wrong, as I earlier mentioned, that sanitizers are not only alcohols, in fact, the 'carbomer' present acts as a stabilizer to emulsify/dissolving all the components in it. These stabilizers prevent the alcohol to evaporate immediately and allow it to disinfect our hands fully. Also, after applying sanitizers, many of you might feel hands becoming sticky it is because of the presence of Glycerin and other cosmetic products used in it which act like a moisturizer. So generally it takes about 30-40 seconds for all(mostly) the alcohol to evaporate and sustaining a risk to be inflammable for a long time.
A thing to keep in mind is that the residue (mostly Glycerin) is also flammable(not that much, it's flammability rating is one) so after the alcohol has evaporated, you aren't fully safe from catching fire.
If your are working in kitchen, there is no actual need of using sanitizers as you can always wash hands in sink with soap and water.
I hope this answer cleared some of your doubts, stay safe :).

thanks. If you can add the references also

In this answer, I don't think I have added something which needs a reference, the enthalpy data is the standard NIST data, and everything else is rather applications of organic chemistry, carbomer being a stabilizer due to it's high molecular mass etc.

Water is also a common ingredient in hand sanitizer.

@SE-stopfiringthegoodguys thank you for the comment, yes, water is also a common ingredient in sanitizers. I have updated my answer.

Isn't the carbomer just a thickener? I've made hand sanitizer (no thickener though) and can confirm that no emulsifiers are used or needed. The WHO formulas are just water, alcohol, glycerin, and optional peroxide.

@piojo thank you for your comment. Comment part 1 You are absolutely right, carbomer(or some other chemicals) can be formally called as a thickener but if you compare the time taken for alcohol to evaporate of the sanitizers you will see a drastic difference,.  the WHO sanitizer evaporates way to fast, but note that it does not decrease it's potential to sanitize (notice the volume you take of your WHO sanitizer and compare it with the branded sanitizer).

@piojo Comment part 2 Using a thickener/solvent just makes it more efficient as most of the alcohol is used up after sanitizing hence it works with very small amounts, while the WHO sanitizer which is somewhat thinner is used in comparatively a bit more amounts at a time works just the same. (The comment was exceeding the word limit so had to answer it in 2 parts ;-))

Thanks for elaborating! I noticed I had to use a large amount of WHO sanitizer, but didn't realize commercial sanitizer might work as well with a smaller quantity. I'm gonna give carbomer a try next time I mix up a batch.


text:
Relative intensities are basically independent from the absolute ion abundances registered by the detector?
I am currently studying the textbook Mass Spectrometry - A Textbook, third edition, by Jürgen Gross. Chapter 1.4.4 Mass Spectrum says the following:

The most intense peak of a mass spectrum is called base peak. In most representations of mass spectral data the intensity of the base peak is normalized to 100% relative intensity. This largely helps to make mass spectra more easily comparable. The normalization can be done because the relative intensities are basically independent from the absolute ion abundances registered by the detector.

I don't understand how it makes sense to say that the relative intensities are independent from the absolute ion abundances. It seems clear to me that the contrary is true; that is, that the relative intensities are derived from the recorded absolute ion abundances registered by the detector, and so the relative intensities are dependent on the absolute ion abundances. After all, the detector records the absolute ion abundances first, and then uses these values to calculate the relatives intensities, right? I'm not sure that it would make much sense to say that it was the other way around.
I would greatly appreciate it if people would please take the time to clarify this.

I think this is simply sloppily written/translated. The pattern (positions & intensities) for a single substance is always the same if you use the same ionisation technique (esp. same voltage) and measure long enough. The absolute intensity is only very loosely correlated with the abundance of the compound in your sample. Ionisation efficiency varies all over the place.

Some detectors may saturate at high count rates, or you get pulse pileup issues, or whatnot, but if you are below the total count rate that causes that, you should be recording the same spectrum regardless of count rate.

@JonCuster Oh, I think I understand what the author is trying to say. He doesn't mean that the absolute abundances are not used in deriving the relative intensities; what he means is that, regardless of the varying levels of absolute abundances, the relative intensities pattern will always be the same for the same sample. Is this correct?

That´s what I meant by "measure long enough". You count fragments reaching the detector. The pattern develops purely statistical.

@Karl Ok, thanks for the clarification.


text:
why the terminal Br/Cl is an electrophile in AlBr5/AlCl5 complex?
In halogenation of benzene, e.g. bromination, $\ce{AlBr3}$ and $\ce{Br2}$ is added in order to make $\ce{Br+}$ by forming delivery agent $\ce{AlBr5}$ (step 2 of picture).
Why is the terminal (outermost) Br atom serve as an electrophile instead of bromine with positive formal charge? I don't get it because the terminal Br atom has an octet and should be stable and not as electron-deficient, why is it bears a partial positive charge?
another question related is that can use the combination of $\ce{AlCl3}$, $\ce{Br2}$ to achieve bromination? Is the mechanism the same as adding $\ce{AlBr3, Br2}$ if it is possible? Will the electrophilicity of terminal halogen be altered and prevent the reaction?

I don't get it what a+1 formal charge means, and yes, I think AlCl3 can act as catalyst, as it can still act as a Lewis acid for Br2. Only the rate of reaction my be altered.

Consider how large the FeBr4 complex is

@Waylander Sorry I don't understand why size is related. I am thinking is it possible that the Br with positive formal charge is partially negative due to high EN difference with Fe, hence terminal Br is induced to become partially positive, is this a possible answer or induction effect falls quickly as size (distance) increase? Thank you


text:
Is is possible to add NO2X, AlX3 to achieve nitration of benzene?
According to Organic Chemistry as Secondary Language (2nd semester), for nitration of benzene, the general way is to add $\ce{HNO3}$ and $\ce{H2SO4}$ and it said to make $\ce{NO2+}$. $\ce{(AlBr3 + NO2Br)}$ is impossible since it is a 'nasty' stuff? Why is it 'nasty'?
I searched online, there is no info about $\ce{NO2Br}$. Is it too unstable? If yes, may I ask why? However, there is $\ce{NO2Cl}$ present, so is it possible for it to react with benzene ($\ce{+AlCl3}$) to achieve nitration?

http://www.sciencemadness.org/talk/viewthread.php?tid=23230: this forum says that NO2F, NO2Cl is more stable where NO2Br tend to decompose and attain equilibrium. Does this means using NO2Cl is a viable option?

Nitration using nitryl chloride($\ce{NO2Cl}$) was studied extensively by Price and Sears where they found $\ce{AlCl3}$ to be a suitable reagent for the reaction. They also noted that nitration was difficult for deactivating aromatics such as halobenzene and benzotrifluoride and hence this method was considered of limited value.
On further experimentation, it was found that $\ce{TiCl4}$ was the best suitable catalyst. Other reagents used were $\ce{FeCl3,ZrCl4,AlCl3}$ and $\ce{AlBr3}$ but the reaction conditions were difficult to control. $\ce{BCl3}$ gave smaller yield with possible ring chlorination and $\ce{SbCl5}$ were effective for only aromatic compounds. The following compounds were nitrated using nitryl chloride and $\ce{TiCl4}$ as catalyst and data was noted:
\begin{array}{c|c}
\mathbf{Compound} & \mathbf{Yield (percent)} \\\hline
\text{benzene} & \text{88%}\\
\text{toluene} & \text{81.5%} \\
\text{ethylbenzene} & \text{79%}\\
\text{fluorobenzene} & \text{91%}\\
\text{chlorobenzene} & \text{89%}\\
\text{bromobenzene} & \text{86.5%}\\
\text{o-dichlorobenzene} & \text{41.5%}\\
\text{benzotrifluoride} & \text{32%}\end{array}
An excess of solvent was used for the reaction. Tetramethyl sulfone was suitable for the reaction. Other solvent experimented with was carbon tetrachloride but many chlorinated byproducts were observed. The possible mechanism for the reaction (with $\ce{AlCl3}$ as catalyst) was noted:

or a possible salt formation mechanism was indicated:
$$\ce{NOCl2 + AlCl3 <=> NO2+AlCl4-}$$
Nitration using Nitryl chloride was once important in making chloronitrosoalkane where carbon-carbon double/triple bond were broken by nitryl chloride and an free radical addition reaction occured but the yield was not that high. Dinitromethane was used as solvent as nitryl chloride breaked into chlorine and nitrous oxide:
$$\ce{NO2Cl -> NO2 + Cl.}$$
$$\ce{CH2=CH2 ->[NO2Cl] Cl-CH2-CH2-NO2(45 {%}) + Cl-CH2-CH2-ONO(35 {%}) + CH2Cl2 + N2O4}$$
$$\ce{C2H5-C#C-C2H5 + NO2Cl -> C2H5-ClC=CNO2-C2H5}$$
As for nitryl bromide, it was found to be unstable. Zuskin et. al. claimed to have synthesized the product$\ce{^{[4]}}$ but were invalidated due to their instability. Many attempts$\ce{^{[5]}}$ of making nitryl bromide were made but failed as they decomposed to nitric oxide and bromine. So, nitryl bromide was observed as a combined entity of (nitryl chloride + $\ce{KBr}$ in sulfur dioxide) at $\pu{-20 ^{\circ}C}$ and using $\ce{TiBr4}$ as catalyst, nitration was done but yields were low(lower than that observed with nitryl chloride) with possible ring-brominated side products. This is due to interference of side products from partly decomposed nitryl bromide.
Notes and References

Across Conventional Lines: Selected Papers by George A Olah, G K Surya Prakash

Chemistry of Energetic Materials by George A. Olah, David R. Squire

Nitrate Esters Chemistry and Technology by Jiping Liu

Bromine vapor saturated with nitrogen peroxide passed through a tube filled with bone-ash at $\pu{200-250^{\circ}C}$.

Halogen exchange of nitryl chloride with $\ce{KBr}$ in liquid $\ce{SO2}$, reaction of $\ce{BBr3}$ with anhydrous niyric acid or dinitrgen pentoxide, reaction of bromosulfonic acid with nitric acid and ozonolysis of nitrosyl bromide.


text:
Is LiNbO3 a perovskite or not?
I am collecting a database of perovskite materials and could not decide if $\ce{LiNbO3}$ is categorized as perovskite or not. $\ce{LiNbO3}$ has been identified as a perovskite [1] as well as non-perovskite [2 , appendix] in literature. Which is correct? Usually the structure type of some compounds are written as ($\ce{LiNbO3}$-type or LN-type), are these perovskites or not?
Relevant publications are also appreciated.
EDIT:
Goldschmidt tolerance factor of $\ce{LiNbO3}$: 0.748 (should appr. be between 0.8 and 1.1 to be perovskite)
Tau factor of $\ce{LiNbO3}$(as proposed here): 8.678 (should be less than 4.18 to be perovskite. Tau factor is proven to be more accurate than tolerance factor)
These tolerance factors indicate $\ce{LiNbO3}$ should be non-perovskite, however, I found some other materials such as $\ce{BiSrCr2O6}$ that pass tolerance factor tests but still form LN-type structure. This is where the confusion comes :(

2 doesn't specifically say that it is a non perovskite

It may have several different known phases...

@Pj30 They have reported a database of perovskites and non-perovskites, as you would find here(https://advances.sciencemag.org/highwire/filestream/210992/field_highwire_adjunct_files/1/aav0693_Table_S1.csv). In this database LiNbO3 was categorized as non-perovskite.

@AchinthaIhalage got it

Perovskites are cubic, orthorhombic, or tetragonal (see https://en.wikipedia.org/wiki/Perovskite_(structure)). Lithium niobate has a trigonal crystal structure. So, not a perovskite.

@JonCuster I am quoting from Figure1 caption of this Wikipedia article, "Pictured is the undistorted cubic structure; the symmetry is lowered to orthorhombic, tetragonal or trigonal in many perovskites". Triagonal is mentioned here?

I guess the real question is how distorted from real Perovskite (the actual mineral) is still a perovskite (general name for similar materials). Orthorhombic and tetragonal at least maintain right angles, trigonal is a distortion along the body diagonal so the sides are no longer rectilinear. What is 'correct'? Strictly speaking, only a true cubic material is Perovskite-like.

Not really, @JonCuster, the perovskite structure implies a 3D structure with corner-sharing coordination octahedra. The concept is still evolving: in the 1970s, people thought only those cubic or quasi-cubic structures with corner-sharing coordination octahedra should be considered perovskite. However, nowadays, it is more relaxed. People even call the "hexagonal perovskites" with face-sharing octahedra perovskite-like.

@QGent - well, I'm old school materials... Of course, by making the term 'perovskite' cover just about any crystal structure does make it much less useful as a technical term, but it looks like that ship has sailed...

LiNbO3 has quite a unique non-perovskite structure (under ambient conditions) and it's referred to as a structural type (LiNbO3-type structure) as a result (see here). This is largely due to Li+ being a very small cation.
The tolerance factor from Victor Goldschmidt is a classifier with moderate accuracy, so it's not surprising that even if a material passes its test, it may still not acquire a perovskite structure.
The other (hidden) assumption that may cause confusion is that these classifiers for perovskite structures are usually trained for conditions at ambient temperature and pressure. So they can fail badly if this condition is no longer valid because a material can be in a non-perovskite phase under ambient conditions, but in the perovskite phase under some other conditions.


text:
Difference between flame test and ability to show photoelectric effect
My textbook says magnesium doesn't give flame test. But it also says that it is a good photoelectric material. Are both concepts using "visible" light to remove or excite electrons?  What is the basic difference between these two and why is magnesium showing only one of them?

It is not the only factor to be thought of to explain the colour for a proper flame test we need to add ${HCl}$ to the compound and heat it to the flame.You may refer this https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__1%3A_The_Alkali_Metals/2Reactions_of_the_Group_1_Elements/Flame_Tests

why do we need to add hcl? What is the differnce caused by it?

I think you have a misconception. Observing an emission spectrum and photoelectric effects are different things. Photoelectric effect requires not only the electron is ejected with some kinetic energy, but it is ejected by light of a suitable wavelength. No heat or flame is involved.
In (Bunsen burner) flame emission spectroscopy, your temperature is not high enough to remove electrons from Mg atoms in the gas phase. Eventually, the flame temperatures can be made high enough that you do start to observe emission from Mg atoms, but it is in the UV range. Your eyes cannot see UV light.

we use visible light for excitation in photoelectric effect right? Or UV rays?

You can use visible or UV light. It only depends on the $work function$ of the metal i.e. how easy it is to eject an electron from a metal.