text:
Why transition metal catalyzed [2+2]cycloaddition reactions allowed?
As you know, cycloaddition reactions which (4n)π electrons are involved in such as [2+2]cycloaddition are forbidden in normal organic reactions, but transition metal catalyzed [2+2]cycloaddition reactions are allowed.
I think the d orbitals of transition metals play important roles in the reactions, but I cannot understand how they work.
I found this article on the Internet, and the reaction is explained.
https://pubs.acs.org/doi/abs/10.1021/ja00734a016#

When metals are involved, the typical rules for thermal/photochemical [2+2] reactions don’t really apply. The metal generally involves a non-concerted (I.e. stepwise) process, so no issue with it


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What happens if you superheat timber in a zero oxygen environment?
I was thinking to myself, what would happen if you were to put a lump of wood in a zero oxygen environment and superheat it? It couldn’t burn because fire needs fuel, heat and oxygen however because wood is a mixture of cellulose + other things, it can’t really just melt... If this were done, what would really happen?

Pyrolysis

A crude version of this process is done to create charcoal. Piles of wood are structured so that burning some creates a great deal of heat in the centre of the pile but little oxygen. The centre of the pile pyrolises leaving charcoal.

Also a name for methanol was Wood spirit. And ants were distilled leading to formic acid.

Similar: Can paper burn without oxygen or air?

@matt_black - indeed, and similarly turning coal into coke.

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How to choose the parameter to scan in finding transition state by using Gaussian 16?
I have problems in finding the TS of reaction dehydrogenation of an organic compound. I read many tricks to find the TS and see scan is an effective method. As I know we can use the product of the reaction as a starting point to scan. But what parameter should I choose if there are two bonds were broken (the bonds in the red circles in the picture)?  I already tried to scan with one broken bond, and for the other one I just elongated it a little.  Unfortunately, the result has not converged. And one more question, does the orientation of the initial structure affect the result of the scan?
I am just a very new person in DFT as well as with Gaussian. Thanks in advance for any help.

By dehydrogenation, do you mean that H2 molecule is being ejected? Have you tried using QST2, or QST3? And also I am not sure what you mean by orientation of the initial structure. The structure can be rotated in any direction, it would not change anything.

@ShoubhikRMaiti yes, in dehydrogenation H2 molecules will be released. I already tried to use QST2 and QST3. But when I used the TS from those calculations to do IRC, the molecule at the forward and reverse side were not as my expectation. About the orientation of the initial structure, when I do the scan (relaxed redundant coord), I scan the bond length of atom 14 and 37 (as the picture), does the result will be affected by the position of atom 38? I mean if I do two scans with 2 different positions of atom 38 in the innitial structures, whether the results will be different?

What are the molecules that you get from the IRC from the TS of QST2/QST3? From what you have written, you are trying to model a concerted mechanism Since both atoms 37 and 38 are involved, different positions of atom 38 will definitely give different results. How much different I can't say.

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How to Test Auto-Ionization of Ammonia
I was wondering how we could test that the auto-ionization of liquid ammonia actually occurs experimentally. The chemical equation is described below:
$\ce{NH_3 <=> NH_4^+ + NH_2^-}$
I was thinking if there might be a substance that reacts with $\ce{NH4+}$ to form a noticeable precipitate, but I couldn’t find anything. Maybe we could test pH? But do pH tests work in liquid ammonia, and what would it tell us?
Any tips would be appreciated!
Thank you!

Autoprotolysis it is usually called.

No, that's not how it all works on more than one level.
First, the amount of $\ce{NH4+}$ out there is small. And by "small", I mean really, really small - about as small as you can imagine, only smaller. Many times smaller, I must reiterate. If some substance was to react with it (which is certainly not outside the realm of possible) and form a product in equally small amount, you won't be able to see it.
I intentionally avoid any numeric estimates here. You must have seen them already. They are too far outside the everyday experience to make an impression. Well, you may look up the auto-ionization constant of ammonia and see how it relates to that of water (which itself is not big). Guess what? It is many orders of magnitude smaller.
Now comes the second point, which kind of negates the first one. If something reacts with that $\ce{NH4+}$ (which, I repeat, is totally possible), then the auto-ionization reaction will produce more $\ce{NH4+}$ to replace the loss, and then more and more, probably until no more $\ce{NH3}$ is left. How so? Well, imagine a crowded theater emptying through one door. Obviously all those people can't fit in the door at once. Not even half of them can. But they don't have to. They just walk through it one by one, and before long, the theater is empty. Much like that, all $\ce{NH3}$ molecules will go through the form of $\ce{NH4+}$ (or $\ce{NH2-}$) to whatever products are there.
In a way, this will prove that the auto-ionization was there in the first place, but tell us nothing about its extent. To measure that, we need other methods; conductometry comes to mind.
As for pH indicators for water, they are definitely no good for liquid ammonia. They all will be far off-scale, much like your cooking thermometer in the middle of Antarctica. To work in liquid ammonia, one needs to develop a whole new set of indicators. Some seems to be known already, but I have no firsthand experience with them, and Google is the same over there as it is over here.
So it goes.

That explains a lot of my confusion. This was asked as an exercise, and knowing that the $K$ value was around $10^{-33}$, I couldn’t think of any experimental methods we had learned so far (from general chemistry... just starting out!). But I guess the trick is to think about it conceptually as an equilibrium problem and do something more conceptual like the setup you mentioned above.

Thank you for your answer though. As a pure mathematics person, I’m still a little rough when it comes to experimental methods. I’ll be sure to research conductometry. We haven’t learned about it yet, but it sounds interesting!


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Can I synthesize an Amide simply by heating an amine and a carboxylic acid (with a vacuum pump)?
I am interested in creating an amide of stearic acid. I thought that simply mixing both of them in some solvent, heating the solution, and using a vacuum pump to distill the liberated water (to drive the reaction forward) would be enough.
But a bit of research showed that this synthesis is normally done with acyl chlorides derivatives or methyl ester derivatives or other more elaborate methods.
Without access to those tools, is it still possible for me to produce a good yield of amide, using only a fatty acid, a primary or secondary amine, a solvent, and heating & vacuum?
(Secondary optional question: If I use ethanolamine, what will be formed preferentially, the amino-ester or the hydroxy-amide?)

What is "some solvent", and how do you separate solvent & water after boiling it off? There is a special piece of glassware to do this with e.g. toluene, another one with chloroform, but you don't have such equipment, right?

And, please: Your "secondary, optional question" must come before the other one. What reaction exactly do you want to do?? Ethanolamine is not just an amine. Methylamine, diethylamine, triethylamine, those are, and they're diverse enough already.

To answer your second question - which should really be a separate question - the amino group is by far the better nucelophile so you'll get the amide.

It can be done but it is not a good reaction. From Lundberg et al. [1]:

Formally, the amide bond is formed through the condensation of a carboxylic acid and an amine with the release of one equivalent of water. This reaction has been considered challenging due to the competing acid–base reaction, which occurs when the amine and the carboxylic acid are mixed. Although the amide bond can be formed from the corresponding ammonium carboxylate salt upon heating, this reaction has generally been considered to be of limited preparative value. Furthermore, the high activation barrier for the direct coupling of a carboxylic acid and an amine can only be overcome using forcing reaction conditions.

Use of borate ester catalysis under Dean–Stark conditions has recently been reported to give good yields without resorting to acyl activation groups [2].
References

Lundberg, H.; Tinnis, F.; Selander, N.; Adolfsson, H. Catalytic Amide Formation from Non-Activated Carboxylic Acids and Amines. Chem. Soc. Rev. 2014, 43 (8), 2714–2742. DOI: 10.1039/C3CS60345H. (PDF)
Sabatini, M. T.; Boulton, L. T.; Sheppard, T. D. Borate Esters: Simple Catalysts for the Sustainable Synthesis of Complex Amides. Sci. Adv. 2017, 3 (9), e1701028. DOI: 10.1126/sciadv.1701028.


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Stabillity of ethyl, iso propyl and vinyl free radicals
Well this might seem like a dumb question but I was just wondering if any other factor would apply here as well?

By simply looking at the number of hyperconjugative structures, I could easily tell that the isopropyl would be the most stable followed by ethyl and the vinyl radical at the last.
But for the vinyl, I didn't find applying hyperconjugation satisfactory.
Would we apply some other concept while comparing the vinyl radical with the other two?

How would we describe the hybrization of the orbital containing the electron in the ethyl radical, the vinyl radical.  Which of those two orbitals would an electron be more stable in?

@ron  I believe the hybridization of the orbital containing the electron in ethyl free radical would be $sp^2$ while in vinyl free radical it would be $sp$ and since sp has more s character so it would keep the electron more closely held. But how would this help us determine whose more stable?

But according to that, the vinyl free radical would be more stable than ethyl free radical?! But that's not true right? Vinyl should be least stable

@ron can we say that since the electron is in sp hybridised orbital in vinyl free radical, it would be pulled more towards the already existing C-C sigma and pi bonds which might lead to its destability. Hence making the vinyl free radical the least stable?

Your thinking is correct.  The orbital containing the lone electron is electron deficient and does not want to be closer to the nucleus.


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What would be the major products of this reaction?

Obviously, A would be 3-Bromoo 3-methyl cyclo hexene:

as the stable free radical intermediate would be formed.
But B is what I am having problems with. I read that $\ce{LiAlH4}$ forms alkenes with three degree halides. So my guess was
2-Methyl-1,3-cyclohexadiene:

But when I checked my book, it said B would just be the Bromine getting substituted by Deuterium in (A)
(I'm sorry I don't know any site that allows making chemical structures with isotopes included so I had to describe it like this.)
Can someone please help me out with this?

How are you introducing Chlorine with N-BROMO-succinimide? Are you aware that LiAlH4 reduces haloalkanes to alkanes?

@Waylander Sorry, that was a mistake. I have made necessary corrections. And yes, I am aware that $LiAlH_4$ reduces haloalkanes to alkanes. But I also read that in case of three degree haloalkanes, $LiAlH_4$ gives alkenes instead of alkanes

There are 4 possible constitutional isomers possible from 2 allylic radicals. Why this one?

According to these incomplete questions https://chemistry.stackexchange.com/questions/125942/how-is-nabh4-able-to-reduce-tertiary-alkyl-halides-to-alkane-but-lialh4-create-a    and  https://chemistry.stackexchange.com/questions/70607/elimination-of-tertiary-alkyl-halide-with-lithium-aluminium-hydride-to-give-alke  you are correct and your book is wrong

Ok thank you so much! Also, the product that my book mentioned would still exist right? Only that it'll be minor while the one I mentioned would be major

@Prajwal Tiwari: A should be 3-bromo-1-methylcyclohexene (not 3-bromo-3-methylhexene; it is cyclohexene, any way). Thus, B should be 3-deutero-1-methylcyclohexene.

@Waylander: Isn't tertiary carbon radical rearranges to give 3-bromo-1-methylcyclohexene as the major product?

@MathewMahindaratne  That rearrangement is possible. This is a bad question given that there is no definitive answer from the literature.

I will give this my best shot. There doesn't appear that there is any conclusive evidence for this reaction since some sources say elimination is favoured while others say substitution is the major product. However, for this reaction, the difference likely resides in the fact that it is an allylic bromide. This is extremely favourable for substitution reactions because the intermediate will be very stable. Just because of that, substitution may be favoured over elimination and result in the major product being Br substituted for a deuterium atom, just as your book mentioned.


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The type of H-NMR spectra observed for S=PF2H and other inorganic molecules similar to this
I was hoping to learn more about the types of splitting observed in inorganic molecules such as S=PF2H. I am familiar with the typical splitting patterns associated with the study of organic molecules studied by C and H NMR. We know that H and F have a spin of 1/2, in reading about the effects of fluorine in on H NMR, it appears that fluorine relaxes in a similar way to protons when pulsed by a radio frequency. Since the two nuclei have the same spin but are different types of nuclei altogether and impact each other according to the Nuclear Overhauser Effect, what kind of signals would be given by an H NMR for the molecule? Would we see signals corresponding to fluorine given that it has the same spin as the proton?

To be honest, I'd say you're overthinking it; for spin-1/2 nuclei relaxation behaviour (and NOEs) don't really matter, at least to a first approximation. Yes you would see H–F coupling in the 1H spectrum. No you wouldn't see 19F peaks unless you specifically looked for them, because 19F peaks appear at a different frequency altogether. It's like how you don't see 13C peaks in 1H spectra, but you could if you went hunting for them and acquired a 13C spectrum.

@orthocresol I think you are right, I probably am overthinking it quite a bit. When I think about coupling, I am typically associating it with the constant that might be found between the distance between, for example, the center-most peak in a triplet and the two smaller peaks to either side. So if I am understanding you correctly, yes coupling will occur, but no, we would not observe signals for the fluorine atoms, so you would probably see a singlet for something like S=PF2H.

No, the proton is a triplet, and all three peaks in that triplet belong to the proton. You will see all of them. No part of the triplet belongs to the fluorine. There's also 31P which is NMR active, but ignore that for now, I suppose....

@orthocresol so suppose we ran an H-NMR experiment with S=PF2H, we would observe a triplet for the single proton due to the two neighboring fluorines, but we would not observe a different triplet due to one fluorine having a hydrogen and a fluorine neighbor?

"we would not observe a different triplet" Yes and no. You indeed wouldn't observe a different signal for the fluorine (unless you were to record a 19F spectrum). It won't be a triplet, though: the two fluorines are equivalent and so you should not see coupling between the two fluorines. It would just be a doublet: fluorine coupled to proton. Note that everything we've talked about so far will in practice be further split by 31P (spin-1/2 too). So the 1H peak is in fact a doublet of triplets.

@orthocresol wow that is really fascinating, I would certainly not have concluded that. Thank you!

The everything OP and  orthocresol discussed in the comment session is in this reference book (Ref.1). I include its PDF file (Ref.1) for benefits of OP and future readers. An example of coupling is depicted in the following figure (from Ref.2):

This spectrum is qualified as other inorganic molecules similar to $\ce{S=PF2H}$ molecule. What I want to show is alike different coupling constants in $\mathrm{^1H \ NMR}$ due to the molecular structure, bond distance, etc., the coupling constants in $\mathrm{^{31}P }$ and $\mathrm{^{19}F \ NMR}$ are also changed as well. For instance, $^1J_\ce{HH}$ is $17$-$\pu{20 Hz}$ in $\mathrm{^1H \ NMR}$ organic compound. However, $^1J_\ce{PH}$ is $\pu{836 Hz}$ for this particular molecule ($\ce{PF2H(^{15}NH2)2}$) while $^2J_\ce{PH}$ is $\pu{11.8 Hz}$ for same molecule (c.f., avarage $^2J_\ce{HH}$ is $\approx \pu{7 Hz}$ for a hydrocarbon with more than 2 carbons).
This particular spectrum consists of 90 peaks due to additional coupling of $\ce{P-^{15}N}$ coupling. If the compound is not labelled with $\ce{^{15}N}$, the number of peaks would reduced to 30 peaks, a relatively simplified spectrum, but still complicated coupling pattern of doublet of triplet of quintet.
References:

Michael O’Neill, In Assigning Inorganic NMR Spectra: a programmed approach to solving NMR problems in main group chemistry; Copyright © 2019 Michael O’Neill (ISBN-13: 978-1-0864-12017) (PDF).
David W. H. Rankin, Norbert W. Mitzel, Carole A. Morrison, In Structural Methods in Molecular Inorganic Chemistry; First Edition, John Wiley & Sons: Chichester, United Kingdom, 2013 (ISBN-13: 978-0-470-97278-6).

Thank you! That is exactly what I needed to see. Examining the splitting diagram is very helpful for these types of problems! I suppose if we considered PF4H for example, we might see a doublet of quintets or a quintet of doublets due to the coupling between the proton and the phosphorous and the subsequent splitting from the fluorine atoms. I worked out the splitting diagram for this and that is what I concluded.


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Reactivity of boron nitride BN, reaction with Cl2
lately I've been interested in chemistry of boron and it's properties.. and I am looking for more information. For example I can't really seem to find lot of information about boron nitride BN other than its possible structures. I know that BN is relatively unreactive, and some stated reations found are with F$_2$ or HF, but are described reations with other halogens? With Cl$_2$? I would appreciate even literature recomandations, to find out more.
Thank you, for your time.

(Some time ago we decided against using $ in titles.) Now to the point. Never heard of such. At high temperature, maybe. Then again, if a compound is so unreactive that somebody bothers to react it with fluorine... well, that's a telling detail.

Cubic or hexagonal BN?

@Mithoron Hexagonal BN


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How to calculate Energy Efficiency of electrochemical production
I'm trying to understand some basic electrochemistry as a non chemist and struggling to find the info online effectively.
Its clear that energy efficiency = Energy(useful) / Energy (input).
So, I know how to solve for energy efficiency (η) when given a quantity of product and quantity of electricity.
η = product combustion energy / (power x time)
However, I'd like some help figuring out what to do with the information that is usually given in papers, so I can calculate η when it is not given.
Most papers report Faraday efficiency ( which I understood to be the moles of electrons that end on desired product/ invested moles of electrons)  (the denominator here can be calculated from the current and F constant).
also given are:
Catalytic surface area.
Cell voltages.
Over potentials.
ohmic drops...
etc.
For instance, I want to calculate the energy efficiency of formate production given in this paper (see image) https://www.nature.com/articles/s41467-020-17403-1/figures/3

Alternatively:
How to calculate the bottom line here:

If you can explain how its done or points me to resources to learn this, that will be super helpful.
Thanks,
D.


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What are quantifier and qualifier ions in mass spectrometry data for GC-MS or LC-MS?
I've read that one should pick an ion peak that is unique to an analyte for quantification purposes and then pick at least 2 qualifier peaks and use their ratios to identify the analyte (or avoid misidentification thereof). How exactly does this work?

This is a general "identification" rule which should not be restricted to mass spectrometry.
In chromatography you may be analyzing very closely related analytes whose masses are very close or even identical. Think of structural isomers (low resolution MS is blind to enantiomers, isotopomers, isotopologues etc.). Structural isomers may very closely related mass spectra and their molecular ion peaks would be identical in mass, but their fragmentation patterns might be slightly different. So the "good" advice is pick not one but more than one line in order to positively identify a target analyte.
Think of the legal consequences of false positives. A sportsman or women may loose their career if the tests show a false positive or a criminal may go scot free. Therefore analytical chemists have a huge responsibility on the society in terms of having a very high degree of confidence in identifying or quantifying anything.
Although most of the people think that mass spectrometer is a "perfect" detector in chromatography, it is nowhere close to being one. Many compounds do not ionize well and hence go undetected. It is extremely useful though.


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DIY lipid nanoparticles
I, and a number of friends, take a bunch of supplements that are polyphenols and often flavin-oids, -ols or -es; also numerous carotenoids.  Examples include quercetin, luteolin, fisetin, curcumin, resveratrol or pterostilbene, apigenin, lycopene and astaxanthin.  Most of them are available in a relatively inexpensive, more or less pure form, and in a much pricier lipid nanoparticle form. The latter is justified because most of these substances are poorly absorbed, and lipid encapsulation has been shown to improve this.
When I look at the ingredients used to produce these lipid nanoparticles, they are generally cheep and easily available. They usually include some lecithin fraction, like phosphatidylserine or phosphatidylcholine, non-toxic organic solvents like propylene glycol or polyethylene glycol, and water or ethyl alcohol or both. You can get all these things from Amazon. So I was looking at this list, and I thought:
Mayonnaise.
I’ve made mayonnaise at home with just a blender. It’s not hard, but it is tricky. Try dribbling the aqueous part into the oil a few drops at a time, rather than the other way around. I’ve done this, and it never emulsifies.
So I have this idea that there should be, or at any rate there might be, a recipe for self-assembling lipid nanoparticles that could be made using kitchen equipment, that could be used to improve the nonavailability of a wide range of polyphenols such as those listed above, at lower cost and with greater satisfaction then having someone else do it and deliver it in capsules. And I was hoping that someone here might know of one, or where I could find one.
Based on my very limited exposure to the literature, I think the model I am aiming for is a colloid of nanoscale balls of one or more polyphenol dissolved in a lipid or a non-polar solvent, with a "skin" of phosphatidyl-something, in a mainly water matrix. I should add that all of these polyphenolic substances are marketed as supplements and considered to be foods under FDA rules.
Of all the things I don't know, I think the thing that would help me the most would be suggested identities for a GRAS lipid or lipid-like solvent that is good at dissolving many polyphenols, under atmospheric pressure and temperatures below the boiling point of water.

It's not really clear what you are asking. What do you mean with "generic"? What is wrong with using a mayo recipe or just (say) olive oil? Is this for home prep or for commercial purposes?

Dear @BuckThorn -- By "generic" I mean a recipe that does not require substantial knowledge of the polyphenol or carotenoid that you are trying to deliver, beyond “keep adding solvent X and stirring until fully dissolved.” I am equally interested in lipid-coated bubbles and in solid lipid (or digestible wax) nano-particles. I'm a member of a small -- call it a club – that pools our research on nutritional approaches to longevity and buys supplements that cost less when purchased in bulk. Some of us share our purchases with our partners. We do not resell anything we buy.

I actually tried a mayo-like approach with a curcumin-quercetin-fisetin blend, but even heated, I could not get much of it to dissolve in olive oil. The result was more like a slurry, and swallowing it was not pleasant.

Welcome to the world of drug formulation. My gut instinct is to try to discourage you. There are ways of getting the stuff into a higher concentration cream, I'm sure but if cooking in the kitchen doesn't work your options narrow quickly. You probably don't want to work with volatile solvents or excipients and end up poisoning yourself. There is a lot of literature on attempts to solubilize curcumin and none look simple to implement in the kitchen. I would stick to curry (a reasonable diet). There is no "magic antioxidant bullet".

Otherwise your question is very interesting and it would be interesting to see if someone here has better advice. You might also try a formulation site for creams (lotions) and the like.

@andrewH Did you take a look at https://onlinelibrary.wiley.com/doi/full/10.1111/ics.12452? It's free access. They can probably point you to a good direction. One of the methods, precipitation, looks relatively simple.

I've tried to poke around the literature, & see the question as resolving into 2 parts. First, what are some GRAS solvents good at dissolving a range of polyphenols? Second, once dissolved, how do you make it form easily absorbed nanoparticles? I assume the answer to the 1st question should be an actual lipid or something nonpolar. On the blender mayonaise model, I add an emulsifier (e.g. phosphatidyl-choline or -serine) to an aqueous base, & drip the dissolved nutrient into the running blender. But all the recipes I've seen have 5 or more ingredients, & I don't know what the others are for

In my last attempt at this (after the olive oil attempt) I just poured a mixed bag of polyphenolic suppliments into a blender with some water, ethyl alcohol, propylene glycol, some polyethylene glycol 40, soy lecithen, and a little DMSO. I believe this by-guess-or-by-golly approach, using all GRAS ingredients, was reasonably safe, but I don't think it gave me nanoparticles. The result was too coarse and seperated too quickly. I think I should have tried to dissolve the polyphenols in the PEG/PG & dripped it slowly into the water/alcohol/lecithin. But the ratios (& DMSO) were complete guesses.

Thanks, @K.Cl Mar ! That is a nice article, and very much in line with my goals. But I am not confident that solvents appropriate for vitamin E, with its single phenol group and long tail, would work on the larger and more ramified polyphenols I am looking at. Do you believe they would be?

Sorry @andrewH, I was only notified now of your comments. Hm, you already tried olive oil, and I was going to suggest other simple oils. You could also look for where these compounds are typically found, and try get your solvent inspiration from that. You should perhaps look at some basic books on the area, I like "Introduction to Colloid and Surface Science" by Duncan J. Shaw as a general introduction to colloids and stability.


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Does the chlorination of 4-methylcyclopentene generate two enantiomers or two diastereomers?
Does the chlorination of 4-methylcyclopentene generate two enantiomers or two diastereomers?

I think it makes two enantiomers (or one pair of enantiomer) since the starting compound is achiral, but the answer key says it's two diastereomers.
When I tried making the products using molecular model kit, I think  I got enantiomers.

Under what conditions?

I'm taking a intro organic chemistry course so sorry if I'm not being specific enough. I'm pretty sure it's just supposed to be the reaction that forms a chloronium ion and undergoes anti addition.

You have a reaction that is not stereoselective. Such a reaction will produce something that is either achiral or racemic on an achiral reactant. Since there are two products here, they must be enantiomers and formed in equal amounts.

It is interesting when there are two diametric answers to a question.[Zhe's comment and KMT's answer] Agreed trans addition to a double bond gives RR or SS configurations about the vicinal carbons. The question is: Is there induced chirality in the carbon with the methyl group? [additional Question what is correct name for this compound?] My first impression is that the methyl carbon has two identical groups either a RR or a SS and is not chiral. [...] Hopefully someone knowledgeable will settle this dilemma and point out the fallacy or correctness in my, ZHe's, or KMT's analysis.

For an anti addition, you get enantiomers (left two structures) in equal amounts. The enantiomers in the orientation shown are related by a mirror plane swapping left and right:

For a syn addition, you get achiral structures that are diastereomers (right two structures) in unequal amounts.

[OP] I think it makes two enantiomers (or one pair of enantiomer) since the starting compound is achiral, but the answer key says it's two diastereomers.

If the question is about anti addition, as the OP mentions in the comments, then you get enantiomers.

[Zhe in comments] You have a reaction that is not stereoselective. Such a reaction will produce something that is either achiral or racemic on an achiral reactant. Since there are two products here, they must be enantiomers and formed in equal amounts.

The syn addition seems to contradict that statement. You start with an achiral reactant, but you get a pair of achiral diastereomers, and not at equal amounts. It is true that in cases where you get enantiomers from non-chiral starting materials (like in the anti addition), it will be a racemic mixture (formed in equal amounts).

[deleted answer] The two products formed will be considered enantiomers if the configuration of all the chiral carbons in the compound are different. But here only the carbons to which chlorine attaches have different configurations and not the carbon to which methyl is attached. Hence they will be diastereomers.

If you compare the first pair of structures in the figure above (the ones labeled anti addition), you might think they are diastereomers just by inspecting the wedges and dashes. However, if you rotate these structures around a vertical axis, the methyl group will now point downward while the chlorines point in the same direction as before the rotation, as the figure below illustrates. So the two structures in the box are identical, and if you compare the left structure with the right one in the box, you can see that they are enantiomers, related by a mirror plane parallel to the paper. Merely comparing wedges and dashes, you might think that the two structures in the box are diastereomers, but in fact they are identical.


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How does one test for lead in dry paint?
What analytical methodologies are capable of identifying lead in peeling paint? I am a hobby chemist owner of a Clinical Research Investigative Site.  Please tell me if this site is only for professional chemists.
If so, I will cancel my registration.

The best way would be by a hand held x-ray fluorescence instrument.

Lead test strips are available in hardware stores where I live and I think they are widely available. I do not know their limit of detection.

This site is not specialized and reserved to professional chemists. On the contrary. A lot of teenagers are using it every day.
Now if you want to identify lead in a dry paint, you may burn it in the flame of a Bunsen burner, and save the ashes. They contain a mixture $\ce{Pb + PbO + Pb3O4}$. Then dissolve the ashes in nitric acid (concentrated $\ce{HNO3}$ diluted with its own volume of water). Filtrate the insoluble part. Dilute the filtrate still more, by adding an equal volume of water. This gives you a lead nitrate solution, with which you carry out the following tests. Add some $\pu{1 M}$ $\ce{HCl}$ to a part of the obtained solution : it should produce a white precipitate of $\ce{PbCl2}$, if the solution contains some lead ions. Or add some $\ce{KI}$ solution : You would obtain a yellow precipitate of $\ce{PbI2}$, if the solution contains lead. The reactions are :
$$\ce{Pb(NO3)2 + 2 HCl -> PbCl2(s) + 2 HNO3}$$
$$\ce{Pb(NO3)2 + 2 KI -> PbI2(s) + 2 KNO3}$$
A further proof is obtained by heating the solution containing the yellow deposit of $\ce{PbI2}$, because this substance gets redissolved at $\pu{100°C}$. And after slow cooling, $\ce{PbI2}$ will precipitate back and it looks like wonderful golden plates, floating in the liquid. It is amazing. If these precipitate do not occur, there is no lead in your paint.


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Why don't 1,2-substitued alkenes polymerize as much as mono or disubsituted alkenes?
Ethylene, propylene, PVC, acrylates. All of these monomers that are industrially produced seem to only have substituents on one of the carbons, rather than both. Why don't we usually see 1,2-substituted alkenes undergoing polymerization? Sterics? Electron cloud interference?

It is true that we do not see the products made by the polymerization of 1,2-substituted alkenes or 2-olefins (even natural rubber does not fall in this category). Although a number of 2-olefins such as 2-butene in large amounts is produced by petroleum industry (naphtha cracking), they have not been utilized as monomers to obtain commercial polymers in daily life because of their low reactivity toward commonly used polymerization catalysts, probably due to steric effects. Perhaps, if 2-olefins are reactive as much as ethylene towards polymerization to give high molecular weight polymers, there would be a large number of new polymer materials in daily use.
Nonetheless, a few studies on the polymerization of 2-olefins by novel transition metal complexes, specifically using trans-2-butene as the monomer, have been reported (Ref. 1-3). According to all these three references, the monomer-isomerization has been occurred during polymerization process, indicating the steric effect. For instance, both cis- and trans-2-butene give high molecular weight polymers with a recurring unit of the corresponding 1-olefins (after monomer-isomerization by metal catalyst prior to the polymerization), which consists of $\ce{–CH2–CH2–CH(CH3)\!–}$ units instead of $\ce{–CH(CH3)–CH(CH3)\!–}$ units (Ref.1 and 3). Thus, it is unlike that highly condensed polymer can be produced from 2-2-olefins until the discovery of a novel metal catalyst, which prohibit the in situ monomer-isomerization.
References:

Mark D. Leatherman, Maurice Brookhart, "Ni(II)-Catalyzed Polymerization of trans-2-Butene," Macromolecules 2001, 34(9), 2748–2750 (DOI: https://doi.org/10.1021/ma002142j).
Mark D. Leatherman, Maurice Brookhart, "(E)−(Z) Selectivity in the Polymerization of 2-Butene Promoted by Ni(II) Brookhart-Type Catalysts," Macromolecules 2005, 38(6), 2072–2075 (DOI: https://doi.org/10.1021/ma048373y).
Kiyoshi Endo, Yoshiyuki Kondo, "Polymerization of trans-2-Butene with ($\alpha$-Diimine)Ni(II) Complex in Combination with $\ce{Et2AlCl}$," Polymer Journal 2006, 38(9), 1160–1164 (DOI: https://doi.org/10.1295/polymj.PJ2006058).


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What qualifies to be considered as a lone pair-lone pair repulsion?
Referring to, NCERT Chemistry Part I, Textbook for Class XI[1].

Here the text puts forward three structures of $\ce{ClF3}$.
My teacher said to me that (b) structure is unstable due to lone pair lone pair repulsion while (a) is the most stable of the given structures as there are minimal lone pair and bond pair repulsions and no lone pair lone pair repulsion.
That got me thinking what actually qualifies as a lone pair lone pair repulsions. Is it when the lone pairs are at an angle equal to or less than the 90° or is it something else?

Reference:
[1]: Chemical Bonding and Molecular Structure. NCERT Chemistry Part I, Textbook for Class XI, First Edition(October 2019 Reprint); Uppal, Shveta, Banerjee, Binoy, Eds.; National Council of Educational Research and Training: New Delhi, India, 2006; pp 113.

Regarding the structure of $\ce{ClF3}$, it has more to do with Bent's Rule than lone pair repulsions.  See this earlier SE Chem Q&A for an explanation. What is the favoured geometry of chlorine trifluoride?

When applying VSEPR, imagine that the various electron pairs occupy "orbitals", in essence volumes shaped like balloons, protruding from the central atom (in this case chlorine). The electrons ("orbitals") repel each other.
Which repulsive interaction is strongest?
One assumes that, placed in equivalent relative positions (angles), LP-LP (LP=lone pair, BP=bonding pair) repulsion is strongest, followed by LP-BP and then BP-BP. The idea is that bonding electrons are more distant from the central atom, whereas nonbonding (LP) electrons are held tightly by the central atom and therefore represent a more "concentrated" charge, and therefore repel other electrons most strongly. The repulsion is also stronger the smaller the angle between the electron pairs.
The strength of repulsion determines the order in which to prioritize the interactions, but this can be a little tricky, as your example shows, because it is a "non-linear" problem. When more than one pair of LP is involved, the geometry should minimize LP-LP repulsions followed by LP-BP repulsions, but there is a tradeoff. While (c) places both LPs on opposite ends ($180^\circ$) and would seem to minimize the repulsion, there is a bit of a tradeoff between LP-LP and LP-BP repulsions. Geometry (c) results in a lot (6) of LP-BP repulsions at right angles ($90^\circ$), whereas geometry (a) has 4 LP-BP repulsions at right angles. This leads to positioning of the LPs in equatorial (!) positions, even though LP-LP distance is not maximized.
Such arguments are not transparent, they are based on a crude method that only approximates the "real" electron distribution. The experimental geometry is in fact a distorted version of (a), with the LP-BP angles slightly greater than right angles ($>90^\circ$).
My practical advice is to simply remember that according to VSEPR lone pairs are always placed in the equatorial plane in trigonal bipyramidal electron pair geometries (those with 5 pairs). See for instance the following table:

So like how are we comparing, what's the criteria so as to call the repulsion interaction as not much as destabilising in comparison like in (a) lp lp is at 120 degrees which is less than that of 180 degrees in (c) but still (a) is being called as more stable.

@Rishi They are all repulsions, but as I try to explain the interaction doesn't change linearly, which makes your question good. It is difficult to add up the total repulsion if you can't say "an LP-LP repulsion of 180 degrees is so-and-so much energy" and "an LP-LP repulsion of 120 degrees is so-and-so much energy". Maybe there is a table but I haven't seen it. It is in my opinion confusing and that is why I give my advice in the end: when you have 5 electron pairs, always place lp in equatorial positions.


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Why is there no such substance as p-phthalic anhydride
Its formula is C6H4(CO)2O.  It is an organic molecule that doesn’t have any branching or functional groups on carbon 4 or 1.

First lets take a look at o-phthalic acid,

and it's anhydride;

Now as you can see the ring formed in case of o-phthalic anhydride is sterically feasible. The angle strain is not very high and the aromaticity is still intact. Now, take a look at p-phthalic acid;

Now, if you try to dehydrate this molecule as you see there is a very big steric problem. To accomplish this dehydration, the molecule must take a boat form, which would look something like this,

(I am using image of cyclo-hexane just to show you the structure, benzene ring will never do this)
But to accomplish this conformation the benzene ring has to loose it's aromaticity and also create a very unstable system with the pi bond densities repelling each other. So, as this conformation is not Thermodynamically feasible, anhydride formation doesn't take place.

Would p-phthalic acid dehydrate to form a polymeric anhydride (possibly a goo) or decompose? it does form polymers with dialcohols.

@James Gaidis I would assume yes as it gives very similar reactions in the production of PET(polyethylene terpthalate) and PBT(polybutylene terpthalate).


text:
Depression of the Freezing Point
Recently we learnt that there is a depression in the freezing point when a non-volatile solute is added to the solvent. Our teacher explained this phenomenon with a graph where the vapour pressure of the solution and pure solvent intersects with the vapour pressure of the solid solvent. my question is: wouldn't the vapour pressure of the solid solution be lower than the vapour pressure of the solid solvent(that would make another curve in the graph)? Because, the non-volatile solute is still blocking the surface of the solvent, thus reducing the vapour pressure.
The solution is of a solid non-volatile solute in a liquid solvent

Blocking the surface is not a part of the picture at all.

In chemistry we assume a mixture of two solids as two different phases, which do not interfere with each other and thus, the solid solute in the solid mixture will not interfere with the solid solvent. So, the vapour pressure of the solid solution and solid solvent will always be same. In a similar way, when two immiscible liquids are mixed together, the total vapour pressure is counted as the sum of Vapour pressures of the liquids if kept separately.
Hope this helps!!

Unless the compounds form solid solutions, if they have similar enough properties. But that is rarely the case for the volatile and non-volatile components.

@Poutnik Thanks for adding the value!!


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Adding aluminum sulfate to neutralize an aqueous solution of sodium silicate with pH of 12.6
How much aluminum sulfate do I need to add to an aqueous solution of sodium silicate to neutralize the solution (pH from 12.6 to 7)?
For a water treatment school project, we remove sodium silicate from water by adding aluminum sulfate to the mixture and filtering the precipitating products by reverse osmosis. Using the following equation along with the molar masses of both molecules:
$$\ce{3Na_2SiO_3 + Al_2(SO_4)_3 + 6H_2O = 3Na_2SO_4 + 2Al(OH)_3↓ + 3H_2SiO_3↓}$$
We calculated that it would take $\pu{51.5 g}$ of aluminum sulfate to react with $\pu{55 g}$ of sodium silcate. In theory, after filtering the precipitants, the $\mathrm{pH}$ of water SHOULD be equal to 7, since sodium sulfate is a neutral salt. However, in practice, we found that it takes a lot more than $\pu{51.5 g}$ of aluminum sulfate (specifically, $\pu{83 g}$) to bring down the $\mathrm{pH}$ of the water after filtration to 7. Chemistry isn't my strongest subject and I really have no idea why $\pu{51.5 g}$ aren't enough. Is this a reversible reaction and after the species get to a certain concentration it simply reaches equilibrium? If so then how do I find the equilibrium concentrations and the equilibrium constant etc. etc.? Otherwise why did we need to add more than $\pu{51.5 g}$ to fully neutralize the sodium silicate particles?

Sodium silicate is never pure in solution. It is usually  a solution containing nearly as much $\ce{NaOH}$ as sodium silicate.

That makes much more sense. I'm not sure if this is the case though, as we added sodium silicate powder to water.

Did you make $\pu{55 g}$ of $\ce{Na2SiO3}$ solution for this experiment? If water sample is unknown, it may contain more than $\pu{55 g}$ of silicate.


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Why is cyclopropenium aromatic but not cyclodecapentaene?
An $\ce{sp^2}$ hybridized carbon prefers a bond angle of $120^\ce{o}$. It would be willing to deviate if it acquires extra stability, for example through aromaticity. For this reason, the carbon atoms in cyclopropenium cation are happy with a bond angle of $60^\ce{o}$, which deviates from the standard angle of $120^\ce{o}$ by $60^\ce{o}$.

My question is, why doesn't cyclodecapentaene acquire a planar shape and become aromatic? I am specifically talking about this isomer:

In a regular decagonal structure, each internal angle would be $144^\ce{o}$. This deviates from the standard trigonal angle by $24^\ce{o}$, which is much less than the deviation for the cation.
My guess is that even though the deviation from the standard angle is much lower in the case of cyclodecapentaene, the number of carbon atoms experiencing the strain is far greater. This increases the overall strain in the molecule. Is my thinking correct?

Yes one can think this way. All boils down to consider that even in benzene what drives aromaticity is the sigma skeleton. In other words, the pi electrons are those being forced to delocalise,and not that delocalisation drives the skeleton. Of course this must be taken with caution, but it is easy to be seen in benzene as everithing fits the geometry of sp2 carbon. Note that for cyclopropenium there should be a banana character, as far the first part of your question is concerned.

Cyclopropenium can't deform to lower strain, while big molecules have plenty options to do it.

@Alchimista Thanks for making the clarification in the first part of your comment. Could you elaborate on this, though: "...for cyclopropenium there should be banana character."

@RayBradbury https://en.m.wikipedia.org/wiki/Bent_bond


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How can change in Enthalpy for a reaction be written like this?
If a gas expands in a container with constant external pressure then:
$W=-P_{ext}\Delta V$
Instead of $P_{ext}$ if we use $P_{int}$ then we will not get the same value because  $P_{int}$ is constantly decreasing from a certain value until equilibrium is attained, while the change in volume remains the same.
This is the reason why we use $P_{ext}$ here:
$\Delta H= \Delta U +P_{ext} \Delta V$
But then, $\Delta H$ for a reaction is:
$\Delta H= \Delta U +\Delta n_{g}RT$ ($n_{g}$ is for change in number of gaseous moles as the solids and liquids are ignored)
where $PV=nRT$ is used. However, the $P$ in the ideal gas equation is $P_{int}$.
So my question is how can $P_{int}$ be substituted in the place of $P_{ext}$ if the values of work done change when we do so?

Who says the change in enthalpy is $\Delta U+P_{ext}\Delta V$?  Please provide a reference.  This is not correct.

I've seen $\Delta H=\Delta U +P \Delta V$ in text books and when showing that $\Delta  H=q$ under constant pressure the work from $\Delta U = q+w$ gets canceled out with $P \Delta V$ from $\Delta H$  so I thought that for it to cancel out with work the $P$ had to be $P_{ext}$.Hence, the $\Delta H=\Delta U +P_{ext} \Delta V$

How would that work for an adiabatic irreversible expansion at constant external pressure (for which q = 0)?

@ChetMiller I don't see why it woudn't work. Wouldn't $\Delta H=0$ for adiabatic irreversible expansion.

Does the temperature change?  Is the enthalpy of an ideal gas a function of temperature?

@ChetMiller Temperature does change, therefore $\Delta H \neq 0$ as $\Delta H$ is a function of temperature for ideal gases. Ok, then what would be the correct expression for change in enthalpy that would work all the time.

The correct equation is always $$\Delta H=\Delta U+P_fV_f-P_iV_i$$Is the pressure of the gas in the initial thermodynamic equilibrium state of the gas $P_i$ in this irreversible expansion equal to the gas pressure in the final thermodynamic equilibrium state of the gas $P_f=P_{ext}$?

But the gas pressure in the initial state is not equal to final pressure in the irreversible expansion.  $\Delta H=q$ only for a reversible isobaric process (for which the final and initial equilibrium pressures of the gas are the same).

@ChetMiller yes, $P_f=P_{ext}$.So,$P_f$ and $P_i$ are internal pressures right. Then will we get the expression $\Delta H=q$ for a process at constant pressure?$\Delta U=q+w=q-P_{ext}\Delta V$ and $\Delta H = \Delta U+\Delta (PV)$ at constant external pressure. Substituting $\Delta U$ will we be able to cancel out some terms to get $\Delta H=q$

Let us continue this discussion in chat.

Only if the initial and final equilibrium pressures are equal, which is only the case if the isobaric process is reversible.  Otherwise, for your case, $\Delta H=-(P_i-P_{ext}) V_i$

@ChetMiller But, I have seen $\Delta H$ being interchangeable used with heat even when the inital and final equilibrium pressures are not equal .
Like to determine whether a process is spontaneous or not.$\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr} \geq 0$. here $\Delta S_{surr}=-q/T$ which is written as $-\Delta H/T$ but if a process is spontaneous it certainly isn't reversible.

For an ideal isothermal reservoir, the fluid in the reservoir is assumed to be incompressible, in which case the changes in enthalpy and internal energy of the reservoir are both equal to the heat received by the reservoir. Which $\Delta H$ does your entropy equation refer to, the enthalpy change of the reservoir or the enthalpy change of the system?

@ChetMiller The system.

What was the process that the system experienced?

@ChetMiller Say for a chemical reaction that takes place under constant temperature and pressure.

In that case, the initial and final pressures of the system are the same, and equal to the external pressure of the surroundings.  So...?

Are you asking why the standard change in entropy of a reaction is not equal to the standard change in enthalpy divided by the standard temperature?

Thanks, I think I have satisfactorily cleared up my doubts.

Your claim that $\Delta H = \Delta U + P_{ext} \Delta V$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $\Delta H = \Delta U + \Delta (PV) $. Here as you can see even when you compare it with $\Delta H = \Delta U + \Delta n_gRT$ there is no problem as we don't have to replace $P_{in} $ with $P_{ext} $.


text:
Why is phenyl carbanion aromatic?
In phenyl carbanion, one carbon atom has a lone pair on it:

This answer claims that the anion is aromatic, because "the lone pair on the benzene is in the plane of the ring and hence orthogonal to the conjugated $\pi$-system so that it cannot interact with it."
Why is the lone pair perpendicular to the plane of the molecule? It seems clear to me that the carbon in question has three $\mathrm{sp^2}$ orbitals, perpendicular to which is a pure $\mathrm{p}$-orbital containing two electrons. Assigning one extra electron (i.e., a negative charge) to an uncharged carbon in a benzene ring shouldn't rotate its $\mathrm{p}$-orbital by $90^\circ$.
The $\mathrm{p}$-orbitals in carbon are, of course, always perpendicular to each other. The $\mathrm{p}$-orbital containing the lone pair should, therefore, be perpendicular to the hybridized orbitals.

No, the lone pair is in one of the three sp2 orbitals, not the p orbital.

@orthocresol Whoops, the hydrogen was still tacked on to the carbon in my mental picture. Thanks.

The lone pair of electrons will be in a $\mathrm{sp^2}$ hybridized orbital and not in  a pure p-orbital. Thus, it wil be in the same plane as the molecule. Only the lone pairs which are in an orbital perpendicular to the molecular plane can take part in resonance, so those $\mathrm{sp^2}$ electrons will not be counted in Hückel's rule.


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What happens when hydrogen bonds stretch?
I'm researching about the Mpemba effect and I came across an explanation saying that the effect occurs because at higher temperatures, the hydrogen bonds between the water molecules stretch and store energy and when the water starts to cool, the hydrogen bonds go back to normal and release the stored energy on top of the normal cooling process, that's why hot water freezes faster than cold water.
I'm quite new to hydrogen bonds, so I don't really understand how hydrogen bonds can store energy. I thought bonding was something like a magnet attracting something, the further you get, the weaker the attractive force becomes. Or am I understanding it wrong?
The article that I saw this in is Hot water freezes faster than cold water; physicists solve the Mpemba effect

Hydrogen bonds are like other chemical bonds. They can stretch and then get contracted. It is like a teeter-totter, or like a spiral spring, or a leaf spring. If you give them some energy they will stretch far away from the equilibrium position, and oscillate later on. .

@Maurice What do you mean by oscillate?

Even if you assume that, while returning to their low energy state they will have to release energy working against freezing. The only way could be something relegated to kinetics....

@Brien Lim. Oscillation is like a vibration. A bond AB between two atoms A and B is continuously vibrating between a short and a long separation distance AB. At low temperature the amplitude of the vibration (= difference between longest and shortest AB distance) is small, maybe a couple of percent of the average AB bond. If energy is transmitted to this bond, it starts vibrating or oscillating more. The amplitude of the oscillation is larger.

We should read the original paper, at least about the freezing effect. H-bond vibrate as all the other bonds, with the difference that they have lower strength and are less defined (in the sense that a sample will have a multitude of slightly different ones).

Yeah this question is actually surprisingly difficult to to answer.  But let me explain it this way.   In thermodynamics you can't have negative temperature.  Thus anytime you are getting heat, you are actually gaining electrons.  When you gain electrons these chemical bonds change and rearrange themselves, which is how they can store energy.
We do not think about this normally because we look at hydrogen and we just see a proton and electron.  But in reality, unless hydrogen is in an extreme environment, it likely has absorbed energy in some manner.
For instance,  if you put a glass of water under concentrated sunlight it will absorb photons, these photons will be absorbed by the moving electrons inside the atom. Which we understand as "heat", moreover If there are more electrons than water molecules than it will start to evaporate because they run out of room.


text:
Why iron(II) spectrometry is not measured in distilled water and it requires a complexing agent like o-Phen?
We know that iron(II) ions in water have a green colour. Why we cannot draw the spectrum of iron(II) directly in distilled water to measure the absorbance then determine the concentration of iron(II)? We generally use ortho-phenanthroline as a complex agent to form a red-orange coloured complex with iron(II), then measuring the absorbance and the concentration. Why we Cannot use just distilled water as complex agent knowing that it gives green colour with iron(II)?

The complex is much more strongly light absorbing than the iron ions themselves. So you are putting a bell on the cat: much easier to hear when you startle the cat with light!

Iron (II) has no color to speak of. You can't see it until you have a lot of it.

The last photo here show how faint the green color was. So the idea is to have a strongly colored complex and have the target species and complexing agent be colorless or only lightly colored.

The o-phenanthrolin is extraordinarily sensitive to the smallest amount of $\ce{Fe^{2+}}$. A concentration as low as of $2$ microgramms $\ce{Fe^{2+}}$ per liter is still colored and its concentration can be measurable by optical absorption. As a comparison, a concentration of $\pu{30g/L}$ $\ce{Fe^{2+}}$ is the minimum measurable by Beer's law in water.

The answer would be greatly improved if you gave the detection limits for just an aqueous solution of $\ce{Fe^{2+}}$. Or perhaps you could compare the molar absorptivities.

I'm sure OP get the answer by reading all comments and the answer elsewhere. However, I'd like to point out few things on this spectrophotometric determination of $\ce{Fe^2+}$ in drinking water (it is not distilled water):

Iron can exist in natural waters where it originates from the dissolution of minerals. Even though most iron containing minerals are not very soluble in water as evident by the solubility products of iron-containing salts. However, a trace amount of iron can still dissolve in water. The dissolve iron is predominantly found in the form of $\ce{Fe^3+}$, because dissolved oxygen in water oxidizes $\ce{Fe^2+ -> Fe^3+}$.
Based on above fact, water cant be green, but probably yellow. But, it can be still determined by UV-vis spectroscopy, only if you have a lot of $\ce{Fe^3+}$ ions in your unknown drinking water sample to do so (see Ivan Neretin's comment and Maurice's answer), which is not possible.
On the other hand, 3 molecules of 1,10-phenanthroline (ortho-phenanthroline, a bidentate) chelate with one ion of $\ce{Fe^2+}$ to give highly intense orange-red $\ce{[Fe(o-phen)3]^2+}$ complex ion in water (Ref.1). The equilibrium constant for this reaction $\left(\ce{Fe^2+ + 3 o-phen -> [Fe(o-phen)3]^2+}\right)$ is $2.5 \times 10^6$ at $\pu{25 ^\circ C}$  in the $\mathrm{pH}$ range between  $\mathrm{pH} = 3$ and $\mathrm{pH} = 9$. Thus, a $\mathrm{pH}$ of about $3.5$ (using acetate buffer) is recommended to prevent precipitation of iron salts.
The molar absorption coefficient (ε) of this ferrous complex, $\ce{[Fe(o-phen)3]^2+}$ is $\pu{11,100 dm3 mol-1 cm-1}$ at the wavelength of maximum absorbance, $\lambda_{max} = \pu{508 nm}$. Thus, spectrophotometric absorbance of this complex can be easily measured in $\pu{10^{-4} M}$ to $\pu{10^{-5} M}$ or even lower levels at $\pu{508 nm}$, which is compatible with $[\ce{Fe^3+}]$ levels in drinking water (usually well water).
However, there is a problem: 1,10-Phenanthroline complexes quantitatively only with $\ce{Fe^2+}$ ions. But that can be solve using a reducing agent such as hydroxyl amine ($\ce{HCl}$ salt, which is soluble in water):
$$\ce{4Fe^3+ + 2HO-NH2.HCl -> 4Fe^2+ + N2O + 6H+ + 2Cl- + H2O}.$$ The reducing agent should be used in excess to prevent $\ce{Fe^2+}$ oxidizes back to $\ce{Fe^3+}$ by dissolve oxygen.

References:

R. Belcher, "The application of chelate compounds in analytical chemistry," Pure Appl. Chem. 1973, 34(1), 13-28 (DOI: http://dx.doi.org/10.1351/pac197334010013).

You forgot HCl in the right hand side of the last equation.

@: Thanks for the notation. I corrected it.


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API formulations - are excipients required
I have the following API's
Itraconazole, Ivermectin, Fenofibrate, Pyrvinium Pamoate,  Sulfasalazine, Artesunate,  All-Trans Retinoic Acid, Doxycycline,  Atovaquone,  Extremestane, Syrosingopine and Mebendazole
Why is it that API's require excipients?
Are they necessary for the active pharmaceutical ingredients bioavailability?
Or are the excipients used more as stabilises, preservatives, flavourings, binders and fillers so that as a finished pharmaceutical drug, it can hold its integrity/ shelf-life / taste etc.?

How can you weigh out your dose of API? It would be very easy to overdose if you cannot do this accurately

Very true, lab scales with 1mg accuracy would have to be used. My main curiosity around the subject was more around the bioavailability issues lots of tablets have - I thought why couldn't API's be directly applied sublingual for improved bioavailability, but i guess the consumer wouldn't take the drugs if they tasted like raw API's - maybe nanoparticle encapsulation of raw API could be sublingual without the taste

The ICH guidelines give some details about the presence of excipients, and the forms they may take (along with acceptable types/classes of excipients);
“The excipients chosen, their concentration, and the characteristics that can influence the drug product performance (e.g., stability, bioavailability) or manufacturability should be discussed relative to the respective function of each excipient. This should include all substances used in the manufacture of the drug product, whether they appear in the finished product or not (e.g., processing aids). Compatibility of excipients with other excipients, where relevant (for example, combination of preservatives in a dual preservative system), should be established. The ability of excipients (e.g., antioxidants, penetration enhancers, disintegrants, release controlling agents) to provide their intended functionality, and to perform throughout the intended drug product shelf life, should also be demonstrated. The information on excipient performance can be used, as appropriate, to justify the choice and quality attributes of the excipient, and to support the justification of the drug product specification (3.2.P.5.6).”
The major use of excipients is often just as bulk material in tablets. Many modern drugs are active at doses of 10-100 mg — these quantities are often so small that fillers are needed to make a tablet of acceptable size to be acceptable to the patient (compare to some older drugs where doses were often in the gram range).

Thanks NotEvans - it makes sense, I guess the main reason they use excipients to finalise the product mostly because it allows ease of administration for the consumer, by being in neat tasteless capsules as apposed to raw API powders which would require precise dosing, and in doing so, they then add further ingredients to prolong the shelf life once formed.


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Universal symbols for molar mass, molarity, and molality
This is a really basic (perhaps trivial) question, but I just want to confirm:
What are the SI symbols for molar mass, molarity, and molality? For molality and molarity, there is a difference, but I’ve found that the symbol is the same for molarity and molar mass, i.e. M (a big m).
Also, what are the meaningful differences between molarity and molality? Does it just have to do with mass (i.e. difference in mass between solute and solvent) vs. concentration (i.e. within the solution)?
I get the sense that molarity is defunct? I read this here -
https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-8

Have you heard of the pigeon hole effect? There are more quantities than English or Greek letters. You will always see some overlap. It is the context which is important.

That is such a good point, @M.Farooq. That is not clarified or emphasized enough. I get the sense that you can use M for molarity or molar mass; just specify beforehand, e.g. molar mass M of NH3 (or molarity M of NaCl). I see no reason to not use both, as long as you make it clear what the context is.

Exactly. I check IUPAC gold when I wish to confirm. You can see $M_r$ for relative molecular mass. If someone sticks to IUPAC recommendations then it is very clear who is who.

https://goldbook.iupac.org/terms/view/R05271

Related on meta: Quantities and units

According to IUPAC, the official name for molarity is amount of substance concentration, and its symbol is $c$. The symbol for molality is $m$ or $b$, and the symbol for molar mass is $M$.

what are the meaningful differences between molarity and molality?

Molality does not change with temperature because mass and chemical amount is temperature-independent. The amount of substance concentration (molarity) does because substances expand (mostly) with in increasing temperature. So for most applications where the temperature changes, molality is the better choice.
On the other hand, for room temperature applications where volumetric measurements are convenient, it is common to use molarity as the measure of concentrations.

See also https://chemistry.stackexchange.com/questions/115196/how-are-lowercase-m-and-uppercase-m-used-in-general-chemistry-courses/115197


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Are Group 2 metal fluoride salts less soluble in organic solvents than Group 1 fluorides?
It is clear that Group 1 metal fluoride salts DO NOT readily form solutions in organic medium. The amount of solvation that can occur leading to the eventual release of a fluoride ion depends on many factors, one is the polarity of solvent (described by dielectric constant). Most examples seen in the literature seem to evolve around using $\ce{CsF}$ or $\ce{KF}$ in the vicinity of 18-crown-6 ether as common examples for the release of fluoride ions (not considering TBAF). However, I cant seem to find any description for $\ce{MgF2}$ or $\ce{CaF2}$ salts and how their solubility in organic medium affects the release of fluoride ions.

Don't get your hopes up.  Wikipedia identifies magnesium fluoride as insoluble in ethanol.  Calcium fluoride fares no differently in acetone.  While these data are hardly exhaustive, they suggest strongly that Group 2 fluorides are not a good way to get fluoride ions into solution.  Of course, we would not expect Group 2 ionic compounds with their doubly charged ions to get very far in organic solvents generally.
Getting fluoride ions into solution with organic solvents is typically done with bulky 1+ cations, such as N,N,N-trimethyl-(2,2-dimethyl-prop-1-yl)ammonium ion, $\ce{C_{8}H_{20}N^+}$.  See Reference [1] as an example.  The use of cryptand-complex alkali-metal complexes fits this paradigm because such ions are also bulky.
Reference
1.
Victoria K. Davis, Stephen Munoz, Jeongmin Kim, Christopher M. Bates, Nebojslš a Momčilović, Keith J. Billings, Thomas F. Miller III, Robert H. Grubbs and Simon C. Jones, "Fluoride-ion solvation in non-aqueous electrolyte solutions", Mater. Chem. Front.,
2019, 3, 2721


text:
How does the anti-bonding eg orbital (eg*) affect the bonding of ligands?
In the molecular orbital for octahedral metal-ligand complexes, there is an eg orbital and an eg* orbital. When the d orbitals on the metal split, it will fill up some of the eg* orbitals I believe. However, my question is that won't these orbitals cancel out with the eg orbitals and make fewer bonds? For example, $\ce{Ni^{2+}}$is a $\ce{d^8}$ metal and will split in the presence of ligands and will partially fill the eg* orbitals. The $\ce{Ni^{2+}}$ metal can then form complexes like $\ce{[Ni(H_2O)_6]^{2+}}$ with 6 ligands attached. My question is how would the filling of the eg* orbitals affect the bonding and chemical properties of the compound? Here is the diagram I am referencing to:

It might be more clear if you indicate that the t and e levels represent 3 and 2 orbitals, respectively, even though there is only one line. Thus a d8 metal has the e* level only half-filled.

Short answer: slightly reduced bond order. As for effect on chemical properties, what is your reference point for comparison?

As you would expect from increasing occupancy of antibonding orbitals, the result is weaker metal-ligand bonding. This is one reason why the octahedral geometry is not as strongly favored for metals with greater numbers of d electrons. For a d8 metal such as the Ni(II) in the example, square planar and octahedral complexes are both common, and planar structures tend to be favored if the ligand is strong enough to result in a low-spin planar complex (as is often the case with Ni(II)). Only a weak field ligand like H2O will tend to form octahedral complexes with Ni(II).
A semi-quantitative approach to this issue is found in "structure energy" approximations and determination of "structure preference energy". Although that approach is not rigorously quantitative, it will show you trends.
It is important to note, though, that even with full occupancy of the eg antibonding orbitals (for example in a d10 metal like Zn(II)), octahedral complexes are possible. One can think of the sigma bonding in that case as being essentially entirely via the s and p metal orbitals, similar to a p-block central atom such as S in SF6.


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What is the heaviest possible water using oxygen and hydrogen isotopes, and what might it's physical properties be?
After a cursory internet search, it appears to me that Hydrogen-3 (a part of Tritiated Water) is the heaviest stable* hydrogen isotope. Oxygen-21 is the heaviest isotope of Oxygen with a half-life of seconds, and Oxygen-16, -17, and -18 are the heaviest stable isotopes (again, this is based on an internet search and may not be accurate).
So could (3H2)(21O) conceivably be a thing? If not, what is the heaviest water possible, and what would some of its theoretical physical properties be? Would the mass overcome surface tension? Would there be stronger or weaker hydrogen bonds? And so on.

No, water with $^{21}\rm O$ is not much of a thing, and the heaviest realistically possible water would look just like the ordinary heavy water (which in turn you won't tell from ordinary water without instruments), only a bit heavier and radioactive.

$\ce{^3H2^{21}O}$ would be tough to synthesize and study since the half-life of $\ce{^{21}O}$ is only 3.4 seconds.

@MaxW - better than a few picoseconds for $^{26}$O and a few $10^{-23}$ seconds for $^{7}$H

@IvanNeretin - one of my favorite short papers on isotopes, "Concerning the Taste of Heavy Water" by Urey and Failla, Science 81 (2098) p. 273 (Mar. 15, 1935). Blind taste test of distilled vs pure heavy water. No difference detected.

The heaviest water obtained with stable isotopes would be $\ce{D_2^{ 18}O}$.  This compound is not well-charactetized itself, but Ref. [1] indicates that the melting point is only about 0.28°C above that of ordinary water.  The implication is that overall physical properties of $\ce{D_2^{ 18}O}$ are close to those of heavy water with oxygen-16($\ce{D_2^{ 16}O}$); apart from density the isotopic effect of $\ce{^{18}O}$ versus $\ce{^{16}O}$ is much less than that of deuterium versus protium/"hydrogen".
References

F. Steckel and S. Shapiro, "Physical Properties of Heavy Oxygen Water.  Part 1.  Density and Thermal Expansion", Trans. Faraday Soc. 1963, 59, 331-343.

I can see how an increase in electron delocalization could increase van der Waals forces, but it seems that the lowering of electron energy levels you describe for the hydrogen would have the opposite effect on hydrogen bonding (weakening it), since this would tend to make hydrogen's electron more attracted to its nucleus.

Is it possible to obtain \ce{T_2^{ 18}O} ? Any idea what it would be like? Edit: I have no idea how to format the way you do, lol


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Why is negative charge more stable on sulphur than on oxygen?
Consider this problem:

Which is more acidic, $\ce{MeOH}$ or $\ce{MeSH}$?

According to my teacher:

To determine which is more acidic, figure out which forms the more stable conjugate base. In this case, $\ce{MeS-}$ is more stable than $\ce{MeO-}$. In general stability reduces in the order: $$\ce{RS- > RP- > RO- > RN- > RC-}$$

Why is the compound with $\ce{S-}$ or $\ce{P-}$ more stable than that with $\ce{O-}$? I know that oxygen is more electronegative than either sulphur or phosphorus. Shouldn't the descending order start with $\ce{RO-}$?

Consider the length of the RS-H bond vs the RO-H bond, this makes the RS-H bond weaker than the RO-H bond

@Waylander I'm guessing RS-H is longer, but I don't know why. I saw your edit: why is RS-H longer than RO-H?

Sulfur is a larger atom, the outermost electron shell is further from the nucelus

@Waylander Isn't the outermost electron shell even farther in phosphorus?

For phosphorus, it's a clear mistake, AFAICT - phosphine would probably be a bit more acidic then amine, but not even close to alcohol.

There is more to it than meets the electronegative eye.
Get into the habit of transferring attributes into energy.  In this case the relevant energies are the electron affinity of the base money ($\ce{RO}$ vs $\ce{RS}$) versus the bond energy of the moiety to hydrogen.  A strong anionic Bronsted-Lowry base will have a low value for the first versus a high value for the latter.
Surprise!  Sulfur has a higher electron affinity than oxygen!  Add the fact that the oxygen-hydrogen bond is stronger than the hydrogen-sulfur bond and by both measures, $\ce{RO^-}$ appears as the stronger base.
The higher electron affinity of sulfur and even selenium, compared with oxygen, has another curious effect.  We know about the aromaticity of certain ring ketones in which the electronegative oxygen atom imparts a positive charge to a three- or seven-membered carbon ring.  It turns out that despite what an elctronegativity table might suggest, the higher electron affinity of sulfur and selenium versus oxygen causes this characteristic to carry over to the corresponding $\ce{C=S}$ and $\ce{C=Se}$ compounds.

Your explanation makes sense for $\ce{RS- > RO- }$, but is $\ce{RP-}$ more stable than $\ce{RO-}$ (I realize I'm asking a slightly different question now)? The figures in the electron affinity table you linked and the fact that phosphorus is larger than sulphur, would suggest that $\ce{RP-}$ is the stronger base.

My bond-energy table does not include phospgorus-hydrigen, but if the latter bond is weaker than sulfur-hydrogen it would compete against your arguments.


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What does Molecular Orbital Theory allow you to predict?
Molecular Orbital theory reveals the outcome when two atomic orbitals overlap. Almost all of the time we draw a Y axis with the notation of energy that can be described using electron volt units (eV). However I don't recall in university lectures or in textbooks any actual values associated with either atomic orbital or molecular orbital states. For example when two Hydrogen atomic orbitals come together. Are there actual energy values associated with these and if so how would they relate to reactivity. In relation to this does Molecular Orbital theory allow to predict reactivity of common nucleophiles and electrophiles seen in Organic Chemistry.
Thanks

It is a long story. But yes, as far one can go with calculations. But one can predict a lot of properties, at least in principle. First of all energetic levels to be seen via spectroscopy. *note that even the principle predicts a degree of approssimazioni, as orbital are solved for mono electronic systems.

For organic chemistry just have a look at Woodward Hoffman rules and Theory of Frontier Orbitals.

For the specific question of how energy values are associated with orbitals, look up photoelectron spectroscopy and also the basic concept of the Schrodinger Equation, even if the details of the math are beyond what you have learned.

When you say the Schrodinger Equation are you implying the Wave function of a atomic orbital and how we can apply the potential energy operator to get a value? Would this also relate to the particle in a box idea?


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Why Heating of certain salts like Lead salts produce crackling sound?
I recently discovered that heating of certain salts produces crackling sound and that is called Decrepitation. But I couldn’t figure out the reason why that happens moreover I couldn’t find anything satisfactory on the internet.

Are they hydrated?

One cause is water.

I read about lead nitrate

Frankly speaking, you're reading too much into this phenomenon. Have you ever heard the satisfying crackling noise when you burn logs of wood in a bonfire? I'm sure you must've got the answer by now:
Decrepitation is more of a physical process rather than a chemical process.
When we write chemical equations, we treat compounds as single molecules, but in reality this is not so.
You specifically mentioned lead nitrate. You must be aware that it is a  white/colorless crystalline salt. To expound further,  solid lead(II) nitrate exists in a face centered cubic structure. Coming back to your question, when we heat Pb(NO3)2, we don't heat a single molecule, we heat a sample of the salt.
When we do so, the crystal structure breaks as the entropy increases. The deformation of the crystalline structure causes sound energy to be evolved.
TL;DR: Breaking down of large crystals into smaller crystals and/or intermediates causes the crackling sound known as decrepitation.


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Effect of anisotropy on electrical conductivity of crystalline solids
I was reading through a chapter named 'Solid state' in the NCERT Chemistry book.
I learnt about anisotropy of crystalline solids.
And what I understood was, you measure a certain property of that solid in one direction and it could be X while you measure it in the other direction it could be Y.
I then learnt about how a solid could show the property of birefringence.

My question : So does that mean, even electrical conductivity isn't the same in all directions? Does that mean a copper wire is going to conduct differently in different directions?
How is that possible? It feels impossible?
Does that mean different copper wires conduct differently?

I understand the last part is like a chunk of questions altogether and our community doesn't like that, but then it is basically having the same origin with just a little counter questions kinda? so I just let that be. I believe the contributors can understand what confusion I'm dealing with if I ask it like that.

Copper is a bad example: it has cubic structure, hence the tensor is spherical and conductivity the same in all directions. Better think of graphite.

And as noted, single crystal graphite has different conductivities in plane vs out of plane. Hcp metals not so much.

Note that deletion of questions with  answers, where the responders have spent their effort on,  is disrespect to users and may highly discourage from posting answers to the questions of the same author. Consider your actions.

@Poutnik Oh, I apologize for that. Didn't mean to disrespect. I thought my question was trivial. And you mentioned the same too? I once came across a chat from the moderators where they would delete the questions that seemed trivial/were AMiRite type? I shall keep your warning in mind. Don't worry :) I appreciate your effort though. No disrespect.

Ok, apology accepted. It is good to write such a comment about deletion decision, before deletion itself, so users can read it. ( It is not really deleted, just hidden from junior users). BTW,nobody is safe from occasionally asking trivial questions, not looking so at the asking moment.

There are indeed compounds with very different conductivity in different directions
Depending on the structure of the crystal, it is possible that a compound has very different properties in different directions. Some crystal structures cannot show this as they are essentially the same in all spatial directions, but others can. Metals don't usually have anisotropic conductivity due to the nature of the conduction bands but other conductors can show it.
The archetypal compound that shows this property is graphite. Graphite consists of multiple flat layers of covalently bonded C-6 hexagons in an extended plane stacked on top of each other by much weaker forces. The delocalised electrons in the plane can conduct electricity well but there is little scope for electrons to move freely perpendicular to the plane. So the conductivity is about 1,000 times higher along the plane than it is perpendicular to it (see Electrical conductivity of graphite).
So it is certainly possible that electrical conductivity can be different in different crystal directions. But this requires particular structural features. Other forms of anisotropy are far more common (like birefringence).

For organic materials (with a much greater likelihood to crystallize in lower symmetry space groups of the monoclinic and triclinic class, than inorganic materials [based on counting the entries in the CSD and ICSD database per space group]) I rather would support for them the perspective isotropy is a specialty when it comes to vector / tensor properties.  On the other hand, e.g., in organic  electronics (like solar cells, OLEDs), there are instances where crystallinity / a regular spatial arrangement of the molecules is a hindrance to the devices built and thus not wanted at all.


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How to find enthalpy of solvation for gas phase cation using Hess's law?

Find enthalpy of solvation $\Delta_\mathrm{solv}H$ using Hess's law, and show the expression that allows to calculate the enthalpy of solvation.
$$\ce{Ca^2+(g) -> Ca2+(aq)}$$

I am not told which enthalpies of reaction to use. I don't need the exact value, but I do need to show the sum of each enthalpy. I managed to find the enthalpy of the following reactions:
$$
\begin{align}
\ce{Ca(g) &-> Ca+(g) + e-} &\quad \Delta H &= \pu{590 kJ mol-1} \\
\ce{Ca+(g) &-> Ca^2+(g) + e-} &\quad \Delta H &= \pu{1145 kJ mol-1} \\
\ce{Ca(s) + 2 H+(aq) &-> Ca^2+(aq) + H2(g)} &\quad \Delta H &= \pu{1925.9 kJ mol-1} \\
\ce{Ca(s) &-> Ca(g)} &\quad \Delta H &= \pu{178 kJ mol-1}
\end{align}
$$
I nearly have the reactions I need, but I am still needing
$$\ce{H2 -> 2 H+ + 2 e^-}$$
I haven't been able to find the answer.
P.S. I am aware of the enthalpy of formation of $\ce{H2}$, but don't know how/if I can use it here.

By definition, the reaction $\ce{H2 -> 2 H+}$ has an enthalpy of reaction equal to zero  kJ/mol

One approach to compute the enthalpy change for the reaction
$$\ce{H2(g)->2H+(aq) + 2e-}$$
is to combine the bond dissociation enthalpy and hydrogen ionization energy (both refer to gas phase species) and the hydration enthalpy of the hydrogen ion:
$$\ce{H2(g)->2H(g)}~~~~~~~~{\Delta H = \pu{439 kJ/mol}  }\\
\ce{2H(g)->2H+(g) +2e-(g)}~~~~~~~~{\Delta H = \pu{262 kJ/mol}  }\\
\ce{2H+(g)->2H+(aq)}~~~~~~~~{\Delta H = \pu{-2260 kJ/mol}  }$$
If you add up the enthalpies you obtain as the solvation enthalpy of $\ce{Ca^{2+}}$ the value $\Delta_{\text{solv}} H = -1546 \pu{kJ/mol}$ which is quite reasonable (see eg a table here)
The result is clearly very different from my earlier answer which I am still pondering:
The catch is that $\Delta H^\circ=0$ for this reaction. The reason is because $\Delta G^\circ=0$ for the reaction (by definition, see standard reduction potential) at all temperatures, and then, since
$$\Delta H = -T^2\left( \frac{\partial (\Delta G/T)}{\partial T}\right)_p$$
it follows that  $\Delta H^\circ=0$. I suspect the problem with this approach is due to a missing term involving enthalpy of solvation of the electrons (or a similar term involving the electrons) but I haven't put my finger on it.

Hm, but is not the standard Delta G = 0 the trivial consequence of considering the same but the opposite half reactions ? The each half reaction has non zero but the opposite standard Delta G and Delta H. The above reaction would not have zero reaction enthalpy just because chemists decided SHE to  have the reference potential.

I would also like to ask, does that also apply to the change needed? I didn't add symbols, but the reaction I need for the hydrogen isn't in a single phase. To make it more accurate to what I need, it'd be this:
$$\ce{H2_{(g)}->2H^+_{(aq)} + 2e^-}$$

That's what I'm saying in my answer, you should just use $\Delta H = 0$ for that reaction. @Poutnik what else would you suggest?

@Buckthorn I say "Why should be delta  G and Delta H of that reaction accidentally zero ?“ It cannot be zero by convention. If we created convention 235U fission has zero enthalpy, it would not have any effect on real enthalpy of fission. That Delta G=0 is for reaction 2H+(aq) + 2e- + H2(g) -> H2(g) + 2 H+(aq) + 2e-. Similarly H2 formation enthalpy is zero, because it relates to  another trivial reaction H2 -> H2.

@Poutnik It may seem trivial in the same way that x-x=0-0, but it is part of the definition. In any case, I agree that it seems odd and I may yet alter my answer.

@BuckThorn By other words, evaluation of Delta G of redox reaction assumes having one of half-reactions from SHE. Evaluation H2(g)->2 H+(aq) + 2e-   is evaluating 2 SHE electrodes together, for which Delta G =  Delta G_forward + Delta G_backward = obviously 0, whatever value the single terms have.


text:
Electrolysis of copper sulphate using copper electrodes: Mass lost from anode is greater than mass deposited on cathode
In short, the aim of my experiment is to determine how the concentration of sulphuric acid in solution affects the mass liberated and deposited at copper electrodes, when copper sulphate is electrolysed.
Over all data points, what I observed is that the mass deposited at the cathode was lower than the mass liberated at the anode. Theoretically, the two should be equal, and even experimentally I expected the opposite to be true, due to oxygen production taking place at the anode which could disrupt the liberation of copper. The voltage was kept constant at 3V, so I monitored the current over the duration of the experiment. I observed that the current increased, which further suggested that more ions were released at the anode than deposited on the cathode, as  that represents an increase of positively charged particles in solution and thus an increase of current density.
Now, perhaps the liberation of oxygen facilitates the oxidation of the copper anode? Is this my misunderstanding?
Additionally, I was unable to find conditions which would stop the production of hydrogen at the cathode. The voltage required to electrolyse water is 1.23V, but if I set my voltage to that low, no plating took place either. Is hydrogen also expected to be produced alongside the plating process? Maybe, the hydrogen production was disrupting the plating? In that case, what is the explanation for why hydrogen was produced at the cathode in the first place? Since copper is lower in the reactivity series than hydrogen,  no hydrogen should be produced.
Can someone point out the flaws in my experiment or where I've misunderstood things. I don't have much experience with electrochemistry. Thank you!

You better forget about the voltage, and only try to check the intensity, which depends on the surface of the electrode, and which should be around $\pu{0.02 A/cm2}$. With such a value, you should not get any Hydrogen on the cathode. The amount of sulfuric acid should not have an effect on the yield. Also try to use a high concentration of CuSO4 in the solution, around $\pu{1 M}$.

I was indeed using a CuSO4 concentration of 1M. What is the theoretical reason for why a lower current density would stop the hydrogen production? Also, I'm not sure what you mean by yield. Increasing the concentration of sulphuric acid increases the current density in solution, leading to a greater current and a greater yield of copper deposited. Maybe it doesn't affect the production of hydrogen. In any case, would oxygen still be produced at the anode?

Also, consider that a) some copper may stay in solution, if you add H2SO4, and b)  look for sludge at the the bottom.

@Poutnik I actually believe that hydrogen production is necessary in order for the sulphuric acid concentration to affect plating. With the reduction of hydrogen ions taking place at the cathode, a constant stream of H ions in the solution is insured, thus increasing overall current. What I observe is that if the voltage is set to too low, the current merely spikes and goes down to near 0. I assume that represents positive ions from the solution being rapidly pulled to the cathode, before the high concentration of hydrogen ions creates a layer around the cathode stopping the plating.

Moved my comments to answer.

$\pu{3 V}$ is too high. Kinetics of deposition of $\ce{Cu}$ is saturated, so another electroactive system kicks on, creation of hydrogen. Similarly for anode. Try to evaluate volt-ampere characteristic of the system and stay on the current plateau before gases start to be created.
Generally, is is more advantageous to use a power source keeping constant current rather than constant voltage.
If gases are being evolved, you cannot expect the dissolved and deposited mass of copper would match, as various portions of the current are spent on gases, instead of copper. The current density that Maurice suggests is a good one.
For low enough voltage, copper deposition + dissolution can absorb the current. At particular current density, one of these reactions is saturated and cannot take more current any more, being diffusion controlled. Then either hydrogen, either oxygen starts to be created, depending on the electrode where it occurs first. When it happens for the other reaction as well, the other gas gets created as well.

Each electrode has a voltampere characteristic given by I=f(E) function, that has a shape of irregular rounded step-wise slope. When a cathodic potential gets low enough value for a given compound/ion, current starts to raise. When reaction gets saturated, i becomes being controlled by ion diffusion and there is zero concentration at the electrode. Then the current reaches a plateou. The plateou ends at a potential where another redox system starts to interact. Similarly for anode.
When an external voltage is applied, electrodes gain such potentials to meet 2 conditions:

The potential difference equals to the external voltage.
Currents of both electrodes, according to the above function, are equal but the sign.

Lower current density means more positive cathode potential, where $\ce{Cu^2+}$ gets selectively reduced. Higher density means lower cathode potential where both $\ce{Cu^2+}$ and $\ce{H3O+}$ may get reduced. It is like if you used mildly reducing reagent, reducing just copper ions versus strongly reducing reagent, reducing also hydronium ions. Similarly for the anode.
The only necessery thing is the cathodic current matches the anodic current. Processes on both electrodes are independent. There is no "quantum entanglement" between these processes. If $\ce{Cu^2+}$ diffusion and $\ce{Cu}$ deposition can absorb all the cathodic current, there is no need of hydrogen creation.
Try to measure the steady current as a function of the applied voltage. You can use a small resistor is a serie with a cell, measuring voltage on it, which is equivalent to the passing current as $I = U_\mathrm{R} / R.$ Find the first current plateau. Or the "like to be a plateau, just if the 2 voltage steps were not so close together."
Choose the voltage from this plateau, but be aware it may change with the change of the cell geometry or electrolyte composition.

Even in cased hydronium ions do not participate on electrode reactions directly, they can participate indirectly by influencing I=f(U) characteristics of the cell via:

Keeping or affecting chemical state of involved ions. ( E.g. pushing equilibrium $\ce{[Cu(H2O)4]^2+ + H2O <=> [Cu(H2O)3(OH)]+ + H3O+}$ to the left.
Balancing the local gradient due charge disbalance
Increasing conductivity and therefore the solution potential difference, that subtracts from the applied voltage.

I really appreciate the answer. Clears things up a lot. But as I stated previously, if the hydrogen ions from the sulphuric acid do not participate in the flow of charge, then they don't have any effect on the current, right? If they were to participate in the flow of charge, then current will increase because I = Q/t. If the concentration of hydrogen ions is high, but the voltage is low, the ionization of the solute is inhibited, so the sulphuric acid concentration will only disrupt the reaction rate.

What do you mean by "the ionization of the solute is inhibited" ?

In concentrated solutions of strong electrolytes (such as sulphuric acid) there are strong forces of attraction between the ions of opposite charges — there are inter-ionic forces. Due to these inter-ionic forces the conducting ability of the ions is less in concentrated solutions. I guess I should have said that this occurs when the concentration of both hydrogen and SO4 ions is too high. This is highlighted in this thread: https://chemistry.stackexchange.com/questions/26781/conductivity-as-a-function-of-acid-concentration

Yes that is true. Your prior formulation is confusing. But it would apply rather to unreasonably high  acid concentration, and would be applicable to any soluble enough ionic compound.


text:
How to know if a reaction forms H3O+ ions or OH- ions?
How would you know when an acid or base is paired with $\ce{H2O}$ that it will form an $\ce{OH-}$ ion or a $\ce{H3O+}$ ion? I just started the acid and base equilibrium unit, and I'm just confused on what conditions $\ce{OH-}$ or $\ce{H3O+}$ ions form.

Why, that's simple: acids give H3O+, bases give OH-.

Acids are hydrogen ion donors.  When the react with water, they can give a hydrogen ion to form $ \ce{H3O+}$.
For example:
$ \ce{HCl(aq) + H2O(l) \rightarrow  Cl-(aq) + H3O+(aq)}$
Simple acids, such as $ \ce{HCl}$ or $\ce{H2SO4}$, can be recognized as acids by the H at the start of the formula.  Other more complex acids may be written with $ \ce{COOH}$ at the end of the formula, which denotes a carboxylic acid.  An example of this is acetic acid: $\ce{CH3COOH}$.  Otherwise, the name of the compound should give it away.
Bases are hydrogen ion acceptors, and generate $\ce{OH-}$ when they react with water.  Strong bases are generally soluble metal hydroxides, such as $\ce{NaOH}$.  You should be able to recognize them from their formulas.
Simple weak bases often contain nitrogen, like ammonia: $\ce{NH3}$.
$\ce{NH3(aq) + H2O(l) \rightleftharpoons NH4+(aq) + OH-(aq) }$

H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative covalent bond with the $\ce{H+}$ ion.
So like Ivan said, whenever you see an acid in an aqueous solution you can consider $\ce{H+ \equiv H3O+}$. The key there though, is that it must be in aqueous solution so that it can dissociate (and the $\ce{H3O+}$ form in the first place). That is the condition that is needed to form it. This is the same for $\ce{OH-}$, which requires an aqueous solution to form and thus release the ions in say NaOH, forming $\ce{Na+}$ and $\ce{OH-}$.


text:
Can there be an aqueous solution with a convex meniscus?
This question is motivated by a quote from a TV show:
A detective is looking at a small vial (really a glass tube) that supposedly contains poison. Then the detective says:

You don't have to take AP Chem to know that, in a glass container, the meniscus of a liquid containing poison, like baneberry, curves downward, but this meniscus curves upward because it's just water, right?

The detective proceeds to drink the content of the vial to convince everyone that it contains water and not poison.
The "vial" in question really was a tube.
Surely the "poison" itself was an aqueous solution. So the detective is implying that there exist some aqueous solutions with a convex meniscus. I've searched the internet looking for examples of convex menisci, and the only example I could find was mercury. Mercury in thermometers is such a well-known example of convex meniscus that every written material I could find used it as the example of convex meniscus.
Hence the question:
Does there exist an aqueous solution with a convex meniscus?

If walls are rather hydrophobic, even water can do that.  If cheater is in a narrow plastic tube, pumped up and down, meniscus shape can be seen pumping between concave and convex.

Also, tiny amounts of detergents or oils, particularly silicones, in or floating invisibly on top of the water drastically affect the meniscus. The detective likely would not have survived long in Derek Lowe's classes, e.g., https://blogs.sciencemag.org/pipeline/archives/2013/05/08/things_i_wont_work_with_dimethylcadmium

I agree with both Poutnik and DrMoishe Pippik, and the remarks. :-) Changing the surface forces (e.g., capping $\ce{Si-OH}$ bonds) make the meniscus change regardles of it's aqueous or not. I have capped them in test tubes by exposing to TMS-chloride and get a flat meniscus. By the way, the detective have drank something other than water.

This has been asked already.

Thank you for the comments! I am avidly reading that link to Derek Lowe's article. @Alchimista: could you please be so kind as to point to the question which has already been asked? I searched quickly before posting and didn't find it.

@Stef look here in Chemistry or in Physics SE... It was likely the same show...


text:
Is 2-bromo-3-methylbutane chiral?
Is 2-bromo-3-methylbutane chiral?

I can't see what the molecule is from your diagram, but in your first comment you state "It is a ethane (2 carbons) with 3 methyls on the 1st carbon, and a bromine and a methyl on the 2nd carbon"
If you draw out the structure you describe above I see four consecutive carbons. Since one of the methyls attached to the "1st carbon" makes the main chain longer and the methyl adjacent to the bromine on the "2nd carbon" also makes the chain longer. Thus, you have a butane as the main chain.
You have two constituents remaining the bromine on carbon 2 and the di-methyls on carbon 3. The name is thus 2-bromo-3,3-dimethylbutane
Also the molecule is chiral since the bromine makes the molecule nonsuperimposable.