text:
ortho/para-Selectivity in the Reimer-Tiemann Reaction
I can't figure out why the dichlorocarbene molecule produced in the Reimer-Tiemann reaction gets attached to the ortho position of phenol.
The ortho position is sterically slightly more hindered due to the oxygen, so why dosen't the carbene attach to the para position?

Does this answer your question? In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position?

No, because this answer was a bit more simple and was straightforward

@EVO Answers can be simple and straightforward. If they answers your question, then yours is a duplicate of that.

I don't think this is a duplicate; the linked duplicate asks about C2 attack vs O attack, not C2 vs C4.

The Reimer–Tiemann reaction will give a mixture of products unless the formation of one product is highly disfavoured. The Kürti-Csakó notes in its introductory paragraph on the Reimer–Tiemann:[1]

the regioselectivity is not high, but ortho-formyl products tend to predominate;

The statement already hints at a product mixture being obtained. To further show that both ortho and para attacks are possible, the 1960 review published by Wynberg contains the following scheme as its very first scheme in the introduction:[2]

So while both products are obtained, a back of the envelope calculation already shows that the distribution is not a perfect $2:1$ ratio. This can apparently be explaned by the effect of positive counterions, as Hine and van der Veen report:[3]

The ratio of ortho to para product was found to be 2.21 [under high base concentrations], showing that the tendency towards o-substitution is indeed increased under conditions where ion-pair formation is encouraged. One factor that would certainly be expected to be present and that would tend to favor o-substitution is an electrostatic effect. When a dichloromethylene molecule attacks the o-position of a sodium phenoxide ion-pair to yield the probable initial product, there is less separation of unlike charges than when the analogous para product is formed.

So while the steric hindrance of a single oxygen atom is not much more than that of a hydrogen atom and thus does not play a huge role, the presence of positive counterions – although increasing steric congestion – may serve to enhance the formation of the ortho-product through favourable electronic effects.

References:
[1]: L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 378.
[2]: H. Wynberg, Chem. Rev. 1960, 60, 169–184. DOI: 10.1021/cr60204a003.
[3]: J. Hine, J. M. van der Veen, J. Am. Chem. Soc. 1959, 81, 6446–6449. DOI: 10.1021/ja01533a028.

In Reimer-Tiemann Reaction, a mixture of ortho and para isomers is obtained in which the ortho isomer predominates (it is not the sole product).
If one of the ortho positions is occupied, the para-isomer is the major product.
The two isomers can be separated by fractional distillation, in which the unreacted phenol and the ortho-isomer distil over leaving behind the para-isomer.
Ortho product is major mainly due to 2 reasons-

Probability factor ( there are 2 ortho positions available Vs only 1 para position )
H- bonding in the final salisaldehyde.(There is a formation of 6 membered chelated ring which increases the stability of this product.)

Is the Reimer-Tiemann reaction thermodynamically controlled, such that the product stability determines the outcome? To me, it seems more likely that it's kinetically controlled, where transition state stability is the key.

Although it is run at refluxing chloroform conditions, I want to agree with orthocresol on this one and think that thermodynamic control is slightly unlikely.


text:
How to find entropy of chemical network reaction?
I have the following system of reactions:
$$
\begin{aligned}
\ce{X_1 + Y_1 &<=>[$k_{21}$][$k_{12}$] X_2} \\
\ce{X_2       &<=>[$k_{32}$][$k_{23}$] X_3} \\
\ce{X_3       &<=>[$k_{13}$][$k_{31}$] X_1 + Y_2} \\
\ce{X_2 + Y_3 &<=>[$k_{42}$][$k_{24}$] X_4} \\
\ce{X_4       &<=>[$k_{34}$][$k_{43}$] X_3 + Y_4}
\end{aligned}
$$
Here $\ce{X_1},\ldots,\ce{X_4}$ represent the state of the molecule that catalyzes the reaction, and $\ce{Y_1},\dots,\ce{Y_4}$ represent a reactant or a product that is consumed or produced by the reaction.
I need to find the free energy flow $J_G$ with respect to $\ce{Y}$ and the entropy generation $R_S$ with respect to $\ce{X}.$
Could anyone recommend the approach to this problem?

First, you should clear up the basics: the network is not in line with your description of X and Y you gave. Second, if you share how far you got it helps us give a meaningful answer. Have you tried to apply Hess's law?

Dear Greg, thanks for your comment. I made a mistake with the indexes for X and Y, corrected it. Does it make sense now? Everything I provided here is all the information I have for this task. I need to find J_in and J_out to get the free energy flow and then find entropy. I think that they must match in this reaction network but I am a bit lost.

OK, so have you tried to apply the Hess's law to the reactions? I do not know what do you mean by the free energy flow for the whole network, but for if you need it for the X1 --> X4 reaction, that would be just the straight application of Hess's law.

source: Stack Exchange

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text:
When can you use the direct flame of a Bunsen burner, and when must you boil a beaker of water first?
I'd like to ask when you would do the following: use the direct Bunsen flame to heat a substance; place the substance in a test tube, which is placed in a beaker of water, then boil the water with the flame while the test tube is in it?
Specifically, I'm doing an experiment where I synthesise aspirin by reacting salicylic acid with acetic anhydride using an 85% phosphoric acid catalyst. Is it potentially harmful to use a direct flame in this instance?

For safety's sake, use the water bath.

SAFETY - The water bath gets the reactants hot enough to cause the reaction without the risk of causing the contents of the test tube to boil thereby turning the test tube into a cannon.

Also, most organic chemicals are flammable, so if you spill any out of the test tube, that might be a fire / explosion hazard, whereas if it were to land in the water bath, it's not as significant a danger.

Bunsen burners are not safe to use–even as the source of heat for a water bath–if you are working with chemicals that are flammable. There are plenty of ways to heat vessels that don't–at any point–involve naked flames. If you are, for example, working with diethyl ether, there should be no naked flames anywhere in the same laboratory, never mind being used to directly heat the vessel. Acetic anhydride might be less volatile than that but it is still wise to avoid any naked flames and to stick with electrical heaters.

Besides safety concerns. Sorry but the answer should be quite obvious. This, in my opinion has nothing to do with chemistry. The Temperature of a body immersed in any bath cannot exceed that of the bath itself (when the latter is compatible). The T of the bath will be rather omogenous and and always below that of the heating source. Of course if you want to boil a water based solution, a water bath won't be suitable. Still, unless we are speaking of calcination, decomposition, or small test tube scale, you will alway use something to spread the flame/heat. Minus 1 to the Q. .

This reminds me of an incident that occured in my school lab: The lab assistant had set up an electrically heated water bath in a corner of the lab (Which was the only place where an electrical socket was available). Since this was a large lab, meant to accomodate 40+ students, it was a long walk to the bath. You had to place your test tube in the bath after marking it, wait for a few minutes and then pick it up again.
A few guys who were located in a corner of the lab didn't feel like walking the distance and waiting, so they turned on the bunsen burner and used it to heat the organic chemicals (Acetone and Formaldehyde) that they had to identify the functional groups for. While heating, the chemical started to boil so vigorously that a small amount 'cannoned' out of the mouth of the test tube and got stuck onto the ceiling. Needless to say, the students spent their afternoon in the headmaster's office.
TL;DR: While working with volatile chemicals, always use a water bath. It ensures slow, safe and even heating and as a bonus, the temperature does not rise above 100° C in the worst case. Most electrical water baths are temperature-controlled as well, meaning a thermostat senses the temperature and shuts off the heater coil if the temperature rises above a certain limit.

Would it make a difference if I placed the reaction mixture in a test tube, which I place in a beaker of water, and heated the water with the bunsen burner? Here's a picture of what I mean: [https://www.google.com/url?sa=i&url=http%3A%2F%2Fwww.smartlearning.in%2Fnew_stsquestion.php%3FQuestionID%3D77223%26page%3D1&psig=AOvVaw0TtlVHYs1Q3Oqz0nsHZAPF&ust=1596613210543000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCOiiwO2FgesCFQAAAAAdAAAAABAD]
The reason I'm considering using a bunsen burner even if a water bath is safer is because my experiment is to test different heating methods in making aspirin.

that would be safer than directly using a bunsen burner, but still risky. You should read @matt_black 's comment on your question.

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text:
Differences in Novichok agent mass spectra
As we have all heard, in 2018, a series of poisonings with a nerve agent alleged to be of the Novichok series was carried out on British soil. It is claimed that the agent used was A-234. However, a novichok agent was also previously used in the poisoning of a banker named Ivan Kivelidi in 1995.
The Russian Wikipedia page on Novichoks states that not only was the mass spectrum of A-234 published in NIST by someone affiliated with Edgewood Arsenal, but also provides the mass spectrum of the agent used in the Kivelidi case.

Top: Mass Spectrum of chemical agent used in the Kivelidi case. Bottom: Mass Spectrum of A-234 as published by Edgewood.
As one will notice, the spectra almost match, except for the fact that the top spectrum contains a peak with a mass of 42.The bottom spectrum has no such peak. What is the significance of this peak? How would the structure of the agent in the Kivelidi case differ from A-234? Is the A-234 spectrum correct?

Can you link to the NIST entry? It would be useful to know the instrument settings used for the Edgewood spectrum. Could be that m<50 was excluded from spectrum

@Andrew The NIST entry was deleted sometime after it was published. This is all that remains: https://archive.org/details/Edgewood

Do you know how the pattern of mass spectra develop, via the fragmentation of the ionised original molecule?


text:
I am having trouble with this question on net ionic equation
Convert into molecular equation:
The reaction between potassium dichromate, ferrous sulphate and sulphuric acid:-

Cr2O7^2- + 14H^+ + 6Fe^2+ ---> 6Fe^3+ + 2 Cr^3+ + 7H2O.
I have no idea how to solve this problem, can someone please help. Thanks!

What is the issue you are having, where are you stuck?

You should simply add the spectator ions on the left and on the right hand side, namely the ions $\ce{K^+}$ (and they are 2), and the $\ce{SO_4^{2-}}$ (and they are 6+7 = 13), and make molecules on both sides with them. That's all.

The only molecule in here is water. What kind of a teacher asks questions in such sloppy terminology?

You showed the correct balanced net-ionic equation:
$$\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 6Fe^3+ + 2Cr^3+ + 7H2O}$$
They asked for the molecular equation (I agree the terminology is horrible), which nowadays is taught as putting the ions in their "molecular" form:
Potassium Dichromate ($\ce{K2Cr2O7}$), Ferrous Sulphate ($\ce{FeSO4}$), and Sulphuric Acid ($\ce{H2SO4}$)
You need to convert the number of ions that you have into the number of "molecules" (really formula unit of the solid). For example, 14 moles of $\ce{H+}$ turns into 7 moles of $\ce{H2SO4}$ since they told your that your acid is sulphuric acid.
Repeat this on both sides so that you get a formula like so (with no free ions):
$$\ce{K2Cr2O7 + 6FeSO4 + 7H2SO4 -> Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O}$$


text:
Is reverse osmosis a fancy term for filtration?
In Reverse Osmosis (RO) systems, the water to be purified is pushed against a semipermeable membrane and comes cleaner the other side, the path of water is the reverse of regular osmosis, hence the name.
However, it seems to me that this process is pretty similar to a regular filtration.
So, is RO just a fancy term for a fancy filtration?
What is it that justifies it not being called filtration, ultrafiltration or some variation of that?

The wikipedia article on Reverse osmosis discusses this in the paragraph before the “History” subsection of the article.

Size of the holes.

Do you prefer to call osmosis "reverse filtration" ?

If the size of the holes are the important factor, could it be called ultrafiltration? If the pressure difference is the important factor, could vacuum filtration be called reverse osmosis? I find it hard to understand the wikipedia text."[RO] differs from filtration in that the mechanism of fluid flow is by osmosis across a membrane", this doesnt seem right as osmosis is not that.

@LocoAsker There is another difference in that in vacuum filtration, the pores are usually big enough that only precipitated particles are separated from the solution, whose dissolved composition (unlike with RO) remains unchanged. "Filtering" away the disolved portion of a solution is a very specific kind of "filtration."

Ultrafiltration is ongoing at any pressure, even if slowly. Reverse osmosis is ongoing at pressure higher then osmotic pressure. At pressure lower than osmotic pressure, normal osmosis occurs. At ultrafiltration,, liquid always goes to the side with lower pressure. At reverse osmosis, it can flow to side with higher pressure,

I think I got it now. Could someone write an answer so I could accept it?

Ultrafiltration is ongoing at any pressure, even if slowly and liquid always goes to the side with lower pressure.
Reverse osmosis is ongoing only at pressure higher then osmotic pressure. At pressure lower than osmotic pressure, normal osmosis occurs and liquid flows to the side of higher pressure.


text:
In the question given below

in the question if we go in accordance with+M effect instead of+I effect we will get the required answer but why are we going in accordance with +M effect and when do and how do we choose which phenomenon to consider ?

+M has priority over I effect

Please try to find a reasonable title for your question. While being as concise as possible, it shall indicate, without ambiguity, the subject matter of the question in such a way as to distinguish it from that of other questions, without going into unnecessary detail.

Instead of a blanket statement like "resonance is more important than inductive effects", I'd offer this version: generally, resonance effects are observed to be more effective at stabilising carbocations than inductive effects.

CH3 does not show resonance effects noticeably.
In case of solving problems like this, the order to consider is +M >+H >+I,
where +M refers to the resonance effect, +H to the hyper conjugation effect and +I to the inductive effect.
Note that in the above problem, if you consider the number of alpha-hydrogen atoms involved in hyper conjugation, you will get the correct order.

+M or +R dominate over inductive. That's because+M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore+M effect has priority over +I.

Will the statement always hold true , means we must first check for +M effect before+I

Yes. +M must be checked before +I. The order to be followed is M > H > I.


text:
Is 2s a subshell or an orbital?
My textbook says

They will then fill the $2s$ subshell, and then the $2p$ subshell...

Yet half of ChemSE refers to $s,p,d,f$ as "subshells" and the other half seems to refer to $1s, 2p...$ for example, as "subshells" as well, even in two answers to the same question.
So is, for example, $s$ a subshell, or $2s$ a subshell?

Am I Ivan or Ivan Neretin?

@IvanNeretin You are ИСН:)

Terms a shell and a subshell are rather obsolete.2 means the orbital with quantum number n=2, p means the orbital with quantum number l=0. By the particular subshell were considered orbitals with the particular quantum numbers n and l.

@andselisk You knew!

Terms a shell and a subshell are rather obsolete.
2 in 2p means the orbital has the quantum number n=2, p in 2p means the orbital has the quantum number $\ell$=0.
By the particular subshell were considered orbitals with the particular quantum numbers n and l.
So s, p, d, f are particular groups of "subshells", each sharing the common value of the quantum number $\ell$=0,1,2,3,  while 1s, 2p,.. are the particular "subshells".


text:
What is the magnetic moment of tris(oxalato)nickelate(IV)?
I'm really struggling with finding the magnetic moment of this complex:

In the solutions, it says that it is 2.83 BM (which corresponds to 2 unpaired electrons). However, in drawing out the d orbital diagram (using crystal field theory) I found that there are actually 4 unpaired electrons (since there are 6 d electrons and the complex will be octahedral and high spin). Am I missing something or is the solution wrong?

If it's high-spin, you'd be correct. I wouldn't be entirely surprised if it's low-spin (hence 0 unpaired electrons), though: Ni(IV) is a pretty high oxidation state, and d6 low spin is nicely stabilised by CFSE (or LFSE, whatever you prefer to call it), just like in Co(III) which is essentially exclusively low-spin...

(Are you or the question writer sure that it's not meant to be $\ce{K4[Ni(ox)3]}$?)

Thanks very much...yes I suspect it should be K4! Thanks for confirming my thinking about the d6 configuration though :)

Just to reinforce orthocresol's point with the exception of some Fe2+ complexes, and one or two fluoro complexes of Co 3+ all d6 is low spin

If the compound is really nickel(IV), then there are only two choices: a high-spin $\mathrm d^6$ or a low-spin $\mathrm d^6$ configuration, corresponding either to 4 or 0 unpaired electrons. As higher oxidation states make low-spin configurations more likely (due to the lower energy of metal orbitals which means the energy is more similar to the corresponding ligand orbitals leading to greater stabilisation and destabilisation of bonding/antibonding orbitals; $\mathrm{t_{2g}}$ is nonbonding while $\mathrm{e_g^*}$ is antibonding if π effects are ignored, so the energy difference increases), I would expect a nickel(IV) compound to be diamagnetic rather than paramagnetic.
However, oxalate is easily oxidised to carbon dioxide and likewise transition metals in unusually high oxidation states (which is true for nickel(IV)) are typically very good oxidation agents as their reduction is preferred. Therefore, I highly doubt this complex would be stable and suggest it immediately decomposes to carbon dioxide and a nickel(II) or nickel(III) complex. To quote Wikipedia:

Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

Thus, I am inclined to suspect some kind of typo. Two unpaired electrons would correspond to the well-known nickel(II) $\mathrm d^8$ system and tris(oxalato)nickelate(II) would seem much more likely.

I agree that it's chemically unlikely, but been wrong enough times to not want to say it. :-)

@orthocresol At least this absolutely terrible source gives $\pu{-0.386V}$ to $\pu{-0.49V}$ for the oxalate oxidation. Finding a number for nickel(IV) is hard but $$\begin{align}\ce{NiO2(s) + 2 H+ + 2 e- &<=> Ni^2+ + 2 OH-}&E^0 = \pu{+1.59V}\end{align}$$ (Wikipedia) That seems to prove my hunch that oxalate and nickel(IV) can’t be in the same solution.


text:
Which Additives May be Present in Cellulose Filter Paper?
I'm wondering what, other than cellulose, might be contained in filter paper (the kind routinely used in Buchner filters / normal funnels).
A search here on Chemistry Stackexchange shows that at least one  additive is used to modify the visual appearance of some filter paper (for improved "whiteness" / "fluorescence").
What additives may be contained in filter paper? What are their properties (solubility and potential chemical interference)? And how could one get rid of them prior to use, without consequence on the paper's filtering properties?
When used to filter away precipitated contaminants I usually rinse my filters in hot purified water before use (more with residual acid traces in mind). How would you guys prepare them for extreme minimal contamination?

Depends on which ones you buy. What are you concerned about? https://www.mn-net.com/filtration/cellulose-filters-information/

The manufacturer in the link seems like a consciencious one at reasonable pricing. I am concerned about possibly (a) unknowingly (b) contaminating my products (c) with a set of substances whose nature I ignore. I must write that given the intended use cases I wouldn't have expected manufacturers to add anything to paper filters other than to improve their filtration characteristics (e.g. mechanical resistance when wet), and I would have expected them to disclose any additives on the package.

I don't care how shiny my paper filters look like (ceteris paribus), but I really care about them containing unstated substances that might end up in my products.

Have you considered glass fibre or a dual piece setup,  usually for mobile phases using Teflon coating which has minimal if any, leaching

Are you filtering at very high/low pH, high temperature, unusual solvents? Your filter paper package certainly contains statements about ash content etc. And the statement c) in your comment is ludicrous, sorry. What is your product, and what kind of contaminations would be critical?

@Beerhunter I've considered glass fiber disks but they are very expensive and unsuitable for strong bases. Though I'll definitely start using them as some point for stuff incompatible with organic matter.

@Karl Saturated solutions of acids and bases at times, -30 to ~190C. No unusual solvents. Purified water / ethanol mainly, triglycerides sometimes.

If the filter paper has been bleached, with say a hypochlorite (OCl-), possibly a small amount of chloride, and if using NaOCl, test for sodium as well. Now, if bleached with say H2O2, check for residual amounts of commonly employed stabilizers (like H3PO4) to lower the pH of the H2O2 and increase shelf-life. If ClO2 used, again chloride.

If additives would leech from the filter paper, washing the paper before filtration (and collecting the wash separately for potential analysis) should remove any contaminants. I routinely do that with cotton woll, when I use that as a filter: prepare filter, wash with a couple of ml of solvent, discard, filtrate my mixture.


text:
Nomenclature of halogenoalkanes

Why is the following compound called "1,1,1-trifluoro-2-chloro-2-bromoethane"?

Why is this not called "1-bromo-1-chloro-2,2,2,trifluoroethane" alphabetically? (the name, 1,1,1-trifluoro-2-chloro-2-bromoethane was in my CGP A level chemistry textbook :))

Both proposed names, "1,1,1-trifluoro-2-chloro-2-bromoethane" and "1-bromo-1-chloro-2,2,2-trifluoroethane", are not in accordance with IUPAC nomenclature.

Indeed; neither name is correct.

thank you!! So the first name from CGP just needs to be written alphabetically as 2-bromo-2-chloro-1,1,1-trifluoroethane and then it would be correct. Please could you explain the rule about the [1,1,1 vs 2,2,2] numbers and why I couldn't write "1-bromo-1-chloro-2,2,2-trifluoroethane"? thanks :D

It's something called the 'lowest locant set', i.e. (1, 1, 1, 2, 2) is considered lower than (1, 1, 2, 2, 2): "The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference." If you search Stack Exchange for that phrase, or for "P-14.3.5", you'll find more explanations.


text:
Material properties differing between vacuum or pressure filtration
During vacuum filtration of a very fine suspension, all solids that need to be removed are, but there is minimal solids loss. If the same material is pressure filtered (1-2 bar), desired solids removal is effective but the precipitated material matrix retains more liquid and other components entrained within. The thin slurry was tested and shown not to be a colloid.
What is the rationale for the filtered solids differing behaviour?

Are you sure you've got that the right way around?  One problem with filtering fines under vac is that the solvent (if a low boiler) starts to evaporate at the underside of the filter and precipitates impurities from the liquors and blinds the filter paper/cloth.

The solvent doesn't have that much time to evaporate in the filter under vacuum, especially if it's cold (this is normally carried out at zero degree Celsius or below). Both situations give rise to blinding eventually.


text:
Confusing definition of mol when considering second order rate equations
I am confused by "$\text{mol}$".
I denote the unit of a physical quantity with square brackets (e.g. $[V]=l$), and the value of a physical quantitiy using curly brackets (e.g. $\left\{ V\right\}$ = 0.01).
A simple second order rate equation for number concentrations $n=N/V$ is given by,
$$\dot{n} = k^*_{on}n^2,\;\;[1]$$
where the $[k^*_{on}]=l/s$.
The same second order rate equation but for molar concentration $c=\frac{N}{N_AV}$, is given by
$$\dot{c} = k_{on}c^2,\;\;[2]$$
where $[k_{on}]=\frac{l}{\text{mol}~s}$.
When transforming equation $[1]$ into equation $[2]$, we get the relation
$$k_{on} = \left\{k_{on}\right\}\frac{l}{\text{mol}~s} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{\left[N_A\right]}s}$$
Hence $\frac{1}{\left[N_A\right]} = \text{mol}$?
But this would be in contraction to the definition I find on wikipedia where it is written that:

In short, for particles 1 mol = $6.02214076×10^{23}$.

Hence, $1~\text{mol} = \left\{N_A\right\}$.

I think the sentence cited above in the wikipedia article is very confusing.
But the wiki article about the Avogadro constant helped me resolve my confusion, which defines $mol$ via the Avogadro constant:
$$N_A = \left\{N_A\right\}[N_A] = 6.022\times10^{23} \text{mol}^{-1}.$$
Hence
$$k_{on} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{[N_A]}s}=\left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{mol~s}$$
This definition is also given on the hompege of the National Institute of Standards and Technology (NIST), see here.

The Wikipedia article was edited by someone who wants to change the definition of mole, and has written an article about it. The version of the Wikipedia article you saw did not have a neutral point of view because it did not reflect the current official definition of the unit.

I think this definition is right. It got estrablished new since 19. May 2019. See here: https://physics.nist.gov/cgi-bin/cuu/Value?na

Yes, I edited the Wikipedia article to reflect the current official definition.

Nice! Good job @KarstenTheis


text:
Phenoquinone vs benzoquinone
Are phenoquinone and benzoquinone the same thing? When I searched for the colour of benzoquinone I found that it's yellow. But when I searched oxidation of phenol in air. It states that it turns pink/red due to formation of phenoquinone. Also when aniline is oxidised with $\ce{K2Cr2O7}$ quinone is formed (perhaps it is aniline black). Is it the same compound everywhere?

& what about aniline. It also forms quinone on oxidation but it is aniline black. It it the same case as a complex?

Benzoquinone($\ce{C6H4O2}$) is a quinone with a single benzene ring. There are two types of benzoquinone:

1,4-Benzoquinone also called p-benzoquinone. More common type of
benzoquinone. It is the main product of oxidation of phenol.
1,2-Benzoquinone also called o-benzoquinone. Less common type of benzoquinone.

Phenoquinone is a crystalline complex made of one quinone molecule and 2 phenol molecules. It is most likely formed from the polymerization of benzoquinone, hence the color change from yellow to reddish pink. See the mechanism below, taken from this answer:

It is formulated as $\ce{C6H4O2.2(C6H5OH)}$. See: http://scripts.iucr.org/cgi-bin/paper?S0567740868002451

Such adducts are not hydrogen bound, but charge transfer complexes.

@Mithoron From ref 4 of my answer: "Different molecules in each complex are connected by hydrogen bond between the hydroxyl bond and carbonyl bond [...] In phenoquinone, one quinone molecule is placed between two phenol molecules to form a short complex chain. In spite of difference of complex chain, no significant difference was observed between complex due to short interatomic distance". I changed the "adduct" to "crystalline complex" but there is indeed a hydrogen bond between them.

What "ref 4"? I've seen this misconception earlier and it's quite obvious.

Here is a bit about CT complexes of quinones and how they are related to color. Thing is h-bonds don't change color  to any significant extent and CT complexes do.

While then sure can be (even should be!) h-bonds between individual phenol-quinone complexes, color change cannot be attributed to them, but to change transfer between the rings stacked one over another.

Here's a simplified pic for adduct with hydroquinone https://files.askiitians.com/cdn1/images/2014825-141131979-6452-ox11e.gifpsig=AOvVaw1Sl6zvV-SZCgg45wC6d5m4&ust=1593453995357000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCNjv5O2MpeoCFQAAAAAdAAAAABAU


text:
Can silicon dioxide melt at 20 °C?

Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$?

I have searched the web for $\ce{SiO2}$ phase diagram, but it seems to me that almost all the graph I can find have the temperature axis where the minimum temperature is in the hundreds degrees Celsius.
Is it theoretically possible that with an high enough pressure I can melt $\ce{SiO2}$ in a temperature range like $\pu{-30 ^\circ C}$ to $\pu{50 ^\circ C}$? Or is it something theoretically simply impossible?

High pressure has tendency to cause solids to melt at higher, not lower temperatures. Ice is exception, as ice/water density ratio is exception as well.
Generally, substances increase volume when melting and high pressure  would defiinitely not to make melting easier - See Le Chatelier's principle.

@Poutnik Thank you. I was reading at https://en.wikipedia.org/wiki/Silicon_dioxide#cite_ref-16 that Molten silica exhibits several peculiar physical characteristics that are similar to those observed in liquid water: negative temperature expansion, density maximum at temperatures ~5000 °C, and a heat capacity minimum. Is this of some help?

but in the same article: silica density 2.648 (α-quartz), 2.196 (amorphous) g·cm−3, molten silica density 2.08 g/cm3 at 1950 °C to 2.03 g/cm3 at 2200 °C. So there is significant melting expansion.

... have the temperature axis where the minimum temperature is in the hundreds of Celsius degree. I guess that means that nothing interesting happens below a few hundred degrees. :-)

The diagram proposed by Alessandro Jacopson is not related to $\ce{SiO_2}$. It is the diagram of water $\ce{H_2O}$.

@Maurice Thank you, I did not add the image, see https://chemistry.stackexchange.com/revisions/135879/2

I just deleted it - even with proper context it could be misleading.

Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$?

According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1:

Reference 1 states that:

An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{15 GPa}$ (α‐quartz, β‐quartz, tridymite, cristobalite, coesite, and stishovite) and the liquid phase is presented. The data set was produced through a computer‐based assessment of the properties in which the available thermochemical (calorimetric), physical (bulk modulus and thermal expansion), and solid‐state and melting transition data (including some newly determined data on the high‐pressure polymorphs cosite and stishovite) were considered. The data set can be used to calculate phase relations at pressures of $\pu{0.1 MPa}$ to $\pu{15 GPa}$ and temperatures of $300$ to $\pu{3200 K}$. The calculated phase diagram using these data agrees quite well with the phase equilibrium determinations except for the high‐temperature part of the coesitestishovite boundary. The properties of the liquid phase obtained are also in good agreement with the available data.

Keep in mind that $\pu{0.1 MPa} \approx \pu{1 atm}$. According to the phase diagram, cristobalite form is in liquid form at temperature grater than $\pu{1750 K}$ in pressure range of $\pu{0.1-0.7 MPa}$. The extrapolation of that boundary would show it would never cross $\pu{20 ^\circ C}$ point on $x$-axis without going to negative pressure. Therefore, it is impossible to have any form of $\ce{SiO2}$ in reduced or high pressusituations at room temperature or below.
References:

V. Swamy Surendra, K. Saxena, Bo Sundman, J. Zhang, “A thermodynamic assessment of silica phase diagram,” Journal of Geophysical Research 1994, 99(B6), 11787-11794 (https://doi.org/10.1029/93JB02968).


text:
How to find a symmetry group of a system if all the symmetry transformations do not obey closure and don't form a group?
For instance, consider a system with $p_x$ and $p_z$ orbitals at the vertices of a square (on xy-plane). A square by itself would have $D_4$ symmetry. However, because of the $p_x$ orbital; the $90^\circ$ rotation ($C_4$) and $270^\circ$ rotation ($C_4^{-1}$) are no longer symmetry operations. Now the rest of the $D_4$ members do preserve the symmetry, but do not form a group since the subset is not closed.
How does one go about formulating a symmetry group for cases like these?

Never saw a rotation denoted with $r$. Normally, 90° and 270° rotations are denoted with $C_4$ and $C_4^{-1}$ (or $3C_4),$ respectfully. This should give you a hint when you look up the point group using a character table. $r$ and $r^3$ in your context would refer to the distance and volume, respectfully.

Orbitals are not relevant to molecular symmetry. Only atoms are. Are the atoms at the four corners of the square identical? If so, then 90 degree rotation does return the original molecule.

@andselisk Sorry about the issue with notations, I've fixed it. Thanks!

I am mildly confused: a square should be $D_\mathrm{4h}$, surely?

@Andrew Actually this is sort of a hypothetical problem from physics. It's not really a molecule, and I'm considering a system that contains hypothetical atoms with only $p_x$ and $p_z$ like orbitals at the four vertices. I'm afraid a $C_4$ rotation will turn $p_x$ into $p_y$ breaking the symmetry.

The symmetry transformations do obey closure and do form a group, otherwise they are not symmetry transformations.

If what you have in mind is something like this https://i.stack.imgur.com/oZTV7.jpg, then it appears to me to be $D_\mathrm{2h}$.

@orthocresol Thanks for pointing out the typo, I actually meant $D_4$ (Dihedral_4 group). I have learned group theory as a topic in mathematics and physics mostly, and I'm a beginner in crystal symmetry and point-groups etc. I think perhaps the Dihedral_4 group of the square is called D4h in the Schoenflies notation (..?)

@orthocresol Thank you so much! I indeed meant what you mentioned in the image. I'm going to look into $D_{2h}$ point group.

"I'm afraid a C4 rotation will turn px into py breaking the symmetry" My point was that atomic orbitals cannot "break symmetry" because they are not considered when determining the symmetry group of a molecule. If you make a molecular orbital out of a combination of atomic orbitals, then you need to ensure that the molecular orbital is sym or antisym w.r.t. one (but not necessarily) all of the operations of the determined group.

Hmm, so I'm not entirely sure what notation you're using ($D_4$ sounded like Schönflies to me), but I'm sure you can figure out the equivalent.
In general I agree with Andrew's point that molecular symmetry is determined by atomic positions and not orbitals. But at least to me as of now, this seems to be a thought exercise more than anything (after all, no real atom has only two p-orbitals), so I'll play along. In Schönflies notation your system would be labelled $D_\mathrm{2h}$. The three $C_2$ axes and three mirror planes are illustrated below. I left out the identity operation $E$ and the centre of inversion $i$.

+1, orthocresol, What software did you use to draw these pictures? Is it a pen used PowerPoint?

@M.Farooq, stylus on tablet, more specifically Apple Pencil + iPad Pro using the Notability app (starting to sound like an Apple commercial so I had better stop here...)

Figure looks nice! I feel a common stylus in PowerPoint should also be able to draw similar figures. Don't have these Apple products.

Yep, I've done similar things in PowerPoint before on my iPad, so I don't think special software is necessary. As far as hardware goes, there are many on the market nowadays.


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Calculating dissociation constant at different pKw values
What are the standard approaches to calculating how dissociation constants for acids and bases, $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$, change as the ionic product of water, $\mathrm{p}K_\mathrm{w}$, changes?

They primarily change with temperature, but they still follow pKa + pKb = pKw

Is the change generally "linear", if pKw decreases, do pKa and pKb both decrease proportionally (pKa + pKb = pKw)?

@Asker123: Show first what you think in details.

I don't really know how you'd think anyone could show more what they think than "if the equation is pKa + pKb = pKw, will both acid and conjugate base disassociation constant generally change "linearly""

The equation
$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}\tag{1}$$
has two degrees of freedom, so two values are independent on each other and the third one depends on the other two.
Reaction equilibrium constant $K_\mathrm{a}$ for
$$\ce{HA + H2O <=> H3O+ + A-}\tag{R1}$$
is chemically independent on $K_\mathrm{w}$ for reaction
$$\ce{2 H2O <=> H3O+ + OH-}\tag{R2}$$
OTOH, the equilibrium constant $K_\mathrm{b}$ for the conjugate base is derived from the above two.
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{K_\mathrm{w}[\ce{A-}]}{[\ce{HA}][\ce{OH-}]} = \frac{K_\mathrm{w}}{K_\mathrm{b}}\tag{2}$$
$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$$
Another thing is that both $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{w}$ depend on temperature. It is not dependence
$$\Delta T \to \Delta \mathrm{p}K_\mathrm{w} \to \Delta \mathrm{p}K_\mathrm{a},\tag{3}$$
but
$$
\begin{cases}\tag{4}
\Delta T \to \Delta \mathrm{p}K_\mathrm{w} \\
\Delta T \to \Delta \mathrm{p}K_\mathrm{a}
\end{cases}
$$
For the temperature dependence of equilibrium constants, there is van 't Hoff equation, which calculates the constant value changes from the reaction enthalpy:

$$\ln \frac{K_2}{K_1} = \frac{-\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right),$$

from which:
$$K_2 = K_1 \cdot \exp{\left( \frac{-\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\right)}\tag{5}$$


text:
how to calculate the dihedral angle
I need to calculate the dihedral angle between C20 and C19, as well as between C19-C18.
 Can someone help me choosing the right combination? I have used the following one: C18C19C20C23 for the first angle and C20C19C18N7 for the second one. (I am using Maestro software and I am trying to make an energy plot for chi and psi angles). However I am not sure if I should use instead the following combinations: C23C20C19C18 and N7C18C19C20. Does it make any difference?
Thank you!

It makes a difference. Phi and psi are defined for polymers of alpha-amino acids, which this molecules is not. "The phi-angle (φ) - around the N-Cα bond. The psi-angle (ψ) - around the Cα-C bond" (https://www3.cmbi.umcn.nl/wiki/index.php/Torsion_angle).

it is not about phi and psi angles. this is not a protein. (I mentionned psi and chi angle). this is nelfinavir

The picture shows how the dihedral angle $\psi$ is defined. You will need to be familiar with vectors to be able to calculate this angle; $\cos(\psi)=m\cdot n/(|m||n|)$ where the vectors are shown in the figure, vector $m$ is perpendicular to $a$ and $b$, $m=a\times b=|a||b|\sin(\gamma)$ and $n=b\times c$.

Thank you a lot!Fortunately I am using a software, so I will not make the calculations by myself. I just need to select the right 4 atoms. Could you tell me your opinion about the selection of the atoms? I have written in the question the atoms that I have selected. Thank you!!

It is up to you but its either 18-19-20-22 or 18-19-20-23 and the second 7-18-19-20

Thank you so much for your answer!!  I really appreciate it!

hoowever I cannot understand why in the the angle C20-C19 we do not begin from C22 and finish to C18 and in the angle C19-C18 we do not begin from C20? Thank you!

you could start at either end.


text:
Will a dangerous amount of lead leach into soil from a galvanised corrugated steel raised bed?
I have searched, found similar questions but none answer my specific need.
I have built a raised garden bed to grow vegetables (ironically due to a high lead content in our garden soil - planning to put new, lead-free soil in the raised bed).
I used galvanised corrugated steel panels for the sides. I believe any zinc leaching would be beneficial, not a hazard.

However, now I learn that the galvanising process often involves lead. When something is "hot-dipped", apparently adding lead to the process helps the zinc run off. My understanding is limited. A crude 3M home swab test indicates lead, possibly a high content (it went very dark red).
https://en.wikipedia.org/wiki/Hot-dip_galvanization
https://galvanizeit.org/knowledgebase/article/why-is-lead-added-to-the-galvanizing-bath
My chemistry isn't very good, which is why I came to the experts. Here are my concerns:

How much lead would be likely to leach into the soil?
Would this lead be in a form that could be absorbed by lead-absorbing plants (e.g. lettuce)?
Would this lead be in a form that might endanger small children "helping" with the vegetables?
Would this lead be in a form that might be absorbed by the body when eating these vegetables, if not carefully washed?

Thanks in advance!

Could you please provide a reference for how lead is associated with galvanization? The wikipedia article says nothing about lead contamination of the zinc, but it does say that acid rain may be a problem with outdoor galvanized steel. Hypothetically, if the zinc did contain lead at low or trace levels, then it will get into the soil as the zinc sacrifices itself to protect the steel it coats. Then the soil would be a problem. But you already have soil with elevated lead levels, if I read you right, so that is already an issue with kids because kids eat dirt.

@EdV - thank you, added some links to the "hot-dip" galvanising process. Planning to use new, lead-free soil in the raised bed, and mulch / cover the rest of the yard to lock in the lead.

If it was me (and it wouldn’t be since I couldn’t grow lichens on a rock), I would paint the panels, as per Galvanization and lime the soil underneath your raised garden bed so calcium is in relative abundance relative to soil lead content.

Excellent, thanks for the link! I had no idea that lead was deliberately added, especially nowadays when people go nuts about lead. The mulch idea is excellent, but adding lime is also an inexpensive additional step.


text:
Effect of magnetization on oxidation (rusting) of iron
Can magenetizing a piece of iron bar slow-down (or speed-up) the oxidation (rusting) process? In other words have any influence on it ?
From what I've looked up( wikipedia, quora), it appears that the property of magnetism depends on 'spin' and also the creation of bonds does have an impact on the 'spin' (apparently has something to do with pauli's exclusion principle).
PS: Tried to ask the question on physics.stackexchange.com but did not find all the tags so decided to ask here.

Interesting query, never thought about it. Anyway, use Google Scholar, and search Effect of magnetization on corrosion. You will get a lot of useful results. Quora  or Wikipedia should be your last resort in reading about some thing. Wikipedia is relatively more reliable but ...still.

Why not try an experiment? Remove oil from some nails, place partially in water, and add a strong magnet outside the container (to avoid electrolytic action), and set up a control without magnet (being sure to arrange nails the same way).

@DrMoishePippik - That was my first thought but was not sure if I could get enough free time but in retrospect may be it will be worth it.

OK, those who say magnetic fields have no effect can buy a relatively inexpensive Magnetizer (as I once did) and try performing a controlled experiment (as I attempted) with the rusting of iron. There is some apparent evidence that alignment of magnetic fields can promote radical activity.
This UK study on the effect of a magnetizer in reducing the amount of chlorinating pool chemicals required, is an example of a successful application. Reference, see Magnetic treatment of swimming pool water for enhanced chemical oxidation and disinfecting.
However, with respect to possible applications and significant results, apparently electrochemistry applications are at the top. For example, see Zou, P.; Leddy, J., Magnetized Nickel Electrodes for Improved Charge and Discharge Rates in Nickel Metal Hydride and Nickel Cadmium Batteries, in Electrochemical and Solid-State Letters 2006, 9, A43-A45. To quote from the article:

...magnetized electrodes yielded higher currents than the corresponding nonmagnetic electrodes by 70 and 230 %, respectively.

Source material for background includes Magnetic field induced motion behavior of gas bubbles in liquid, by Keliang Wang, Pucheng Pei, Yu Pei, Ze Ma, Huachi Xu & Dongfang Chen, in Scientific Reports 6, Article number:21068 (2016), doi:10.1038/srep21068 ), interesting article comments include:

the kinetics of oxygen bubbles is mainly proportional to intensity of the electromagnetic field. The magnetic-field induced rotational motion of oxygen bubbles in a square electrolyzer can increase liquid hydrodynamics, thus solve the problems of oxygen bubbles coalescence, and uneven distribution of electrolyte composition and temperature. These types of oxygen bubbles movement will not only improve energy saving and metal deposition for energy storage and metal refinery, but also propel object motion in application to medical and martial fields.

Note, with my experiment reported above with iron rusting, I observed significant more frothing/air bubbling activity around iron in the presence of the Magnetizer. One could look at my experiment as a crude rendition of an iron metal/air battery cell with the Magnetizer forming a magnetic iron electrode.
Finally, a link to a prior StackExchange comment on the topic here.

Did take a quick look but will need a more detailed readup into it. Also am looking for a scenario where the metal (like iron filings in you case) itself is magnetised rather than just being subject to a magnetic field. So not sure if the two scenarios are equivalent.

As far as my sadly neglected physics knowledge goes It shouldn't. Oxidation is a chemical reaction process - electron exchange; and magnetism is spin dependent. Oxidation of metal will affect magnetism because it affects the physical structure of the material, but the inverse should not apply, as the energy drivers are order level lower for magnetization.
That said, I haven't seen experimental proof ...

I propose an experiment that does not involve sub-atomic effects: take a very highly magnetized iron (or perhaps magnetic stainless steel) and immerse it into a concentrated ferric chloride solution. The metal will corrode, producing ferrous ions which combine with ferric ions to give magnetite $Fe_3O_4$, which largely adheres to the iron/steel. (I know this occurs, for a fact.) Corrosion continues until the metal is consumed.
I have not done this with magnetized metal. Now, magnetite is magnetic; if it is bonded more tightly to the magnetic iron/steel than to non-magnetic metal, it might impede corrosion by preventing easy access of the corrosive ferric chloride to the body of the metal. Sort of like a passivation layer. But probably not nearly as effective.
One way to measure the effect is to do the experiment for a period of time and weigh the metal (before and after) to determine how much metal was lost. Another possible method would be to measure the electric potential of the iron/steel electrode as it was dissolving. Any data would surely be worthy of much explaining.


text:
Convention in electrochemical cell notation
While writing the cell notation for a cell, doesn't one have to write the two states in which a species exists in a sequence appropriate to the reaction in which they participate?
For example, in the Daniell cell reaction, shouldn't it always be written $\ce{Zn}|\ce{Zn^2+}$, and never, $\ce{Zn^2+}|\ce{Zn}$?
I didn't see anything in the definitions, and I saw a question asking for the cell reactions for this;
$$\ce{Pt,Cl2}|\ce{Cl^-(aq)}||\ce{Ag^+(aq)}|\ce{Ag}$$
Obviouly, chlorine doesn't react in that direction, but gets oxidised.

Good question. You can remember the cell notation with the mnemonic "reduction on the right."
So, on the right hand side, you should reduction process and on the left hand side you should show the oxidation process.
For example, for a Daniel cell, one can write
$$\ce{Zn|Zn^{2+}, SO_4^{2−}}||\ce{SO_4^{2−},Cu^{2+}}|\ce{Cu}$$
Now coming to your cell, $$\ce{Pt,Cl_2}|\ce{Cl^-(aq)}||\ce{Ag^+(aq)}|\ce{Ag}$$ is indeed written incorrectly because the writer seems to have the misconception that even in the cell notation, half cells must be written as a reduction.


text:
Reaction of 2,4,6-triaminobenzoic acid in a protic medium and heat

The given answer is (A). At first I thought that it was a simple decarboxylation (due to high heat).
But that does not seem feasible at all as the ring has no electron withdrawing groups which stabilise the carbanion; infact it has electron donating groups which rules out decarboxylation. What else could be a possible mechanism here?
Edit:In the answer posted below could anyone explain how the hydrolysis of triamminobenzene occurs?

Redraw the starting material as its imino tautomer and maybe it will become clearer

It is normal decarboxylation - pericyclic. Carbanion isn't usual intermediate.

My initial guess for the product was Phloroglucinol, or 1,3,5-trihydroxybenzene1,2

Belskaya et al.3 says that 1,3,5-triaminobenzene can be formed, but it can eventually be hydrolyzed to Phloroglucinol

The sources mention decarboxylation as the mechanism. Note that it is the trihydroxylamine 10 that is getting decarboxylated here. The mechanism becomes clearer if you consider the imino tautomers 6 or 7 with the acid group attached and then dehydrate.
In short, Phloroglucinol should be the final product. If that is not one of the options, then 1,3,5-triaminobenzene is the answer you should select.
References:

“PHLOROGLUCINOL.” Organic Syntheses, vol. 9, 1929, p. 74. doi:10.15227/orgsyn.009.0074
Tewari, K.S and Vishnoi, N.K : A textbook of organic chemistry, 4th edition.
Olga B.Belskaya, Valentin P.Talsi, Roman M.Mironenko, Vladimir A.Rodionov, Sergey V.Sysolyatin, Vladimir A.Likholobov, "Transformation pathways of 2,4,6-trinitrobenzoic acid in the aqueous-phase hydrogenation over Pd/C catalyst", Journal of Molecular Catalysis. doi: 10.1016/j.molcata.2016.04.014

Did you notice it's "side reactions" there? It's not usual at all. Also catalytic hydrogenation environment can influence things.


text:
Temperature required for E1 elimination reaction
I wanted to ask what's the temperature required for carrying out elimination reactions?
Wherever I read about elimination reactions, they just mention that elimination reactions occur at high temperature and substitution reactions are favoured at low temperatures.
So how do I decide which is high temperature and which is low, because we always talk in relative terms
Does it mean room temperature? If so, then will elimination reaction be favoured at $40^{\circ}$C for instance?

This is a question that is answered by careful experimentation [Possibly the experiment has been done already]. Primary and secondary require high T to dehydrate usually about 150-200C, intermediate temperature can give ether formation especially for primary; Usually conversion to a halide or sulfonate followed by base E2 reaction is superior. Tertiary or benzyl will dehydrate more easily. My experience with reacting substituted benzyl alcohols to make chlorides and bromides [more difficult] with thionyl chloride and PBr3, both very acidic reagents, gave very little E1 reaction at room temperature for most compounds. However, methoxy substituted phenyl compounds would eliminate almost immediately; I made some but they were thermally unstable probably because I couldn't purify them. My point is that one doesn't know until one tries. The text book pat answers are guides.

25°C (approx) is considered to be low (room temperature). Substitution reactions are favoured. And we consider 50°C
(approx) to be high and elimination reactions are favoured at this temperatures.


text:
Why do old buildings in the UK smell like ammonia / boar taint?
As a US resident who visited the UK, I was surprised that many wooden buildings built before say 1800 smelled strongly of pork, an ammonia-like smell but most precisely like boar taint. I have never smelled this in old buildings in other countries.
What is the smell? Is it residue from centuries of cooking pork? Is it from an old type of varnish? Decaying wood?

Probably related to specific fungal metabolites. This seems to be an interesting question and I'm not sure why it's getting downvoted and marked as unclear. I suppose every person who ever visited a museum of wooden architecture or just an old village log house in temperate climate zone would know what OP is talking about.

The What causes the old book smell? question and answer were very well received, so maybe someone will provide an answer with corroborating literature references. I hope it does not turn out to be centuries of sweat. ;)

@andselisk, This is exactly what I am wondering. What is wrong this with this genuine question! Above all "-3" votes. What is wrong folks?

@M.Farooq I gave it an upvote, but maybe the downvotes were due to something like assuming answers would be predominantly just opinions. That is just a guess. And maybe andselisk nailed it with his fungal metabolites hypothesis. Would have been nice to see a demonstrably valid answer.

@EdV, it will be a great gas chromatography-mass spectrometry project for undergraduates for summer projects. May be you can also vote to "reopen".

@Fritz, Search the paper "Smell of heritage: a framework for the identification, analysis and archival of historic odours" by Cecilia Bembibre* and Matija Strlič. It is open access and probably you will find the answer.

@M.Farooq I would vote to re-open if I could, but that requires 3000 or higher rep. ;(

@EdV Congrats on reaching 3000 rep points, feel free to vote:)

I would imagine the type of heating and cooking fuels e.g. wood/peat/coal/dung would also be a major factor.  I find it surprising that there haven't been any answers. Isn't there some historian/archeologist who also knows their chemistry and can give us an answer?

It might benefit possible answerers if you mention one or more examples of the buildings in question. This seems a very difficult question to answer without other than sensorial anecdotal information. Still, the smell you describe seems pretty specific (and rather odd). The link provided by Farooq seems to provide a narrow set of possible sources for the smell.

@BuckThorn The strongest smelling examples I can recall are (1) St. Mary's Guildhall in Coventry, build in the 1300s, and (2) the ground floor wooden hallways of Warwick Castle, I believe built in the 1600s. Many other buildings smelled similarly.

From your description it sounds like it might come from food preparation. Ventilation was often poor and kitchens used wood for fire. I am not sure what candles were made of, but I would not think lard. In later times lanterns were fueled with different kinds of oil of both animal and vegetable origin.

From the wikipedia: "Tallow, fat from cows or sheep, became the standard material used in candles in Europe. "


text:
Absolute Configuration in Stereochemistry
I got all the rules of writing RS nomenclature for Perspective formula of chirality center only problem being, I m confused that given compound can have any configuration depending on how I write the perspective formula.
 It's basically Same Molecule

You need to get a physical model ... match sticks and blue-tac with paper labels / straws and blue-tac with paper labels build it and see if you can see it ... all your angles are supposed to be 109 ° are they in the second diagram?  (You'll see when you build it).

Not necessarily the same molecule.

the problem is that u are focussing on the theoritical aspect of it.u r just using wedge-dash concept.try converting it into fischer form(in the correct way), and u will see that there is only 1 way doing so, and hence the same molecule.

Keep bonds of the same type together. In structure 1, the Br and its bold bond should be flipped to the right just above the dotted hydrogen, similar to structure 2.

@user55119 is that a rule?

@granthium i want to apply rules to perspective formula only as i mentioned in my question

@MathewMahindaratne can you elaborate

@user1945827 i know what u are sayin, but question is can i apply rs rules to any compound or there is some kind of restriction

It is not a rule but it surely makes the picture more understandable. The structures in the answer are good models.

You're right, the two compounds you depicted are basically the same molecule except for in their chirality. They are, essentially, enantiomers of each other.
In the slightly clearer depiction below,

by the Cahn-Ingold-Prelog convention we would call the molecule on the left S-bromochlorofluoromethane, while the one on the right would be its R-enantiomer, i.e. R-bromochlorofluoromethane.

i get this , this wasn't exactly the question. question is that i can change the configuration of same molecule by adjusting how i draw the solid wedge

@ManRow what software do u use to create such structures?

@ManRow didn't create it . He just used Wikipedia.


text:
Would any alkene possibly aromatize with proper conditions?
I was reading about 2 specific reactions: polymerisation of alkene and aromatization of alkane (of a sufficient size).
I thought why can't, having a condition that favored both these reactions be created, there exist a chain reaction where an alkene starts polymerizing and when the chain becomes long enough, it aromatizes.

It's just a thought that I had but I can't figure out if it IS possible.

In incomplete combustion things like that happen all the time, but wording of your question makes it somewhat unclear / too broad.

@Mithoron Yes I felt too while typing it that it could be too broad. But I felt narrowing it down more could contradict this idea if I narrowed it down too far. I wanted to know if an ideal case is possible where a significant amount of an aromatic compound is generated this way? And what conditions would it require?

Your thought is actually happening in real life. Petrolium industry uses the same concept except for polymerizing ethene (may be that also used to get higher alkanes). Fluid Catalytic Cracking (FCC) is one of the most important conversion processes used in petroleum refineries. FCC (see the schematic diagram below)) is widely used to convert the high-boiling, high-molecular weight hydrocarbon fractions of petroleum crude oils into more valuable gasoline, olefinic gases, and other products such as aromatics (Ref.1):

For example, a conversion reactor is used to carry out several reactions. A few of them are:
$$\ce{n-C26H54 -> C6H12 + C6H14 + C10H20 + C4H10}\tag1$$
$$\ce{C10H20 -> C6H12 + i-C4H8}\tag2$$
$$\ce{C6H12 + 3 i-C4H8 -> 3C4H10 + C6H6}\tag3$$
The equation $(1)$ (cracking reaction of paraffin) shows the formation of cyclohexane from a long-chain paraffin. The equation $(3)$ (hydrogen transfer reaction between cyclohexane and isobutene) shows the conversion of cyclohexane to benzene.
References:

A. Kayode Coker, In Petroleum Refining Design and Applications Handbook, Volume 1; John Wiley & Sons, Inc.: Hoboken, NJ, 2018 (ISBN: 978-1-118-23369-6).


text:
Does plant material oxidize from 02 gas (air or solved in water) and/or oxygen in H20?
https://en.m.wikipedia.org/wiki/Curing_of_tobacco
Hi
This is my first time growing tobacco for fun, pleasent green at home, to make snuff for personal use and maybe save som money.
Ive been reading a bit on wiki, it says oxidization is what makes tobacco cure/age/mature take out the harsh.
Im wondering about if the oxidization of caratenoids and chlorophyll in tobacco curing is from O2 in air (or solved in water) and/or from O in H2O itself?
From what i remember, from school and various archeology programs on tv, wood in sea bed with no O2 solved in it doesnt rot but metals will rust. ?
At what relative humidity at room temperature will tobacco leaves sealed in a jar (in fall and winter the relative humidity in my apartment will drop too much for curing tobacco, i guess) cause them to cure? If air is required, how much?
/Johan

Oxidation is done by oxygen O2. H2O cannot oxidize anything. H2O is the result of the oxidation of hydrogen.

Ok, thanks, im thinking about going with sealed yar with lots of air space. Venting once in a while with a light spray of moist since O2 is needed (The moist is needed at least because otherwize it taste horse food - every vegetable ive dried without a light boil taste and smell like hay). Is water required for oxidization?

I would flip that question around: figure out the humidity you need and then figure out how to achieve it by asking for information on that here or elsewhere. It seems to me you might also get answers by looking for a website specialized in growing and curing tobacco and such (like https://dengarden.com/gardening/Tobacco-Growing-and-Curing-at-Home)

This link says "In order to have the best quality tobacco leaves for further processing, the leaves should have a moisture content between 12-16%, so an ambient relative humidity of 60 to 68%"

Here is an interesting experiment to possible accelerate O2 activity. Water two plants, but on one sprinkle on iron filings from a magnetic, or purchase an inexpensive Magnetizer and mount it close to the leaves. Per an old experiment of mine and a recent question on this forum and my answer relating to increased O2 activation, there could be some related growth benefit. See https://chemistry.stackexchange.com/questions/135898/effect-of-magnetization-on-oxidation-rusting-of-iron and references cited,

OK, for the disbelievers of my last comment, to quote a source: "The results showed that magnetism had a significant positive effect on plant growth. Plant seeds under the influence of the magnetic field had a higher germination rate, and these plants grew taller, larger, and healthier than those in the control group. No adverse effects of magnetism on plant growth were noticed." Link: https://www.ysjournal.com/wp-content/uploads/Issue11/The-effects-of-magnetic-fields-on-plant-growth-and-health.pdf

Okay first, it's the oxygen in the air that causes oxidation, also dissolved oxygen in water, but if you let your tobacco get wet enough for the latter to be significant you will be getting more rot than curing ...
Anyway, I don't think a sealed jar would be a good idea. Ammonia build-up, wet rot, etc would tend to be an issue.
You could build a proper curing/fermenting kiln to cure the leaf, there's DIY instructions here: https://www.seedman.com/wkiln.htm for one type.  But if you're using the finished product for dry snuff rather than smoking or chewing, you might not be too bothered about the taste/smell so careful fermenting is probably not that important to you. Then easiest is to just hang them somewhere like the garage and spray them with a plant mister when they get too dry like this guy: https://dengarden.com/gardening/Tobacco-Growing-and-Curing-at-Home or if it gets too cold where you live to use the garage, in a closet, wooden box or linen chest.

Sorry for late answer. I confirm what you say by having tried it except kiln. The best result for my palette was to let leaf dry a day so that it "dehydrate collapse wilt" then put in blown up plastic bag and like you say it will rot and get too sour if left for too long. After a couple of days it smelled nice like cigars to me although i rarely smoke. The snuff was good but i used too much pot ash.

Sounds good more than good enough. You really don't need anything fancy. After all, the improvement to taste caused by fermentation was supposedly discovered by someone transporting bales of tobacco somewhere on the open deck of a barge/boat/ship, where they they got splashed, rained on, etc while drying in the sun ...

OK, I have a different opinion on the curing of tobacco step from my colleagues.
Water in contact with vegetation (likely rich in transition metals) with dissolved O2 and CO2 likely produces (especially in direct or diffused sunlight) powerful radicals that slowly removes/decompose organics. This is part of the science cited in the advanced oxidation process (AOP)  literature.
See, for example, this review reference and also, especially this, A review on advanced oxidation processes for the removal of taste and odor compounds from aqueous media.
Note, per the supplied Wikipedia reference: Curing involves, to quote:

...the slow oxidation and degradation of carotenoids in the tobacco leaf...

So water, oxygen, and metal oxides are likely instrumental in this process, all fostering radical formation and subsequent targeted 'favorable' alterations.

Water/humidity is important in the tobacco fermentation process. If the leaf gets too dry, the fermentation process inside the leaf will stop. But you don't want the leaf to rot - the surface has to remain intact - so the leaf should not get too wet or stay that way for too long.

interesting. It turned out that i asked for too much for my first curing. it went ok without understanding why. although i keep in mind for next years grow and cure


text:
Order of nucleophilicity of ammonia, hydrazine and hydroxylamine
I was reading about nucleophilicity and basic strength orders. I came across this interesting set of three compounds $\ce{NH3},$ $\ce{N2H4}$ and $\ce{NH2OH}$ where I am not able to apply the rules I have learnt.
The correct order of basic strength is given to be:

$K_\mathrm b$ value: $\ce{NH3} > \ce{N2H4} > \ce{NH2OH}$

The rules I know are:

If the donor atom is same, then order of nucleophilicity is same as the order of basicity.
Alpha effect: if the adjacent atom attached to the donating atom has lone pair of electrons, then nucleophilicity is enhanced.

When I try to conclude the nucleophilicity order, I am getting contradicting results with the above rules. Also, I wonder whether the −I effect due to the presence of highly electronegative atom may have a role to play.
So, what is the correct order of nucleophilicity of $\ce{NH3},$ $\ce{N2H4}$ and $\ce{NH2OH}?$
Is it theoretically possible to conclude the correct order based on deciding which factor will dominate in a given situation?
CLARIFICATION:
My doubt is pertaining to specifically these three compounds.Many have said that this question is a duplicate of Is there a general consensus on the causes of the alpha-effect?
But, I am not able to get the order of nucleophilicity that I need from the answers to that question.I am not able to find the information on order of nucleophilicity of these three compounds in the books I use.So, it would be helpful if someone clarifies the position of $\ce{NH2OH}$ in this.

Does this answer your question? Is there a general consensus on the causes of the alpha-effect?

@Mithoron I have already seen that question.But, I am not able to get the answer to my question over there

Well, nucleophilicity has little to do with basicity, that's why alpha effect was coined.

But what would be the correct order then?I saw this somewhere:N2H4 >NH3>NH2OH first one eventhough Kb is less than ammonia due to lone pair-lone pair repulsion and larger size it becomes a better nucleophile and -I effect of oxygen dominate over alpha effect.Is this correct?

https://chemistry.stackexchange.com/questions/83072/is-hydrazine-a-better-nucleophile-than-ammonia

Yeah, I have seen this too.Still, the position of NH2OH isn't clear to me.Can you clarify that?

You can only come up with a rough trend. Nucleophilicity of substances differs in different contexts.

@Zhe Got your point.Still would like to know the rough trend.Thanks

@Chem-Learner Why do you keep on insisting on [tag:organic-chemistry] tag? There is not even a single carbon atom in any of the compounds you are asking about.

I was reading this in my organic chemistry course under the topic effect of Nucleophiles.So, I thought I may get a related answer if I include this tag in my question.


text:
"e and a" as combinations of pi bonds
I am trying to get a better intuition on solid-state physics but I am a chemist by degree. I came across this article that attempts to bridge the gap between solid-state physics and molecular chemistry. However, I am unsure what the author is referring to when he says "a and an e combination". I believe it is talking about the symmetry of the pi orbitals but I really have no clue.

A is symmetric with respect to the principal axis, e is degenerated. You may want to look at character tables.

They are called Mulliken symbols or Mullikan notation if you try read more about it.


text:
If I break a crystal, would the "broken" pieces also have the same crystal shape?
If I've a crystal of some particular geometrical shape and i hammer it and it breaks into small pieces. Will the small pieces have the same shape as of the parent crystal?

No, it won't. $\ce{}$

What M. Farooq said is this site.

What is "the same shape", exactly? Preservation of crystal symmetry, morphology, linear/angular parameters of the bulk sample, their ratio, or something else?  OP might want to have a look at the old video by Alan Holden demonstrating what cleavage is (if that's what the question is about, and sorry, no hammering there).

Having worked with many highly perfect crystals, in the form of silicon wafers for semiconductor fabrication, I can tell you that I have never dropped one and had beautiful round pieces result.

In most cases, a random hammering on a crystal will smash the later into randomly shaped, even irregular pieces of matter.  As shown in the video suggested by @andselisk however, if you aim parallel to the cleavage planes you may obtain fragments which are similar to each other.  This similarity however is not necessarily about the shape of the objects (as in «a cube yielded smaller cubes»), nor in the ratio of the side lengths of the smaller fragments.  The similarity you may find is that the constant angle two faces enclose for one crystallographic phase of a given compound as illustrated below:

(adapted from enter link description here, part of this)
This form of regularity was identified by the Danish Niels Steensen, better known as Steno's law of the constancy of interfacial angles, published in 1669 which you may read digitized e.g., on archive.org.
The identification of the same surfaces among crystals stays of relevance today describing the habit of a crystal with Miller indices

(credit)
for example because the speed of propagation of light and refractive index generally depends on the direction in respect to the crystal's coordinate system.  Knowing the correct orientation helped you to pick the right face to look at while using the sunstone (described further here).
Addition:  The video shows how to split inorganic crystals with hammer and razor blade, which may appear as brute.  But it is not.  Indeed, the technique equally is used today in crystallography to «cut» needle-shaped crystals of organic matter into specimen around 0.3 x 0.3 x 0.3 mm (or below) before mounting them on a diffractometer.  Not only that this may be a delicate act dangerous to the sample quality (which you estimate rapidly while recording the data), often, you need only a very gentle tap with the razor blade along the cleavage plane to perform it.

(+1) Excellent answer! I just took a couple of pictures of the some of my calcite (Iceland Spar) and my faceted Iolite (Cordierite aka sunstone aka water sapphire). My photos are here for a couple of days. Feel free to use (or entirely ignore) them. The red arrows, on the picture with the raw calcite crystal (from Mumbai, India), show where the single edge razor blade could be placed. I actually cleaved off a small piece that way and very little force was required.


text:
Reaction of cinnamic acid with Br2/Na2CO3

The first reaction produces benzaldehyde, and the next one (perkin's condensation)produces Cinnamic acid.(X)
Now the treatment of X with $\ce{Br2/Na2CO3}$ is whats troubling me.  $\ce{Na2CO3}$ being a base, abstracts the hydrogen from the  $\ce{COOH}$ group.  $\ce{Br2}$ reacts with the alkene portion to yield a cyclic intermediate. What next?
The solution claims that somehow The  $\ce{CO2-}$ group leaves and a bromoalkene forms. The addition Of moist KOH in the next step results in  $\ce{E2}$ elimination to yield option (c).
I cant quite digest the $\ce{CO2-}$ group leaving. I thought of a mechanism, which is similar to the syn-elimination in esters: but that doesnt work out well..
Any hints/ insights will be appreciated. The correct answer is (c).

Br2 /sodacarbonate gives just anti addition to vicinal bromides. Just look carefully,the reaction is similar to treatment of vicinal bromide with soadamide(this rxn requires very strong base),KoH being less basic than sodamide ,thus additional heat has been provided. Thus 'C'would be formed.

The sodium carbonate is just there to mop up any $\ce{HBr}$. The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide.
The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.

I agree with answer (c) but, possibly,  a different mechanism. The post says that a bromoalkene is formed first. Dibromocinnamic acid in base concertedly loses CO2 and bromide to give (Z)-(2-bromovinyl)benzene. This reaction is known and is stereospecific using amines as the base. [Note: Bromination of methacrylic acid followed by heating the dibromide in pyridine give 2-bromopropene]. KOH then effects E2 elimination of the vinyl bromide to give phenyl acetylene. I think this is a made-up problem and I think your mechanism, in the presence of KOH, would prevail in the flask!

...continued. Reference for the stereospecific decarboxylation: DOI:10.1016/j.tet.2005.02.043

Just a doubt, in our lectures the difference between alcoholic KOH and aqeuous KOH was taught. I guess it's aqueous KOH here with 200°c. It was taught to us that aqueous KOH can only do substitutions not E2.

@vanshitarawat  note the temperature at which the reaction is - 473K - there is no KOH in solution as there no liquid water present.

An alternative to @Waylander 's explanation.  First, we brominate cinnamic acid to form cinnamic acid dibromide and because of the base $\ce{Na2CO3}$ the cinnamic acid is deprotonated, and thereby both CO2 and Br are eliminated  leading to the formation of bromostyrene and then KOH does its work.
Note: The answer may seem similar as of user55119 but here the reagent $\ce{Na2CO3}$is used and reaction conditions are similar to that of the author's question.

Just trying to share my 2 cents here.
Firstly, brominate the cinnamic acid to furnish a dibromide(trans).
Then using Sodium carbonate, de-protonate the carboxylic acid.
The anion would conventionally prefer to form a ring but, on close observation we see that the ring ,if formed, would be a 4 membered one which is known to be an unstable form.(5 and 6 membered rings are preferred as products)
Hence the anion would cause CO2 and -Br to be eliminated from the molecule.
Following a treatment by KOH, we would get the desired product.


text:
What is the derivation of the formula for finding number of pure and hybrid orbital?
I am studying Organic Chemistry. Recently I came up with the formula for finding number of pure and hybrid orbital that can be present in a organic compound. That is:

Number of Hybrid  Orbital = number of "Carbons" $\times$  (number denoting the hybridisation of carbon) and
Number of Pure Orbital = number of Hydrogens present $+ 2 \times$  (no of $\pi$-orbitals present)

Example Benzene:

number of hybrid orbital $= 6\times (3) = 18$ (Hybridisation is $\mathrm{sp^2} = \mathrm{1s} + \mathrm{2p} = 3$)
number of pure orbital $= 6 + 2 \times 3 = 12$

which tells that there are 18 hybrid orbitals and 12 pure orbitals present in benzene.
My question is, What is the derivation of this formula?

All we are doing is counting orbitals, and these formulas are generalizations of counting the same way that multiplication is generalization of repetitive addition. The derivation is exactly as you present it. For the first formula, we are multiplying the number of hybrid orbitals per carbon atom by the number of carbon atoms. For the second formula, we are multiplying the number of non-hybrid orbitals per carbon atom by the number of carbon atoms and adding the number of hydrogen atoms (since each hydrogen atom only has one orbital that cannot hybridize**).
If you took benzene $\ce{C6H6}$ and counted all of the hybrid orbitals, you would count 18. Each carbon atom is $sp^2$-hybridized, so each has three hybrid orbitals. We could count each individually:
$$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 18$$
Or you could group by each of the six carbon atoms, which eventually simplifies to $3\times 6$
$$(1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) = 18\\
3 + 3 + 3 + 3 + 3 + 3 = 18\\
3 \times 6 = 18$$
Your formula has some limitations. First, it only works for hydrocarbons (compounds with only carbon and hydrogen atoms), and it assumes all carbon atoms have the same hybridization. Here is a hydrocarbon where the carbon atoms do not all have the same hybridization. This is hex-1-en-5-yne and it has two $sp^3$-hybridized carbon atoms, two $sp^2$-hybridized carbon atoms, and two $sp$-hybridized carbon atoms.

Your formula could be extended to:
$$N_{\mathrm{HO}} = 2n_{sp} + 3n_{sp^2} + 4n_{sp^3}$$
Where $N_\mathrm{HO}$ is the number of hybrid orbitals, $n_{sp}$ is the number of $sp$-hybridized carbon atoms, $n_{sp^2}$ is the number of $sp^2$-hybridized carbon atoms, and $n_{sp^3}$ is the number of $sp^3$-hybridized carbon atoms. For hex-1-en-5-yne, we could count:
$$N_\mathrm{HO} = 2\times 2 + 3\times 2 + 4\times 2 = 18$$
There is a more general formula for hydrocarbons. Let's look at it this way. Each carbon atom has four orbitals (either hybridized or non-hybridized). If there are no $\pi$-bonds, then the number of hybrid orbitals is four times the number of carbon atoms. Each $\pi$-bond requires to non-hybridized $p$-obitals, so each $\pi$-bond reduces the number of hybrid orbitals by 2. Here is a general formula for hydrocarbons:
$$N_\mathrm{HO} = 4n_\mathrm{C} - 2n_\pi$$
Here $n_\mathrm{C}$ is the number of carbon atoms and $n_\pi$ is the number of $\pi-bonds$. For benzene, $n_\ce{C} = 6$ and $n_\pi = 3$, so
$$N_\mathrm{HO} = 4\times 6 - 2 \times 3 = 24 - 6 = 18$$
For hex-1-en-5-yne, we have the same number of carbon atoms and $\pi$-bonds, so we get the same result. For the number of non-hybridized orbitals ($N_\mathrm{NHO}$), we have a general formula as well. Each hydrogen atom contributes one non-hybridized orbital ($n_\ce{H}$), and each $\pi$-bond contributes two.
$$ N_\mathrm{NHO} = n_\ce{H} + 2n_\pi$$
For benzene and hex-1-en-5-yene, $n_\ce{H} = 6$ and $n_\pi = 3$, so
$$ N_\mathrm{NHO} = 6 + 2\times 3 = 12$$
This is the same as your second formula.
** I expect one of my physical/theoretical chemist colleagues to come along and inform us that under certain circumstances, hydrogen atoms may in fact act as if they have hybrid orbitals, but for most organic compounds we can ignore this behavior.


text:
What is the difference between pig iron and wrought iron?
I mean, I know the difference in the properties.. but how are they produced? what is the difference in the method of production?

Did you search the web?

Very little similarity except they are both mostly iron. Pig iron is from an archaic process of steel making ; The product of the blast furnace, very high carbon and silicon. Poor physical properties ,about the only direct use would be weights. Wrought iron is essentially "pure" iron, soft,ductile . Not much direct use ,it can be drawn into wire which hardens and gives it some strength. However, many products today are called wrought iron , they are actually (low) carbon steels . Today very little actual wrought iron is made ,I expect the furnace ( eg. oxygen converter )just blows oxygen ( and Argon ?) until carbon is down to 0.02 %+/-, adding no silicon or manganese. A pilot plant size operation.


text:
How to remove Acetic Anhydride?
I am trying to form 2,2-diacetoxy acid using acetic anhydride and glyoxylic acid monohydrate (Scheme below)

I could not find a suitable synthetic protocol for this, so I just added 10 equivalents of acetic anhydride to 1 equiv. of glyoxylic acid monohydrate and refluxed the mixture at 140 °C. Now the limiting agent (glyoxylic acid monohydrate) is gone, but I am not sure with how to remove excess acetic anhydride; its boiling point is too high (139.8 °C) and the pressure of the rotovap in my lab isn't that strong (I tried removing toluene before, it only worked when I increased the water bath temperature to 70 °C).
Yes, I am aware that I can make an azetrope by adding toluene, which will have a lower boiling point, but that would still require the water bath temperature to be quite high and I'm scared that my product might be broken down or perturbed at such a high temperature. Should I add ~2 N HCl and extract my product with DCM? Any other tips or guidance? I'd appreciate any help.

If you have access to a good vacuum pump (~30-50 torr) you probably should try fractional vacuum distillation.

You refluxed the reaction mixture at 140C and you are now worried about a high water bath temperature?

@Waylander you are right, wasn't thinking straight

@Waylander I made an azetrope with toluene but still can't distill it off. Would adding ~2 N HCl followed by extraction with DCM do?

I think there is every chance the 2N HCl would hydrolyse the acetal. Can you get an vac pump?


text:
Electron defficient molecules
Electron defficiency is a common feature in some class of compounds of Group 13 elements (B,Al,Ga). In some covalent compounds of Be too the same nature prevails. To compansate this electron defficiency they undergo dimerization forming bridges('3c 2e' or coordination) or in some cases they get stability through backbonding (halides of group 13 elements). Hydrides of them are known to stay in dimeric form through bridging.
Halides of B are found to remain in monomeric form due to backbonding where as 'Be, Al'-halides highly favour bridging structures through coordination of halide. Why is such difference? In Be,Al, vacant p orbital is present too to favour backbonding.
Then why Be,Al favour dimerization (or polymerization) over backbonding in their halides?

Why my questions are getting downvoted everytime? What is the reason?, I can't get it. What's wrong with this question?

Generally because there are some people around with nothing better to do than downvote questions and answers anonymously. The people who actually have a good reason for their votes and who want to help others rather than put them down will post a comment with their reasons for downvoting and suggestions on how to make your question or answer better. The rest should be ignored as the bottom feeders they are.

@Gwyn That is a very wild accusation you throw out there. This question has probably gotten a down-vote (and close votes), because it does not really fit the scope of this site. There are plenty of excellent textbooks, which explain this very well, see Resources for learning Chemistry  for a list or recommendations.

@Martin : I stand by my comment. Doing someone the courtesy of explaining a down vote is simple good manners (what us fossils would call proper 'netiquette). How can they learn if the just get slapped down with no explanation. And really? Doesn't fit the scope of the site? I would have thought voting to close as a duplicate would be more likely, since you are one of those who once spent ages debating a very similar question at:
https://chemistry.stackexchange.com/questions/34842/why-is-bcl3-a-monomer-whereas-alcl3-exists-as-a-dimer

Does this answer your question? Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

I believe this may be a duplicate of this question:
Why is BCl3 a monomer whereas AlCl3 exists as a dimer?
From what I remember, the stronger the backbond, the smaller the tendency to undergo dimerisation, however, way back in the dark ages when I was a student and asked my second-year chem prof, his answer was that the simple reason why none of the tri-halides of boron undergo dimerisation is steric hinderance - i.e. the boron atom is just too small to accommodate the fourth larger Cl/Br/I atom. He then pointed me to several sets of further reading.
However, you will probably find a better answer at one of the following links:

Mulliken J. Chem. Phys. 3, 635 (1935); https://doi.org/10.1063/1.1749567
Lappert, et.al. Chem. Soc., Faraday Trans. 2, 1976,72, 539-55 https://doi.org/10.1039/F29767200539

That is a very crude oversimplification.

@Martin:Yup. Still better than nothing.


text:
Reversibiliy and equilibrium
My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".
The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $\ce{Tsys=Tsurr}$ at every instant of the process.
Atkins-Physical Chemistry, page 77 states, while explaining the entropy change of the surroundings: "furthermore, because the temperature of the surroundings is constant whatever the change, for a measurable change,$\ce{\Delta}Ssurr=\dfrac{q_{surr}}{T_{surr}}$".
But then this treatment seems to contradict the fact that the system and surrounding are in thermal equilibrium! Since $\ce{Tsys=Tsurr}$ implies that $\ce{Tsurr}$ is not constant as long as the process isn't isothermal.
I therefore believe that my first statement is incorrect.If it isn't incorrect,then what am I misinterpreting?

I don't see the problem. The temperature is the same and is constant in the system and in the surroundings in all your statements. So they are all correct. There are no contradictions.

I really dont get what you mean

One paragraph (referencing atkins) clearly states that the surrounding temperature is constant.

While Tsys=Tsurr means that Tsurr is not constant. If the temperature of the system varies, This equation ensures that the temparature of the surroundings will also vary.

@Maurice would you please elaborate upon your comment?

The equation does not ensure that the temperature of the surroundings will also vary.  They are modeling the surroundings as an ideal (infinite) reservoir, whose capacity to absorb heat without its temperature changing is infinite.  This would approximate the behavior of a very large temperature bath with very high heat capacity and thermal conducivity, or a bath in which a phase change (e.g., an ice/water bath) is occurring.

Tsys=Tsurr. If Tsys changes, Then Tsurr changes. What's incorrect in this?

This equation for the entropy change of the surroundings assumes that you are dealing with a single reservoir at a constant temperature.  If, in a given reversible process, both the system and the surrounding temperatures are changing (in temdum), then, to use that equation, you need to use not just one reservoir for the suurroundings, but, rather, a (continuous) sequence of constant temperature reservoirs, each at a slightly different temperature.

If Tsys changes, the reaction is not reversible.

@Maurice  That's not correct.  Tsys can change in, for instance, adiabatic reversible reactions.  In addition, you can have, as Chet Miller described, a reversible process with diathermal walls in which Tsys changes.

@theorist. You are right. I was thinking of processes which are not adiabatic, where heat can get in or out of the system.

When a system is in thermodynamic equilibrium, there is no tendency for spontaneous change. In other words everything stays the same, the composition, temperature, pressure do not change over time, etc.
Therefore:
Tsurr is constant
and
Tsys is constant.
If you make an infinitesimal change, the system will tend to revert to the equilibrium.
So there is no issue.
When you are talking about a reversible process in thermodynamic terms, you're talking about a system which can be taken from its initial state to another state, and then back to the initial state without any net change to either the system or the surroundings.
The idea being that you make an infinitesimal change to some aspect of the surroundings, the system changes to a new state which is again in equilibrium with the surroundings, lather, rinse, repeat. Of course it would take infinite time for the process to be completed.

My textbook states "A system is in thermodynamic equilibrium if it is in Thermal, Mechanical, and chemical equilibrium simultaneously".

Since you are talking about thermodynamic equilibrium with the surroundings, I would replace that statement (which is fine by itself, though it is not comprehensive) with something like: "A system is in thermodynamic equlilibrium with its surroundings when it is in equilibrium with respect to all forms of work or heat flow that can move across its boundary".
Thus if the boundary is rigid, even a system is in thermodynamic equilibrium, it would not be in mechanical equilibrium with its surroundings, because it is not communicating with its surroundings mechanically.  If the boundary is adiabatic, a system at thermodynamic equilibrium would not be in thermal equilibrium with its surroundings, because it is not communicating thermally with its surroundings (e.g., if the walls are adabiatic, the system could be in thermodynamic equilibrium at, say T = 300 K, while the temperature of the surroundings could have any value (yes, above absolute zero, of course).

The Wikipedia article on "reversible processes" states that the system is thermodynamic equilibrium with the surroundings throughout the process. This means, as per my initial statement, $\ce{Tsys=Tsurr}$ at every instant of the process.

Correspondingly, the above is not generally true; it's only true for isothermal reversible processes (you may be misrepresenting the content of the Wikipedia article here by not including the isothermal restriction in your statement—I don't know, since you didn't provide the reference).  In an adiabatic reversible process, by contrast, the system is not in thermal equilibrium with the surroundings and, typically, $\ce{Tsys \ne Tsurr}$ (except perhaps at one point, or in the case that you have an adiabatic process with no temperature change, such as the free expansion of an ideal gas).

Since $\ce{Tsys=Tsurr}$ implies that $\ce{Tsurr}$ is not constant as long as the process isn't isothermal.

You have it backwards.  For $\ce{\Delta}S_{\text{surr}}=\dfrac{q_{\text{surr}}}{T_{\text{surr}}}$ to apply, the surroundings must be an infinite heat bath. I.e., the surroundings are so vast that their intensive properties (e.g., temperature) are unaffected by exchange of heat, work, or matter with the system. Thus, for isothermal processes in a system that is in thermal equilibrium with its surroundings, it is the temperature of the surroundings that determines the temperature of the system, not the other way around.

The process described in Atkins must refer to an isothermal process, since otherwise the appropriate expression to use in computing the change in entropy of the surroundings is
$$ \ce{\Delta}S_{surr}=\int\dfrac{dq_{surr}}{T_{surr}}=-\int\dfrac{dq}{T_{surr}}$$
where $dq$ represents an inexact differential.
This becomes equal to the statement you quote only if the temperature of the surroundings remains constant during the process.
The entropy change during a reversible process, during which there is thermal balance between system and surroundings at each step ($T_{surr}=T_{sys}=T$), is given by
$$ \ce{\Delta}S_{sys}=-\ce{\Delta}S_{surr,rev}=\int\dfrac{dq}{T}$$

The apparent contradiction comes from the fact that in the case of Atkins' Physical Chemistry there is no assumption of equilibrium, just that the surroundings have a temperature that is not affected by whatever happens to the system.