text:
Reaction of 2,3-dichlorobutanoate with a decarboxylating agent followed by hydrochloric acid and then aq. potassium hydroxide
Question

$$\ce{CH3-CH(Cl)-CH(Cl)-COO- ->[-CO2]A->[HCl]B->[aq KOH]C}$$ Product C is:
a) $\ce{CH3-CH2-CHO}$
b) $\ce{CH3-CH(OH)-CH2-CH2-OH}$
c) $\ce{CH3-CHO}$
d)$\ce{CH3-C(=O)-CH3}$

Answer

 The answer given is (a)

My attempt

Where did I go wrong?

the addition of HCL youve done to alkene is wrong , it will be geminal not vicinal there by making ans (a)

The answer is a) as long a propionaldehyde can avoid an aldol condensation in the presence of aqueous KOH. c) can't be the answer. Do the math on the number of carbons.

Yeah the answer is A my bad( i will edit it). But why where would the H+ attack(in the 2nd step)? I thought as i did because Cl- would make C+ unstable?

Also can someone advice me how to convert pictures to text like the above text ? I will try to make my questions better.

Shreya why would it be geminal ?

@AdityaSURESH check this out-https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here  for writing molecules on here or else you can use online softwares to draw molecules

See here and here. For a more detailed MathJax guide, look here, minor other details

(continued) To draw chemical compounds, this should help

Considering the follow up doubt, why geminal and not vicinal?

the carbocation formed is more stable because of $\ce{Cl}$ +M.

also make sure you dont use this logic anywhere else, this is an observation specific as theres +M and -I continuously fighting.


text:
How toxic chemically is plutonium (Pu), neglecting the radioactive damage?
In Rhodes' The Making of the Atomic Bomb, he says that, while Pu is not that radioactive (which is surprising -- maybe he means compared with radium and some other elements), it is very toxic.
I would guess it could not be much more toxic than uranium (U) or lead (Pb) -- I believe thallium is more toxic than U and Pb and so is mercury (Hg). If it is more toxic than lead, do we understand the mechanism?

Since we don't have any non-radioactive plutonium it is probably impossible to determine whether it has other forms of toxicity. Rhodes may just have been (badly) describing the difference of lumps of plutonium outside the body and small amounts inside the body.

I suppose one could perform experiments with different Pu isotopes (with different half lives and decays), but since most folks don't encounter Pu it probably is not particularly cost effective.

I think Mr. Rhodes has just fallen for an urban myth, or to put it another way, he´s been reasoning by analogy. The statement is just completely pointless.

Apparently, it's supposed to be less toxic than caffeine. See last paragraph in  https://en.wikipedia.org/wiki/Bernard_Cohen_(physicist)#Scholarly_achievements

You should read this: https://pubs.acs.org/doi/pdf/10.1021/acs.organomet.7b00605 - about how it is not easy to generalise the toxic properties of heavy metals

I have attended lectures given by actinide chemists who remarked that Pu has very high chemical toxicity, in the sense of rapid effect and a small LD50. That is consistent with Rhodes's remarks, but I cannot personally attest to it or provide a citation. The half life of the fastest-decaying Pu isotope is over 14 years, and this isotope does not occur naturally. The next smallest is nearly 90 years, and the rest are in the thousands of years or more.  Most of them are a lot shorter than any of Uranium's, but much longer than, say, Np-238's.

Thanks for asking this question. I'd heard before that Pu was actually more chemically toxic than its toxicity due to its radioactivity, but had never followed up by checking out this claim in detail.
tl;dr: Plutonium is very safe unless you grind it up into dust and inhale it, in which case the hazard is probably from radiation, not chemical toxicity. There is probably no way of knowing for sure, because we don't know enough about things like the dependence of harm from radiation on the dose rate as opposed to the total dose.
I'm going to assume we're talking about 239Pu, which is what is used in bombs. It has a half-life of 24,000 years, and it decays via alpha emission to 235U, which is effectively stable (half-life of almost a billion years). Note that it doesn't decay by fission (that only happens in bombs), so comparison with nuclear fallout is not relevant.
To get a comparison of chemical toxicity to toxicity due to radiation, I think all we need is a very rough order-of-magnitude estimate of chemical toxicity. Organic mercury is extremely toxic (about 10 times worse than arsenic), and it has an LD50 of about 0.1 g, so that's probably a reasonable upper limit for any other heavy metal such as plutonium.
239Pu's radioactivity is in the form of alpha particles, which are only dangerous if emitted internally, because they can't penetrate the epidermis. The hazard then seems to me like it would depend crucially on how the internal exposure occurred.  If you eat an alpha emitter, then it will stay in your body until it's either excreted or decays. The half-life for excretion of organic or metallic Hg is on the order of months, so that's probably a reasonable order-of-magnitude estimate for other heavy metals. On the other hand, there are people at nuclear weapons labs who do machining of plutonium pieces, and this kicks up plutonium dust. They do this machining inside glove boxes, and I assume the chips and dust are swept up very carefully. The dust can be breathed in, and microgram particles that get into your lungs are likely to remain there for the rest of your life (Lenntech, ATSDR 2010). From the lungs, it can also migrate to the bones or liver. ("Much less than 1%" of ingested Pu would do so.)
For internal exposure to alpha emitters, we have good data on 210Po, which was used in the Litvinenko assassination. The LD50 for this substance is believed to be about 1 μg. It has a half-life of 138 days, and although I don't know its half-life for excretion, as described above I'd guess it's on the same order of magnitude, so to within a factor of order unity, we can probably pretend that all of it decays within the body. But if you eat 239Pu, then assuming it gets excreted within ~100 days, its probability of decaying within your body is only about $10^{-5}$. Therefore we would expect the LD50 for ingested 239Pu to be on the order of $10^5$ times greater than that for 210Po, or 0.1 g. This is on the same order of magnitude as the chemical toxicity. Well, this would be of interest if any human being (or lab animal) were ever going to be exposed to this much of the stuff, and then maybe we would be motivated to work out the biochemistry in more detail, do studies, and so on. But in reality, nobody has ever eaten or will ever eat this much plutonium, so it's of no interest.
For inhalation, the balance shifts. If someone lives for 50 years after exposure and carries the dust in their lungs for that whole time, then the probability of decay within the body is $\sim10^{-3}$. That means that in terms of radiation damage, the LD50 should be about 1000 times higher than for 210Po, i.e., about 1 mg. This would make it about 100 times more toxic in terms of radiation than any plausible chemical toxicity.
You could argue that plutonium might be less lethal than implied by this estimate because a radiation dose that would be lethal if suffered all at once can be less dangerous or almost harmless if spread out over a period of years. This probably reduces the radioactive toxicity of plutonium by a big factor relative to the estimate above, but I don't think it can reduce it by a factor of 100, which is what we would need in order to make the radioactive toxicity comparable to any realistic level of chemical toxicity.
So to summarize:
For external exposure: Zero hazard from radiation, some hazard from chemical toxicity but probably no worse than the (very small) risk from external exposure to substances like lead or mercury.
Ingestion: Chemical toxicity is not relevant because nobody will ever ingest enough to be in danger. On a population basis, radiation could conceivably cause some excess cancers, but I doubt that there is very much unreacted plutonium in fallout, and the population risk does not seem to have been considered sufficient to merit discussion in Simon et al.
Inhalation: The hazard from radiation is probably 1-100 times greater than the chemical toxicity is likely to be, even if the chemical toxicity is as bad as organic mercury.
References
ATSDR 2010, "Public Health Statement for Plutonium," https://www.atsdr.cdc.gov/PHS/PHS.asp?id=646&tid=119
Lenntech Water Treatment Solutions, "Health effects of plutonium," https://www.lenntech.com/periodic/elements/pu.htm
Simon et al., "Fallout from Nuclear Weapons Tests and Cancer Risks ," https://www.cancer.gov/about-cancer/causes-prevention/risk/radiation/fallout-pdf

The toxicity is primarily due to radioactivity and to absorption by the body, where that radioactivity can act internally. There is, "significant deposition of plutonium in the liver and in the "actively metabolizing" portion of bone," according to Miner and Schonfeld.
Many $\ce{Pu}$ isotopes are primarily alpha-emitters, with "high energy radiation with low penetration and thereby requires minimal shielding. A sheet of paper can be used to shield against the alpha particles emitted by plutonium-238," from Wikipedia. The risk of holding a small amount of plutonium, if it is sealed inside almost any container, is negligible, because the the heavy, positively-charged alpha particles are stopped by the container in a very short distance.
By that same token, when absorbed, the plutonium migrates to rapidly dividing liver tissue and bone's hematopoietic cells. The alpha particles are stopped in a very short distance, depositing all their energy in the few nearest cells and massively interfering with cellular reproduction. Small amounts can cause various types of cancer, such as leukemia, a cancer of the previously-mentioned blood cell producing tissues.
That said, plutonium is not quite as toxic as one might think: "The U.S. Department of Energy estimates that the lifetime cancer risk from inhaling 5,000 plutonium particles, each about 3 µm wide, to be 1% over the background U.S. average... no human being is known to have died because of inhaling or ingesting plutonium, and many people have measurable amounts of plutonium in their bodies."
Perhaps what Rhodes means by "not that radioactive" is that the radiation from plutonium is not very penetrating. The isotope used in nuclear weapons, 239Pu, has a half-life of about 24,000 years, comparable to radium, or to 231Pa. If I've done my math  correctly, one mole of 239Pu, about 6e23 atoms, with  half-life of 8e11 seconds, should emit about 400,000,000,000 alpha particles per second. Of course, if it were a sphere, most of those particles would be absorbed inside the mass of metal, releasing energy as heat. However, spread through a human body, each particle has a much greater chance to cause damage. And 238Pu, used in radioisotope thermoelectric generators, is far more radioactive!

My sense from Rhodes is toxicity not due to radioactivity just as lead is not due to radioactivity but rather, as i understand it, displacement of vital minerals in the body. It sounds like you are saying that it is only or primarily due to radioactivity so i am still puzzled about what Rhodes meant.

Interesting - I'm not sure I buty the 'no human ... is known to have died because of inhaling' - perhaps not directly, but plutonium-caused lung cancer leading to death certainly is a thing (similarly various leukemia varients from Pu inhalation/ingestion as well).

@JonCuster, Agreed! There were many children downwind of the US nuclear test sites who died of leukemia (the Wikipedia article, Downwinders, was cited). The statement, I believe, is trying to show that Pu alone has not been demonstrated to have caused deaths... a quibble.

Another way to do the calculation (sorry for not using SI units): The activity of 1 g Ra-226 is (by definition) 1 Ci. As Pu-239 has a half life which is a factor 24000/1600 longer than that of Ra, you'll need 15 g of Pu to give 1 Ci.

It would still be interesting to know about the chemical toxicity (heavy metal), e.g. have it quantified. What is the ratio of  LD50s? Is it 1:10, 1:1000, or 1:1000000000000? Wouldn't acute chemical poisoning dominate at a high dose?

What you say in this answer is pretty much true, and yet it doesn't do anything to answer the question, which asks about the chemical toxocity.

@DrMoishePippik Those children who died of leukemia likely died due to inhalation of fission byproducts, not plutonium itself. Typical fission byproducts tend to be much, much more radioactive than plutonium. Pu-239 has a half-life of 24,100 years, which makes it not very radioactive in the grand scheme of things.

Actual toxicity other than radioactivity is not, as far as I know, very well studied. Quite simply, most of the danger is the radioactivity in general, as well as the toxicity of decay products (uranium and americium).
Basic knowledge of biochemistry though suggests that it should be toxic for the same reasons that most high atomic weight metals are, namely that most biochemical processes do not differentiate well between elements beyond the ionization charge (this is why rubidium and cesium salts are toxic, the large ion size for both of them combined with the +1 ionization state causes them to get stuck in and block potassium and sodium ion transfer channels in cells). The tendency of plutonium to concentrate in the liver and areas of bone growth supports this, as both are areas where ions with similar charges tend to concentrate (note that plutonium ions can be anywhere from +2 to +7, with +4 being the predominant form and +7 being exceedingly rare).
Also, for the record, Rhodes is correct that weapons-grade plutonium is comparatively not very radioactive. There are five major isotopes of plutonium, with atomic weights from 238 to 242. Plutonium-239 is the isotope used for weapons production, and decays via alpha particle emissions, which means it's not really all that dangerous except internally, especially since it decays rather slowly (it's half-life is 24100 years). Notably, other than the spontaneous fission of Plutonium-240, all the isotopes except Plutonium-241 (which undergoes beta decay to Americium-241) only undergo alpha decay, and as such are relatively safe with minimal shielding (ignoring the high decay-heat of Plutonium-238 which is utilized to power RTGs).

The radioactive aspect of toxicity is not that interesting; the idea that any element (as opposed to chemical compound) can be extremely toxic is interesting. I imagine that non-radioactive effects of an element would be "swallowed up" as the element combines with cells in the body so that there is a limit on how dangerous a very small amount of anything can be although I think microgram-range amounts of something like botulism toxin can be fatal.

@releseabe re toxic in small quantities, see, e.g., https://en.wikipedia.org/wiki/Dimethylmercury

even dimethyl mercury requires more than micrograms.

@releseabe It kind of depends on the context of the exposure how toxic/dangerous something is. Ingesting a few mg of pure zinc can easily kill you for example (due to the formation of a zinc-chloride solution in the stomach, which will eat through your stomach lining) despite zinc being largely non-toxic. You're generally right though that pure or ionic elements tend to require non-negligible concentrations to show actual symptoms of toxicity, though they often are still toxic in very small quantities.

The word "toxic" as applied to a radioactive element always troubled me: is molten iron "toxic" or just very hot? You would not attribute heat stroke to solar toxicity. Are acids toxic or just corrosive? (Although hydrofluoric acid is indeed also toxic -- hydrochloric acid is not to me toxic although you can die from it just as gulping hot chowder might kill you by burning your esophagus.)

@releseabe Agreed, it's kind of hard to determine at times what constitutes 'toxic' in some cases versus just 'damaging'. The usual differentiation I've seen is that things that are 'toxic' actually interfere in some way with normal biological processes (in contrast to things that just outright destroy organic matter, as is the case with most caustic solutions, such as zinc chloride).

I guess this the the only answer which attempted to answer the question. The asker specifically said "Neglecting the radioactive damage". The other 2 answers are explanations about radiation, which the question did not ask about.

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text:
What specific toxins are formed during deep frying? And how (chemically) do they attack the body cells?
From what I understand, deep frying causes the oxidation of lipids into advanced lipid oxidation products. When these are consumed and metabolized by the body, they release ROS or free radicals which damage DNA and other cellular components, paving the way for age related decline, chronic inflammation and terminal illnesses such as cancer, Alzheimer's etc.
What are the names of some of these carcinogens from frying?
Here's an arbitrary example: Lets say we had some chicken fingers and french fries. What specific carcinogens are formed when the chicken and potato are kept in hot oil for a long period of time?
And what are the biochemical mechanisms by which they harm the body?

This question could only be answered after someone finishes a PhD and a postdoc on this topic. Believe it or not it might take 7-8 years to find answers to those questions and even more.

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text:
Definition of throughput
I wonder what's the exact meaning of throughput in chemistry synthesis, eg. in organic synthesis or production of pharmaceutical products. For instance, what are their units? How are they measured?
I've read one paper mentioning their production methods have a high throughput and production rate (https://pubs.rsc.org/en/content/articlelanding/2013/sc/c3sc50859e#!divAbstract).
Are these two concepts the same thing?
Share with me your thoughts. Thanks a lot!

Usually they're synonymous.  Throughput would have the same units e.g. tonnes/day, kg/h.  Basically,  mass or volume per unit time.


text:
Is nitrogen with 3 different substituents and lone pair a chiral center? Does it make the whole compound chiral too?
Is nitrogen considered a chiral centre or stereocenter when it has three different substituents attached to it along with a lone pair.
By definition LG Wade says

An  asymmetric  carbon  atom  is  the  most  common  example  of  a  chiral  center  (or chirality  center),  the  IUPAC  term  for  any  atom  holding  a  set  of  ligands  in  a  spatial arrangement  that  is  not  superimposable  on  its  mirror  image.  Chiral  centers  belong  to an  even  broader  group  called  stereocenters.  A  stereocenter  (or  stereogenic  atom)  is any  atom  at  which  the  interchange  of  two  groups  gives  a  stereoisomer.*  Asymmetric carbons  and  the  double-bonded  carbon  atoms  in  cis-trans  isomers  are  the  most  common types of stereocenters.

So by the definition of chiral centre and stereo center , if here I interchange two of the substituents then I get a non super-imposable structure which is the mirror image of original compound. Sure, it can undergo inversion, but the definition only says about non super-imposability of mirror image and getting a new stereoisomer on interchanging 2 substituents.
Also if we are calling the nitrogen as chiral centre will be also called the compound to be chiral??
Here The answer says it can only be chiral if we slow down the inversion... But the definition doesn't have any condition about how fast the inversion is , it's just checking super-imposability of mirror images
Note : I'm not just asking about the compound being chiral, but the chiral center. Kind of like (R,S)-2,3 dichloro butane is not chiral molecule, but has a chiral center.
It's not a duplicate as by definition of IUPAC Chiral center is the term for any atom holding a set of ligands in a spatial arrangement that is not superimposable on its mirror image. They don't talk about how fast the inversion is at all

If inversion takes place, then the molecule is similar to an sp2 Nitrogen and so doesn't show chirality, however if you can restrict the rotation by attaching it to a ring or attaching bulky substituents then it is chiral if th 3 substituents are different. Inversion happens at very high orders of rate and so the nitrogen seems to be as in its sp2 transition state

Well it will not be isolatable, into seperate constituents, but that shouldn't mean it's not chiral

It keeps interconverting between the two enantiomers and so avg. rotation would be zero. Hence it would not be chiral. This is a similar reason to why butane has asymetrical conformers but does not show chirality.

The accepted answer in the the linked question answers this very well. It talks about what I just said

Please read the definition that LG Wade and IUPAC gives . Chiral cneter is the term for any atom holding a set of ligands in a spatial arrangement that is not superimposable on its mirror image

@Safdar Not super-imposable on mirror image is the only condition that they demand for being a chiral center

@ManishMittal An amine would have been chiral if it only existed in one form. But since it undergoes inversion rapidly and establishes a chemical equilibrium, irrespective of whether you started with x moles of the R form or the S form, at equilibrium, you will have x/2 moles each of the R and S form of the amine each. Since the solution has equal amounts of enantiomers rotating light in the opposite directions, a racemic mixture will be formed and the net rotation of the amine solution will be zero, hence making it achiral. Does this answer your question?

@YusufHasan I think you can write that as an answer.

Basic idea being that you can't isolate a chiral amine due to racemization, as the solution becomes optically inactive. In a sense you can call this situation "achiral" or more technically "optically inactive". But since we really can't isolate chiral amine molecules, so it is generally stated that these chiral molecules as such don't exist as independent entities, and  the solution as a whole is "achiral"

I'm not sure why you say it's not a duplicate: you say you're asking about whether it's a chiral centre, not whether the molecule is chiral. Firstly, I don't think there's any difference between the two in this case (the meso compound you raise is a different matter altogether). If the N is a chiral centre, then the amine is chiral, and if the amine is chiral the chirality has to come from somewhere, which in this case is a chiral centre. And anyway, secondly, that question and answer explicitly discuss chiral centres.


text:
Reasons for antiperiplanar arrangement in E2 reactions
From what I have learnt, there are 2 reasons that synperiplanar arrangement cannot give E2 reactions:

In Newman projection, synperiplanar arrangements is in a eclipsed form (very high in energy), so less likely to be in syn formation than anti. However, I am thinking that as the C-C sigma bond is always rotating, there is still a little chance for synperiplanar to undergo E2

In synperiplanar arrangement, the base (usually rich in electron density) will face a high electron repulsion with the leaving group when approaching the beta-hydrogen, hence reduce chance of successful E2.

However, these reasons only reduce the chance of synperiplanar arrangement from going E2, but cannot comepletely eliminating it. So I want to ask whether there are other reasons (e.g. cannot form / break the bonds if in synperiplanar during E2), in line with the experimental results that only anti can gives E2?

I'm fairly sure synperiplanar E2 reactions can happen; they're just less favourable. There is a classic case with norbornyl bromide where the ring system forces a syn elimination https://pubs.acs.org/doi/abs/10.1021/ja01067a018


text:
Water solubility comparison of stereoisomers of bicyclo [4.4.0] decane-3,4-diol

I think, (I) must be more soluble in water. It is due to trans configuration and thus, forming more effectively inter-molecular hydrogen bonds with water molecules.


text:
How did Mendeleev know elements from compounds or mixtures?
I read that Mendeleev initially thought didymium was an element, but it was actually a mixture.
How did he know the rest of the chemicals in the periodic table were elements and not compounds, especially since back then most elements would have been incredibly difficult to isolate and get a mono-atomic version?
It seems like an impossible task in the 19th century.

Element was what you can not split experimentally. For example, you can split rust to iron and oxygen if you really want to, but you can not split iron to other compounds (elements) by any chemical means. Yes, it was hard to decide, what is an element at the time. Playing with a lot of compounds, elements, checking also the mass ratio, made slowly clear the "atomic theory" (which was not very widely accepted even until the early XX. century!).

It is impossible to give a simple answer in this forum. It was a difficult problem and each element is different. So a long discussion would be needed to deal with each element, from the lightest to the heaviest. It would need at least one page per element. And there are more than 100 elements ...

Please do not underestimate the scientists of 19th century. They were as creative, intelligent and perhaps more genuinely dedicated to science than the scientists of the 21st century. Spectroscopy was the tool of the trade to identify and verify that a given substance is not a mixture.
The original reference which established that Didymium was a mixture is in German [1] (English title: "On the decomposition of didymium in its elements").
I don't know enough German to read it conveniently, but it is clear that the authors did fractional crystallization of a lot of salts, made measurements absorption spectra, did emission spectroscopy and finally led to a beautiful result. The original paper has plenty of spectra! From [1, p. 490]:

Da sonach die exacte Zerlegung des Didyms in mehrere Elemente realisirt ist, so schlage ich vor, die Bezeichnung Didym nunmehr ganz zu streichen und beantrage, für das erste Element, entsprechendder Grünfärbung seiner Salze und seiner Abstammung die Benennung:
Praseodym mit dem Zeichen $\ce{Pr}$
und für das zweite, als das „neue Didym“, die Benennung:
Neodym mit dem Zeichen $\ce{Nd}.$
Aus den Atomgewichtsbestimmungen, die in allen Einzelnheiten nach Bunsen ausgeführt wurden, ergaben sich als vorläufige  Werthe für Praseodym = 143·6, für Neodym = 140·8, wobei dem entsprechenden Oxyd die allgemeine Formel $\ce{M2O3}$ zukommt.

English translation with assistance of DeepL:

Since, therefore, the exact division of the didymium into several elements has been carried out, I propose that the name 'didymium' be deleted altogether and that the first element be given a name corresponding to the green colour of its salts and its origin:
Praseodymium with the sign $\ce{Pr}$
and for the second, as the "new didymium", the naming:
Neodymium with the character $\ce{Nd}.$
From the atomic weight determinations, which were carried out in all individual units according to Bunsen, it resulted as provisional values for praseodymium = 143·6, for neodymium = 140·8, whereby the corresponding oxide is assigned the general formula $\ce{M2O3}.$

Anyway, historically, it is not Mendeleev who had to decide whether the elements were actually elements. You may ask the right question, how did the element discoverer decided that it was an element in Mendeleev's time? The answer is that each element has a story to tell and that is why Isaac Asimov wrote a beautiful book called "The Search for the Elements". Go ahead and get hold of it.
Reference

v. Welsbach, C. A. Die Zerlegung des Didyms in seine Elemente. Monatshefte für Chemie 1885, 6 (1), 477–491. DOI: 10.1007/BF01554643.

"Please do not underestimate the scientists of 19th century. They were as creative, intelligent and perhaps more genuinely dedicated to science than the scientists of the 21st century." Ain't that the truth.

To anyone interested, I strongly recommend this article as a window into the exhaustive work involved in the isolation of an element, even in the early 1900s. As the author states: The most important point proved by this work is that the element, giving the characteristic absorption bands of thulium, cannot be separated into simpler substances. After about 15,000 operations the absorption spectrum underwent no change.

@NicolauSakerNeto, I was teaching a lab, and a student was collecting a Raman spectrum, which took exactly one min for acquiring the data. She was quite annoyed that it was taking too long!

@M.Farooq This is why it's so important to know the history of our field! It blows my mind to think of the conceptual leaps and bounds that were made with flame and flask, yet the first time I heard of the 1800s obsessions with balloons, I myself thought it funny. The very fact that many of us initially take their work for granted/as obvious is testament to how well they did it.

@jezzo, I completely agree! That is why I wanted to learn functional German. It comes to use in these historical cases! If you really like history, search "Ask the Historian" William B. Jensen. It is book of historical chemistry question and answers whose answers are not available in textbooks. For example, why R is used for universal gas constant, and so on.

@M.Farooq I look forward to the read!


text:
Molality of solution to be used when finding water separated from solution as ice
Question

$\pu{1 kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?
$K_\mathrm{f} \ \text{(water)} = 1.86\ \mathrm{K\ kg/mol}$

Solution given:

$\Delta T_\mathrm{f} = 0 - (-4)=4$
$i = 1$ for sucrose.
Therefore, since $\Delta T_\mathrm{f}$ is defined as
$$\Delta T_\mathrm{f}= K_\mathrm{f}\cdot i \cdot \frac {\text{amount of sucrose in } \pu{mol}}{\text{weight of water left in }\pu{kg}}$$
[...]

My doubts
$\Delta T_\mathrm{f}$  is the absolute value of the difference of freezing point of pure solvent and the solution with the non volatile solute. Here, the freezing point of the solution is not $\pu{-4 ^\circ C}$. So how can we say $\Delta T_\mathrm{f}$ is equal to $4$?
Why does the molality term contain amount of remaining water in the denominator? In the original equation shouldn't it be the original molality here?

This is because we say the temperature is fixed at -4 when actual freezing point is zero..

Go on, plug the original molality and find the freezing point of the original soultion (which is not $-4^\circ\rm C$, that's true). Think about what happens next.

The question: $\pu{1kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?
Given: $K_\mathrm{f} \text{ (Water)}= \pu{1.86 K kg/mol}$

The solution is given assuming you understand the concept. It seems you did not completely understand the colligative properties of solutions. Thus I'm giving some clues you how to approach this problem.
The textbook equation you are going to use is:
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times \text{molality of the solution} = K_\mathrm{f}\times i \times \frac {\text{amount of sucrose in }\pu{mol}}{\text{weight of solvent in }\pu{kg}} \tag 1$$
where $i = 1$ for sucrose (sucrose does not ionize in water).
The solution is kept at $\pu{-4 ^\circ C}$. The freezing point of pure water is $\pu{0 ^\circ C}$ ($\therefore \ \Delta T = 0 - (-4)=4$). First what you should do is find the freezing point of given solution using equation $(1)$. If $\Delta T_\mathrm{f} \gt 4$, the game is over. That means, this solution does not freeze at $\pu{-4 ^\circ C}$. If $\Delta T_\mathrm{f} \lt 4$, the game is on. That means, some water will freeze out until the final molality reached the molality of solution, which freeze at $\pu{-4 ^\circ C}$.
For given solution (no units was used):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m = 1.86 \times 1 \times 0.75 = \pu{1.395} $$
Thus, this solution freeze at $\pu{-1.395 ^\circ C}$.
Now, you have to find out at what molality of a solution would be just about to freeze at $\pu{-4 ^\circ C}$. That means, $\Delta T_\mathrm{f}= 4$ (still, $i = 1$):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m \ \Rightarrow \ m= \frac{\Delta T_\mathrm{f}}{K_\mathrm{f}} = \frac{4}{1.86} = 2.15$$
This means, water in original mixture freeze out until the molarity of solution becomes $\pu{2.15 mol/kg}$.
Put it this way:

The original solution has $\pu{0.75 mol}$ of sucrose in $\pu{1 kg}$ of water. The amount of sucrose would not freeze.
Thus, how much water would remain with $\pu{0.75 mol}$ of sucrose to keep molality at 2.15?

I'm sure it is given in your solution manual. Regardless, I'll show you the calculations. Suppose the weight of water is $W \ \pu{kg}$ (ignore the units in equation}:
$$ \frac{0.75}{W} = 2.15 \ \Rightarrow \ W = \frac{0.75}{2.15} = 0.349$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.349 kg}$ (the amount of sucrose is still $\pu{0.75 mol}$).
However, this calculation is in significant deviation from actual amount since the given amount of original solution is $\pu{1.0 kg}$. That means it does not contain $\pu{0.75 mol}$ of sucrose. Now, molecular weight of sucrose is $\pu{342.3 g/mol}$. If amount of sucrose in $\pu{1.0 kg}$ of original solution is $x$, then:
$$\frac{\frac{x}{342.3}}{(1000-x)\times 10^{-3}} =0.75  \ \Rightarrow \ 1.257x = 0.75 \times 342.3   \ \Rightarrow \ x = \frac{256.725}{1.257} = \pu{204.2 g}$$
Therefore, the amount of water in that solution is $\pu{1000-204.2) g} = \pu{795.8 g}$ or $\pu{0.796 kg}$. Thus, amount of sucrose in the of original solution is: $\frac{\pu{204.2 g}}{\pu{342.3 g/mol}} = \pu{0.597 mol}$.
Using these values, we can recalculate the water amount in final solution. Suppose the weight of water in final solution is $W_f \ \pu{kg}$:
$$ \frac{0.597}{W_f} = 2.15 \ \Rightarrow \ W_f = \frac{0.597}{2.15} = 0.278$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.278 kg}$ (the amount of sucrose is $\pu{0.597 mol}$ making it $\pu{2.15 molal}$).
Thus, the amount of water freeze out from the original $\pu{1.0 kg}$ of solution is:
$$\pu{(0.796-0.278) kg} =\pu{0.518 kg}$$


text:
Questions on formation of cyclic ethers using Williamson Ether synthesis
question 1:

In the synthesis, both intra- and intermolecular SN2 can undergo, in which only intramolecular reaction gives the cyclic ethers. However, both reactions happen in the reaction mixture. How can I alter the reaction conditions / add other reagents such that the intramolecular reaction becomes more favorable?
One solution I found is diluting the solution since the rate of 2nd order (intermolecular) reaction drops faster than 1st order (intramolecular) reaction. However, I wonder is it practical to dilute the solution to a large extent for purity, compromising the increased reaction duration and is there any faster and simpler reactions?
question 2:
This synthesis involves SN2, so I am thinking if there is a tertiary carbon bonded to the leaving group, it is impossible to form a cyclic ether. Therefore, are there any other ways I can synthesis cyclic ethers with the starting material as monohaloalcohol?

This is not an epoxide that you are making

High dilution is a well-recognised technique for favouring intramolecular cyclisations over inter, particularly in the cases of forming large rings

@Waylander Thanks for reminding. But may I ask under high dilution, what is the typical duration for the reaction?

Suspended NaH in refluxing diethyl ether. Add bromoalcohol in ether dropwise to the base. You will have to separate ether from the cyclic ether (THF). Notice I didn't use THF as a solvent.;)

High dilution reactions can be many hours with the reactant added by syringe pump.

Thank you all! I just found out I was drawing THF as the product...


text:
In the absence of a physical model, how can I use free software to visualize stereochemistry?
In a different question, the task was to turn a wedge-and-dash representation into a Fischer projection. This involved rotation of a fairly complex molecule - either as a physical model or a mental model - to orient each chiral carbon in preparation for drawing the Fischer projection. How can you use free software to visualize this process?

This might be useful if kept in meta since most of formatting tricks are kept there.. +1

There is a lot of free 3d molecular viewers - Maestro for academics, VMD, Pymol, Avogadro... Some of them have option to optimize/clear structure

I used Jmol to make the following animated GIF:

First I used NCBI's Pubchem Sketcher to get the SMILES code for the molecule: CC(C(C(C)F)(Br)F)O
Then, I loaded it into JSmol running here using the load molecule by name button.
Internally, this triggers the command
load /*file*/"https://cactus.nci.nih.gov/chemical/structure/CC(C(C(C)F)(Br)F)O/file?format=sdf&get3d=true
I fixed the stereochemistry by clicking on the carbons with incorrect chirality after issuing the command "set picking INVERTSTEREO". To rotate around the bonds, I used "rotate branch @3 @2 180 180" and "rotate branch @3 @4 180 180". Finally, I chose the viewing orientations, displayed the orientationwith "show moveto" and issued the resulting moveto commands before and after rotating around the single bonds.
A more elaborate example with glucose (4 chiral centers) can be found on Proteopedia.


text:
Clar's rule of polycyclic aromatic hydrocarbons
Clar's rule  states that the Kekulé resonance structure with the largest number of disjoint aromatic π-sextets, i.e., benzene-like moieties, is the most important for characterization of properties of polycyclic aromatic hydrocarbons (PAHs). What is the reason for the validity of Clar's rule?
I mean why does polybenzoid structures follow Clar's rule (what is the mechanism lying behind it) and what is the reason for the stability of these structures?

And what reason other then "it works" would you like? These are just some rules of thumb not a Relativity Theory.

what is clar's rule? Please add more detail into your question.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3982536/

@Mithoron i just wanted to know why it works, the mechanism behind it, and not to prove its validity

@NilayGhosh i saw this article but it doesn't address the mechanism behind the rule.

@AniruddhaDeb i edited the question

OK, we''ll see if it's reopened. I guess MOT can give some more insight then simple rule about this.

related https://chemistry.stackexchange.com/questions/16562/is-pyrene-aromatic-despite-failing-h%c3%bcckels-rule

You might agree that it is a good and reasonable starting point.


text:
What drives the 'internal return' in SNi?
I just learnt that alcohols can undergo $\mathrm{S_N}i$ with $\ce{SOCl2}$ (without pyridine) to achieve retention:

However, what I really do not understand is the 'internal return' step (forming 'intimate ion pair'). What drives the departure of oxygen + attack from chloride? Electronegativity of oxygen is more negative than chlorine, and bond strength of $\ce{C-O}$ is $\pu{358kJ/mol}$, $\ce{C-Cl}$ is $\pu{339kJ/mol}$. It seems to me that it is more stable to stay at $\ce{C-O}$ bond then forming $\ce{C-Cl}$ bond.
Moreover, what causes $\ce{Cl}$ to attack the carbocation on the same face? Like $\mathrm{S_N1}$, it can attack from the front and back (both sides of empty $\mathrm{p}$-orbital), what cause this to differ from $\mathrm{S_N1}$? Also, is primary/methyl alcohol unable to undergo $\mathrm{S_N}i$ due to carbocation formation?

Here's another link, https://en.wikipedia.org/wiki/SNi which suggests that part of the driving force is the formation of sulphur dioxide, this is thought to occur at the step you lable chlorosulfite with a greater degree of concertedness than you draw.  "what causes Cl to attack the carbocation on the same face?"  This is the idea of the intimate ion pair, the species are bound togethre with their ionic charges which restricts their ability to migrate.  It would be interesting to model ...

The solvent cage effect. https://sites.google.com/site/mechanismmordor/first-year-mechanisms/nucleophilic-substitution-1

I see, so SNi (using SOCl2)  is unavailable for methyl / primary alcohol since it forms a carbocation, unlike the usual route SN2 (using HCl)?


text:
How is bond angle of a molecule measured?
Simply, how are (average?) bond angles for simple (or complex) molecules such as methane or ethene measured (determined)?
I'm guessing that X Ray crystallography can be used to determine bond angles for, well, materials that have a regular crystalline form. However, is there some other technique for determining bond angles in non crystalline  molecules?
I have researched on Chemistry SE and on the wider web and I can't find this question addressed - which I find strange as it seems to me an elementary question. Apologies if this question has already been addressed in Chem SE - I just wasn't able to find it using my search terms. I can find many resources about predicting bond angle and none on measuring them.

One method is rotational spectroscopy: https://en.wikipedia.org/wiki/Rotational_spectroscopy

Related: https://chemistry.stackexchange.com/q/18777/72973

Also related: https://chemistry.stackexchange.com/questions/57784/has-anyone-even-taken-a-picture-of-a-molecule-to-confirm-the-geometry-predicted/57818#57818 (I search "X-ray bond angle" on this site).

Thanks to both  for such quick, relevant and helpful links. @buck if you can post as an answer I will accept. karsten, as you have shown, most of the trick in getting the right answer is asking the right question. In other SEs (ahem , maths, ahem, CS) you get your knees cut off for asking anything other than a perfect question.

Of course for methane, there is no need to measure anything ... ;-)

@Karl. But surely that's exactly the point. Without evidence that C-H bond angles (or maybe bond lengths) are all the same then we haven't any evidence that orbital hybridisation has occurred.

@CliveLong Also without single crystal structure analysis, there is overwhelming evidence that methane has four chemically identical H atoms in a tetraedrical arrangement, resp. that excludes any possible alternatives. Rotational and vibrational spectroscopy, NMR, heat capacity, powder diffractometry, isotope substitution experiments ...

@karl. Thanks for listing all the other methods/techniques.

Your question was sth else, however: What are the methods (apart from theoretical calculation) to determine absolute angles / distances in molecules that do not crystallise. Rotational spectroscopy applies to very simple gaseous molecules, which could also be crystallised. Right?

@CliveLong Orbital hybridisation doesn’t occur; it’s just a mathematical convenience we’ve adopted in order to describe the tetrahedral geometry. You don’t actually need sp3 orbitals to explain why methane is tetrahedral.

The most obvious method is single crystal xray crystallography, of course, and for simple molecules, IR and microwave spectroscopy in the gas phase can be used.
For many complex (e.g. bioorganic) molecules, this fails, but there are two (afaik) last hopes to help solve/prove structures if they don´t crystallise and the usual NMR spectroscopy with added computational chemistry is not enough.

Single molecule xray diffraction, a very new approach that can only just be done on the one most brilliant xray "laser" source in the world, the XFEL in Hamburg. It cost 1.2 billion euros to build, and this is the "killer" application it is supposed to prove feasible.

RDC (residual dipolar coupling) measurements with NMR, a very fancy method where the interesting compound is trapped in an oriented (sort of "stretched") gel. If the molecules of the compound have a preferred orientation in that gel (they usually do, because very few biomolecules are spherical), you observe dipolar splittings in the spectra, which would normally be averaged out due to the free tumbling of the molecules. These couplings can be used to calculate relations between distances and angles between the magnetic field and the direct connecting line between the atoms involved.

No 2 of course relies heavily on additional input from conventional (spectroscopic) methods and computer simulations.

I tried finding data on 1H-13C NOEs for methane but had a hard time. That should be a pretty good way to obtain bond distances I would think. For simple molecules simulation (I am thinking of RDCs) should not always be necessary, depending on the amount of data. Motional averaging is always an issue, but presumably also for rotational spectroscopy.

@BuckThorn interesting thought - AFAIK, heteronuclear NOEs are not all that popular and have mostly focused on abundant nuclei like 1H/19F. Even then, going from a NOE intensity to an actual distance is not trivial; for homonuclear NOEs you need a “reference” peak for which you already know the distance (eg a geminal CH2 which can be accurately calculated computationally). In methane this would not be possible as there is only one 1H/13C NOE. Not sure if anybody’s actually tried doing it on a labelled protein, though. (see eg Craig Butts’ work for the small molecule stuff I mentioned earlier.)

@orthocresol Heteronuclear NOEs are often measured for specific applications in protein NMR and since 13C/15N labelling is commonly performed their measurement has become fairly routine. For something like methane you probably have to go back a half century to find studies in the literature on the determination of its bond length from NMR. The catch is that you have two unknowns, the bond length and the rotational correlation time, but I'm sure it's been done (or rather I'd be extremely surprised if it hasn't).

@BuckThorn good to know, thanks. :-) Yes the two unknowns issue is why reference peak is needed in that procedure. Going very offtopic here, but do you happen to know how one might otherwise determine $\tau_c$?

@orthocresol For protons in methane this does not appear to be simple because there are a number of different relaxation mechanisms. A seminal ref is J. Chem. Phys. 39, 552 (1963). It might be complicated even for 13C but I'm still trying to find the relevant ref.

@karl. A very thorough and actually understandable answer. Thanks to you and all other contributors.


text:
How to calculate the pH of solutions of amphoteric salts?
I've been trying to understand how to calculate the pH of solutions of amphoteric salts, but I just can't figure it out. So, I've seen that it is commonly explained like this (example for a solution of NaH2PO4):
The salt completely dissociates like so
$\ce{NaH2PO4}$-->$\ce{Na+}$ + $\ce{H2PO4-}$
Then the H2PO4- ion reacts with water to form HPO4 2- and H3O+.  (reaction 1)
$\ce{H2PO4-}$ + $\ce{H2O}$ <=> $\ce{HPO4  2-}$ + $\ce{H3O+}$ _______   $\ce{Ka2=6.2 x 10^{-8}}$
And then it reacts with H3O+ to form H3PO4 and H2O.  (reaction 2)
$\ce{H2PO4-}$ + $\ce{H3O+}$ <=> $\ce{H3PO4}$ + $\ce{H2O}$   ________   $\ce{K=Ka1^{-1}, Ka1=7.52*10^{-3}}$
The autoionization of water is negligible.
Now, I imagine this process like this:
The concentration of the $\ce{H2PO4-}$ ion in the first reaction is equal to the concentration of the salt in the solution because the salt dissociates completely. As it reacts with $\ce{H2O}$ to form $\ce{HPO4 2-}$ and $\ce{H3O+}$ (reaction 1), its concentration decreases for a certain value of x. The equilibrium concentration of $\ce{H3O+}$ and $\ce{HPO4 2-}$ in this reaction are the same and equal x. The equilibrium concentration of $\ce{H2PO4-}$ is equal to the concentration of the salt minus x.
$\ce{[H2PO4-]1=c(salt)-x}$
$\ce{[H3O+]1=x}$
$\ce{[HPO4 2-]=x}$
$\ce{Ka2=\frac{[H3O+]1*[HPO4 2-] }{[H2PO4-]1}}$
Then, what is left of the $\ce{H2PO4-}$, that is the equilibrium concentration of $\ce{H2PO4-}$ from the first reaction reacts with the equilibrium concentration of  $\ce{H3O+}$ from the first reaction to form $\ce{H3PO4}$ and $\ce{H2O}$. The concentrations of $\ce{H2PO4-}$ and $\ce{H3O+}$ decrease for a certain value of y. So, in the equilibrium of this reaction, reaction 2, the equilibirum concentration of $\ce{H3PO4}$ will equal to y.
$\ce{[H2PO4-]2=[H2PO4-]1-y=c(salt)-x-y}$
$\ce{[H3O+]2=[H3O+]1-y=x-y}$
$\ce{[H3PO4]=y}$
$\ce{Ka1^{-1}=\frac{[H3PO4] }{[H2PO4-]2*[H3O+]2}}$
So the final concentration of $\ce{H3O+}$ is equal to the concentration of $\ce{H3O+}$ in the equilibrium of the second reaction.
$\ce{[H3O+]final=[H3O+]2}$
We can see that the final concentration of $\ce{H3O+}$ is equal to x-y. Because $\ce{[HPO4 2-]=x}$ and $\ce{[H3PO4]=y}$, we write:
$\ce{[H3O+]final=[HPO4 2-]-[H3PO4]}$
We can express $\ce{[HPO4 2-]}$ and $\ce{[H3PO4]}$ using the equilibrium constants, concentrations of $\ce{H3O+}$ and $\ce{H2PO4-}$.
$\ce{[H3O+]final=\frac{Ka2*[H2PO4-]1 }{[H3O+]1}-\frac{[H2PO4-]2*[H3O+]final }{Ka1}}$
If we assume that the initial concentration of $\ce{H2PO4-}$ is so large that the change of x and y are negligible, we can write that:
$\ce{[H2PO4-]2=[H2PO4-]1=c(salt)}$
If we assume that the concentration of $\ce{H3O+}$ is large enough so that the change of y is negligible, we can write:
$\ce{[H3O+]final=[H3O+]1}$
With those assumptions we can get this commonly used expression for explaining this problem
$\ce{[H3O+]final=\frac{Ka2*c(salt)}{[H3O+]final}-\frac{c(salt)*[H3O+]final }{Ka1}}$
and with a little bit of rearranging, we get this expression:
$\ce{[H3O+]final=\sqrt{\frac{Ka2 * c(salt)}{1+\frac{c(salt)}{Ka1}}}}$
What I don't understand is the fact that we were able to get the last equation because we assumed that the final concentration of $\ce{H3O+}$ ions is equal to their concentration in the equilibrium of the first reaction. But at the beginning, when we established the first equation for the final concentration of $\ce{H3O+}$, we assumed that the concentrations of $\ce{H3O+}$ ions in the equilibrium of the first and second(final) reaction aren't equal. If in the end we assumed they are equal, why can't we calculate the pH just using the first reaction? That way the $\ce{H2PO4-}$ wouldn't be treated as an amphoteric, which obviously isn't correct, so I'm wondering what is wrong in my approach?

Related: https://chemistry.stackexchange.com/questions/114488/find-the-ph-of-an-amphiprotic-salt-dissolved-in-water-at-room-temperature


text:
What chemical properties of ethanol make it usable for drinks as compared to that of methanol?
Methanol ($\ce{CH3OH}$) and ethanol ($\ce{C2H5OH}$) both are the organic compounds having an alcoholic group. The alcoholic beverages (liquors and spirits) for human consumption (albeit injurious to health) contain a certain percentage of ethanol but nothing of methanol.
What chemical properties of ethanol ($\ce{C2H5OH}$) make it usable in beverages as compared to those of methanol ($\ce{CH3OH}$)?
EDIT: I am looking for a comparison of chemical properties of ethanol and methanol with respect to suitability for drinks i.e. comparison of their mechanism upon consumption.

Did you read, for example, https://en.wikipedia.org/wiki/Methanol_toxicity?

Thank you, but for ethanol what are those properties to make it drinkable? I want to know comparison of related chemical properties w.r.t. suitability for drinks.

Does this answer your question? Why is methanol toxic?

@Mithoron: No it doesn't completely answer my question. orthocresol has already given a similar link above about toxicity of methanol. But what about the ethanol? What chemical properties make ethanol suitable for beverages?

The biochemical breakdown paths are quite analogous, but the toxicity of the various steps are quite different. Just why is more an issue of biology than chemistry.

The problem arises from the metabolized products of methanol. Methanol oxidizes in the liver by an enzyme called alcohol dehydrogenase to formaldehyde which is further metabolized to formic acid by another enzyme called aldehyde dehydrogenase. This formic acid is the source for acute toxicity associated with methanol poisoning. Accumulation of this chemical in the blood deprives cells of oxygen by inhibiting the enzyme cytochrome c oxidase in their mitochondria, a key element of the respiratory electron transport chain.
Formic acid, together with formaldehyde, are responsible for nerve damage, blindness, and other unpleasant effects associated with methanol poisoning.

Note that ethanol is also metabolized in the same way by the same pair of enzymes to ultimately form acetic acid but human can tolerate acetic acid to an extent because it is less toxic compared to formic acid and thus can be consumed like for example, vinegar. In fact, ethanol is used as a remedy for methanol poisoning as it acts as a competitive inhibitor by more effectively binding and saturating the alcohol dehydrogenase enzyme in the liver, thus blocking the binding of methanol with the enzyme rendering it useless.
References

https://theskepticalchemist.com/methanol-toxic-ethanol/ (above reaction source)
https://metode.org/metodes-whys-and-wherefores/why-can-we-drink-ethanol-but-not-methanol.html
https://en.wikipedia.org/wiki/Methanol_toxicity

I would also like to add something that I saw in a comment of the accepted for the question Why is methanol toxic? which I believe gives a nice intuitive feel to the idea of methanol toxicity. As this answer states that methanol gets metabolized into formic acid into the body. Formic acid is also basically the main component of ant poison. If a sizeable amount of methanol metabolizes into formic acid, you can imagine the pain like being stung by a 100 ants...

Nilay's answer, as well as the comments, do a great job of explaining why methanol is toxic (and ethanol is comparatively less toxic). I'll take a different approach here: I'll explain why ethanol is the major component of alcoholic drinks. To start off, ethanol is not the only alcohol in alcoholic drinks1. Other alcohols such as Glycerol, Tryptophol, Tert-butyl alcohol, Tert-amyl alcohol, Phenethyl alcohol, Propanol, etc are also present, though in smaller amounts.
The main requirement of an alcohol in a drink is to produce the feeling of intoxication: The most common Alcohols satisfying this are Ethanol and Propanol (which is 2-4 times more potent than ethanol). While these alcohols give you a tour of the skies, Toxic alcohols take you straight to heaven, and are a strict no-no. Other requirements include solubility: due to a larger hydrocarbon portion, propanol is less soluble than ethanol in water. This is not a problem for beer, but can be a problem for spirits (which may have up to 40% ethanol).
The simplest reason for the usage of ethanol over propanol would be because nature has willed it so. Zymase is the enzyme that fermentation bacteria use to convert glucose to ethanol, and there is no common analogue that creates propanol from glucose. Ethanol has been used for beverages since the time some farmer left some foodgrains in water for a long time, and later got a good feeling on drinking that water. There is no intertia to change, ergo we continue to use the same process for creating drinks even today.
References:

IARC Working Group on the Evaluation of Carcinogenic Risks to Humans. Alcohol Drinking. Lyon (FR): International Agency for Research on Cancer; 1988. (IARC Monographs on the Evaluation of Carcinogenic Risks to Humans, No. 44.) 3, Chemical Composition of Alcoholic Beverages, Additives and Contaminants. Available from: https://www.ncbi.nlm.nih.gov/books/NBK531662/

"zymase: is not a single enzyme. It is a mixture of enzymes. Propanol (both 1- and 2- isomers) is also a known fermentation product of bacterial and yeast fermentations. It just is not produced at the same concentrations or with the same high rates as ethanol, so it is not a major component (although it is a minor component) in fermented alcoholic beverages.

So there's no qualitative difference, just a quantitative difference in the level of intoxication ...


text:
Database of 3D crystal models
I'm looking for a database of very basic crystal models for all the various types - triclinic, monoclinic, orthorhombic, tetragonal, rhombohedral, hexagonal, and cubic.
A basic example would be a CAD file for a cube, to represent the cubic structure. I don't want like a packing model which shows the individual atoms because I'm trying to represent macroscopic crystals.
Any ideas where I could find what I'm looking for? Not a 3D graphic designer.

Crystallographic database. CAD file formats are not used in crystallography (maybe unless you are an architect building another Atomium), look for CIF instead.

There are exactly seven of them, all of which you already listed in your question. Why a database? And what do you want to do with them, with what software?

sounds like you need a database of polyhedra, not chemicals

for example, something like : https://www.georgehart.com/virtual-polyhedra/vp.html

Joining @Karl's comment on «what do you want to do with them?» the work to present point groups is work already solved.  For off-line teaching, the freely accessible collection of the 32 point groups by Casas and Estop by (UB Barcelona) may suit your needs:

The model data are provided as interactive 3D .pdf files which a reader like Adobe Acrobat 9.0 may let you toggle on/off the axes / planes, etc.
Beside some background provided here, the authors outline the workflow and tools used (SolidWorks, Adobe Illustrator CS3, and Adobe Acrobat 9 Pro) in an accompanying publication (doi 10.1016/j.cageo.2013.09.004, paywall).  Later, the concept was extended to cover the space groups as to complement the International Tables of Crystallography, too (title page, paywall).


text:
Reaction of styrene oxide with hydrogen iodide
The question is to find the product of the reaction of Styrene Oxide with Hydrogen Iodide.
According to me, this should be the mechanism-

As the benzylic carbocation will be more stable, therefore the product made from A should be major and the other one should be minor. But, in my book the answer to this question is the opposite of mine. Where did I go wrong?

Did your book mention cyclodextrin? In its presence iodide opening at the primary site.

No, only the structure of the two reactants was given.

Your answer is the correct one. Acid-catalysed hydrolysis of epoxides proceeds exactly as shown. In the specific case of styrene oxide, nucleophiles attach themselves to the benzylic carbon1,2, which means that A should be the major product. I'm attaching an image from (1) which gives the products of reaction of methanol with styrene oxide:

If you want to, you can try to drop a note to the authors of your book about this error.
References:

Mohseni, Sharareh, et al. “Theoretical and Experimental Studies on the Regioselectivity of Epoxide Ring Opening by Nucleophiles in Nitromethane without Any Catalyst: Nucleophilic-Chain Attack Mechanism.” Progress in Reaction Kinetics and Mechanism, vol. 39, no. 1, 2014, pp. 89–102. doi:10.3184/97809059274714X13874723178403.
Morrison, Robert Thornton, and Robert Neilson Boyd. Organic Chemistry. 5. ed, Allyn and Bacon, 1987, pp. 780-785.


text:
Is the Hartree-Fock determinant always the minimum energy single Slater determinant solution to a molecular Hamiltonian?
Is it always true that the Hartree-Fock slater determinant is the minimum energy single slater determinant solution to a given molecular Hamiltonian $H$?
By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals). The Hartree-Fock slater determinant $|\Phi_0\rangle =|\chi_1\ldots\chi_N\rangle$ is constructed by choosing the lowest $N$ energy solutions of the Hartree-Fock equation.
Now, $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ is an eigenstate of the Hartree-Fock Hamiltonian $H^{HF}$ with energy $\sum_{i=1}^N \epsilon_{i_j}$, so constructing $|\Phi_0\rangle$ from the $N$ lowest energy solutions of the HF equation clearly minimizes the energy of $H^{HF}$.  But the molecular Hamiltonian $H$ is given by
$$H = H^{HF} + \sum_{i=1}^N\sum_{j>i}^N \frac{1}{r_{ij}} - \sum_{i=1}^N v^{HF}(i)$$
where $v^{HF}$ is the HF potential.  This satisfies
$$\langle{\Phi_0}|H|\Phi_0\rangle = \sum_{i=1}^N \epsilon_i - \frac{1}{2}\sum_{i=1}^N \sum_{j=1}^N \langle ij|\,|ij\rangle $$
where
$$\langle ij|\,|ij\rangle:=\int dx_1 dx_2 \chi_i^*(x_1)\chi_j^*(x_2)\frac{1}{r_{12}}(\chi_i(x_1)\chi_j(x_2) - \chi_j(x_1)\chi_i(x_2))$$
Due to the $\langle ij|\,|ij\rangle$ term, it is not clear to me that constructing $\Phi_0$ from the $N$ lowest energy solutions of the HF equations still results in the lowest energy single slater determinant for $H$.  An excited determinant would have a higher energy contribution to the $\sum_i \epsilon_i$ term, but could this is offset by a greater deduction from the $\langle ij|\,|ij\rangle$ term?

By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals).

This is not true: only the converged Hartree-Fock determinant $|\Phi_0\rangle = |\chi_{i_1}\ldots \chi_{i_N}\rangle$ has stationary energy with respect to orbital perturbations.
Stationarity of energy requires all orbitals in the determinant to fulfill the Hartree-Fock equations
$$
\hat f \chi_i = \varepsilon_i \chi_i
$$
where the Fock operator
$$\hat{f} = \hat h + \sum_j^N(\hat{J_j} - \hat{K_j})$$ depends on the $N$ occupied orbitals $\{\chi_j\}$ included in the determinant.
The Hartree-Fock determinant $|\Phi_0\rangle$ fulfills this condition and is therefore stationary with respect to orbital perturbation, because the constituting orbitals are converged (in SCF iterations) solutions to these equations, where the Fock operator is calculated from the $N$ orbitals with lowest energies.
Any other (excited) determinant $|\Psi\rangle$ is generally not stationary with respect to perturbing the orbitals. Since other orbitals are occupied in this determinant, it corresponds to a different Fock operator $\hat {f'}$ where the sum is over the orbitals in the determinant $|\Psi\rangle$ which differs from the set of orbitals included in $|\Phi_0\rangle$.
Using this Fock operator $\hat{f'}$ in subsequent SCF cycles, the total energy of the excited determinant is again minimised by tweaking the orbitals. In this course, I suppose, it is quite likely that the excited determinant relaxes back to the ground state Hartree-Fock determinant  $|\Phi_0\rangle$. The iterations might also converge to a state with higher energy, but one-determinant representations of excited states are rather inaccurate anyway.
In principle, in should be possible for an excited determinant to converge to a determinant with lower energy if and only if the initial converged solution is merely a local minimum. However, this is very unlikely if using a reasonable initial guess.


text:
Reason for same frequency of spectral line in infrared and visible region
I encountered this question while trying to understand the basics of spectroscopy and atomic structure.

Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n=6 to n=4 transition in Helium positive ion?

I am aware that lines of Balmer series correspond to visible region of electromagnetic spectrum and that to Brackett series correspond to infrared region.
How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

The Balmer series is said to be visible in case of Z=1 $\mathrm{n_1 = 2}$, The spectral lines are not in different regions.

What do you mean by 'the spectral lines are not in different regions'? Spectral lines have different wavelengths and therefore should belong to different regions.

But that is if we consider them to be emitted from the same atom. Here you are asked to compare between two different atoms.

According to Bohr's theory of the structure of the hydrogen atom, the atom is defined as a system with a positively charged nucleus that has electrons revolving around it in discrete orbits of fixed size and energy and stated that the orbits were held stationary using electrostatic forces.
Upon calculation of the various parameters of the model, we get the energy of the $\mathrm{n}^{\text{th}}$ orbital in the hydrogen atom to be:
$$E=\frac{\mathrm{-13.6}}{\mathrm{n^2}} \mathrm{eV} \tag{1}$$
If we state that the atom to be considered were to be made mono-electronic and had an atomic number of Z, satisfying the conditions for the Bohr's model to work, we get that the energy of the $\mathrm{n}^{\text{th}}$ orbital to be:
$$E= -13.6 \cdot\frac{\mathrm{Z^2}}{\mathrm{n}^2} \mathrm{eV} \tag{2}$$
A spectral line is observed when the electron de-excites itself from a higher energy state to a lower energy state that is more stable. This is done by emitting a photon whose energy is the same as that of the difference of energy between the two levels. Applying $\frac{\mathrm{hc}}{\lambda} =E$, and $\frac{\mathrm{c}}{\lambda} = f$ we get
$$f = \frac{\mathrm{c}}{\lambda} = \mathrm{c} R_H\big[\mathrm{\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]Z^2} \tag{3}$$
Here $R_H$ is the Rydberg constant whose value is $\approx\pu{109,678 cm^-1}$
In the question, you have been asked to figure out which line in the Balmer series($\mathrm{n_1} = 2$) of hydrogen would have the same frequency as that of the $3^\text{rd}$ line in the Brackett series($\mathrm{n_1} = 4$). I believe you would know how to solve the question and if not would now be able to via equating the frequencies of the two spectral lines.
Moving on to your doubt which is based on a misconception.

How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

The issue with this statement here is that you are assuming that the Balmer series is always visible for all atoms. This is only true for hydrogen as the the wavelengths provided for the different series are based on measurements taken on the hydrogen atom. That is, the Paschen series is infrared for hydrogen and this may change if the atom changes. As the atomic number increases, the frequency becomes greater as you can see from equation ($3$). This implies that at higher atomic numbers, there is a greater shift towards the UV region.
TL;DR: What you say is correct, frequency is a characteristic property unique for a certain electromagnetic radiation. However, the Balmer series is defined to be visible specifically for the hydrogen atom and not for any other mono-electronic atom/ion. As the atomic number becomes greater, a shift towards the UV region takes place.


text:
Overview of material characterization methods
I have been searching for quite some time now, but I feel like it is impossible to find some sort of chart where I can determine which characterization method (microscopy, spectroscopy, electrochemical, DMA etc.) is the best in which situation. There are so many methods and an overview would really help me to understand them better.
Does anyone know if a book or a site provides such an overview? I am pretty desperate, that's why I'm asking it here.

The entire scientific literature contains that information. Scientists ask a question, and they choose multiple methods to get closer to an answer. So every paper would be a row in your table, the cell with the "situation" would be the entire introduction to the paper, and the methods section would not tell you which single method is the best, but which set of methods they decided to use (based on expertise, availability, cost and expected outcomes).

Yes, I understand that. I have been reading a lot of papers to gain some insight. But I mean it would be nice to have some kind of overview of the methods and a flowchart that determines what would be a good option. I can make it myself, and probably will, but it seems odd that there isn't already such a thing in the literature.

For molecular biology, there is a series often just called Methods. (see https://www.springer.com/series/7651). I don't know whether material sciences has something similar. The corresponding wikipedia article has a couple of references that might be useful: https://en.wikipedia.org/wiki/Characterization_(materials_science)

You might find such a table for a more focused set of materials, such as https://www.researchgate.net/profile/Venkata_Narayana_Kalevaru_Dr/publication/303790692/figure/fig3/AS:369912011870209@1465205304082/Summary-of-some-selected-characterization-techniques-of-MNPs_W640.jpg

There is a collection available at Libretexts.org, by students and faculty from Franklin & Marshall College. The intent is to provide a quick overview for readers of the literature (rather than practitioners doing the experiments), so the focus is the type of result each method yields, and how to interpret it.


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Software for modelling decomposition reactions
What are some good softwares that can model a decomposition reactions?
I have a reaction of H2O2 decomposing on metal exchange hexaaluminate catalyst. The catalyst is present in a packed bed thruster. I have done the micro-kinetic modelling of the diffusion and reaction that might be happening on the catalyst on paper.
I need a software that can model the mass transfer and other transport phenomena that might be happening inside the thruster. Also are there some other softwares that can model the surface reaction and the adsorption/desorption steps that be happening on the surface of the catalyst.
I have the following data that could serve as the input to the software-

Experimental rates (average rate and instantaneous rate)
Some assumptions of the kinetic rate and equilibrium constants derived using non linear regression and least sum of squares method
All physical dimensions of the thruster bed and the catalyst pellets.

Any suggestions will be welcomed. I work mainly on the experimental chemistry side ,and this is the first time I will be working on the computational side.
P.S please ignore grammatical errors (english is not my first language)

Actually, I recall a paper citing numerous (perhaps 6) ascribed possible decompositions pathways involving H2O2. On your select catalyst, some (as in more than one simultaneous occurring path) may be applicable. It would be interesting to compare actual and model predicted.

Here is a paper, for example, citing 3 paths for H2O2 decomposition with Cupric ions at https://www.sciencedirect.com/science/article/pii/S0162013403004392 .

@AJKOER Thanks for the input .Yes model validation is of immense importance to me

Curious, what standards do you ascribe to ascertain an answer to the question: "What are some good softwares that can model a decomposition reactions..."?

So I was wondering if there are any softwares ,so if I input to them 1)Chemical and physical properties of the reactants and products 2) The physical properties of the catalyst and the thruster/reactor 3)any additional info it needs ,will it be able to model the transport phenomena going on inside

So I know I can model the transport phenomena using experiments ,but due to covid my lab is closed ,so was wondering if computational chemistry could help in this matter ,so as to not waste time


text:
The best chromatographic technique to analyse a solution of benzene, toluene, and ethylbenzene in n-hexane

What is the best chromatographic technique to analyse a solution containing benzene, toluene, and ethylbenzene in n-hexane?

Is gas chromatography (GC) the best way to analyse this solution? Or is using liquid chromatography with a non polar eluent and column better than GC in this case?

Seems tailormade for GC with flame ionisation

Define better :-)

What you want to analyze? to separate? percentage of a mixture? or anything else? You need to elaborate a little more.

Best is relative. Fastest and easiest would be a direct GC liquid injection. That run would be less than 10 minutes on a 624 column or other non-polar column. HPLC is much more complicated than GC with the different mobile phases and solvents.

The choice of analytical technique is based on three simple factors: what is available, and how low concentration do you want to go, and how much time do you have? Do you have a gas chromatograph with flame ionization detector or a UV high performance liquid chromatograph or both? This type of mixture is readily amenable to both GC and reversed phase HPLC. The only problem is hexane in reversed phase HPLC, but if less than 1 uL is injected it should be fine.


text:
Why aren't the hydrogen bonds in carboxyllic acids like this?
Why don't carboxylic acids form H-bonds in this formation with the red dots (diagonal) instead of the black dots (horizontal)?

I was always taught that H-bonds formed between partially negative and partially positive atoms. I agree that the $\ce{=O}$ carboxyl oxygen is partially negative, so there should be a H-bond there from the partially positive H.
However, the $\ce{O}$ in $\ce{OH}$ is also partially negative. So why aren't the H-bonds denoted with red dots formed?

And nobody told you that it goes straight to lone pair?

So, while bifurcated hydrogen bonds do exist, it's not the case here.

Does the red-dotted line from the hydrogen lead to electron pairs on oxygen? I think not.

You need some degree of orbital overlap. Pretty challenging to overlap at that kind of angle.

In reality in carboxylic acids, there are no double bond and single bond on oxygens. It is a hybrid (e.g., where is the $\ce{^{13}C}$-signal for carboxylic carbonyl in NMR?).

Often times you ignore the stuff that is not as important. It's a simple cartoon after all.

@MathewMahindaratne Isn't that just an issue with fast exchange in NMR? I would think that the oxygens are inequivalent (in a carboxylic acid in general; in the picture shown here for paired groups the inequivalence of the oxygens is less clear).

H-bonds are indeed bonds, and your black dotted lines give a 6-ring, and with the red, it´s a 4-ring. If you want to know for sure, check the crystal structure of any carboxylic acid.

@Mithoron but isn't there a lone pair on the red bond oxygen?

And what does "straight to" mean? - Lone pair is no the other side of oxygen.

https://chemistry.stackexchange.com/questions/60640/why-cant-alcohols-form-hydrogen-bonded-dimers-like-carboxylic-acids


text:
Solving the Schrödinger equation for a rotating triatomic linear molecule
This source is showing that solving the Schrödinger equation for a triatmoic linear molecule yields the same formula for the rotationaI quantum states $BJ(J+1)$ as for dipoles.
For dipoles, the total rotational energy can be expressed as the total length of the molecule $R$ and a particle with reduced mass $\mu$ which makes it possible to express the coordinates in the Schrodinger equation specifically for only one particle with that reduced mass $\mu$.
I am clueless how the Schrödinger equation is solved for triatomic linear molecules because there are more variables that can not be reduced the same way. I would therefore expect to be seperate coordinates for each atom in the Schrodinger equation, as well as more than one bond length.
How is it derived and is it actually possible to do so?

See this answer https://chemistry.stackexchange.com/questions/135539/expression-of-the-potential-in-the-hydrogen-atom-while-solving-radial-part-of-th/135547#135547 . The 'trick' is to change to polar coordinates and then in your case to make the potential energy $V(r)=0$ and then to solve the Schroedinger equation. Its complicated but standard maths.

@porphyrin I indeed know that polar coordinates has to be used just like with dipoles. But for dipoles, there's only one particle with reduced mass and one bond length $R$. In the case of a triatomic linear molecule, one would have to use polar coordinates for each of these atoms, each having a different distance to the center of mass. Is there a derivation that shows how all these variables can be reduced to solve the Schrödinger equation?

do you mean diatomics in the place of dipoles?

@AntoniodeOliveira-Filho I mean triatomics, a linear molecule with three atoms (or more).

You use still use the reduced mass to form the moment of inertia as the molecule rotates about it s centre of mass.

@porphyrin Last question, how is the distance to the centre of mass formulated if there are three different distances for each
of the 3 atoms that are expressed as a reduced mass?


text:
Can the inner amine groups of Bacteriopheophorbide A become positively charged in aqueous solutions without the use of very strong acids?
In an attempt to calculate the pKa of the carboxylic acid group of Bacteriopheophorbide A (BPheo, Pubchem), I generated the structure in MarvinSketch (ChemAxon product) and I came across an interesting result. According to the software, the inner amine groups which are part of the resonance structure can become positively charged at relatively high pH. The photo below shows the calculated pKa of the groups in blue (groups of interest are highlighted in yellow). According to this software, this means that at pH 5.5 the groups will both be positively charged, whereas I was under the impression that these groups can only be altered using very strong acids such as TFA due to the aromatic ring.

Is this realistic? Or is this simply an issue with the software. I've tried looking into this myself but the literature on BPheo seems rather sparse.
Thank you in advance for your input.

The one on the right in your image makes sense, as the basic line pair is perpendicular to the pi system. Compare to pyridine. The one on the left doesn’t make sense to me, unless the 3D conformation breaks the conjugation.

Also, this structure is a form of porphyrin ring, which will chelate metal ions very readily. Perhaps the strong acid experiments you are thinking of were needed to displace a metal, not just protonate a free porphyrin?


text:
What is the molecular weight of bilirubin?
There is a discrepancy among different sources on the molecular weight of bilirubin, $\ce{C33H36N4O6}$:

584.673 (Fisher Scientific);
584.66 (Sigma Aldrich).

What is the correct value and why?

This may be above my pay grade, but try 33M_C + 36M_H + 4M_N + 6M_O ;)

And that's supposed to be a different value? 0.01 difference? They could be both right! Isotopic content isn't constant.

I would get 584.66 with this formula. https://www.lenntech.com/calculators/molecular/molecular-weight-calculator.htm But wiki and fisher scientific both say 584.673

I understand that this difference of 0.013 is trivial. I just would like to know why some would get 584.673, and some would get 584.66. How come nobody gets 584.669 or 584.671, etc.?

Huh, I guess it's valid question,but you should elaborate more in this vein in the body of question.

The discrepancy is simply due to round-off error.  The atomic masses of these elements are usually given as follows.  These are known as the standard conventional form, or standard formal short form, since they are given without an interval indicating the uncertainty in their values (see  https://en.wikipedia.org/wiki/Standard_atomic_weight):
C = 12.011
H = 1.008
N = 14.007
O = 15.999
Hence 33 C + 36 H + 4 N + 6 O = 584.673
However, if someone were to round off all the masses to four significant figures, but then report the result to five significant figures, we would have:
C = 12.01
H = 1.008
N = 14.01
O = 16.00
And now 33 C + 36 H + 4 N + 6 O = 584.658 = 584.66

Which shows that it makes sense to leave a "guard digit" on the numbers in the calculation, and wait to round until you have the answer (This would give the molar mass as 584.67 g/mol at two decimal points).

Yes, it's best practice to leave the rounding to the end (including not rounding the results of intermediate calculations).  Though note that, for some calculations, a single guard digit isn't always sufficient.

Here's a toy example:  Suppose your data is {1.214, 1.224, 1.214}, and you want to express the sum of these numbers to the nearest tenth.  Then, with one guard digit, you would have:  1.21 + 1.22 + 1.21 = 3.64 = 3.6.  Without any rounding, you would have:  1.214 + 1.224 + 1.214 = 3.652 = 3.7.


text:
Alternate pathway of Gabriel synthesis to prepare secondary amine
Can modifying the Gabriel synthesis, which is generally restricted to synthesis of primary amines, be used to synthesise secondary amines?
I have attached my idea regarding the synthesis (the final product obtained is a phthalic acid):

You cannot do the 2nd reaction using alcohol as solvent. It will react with the acyl chlorides/cyanides to give the diester

Some of these intermediates feel too unstable... Is there precedent for adding HCl across an acyl cyanide twice? Practically, reductive amination is probably a simpler way of getting to secondary amines.

The benzocyclicamide (just before adding $\ce{Mg(OH)2}$), is not stable. $\ce{HCl}$ would leave quickly, giving cyclic carboxyimine product.

W-K reduction does not work for formamides. This looks like a lot of trouble to go to when reductive methylation procedures are well precedented, work very quickly and in high yield.

@orthocresol I added HCl twice to display the intermidiate. I wanted to see if somehow, even if with difficulty, secondary amines could be prepared by Gabriel synthesis. Hence I proposed a mechanism, though I have realised that it is not efficient enough to be implemented practically.

@Waylander If some polar aprotic solvent, like DMSO, is used, can the succesive product be formed by reacting cyanide ion with the carbonyl carbon of the acid chloride?

DMSO reacts with acid chlorides

Why did alc. KCN replaced one chlorine atom from phthaloyl chloride?


text:
Difference between spin-orbit coupling and the Russell-Saunders Effect?
The Russell-Saunders effect is the same thing as 'spin-orbit interaction, correct?
The reason I am asking is because I was reviewing the Wikipedia page on 'spin-orbit interaction' and it does not mention Russell-Saunders at all...

I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will.
The answer is... yes and no.
The word "coupling" refers to the coupling of several sources of angular momentum, namely the spin component and orbital component. Now, the issue is that in an atom you typically have many electrons, and every electron has its own orbital angular momentum $\vec{l}$ and spin angular momentum $\vec{s}$,* so you have many, many sources of angular momentum.
The challenge is to bring all of these together in a way that allows us to describe the electronic state of an atom using its angular momentum properties (for example, this is what a term symbol does). There are two approaches towards coupling all of these angular momenta together:

(Russell–Saunders or LS-coupling) Couple all the individual $\vec{l}$'s together to form one gigantic orbital angular momentum $\vec{L}$, and couple all the individual $\vec{s}$'s together to form one gigantic spin angular momentum $\vec{S}$. Then couple these two together to form the total angular momentum $\vec{J}$.
If you have studied term symbols before it is probably using the Russell–Saunders scheme, where you calculate $L$, $S$, and $J$, then write the term symbol $^{2S+1}L_J$.

(jj-coupling) For each individual electron, couple $\vec{l}$ and $\vec{s}$ together to form the total angular momentum $\vec{j}$ for that one particular electron. Then bring all the electrons' total angular momentum together to form $\vec{J}$.
Note that we haven't ever mentioned $L$ and $S$ here, so the term symbols under this coupling scheme are different. Instead, you'd label the term symbols with the individual values of $j$ for each electron. For an example see e.g. Atkins Molecular Quantum Mechanics.

Now, which you use depends on whether electron-electron repulsions or the spin-orbit coupling is a "bigger" term. If spin-orbit coupling is very significant, then it means that the spin and orbital angular momenta on their own (i.e. $\vec{L}$ and $\vec{S}$) are not very useful quantities,† since the interaction between them is large. In this scenario, jj-coupling is a more appropriate way of describing the effects of spin-orbit coupling on the electronic state.
On the other hand, if the spin-orbit coupling is relatively small, then $\vec{L}$ and $\vec{S}$ are useful quantities which are still applicable to the atom's electronic state, so the Russell–Saunders scheme is appropriate.
And of course, sometimes we get stuck in the middle ground where neither scheme is fully appropriate.
TL;DR The Russell–Saunders and jj schemes are both methods which can be used to describe the effects of spin–orbit coupling, but they are not the same thing as spin–orbit coupling.‡

* Well, sort of, anyway: the electrons are indistinguishable, so it's more accurate to say that the $n$ electrons in the atom have $n$ orbital angular momenta $\{\vec{l}_1, \vec{l}_2, \cdots, \vec{l}_n\}$ and $n$ spin angular momenta $\{\vec{s}_1, \vec{s}_2, \cdots, \vec{s}_n\}$.
† To be precise, $L$ and $S$ are not "good quantum numbers" because the operators $\hat{L}$ and $\hat{S}$ don't (approximately) commute with the total Hamiltonian $\hat{H}_0 + \hat{H}_\text{so}$, where $\hat{H}_\text{so}$ is the spin-orbit coupling Hamiltonian and $\hat{H}_0$ is the rest of the Hamiltonian (which does commute with $\hat{L}$ and $\hat{S}$).
‡ If you read the previous footnote, then the spin–orbit coupling itself is represented by the Hamiltonian $\hat{H}_\text{so}$. The two coupling schemes can be thought of as ways of dealing with this term as a perturbation to $\hat{H}_0$. In Russell–Saunders the perturbation is small, and consequently, the "good" quantum numbers are similar to those of $\hat{H}_0$. In jj the perturbation is large.


text:
How do you purify gases formed in electrolysis with a potassium hydroxide electrolyte?
I am interested in the $\ce{KOH}$ vapor that is released with my gasses during electrolysis.
I have two questions.
What would be the best place to start research on the $\ce{KOH}$ vapor produced in electrolysis and how to separate the $\ce{KOH}$ vapor formed.
I was thinking about filters, or bubblers that require constant water replacement. For me, a bubbler is a jar with a gas inlet at the bottom and gas collection at the top.
Also I have no clue on what type of filter to use since I don't know where to start research on $\ce{KOH}$ vapor.
My goal is to make the gases safer by eliminating most of the $\ce{KOH}$ vapor. I will probably use a pH testing paper to test the air to see if $\ce{KOH}$ vapor is present.

What type of electrolysis are you talking about? Aqueous KOH or molten KOH? If you talking about aqueous, then KOH vapor can be called nonexistent, on the other hand molten KOH well that is another story.

aqueous KOH, I am not even heating up the electrolysis. Thank you for the info I could not find it anywhere. I will still use 2 bubblers and replace water frequently. As well as testing the air for KOH with ph paper.

To be precise, i have no proof of what i said, but is an intuition from my experience and sense. It could be possible for KOH vapors to escape, but based on its hygroscopic nature it is neglectable. It also could be predicted that if a little KOH vapor would escape it would react with atmospheric CO2 to produce K2CO3 and water, decreasing its basicity in orders of magnitude rendering it more safe.

There are plenty of examples on youtube about hydrogen and oxygen (HHO) production from electrolysis. Mostly they use baking soda, NaOH rarely KOH for the electrolyte. In every video there are instruction that you have to attach a water bubbler to secure burnback. The water bubbler may probably filter the NaOH and KOH vapors if there any measurable quantity at all present.

I very much so appreciate the help. Your expertise is much better than my small chemistry knowledge. My design is all about safety and i though koh being a better electrolyte would allow for less to be used. there for less to be cleaned up in a spill and less danger overall. Should I be worried that Everyone else is using NaOH or did I just choose a strange electrolyte. If this is too much to ask you have been an amazing help and thank you very much. I was very worried about koh mist. I will still proceed with caution.

I think NaOH is just more commercial than KOH simple as that, it makes no difference. In theory you would be better off with NaOH since its molecular weight is less, so you need less NaOH for the same conductivity than KOH. (Also the conductivity is heavily factored on the OH- ion, not the cation, thats why NaHCO3 and Na2CO3 conduct less, since they are far less basic = less OH- molecules, but safer for sure)

If you want to take it further for better conductivity in exchange for lower shelf life I reccomend sulfuric acid, since H+ has the highest mobility (higher than OH-) yielding higher conductivity. The downside is that there will be surely sulfuric acid vapors escaping and generally speaking more corrosive to metals than alkali hydroxides.

My first electrolysis prototype was sulfuric acid. It ended up being way too expensive and unsafe. Finding materials that Sulfuric acid wouldn't destroy was difficult. I choose hdpe and couldn't make an airtight seal due to the same reason I picked it. The chemical inertness stopped all my adhesives from working. Even the poly welds failed. Thank you for all the help Andrew. I am close to being done and am just hoping at this point I don't run into any major design flaws. Success or not the knowledge and experienced I have gained is more valuable than the hydrogen generator.


text:
A dataset of drugs or molecules that failed the FDA approval process
I am looking for a dataset of bad drugs, the opposite of FDA approved drugs.
There is this database of withdrawn drugs, but these drugs were withdrawn after making it into the market.
Is there any such dataset of drugs that failed the FDA approval process because of their toxicity or poor pharmacokinetics properties?


text:
Formation of toluene from (cyclobutane-1,2,3-triyl)trimethanol using conc. sulphuric acid
In my book this reaction is given without any mechanism:

I thought that simple dehydration reaction is going to take place with rearrangement of the carbocation but I have no idea as to how toluene turned out to be the product.

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated.

The mechanism (on paper) seems to be as follows
Dehydration $1$
For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows:

Dehydration $2$
Similar to the first dehydration, a ring expansion occurs to stabilize the carbocation by forming a 6-member ring:

Dehydration $3$
In the third dehydration, a hydride shift takes place concomitantly forming a tertiary carbocation stabilized by conjugation with the double bonds. The double bond is then formed giving the compound aromaticity.

The mechanism should provide for the successive processes of carbocation formation, rearrangement and proton loss.
Observations(Points to consider):

The tendency to attain stablity by aromaticity.
The high ring strain in the cyclobutane ring.

The first step step should involve protonation of each hydroxy group attached. Protonation produces hydronium ion from hydroxide, which is a good leaving group. It leaves forming carbocation.
The carbocations, although formed in a single step, can be dispayed using successive steps for ease.
Each step of hydronium ion loss will produce a primary carbocation, which is unstable(due to lack of any prominent stabilising effect like resonace or  hyperconjugation). Hence, rearrangement is the only method emplyed for efficient stablization of catbocation.
The neighbouring migration may be of -CH or -H. Generally, the migratory amplitude of -H is higher than -CH. But hydride migration will not solve the problem of enormous ring strain. Therefore, to ease ring strain, it is -CH which migrates.
Clearly, now:

The carbocation is secondary and more stable than the initial primary.
The extreme strain in the ring is also relaxed.
This will be followed by a deprotonation step, forming double bond.

This migration of -CH group and successive step of deprotonation occurs twice, in two -OH losses.
In the third -OH loss the ring is expanded enough to ease ring strain. Thus, hydride, having lower migratory amplitude, is preferentially, migrrated to transfer positive charge. But this does not cause any ring size change. Next deprotonation step causes, double bond formation, to generate toluene. This is also enhanced due to the tendency to acquire aromaticity, as multiple double bonds are quite unstable.


text:
Ice geometrical shapes
https://en.wikipedia.org/wiki/Ice
https://www.google.com/search?q=ice&prmd=inmv&sxsrf=ALeKk028fJvrUcWLWu8EV-KsRGuiw0iWWw:1596098668929&source=lnms&tbm=isch&sa=X&ved=2ahUKEwiqjcDxyvTqAhWZaCsKHY36BIYQ_AUoAXoECBMQAQ
Can Ice formation creation possible for all geometrical shapes viz Rectangle, Square, Circular, Cylinder, Kite, Cone, Rhombus, Sphere, Cube etc?
If yes, What is the process?
If no, Why?

Do you mean with the lattice structure or just the shape of the ice block? If it is just shape then it only depends on the mold you use

Pour water into any shape, and it will freeze in that shape. There is no chemistry behind it.

With the Lattice structure. Can we create a Cone or Cylinder structure of Ice?

https://www.google.com/search?tbm=isch&sxsrf=ALeKk00nBSln87IXevZplyFbiNzikRyRWg%3A1596100051626&source=hp&ei=040iX-qhJI-T4-EPmJuMwAY&q=lattice+structure+of+ice&oq=Lattice+structure+of+ice&gs_lcp=ChJtb2JpbGUtZ3dzLXdpei1pbWcQARgAMgIIADIGCAAQBRAeOgcIIxDqAhAnOgQIIxAnOggIABCxAxCDAToFCAAQsQM6BAgAEAM6BAgAEBhQqgxYoGZgrXBoAXAAeACAAasFiAHJJpIBDDAuMTcuMS4xLjEuMpgBAKABAbABBQ&sclient=mobile-gws-wiz-img  https://www.stlfinder.com/3dmodels/lattice-structure-of-ice/

The answer of this question does not depend on the scientist's will. It depends on the nature. The crystalline form cannot be obtained by calculation. It is an experimental observation. Nobody knows why ice is cubic or orthorhombic in a given domain of temperature and pressure. Nobody can predict the limit of these domains. It cannot be predicted.

Lattice structure is not a relevant or useful concept when making ice cubes. Stop thinking about it.

I think you wish to refer to the fact that different forms of ice called ice phases of water are available. Like if under great pressure, in some gas giants like neptune, ice VII may form which has a cubical structure.
For more information on the different ice phases, Water Structure and Science would be a good read.


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Boilling colloidal silver?
I have some colloidal silver that i bought from a shop. It is 18ppm, however, I need a higher PPM and I can't seem to buy it anywhere where I live.
If I boil this colloidal silver, would it make the final concentration of silver go up?
For example, If I had 1 liter of 18ppm colloidal silver, and boiled it until I only had 500ml of colloidal silver, would that mean that I now have 500ml of 36PPM colloidal silver?

Better evaporate under a vacuum !

Boiling will probably destroy it.

It may cause silver aggregation and loosing colloidal properties.