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Enthalpy Changes
0.50 g of Mg is added to a 50 mL HCl solution. The temperature increased by 5.2 degree Celsius. Calculate the enthalpy change for the reaction per mole of Mg that reacted.

What is the concentration of the HCl solution? Did all the Mg react?

We assume all the Mg reacts, which means that $\frac{1}{48}$ mol of Mg reacts with 50 ml
The enthalpy change for $\frac{1}{48}$ mol is given by $\Delta H = mC\Delta T = 50 \times 4.186 \times 5.2 \space J$
Therefore, the enthalpy change per mole is $48 \times \Delta H = 52.241 \space kJ \space mol^{-1}$


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Why does snow sublime even when the surrounding temperature is below 0 °C?
Snow in winter below 0 °C, still sublimed very slowly over time. If I’m not mistaken, sublimation is an endothermic process. Where is this absorbed energy coming from when the surrounding temperature is below 0 °C?

Water vapor has a finite value even at temperatures below 0 C, thus there is an equilibrium between water's vapor pressure and the temperature of the ice.

As long as the temperature is above 0 K and the air is sufficiently dry, it’s fine.

Above 0°C or below 0°C, sublimation (as well as melting) is endothermic all the same. What's so special about 0°C?

Evaporating liquids and sublimating solids have temperature lower than the surrounding.
That is because the most energetic molecules escape to gaseous phase, cooling down the condensed phase, as it's average energy per molecule decreases.
The standard  psychrometric method of air humidity measurement, used in  meteorology/climatology, is based on this effect, evaluating the temperature difference between "dry" and "wet" thermometers. Note the the dry/wet temperature tables to calculate the humidity are provided even for the frozen "wet" thermometer".
Therefore, during ongoing sublimation or evaporation, the surrounding continuously provides thermal energy to sublimating/evaporating matter, as it is permanently colder ( unless 100% vapour saturation is reached and sublimation/evaporation stops )
If you have ( in interior ) a big enough freely hanging piece of washed up wet cloth, put your hand just below it. You may feel very slight, somewhat colder downward air flow, as air near cloth surface gets little colder and flows down.

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Need of refining aluminium and Hoope vs Hall–Héroult processes
Recently I was taught how aluminium is commercially extracted. The ore is first concentrated by leaching either by Bayer's process for red bauxite (impurity: $\ce{Fe2O3}$) or by Serpeck's process for white bauxite (impurity: $\ce{SiO2}$).
Then alumina is reduced electrolytically via Hall–Héroult's process directly to 99% pure aluminium. But for further purification to remove remaining copper or iron impurities, Hoope's process is executed to get 99.99% pure $\ce{Al}$.
Since both processes involve electrolysis, and the materials added are similar in both (except $\ce{BaF2}$) with same graphite electrodes, shouldn't we get the same purity in both process (thus not requiring a further refinement)?
Also, the arrangement of the electrodes are exactly opposite to each other. Hoope's argues that pure aluminium is lighter and thus floats so cathode is kept at top, but why doesn't that logic work in Hall–Héroult's? The difference in purity varying by 0.99% and the electrolyte used being same.
Any insight into this matter is helpful because I haven't been able to come to a convincing conclusion.

But why doesn’t that logic work in hall-Heroult? : You answered it yourself - the $\ce{BaF2}$ mixture is much more heavier than $\ce{Al}$.

Ok so you seem to be right that BaF2 is heavier than Al, but Na3AlF6(cryolite) is not. So it should possibly float over the pure molten Al in Hoope's process? Making the conditions same as for Hall-Heroults.

Moreover I feel that adding cryolite is lot more necessary than BaF2, and Wikipedia says that "https://en.wikipedia.org/wiki/Hall%E2%80%93H%C3%A9roult_process" molten Al is heavier than cryolite, so it should sink down both in Hall-Heroult's and Hoope's processes. I don't think there's anything to argue on that.

From the wiki link, molten Al is denser than molten cryolite at the H-H operating temperature, so Al is on the bottom. In the Hoope’s process, as per wiki, there are 3 molten layers: a Cu-Al-Si metal anode, a mix of cryolite and barium fluoride middle layer, and molten Al as the cathode on top. Barium fluoride is much more dense than cryolite. So the layering makes sense if enough barium fluoride is used. What is especially interesting is how the Hoope’s process came to be invented: I never would have guessed it would work! Thanks for asking the question!

@EdV Yeah I found the idea pretty intriguing, and since a human being must have invented these processes so there must have been some sort of research and evidence involved. In Hoope's I couldn't understand how the pure $\ce{Al}$ migrated and was collected at the top, since cryolite must float over it. Also the both cells are nearly same but still Hoope's is run after Hall-Heroult for better purity.

I just upvoted your question and realized a mistake I made: “Hoope’s” is incorrect, since the process is named for its inventor “Hoopes”. The wiki blurb says he invented the process at ALCOA and it was patented in 1925. Probably not worth looking up the patent, since they can be pathetically lacking in important and proprietary details. Anyway, excellent question and maybe you or someone else will post an answer (other than the current answer)!

@EdV Thanks. Yeah the wait seems to be long. Nice update on Hoopes!

I think the difference is due to the way eddy currents form Magnetohydrodynamics.
As the current flows through the conductive liquid, a magnetic field is produced. This in turn causes the liquid to be pushed (at right angles to the current and field).
In one version of the cell, this current causes stirring of the aluminum, alumina, flux mix. This is more efficient from a time and electrical point of view. The trade-off however is that some of the impurities which would otherwise get captured in the slag (waste), end up mixed into the final product.
Edit: Thinking about it some more, the reason for the mixing may be due to where the $\ce{CO2}$ gas is formed. ($\ce{Al2O3 + C_{electrode} -> Al +CO2}$) Bubbles formed a the bottom of the cell would rise up and cause more stirring than if they formed near the top of the cell.

Thanks for providing a new perspective, I studied eddy currents recently but little did I know that its effects are such profound. But according to Ellingham diagram for oxidation reactions of metals and non-metals, its pretty clear that the thermodynamical feasibility of the above mentioned reaction is less. Also if we are going to really account eddy currents, then don't you feel it'd be likewise applicable in Hoope's process?

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Representation of bromelain molecule in two dimensional diagram
If I'm not wrong, I learned in 2011 introduction to neuropsychology course that all enzymes are proteins.
As far as I know, any protein, not even the smallest of the TAL group (which I understand to be the smallest currently known protein group), can be represented on a common-size two dimensional diagram (say, A4 page size) simply because it's too big in comparison to other molecules (such as methylcobalamin).
If bromelain is an enzyme which is larger than, for example, TAL proteins, how come it is represented on a common-size two dimensional diagram in PubChem?

Well, just like that. Look at that 2D diagram, it is all before you. Which part of it you don't agree with?

@IvanNeretin hello. Sadly I don't understand your comment - if it is sarcasm, this is a case in which I misunderstand the purpose of the sarcasm...

No, it is a genuine incomprehension on my part. You seem to be seeing some kind of mistake or contradiction where I see none.

@IvanNeretin thanks, I have edited the question to hopefully make it better.

I was forced to look at the wikipedia entry for enzyme, which states:

Enzymes /ˈɛnzaɪmz/ are macromolecular biological catalysts that accelerate chemical reactions. [...] Most enzymes are proteins, although a few are catalytic RNA molecules. The latter are called ribozymes. [Enzymes' specificity comes from their unique three-dimensional structures.]

The definition encompasses a rather broad range of molecules (macromolecular biological catalysts), is more of a functional rather than a structural definition, and requires clarification of what is meant by macromolecular. Also, not everything the wikipedia says should be accepted as consensus, of course.
This is an interesting question. For one thing, it challenged my biased view of enzymes as proteins (even if I know of ribozymes). I don't usually think of glycoconjugates as enzymes, but there you go, you have presented such a case. However, this is the structure shown in pubchem:

I see a couple of amide bonds, but mostly I see sugar rings, consistently with the description of the molecule:

Alpha-L-Fucp-(1->3)-[alpha-D-Manp-(1->6)-[beta-D-Xylp-(1->2)]-beta-D-Manp-(1->4)-beta-D-GlcpNAc-(1->4)]-D-GlcpNAc is a branched amino hexasaccharide consisting of a GlcNAc residue at the reducing end with a Man-alpha(1->6)-[Xyl-beta(1->2)]-Man-beta(1->4)-GlcNAc moiety attached via a beta-(1->4)-linkage and a Fuc residue attached via an alpha-(1->3)-linkage.

And yet the wikipedia says that "Stem bromelain (SBM) (EC 3.4.22.32), is a proteolytic enzyme". That's because I also missed the fine print: the pubchem entry is for MUXF, carbohydrate moiety of bromelain (hat tip to Ivan). Bromelain itself is a glycoprotein. And I am still looking for an experimental 3D structure to go with the molecule. Heading down another rabbit hole, the best I could do for now is find a link to a theoretical structure (Ref. 1):

A polysaccharide, attached to a folded protein, is shown in the lower right of the image.
So you opened up your post with a false premise, that all enzymes are exclusively polypeptides. Some of them, such as bromelain, are glycoconjugated proteins. And some such as ribozymes lack a peptide component altogether.
Finally, as to why the hexasaccharide is presented in 2D rather than as protein enzymes usually are, as a folded 3D structure, pubchem explains (see under 3D structure):

Conformer generation is disallowed since too many atoms, too flexible

Reference

Banerjee, N., Banerjee, A.. A three dimensional structural model of stem bromelain: a critical comparative analysis of glycosylation torwards functional structural stability. DOI: 10.5452/ma-c3x5z

I think I got it. The diagram in PubChem is not bromelain molecule, but a tiny fragment of it. This thing is indeed too small to be an enzyme. Also, it has nothing to do with proteins. Look at the formula, there is not a single amino acid residue here. Indeed, scroll down to 3.4.2 and read: "carbohydrate moiety of bromelain". That's what it is, and not the entire enzyme.
If you want to know what the rest of it is made of, then google "bromelain amino acid sequence". It seems to be known for quite a while. No nice pics, though.
On a side note, there is no such thing as "the smallest protein" or even "the smallest known protein", for the border between proteins and polypeptides is rather arbitrary. You are right, though, in that most enzymes are huge proteins, too big to fit on a 2D diagram. So is bromelain.
So it goes.

There is a recent crystal structure of a bromelain precursor protein.
PDB Entry 6U7D   http://www.rcsb.org/structure/6U7D
Usually "precursor" means that the active form of the protein is formed by cleaving off a piece or pieces. I do not know if that is the case for this protein.  However the protein used to grow these crystals was expressed in E. coli, so it does not contain the glycosylation characteristic of proteins produced natively in eukaryotic cells. If you are interested in the sugar residues of the native form, this crystal structure will not help you.


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Literature value for the activation energy of the reaction between hydrochloric acid and sodium bicarbonate
Does anybody have the literature value for the activation energy of the reaction between hydrochloric acid and sodium bicarbonate? If yes, could you please cite the source for it and the value?
I searched through the internet for hours, but the closest I could find was the activation energy between sodium carbonate and HCl, nothing for bicarbonate. I tried the NIST database and could not find anything, and other databases too.

@andselisk How is this a homework question? I think it needs a separate classification

@StackUpPhysics I would even say that this is a canonical homework question as it completely lacks research attempt and asks for the data on specific reaction, which would be of questionable help for other users. This could also be closed as a narrowed-down duplicate of Is there a database on chemical reactions, similar to NIST, but far more complete?. For a decent example of how a similar question could be asked, check out Activation energy of chlorine trifluoride and hydrazine.

@andselisk Got it

I thought homework questions were allowed. I searched through the internet for hours but the closest I could find was the activation energy between sodium carbonate and HCl, nothing for bicarbonate. I tried the NIST database and could not find anything, and other databases too. What makes the question not allowed? Also, if you have an answer I will be eternally grateful.

@IoanaRaducanu If you had a closer look at the link in the closing reason, you find out that homework questions are allowed all right, it's just that you need to demonstrate your thoughts and attempt to solve the problem on your own. I appended part of your last comment to the question and re-opened it. It appears that this question resulted in a Meta post, so I keep all the comments JIC.

@Ioana Raducanu It is rather reaction of H3O+ and HCO3-.


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Software platforms with integration of multi vendor mass spectrometer device data?
I know about ACD/Spectrus Processor which is a software platform to integrate analytical chemistry data from multiple mass spectrometer device vendors (e.g. Agilent Technologies, Bruker, Waters) to allow data processing. Are there comparable software platforms out there which enable the integration of MS/chromatography data of different vendors for data processing?
BTW: I know that questions of this kind usually fits better to Software Recommnedations. However cause it's specific to the chemical domain I think it's more suitable to ask here.

I'd recommend first converting all your data to an open format such as *.mzML and then using MS-DIAL, mzMine2, Open-MS, or any other common open-source softwares that work with this data.  All different sorts of processing workflows are available.

Some of the MS vendors made their software work with other vendors as well. E.g. ThermoFisher's Chromeleon, Waters' Empower seems to have Data Converters for this. If you already have software bought from your vendors it may make sense to research their cross-vendor capabilities.
There are also pure software companies that build vendor-agnostic tools:

As you mentioned - ACD/Spectrus which has a wide range of vendors supported.
Virscidian's Analytical Studio.
Elsci's Peaksel (disclaimer: I'm one of the developers).

The last two have good support for handling libraries (24, 96, 384, 1536 or more samples per experiment).
There are also open-source packages and libraries:

OpenChrom, if you don't like something you can contribute or just change it to your own needs.
MZmine2. There's also a repo with MZmine3, but I don't know what's the status there.


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Can NaCN or HCN be used as sources of cyanide ion in substitutions?
When cyanide ions act as nucleophiles in substitution reactions, why is KCN the reagent of choice? Would some other source of cyanide such as NaCN or HCN work equally well in the reaction, since they should dissociate to give cyanide ions as well?

KCN and NaCN are essentially interchangeable as reagents and very much easier to use as nucleophiles than generating CN- from HCN.

I don't, however, see why this question should be downvoted.

@orthocresol Poor formatting, lack of research?

@Mithoron You could take 2 minutes to improve it, instead of finding a duplicate that isn't a duplicate because it doesn't actually cover HCN.

NaCN is quite alright. The main problem with HCN, on the other hand, is that it's a relatively weak acid which is barely dissociated (pKa 9.71). So, trying to use HCN as a source of CN− is like trying to use H2O as a source of OH− ions, which is not likely to work. On top of that, it's gaseous, and getting it into your reaction without killing yourself can be quite tricky.

Well, then why actually NaCN, or other salt isn't used? Why is this one of these cases that potassium salt is common and not sodium? BTW if OP did his research he wouldn't ask about HCN.

@Mithoron not sure what you're aiming for. NaCN is used quite regularly http://commonorganicchemistry.com/Common_Reagents/Sodium_Cyanide/Sodium_Cyanide.htm

@Mithoron Maybe the OP tried to research and couldn't understand as the literature he/she may have found wouldn't have been very easy to understand. Also I believe SE aims to answer questions which the users face trouble with (If they aren't very basic but that again is subjective from person to person) and all question which are asked aren't necessarily open questions or award winning questions. At different stages of ones studies one can have different understanding and maybe a beginner might find the same page much difficult to read than someone who does so on a regular basis.

Well, it sure is used, but I think KCN is the stereotypical cyanide.


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What is the function of toluene in this reaction?
I'm trying to find the mechanism for this reaction:

My (tentative) thought for a possible mechanism is:

EDIT: New thought for mechanism:

But this leaves some loose ends that I haven't been able to figure out. How does the H on OH get removed? (Does it get removed at all?) Most importantly, what is the function of the toluene in this reaction? Why is there a time and temperature given? Any help is greatly appreciated.

Toluene is the solvent, that's all. The time and temperature are standard parameters that accompany any reaction description to assist anyone who might want to replicate the reaction. You appear to have a mixture of a radical and ionic mechanism, you might want to decide which it is.

Thank you for your comment. I now see that the first step of my mechanism doesn't make sense, but even if Br is not a radical, and it forms a bromonium ion on the alkene, I don't see why the OH group suddenly loses its hydrogen and resolves the bromonium positive charge by bonding to the carbon.

What is the other component present if Br+ is formed from NBS?

Oh! I think I get it. The nitrogen in NBS now has a lone pair which is basic enough to dehydrogenate the OH group.

Exactly so, the product is succinimide which precipitates out of the reaction mixture.


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How would one determine an element simply by looking at its binding energy?
I am self studying MIT OCW chemistry 5.111 2014, one of the lecture questions states the following:

Consider a neutral atom with 8 distinct electron binding energies: −14 eV, −28 eV, −94 eV, −218 eV, −293 eV, −1730 eV, −1921 eV, and −14326 eV.
  (a) Name all of the possible ground state atoms that could have these binding energies (without looking up any values).

It seems like the values are quite arbitrary without having a reference place to begin from i.e. electron configuration, quantum numbers, period and so on.
I thought maybe to use equation
$$E_{nl} = -\frac{Z_\mathrm{eff}^2R_\ce{H}}{n^2}$$
as it has an energy parameter, but it appears I have one equation and two unknowns.
Any ideas on how to go about this?

hint: consider the Aufbau principle, https://en.wikipedia.org/wiki/Aufbau_principle#/media/File:Aufbau_animated.png

@BuckThorn Thank you for your response although I'm not sure I understand. Could you please help me reason through it? The Aufbau principle pertains to filling in sub-shells to create the electron configuration of an atom. How could I combine that principle with knowledge of only the binding energy to determine the identity of an unknown atom?

Look at the binding energies of any known element. Look how do they grow when you are done with stripping one shell and are starting a new one.

I think some context is missing in the question. They should have discussed the X-ray spectra in the lecture. Have they? Forget Bohr's equation for this question, it fails miserably after hydrogen. The answers are provided by the online answer key. See one possible answer and its X-ray spectra notation https://xdb.lbl.gov/Section1/Periodic_Table/Ga_Web_data.htm. If they did not discuss X-ray spectra then I don't know how to address it.

@IvanNeretin Thank you for your response, it is my understanding that the binding energy is stronger the lower the energy level as they are closer to the nucleus so stripping electrons while maintaining the same amount of protons in the nucleus should result in a stronger binding energy. I am unsure how to quantify this in terms of energy though.

@M.Farooq X-Rays were not discussed beyond a general discussion of waves and electromagnetic waves as a whole, this is the assignment in full that I am referring to https://ocw.mit.edu/courses/chemistry/5-111sc-principles-of-chemical-science-fall-2014/unit-ii-chemical-bonding-structure/MIT5_111F14_Lec08Prob.pdf and these are all of the lecture related materials https://ocw.mit.edu/courses/chemistry/5-111sc-principles-of-chemical-science-fall-2014/unit-ii-chemical-bonding-structure/lecture-8/

@M.Farooq As well here is the link to the handout for the specific lecture to which this assignment corresponds https://ocw.mit.edu/courses/chemistry/5-111sc-principles-of-chemical-science-fall-2014/unit-ii-chemical-bonding-structure/lecture-8/MIT5_111F14_Lec8.pdf

There are a few clues on how to solve this question:

you are given binding energies. The energies in question are just enough to liberate electrons from their bound state.
The equation you've written $$E_{nl} = -\frac{Z_\mathrm{eff}^2R_\ce{H}}{n^2}$$ describes the binding energy. The important part is that it can be assumed to be only a function of n and l, that is, of the subshell to which an electron belongs ($Z_\mathrm{eff}$ is a function of n and l quantum numbers. Unfortunately this is probably a somewhat complicated function but that is not very important just now)

Finally consider the Aufbau principle, outlined in this libretext:

In essence, you can think of electrons filling in "n+l " order. If you have an equal "n + l"  value for two orbitals, then electrons will occupy the orbital with the lower "n" value.

which is followed by a nice illustration (shown below). The number of distinct binding energies equals the number of subshells. The energy for emission from each subshell decreases in the same order in which the subshells are filled. All electrons in a subshell share a common binding energy. Following the Aufbau pattern you can identify the highest-energy occupied (valence) subshell, which corresponds to the smallest binding energy. I count 8 subshells and assign the minimum $E_{nl}$ to the 4p subshell (the electron configuration is $\ce{[Ar] 3d^{10} 4s^2  4p^?}$). This means there are six possible atoms that match the number of binding energies, namely Ga through Kr.
If we cheat and check a table of ionization energies we see that the lowest binding energy (-14 eV) matches that of Kr (other values in the table don't match those in the problem since the table lists the ionization energy for multiple electron emissions whereas the problem lists energies for all possible single electron emissions $\ce{A->A+ + e-}$, starting from the neutral atom, an emission from one subshell corresponding to one energy). You should note the pattern of decreasing binding energies (magnitudes) starting from the 1s electrons (−14326 eV) up to the valence shell (-14 eV). The following diagram adapted from the libretext quoted above illustrates how this works:

Thank you so much for your very detailed answer, so simply by knowing the number of potential binding energies we are able to deduce the block on the periodic table that these binding energies must belong to. In this case it is not one atom with 8 binding energies at different levels but rather 8 different elements with differing ground state energies. But if it is the case that there are 8 differing elements with 8 differing ground states how could we know their respective energies?

@TheRedCamaro3.03.0 I find there are 6 elements that match the emission pattern. I tried to explain more clearly in the answer.

Thank you so much truly for your detailed answer it makes it much more intuitive. I believe my confusion was in the wording of the question suggesting that we're looking for a single element with 8 ground states vs 8 elements with 8 distinct ground states when we were actually looking for potential elements with 8 potential sub-shells.


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Eletrolytic rust removal
I've seen quite a few youtube videos about electrolysis for rust removal. The process seems to be effective. My question is: how does it work?
I get that the water is ionised so we have O- and OH+ ions. I get that O2 is liberated at the anode. But what is going on at the cathode to release the rust? Is the rust chemically changed or is the rust somehow chemically released from the parent metal?
I'm a (rusty - graduated 1981) physicist, so don't hold back on the answer.

XRD of rust shows Fe3 O4 -H2O mostly ,and several similar compounds . Very rarely Fe2O3  ; at ambient temperature it is diagnostic of fretting corrosion.

@DrMoishePippik rust ≠ $\ce{Fe2O3}$.

You mean $\ce{H+ and OH-}$ ions, right? Ideally, electrolysis would reduce rust to iron, and the iron would be in the same spot it was before rusting. However, nobody has figured out how to do that, yet.

Electrolytic rust removal appears more like a semi-hoax to me. I think it is more like a mechanical removal of rust layer. As we dip a rusty object in sodium bicarb solution and you pass cathodic currents, we would see bubbling of hydrogen gas from the rusty object. These bubbles apparently help to loosen up the tight rust layer (bottom up). The cathodic current may also reduce some iron oxide.
The anode, which most videos suggest should be made of iron, is also useless in a sodium bicarbonate solution. Any inert conductor would have worked like a graphite anode. The reason is that the iron from the anode must go into the solution, if it were a true electroplating. However, in most videos iron oxides from the anod float as a precipitate of iron hydroxides. The whole solution turns rusty. Finally, people suggest a mechanical brushing or polishing. I am sure if a rusted object is polished with an abrasive, the results would be the same as electrochemical "rust" removal.
Now some over-the-counter rust stain removers are quite effective. They contain powerful reducing and perhaps some chelating agents. No, it is not oxalic acid, if that is coming to mind. These reducing agents immediately remove the rust by chemical means, but the iron surface is no longer shiny (greyish black). One has to polish it again to look shiny.


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Historic Importance of Tin(Sn)
How important is tin for a bronze age society when there's many copper alloys including arsenical bronze. If you're out of tin whats the next best copper alloy(s) for making shortswords and axe blades and is it such a downgrade. This is a chemistry and history question.

It's answered by dating back copper artifacts. At a glance I would answer as: arsenical bronze if you have the right ore. I'll keep digging and discuss it somewhere else.

Hardness is easy to look up; this ended up being way more history leaning than i expected. Whatever they used when out of tin was what worked for them; this is barely a chemistry question at all.

Some tools were made of "pure" copper. But tin was the best and only available addition to copper for strength. At the time tin was the only type of bronze. Today a variety of copper alloys are called "bronze"  that do not contain tin. Lead could possible have been available depending on time and place but it does not improve strength of copper and could actually weaken it. I see in alloy listings there are silicon bronzes ( today) that contain 1 to 2 % . Silicon was available, but it takes modern technology to alloy it into copper. Silver and gold could be added to copper ( they are all mutually soluble ) but make little increase in strength. There is a copper alloy today that contains about a quarter % of each phosphorus and arsenic ( C14200) , it is a wrought product so not available in the bronze age. The Copper Development Association  presumable has some information on the internet.

Arsenical bronze is strong and good for edged tools; it dates back pretty far although it was only made from certain ores until they could isolate arsenic as an additive.

My answer is in modern ( last few hundred years) context. The Wiki article on archaeological bronze is interesting , much higher levels of As were used than the one modern alloy. But consistent as it notes the high strength of As bronze is achieved by forging ( wrought ); which is not necessary with tin.  Traces of As are used for corrosion resistance in some modern brasses. And as noted, unrefined copper often contains As as a tramp element.

And , roughly 100 yr ago arsenic copper was used for tubes in low temperature boilers for improved corrosion resistance and better elevated temperature strength than pure copper. It was made from high As ores and sometime called "tough lake copper".


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Is fluorinated steel corrosion resistant from normal oxidation?
I've recently learned of a compound named Chlorine trifluoride a powerful fluorinating agent. The common way to store it seems to be in regular steel drums where it flash oxidizes the inside of the drum to make a metallic fluoride. This layer of metallic fluoride keeps the substance from reacting any further with the steel drum and passivates the steel from further oxidation by fluorine. My question is if steel that has been fluorinated comes in contact with water and oxygen(or even worse a hydrogen peroxide water mix) is the metallic flouride layer fully passivated or is it pourous like the magnetite layer used in gun bluing? Would this result in any better corrosion resistance than chrome plating, aluminizing, bluing or galvanizing? I can't seem to find any studies on the mechanical properties of the fluorine passivated layer of steel and I am curious if anyone has any experience with working with either flouring gas and steel or even chlorine trifuoride itself.

I am very dubious of the premise that some iron fluoride compound is protective.  HF has been used in alkylation of hydrocarbons for many years; It is less common now because costly alloys ( Hastelloys , Monels, etc) are necessary to control corrosion. Keep in mind the competing process uses concentrated sulfuric acid as it is less corrosive to steel.


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why Ammonia is proton acceptor
I understand that this is an elementary question, but I tried to find answers which I could not get.

I am trying to understand why Nitrogen in Ammonia is a proton acceptor.
Now as I can see from the structure the octet is complete hence should be stable as it is.
But couple of points here:
1) Nitrogen is more electronegative than hydrogen hence $N$ will pull electrons towards itself. Hence not a good candidate to donate electrons! In addition, the lone pair is in 2s orbital.
2) The other point is the presence of lone pair, it is claimed that this lone pair is more stable if its part of some bond, hence it accepts proton. But if this is the case then lets take the case of Oxygen $O=O$, here each Oxygen atom has 2 lone pairs, and going by the above logic these lone pairs should also participate in bond formation and create an Oxygen polymer(!!), but they do not and Oxygen is quite stable.
Hence with the above points I still could not see why Nitrogen has to accept proton.
The only explanation that I can think of is the lone pair is in 2s orbital and the spread of this electron cloud is larger than electrons participating in bonds. Hence to be stabilized further it accepts a proton $(H_+)$ to form a bond!!

NH3 is quite stable as it is. Also, when given a chance, it readily accepts protons and occasionally reacts with some other things. You seem to perceive stability as some universal definitive ordering of all compounds. This is very much not the case.

NH3 is octet-stable as is, but the hydrogen ion could do with a pair of electrons. The nitrogen lone pair is more available than an oxygen lone  pair, generally speaking. Forming the bond with hydrogen keeps the octet on nitrogen. So even with this simplest octet model, you can sort of rationalize the observations.

@IvanNeretin exactly, why it should readily accept protons. I know the effect, I am looking for the cause, the exact cause.

@KarstenTheis "forming the bond with hydrogen keeps the octet ... " - the octet is already present and the system is stable enough, so why it should go the extra trouble to bond with hydrogen. The only explanation that I can think of is if the system  goes into more stable state by releasing energy!

NH3 is perfectly content and doesn't want any trouble. H+ is a different story, though.

Huh, I have a feeling your fallacy is that you're for some weird reason completely missing acid in acid-base reaction - completely not taking into account what acid "has to say" in reaction. If acid is significantly stronger then NH4+ then NH3 will be mostly protonated and protonation is gonna have negative Gibbs free energy.

Nitrogen is more electronegative than hydrogen hence N will pull electrons towards itself. Hence not a good candidate to donate electrons!

Yet it makes bonds with hydrogen, as do other electronegative elements (think of hydrogen chloride and water). So polar bonds form in all kind of molecules.

lets take the case of Oxygen O=O, here each Oxygen atom has 2 lone pairs, and going by the above logic these lone pairs should also participate in bond formation and create an Oxygen polymer(!!)

The concept of stability makes sense when comparing two states with the same number of atoms (e.g. ammonia and hydrogen ion vs. ammonium ion). For elemental oxygen to polymerize, each $\ce{O2}$ molecule would bring in electrons. For a molecule to act as base, a lone pair turns into a bond with hydrogen (where hydrogen provides zero electrons and the base provides two). This does not "break the octet" of the base, and give the hydrogen access to two electrons (completing its duet).

Hence with the above points I still could not see why Nitrogen has to accept proton

It does not have to. It is a weak base. If you add ammonia to neutral water, most of it remains undissociated (the pH goes up, though, because some of it does dissociate).

[devils advocate question:] How can I use electronegativity to predict something about acid/base chemistry?

The trick is to compare green apples with red apples, not oranges with bananas. The easiest comparison is isoelectronic species, such as:
$$\ce{HF vs OH-}$$
Hydroxide is a better base than hydrogen fluoride.
$$\ce{H2O vs NH2-}$$
Azanide (had to look up the name) is a better base than water.
$$\ce{H3O+ vs NH3 vs CH3-}$$
The methyl anion is a better base than ammmonia, which is a better base than hydronium.
Or you can compare compounds by exchanging O for NH or CH2.
$$\ce{H2O vs NH3 vs CH4}$$
$$\ce{CH3-OH vs CH3-NH2 vs CH3-CH3}$$
$$\ce{CH2=O vs CH2=NH vs CH2=CH2}$$
This is more complicated because the carbon atoms lack a lone pair while nitrogen and oxygen have one. So you can compare nitrogen and oxygen using an electronegativity argument, but have to exclude carbon because it can't act as a base without the lone pair (well, $\ce{CH5+}$ does sort of exist but is called a super-acid, so $\ce{CH4}$ is hardly a base).
Hydrides
Here is a set of examples from OpenStax Chemistry 2e showing some of the trends across the periodic table. Again, $\ce{CH4 and SiH4}$ don't have lone pairs, so they can't be bases, even though the trend suggests they might be the strongest base in the series. The reason that water is described as neutral is not some special property of water but because the pKa is typically defined in aqueous solution, and a neutral pH references the hydroxide and hydronium concentration in pure water. If liquid ammonia were our solvent of choice, that would be the "neutral" case.

"I  am trying to understand why Nitrogen in Ammonia is a proton acceptor."
I am going to modify your question slightly: it is a proton acceptor in aqueous solution.  Water, being the solvent of life and being very common on Earth, is generally the most pertinent solvent.  Ammonia is a proton acceptor, or if you prefer an electron-pair donor, in water.
"What about the oxygen in water?" you ask.  It too, is a proton acceptor, having two exposed electron pairs.  In water solution, it is a weak acid, generating a hydrogen ion concentration of 10^-7.  Water is a very strong base, or proton acceptor, in sulfuric acid solution. So strong that there is a strict rule against adding water to any strong base.
There is plenty of internet material on  Classical (or Arrhenius), on Bronsted-Lowry; and   on Lewis acids and bases.  Also quite a bit on autoprotolysis (or autoionization) of water and other solvents.  Khan Academy has clear, comprehensible writing on these subjects.


text:
Method of separating product and conserving the catalyst
I'm looking for a way to separate reaction product of iodobenzene with ethyl 2-methyl-3-oxopentanoate in 1,4-dioxane (toluene is an alternative as a solvent). The catalyst being used is $\ce{Pd2(DBA)3}$ with $\ce{Ph3P}$ as a ligand and $\ce{K3PO4}$ as a base.
I was thinking about fractional distillation under partial vacuum because there's a risk of explosion if there's some leftover unreacted iodobenzene heated to high temperatures.
Provided the liquids distil off, the reaction flask should be left with some solid junk containing $\ce{[Ph3P + Pd2(DBA)3]}$ and $\ce{K3PO4}$. Then this mixture could be washed with water to get rid of $\ce{K3PO4}$, $\ce{KI}$, water-soluble side products and dry the catalyst + ligand under vacuum and use it again in another run.
So, is it possible for me to separate the product from the liquids using distillation and to reuse catalyst with the procedure I proposed?

I don't see any problem with the procedure you describe for re-isolating the Pd-containing stuff, but I would question how worthwhile it is. I'd imagine by the end of the reaction the majority of the Pd is in the form of palladium black, and no longer catalytically active.

On what scale is this?

using 1kg of iodobenzene would mean 50g of palladium catalyst is needed.. given the price for 1g is 20$, this synthesis gets quite expensive, so is there any way how to recycle the catalyst?

Could try using Cu as catalyst as here:  https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201106719

What is your final target? If cost is a major consideration then starting with ethylphenylacetate (just over £60/kg) may be a better route.

I'm in a hurry now, so I'll get back to your suggestion about ethylphenylacetate later, but I've found an article using Cs2CO3 (which would be the limiting factor for me in this reaction as it consumes a 4 eq).
https://www.organic-chemistry.org/abstracts/lit1/009.shtm - is it recoverable?

The answer is different depending on whether this is for a commercial process, a custom synthesis or proof of concept total synthesis confined to the lab.


text:
Chemical equation for the reaction of Allura Red AC with NaClO
I am trying to figure out the chemical formula for the reaction between Allura Red AC and $\ce{NaClO}$ (bleach). I have never met such reactions before in my study, and I was unable or maybe looked in the wrong places, to find the chemical equation for the reaction. The closest thing I have found is an image with the reaction but without the chemical equation. Could somebody please help me figure out the chemical equation for the reaction in the image?

Thanks for the help in advance.

The structure of Allura Red AC appears to be incorrect. The methyl group should be para to the methoxy group in the benzene ring.

And the products are a problem. The nitrogen atoms are each missing a valence, and filling this in with a hydrogen atom is not consistent with an oxidation reaction.

@Maurice has pointed out the shortcomings of the proposed colorless products from the NaOCl oxidation of Allura Red AC (1, including the red atoms). A study has been reported[1] on the kinetics of the oxidation of Sunset Yellow (1, without the red atoms) with stoichiometric Chloramine-T (sodium N-chloro-4-methylbenzenesulfonamide) in alkali in the presence of catalytic Os(VIII). The structural similarities of the two dyes suggest that the products of oxidation are similar in the two cases. The authors were able to identify, after acidification, the sulfonic acid of o-naphthoquinone 3, benzenesulfonic acid 5 and nitrogen. The products from the oxidation would be expected to be the "red" analog of 5 and o-naphthoquinone 3.
The decomposition of 2 is effected by alkaline hydrolysis and elimination of the chlorohydrazone functionality. The authors propose a similar mechanism but they do not invoke a role for Os(VIII). It is likely that in their case, Os(VIII) is the oxidant and Chloramine-T is the regenerative reagent. Nonetheless, the Chloramine-T mechanism would apply to NaOCl. The OP's post claims that the products of oxidation are colorless. o-Naphthoquinone is yellow.

K. N. Vinod, Puttaswamy and K. N. Ninge Gowda, Inorganic Chimica Acta, 2009, 362, 2044.

The two molecules created by breaking the $\ce{N=N}$ bond have something wrong. They both have a final bond $\ce{-N-OH}$. This is impossible. It may be at least $\ce{-NH-OH}$. Maybe the $\ce{N2}$ group is removed, and a phenol or naphthol group is formed.


text:
Finding the central atom of N2O
Why do all structures of $\ce N_2\ce O$ have Nitrogen as their central atom? Can we not draw a lewis structure with Oxygen as the central atom with 2 double bonds to the two nitrogens? This will still satisfy the octets of each atom.

Unhelpful answer: that's just how it is, really. And no, oxygen can't form two double bonds.

I guess we could draw a Lewis structure with O as the central atom that would satisfy all octets. The reason we don't do so is simply because the real-world molecule is not like that.

An all-octet Lewis structure for $\ce{NON}$ would look like this:
$$\ce{\overset{-}{N}=\overset{2+}{O}=\overset{-}{N}}$$
That’s all atoms bearing formal charges, the charges applied contrary to electronegativity (oxygen having the positive formal charge), and oxygen carrying a double formal charge. This violates practically every single rule of thumb that students get to decide whether structures are stable.
Compare to $\ce{NNO}$:
$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O}$$
This is a much more conventional structure. There is only one pair of formal charges and the positive charge never goes on the oxygen. Furthermore, you can draw resonance structures which typically provide additional stabilistion (no significant resonance structures are possible in the $\ce{NON}$ case). It can be expected that the left structure contributes slightly more to the overall picture by virtue of the oxygen being formally negatively charged.
Thus, even without knowing a thing about the real-world molecule the central oxygen atom does not lend itself well to any axioms about favourable structures.


text:
How are London Dispersion Forces generated?
While talking about the gaseous state of matter we came to the topic of London Dispersion Forces caused by the generation of a dipole in one atom which induces a dipole in another. While talking about the cause of such interactions I stated that, since Schrödinger's equation for multi-electron atoms is time-dependent, at some moment a dipole may get generated (due to the increase in the probability of electrons in that region). Though now I am thinking that what I said was quite vague and may not be the reason for it.
So what is the reason for the generation of dipole moment in one atom to cause London forces?

Let me crash the party here.
TL;DR: The classical explanation of induced dipole attractions from electron densities "evading" each other does, by itself, not adequately or intuitively explain the actual charge density patterns that arise in these situations.
(I know that this must seem like an outrageous statement, and surely downvote fingers are itching now. Bear with me.)

Based on the Hellmann-Feynman theorem, it is known that the forces acting on a nucleus arise from two purely coulombic sources: Its attraction to its surrounding electron distribution, and its repulsion with other nuclei. Hence, the observation that e.g. the two atoms in a rare gas dimer are attracted to each other immediately implies that there is a concentration of electron density in between the nuclei, so that the resulting net forces pulls them "inwards" towards each other. As Feynman put it himself in 1939 (emphasis in the original):

... the charge distribution of each is distorted from central symmetry, a dipole moment of order $1/R^{7}$ being induced in each atom. The negative charge distribution of each atom has its center of gravity moved slightly toward the other.  It is not the interaction of these dipoles which leads to van der Waals' force, but rather the attraction of each nucleus for the distorted charge distribution of its own electrons that gives the attractive $1/R^{7}$ force.

This is the complete opposite picture of the momentary effects in the induced-dipole explanation, where the electron densities "evade" each other by simultaneous displacements in the same direction to create attractive dipole interactions. The Feynman explanation is not very popular, but there are in fact some authors who have picked it up and commented on it, most notably perhaps Richard Bader (of Atoms-In-Molecules fame). Politzer and Murray have written a nice article on the topic.

But, is Feynman actually correct? If so, then we should be able to observe some accumulation of charge density between two neutral atoms that are bound by dispersion interactions, right? Indeed we do. The below image comes from a publication by Thonhauser et al. and shows the difference in charge density that arises when the dispersion interaction between to Ar atoms is "switched on" (highlight mine):

So, this may hold true for atoms, but what about entire molecules? Luckily, Hunt has shown in a very laborious 1990 paper that Feynman's picture holds true in that case as well:

... Feynman's statement concerning forces between well-separated atoms in S states generalizes to interacting molecules A and B of arbitrary symmetry (with A and B in their ground electronic states). To leading order, the dispersion force on each nucleus I in molecule A results from the attraction of I to the dispersion-induced change in polarization of the electronic charge cloud on molecule A itself.

Obviously, the "inwards polarization" effect must seem counter-intuitive at first. Why would the negatively charged electron clouds actually want to approach instead of evade each other? Thankfully, a straightforward rationalization of this effect for a rare gas dimer comes from another paper by Clark, Murray and Politzer:

What causes the polarization of the electronic densities toward each other (...)? This can be easily understood when it is noted that the electrostatic potential produced by the nucleus and electrons of any free ground-state atom is positive everywhere; the effect of the nucleus dominates over that of the dispersed electrons. This positive potential is what each atom “sees” of the other atom (...).

Of course, the astute reader may also voice another point of protest: "The fluctuating dipoles are variable in time, whereas Feynman's deformed charge density explanation is entirely static. How do we even compare the two? And what is the effect of the fluctuating dipoles when averaged over time?"
As it turns out, the two explanations are apparently consistent with each other, as detailed by several authors; Hunt herself in the paper mentioned above acknowledges fluctuating dipoles as a possible starting point, and a paper by Clark dedicates a full paragraph to the seeming dichotomy of the two pictures. At its core, however, the suggestive "electron-evading" nature of this explanation is very much misleading in light of the observable static effects of the "inwards" charge redistributions -- which, again, are actually required to create the resulting attractve interactions.

It requires some thought to follow your argument but it certainly shouldn't be down voted.  My impression of VDW forces is simply on close contact the attraction of the nuclei for electrons on the "bond axis" is slightly greater that the repulsion of the perturbed outer electrons.  These slightly inelastic collisions might also account for blackbody radiation.

@jimchmst Thanks for your comment. Yes, the nuclei primarily follow the electron density, which directly implies its accumulation between the fragments. I've added an image that demonstrates this. -- After some more literature research I also have to concede that, in the limit of full mathematical rigour, the fluctuating-dipole picture is apparently consistent with the "Feynman" one. Still, I consider its simultaneous electric dipole attractions to be utterly misleading, and it also doesn't offer any straightforward way to deduce the observable static charge shifts that ultimately arise.

You are correct that simply stating that the Schrodinger equation is time
dependent does not really answer the question.
In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions.
It's the same idea as any of the other intermolecular forces (coulomb, dipole-dipole, etc.) where some interaction between particles leads to a lower energy state than the free particles.
Generally, the London Dispersion interaction potential is not derived in a general chemistry class because it is purely quantum mechanical and rather involved. These slides provide a clear and thorough (albeit advanced) derivation, but I will explain the main points below.
Derivation Explanation
Let's imagine for this model that a polarizable atoms can be thought of as a simple harmonic oscillator with a positive charge ($+q$) on one end of a spring and a negative charge ($-q$) on the other end. The spring has an equilibrium length of 0, so that the atom has zero average dipole.
This model of an polarizable atom captures the physics we want because applying a force of some kind to the spring will induce a dipole in this atom  ($\vec{p} = q \vec{d}$).
So far we have only considered one of these atoms. What if we move to consider multiple of them together?
Remember we're really interested in the energy difference between two non-interacting atoms and two interacting atoms, so let's start by finding the non-interacting ground state energy of two atoms. We'll let the two atoms have two different spring frequencies $\omega_1$ and $\omega_2$ respectively.
Setting $z_1$ and $z_2$
as the displacements of the springs from equilibrium for atoms 1 and 2 respectively and $m$ as the mass of an atom,
we can write the total potential energy of the system as:
$$
U_{\text{non-interact}} = \frac{1}{2}\left(m \omega_1^2z_1^2 + m \omega_2^2z_2^2\right)
$$
Solving the Schrodinger equation for this potential yields the ground state energy for the two non-interacting harmonic oscillators:
$$
E = \frac{\hbar}{2} (\omega_1 + \omega_2)
$$
Now, we turn on the interaction and the problem becomes harder more interesting because the positions of the springs affect the magnitude of the interaction. Physically, because the dipole-dipole interaction can lower the energy, we expect that the interacting ground state of the two atoms will be a balance between minimizing the stretching of the springs (which increases the energy) and maximizing the amount of dipole-dipole interaction (which lowers the energy).
Setting $z$ as the distance between the two atoms, with some algebra one can show that the interaction energy between these two dipoles is given by
$$
U_{\text{interact}}(z,z_1,z_2) = \frac{q^2}{2\pi\epsilon_0}\frac{z_1 z_2}{z^3}
$$
Now, our total potential energy for the system is given by:
$$
U_{\text{total}}(z,z_1,z_2) = U_{\text{non-interact}}(z_1,z_2) + U_{\text{interact}}(z,z_1,z_2)
$$
While intimidating looking, this potential can also be solved exactly by changing variables to $z_+ = (z_1 + z_2)$ and $z_- = (z_1 - z_2$) and rearranging terms to make it look like the non-interacting potential energy. Once that is done the ground state energy can be found just as easily as in the non-interacting case.
With this done, we can find that the (one dimensional) energy stabilization is given by:
$$
U(z) = \frac{-\alpha^2 I}{2}\frac{q^2}{(4\pi\epsilon_0)^2}\frac{1}{z^6}
$$
Where the final step was to replace some of our less measurable parameters with their atomic counterparts: polarizability ($\alpha$) and ionization potential ($I$)


text:
How dangerous is the Mercury-Thallium filled thermometer?

What's the amount of Mercury and Thallium in a typical low-temperature thermalmeter?
How poisonous is Thallium when bounded with Mercury?

The lowest eutectic in the Hg-Tl system occurs at 8.6 at.% Tl at 213K (see W. Gierlotka et al., CALPHAD 30425-430 (2006)). This is about 21K lower than the freezing point of pure Hg.
For toxicity, refer to an SDS (MSDS) or ask you local ES&H or industrial hygiene professional.


text:
About the use of pascal or atmosphere in chemistry
I was wondering if there are specific formulae in chemistry for which it is "compulsory" or convenient to use atmosphere (atm) rather than pascal (or even mmHg) as a unit of measurement.
Of course, we can convert one into the other, but the choice of using one of the two also conditions other units, such as in the case of volume.
Can someone provide me with a good picture of the relationship between these units and the others that we may need to use in the same context? (e.g. we use L  with atm; $\mathrm{m^3}$ with Pa).

Do you mean something like this? https://en.wikipedia.org/wiki/International_System_of_Units

@J.Ari Unfortunately in your site there's no reference to atm as a unit. I perfectly know that the SI unit of pressure is the pascal, but I'm asking something different, like hints or references to the main scenarios where to use preferably the one or the other, and how other units modify if I choose one of the two

Chemistry, and other sciences in general, have been trying for years to switch to the International System of Units exclusively. Older books don't adhere to the convention of course. So the older books may use pressure units such as torr, atmospheres, bar and psi.

I first started studying chemistry in 1968. All heat analysis was done with the calorie. I did so many problems with calorie that I think calories. I can work the problem out in Joules, but I don't really have a "feeling" for joules like I do calories.

@Shootforthemoon I am struggling to understand your question, mostly because I've been thinking of units my way for about 10 years and am a little stuck in my understanding. Is there a specific situation/problem that has prompted this question? Maybe that will be easier to answer lol

@J.Ari yes, there is; actually my doubt came up from more than one situation. For example, standard pressure for chemical reactions is usually stated in atmospheres, as 1 atm, and in school we also used atmospheres to calculate the equilibrium constant for a gaseous homogeneous reversible reaction. In the ideal gas law pV=nRT, we were taught to use atmospheres. Sometimes, instead, like for thermodynamic transformations, I see that my book uses Pascal.

@J.Ari I see that the two units are of course interconvertible, but is there some convention for which for instance pressure is usually in Pa in thermodynamics, while in atm for other fields? If not, what should I consider when using one of the two, since I saw that P in atm, when combined with Volume in liters, is equal to P in Pa with volume in m^3 (if I'm not wrong)? And what if I used mmHg? Hope to clarify a little my question, sorry for being so contorted lol

what should one take into account when choosing his pressure unit?

Well when I learned pV=nRT, i was taught pressure was in torr. If you look up the ideal gas constant, R, you can see that it has many values depending on the units. Note that even with all the values listed temperature can be measured in grad, time in fortnights, and length in leagues. So there are a lot of other values "missing" in the table.  Having countless values is exactly the situation that the International System of Units tries to avoid.

Are you thinking something like the cigs system? There are lot of inconsistencies between fields, but few has good arguments to do it

Choose either the unit you are asked for (as in hwk or exam) or what is useful. For instance if you want to compare pressures under different circumstances then you want to use the same units. If the pressure is large you might use kPa, if small, mTorr, etc. You may seek to avoid dealing with large multipliers (ie 3 bar rather than $3 \times 10^5$ Pa.

See also https://en.wikipedia.org/wiki/Atmosphere_(unit)

@Greg Yeah, exactly

Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units.
For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation
$$v/(\mathrm{km/h})=3.6\frac {l/\mathrm m}{t/\mathrm s}$$ or $$\left\{v\right\}_{\mathrm{km/h}}=3.6\frac{\left\{l\right\}_{\mathrm m}}{\left\{t\right\}_{\mathrm s}}$$
you should write “speed $v$ is the quotient of distance $l$ by time $t$”, preferably in form of the quantity equation
$$\displaystyle v=\frac lt$$
Generally, you can work with any units you like; however, this requires the use of conversion factors. The use of conversion factors is never required when only coherent SI units are used; therefore, the use of coherent SI units (the SI base units and the SI coherent derived units, including those with special names) are preferred.
In order to avoid large or small numerical values, decimal multiples and submultiples of units may be formed with the SI prefixes. However, when prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor in the expression for the unit in terms of the base units.
Furthermore, various non-SI units are accepted for use.
In the International System of Quantities, pressure $p$ is defined as
$$p=\frac{\mathrm dF}{\mathrm dA}$$
Where $\mathrm dF$ is the force component perpendicular to the surface element of area $\mathrm dA$.
The coherent derived units in the SI for the pressure has the special name pascal and the symbol $\mathrm{Pa}$. According to the definition of pressure, the pascal can be expressed in terms of other SI units as $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. The pascal can also be expressed in terms of SI base units as $1\ \mathrm{Pa}=1\ \mathrm{m^{-1}\ kg\ s^{-2}}$. The last form is especially useful to check whether units in equations and calculations are converted correctly.
The SI prefixes may be used with any of the names and symbols, but when this is done the resulting unit will no longer be coherent.
The bar (symbol $\mathrm{bar}$) is included in the list of non-SI units that are used by special interest groups for a variety of different reasons. Although the use of SI units is to be preferred for reasons already emphasized, authors who see a particular advantage in using these non-SI units should have the freedom to use the units that they consider to be best suited to their purpose. In particular, the bar is included because since 1982 one bar has been used as the standard pressure for tabulating all thermodynamic data. (Prior to 1982 the standard pressure used to be the standard atmosphere, equal to $1.01325\ \mathrm{bar}$, or $101\,325\ \mathrm{Pa}$.)
The following other non-SI units can be used, which requires the use of conversion factors. However, the use of these units is deprecated.
$$\begin{array}{lll}
\hline
\text{Unit} & \text{Unit} & \text{Conversion} \\
\text{name} & \text{symbol} & \text{factor} \\
\hline
\text{standard atmosphere} & \mathrm{atm} & 1\ \mathrm{atm} := 101\,325\ \mathrm{Pa}\\
\text{kilogram-force per square metre} & \mathrm{kgf/m^2} & 1\ \mathrm{kgf/m^2} = 9.80665\ \mathrm{Pa}\\
\text{technical atmosphere} & \mathrm{at} & 1\ \mathrm{at}=1\ \mathrm{kgf/cm^2} = 98\,066.5\ \mathrm{Pa}\\
\text{conventional millimetre of water} & \mathrm{mmH_2O} & 1\ \mathrm{mmH_2O}:=10^{-4}\ \mathrm{at} = 9.806 65\ \mathrm{Pa}\\
\text{conventional millimetre of mercury} & \mathrm{mmHg} & 1\ \mathrm{mmHg} \approx 13.5951\ \mathrm{mmH_2O} \approx 133.3224\ \mathrm{Pa} \\
\text{torr} & \text{Torr} & 1\ \mathrm{Torr}:=\frac1{760}\ \mathrm{atm} \approx 1\ \mathrm{mmHg} \approx 133.3224\ \mathrm{Pa} \\
\text{pound-force per square inch} & \text{psi}, \mathrm{lbf/in^2} & 1\ \mathrm{lbf/in^2}\approx6\,894.757\ \mathrm{Pa} \\
\hline
\end{array}$$
Note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted.
For example, write “the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”.
(The symbol $p_\mathrm e$ is recommended for gauge pressure, defined $p-p_\mathrm{amb}$, where $p_\mathrm{amb}$ is the ambient pressure. Thus the gauge pressure is positive or negative depending on whether $p$ is larger or smaller than $p_\mathrm{amb}$.)

Choose either the unit you are asked for (as in hwk or exam) or what is useful. For instance if you want to compare pressures use the same units. If the pressures are large you might use kPa, if small, mTorr, etc. You may seek to avoid dealing with large multipliers (ie 3 bar rather than $3 \times 10^5$ Pa).
Some equations in physics imply use of a particular system of units e.g. in electromagnetic theory, where equations can change depending on whether SI or Gaussian units are used. However, I am not aware of a similar situation where equations change depending on the selection of particular pressure units (perhaps when using Planck units in cosmology?). Of course the numerical value of the gas constant R depends on the units in which it is expressed, but it remains constant. Also, you'll find that most common pressure units can be related to the SI unit system by fixed constant factors (or ratio). The exception might be mm Hg (I now understand that both 1 Torr and 1 mm Hg are fixed quantities, see eg this post, but not exactly equal).
Basically different units are more convenient in different contexts depending on the magnitude of what you are discussing, but the same rule always applies: be  meticulous and consistent. This means simply that you keep track of the units in use, and perform conversions as required when evaluating the result of an expression.

In the formula Delta G° = - RT lnK, the numerical value of the equilibrium constant K must be in atmospheres.

I'm afraid this is not correct and addresses the question only tangentially.

I’m afraid this is not correct although if it were it would be a full answer to the question.


text:
How does the sec-pentanyl group look?
I'm starting the very basics of hydrocarbons and I'm having a hard time at understanding when to use the common/trivial name prefixes (sec-, iso-, neo-, etc.).
One particular problem is the difference between iso and sec.
I seem to get a grasp of when iso is used. But I'm having a difficult time with sec.
It's clear with butane, for example:

secondary butyl group: $\ce{CH3-C(\bond{...})H-CH2-CH3}$
iso-butyl group: $\ce{CH3-C(CH3)H-CH2\bond{...}}$

However, when it comes to pentane, how can I find the secondary pentyl group? The isopentyl one is clear: $\ce{CH3-C(CH3)H-CH2-CH2\bond{...}}$.
But with secondary pentyl group (of normal pentane, $\ce{CH3-CH2-CH2-CH2-CH3}$), which one is the secondary? Does the common naming fail here and we need the IUPAC system or am I missing something?
If someone can sum up when to use the prefixes as well, I'd be very grateful.

Related: When to use iso and sec while naming compounds in alkyl halides?

Summing up: Don't use them. They are ambiguous, use the proper IUPAC recommendations.

See, let me explain iso means a type of isomer of the existing carbon chain for eg when you say iso pentane the srtucture would be CH3--C(CH3)H--CH2--CH3  as you can see from here that in the normal pentane all the carbon atoms come in a straight long chain without branching. iso- i.e. isomer having same molecular formula and different structural formula. Here the structure that i have made is an isomeric form of pentane.
And when we talk about sec it means secondary carbon atoms carbon atom joined with two other caron atom eg CH3-CH2-CH3
                                       |
The carbon in the center is a sec carbon at since it is attached with 2 other carbon atom if it was attached with only one carbon atom like the first C atom then it would be called a primary carbon atom. Again if you look at the first structure i.e. iso pentane you would notice that the 2nd carbon is attached with 3other carbon atoms which is called a tertiary carbon atom.
Again on coming to neo pentane if one carbon is surrounded by 4 other ch3 it becomes neo pentane again different structure of the same molecule.
Iso,neo are sort of convention created to differentiate between different isomer of the same molecule. Draw the structure and identify each Carbon and the number of carbon they are attached with slowly and steadily you would come to know.
I have tried to explain this but i would suggest you to first read from a chemistry book or use any reference or ask a teacher or you may even use youtube to know about it and then again read this for better understanding.
Happy learning

The prefix iso- does hold a very specific meaning, even though it is no longer recommended and its usage is basically deprecated. Similarly the sec- prefix has a specific meaning and does not mean secondary carbon.


text:
Counterion exchange
I have a quick question about counterion exchange. I have this very practical problem at my hands that I have a ethyl sulfate $(\ce{EtSO4-})$ counterion for a $\ce{R4N+}$ moiety ($\ce{R}$ = 3 different substituents).
I need to change the counterion to something like a $\ce{BF4-}$ or a $\ce{NO3-}$. What would be the best approach? What role does the solubility of my $\ce{R4N+}$-moiety play?
(good solubility in: $\ce{EtOH},$ $\ce{MeOH},$ $\ce{DMSO};$ medium in: water, $\ce{CH3CN},$ choloroform; insoluble in: $\ce{EtOAc})$
I thought about a classical silver-halide approach where I introduce a silver salt with the fitting counter-ion, but I am afraid that the $``\ce{AgEtSO4}"$ might be too soluble and that I get silver contamination. ($\ce{Ag2SO4},$ for example, has a solubility product of $10^{-5},$ whereas $\ce{AgBr}$ has $10^{-13},$ so there is a big difference.)
Any tips are appreciated!

Look up ion-exchange resins and papers on ionic liquids. They do a lot of exchange of ions.


text:
Can I make a buffer with a weak acid without adding it's conjugated base, when the desired pH of the solution is exactly the pKa of the weak acid?
An acid will be 50% dissociated when the solution it's in have a pH equal to the acid pKa ('afirmation 1', for later reference).
We also know that a good buffer solution have equal concentrations of an acid and it's conjugated base.
That's why to make a good buffer we look for an acid that have a pKa close to the pH we want to maintain.
So, suppose I have a solution with a pH 6.2, and I want to add a buffer to maintain this pH at 6.2.
Suppose also, that I have a ideal weak acid for this situation: with a pKa of exactly 6.2.
Using the Henderson–Hasselbalch equation, we know that we need to have the same concentration of the acid and it's conjugated base in the buffer. To add the conjugated base, we use a salt of the refereed acid.
So my question is: If we just throw the acid at the solution without the salt, since the solution pH is exactly the pKa of the acid, the acid will dissociate 50% (afirmation 1), and we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.
So, in principle, it's possible to make a buffer without adding the salt?
Is this logic correct? If not, what error I'm making?
Thanks

For a monoprotic acid, HA, there are two variables. First the pKa and second the concentration. There must be a function of the two which would yield a solution with 50% HA and 50% A-.

The acid alone solution will have pH approximately 3.1 - 0.5.log(c). Why do you think adding it to solution with pH 6.2 would help you to keep  pH 6.2 ?

So your solution is already at a pH equal to the pKa of the weak acid you are about to add. Your hunch is that the pH remains the same after adding acid to a solution.

[...] we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

It depends on the concentration of what is already in solution and of what you are adding. If you add a weak acid at very low concentration (e.g. an indicator) to a buffered solution, the indicator's protonation state will be governed by the pH of the solution. On the other hand, if you add a weak acid at very high concentration (e.g. a substance that is supposed to act as buffer) to an unbuffered or hardly buffered solution, the buffer will govern the pH.
In your case, the pH will be off (too acidic), and most of the buffer substance will be in the acid (protonated) state. You would have to adjust the pH by adding a strong base to the amount that deprotonates half of the just added acid to get the pH to match the pKa again.

So, in principle, it's possible to make a buffer without adding the salt?

Yes, but not in the scenario you suggested. Most buffers are made by starting with a weak base (pH is much higher than pKa of conjugate weak acid) and adjusting the pH with strong acid, or by starting with a weak acid (pH is much lower than pKa of the weak acid) and adjusting the pH with a strong base. In both cases, you effectively get a mixture with the "salt", i.e. you introduce some spectator ions (often chloride or sodium ions).

So, in principle, it's possible to make a buffer without adding the salt?

Yes and no....
The math is pretty simple. Let's say that there is a monoprotic acid $\ce{HA}$.
$$\ce{HA <=> H^+ + A-}$$
$${\mathrm{K}_\mathrm{a}} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}$$
let $c$ represent the overall concentration of HA species. If $$\ce{[A-] = [HA] }$$
then
$$\ce{[A-] = [HA] = \frac{1}{2}}c$$
and if we neglect the self-hydrolysis of water then
$$\ce{[H+] = \frac{1}{2}}c$$
So
$$\mathrm{K}_\mathrm{a} = \frac{1}{2}c$$
or
$$c = 2\mathrm{K}_\mathrm{a}$$.
However to be a buffer there must be some buffer capacity. For an acid with a pKa = 6.2 we couldn't ignore the autoionization of water, so the calculation would be a bit more complicated. So although there would be a concentration of acid  which would yield a solution with a pH of 6.2, the buffer capacity would be so low that the solution couldn't really be considered to be a buffer.


text:
Why is the solution of (NH4)2[Ce(NO3)6] used as an oxidizing agent?
In an MCQ, the correct statement regarding the abovementioned compound is that its solution is used as an oxidizing agent.
I do not understand why that is so.

calculate the oxidation state of Ce in the compound.

I think it's +4. Which means it has achieved Xe configuration.

Most lanthanides display stability in their +3 oxidation states. Cerium does the same, which is why it has a very high reduction potential. The reduction potential for the $\ce{Ce^{4+}|Ce^{3+}}$ couple is 1.61 V (for comparison, $\ce{Cl2}$ has a potential of 1.32 V)
Cerium in the form of ceric ammonium nitrate is also quite stable compared to other oxidizing agents and can be stored on the shelf for a longish period of time. It's also quite usable: the bright orange colour is decolourized to a pale yellow on reduction of $\ce{Ce^{4+}}$. This is why it's used as a common oxidizing agent.


text:
Does the Boltzmann distribution hold for supercritical fluids?
Above the critical temperature of a real gas, if we compress it, to whatever extent, it doesn't get liquefied (no gas liquid interface). However, it does behave like liquids , esp. in case of very high pressures (upto 300 atm) it's physical properties are almost like liquids due to decreased intermolecular distance and increased intermolecular forces, yet is not a liquid but condensed vapors. Now , the Boltzmann distribution assumes negligible intermolecular forces so that the total energy of the system consists of only individual kinetic energies of the molecules, $E_{\text {total}} = \Sigma_i K_i$. So, in the aforementioned case of supercritical fluids, does the Boltzmann distribution still hold true, with only an increased number of energy states , including the intermolecular potential energies, i.e. $E_{\text {total}} = \Sigma_i K_i +\Sigma_{i≠j} V_{ij}$? Or will we have to search for another statistical distribution? Any help/suggestions please.


text:
Acidic strength order in gaseous phase
I got this question in my exam:

Arrange the following in increasing order of acidic strength in gaseous phase
(i) $\ce{CH3CH2COOH}$ (ii) $\ce{CH3COOH}$

I felt that increasing order must be (i) $<$ (ii). But answer given is (ii) $<$ (i). Why will the order change in gaseous phase?

Just guessing that tendency of creation of a dimer may be stronger at acetic acid.

It's in the gaseous state: IMO dimerization is more likely in aqueous medium, where the molecules are much closer to each other.

@AniruddhaDeb In aqueous medium, water competes as hydrogen bonding partner (and is present at higher concentration). In the gas phase, the only available hydrogen bonding partner is another acetic acid molecule.

@Aniruddha Deb The acid dimerization is typical for the gaseous state and nonpolar solvents like benzene. As the result, Acetic acid distribution constant for benzen/water is highly dependent on concentration, as higher concentration of acid in benzene supports the dimerization. ( I remember doing it in past as a physical chemistry lab experiment )

Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists.
In aqueous solution, $\mathrm{p}K_{\mathrm{a}(\ce{CH3CO2H})} = 4.76$ while  $\mathrm{p}K_{\mathrm{a}(\ce{CH3CH2CO2H})} = 4.87$, thus acidity of acetic acid is greater than that of propanoic acid. Keep in mind that solvent plays a huge role in $\mathrm{p}K_{\mathrm{a}}$ measurements. For example, compare $\mathrm{p}K_{\mathrm{a}}$ measurements in $\ce{H2O}$ and in $\ce{DMSO}$:
$$
\begin{array}{c|cc}
\hline
\ce{HA} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{H2O} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{DMSO}  \\
\hline
\ce{CH3CO2H} & 4.76 & 12.3  \\
\ce{H2O} & 15.7 & 31.2 \\
\ce{CH3OH} & 15.5 & 27.9 \\
\ce{C6H5OH} & 9.95 & 18.0 \\
\hline
\end{array}
$$
However, the proton affinity (PA) measurements of the conjugate bases of these acids in gas phase tell a different story. Proton affinity is a thermodynamic measurement conducted in the gas phase. These measurements give a pure view of base/acid strength, without complications from solvent effects such as salvation, thus take the solvent out of the equation.
Cardwell, et al. (Ref.1) has measured the proton affinities of various $\ce{R-CO2H}$ acids, including ethanoic and propanoic acids ($\ce{R}=\ce{CH3}$ and $\ce{CH3CH2}$, respectively). They conclude that:

The acidity increment (with the size of $\ce{R}$-group) is relatively small and can be attributed to the stabilizing effect of the polarizability of the alkyl group on the negative ion.

References:

Gary Caldwell, Richard Renneboog, Paul Kebarle, “Gas phase acidities of aliphatic carboxylic acids, based on measurements of proton transfer equilibria,” Canadian Journal of Chemistry 1989, 67(9), 611-618 (https://doi.org/10.1139/v89-092).
Also read: Diethard K. Bohme, Edward Lee-Ruff, L. Brewster Young, “Acidity order of selected Brønsted acids in the gas phase of $\pu{300  K}$,” Journal of American Chemical Society 1972, 94(15), 5153-5159 (https://doi.org/10.1021/ja00770a002) and references there in.

One possible explanation for this is hyperconjugation due to the attached alkyl chain, as shown below:

since the acetate ion has one more $\alpha$-H atom than the propionate ion, the acetate ion is more stable than the propionate ion in the gaseous state and has more tendency to donate the hydrogen atom, making it more acidic.

Please cite the source, so that I may believe it


text:
Predicting color of a TMB derivative affected by peroxidase
The ELISA test on HIV works by verifying the presence of peroxidase within samples. The verification process is done through the enzymatic reaction that said peroxidase has on TMB (3,3',5,5'-tetramethylbenzidine) in an environment that contains $\ce{H2O2}$. This reaction leaves the enzyme untouched and transforms the $\ce{H2O2}$ into $\ce{2H2O}$ by adding two hydrogen atoms taken from the TMB. This also transforms the TMB into another compound, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which has a blue color in that specific case.
Therefore my question is, how could one predict from the structure of the one product that it is going to have a blue color in that situation? I know that it probably comes from the conjugated double bonds but how do their count relate to the absorbed wavelength(s)?

Your question: How could one predict from the structure of the one product that it is going to have a blue color in that situation?

This is a very broad question. OP has to go to chemistry fundamentals to learn, how one can predict the absorbance wavelength of a compound by analyzing the structure of the given compound. One easy subject for this study for OP is basic fundamentals of UV-visible spectroscopy in organic chemistry. For example, how absorbance wavelength increase with each additional conjugated double bond (Absorption in the ultraviolet and visible regions is related to the transition of electrons). Look following table:
$$
\begin{array}{l|cr}
\text{Substance} & \lambda_\mathrm{max}, \ \pu{nm} & \epsilon, \ \pu{L mol-1 cm-1} \\\hline
\ce{CH2=CH2} & \pu{180nm} & 10000  \\
\ce{CH2=CH-CH=CH2} & \pu{217nm} & 21000  \\
\text{Vitamin A (5 conjugate double bonds)} & \pu{328nm} & 51000  \\
\text{Benzene} & \pu{255nm} & 180  \\
\text{Naphthalene} & \pu{286nm} & 360  \\
\text{Anthracene} & \pu{325nm} & 7100  \\
\hline
\end{array}\\
\epsilon = \text{Molar Absorption Coefficient}; \  \lambda_\mathrm{max} = \text{Wavelength of maximum absorption peak}
$$
Thus, one can suggest diimine, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which contains 5 directly conjugated double bonds (two of them are imines), may absorb wavelengths of lights way larger than $\pu{328  nm}$ given for Vitamin A. That's because it involves transition of $n \rightarrow \pi^*$ in addition to $\pi \rightarrow \pi^*$ in Vitamin A (e.g., $\pi \rightarrow \pi^*$ in acetone is $\pu{195 nm}$ in addition to $n \rightarrow \pi^*$ transition at $\pu{274 nm}$, when compared to $\pi \rightarrow \pi^*$ transition of ethylene at $\pu{180 nm}$). Experimental evidence has shown that 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine absorbs light at $\pu{450 nm}$, giving its characteristic complimentary color $\color{blue}{\text{Blue}}$ (Ref.1).
References:

P. D. Josephy, T. Eling, R. P. Mason, “The horseradish peroxidase-catalyzed oxidation of 3,5,3',5'-tetramethylbenzidine. Free radical and charge-transfer complex intermediates,” The Journal of Biological Chemistry 1982, 257(7), 3669-3675 (http://www.jbc.org/content/257/7/3669)(PDF).
For basic learning, read: Master Organic Chemistry.


text:
Solubility of Lead in water from leaded pipes in Roman aqueducts
Lead can undergo the following reactions when cold and hard water passes over it:
$$\ce{Pb^2+ + 2Cl- -> PbCl2}$$
$$\ce{Pb^2+ + SO4- -> PbSO4}$$
All these salts are sparingly soluble in water and have a tendency to stick to lead, rendering it passive and not allowing plumbosolvency to occur:
$$\ce{2Pb + 2H2O + O2 -> 2Pb(OH)2}$$
My question is:
How did lead dissolve in the water flowing through the aqueducts if most of it's salts are sparingly soluble?
I have considered erosion of the salt layer formed on the surface but according to Wikipedia, the aqueducts had a very gentle slope

Vitruvius recommends a low gradient of not less than 1 in 4800 for the channel, presumably to prevent damage to the structure through erosion and water pressure. This value agrees well with the measured gradients of surviving masonry aqueducts. The gradient of the Pont du Gard is only 34 cm per km, descending only 17 m vertically in its entire length of 50 km (31 mi):

Wikipedia also has the following to say regarding lead concentration in the water:

Where lead pipes were used, a continuous water-flow and the inevitable deposition of water-borne minerals within the pipes somewhat reduced the water's contamination by soluble lead. Nevertheless, the level of lead in this water was 100 times higher than in local spring waters.

Any insights would be much appreciated!

Let me give an extreme example of chemical analysis. It is an anecdote. Some analytical chemists were using a very sensitive technique to analyze metals in blood. It is called neutron activation analysis. What they saw was that a very small quantity of molybdenum (Mo) was present in small quantities persistently. Nobody expected that. It turned out after a lot of hard work that Mo was coming from the stainless steel syringe needle used for drawing blood. Lesson of the story: There is nothing which is perfectly insoluble in water. It all depends on the analytical technique. If you try to weigh a hair on a "weighing machine" it will tell you that the mass is zero kg, but if you use a fine analytical balance it will tell you that hair has a mass.
The same goes for lead pipes. Natural water has plenty of minerals, dissolved oxygen anions, cations and don't forget the microorganisms. The bacteria do a lot of chemical transformations in water. As a result, Pb would slowly leach into the water at parts per million or parts per billion levels. Don't assume that Romans were drinking lead salt solutions in high concentration ranges. Not only water lead was present in water, it was also present in alcoholic beverages.
Read more here Lead in ancient Rome’s city waters
Lead in water is still a serious problem in several areas of the modern world. It is not an ancient problem.

More anecdote :Admiralty brass ( 70 Cu, 29 Zn, 1 Sn ) is commonly used for heat exchanger tubes in many different waters. Usually a trace of arsenic is added to stop zinc from being  removed by the water ( dezincification ).


text:
Is this a plausible reaction mechanism? (H2SO4 cyclization after Friedel-Crafts)
in my undergraduate chemistry lab we synthesized the substituted benzene on the left in a Friedel-Crafts reaction.
As a next step we're supposed to reflux for 1.5h at 90°C with concentrated sulfuric acid and obtain the product on the right. A bicyclical product.
We have to come up with a plausible mechanism for the ring closure reaction and I am unsure that what I came up with makes sense or not.
Thank you!

edit: this is the FC reaction we did in the first step:

The Steps for this reaction are as follows:
To a Solution of AlCl3 in abs. DCM the following reagents were added:
-3-chloropropionic chloride.
-p-xylene.
The reaction mixture was stirred at rt for 1h.
Then added to a mixture of 1:1 conc. HCl and ice water.
Then extracted with DCM in a separatory funnel.
Washed twice with a sat. NaCl solution.
The organic phases were dried with MgSO4.
The DCM was then removed in the rotary evaporator.
A yellow-orange Solution was left at the end.
It was left to crystallize over night.

Now this is where the second part happens, the H2SO4-cyclisation reaction.
.......
Some comments suggested that there might be a hydrolyzed intermediate with a displaced chloride. Here is my take on how that would look like:

Are you sure you made the starting material shown by the F-C reaction? What went into it?

Yes, I'm reasonably sure that it's the product. We began with para-xylene, 3-Chloropropionyl chloride, AlCl3 as the catalyst. (DCM was the solvent)

OK, that looks to be the correct intermediate. It's just that the H2SO4 cyclisation conditions would be far more appropriate for cyclising a carboxylic acid. What solvent is the cyclisation done in?

H2SO4 has to be added to the starting material which is crystalline. After the reflux step, there's an aqueous workup and an extraction with ether.
Thank you very much for taking the time to answer my question.

That simply does not make sense according to me. If anything, you'd add another aryl to carbonyl.

My thought on that is that the path I chose is the kinetically favored one. Or to put it differently, intramolecular reactions are favored above intermolecular reactions. But maybe the bicyclical product isn't the only one being formed, maybe it's just the preferred one.

And if the aryl were to add to the carbonyl, it would be the double bond of the enolate acting as the nucleophile, right? But isn't it too bulky then?

No, the protonated ketone (bottom left) is a substrate for electrophilic attack by the second aryl. How strong is the sulfuric acid used in the cyclisation step? 98% cH2SO4?

You should investigate the Nazarov reaction.

There's obviously some detail missing here. Is there a work up after the FC reaction or is sulfuric acid simply added in? I'd this isn't 98% then water is going in too which potentially backs up the thoughts around a hydrolysed intermediate being cyclised. Also, to hit 90 degrees, the DCM has to be removed for standard batch reaction.

@Beerhunter There's a work up with a 1:1 mixture of HCl and water.  The DCM is removed with a rotary evaporator. It's just  the conc H2SO4 in the flasc at the beginning of the cyclization. (I've edited my post and added the details of the initial steps).

@Waylander I haven't started the cyclization step yet, so I'm not entirely sure what exact concentration the conc. H2SO4 has. Bit I think it's 98%.

I was thinking along the same lines as @Beerhunter that there could be displacement of the Cl with OH, to give a primary alcohol

I've edited the post with my 2 cents on how the mechanism would work if there were a hydrolyzed intermediate.

@Michael would have been interesting see if you ended up with oligomeric mess or product if you'd added more Lewis acid and a higher b.p. solvent instead of doing the work up.

I would propose the mechanism illustrated below, which would seem plausible in concentrated sulfuric acid:

I'm doubtful about the mechanism of your chloride elimination

your mechanism is in line with the Nazarov Cyclization, interesting!

I think an allyl chloride with an electron donating OH group conjugated in the 3 position could likely solvolyze in concentrated sulfuric acid. Then the electrophilic substitution is intramolecular, therefore favorable.

I had the reaction today and this was more or less the accepted mechanism. Thanks!


text:
Is Hoffman elimination anti or syn elimination?
I read an amazing PDF from nptel on elimination reactions...but I was really confused when it used used $\ce{^+NMe3}$ as a leaving group in the reaction pathway via:

$\mathrm{Ei}$ (syn-elimination)
$\mathrm{E2}$ (anti-elimination)
And also, in an illustration for $\mathrm{E1_{cb}}$.

My organic chemistry teacher also did syn-elimination but in the problem solving book and on various online sites it took the $\mathrm{E2}$ elimination pathway.

Plenty of other online resources to answer this question. It is E2. You usually see a neutral leaving group leave a an anion. Here it is a positively-charged group leaving as a neutral species. BTW, the Cope elimination is syn.

Yeah copes elimination is syn but I said n+me3 as a leaving group...which is Hoffmann elimination. And the problem is that I found examples in which Hoffmann elimination stereoselectivity was anti and syn.

BTW = by the way. It was an additional comment. I believe I addressed your issue.

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

(Taken from Ron's answer, in order to start an explanation)(1)
From Wikipedia:

The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.

So on further thinking about the Hofmann elimination, you notice the following

$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))

Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage

In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^2+}$, $\ce{NR^3+}$ all have the same observation

What's the conclusion?
Of course, these undergo E1cB when heated to a suitable temperature
There are sources which infer the same -

Wikipedia

Organic mechanisms by Xiaoping Sun (3)

Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.

The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion  character (e1cb) for the transition state.

Hence, E1cB is the correct answer

The Hoffman elimination is nost often an E2 elimination (as in your last example), as the comments already mentioned. There are many resources that go into greater depth about the reaction.
The "counterexamples" that you gave can all be explained, but they are not peculiar to Hoffman elimination. The same outcome could be observed with a different leaving group, I believe.
The E1cb elimination happens, because the intermediate carbanion is very stable - it is stabilized by the nitro group and the benzene ring (and also the inductive effect of the trimethylammonium group.

The Ei elimination happens due to the constraints on the geometry of the ring. As you can (hopefully) see on the picture below, the substitutents are forced into an almost eclipsed configuration. Therefore, there can be no anti-periplanar elimination, simply because no bond is anti-periplanar to the C-N bond being broken. The best orbital overlap is therefore reached with the (more or less) syn-periplanar C-D bond, which is therefore the broken one.

Honestly, leaving groups usually do not determine whether it is syn or ante elimination. Hence, although the leaving group is an amine, it does not change how you view the problem. What determines syn or ante elimination is the type of reaction occurring, whether it is E1, E2 or E1cB.
E2 would always go through an anti elimination due to the requirement of the antiperiplanar transition state.
Whereas for E1 and E1cB, there is no fixed answer to whether it is syn or ante. The product that ultimately forms is dependent on the thermodynamic stability of the product.
Now, let us go back to your "E1" example. That, my friend, is not an E1 elimination! That is in fact an E2 elimination. Your E2 elimination has the requirement of having your hydrogen and leaving group be coplanar. Since the D and ammonium quaternary cation are coplanar, though it is fixed in a syn position, they are forced to go through a syn coplanar E2 elimination. The reason it is not E1 is that your leaving group is on a secondary carbon and therefore E1 is unfavourable.

anti not ante. ${}$

Somebody had edited my question as I wrote NMe3+ as nme3+ and stuff like that. They must have mistaken the Ei for E1. In the picture also you can see it's Ei not E1

And I wanted to ask that what is the exact mech. For Hoffmann elimination, Like we decide which reaction to follow in alkyl halides based on solvent, reagent and alkyl halide itself. What's the hard rule here?

See right, $E_i$ elimination occurs when there is a strong base (like alkylithiums) or there the proton is sterically hindered. The base would deprotonate the quaternary ammonium and form a ylide. Then this would go through an $E_i$ mechanism, like this: https://en.wikipedia.org/wiki/Ei_mechanism#Special_Cases_for_the_Hofmann_Elimination
As you can see, your final product is DOH not the deuterated amine. Hence, it goes by E2 and not Ei.
The hard rule is, look at the base, the carbon attached to the leaving group and the carbon attached to the hydrogen (pri, sec or tert).


text:
Isothermal transforamtion diagram - Austentite

A steel alloy is rapidly cooled to $\pu{600^oC}$, held for $\pu{4 sec}$, rapidly cooled to $\pu{450^oC}$, held for $\pu{10 sec}$ and quenched to room temperature. What phases are produced?

Why is the time reset back to $0$ when you cool to $\pu{450^oC}$? Is it because I am only considering the remaining austenite?

The given phase diagram provides some information on the time-temperature relationship for the formation of pearlite (P), bainite (B), and martensite (M) from austenite (A) form, which is known as  known as gamma-phase iron ($\gamma$-$\ce{Fe}$). For example, It would take $\pu{10 sec}$ to 100% austenite to the 50% austenite-to-pearlite transformation to complete.
As following diagram indicate, rapid cooling of eutectoid composition of steel alloy (100% A) to $\pu{600 ^\circ C}$ stays as 100% A. If the temperature held at $\pu{600 ^\circ C}$ for $\pu{4 sec}$, austenite-to-pearlite transformation occur to ~25% (see the point noted at the diagram):

During the second rapid cooling from $\pu{600 ^\circ C}$ to $\pu{450 ^\circ C}$ (time $\approx 0$), there is no additional transformation (see the second point noted at the diagram with red line). At $\pu{450 ^\circ C}$, the time is start to count again at zero second, so  that the time is moved to $t=0$ point (this is also because isotherm conversion will be started with unconverted austenite, which is now 75% of original specimen). While holding at $\pu{450 ^\circ C}$ for $\pu{10 sec}$, approximately 50% of this remaining unreacted 75% A (or ~37.5% of the original specimen) will transform to bainite (see the point noted in the diagram). Finally, upon cooling to room temperature, the remaining 37.5% A (of the original specimen) transforms to martensite. Thus, the final micro-structure of given steel alloy consists of about 25% pearlite, 37.5% bainite, and 37.5% martensite (no austenite).