text:
Synthesize pivaloyloxymethyl ester, but with a substrate that is a terrible nucleophile
I want to synthesize an pivaloyloxymethyl ester from a fluorophosphonic acid. The idea is, to make a compound cell permeable that than be cleaved in the cell by enzymes. Apparently, the acid is a very bad nucleophile, so therefore the reaction with chloromethyl pivalate does not seem to work. I have the reaction running know for more than 26 h with 3 eq. of the chloride, but according to LCMS no product is being formed.

To deal with this problem, my idea was to hydrolyse the chloromethyl pivalate with pure water and let the resulting hydroxymethyl 2,2-dimethylpropanoate react with acid chloride of my starting material. Do you think this is a valid approach? Unintuitively I did not find any similar reactions like this in the literature. What are your thoughts on this?

My thoughts on this so far: I attempted the hydrolysis with water and added ACN for better solubility. However, the product is either volatile or the reaction was not yet complete. TLC was not showing strong coloring with CAM/KMnO4.


text:
How dangerous is CO gas poisoning?
I have learned that there are carbon monoxide canisters, and I can't help but feel discomfort at the thought of how morbidly dangerous those will be in a hobbyist home setting. Now let's imagine that someone would mistake it for nitrous oxide or helium and inhale it out of the canister. I imagine that this will be a death sentence, and there is nothing short of a massive blood transfusion that can be done to save that person's life. Is this accurate?
P.S. Please note that I am not planing a murder or a suicide. I just find some of those morbid topics interesting, from time to time, and I think that it is good for us as people, to give some thought, from time to time, to things that can go horribly wrong

Why would hobbyist need carbon monoxide ? Are you sure it is not carbon dioxide?

I red that there have been instances of planned group suicides where people intentionally produce CO gas. So I thought - wait a minute, there is probably a more sinister way  of going about that. So after finding out that it is a useful reagent, I thought - nooo, there might be fatalities related to poor practices. And what is more dangerous then an improper home setting

I am fascinated at what biests chemical engineers contain behind pressure valves

How dangerous is it? Either you die or you fully recover, afaik. How would you make a transfusion in a time one needs to suffocate? (rhetoric question)

@Mithoron strictly speaking CO does not suffocate. It damages the blood's ability to transport oxygen and is far slower than suffocation, but more insidious as smaller amounts do the job well. and it isn't always something from which a full recovery is possible.

source: Stack Exchange

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sort: votes


text:
Can triethylamine be substituted for pyridine as a base catalyst in acryloylation reaction?
I work in a polymer research lab and I'm writing a procedure to synthesize glycerol 1,3 diacrylate but in almost all the reactions I've been reading, the synthesis occurs at hot temperatures(60 C) and using pyridine.
We already synthesize poly(glycerol-co-sebacate) acrylate and in that procedure, we found that it works best using triethylamine and acryloyl chloride at cold temperatures(0 C) due to higher reactivity, so I'm wondering if pyridine can be substituted with triethylamine at colder temperatures for the glycerol 1,3 diacrylate synthesis.
In my head, I think that it should work since the role for both in their respective reactions is as a base catalyst with the main difference being that triethylamine is much more reactive than pyridine but if anyone else can give their two-sense on the matter that would be very much appreciated.
Thank you!!

Without additional info nobody here can say for sure. The proper way to go is to perform both experiments and compare yield, reaction time and product purity. In general one can say, that pyridine is less nucleophilic and less volatile. The first property is often desired, the second not such much. On the other hand, pyridine can also be used as a polar solvent or as a catalyst for acylation reactions. What this means for your system has be evaluated by yours on the basis of multiple experiments and evaluation of the same.

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text:
Why should higher energy empty orbitals be chosen over lower energy empty orbitals during hybridisation?
Organic chemistry by jonathan clayden states during his explanation of hybridisation using trigonal boron structure (sp2) and tetrahedral boron structure (sp3):

You want to populate the lowest energy orbitals for greatest stability and sp2 orbitals with their greater s character are lower in energy than sp3 orbitals. Another way to put this is that, if you have to have an empty orbital, it is better to have one with the highest possible energy since it has no electrons in it and so it doesn’t affect the stability of the molecule.

I get the first part of the statement; however, the bolded part confuses me.
If a higher energy orbital is empty compared to a lower energy empty orbital, how does it affect the stability of the molecule differently?
If they are all empty shouldn't they all not affect the stability of the molecule?

It's just to add the first thing a different perspective. As you can see in the picture if the orbital was sp3 then others would be too - effect is indirect.

The stronger a bond the lower the total energy. A sp2 orbital has more s character than a sp3 orbital so the molecule with 3 sp2 orbitals is of lower energy because the electrons in the bond are closer to the nuclei. The empty orbitals have no effect until occupied by excitation or addition of electrons.


text:
what's the meaning of the "strikethrough" bond in the skeletal formula of this aromatic compound?

That's taken from https://doi.org/10.1039/A901537J , which is about direct amination of the depicted nitrobenzene; I'm puzzled about the meaning of that --NHR moiety, specifically why it doesn't connect with a carbon atom, but instead the other end of the "bond" ends inside the benzene ring.
Any hints about what's meant with this notation?


text:
DIY method for regenerating activated carbon in air filters
I am interested in regenerating activated carbon used in general house-hold air filters. I am looking for a sufficiently efficient and easy method to do this in home setting. A little tinkering in a workshop is possible, but sophisticated laboratory or industrial protocols are out of reach.
I think the most probable mechanism would be the desorption of attached non-polar molecules by heat, already discussed here. What are the required temperatures to desorb typical gases in houses that attach into the activated carbon, and what heating times are required? If necessary, to avoid oxygen attacking (burning) the activated carbon in high temperatures, I think (super)heated water vapor would be the easiest inert gas to obtain. But is it needed and suitable? And whatever method is used, the activated carbon should not liberate polluting carbon particles itself more than it typically does, and even less carcinogens. I know that activated carbon is a little bit dusty, but the method should not cause excess amount of particles.


text:
Does DIBAL-H reduce acid chlorides?
It's only really mentioned with esters, but acid chlorides are more reactive than esters, so following that logic is this a feasible reaction to convert acid chlorides into aldehydes?

Yes, according to this reduction chart from the Myers group at Harvard it gives the alcohol  https://hwpi.harvard.edu/files/myers/files/5-reduction.pdf    Other sources suggest the aldehyde is the product but this appears more authorative


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Why does epsilon caprolactone not have triplets on 1H NMR?
Looking at the structure of epsilon caprolactone (CL), you'd assume that the methylene protons (CH2) on either side of the ester would give you triplets. However, experimentally those peaks actually look more like inverted triplets (happens in both 300 and 400 MHz NMR machines and in both CDCl3 and C6D6). Anyone know what causes this splitting pattern?

That seems to be the result of second-order effects due to small shift differences relative to the J-couplings.

It has to do something with the seven-membered ring because the linear polymer of caprolactone has the spectrum you would expect.

The following is a simulated 1D $^1H$ NMR spectrum of gamma-caprolactone available from the Human Metabolome Database (entry HMDB0060476) at 100 MHz in D2O.

The following is a blow-up of the downfield multiplet due to the epsilon proton (labelled "a" in the OP spectrum)

The splitting pattern might look elaborate but approximates a standard low-order ddd multiplet (not a triplet) as taught in introductory chemistry.
It might be unexpected however to see a ddd pattern. The origin is the nonequivalence of the geminal protons and of couplings to vicinal protons. The geminal epsilon protons have slightly different chemical shifts and are coupled through a relatively large (for a homonuclear coupling) ~10.6 Hz 2J coupling.
The vicinal delta protons resonate~2 ppm away (200 Hz in the simulated spectrum) and are coupled through 1.3 Hz and 5.4 Hz couplings (these differing between trans and cis protons).
Complex splitting patterns diverging from first-order are common in homonuclear NMR spectra with narrow spectral widths and close proximity of chemical shifts, particularly at low magnetic fields. This might also be the case here and might explain the unexpected multiplet peak intensities.
As explained in this site:

Second order effects occur when the ratio of the chemical shift separation in Hz to the coupling constant is less than approximately 10.

Note that the field magnitude (in Hz) relative to the J-coupling is important. Simpler multiplet patterns are observed when shift differences are large relative to couplings, for instance when J-couplings are heteronuclear or when the applied magnetic field is incremented (also, as suggested in comments, when the symmetry of geminal protons is not broken by a ring structure). Such multiplets are often slanted in appearance (the "roof" effect). This is often seen in spectra otherwise lacking complex splitting patterns.


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When to use Heat of Products - Heat of reactants and Heat of reactants - Heat of products when calculating enthalpy change
In my schoolbook there are multiple questions where I have to calculate the enthalpy of formation, heat of reaction, enthalpy of reaction, etc etc. But sometimes the formula to be used is Hp - Hr, and sometimes it is Hr-Hp. So, when do I use what?
(Hr = Heat of reactants
Hp = Heat of products)

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

Welcome to ChemSE! Please find guidance for MathJaX / $\LaTeX$ as follows and reformat your post - Basics // MathJaX and mhchem formatting for equations and chemical formulae // Guidelines on which should be upright / italics

Also you can refer this guide on how to use MathJaX.

It's not clear without some examples.


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Determining the reduction potentials of elements?
How can we predict the order of reduction potentials for given elements? Consider the reactivity of metals, there is a standard explanation to determine reactivity. Eg. Sodium is more reactive than lithium due its larger ionic radius. Hence we have intuition for the order of reactivity. However, for reduction potentials, I can't seem to find an explanation. Unintuitively, the reduction potential of sodium is actually lower than that of lithium, meaning that reduction potential cannot be determined by reactivity, so there must be something else going on.
The best explanation so far, states that reduction potential is determined by ionisation energy and hydration energy, a mixture of the two? Could this be correct?
Comprehensive explanation of how we can build intuition for reduction potential would be great. Could you provide links or materials to learn about this?
Edit: Answers always explain that the reduction potential is measured relative to the hydrogen potential which makes sense. However, what are the underlying principles? Eg. For reactivity with metals, we could obviously just react them together and determine a reactivity series. However, we have an explanation for this reactivity series, which is what I want for reduction potential. Could reduction potential be determined by ionisation energy and hydration energy?

Do thorough searching, similar questions were asked multiple times on this site. Be aware of kinetic reactivity and thermodynamic reactivity are 2 different things. In Li and Na case, their order is mutually switched. Na reacts faster, but Li releases more energy per molar amount (free molar enthalpy).

"Sodium is more reactive than lithium due its larger ionic radius." That sounds like a proximate rather than ultimate explanation. What definition do you use for reactivity? Oxidation potential is a factor in the definition provided at the Wikipedia: "lose electrons (oxidize) more readily to form positive ions;"

related: https://chemistry.stackexchange.com/questions/87514/why-are-not-all-metals-included-in-the-reactivity-series/87529#87529

I think you are asking about the measurement of Standard Reduction potential. It is done by building a electrochemical cell, where the element under observation is taken as an electrode and the other half cell is the SHE (Standard Hydrogen Electrode). SHE acts as a reference point(its reduction potential is assigned as 0 V) and based on the unknown electrode it will get oxidised or reduced. In this manner we can find Std Reduction potential values.


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Factors deciding whether acidic / basic character dominates in an amphoteric compounds
In my NCERT textbook, in the chapter 'The d- and f-block elements', the last paragraph under oxides and oxoanions of metals goes like:

$...$ $\ce{V2O5}$ is, however, amphoteric though mainly acidic and it gives $\ce{VO4^3-} \space\text{&}\space\ce{VO2+}$ salts $...$

What do they actually mean by the phrase amphoteric though mainly acidic?
Wikipedia clearly states that,

Vanadium(V) oxide (vanadia) is the inorganic compound with the formula $\ce{V2O5}$. Commonly known as vanadium pentoxide, it is a brown/yellow solid, although when freshly precipitated from aqueous solution, its colour is deep orange. Because of its high oxidation state, it is both an amphoteric oxide and an oxidizing agent. From the industrial perspective, it is the most important compound of vanadium, being the principal precursor to alloys of vanadium and is a widely used industrial catalyst.

What could the author mean?

EDIT:
I got information from Poutnik that the phrase in concern means that $\ce{V2O5}$ has more of an acidic character than basic character and thus it is known as 'amphoteric though mainly acidic'.
Now my question is, what are the factors which determine whether the basic or the acidic character dominates in an amphoteric compound?

The Q can be too broad, as you mere ask what determines the strength of an acid or a base.  The rest, related to amphoterism, is just comparative arithmetics.

So now I'll look for answers for factors determining strength of acid / base? Thank you @Poutnik!

General answer is not much smaller than chemistry. But narrowing it to character of metal oxides, the rule of thumb is that with increasing of metal oxidation state the oxide basicity gets weaker and acidity stronger. Compare with Na2O, MnO, Al2O3, TiO2, CrO3, WO3, Mn2O7. The first is basic, the last acidic.

Related: https://chemistry.stackexchange.com/questions/49698/classification-of-oxides-transition-elements/49700 ; https://chemistry.stackexchange.com/questions/70415/classification-of-oxides ; https://chemistry.stackexchange.com/questions/22189/how-to-decide-acidic-or-basic-nature-of-oxides


text:
Analysis of substances
The crystalline substance X completely decomposed without the formation of a dry
residue when heated. As a result of adding a solution of calcium hydroxide to the
solution of substance X, a precipitate insoluble in water fell out. Substance X is: A

ammonium carbonate, B - sodium sulfite, B - ammonium chloride, G - sodium
hydrogen carbonate.
Explain your answer.


text:
What is the energy of sum of ionic equations?
Say energy depends on the relative number of ions in the mixture, and factors P, and T.
$$
u = u (P,T,x_1,x_2,...,x_{n-1})
$$
Mixing two solutions at the same temperature, may change the temperature if there is some interaction between the two ionic solutions. Otherwise,
$$
u_1+u_2 = u (P,T,x_{11},...,x_{2m-1})
$$
A change in temperature is a result of some interaction between ions:
$$
\Delta u = \Delta U (k_{x_{1n}x_{2m}})
$$
We know that change in energy is zero, otherwise we can ignore the change in temperature over long periods:
$$
\Delta u = 0
$$
For two ionizations:
$$
u_{I_1} + u_{I_2} = u(P,T,x_{11})
$$
for $I_1 = I_2$:
$$
2u_I = u_I
$$
since our energy $u$ is an intensive property. This also means that while the total energy is not changing, energy per unit volume may change:
$$
\Delta u_V \neq 0
$$
This simply means volumes of ionic solution matters, for one, it matters in size.
$$
U = U (P,T,V,x_1,x_2,...,x_{n-1})
$$
Is there such a function for $U_1+U_2$?

Hi. It was hard for me to follow, because some symbols seem strange to me. Nevertheless, are you interested in a general function $U$ which describes the internal energy when two ionic liquid solutions are mixed? Is that the last result?

@MetalStorm yeah, figuring out the required variables, I think I overcounted.

I know that it is not of much help, but there does not exist a universal function for the internal energy, because it will depend ultimately on a model. It is possible to arrive at a general differential relation, I can guarantee that. But, some model for the excess Gibbs energy $G^\mathrm{E}$ or other thermodynamic function will be needed.


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Why is de Broglie applied in case of photons
Q: Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s¯¹) of He atom after the photon absorption is (Assume: Momentum is conserved when photon is absorbed)
Use: Planck constant = 6.6 * 10 ^ - 34 , Avogadro number = 6x1023 mol-1, Molar mass of He = 4 g mol¯¹)
My approach:
Calculate Energy of the photon by
 E = hc/wavelength
And then equating it with the change in kinetic energy, however the difference is answer is by a factor of 2^1/2
In the solution, they have done it simply by substituting in de Broglie equation. I mean how can we just substitute the mass of helium and the wavelength of photon in the same eqn? And how is my approach wrong?

Focus on momentum, not energy. For m>0 and v << c, p =m.v. For m=0, p=E/c = hc/(wavelength . c) = h/wavelength

I understand, but why is the energy approach wrong? Am I missing something about energy conservation, is energy also converted into some other form?

Why do you think the whole photon energy is transformed to the atom kinetic energy?

I mean if not K.E. then it will be transformed into which form?

It could be an electron excitation.

You might also be interested to look up  details of the Mossbauer Effect.

Einstein's special theory of relativity is the basis of the expression of de Broglie.
the actual expression was
$ E^2=(pc)^2 + (mc^2)^2 $
In case of photons the rest mass was zero. however this is not the case with helium atom. So, you cannot equate change in kinetic energy of a non zero rest mass object with the energy expression of photon mentioned above.

I don't understand how special relativity is the basis of de Broglie's expression. I think de Broglie was inspired by the photoelectric effect, so he extended the expression $h\nu$ which Einstein applied only to photons to all bodies composed of matter. Am I incorrect?

yes. He argued nature is symmetric and matter also shows wave nature.


text:
Does this synthesis of dimethylmercury require anything else?
I've seen that dimethylmercury can be made using sodium amalgam and methyl iodide. Are those the only chemicals needed or does the reaction require a solvent or application of heat or anything else like that? Or will the reaction take place just by mixing the two? Not trying to make it but concerned about possible accidental contamination.

I don't think we're gonna give you preparative receipt for a serious poison. BTW you cannot mix  sodium amalgam and methyl iodide - reaction takes place on the interface between them. Contact between them is enough for reaction to take place, as mentioned in earlier post.

The link you provided seems to answer my question. I was hoping there was more to it but thanks for the reply.

If you don't want to make it, this should be enough. What temperature, solvent etc. to use for actual synthesis of pure compound with good yield aren't really relevant otherwise.

Why do people keep being worried about such unlikely reaction, seems to be beyond me...


text:
Resonance stabilization : lowering of kinetic energy or electron repulsion?
Resonance stabilization, or delocalization stabilization, occurs when the overall energy of the molecule is lowered compared the localized "resonance structures".
In solid state physics, or even when explaining covalent bond formation, it is stated that delocalization occurs due to a lowering of the kinetic energy, just like the ground state evergy of a particle-in-a-box lowers when the size of the box is increased. This explanation is sometimes brought up to explain delocalization stabilization.
This explanation hold even in a single-particle picture, without considering electron-electron interactions. Of course, a molecule is an ensemble of possibly strongly interacting electrons. Several sources claim that delocalization is due to the lowering of the charge repulsion, lowering the potential energy instead.
Of course, both mechanisms are probably at play. My question is the following : In general, does any of these mechanisms dominate ? If not, what are the factors favoring one or the other ?
I could easily understand that charge repulsion plays an important role for charged species like the nitrate ion, while for conjugated molecules or aromaticity, I wouldn't expect electron-electron interactions to dominate. Is this right ?

Note that lowering an electron energy means increasing its kinetic energy, as  in mean values , Etot = Epot + Ekin = -2.Ekin + Ekin = -Ekin

I guess it is model dependent. You can set one effect as a background and see how the other changes or vice versa. Cannot verify anything experimentally since "resonance structures" aren't real.


text:
Can NaBH4 reduce acid chlorides to alcohols?
My professor says it reduces only aldehydes and ketones, but acid chlorides are generally more reactive than both of them.

@Zhe That question refers to Dibal-H reduction of acid chlorides

This paper reports the reduction of acyl halides to  alcohols with NaBH4/K2CO3 in dichloromethane/water at low temperature under phase transfer catalysis.
This paper reports the reduction of acyl halides by sodium borohydride in dimethyl formamide to aldehydes.
The choice of solvent conditions is critical as acyl halides will react with most of the solvents that are commonly used with NaBH4.


text:
Why does water sometimes split into H+ OH- (H3O* OH-)?
I am aware that this isnt the water splitting into H+ and OH-, and that is is just a transfer of Hydrogen via autoionisation ...
(found this on here: Why does water split into ions? //
"Water does not really split into ions. It transfers protons, reversibly, from one molecule to another:
2H2O = H3O+ + OH−
The ions represent, of course, a less stable state than molecular water, at least under ordinary conditions, so the equilibrium constant is low. Yet this reaction strongly affects the chemical prpoeryies and even some physical properties of aqueous solutions." )
...but I want to know why it is that this happens, as Kw is very small 10^-14, this doesnt happen much, but I wish to find an explaination for why this actually happens, what causes the hydrogen to be transfered to another water molecule?
Thank you to anyone who gets back to me!

This should be covered in physical chemistry under statistical thermodynamics. Have you learnt those?

I'm not sure what explanation would you what. Water is both slightly acidic and basic, and that's the result.

You mere ask why acids sometimes provide a proton and why bases sometimes accept a proton.


text:
Can evaporation of EtOH be slowed by adding dissolved solids to it? Does doubling the dissolved solid slow the evaporation in a predictable way?
I use EtOH to dissolve shellac flakes for finishing procedure.  Does adding dissolved shellac slow the evaporation assuming all else is equal?  What if you double the amount of dissolved shellac.  Would that slow evaporation even more?

It does slow evaporation, but there can be problem with solubility or application issues with high concentration and viscosity.

BTW, if the goal is simply to dissolve shellac flakes, why not do that in a closed container? There would be no evaporation, and if the container is strong, it could be gently warmed to speed dissolution.


text:
In Isobaric process, is external pressure equal to gas pressure?
In which processes are external pressure and gas pressure the same or constant?

Note that questions looking like homework/test/exam/lab tasks, posted without significant attempts to understand/resolve them, are explicitly off-topic on this site and will be closed.
// Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

What exactly do you mean by the external pressure?


text:
Fundamental understanding of electrode potentials
The electrical double layer, is commonly used to describe the origins of electric potentials that is fundamental to electrochemistry. This model is quite outdated from my understanding, but it is still used inside my textbook.
Right now, my understanding is that when a metal rod is placed within a solution of its respective ions, some of those metals atoms change to metal ions and go into the solution leaving a negative charge at the electrode surface. Hence, the metal cations inside the solution move towards the electrode surface. At this electrolyte - electrode interface there is now a seperation of charge, that gives rise to the phenomenon of electric potential. At least this how my cambridge textbook explains it.
Some answers on stackechange reinforce the theory as stated above. However,  my problem lies in understanding why a negative layer of surface charge builds up in the first place. How can metal atoms just convert to metal cations and move into the solution? This doesn't make sense to me. The surface of the metal rod comes into contact with the electrolyte solution, so what is causing the surface to gain this initial negative charge?
The rest seems to make sense, but this part specifically doesn't.

Consider also inert electrodes, where only exchange of electrons happen, like $\ce{Pt|V^2+/V^3+}$ // Double layers are rather the consequence than cause.

Very similar to my question a while ago: https://chemistry.stackexchange.com/questions/181451/do-electrons-really-hang-around-the-electrode-while-ions-go-take-a-swim

"processes which can lead to a surface being charged, including adsorption of ions, protonation or deprotonation"

Consider an electrode immersed into an electrolyte, two things can occur.

Metal ion $M^{n+}$ from the solution may collide with the electrode, gaining $n$ electrons from it, and convert to metal atoms. This means that the ions are reduced.
Metal atom on the surface may lose $n$ electrons to the electrode and enter the solution as the ion $M^{n+}$. This means that the metal atoms are oxidized

Notice that for different elements, the redox equilibrium established is different, either further to the right or further to the left. Hence, depending on the electrode, one of the processes above is going to be more favourable then the other, creating either a positive or negative surface charge.

Negative charge is kept near surface by attraction of excess positive ions in solution. The reason for ions going into solution and electrons staying in metal is electron's De broglie wavelength strongly depending on environment. Creating a bubble larger than intermolecular distance in a liquid made of polar molecules needs energy. That does not happen in the situation stated in the question.
For a nice example of a bubble filled with electron, see https://en.wikipedia.org/wiki/Electron_bubble
There are counterexamples, where both cations and electrons go into solution, such as alkali metals in liquid ammonia. They can do this because in these cases surface electron affinities of the alkali metals are lower.


text:
determine activation energy using the time for maximum temperature to be reached for catalytic decomposition of hydrogen peroxide
when determining the activation energy for the catalytic decomposition of hydrogen peroxide using manganese dioxide,
a common way is timing the collection of a given volume of gas or pressure when carried out at different temperatures.
Is it possible to instead use data measuring the time to reach the maximum temperature for a reaction instead.

Where have you found that the activation energy can be obtained by timing the collection of a given volume of gas at different temperatures ?


text:
Why do we use cis/trans for 1,4-dichlorocyclohexane, but R/S nomenclature for its 1,3 isomers?
1,4-dichlorocyclohexane has two isomers, and 1,3-dichlorocyclohexane has three:

Looking at the substituted carbon, the reason for the isomers is the same: either the chlorine points up or it points down.
Why does IUPAC nomenclature use cis/trans for the former and R/S for the latter?

Those 1,4 carbons on the first one aren't chiral centers, and the images aren't stereoisomers: those two images are not non-imposable mirror images of each other, nor are there 4 different substituents on the 1 and 4 carbons.Thus R/S is not applicable here and cis/trans is. See this post for some more details.

@ToddMinehardt I updated the question, stating "isomers" instead of "stereoisomers". I am unsure whether geometrical isomers are a subset of stereoisomers (same Lewis structure, different 3D structure?), e.g. this figure.

The preferred IUPAC nomenclature doesn't use cis,trans for 1,4-dichlorocyclohexane. The preferred names are (1s,4s)-1,4-dichlorocyclohexane and (1r,4r)-1,4-dichlorocyclohexane, respectively.

Related What does lowercase r-s notation mean?

Geometrical and optical isomerism are both forms of steoreoisomerism, which is isomerism that arises due to difference in orientation in 3D space, and they are not mutually exclusive.
As per IUPAC, the strict definition of geometrical isomerism is

Stereoisomeric olefins or cycloalkanes (or hetero-analogues) which differ in the positions of atoms (or groups) relative to a reference plane: in the cis-isomer the atoms are on the same side, in the trans-isomer they are on opposite sides.

RS Nomenclature may be used to refer to the absolute configuration of a chiral molecule.
1,4-dichlorocyclohexane does not possess a chiral carbon, and you cannot use RS to denote the absolute configuration. It can however have cis and trans forms
1,3-dichlorocyclohexane has two chiral carbons, and 3 stereoisomers, 2 optically active forms which are enantiomers, and 1 optically inactive (meso) form

Using 'cis' or 'trans' here is not wrong, as it does satisfy the necessary conditions, but there are two 'trans' forms.


text:
Enamine Stability determiners
been wondering what are the factors that determine the stability of the enamine structure, like steric hindrance, resonance effects, and stuff like that. Like if I have four compounds, how do I check which one's gonna form a more stable enamine


text:
What type of sugar is produced A. Tutton's Paper on the Synthesis of Glucose?
https://www.nature.com/articles/037007b0
At the end of all the chemical reactions, a sugary substance is identified by what sugar it is is not specified so I would like to know what it may be. (On a side note, if there is a other method of artificial synthesizing glucose please mention it)


text:
Question about regioselectivity depending on the base used
In one of my organic chemistry lessons, it is possible to produce 2-methylbut-2-ene and 2-methylbut-1-ene from 2-bromo-2-methylbutane using a base, using the mechanism E2. The percentage of the two products 2-methylbut-2-ene and 2-methylbut-1-ene depends on the base. Using NaOEt, we obtain 69% of 2-methylbut-2-ene and 31% of 2-methylbut-1-ene. Using t-BuOK, we obtain 28% of 2-methylbut-2-ene and 73% of 2-methylbut-1-ene. The reason mentioned for these percentages is that NaOEt is more nucleophilic and t-BuOK is more basic than NaOEt. But it doesn't really answer my question. Why using these different bases favor one product or the other one? (See picture attached, and yes it's in french)

Consider the steric hindrance affecting the two bases. tBuO- is considerably more hindered than EtO-. To produce 2-methylbut-2-ene requires the removal of a more hindered proton than 2-methylbut-1-ene.


text:
The preparation of ethene from ethanol
Draw an energy profile diagram for the dehydration of ethanol to produce ethene, labeling your reactants, products as well as intermediates.
import matplotlib.pyplot as plt
Define energy levels
reactants_energy = 0
transition_state_energy = 100  # Example value
intermediate_energy = 50  # Example value
products_energy = 20  # Example value
Define labels
labels = ['Reactants', 'Transition State', 'Intermediate', 'Products']
Define energies list
energies = [reactants_energy, transition_state_energy, intermediate_energy, products_energy]
Create energy profile diagram
plt.figure(figsize=(8, 6))
plt.plot(labels, energies, marker='o')
plt.xlabel('Reaction Progress')
plt.ylabel('Energy (kJ/mol)')
plt.title('Energy Profile Diagram for Dehydration of Ethanol to Ethene')
plt.grid(True)
plt.ylim(0, max(energies) + 20)  # Adjust ylim for better visualization
plt.annotate('Activation Energy', xy=(1, transition_state_energy), xytext=(1.2, transition_state_energy + 10),
arrowprops=dict(facecolor='black', arrowstyle='->'))
for i in range(len(labels)):
plt.text(labels[i], energies[i] + 5, f'{energies[i]} kJ/mol', ha='center', va='bottom')
plt.show()


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Definition of adsorption
Wikepdia defines adsorption

Adsorption is the adhesion of atoms, ions or molecules from a gas, liquid or dissolved solid to a surface

Does this make, a cation getting reduced and attaching to the surface of an electrode during electrolysis, an example of adsorption? It seems to satsify the above definiton. However, no matter what I search, I never see this as an example of adsorption.

There must be countless articles on reduction of adsorbed species on electrodes. Adsorption "typically" means that there is NO chemical reaction. Since you also want to adsorb and then do an electrochemical reduction, these are two distinct steps. Any definition of any term cannot encompass all examples in the world. Definitions are limited and man-made.

Yup, that would be adsorption. If it's forming a specific bond to the surface, that would be chemisorption; if it relies on intermolecular forces (e.g., ion-ion, London dispersion forces, etc.) it would be physisorption. Found some papers for reference. > https://doi.org/10.1039/D1SE01760H > https://doi.org/10.1002/elsa.202100183 > https://doi.org/10.1021/jp500425g

Yes, adsorption is one step on the electrode, the next step will be electroreduction. Adsorption's definition does not include electrochemical reduction as the OP is thinking.


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What does a “B” mean on the special hazard area of a safety diamond?
I was passing by a building and noticed a safety diamond on it. I also saw that the specific hazard area has a “B” symbol. I tried to find information about this and the closest thing I could find was “BIO” indicating a biological hazard, but the diamond says “B”, and is on a manufacturing building. What could this symbol mean?

That diamond comes from the National Fire Protection Association (NFPA), based  in the US. According to https://en.wikipedia.org/wiki/NFPA_704, the white division is for specific hazards, and that "B" is not part of the NFPA standard. The Wikipedia article mentions that sometimes jurisdictions will use unofficial symbols there, which have specific meaning within that jurisdiction. I think your best bet for getting an answer would be to contact the fire dept. in that jurisdiction and ask them what "B" means (or contact the company directly). Let us know what you find! I'm hoping for "boom".


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Effect of particle size on yield of copper acetylide
To avoid pure trial and error, I need some examples of experiments already done with (possibly ammoniacal) aqueous solutions, particles/precipitates of copper (both oxidation states) salts, and acetylene gas.
Since sensitively dependent reactions are often misreported, be sure to provide enough experimental details. No need to interpret anything, just pieces of information are OK.


text:
Different method of solving a question giving different answer

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $\pu{1.41 gmL^–1}$ and the mass per cent of nitric acid in it being $69 %$.

I attempted to consider a 1L solution, and since $\pu{Density= \frac{Mass}{Volume}}$, obtained the mass of $\pu{1L}$ of the solution as $\pu{1410g}$.
I then attempted to acquire the mass of $\ce{HNO3}$, which I got as $\pu{972.9g}$ Then the number of moles of $\ce{HNO3}$ is $15.44$ moles. This makes the molarity of the solution $15.44$. Of course, this is far from correct, but I am unable to trace any mistake in my methods.
Any hints and answers are extremely appreciated.

What other methods and what other results?

Wikipedia: The concentrated nitric acid of commerce consists of the maximum boiling azeotrope of nitric acid and water. Technical grades are normally 68% HNO3, (approx 15 molar), while reagent grades are specified at 70% HNO3. The density of concentrated nitric acid is 1.42 g/mL

There is nothing wrong with your calculation (except maybe sig figs).

Dear OP, you are supposed to show your working so that the community can better gauge how it can help you. They are asking molarity from given mass %.

Your answer is correct, although a more direct calculation would be:
$C_i=y_iC=\frac{x_i\require{cancel} \cancel{M}}{M_i}\frac{\rho}{\cancel{M}}=\frac{\left(0.69\right)\left(\pu{1410g/L}\right)}{\pu{63g/mol}}=\pu{15.44mol/L}$
Where:
$\,x_i$ is the mass fraction of $\ce{HNO3}$
$\,y_i$ is the mole fraction of $\ce{HNO3}$


text:
Why does a nucleophile have to be replaced by a nucleophile?
I'm a highschool student for context and I was thinking why does a nucleophile need to be replaced by a nucleophile in a nucleophilic substitution reaction -  why doesn't it just replace one of the $\ce{C-H}$ bonds instead. Asked my teacher and she said, "stability". I have no clue what that means and I would like to understand what exactly this term means.

o.O Nucleophile replaces a leaving group, which has to good enough to be substituted. You thought what would leave in your scenario?

@Mithoron I thought the carbon-hydrogen bond would break and the hydrogen would leave.

Hydrogen anion would leave - it's one of worst leaving groups. No way it would happen, unless something would accept it at the same time.

BTW why didn't you consider that C-C bond could be broken? It's actually more likely scenario, but still a rare thing.

"Stability" in this context refers to a bond's resistance to breaking. A C-H bond is way more stable than a C-X bond, and therefore way more resistant to breaking. In other words, C-X bonds are easier to break because they are more unstable.

How can a nucleophile be replaced by another nucleophile? Something's highly inaccurate here with your terminology. Is it by any chance that you missed out on the term electrophile / electrophilic nature?

It is an ionic reaction. The Original compound has a polar bond with a negative leaving group and a partial positive charge on the carbon. The nucleophile approaches the carbon forming a bond while the bond to the original nucleophile breaks. [SN2 mechanism]. C-H and C-C bonds are usually nonpolar so they do not participate directly in ionic reactions. The actual reaction depends on the bond strengths, concentrations and reaction conditions.

@jimchmst Ty that makes perfect sense.


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Calculating mass of a solution
I am trying to calculate the mass of a specific volume of glucose 5% solution. I calculated the values for the components of the solution and wanted to seperately calculate the mass of the solution based on it's volume as if just the volume and molarity of the solution are known. But I can't figure out why the mass is not equal in both calculations.
I used the weighted average of the glucose and water molar mass to calculate the molar mass of the solution. As well as other combinations of different molar mass and molarity.
I used both the molarity of the solution as moles of solute per litre of solution and the sum of the number of moles of solute and solvant.

How do I calculate this correctly?

The density, g/cm^3, is a property of each, separately. Why are you adding them?? When glucose dissolves in water, the density is somewhere between that of water and that of glucose. 100% water has a density ~1.00, 100% glucose, ~1.54 (according to the question), so a 50% solution would be ~1.27 (give or take a bit). You can figiure out what 5% density is.

Note that volume is not additive during dissolution. Density of solutions is empirical value and cannot be derived from basic principles, nor from densities and ratio of the solute and the solvent. You need to get density of 5% glucose solution to know what volume is 100g = 5 g of glucose + 95 g of water (approx. 95 mL )


text:
Calculating pressure inside a bag using the ideal gas laws
I was doing an experiment to inflate a bag using CO2 produced from a reaction between baking soda and vinegar. I came up with a balanced chemical equation and used the ideal gas law to calculate the amount of baking soda and vinegar I would need to fully inflate the bag. This worked great, but now I need to do something else: find the pressure inside the bag. Here's my process:

Weigh the bag with gas inside it, then deflate it and weigh it again. The difference in weight is the grams of CO2 produced as gas.

This yielded 0.66g. Using the molar mass of CO2, I calculated roughly 0.014 moles of CO2 gas produced. I plugged this into the ideal gas law equation along with gas constant 0.0821, temperature of 295K, and volume 0.75L for my bag.

Solving this yields a pressure of 0.48 atm. That's impossible, since the bag cannot inflate unless it has a pressure greater than the pressure of its environment (around 1.0 ATM).

What did I do wrong here?
(Disclosure - also asked at https://www.reddit.com/r/chemhelp/comments/1bkmm0j/calculating_pressure_inside_a_bag_using_the_ideal/ )

Did you calculate with the bag buoyancy? The difference of the weights is not the mass of CO2.

Note that e.g  440 mg CO2 at atmospheric pressure and temperature has weight(= what scales show, i. e. not mass) not 440 g but about 440 g - 288 g =  152 g.

Unless that thin bag burst, the pressure was approximately one atmosphere (depending on the elevation and weather at the time), because the bag would be too weak to maintain a difference between inside and outside pressure. Therefore, consider that the volume of $\ce{CO2}$ collected was less than the theoretical capacity of the bag, 0.75 L. I.E., the bag was not fully inflated.
Also, as Poutnik mentions, consider that $\ce{CO2}$ in the bag displaces air, a mix of $\ce{N2}$, $\ce{O2}$, $\ce{Ar}$, variable amounts of $\ce{H2O}$, and other trace gases -- about 1.2 kg/m3. Subtract that from the absolute density of $\ce{CO2}$ to account for buoancy.
Do you have time and facilities to perform a further experiment? You could get an idea of the actual volume of gas generated by measuring it by displacement of water.

After repeating the original experiment, carefully tie the inflated bag around a rubber tube, the end of which is immersed in water, underneath a measuring container, also filled with water, as shown above. Squeeze the bag so that the gas inside goes through the tube, bubbling up inside the inverted graduated container. Though you might not have a one liter graduated cylinder, a large, clear measuring cup, or even a plastic bottle that has been roughly calibrated, could be used to measure the $\ce{CO2}$.
Ask your instructor about performing this -- personally, I'd be pleased to have a student enhance an experiment.
Don't expect great accuracy -- some $\ce{CO2}$ dissolves in water, some will be lost transferring and tying the bag, some will remain in the bag and tube -- but it will give an idea of the volume, which should agree better with your calculation.

Here is a thought experiment. Take a bag, fill it with air, close it and measure the mass. Then, open the bag a bit, and measure it again. Then turn it around so that the opening is on the bottom, and measure again. Finally, squeeze the air out of the bag and measure again. What mass changes do you expect? If you have a bag and a balance, do the experiment.
Then, evaluate the protocol you used in point 1.


text:
What are a some of the best crosslinkers for uv curing of benzyl methacrylate?
I'm hoping someone can suggest some crosslinkers for uv curing benzyl methacrylate that are low in viscosity, relatively high in refractive index and most importantly that will impart hardness and rigidity to the produced homopolymer.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

I'm hoping someone can suggest crosslinkers for uv curing benzyl methacrylate that are low in viscosity, relatively high in refractive index and that will impart hardness and rigidity to the produced homopolymer.

Can you please edit your question and include this description? This is lot clearer than the actual text of the question.

Just looking for some free consulting work. For requests such as this a chemist, consultant should be hired given the details and PAID for their work.


text:
Pressure in Clapeyron and clausius Clapeyron equation
The Clapeyron equation is
$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$. Here, is the change in pressure 'dp' actually a change in external pressure that is being applied on the whole system?
Clausius Clapeyron equation:
$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$, where $p_2$ is vapour pressure at $T_2$ and $p_1$ is vapour pressure at $T_1$ respectively.
I am confused as to how can we get from pressure on the whole system to vapour pressure? Am I missing something? Or is Clausius Clapeyron equation derived in such a way from Clapeyron equation so as to relate the vapour pressures with corresponding temperatures?

@Poutnik, I read wikipedia. But I'm still not clear as to why we CAN change from pressure to vapour pressure?

@Poutnik. I read it again. It says that "The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along phase boundaries". Along the phase boundaries where gas is in equilibrium with condensed phase vapour pressure is equal to external pressure. Hence we can put vapour pressure. Is this correct?

So my understanding that CC equation is when gas-liquid equilibrium with vapour pressure being equal to external pressure is correct? When we are not in equilibrium C equation is more appropriate where dp is change in external pressure?

What about when they are not in equilibrium? Is there another equation for that?

There all bets are off. There is infinite number of different possible nonequilibrium states, with spatial and temporal variation of parameters.

@Poutnik, thank you for bearing with me. And I am sorry if I offended you.

Not at all  Everything is OK.

@Poutnik, I'm sorry if there is a mistake from my side but why did you delete your comments? They were very helpful...

I do not remember exactly now, partly because I may have considered them as already obsolete, partly to avoid moving comments to chat. Generally, comments are not supposed to be a permanent info. I apologize if it caused you any trouble.

@Poutnik no, it didn't cause any trouble. I just thought that they were helpful for future reference. Also, just to be sure, in Clausius Clapeyron equation the pressure is vapour pressure which due to equilibrium is already equal to external pressure?

It is vapor pressure, which is constant for given temperature. External pressure must be equal to it, otherwise there is ongoing evaporation or condensation

Thanks again, I get it now

For a single component system, p is the equilibrium vapor pressure and the pressure within the liquid.  For a system with more than one component, if the gas phase is ideal and the Poynting correction for the effect of pressure on the Gibbs free energy of the liquid is negligible, p is the partial pressure of the component in the vapor phase.  The overall pressure in the liquid and vapor can, of course, be higher than this.

The atmosphere is a second component and changes the Phase Rule. It may or may not be negligible.

Like I said, if the Poynting correction is negligible (which is typically the case if the gas phase gives close to ideal behavior), even with a second component present (say air), the partial pressure of the component in the vapor phase is equal to the equilibrium vapor pressure of the pure component at the system temperature.


text:
Inconsistency in balancing of the equation for lead sulfide oxidation by ozone
The following question was asked in JEE Examination :
$\pu{1 mol}$ of $\ce{PbS}$ is oxidized by $X\ \pu{mol}$ of $\ce{O3}$ to get $Y\ \pu{mol}$ of $\ce{O2}$ :
The answer to which is given as 8.
If I balance the equation $\ce{PbS + $X$ O3 -> $Z$ PbSO_4 + $Y$ O2}$ by pure observation I seem to get the answer:
$\ce{PbS +2 O3 -> PbSO4 + O2}$
Thereby $X=2$, and $Y=1$
However, upon using the Ion electron method for balancing the former answer(=8) seems to uphold.
Hence, I was hoping that anyone could help me point out what part I'm overlooking.
TIA for any valuable insights.

Note that titles of both your and the other question suffer by the lack of information.

Chem+Math MathJax formatting:
Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Not exactly, can you please elaborate on why I'm getting two different answers, is it informative that 4 ozone molecules should be required to react with PbS?

You're assumption in $\ce{PbS + 2O3⟶PbSO4 + O2}$ is that each ozone molecule contributes two oxygen atoms. Consider that a simplified path for ozone decomposition is $\ce{O3 -> O2 + O^*}$ (though it is far more complex than that). In that scenario, each ozone molecule contributes only one oxygen to the reaction, while the other two wander away as $\ce{O2}$. Thus, you'd need twice as much ozone.
Also consider that all oxygen in ozone could be used in the oxidation of $\ce{PbS}$, depending on temperature, so this is not a good test question unless temperature is specified. Sulfide ores (or sulphides, as found in UK), have been roasted in air to release all the sulfur as $\ce{SO2}$, instead.

The reaction with ozone is different from the reaction with O2. This illustrates the importance of lab work to ascertain what actually happens. The simple equation is a composite of 2 equations:  reaction with ozone and subsequent reaction with O2. However, O2 seems to be incapable of oxidizing S= to sulfate so the reaction is not happening with O3 present. Limited ozone might give a mix of sulfate and PbO + SO2 It is very possible to write perfectly balanced equations with not even a toehold in reality.

You can write this in many ways:
$\ce{PbS +4/3 O3 -> PbSO4 + no O2}$
$\ce{PbS + 2 O3 -> PbSO4 + O2}$
$\ce{PbS + 8/3 O3 -> PbSO4 + 2 O2}$
$\ce{PbS + 10/3 O3 -> PbSO4 + 3 O2}$
$\ce{PbS + 4 O3 -> PbSO4 + 4 O2}$
The reason is that you are combining two independent reactions:
$\ce{3 PbS + 4 O3 -> 3 PbSO4 }$
and
$\ce{2 O3 -> 3 O2 }$
In these cases, you need one other bit of information (such as the ratio of ozone reacting to dioxygen formed) to do stoichiometric calculations.

The reaction that is being combined is one that does not happen, the oxidation of sulfide or sulfur to sulfate by molecular O2. That reaction is the reaction limiting in sulfuric acid production needing catalysis by nitric oxides or vanadium oxide.

The oxidation of $\ce{PbS}$ requires $4$ times $1$ atom oxygen, according to Dr Moishe Pippik. As each molecule ozone produces only one useful oxygen atom, it is reasonable to think that $4$ molecules $\ce{O3}$ are needed to produce $1$ $\ce{PbSO4}$. In this case the oxidation of $\ce{PbS}$ into $\ce{PbSO4}$ occurs according to : $$\ce{PbS + 4 O3 -> PbSO4 + 4 O2}$$

What would happen were O3 were limited?

Ij th amount of available ozone is limited, the amount of $\ce{PbS}$ being oxidized would be rather low.

I would think that the O2 formed might oxidize PbS to PbO and SO2 No? My point is that they are two independent reactions, possibly different conditions.

Well. I start from Poutnik's hypothesize stating that 1 $\ce{O3}$ produces only one useful oxygen atom, plus one $\ce{O2}$ molecule. It is also Kristen's last proposition. I don't really know if this hypothesis makes sense.


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Exchange of Fluorine to Oxygen for Fluorophosphonic acid by silylating agents or other methods?
[Warning, long post!]
Dear all,
I am confronted with a seemingly easy transformation. I want to remove a fluorine atom at a fluor phosphonic acid salt. The usual method to achieve this is the basic hydrolysis or the hydrolysis with acid. These methods are not applicable here.
See also:  The Hydrolysis of Phosphinates and Phosphonates: A Review https://www.mdpi.com/1420-3049/26/10/2840

Base will not hydrolyze the compound, because it is shielded from nucleophilic attack.
Acid will deteriorate the compound and is also not applicable.
I recently tried to achieve hydrolysis by TMSBr cleavage. I see the intended mass in LCMS, but the compound is likely to be contaminated by other polar species. Also, during the workup traces of acids may emerge.
I did the reaction with TMSBr at 60 degrees. Maybe other conditions would lead to a different outcome, but this is not explored yet.

Because these reactions are neither robust nor is the outcome easily quantified in terms of yield and purity, I tried the attachment of a cleavage group like benzyl, that does not require acidic or basic conditions.

This route works nicely in theory, but in reality, I see almost no product. The concept is elegant, but performing 3 reactions to get the final product will lead to a major loss of the final product.
Therefore I still have not a way to get the structure I want yet in a robust fashion. Waylander from this board recommended the use of Siylating agents as a temporary cleaving group and the concept I find to be interesting.

I already observed with other esters, that in the moment the negative charge is neutralized, the fluorine will leave the molecule eagerly. The advantage here is definitely, that a TMS-Ester may also be applicable for easy cleavage, even after the fluorine left the molecule. I did not have the opportunity to test this strategy yet, but I am a little bit concerned, that the -OTMS will be cleaved before the fluorine leaves, leaving me right where I began. However, I did further research on the topic, and I found some interesting reactions where fluorine itself will act as a leaving group.

If I receive the double TMS ester, I would have an easy time, because this would go easily in the organic phase in aq/DCM workup and could be easily purified from any excess fluoride and inorganic byproducts and then could be hydrolyzed with water to the salt in the next step under basic conditions with ammonia.

Do you think double silylating is applicable? Would it be feasible to use for example a mixture of diethylaminotrimethylsilane and  Hexamethyldisiloxan (for the second silylation step)? I don't care what the intermediate is, as long as everything can be cleanly hydrolyzed to the free phosphonate salt. The reagents themselves must not introduce acid into the system when in contact with water and all byproducts must be volatile.
I found a lot of different silayting agents here, https://technical.gelest.com/brochures/silicon-based-blocking-agents/general-silylation-procedures, but I am not experienced with the usage of such chemicals with individual pros and cons.

I tried today to silyate my compound in excess with HMDS unter reflux conditions. In LCMS I did not find the TMS-Ester or the free phosphonic acid. I guess it might be possible that the TMS-ester, if it had been formed, just  immediately hydrolyses during the analytic run at pH 2.8 for 11 min. I will try Si NMR, I would assume that from my starting material I will only observe one other SI signal and will also do an NMR of HMDS for reference.

I fully believe it @raptorlane

What do you believe?

I think the silylation strategy is interesting, but HMDSO is not a strong silylating agent, so I don't think you'd get your desired fluoride elimination in the last scheme. Have you considered using potassium trimethylsilanolate (KOTMS)? I'm not fully convinced when you say SN2 can't be done due to shielding, because phosphorus is big. You could just be deactivating it from nucleophilic attack after deprotonation. If you silylate the OH/O- and then heat with KOTMS in, say, THF or dioxane or MeCN, you might displace the fluoride and crash out KF.

Another possible silylating agent to consider is hexamethyldisilazane (HMDS). You'd think it might end up making a phosphoramide, but for some reason the ammonia byproduct it generates does not seem to participate in further reactions. It's possible that HMDS plus a catalytic acid like TMSCl/TMSBr or saccharin may trigger the fluoride to leave, but I can't immediately think of how to best leverage it, because you'd need to add a nucleophile to the medium which either ends in your product or is easier to remove later. Maybe chloride/bromide already fits the bill? And workup is just rotovap.

Regarding the substitution of the fluorine: This is more optional and not my primary goal. In the moment the silylether is formed, the fluorine is likely to go off relatively easyl. I read that TMS is normally deprotected with MeOH and K2CO3 (pH 12) while the Fluor in Sarin for example will leave quite fast at pH 9. When this is true, one could achieve a sequential hydrolysis. However, the TMS ester was not observed after refluxing with HMDS, I did columns runs via LCMS as direct injection, so that the compound is only exposed a couple of seconds to the acidic mobile phase.

However, I did not observe the suspected masses. Maybe the compound does not tolerate reflux conditions, an alternative route would be to use TMSEDA at RT for couple of hours.

Indeed, the phosphonic TMS ester is likely extremely sensitive to hydrolysis. I have experience with carboxylic TMS esters (which should be more stable than phosphoric TMS esters) and simply dissolving the pure COOTMS ester in CDCl3 (stored over basic alumina to remove free acid/phosgene) and transferring the solution in air to an NMR tube over the course of ~15 s lead to ~5% desilylation by 1H NMR, presumably from trace moisture. It could be helpful if you run reaction tests in deuterated solvents and do NMR vial reactions, checking the progress directly by 1H and 19F NMR.

Let us continue this discussion in chat.


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Splitting of electronic states in Tanabe-Sugano-diagrams
In Tanabe-Sugano-diagrams, different electronic transitions of term symbols of a given symmetry are shown. However, these electronic states arise due to splitting of term symbols of spherical to octahedral symmetry. What does this mean? (https://en.wikipedia.org/wiki/Tanabe%E2%80%93Sugano_diagram)
Does "spherical" mean the usual term symbol splitting in gaseous metal ions, where all the d-orbitals are degenerate because they have no ligands? And octahedral symmetry the different electronic states that arise due to the d-orbital splitting?

For example, in a gaseous metal ion the state $^{9}G$ would split into the four given states in an octahedral ligand field?

Yes. The (potential) splitting is purely a reflection of the symmetry.

Okay that makes sense


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Norbornadiene + hexachloro-1,3-cyclopentadiene, does norbornadiene react with 1 or 2 molecules of the hexachloro-1,3-cyclopentadiene?
So I was doing an exercise from the book and i saw this question: Norbornadiene + hexachloro-1,3-cyclopentadiene gives aldrin and I had to determine the structure of aldrin. The reagents of course undergoes a Diels-Alder reaction. In the end i came to conclusion that there are 2 options, either 2 hexachloro-1,3-cyclopentadienes get connected to norbornadiene or just one, I checked the answers and it was stated exclusively that only molecule gets connected, and the other option wasn't even mentioned.

Source
So i think there aren't any things making connecting second hexachloro-1,3-cyclopentadiene to aldrin impossible, the double bond is still relatively electron abundant, the chlorines are far away and not withdrawing the electrons from double bond, there isnt any steric strain and stuff like that, so why doesn't norbornadiene react with 2 hexachloro-1,3-cyclopentadienes but with only 1?

Have you tried building a model? Based on intuition, I'd have trouble believing that you can squeeze a second reaction onto the norbonyl skeleton, but I could be wrong...

@Zhe I haven't as I don't have such resources, but when i saw the 3d structure of aldrin (both done by me and those looked up on internet) it looked like one more reaction could happen, but I might be wrong

It is true that norbornadienie and hexachlorocyclopentadiene can give two products: aldrin (1) and a product of another Diels-Alder cycloaddition between 1 and hexachlorocyclopentadiene (2). There are scientific articles showing syntheses of both of these compounds. And published procedures for synthesis of 1 and 2 look as if there is no significant reduction in reactivity in Diels-Alder reaction between 1 and hexachlorocyclopentadiene when compared to norbornadiene. In short, article 1 describes synthesis of compound 1 at conditions at which norbornadiene to hexachlorocycopentadienie are in 3 : 1 ratio, i. e. there is significant excess of norbornadiene - probably to prevent formation of 2. When it comes to synthesis of 2 the norbornadiene to hexachlorocycopentadienie ratio is 1 : 2.4, i. e. with a slight excess of hexachlorocycopentadiene. Both reactions are heated under reflux for similar time and both give 65% yield of respective products in the end.

I don't think that exercise from your book was meant to show that only one product is possible. Seems to me that the intention was just to present a problem regarding Diels-Alder reaction between not so trivial substrates - of which one is bicyclic and overall electronics are somewhat reversed as diene substrate is electron depleted by chlorine substituents.
References

T. A. Robbins, V. T. Vien, J. W. Givens III, D. A. Lightner J. Am. Chem. Soc. 1992, 114, 10799-10810.
https://doi.org/10.1021/ja00053a016
N. R. Lokan, M. N. Paddon-Row, M. Koeberg, J. W. Verhoeven J. Am. Chem. Soc. 2000, 122, 5075-5081. (Procedure for synthesis of 2 can be found in supporting information which is accessible free of charge)
https://pubs.acs.org/doi/10.1021/ja991895m


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Converting FeCl2 to FeSO4
I have FeCl2 crystals and I want to produceFeSO4.7H2O based on the reaction below:
$$\ce{FeCl2 + H2SO4 -> FeSO4 + HCl}$$
I dissolved the FeCl2 crystal in water, added equivalent amount of H2SO4 to react with all Fe (No excess) and increased the temperature to 40 C and 90 C (separate experiments). But the yield at 40 C is 70% and at 90 C is less than 30%. Any idea how I can improve the yield?

So what is the relationship between yield and temperature? You thought here would be appreciated.

This is a solubility issue, ferrous sulfate is more soluble at higher T. Try small excess of sulfuric acid and lower temperature. Add a pinch of bicarbonate to work under a CO2 atmosphere to prevent air oxidation of the ferrous salt to a ferric salt. Check the solubilities of everything to ascertain the limits.

Recrystallizing Fe(II) salts from water is extremely ungrateful task. They are oxidized by air too readily.

@DrMoishe Pippik
At room temperature it did not work, so I am confused.

You better produce $\ce{FeSO4}$ by the reaction of metallic iron powder on sulfuric acid $\ce{H2SO4}$


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Will a reaction happen slower if it has a higher enthalpy change than its counterparts?

Suppose both reactions are conducted in identical conditions and they have the same activation energy, and B has 0 enthalpy. Which one will happen faster? My experiment said reaction A happen slower than B but I have no idea why. Is it because of the enthalpy change difference, or other factors?
Thank you!

Going from left to right both reactions could have the same rate constant as the barrier to overcome is the same and so number of molecules with enough energy to cross is the same, assuming everything else is the same.  So $\Delta H$ does not come into it. Going from right to left,  the barrier is less in A than B so rate constant bigger, everything else being the same.

Unless reactions are far from equilibrium, there must be subtracted backward reaction rates from forward reaction rates, providing net reaction rates. This gives the hint the net reaction rate of A is smaller.

What exactly was your experiment? Did you monitor the reactant concentration, the product concentration, or something different. Was it a real or a simulated experiment?

In Diagram A, one puts in X amount of energy to get "over the hump", and, as the reaction completes, gets out somewhat less than X. That means one has to continually feed in energy to keep the reaction going... some energy is consumed.
In Diagram B, it appears that as much energy comes out of the reaction as was put into it, implying that energy can be recovered for the reaction to continue.
If more energy is released than added, then the reaction is not just self sustaining, but can proceed very rapidly, as more and more heat accumulates. An example would be thermite: $\ce{Fe2O3 + 2Al->Al2O3 + Fe}$.
This is not exactly as simple as stated, though. For example, in Diagram B, perhaps the energy source is heat, but the reaction emits light (chemiluminescence), allowed to escape, so more heat must be added -- some heat is being lost as light.


text:
Labelling cathode and anode for electrolysis
Determining which electrode is the cathode and the other the anode is providing to be quite confusing.
We are given the polarity of the dc supply, a cell, and then we should be able to determine the cathode and anode.

As shown by the diagram above the positively charge anode is on the same side as the positive terminal of the voltage source and vice versa. My problem with this, is that the reality is the current flows from negative to positive, and does not obey the conventional current, hence the anode and cathode should be switched.
However, every diagram I have seen has done the same thing as above. I understand that this this convention, but here we shouldn't be able to do this, if we were to observe this process, there would be a build up on the wrong electrode if we followed this convention.
How do we label it?

https://chemistry.stackexchange.com/questions/68533/which-is-anode-and-which-is-cathode

Anode is always the electrode where oxidation happens, whatever the cell. It is valid both in electrolysis and in a galvanic cell.

Pay attention to the flow of the electrons (or that the spoon is being silver-plated), not the plus and minus signs. Reduction happens at the cathode ("red cat"), oxidation happens at the anode ("an ox").


text:
Why are there no two steps involved in this following reaction?
I was studying the chemical properties of alkynes in which I found the following reaction with alcohols

Ethyne gas reacts with methanol to form methyl vinyl ether.
$$\ce{H-C≡C-H + CH3OH -> CH2=CH-OCH3}$$

Now I was quite surprised at seeing this reaction as in most of the other reactions(such as addition of acetic acid to ethyne). The electrophilic addition process was taking place twice. What I think is that $\ce{OCH3}$ must be a +M group (because of lone pairs on Oxygen atom, I suppose) and as a result the pi bond between the carbon atoms in methyl vinyl ether should break in the following way:

and then undergo another electrophilic addition reaction to form 1,1-dimethoxyethane (IUPAC nomenclature). Where am I wrong?

Under what conditions is this reaction reported to take place?

I checked my notes and I found out that I havent written the condition for this reaction to proceed... neither did our teacher tell us the conditions under which this reaction has to be perfomed-as far as I can remember

According to the literature this is done at high temperature and pressure in the presence of a basic metal catalyst and may be a radical process. It is certainly a high energy surface process and those are difficult to be precise about the mechanism.  see https://pubs.rsc.org/en/content/articlelanding/2016/ra/c5ra25100a


text:
What effect does the weak base have here?
Suppose we reacted enough strong base to fully neutralise a weak acid, clearly at this point (the equivalence point) the $pH$ would not be $7$. This is because the reaction between the strong base and weak acid forms a salt, and this salt can dissociate to form a strong base, which affects $pH$. Now, suppose we added some more of the strong base, the influx of $OH^{-}$ ions would contribute to a higher $pH$. If we calculate the $pH$ based purely off the extra hydroxide ions we have added, ignoring the weak base, would the error be negligible?

Could you provide some insight as to why the effect gets less with adding more of the strong base? What interaction is there between the strong base and the carboxylate ion (weak base)?

Chem+Math MathJax formatting:
Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

The key is the fact that strongly alkalic pH due strong base presence practically suppresses the salt hydrolysis. Unless the acid is very weak.
Imagine there is $\pu{1 L}$ of the solution, containing $\pu{0.01 mol}$ of a monoprotic weak acid with $\mathrm{p}K_\mathrm{a}=5.0$ with added $\pu{0.02 mol}$ of $\ce{NaOH}$.
The resulting $\mathrm{pH}=12.0$.
The salt $\ce{NaA}$ is dissociated to $\ce{Na+}$ and $\ce{A-}$. The latter than undergoes hydrolysis
$$\ce{A- + H2O <=> HA + OH-}$$
with
$$\log{\frac{[\ce{A-}]}{[\ce{HA}]}}= \mathrm{pH} - \mathrm{p}K_\mathrm{a}$$
$$\frac{[\ce{A-}]}{[\ce{HA}]}=10^{ \mathrm{pH} - \mathrm{p}K_\mathrm{a}}$$
$$[\ce{OH-(hydr)}]=[\ce{HA}]=10^{(  \mathrm{p}K_\mathrm{a} -  \mathrm{pH}  )}[\ce{A-}]$$
There would be about $\pu{e-9 mol L-1}\ \ce{OH-}$ resulting from the salt hydrolysis ( $10^{-2} \cdot 10^{5-12}=10^{-9}$) which is negligible.
The salt would have an effect in resulting pH, but for reasons of affecting ionic strength and ion activity coefficients. I assume this is out of scope of your level.

I can work out that there is $10^{-2}$ mols of $OH^{-}$ and $3.16 ⋅ 10^{-4}$ mols of $H^{+}$. Then we are left with $9.68 ⋅ 10^{-3}$ mols of  $OH^{-}$  which gives $pH = 11.986$. However from this can we not say that the amount of mols of $H{+}$ is $10^{-12}$, when before we clearly had more  $H{+}$ than $OH^{-}$, and hence we were left with $0$ $H^{+}$?

I used an approximation which is only valid for when the acid is in water on its own... I see my mistake. Why on Earth does this give me the same answer as your's? If my method was completely off. How did you determine the amount of hydroxide ions resulting from the hydrolysis? I think all of the salt dissolves into carboxylate ions and hydrogen ions. How do either of these result in hydroxide ions?

I see, what about the other part of my comment? You have provided a lot of help so far, thank you a lot

Everything you wrote makes sense. One final question, the equilibrium I am used to seeing for the dissociation of a weak acid is not what you have written, but rather your equilibrium with a hydroxide ion "subtracted" from both sides. Are they the same equilibria?

They are two mutually dependent equilibrii with pKa(of HA) + pKb (of A-) = 14. The stronger acid HA is, releasing H+, the weaker base A- is, releasing OH-.

We are both referring to $\ce{H^+ + A^-(aq) <=> HA}$, right? And in turn, for a weak acid, even though the conjugate base is strong, it is still nowhere near strong enough to have a major effect (as you deduced in your answer).

No, this is reversed  acid dissociation , not salt hydrolysis.

What did you interpret my message of "are they the same equilibria" to be comparing then? I was asking whether the acid dissociation and salt hydrolysis where the same equilibria, to which they clearly aren't.

Exactly, they are not, they are just dependent. You can easily derive that Ka = [H+][A-]/[HA]=Kw[A-]/([OH-][HA])=Kw/Kb. // Ka is the equilibrium constant for HA <=> H+ + A-, while Kb for A- + H2O <=> HA + OH-


text:
Surface Tension at Critical Temperature and at Boiling point
I recently did an experiment on finding the surface tension of a liquid at different temperatures, When I was going through my lab manual, I found the following statement:

As temperature decreases, surface tension increases. Conversely, as surface tension
decreases strongly as molecules become more active with an increase in temperature
becoming zero at its boiling point and vanishing at critical temperature.

What is the key difference between becoming zero and vanishing?

I think it may be a result of a wrong sentence analysing and its confusing formulation.

Can you explain please?

Well, I'm pretty sure you know what b.p. is so check critical point and you should be good :/

The first term IMHO does not relate to surface tension, as the second one, but to temperature.

critical point is set of conditions beyond which the gas cannot be liquified right?

So there is no liquid - nothing to have surface tension!

@Poutnik can you please be a bit clear

increase in temperature becoming zero at its boiling point is not about surface tension

@Mithoron But surface tension is due to intermolecular forces right? At critical point there are no intermolecular forces?

@Poutnikwhat else becomes zero at boiling point?

increase in temperature becoming zero

Surface tension is due to having surface.

@Mithoron Yeah, imbalance of forces at the surface? So the distinction between surface and bulk vanishes at critical point?

You obviously cannot measure surface tension above b.p. at given pressure, but the surface tension for T=Tb < Tc is not zero yet.

For T going to Tc, all respective properties of the liquid and the vapor converge to the same values at TC, where the liquid and vapor become the same thing, so there is no boundary, no surface any more.

"As temperature decreases, surface tension increases. Conversely, as surface tension decreases strongly as molecules become more active with an increase in temperature becoming zero at its boiling point and vanishing at critical temperature. "

What is the key difference between becoming zero and vanishing?

The key difference between

temperature increase becoming zero at the boiling point
vanishing surface tension at the critical point

is this:

The former is about the temperature of the liquid that cannot be increased above the boiling point at the given pressure, assuming equilibrium state.
The latter is about surface tension that decreases with temperature and converges to zero at the critical point where both phases become the same and the phase boundary vanishes together with the surface tension.

Note that if you would preform a cyclical state change of the system around the critical point at the phase diagram, you would be able to achieve repeated evaporation without condensation, or vice versa, depending on the orientation of the cycle.
It is like if there was a beach and a plateau above a cliff, the converge in some direction to the same elevation so the cliff gradually disappears.  If you were following repeated closed path around the point where the cliff disappears, you would be repeatedly ascend the cliff without descending it (or vice versa in the opposite direction of the path.

The beach is like a liquid.
the plateau is like a vapor.
The cliff is the phase transition between them.

Chosen values from Engineeringtoolbox site:

Temperature [$\pu{^{\circ}C}$]
Surface tension [$\pu{e-2 N m-1}$]

0
7.56

50
6.79

100
5.89

150
4.82

200
3.76

250
2.64

300
1.47

350
0.37

374.1
0.00

This zero at the boiling point is confusing a boiling liquid shows bubbles that usually require surface tension to happen. also boiling point T depends on applied atmosphere so surface tension becomes zero at every possible BP?

I have rejected the idea of surface tension being zero at the boiling point at several comments and now the answer. The whole OP question is rather about grammar and stylistics than about chemistry.

So isn't it zero at Boiling point?

I am trying to say it all the time, even in comments before posting the answer. // See also https://www.engineeringtoolbox.com/water-surface-tension-d_597.html

At a substance's critical point, it cannot be liquefied by pressure alone. In contrast, if a substance has reached it's boiling point it can be liquefied with an increase in pressure. The distinction between "zero" and "vanishing" is simply a distinction between pressure's ability to physically change a substance into a liquid past a certain temperature.

what does it have to do with surface tension?

Surface tension can only exist when a substance is a liquid.


text:
Henry's Law vs Raoult's Law Standard States for Activities
I am currently reading McQuarrie and Simon's Physical Chemistry and don't fully understand their discussion of standard states for activities. On page 990 they write (with a few minor adjustments to make it make sense):

If, on the other hand, one component is sparingly soluble in the other, then picking a standard state based upon Henry's Law instead of Raoult's law is more convenient. To see how we define the activity in this case, we start with
$$\mu_j^\text{soln} = \mu_j^*+RT\ln\left(\frac{P_j}{P_j^*}\right)$$
(where the star refers to the quantities for the pure liquid, in this case chemical potential and vapor pressure). Because component $j$ is sparingly soluble, we use the fact that $P_j\to x_j k_{H,j}$ as $x_j\to 0$ where $k_{H,j}$ is the Henry's law constant of component $j$. If we substitute the limiting value $x_j k_{H,j}$ into the above equation for $P_j$, and obtain
$$\mu_j^\text{soln} = \mu_j^*+RT\ln\left(\frac{x_j k_{H,j}}{P_j^*}\right)= \mu_j^*+RT\ln\left(\frac{k_{H,j}}{P_j^*}\right)+RT\ln{x_j}.$$
We define the activity of component $j$ by
\begin{equation}
\mu_j^\text{sln} = \mu_j^* + RT\ln\left(\frac{k_{H,j}}{P_j^*}\right)+RT\ln a_j
\end{equation}
so that $a_j\to x_j$ as $x_j\to 0$. The above expression becomes equivalent to
$$\mu_j^\text{sln} = \mu_j^* + RT\ln{a_j}$$
if we define $a_j$ by
$$a_j = \frac{P_j}{k_{H,j}}$$
and choose the standard state such that
$$\mu_j^* = \mu_j^* + RT\ln\left(\frac{k_{H,j}}{P_j^*}\right)$$
or such that $k_{H,j} = {P_j^*}$. The standard state in this case requires that $k_{H,j} = P_j^*.$ This standard state may not exist in practice so it is called a hypothetical standard state. Nevertheless, the definition of activity involving Henry's law for dilute components is natural and useful.

The mathematical operations done here all make sense and I understand how they arrived at
$$\mu_j^\text{sln} = \mu_j^* + RT\ln\left(\frac{k_{H,j}}{P_j^*}\right)+RT\ln a_j.$$
However, it is after this that I began to get confused. I don't understand what they mean by "and choose the standard state such that
$$\mu_j^* = \mu_j^* + RT\ln\left(\frac{k_{H,j}}{P_j^*}\right)$$
or such that $k_{H,j} = {P_j^*}$." since $k_{H,j}$ and $P_j^*$ are properties of the substances involved and not necessarily as a state. Therefore, I don't see how fixing $k_{H,j} = {P_j^*}$ actually specifies a single standard state. Furthermore, it isn't clear why we can arbitrarily choose to have $k_{H,j} = {P_j^*}$ since the expression for chemical potential is already referenced to a standard state: namely the pure liquid $j$. Finally, it seems that the final expression for chemical potential
$$\mu_j^\text{sln} = \mu_j^* + RT\ln{a_j}$$
is referenced to the pure liquid standard state chemical potential $\mu_j^*$ so I don't see where the "Henry's law standard state" actually factors into this equation. I apologize if this is a very basic question, but I would really appreciate if someone could clear up these confusions.


text:
Electrolysis to refine metals
Why does the cathode get bigger and the anode smaller when we use copper electrodes for the electrolysis of Copper Sulphate? Why does the anode not receive electrons from the anions (e.g Sulphate and Hydroxide) in the solution instead of dissolving? The anode has a positive charge so as a result the Sulphate and Hydroxide ions should be attracted towards it, yet the anode made of copper only dissolves in the solution. Should these ions not attach to the anode?

If a copper plate is dipped into a $\ce{CuSO4}$ solution, your process may occur. Some electrons may be creaated at the surface of the copper plate, which later on will interact with copper ion from the solution and deposit a copper atom. Of course. But simultaneously, the initial copper atom from the plate is ionized : a new copper ion is created which goes into the solution. The only result is that you destroy one copper ion to produce another copper ion. No visible change.

It is much easier to oxidate metallic copper than hydroxide or sulfate ions, what you can check yourself in a table of standard reduction potentials. Dissolving copper ions does not allow raising the anode potential to values leading to hydroxide or sulfate oxidation.
Displacement of ions by their migration toward respective electrodes create very quickly electrostatic counter potential that stops further migration. So, there is not much of extra anions on anodes.


text:
Product of Monochlorination of 2-methylbutane
Question: Find out the number of products of monochlorination of 2-methylbutane including stereoisomers.
The given answer is 6, but why can't it be 7? I am getting 7 products, the one which I am getting but not in the answer is given below:

Other 6 Product,

Hard to tell what that diagram represents. It could be edited to show a clear diagram. See https://pubchem.ncbi.nlm.nih.gov//edit3/index.html for help. . For that matter, create th diagram there, and it can identify the name.

Thankyou for responding sir, I have changed the image

OP, have they asked with stereoisomers / without stereoisomers?

Could you draw all the 7 isomers you obtained

+1 for improving the diagram... but now you need to sow what the other 6 diagrams are. to see the difference. BTW, add all hydrogens, to show that all bonds are satisfied.

I shall answer! This is a confusing question, and requires the use of which are the similar compounds. I figured out the thing, and they are asking the number inclusive of stereoisomers.

@HarikrishnanM Yes Sir, they have asked including stereoisomers.

@ananta Sir,I have included the other 6 products too

@Rudresh don't the first 2 structures you have added match the one you drew originally?

Let us take the base compound, 2-methylbutane, as below.

I shall list down each compound, and explain my reasoning also.
This question is one of the more confusing ones as such, because people can easily leave out / count extra structures which might be same.

1-chloro-2-methylbutane

You must note that the below compound is also the same - the name will be the same.

Note that the 2nd carbon is chiral in nature (marked by the $\color{orange}{\text{orange letter A}}$) and hence we need to count this as 2 isomers - the d-isomer and the l-isomer.

2-chloro-2-methylbutane

Note that this is achiral and thus is counted as 1.
So, we've got 3.

2-chloro-3-methylbutane

Note that the carbon marked 'A' is chiral and thus, this isomer is counted as 2.
That makes it 5!

1-chloro-3-methylbutane: No chirality is present here, and thus it is counted as 1.

That makes it six, and we have come to the end of the list of monochloro derivatives from the reaction.

The reason why you might have got 7 is that (refer point 1) you have counted the 1st compound as 2, but in fact those 2 structures point to the same compound.

To the downvoter: is there any way in which I can improve my answer? As far as I know I am sure that I am correct in my reasoning.


text:
Existence of (anti)perovskite (Cs+)3(Pt2-)(Au-) and (Cs+)(Ba 2+)(Au-)3
I know that there are oxide-aurides of cesium and rubidium from wikipedia and https://www.researchgate.net/publication/243821351_Cs3AuO_the_First_Ternary_Oxide_with_Anionic_Gold and that the component compounds cesium/barium auride/platinide exist. Would a double salt be able to exist with these components?
(may have to get Au- and Pt2- radii from lattice parameters)
(CsAu a = 4.24 angstrom)(Crystal data of Cs2Pt: hexagonal, spacegroup P63/mmc (No. 194), a=567.6(1), c=947.1(3) pm, V=264.3(1)×106 pm3, ρcalcd=5.783 g cm−3, Z=2, μMoKα=39.86 mm−1, F(000)=376, λ=71.073 pm,)

There are certain areas of chemistry where we may say with a good deal of confidence that such-and-such compound exist, even though it has never been obtained. This is not such area.

I'd bet we find caesium(>I) first.

alkaline metal anions do exist, but the cation must be protected by ligands like EDTA


text:
Why can't group 2B metals (Zn, Cd, Hg) form complexes with ligands such as CO, nitrosyl (NO), and olefins but do from complexes with NH3, CN- etc?
Why Group 2B metals (Zn, Cd, Hg) can form complexes with Ammonia ($\ce{NH3}$), Cyanide ($\ce{CN-}$) etc. and not with ligands such as carbonyl ($\ce{CO}$), nitrosyl ($\ce{NO}$), and olefins?
What I think is that M(2B)-$\ce{NH3}$/$\ce{CN-}$ complexes are stable without any pi-bonding whereas $\ce{CO}$/$\ce{NO}$ complexes require pi-bonding for stability.  Backbonding (pi-bonding) is what stabilizes Carbonyl, Nitrosyl and olefin complexes. Group 2B metals (Zn, Cd, Hg) have filled 3d orbital and cannot accept electrons from ligands and do not form stable complexes with aforementioned ligands. Am I correct? Please explain this further if possible.
Now, $\ce{Cu (1)}$, group 1B, also has a filled d-orbital (3d10), but its complexes with Carbonyl do exist such as $\ce{[Cu(CO)n]+}$ where n=1,2,3. Why is it so?

Welcome to ChemSE! Please find guidance for MathJaX / $\LaTeX$ as follows and reformat your post - Basics // MathJaX and mhchem formatting for equations and chemical formulae // Guidelines on which should be upright / italics

Also you can refer this information bit on how to use MathJaX.

Zinc actually has been incorporated into a room-temperature carbonyl complex embeeded into a zeolite matrix [1]. Ordinarily, G12 metal centers have their $d$ orbitals too tightly bound to serve as pi donors to the pi-accepting carbon monoxide, but in the zinc-zeolite compound the oxo ligands provide that function and thus enable the zinc to accept carbonyl ligands.
Reference

Guodong Qi, Qiang Wang, Yueying Chu, Jun Xu, Anmin Zheng, Jihu Su, Jiafu Chen, Chao Wang, Weiyu Wang, Pan Gaoa  and Feng Deng, "Room temperature stable zinc carbonyl complex formed in zeolite ZSM-5 and its hydrogenation reactivity: a solid-state NMR study", Chem. Commun., 2015,51, 9177-9180. https://doi.org/10.1039/C5CC02601F

In addition to Oscar's zinc complex, I want to point out the very first thermally stable cadmium carbonyl complex, $\ce{[Cd(CO)3(C6H3Cl)]4·2H2O}$. It has been synthesized with an overall yield of 35%.
Ref.: Subrata K Dey, Shyamapada Shit, Sankar P Dey, Samiran Mitra, K M Abdul Malik, First Report on Thermally Stable Cadmium Carbonyl Complex Containing an Interesting Chloroaryl Bridge: Isolation and Characterization, Chemistry Letters, Volume 40, Issue 8, 2011, Pages 810–812, DOI: 10.1246/cl.2011.810

FWIW, zinc(II) is also shown to form a series of divinylzinc π-complexes, $\ce{Zn[C(Me)CH2]2}$  and $\ce{Zn[C(H)CMe2]2}$, have been synthesized and isolated as white crystalline solids in 66 and 72% yield, respectively.

Ref.: Unprecedented Alkene Complex of Zinc(II):  Structures and Bonding of Divinylzinc Complexes, Alfred Wooten, Patrick J. Carroll, Aaron G. Maestri, and Patrick J. Walsh, Journal of the American Chemical Society 2006 128 (14), 4624-4631, DOI: 10.1021/ja058700x


text:
Optimizing Saponification Time for Oil Residue Removal from Fabrics with Sodium Carbonate (Na2CO3) at Different Temperatures
community,
I'm delving into the specifics of using $\ce{Na2CO3}$ for cleaning fabrics, specifically aiming at optimizing the saponification process to remove oil residues. Given the nature of the materials I'm dealing with, I'm considering two distinct temperature regimes: 80°C for cotton fabrics, and 40°C for synthetic fabrics, to prevent damage.
My objective is to find the optimum settings for two variables: concentration and reaction time, in order to achieve the most effective saponification process. Avoiding letting the fabric soak overnight in 20% concentration solution if 15 min at 5% were enough.
I'm sure such basic problematic has been covered somewhere in the scientific literature but I'm enable to find some answers using google scholar, maybe because I'm a total noob in chemistry and I might not use the right keywords.

Sodium carbonate (aka "washing soda") is Na2CO3 (resp. Na2CO3 . 10 H2O), while NaHCO3 (aka "baking soda") is sodium bicarbonate (hydrogencarbonate)

I assume you cannot avoid relying on empirical experience from sources related to cloth washing industry, or personal experience.

thanks I edited it


text:
Geometrical isomerism in 1-(chloromethylidene)-3,5-dimethylcyclohexane

Will this compound exhibit geometrical isomerism?

This question was asked in JEE (Mains) 2020, which is an entrance exam held in India annually. The official answer given for this question was "No", but some books say otherwise.
According to me, this would not show GI about the double bond, because even though the groups on one side are different, the cyclohexane ring is symmetrical and has the methyl groups on both sides. The answer might lie in the groups coming out of the plane, but am not quite sure about that.
Question: Will the above compound show geometrical isomerism?

According to the NCERT (which the National Testing Agency follows religiously and accepts no other source), $\ce{XYC=CXY, XYC=CYZ, XYC=CZW}$ kinds of compounds show geometrical isomerism. Going by that alone it is seen that the cyclo ring is symmetric with respect to the doubly bonded carbon, and thus it does not exhibit geometrical isomerism. // However, a point is that you should include the sources which you have seen and quote it (...some books say otherwise... - include which books and quote them) so that the community can help you better, and thus I am downvoting the question now.

@HarikrishnanM By "some books" I meant my institute's book, and the answer just says "Will show GI", and no further explanation is provided, so I don't think there is much to quote.

just be sure by checking the NCERT as that is THE KING in this case. From my personal experience, institute books are good only for questions.

@HarikrishnanM According to NCERT this compound will NOT show GI, so I think I would just follow that. Thanks for the help :)

https://chemistry.stackexchange.com/questions/58630/why-is-1-ethylidene-4-methylcyclohexane-chiral

There are four stereoisomers of the compound (two pairs of enantiomers), shown in the first row.

If you swap the chlorine with the hydrogen (second row, swap from structure right above), you don't get a new isomer. In two cases, you get the other enantiomer, and in two cases you get the same compound. I wouldn't call any of these cases geometrical isomerism. Any pair of isomers is related by switching methyl groups from R to S or vice versa.
Note that without the chlorine, there would only be three isomers:

You could analyze the two hydrogen atoms directly attached to the double bond. Using terms from NMR spectroscopy, the two hydrogen atoms in the left structure are enantiotopic, while the ones in the other two structures are homotopic. Thus, substituting a chlorine on the leftmost structure yields one additional stereoisomer, while substituting a chlorine for the other two structures does not.

Will this compound exhibit geometrical isomerism?

No, not in the sense that we would have to break the double bond to turn one stereoisomer into another.

Thanks @Karsten! Unfortunately I am not familiar with the NMR spectroscopic terms, and would prefer if you could give any source from where I can read (12th grader from India). I got the real logic behind this one.

@HarikrishnanM e.g. here

I get that there is no geometrical isomerism about the double bond, but wouldn't switching position of methyl groups also be a type of geometrical isomerism?

After all, isomers of 1,3-Dimethylcyclohexane are labelled cis and trans, i.e. they show geometrical/cis-trans isomerism:
https://www.stereoelectronics.org/webSC/SC108.html

@Thecoder001 1,4-Dichlorocyclohexane uses cis and trans in its IUPAC name, but 1,3-Dichlorocyclohexane uses R,S nomenclature. For the 1,4 substitution, you can't use R,S because of the symmetry of the molecule. This is not the case for the molecule the OP asked about.

Related: https://chemistry.stackexchange.com/q/181724

According to NCERT, which is the textbook followed by the National Testing Agency (which conducts JEE), this compound will NOT show geometrical isomerism, because both the methyl groups in the ring are identical/symmetrical, so I would just follow this as an answer.


text:
Preference in choosing SODIUM or POTASSIum hydroxide in nucleophilic substitution of haloalkanes
When you substitute a halogenated alkane to form alcohol, why is it with reagents like NaOH or KOH? I understand the need for the OH hydroxide bit, but I am not sure why you need the Na cation or K cation. Can't you use some other cation? It seems to me that use of these cations will react with the halogen atom to produce a soluble ionic product. Is this a desirable outcome? And why?

What other source of OH- would you propose using? If you want OH- acting as nucleophile you need a cation

$\ce{OH-}$ ions do not exist alone. They must be accompanied by a cation, like $\ce{Na+}$ or $\ce{K+}$. But of course, another cation can be used, like $\ce{Ba^{2+}}$ or $\ce{Ca^{2+}}$, provided its hydroxide is sufficiently soluble in water.

Perhaps the question should be rephrased Does Na+ or K+ make a difference as the title suggests.  Yes the cation can make a profound difference depending on the reaction.

OH- doesn't exist on its own. I like to think of the Na+ and K+ as “carriers.” But like @jimchmst said, the way a reaction proceeds can depend on the action.

Any reason for writing element names in all capitals?