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Need Suggestion for Picking the Right Solvent for Column Chromatography
I've completed my reactions, and based on the stoichiometry and TLC analysis, I think my mixture should contain 3-4 compounds:

Ethyl gallate (my product)
Gallic acid (the reagent/starting material)
Dicyclohexylurea (by-product from my reaction)
Pyridine (catalyst that I used)

Here's my TLC result (please look carefully there are 3-4 spots):

Spot No. 1 = gallic acid

Spot No. 2 = my successful mixture

Spot No. 3 = my semi-successful mixture (it's not completely successful because there are still pretty much gallic acid remaining)

Solvent A = chloroform : methanol => 100 : 1

Solvent B = n-hexane : ethyl acetate => 1 : 3

Based on that, I must do purification by doing column chromatography. However, I've very little experience in doing column chromatography and I'm unsure about the appropriate solvent for my column.
My questions:

Based on the TLC result, which solvent is more suitable for my column chromatography to separate compounds? And could you explain why it's more suitable, please?
Do you have any suggestions regarding the mixture ratio for my solvent system? (like should I lower the ethyl acetate etc.)

Feel free to add other suggestions/advice that you've for me.
Note:
I also don't know which spot on the TLC plate corresponds to my product.

Your product is most likely the high running spot on system B. Try running another tlc of 1:1 hexane/EtOAc which should bring your gallic acid SM down to approx baseline and your product to Rf 0.3-0.4

I see, but could you share your thoughts on my TLC result using chloroform:methanol (1:1)? What are the issues with that TLC? Is my product spot (Rf) too low?"

Try another tlc with a bit more MeOH (2%) to try to separate out the product. Note that the DCHU does not really stain with most reagents

Okay, I kind of get it now. Regarding DCHU not staining, could I use KMNO4 or Ninhydrin to make the DCHU spot appear?

@Jonathan You wrote you used 1 Percent methanol. 1:1 would be 50 and at this point most compounds will not show any retention. It would make sense to run a stepwise gradient from 10 to maybe 50 Percent EtAc to purify your product. You can check fractions by running them on a TLC and check which fractions will be pure and and separated from the starting material/ side product.

@raptorlane I agree. I think a gradient hexane/EtOAc ending with neat EtOAc column is likely the way to go.  Ninhydrin will not visualise DCHU, KMnO4 + heating the plate may do but most likely as a white spot against the background

I see, I think I could try that. Thanks for your suggestions/advice.

@raptorlane I have an additional question. The solvent I use to dilute my mixture and starting materials, such as gallic acid, is THF. Is it okay to use THF too as a solvent "for diluting my compounds" for TLC and column chromatography? Because my mixture texture is kind of like a paste.

I would use DCM as a solvent for dilution on TLCs, because it evaporates fast. Before you put your compound on the column, I would evaporate all volatiles as good as possible. THF might cause problems in the purification process. For full pyridine removal you might put the sample on lyophilizer or remove the major amount by water/DCM workup (some acid will pull the pyridine in the water phase).

Oh, okay. I thought the pyridine could be remove through the column too. It seems I was wrong all this time.
One thing I forgot to mention is that gallic acid is insoluble in DCM. So, what should I do?


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Properties of Gibbs free energy
I have been asked to find the correct statement/(s) from the below:
$1.$ In any closed system with P, V work only, $G$ is always minimized at equilibrium.
$2.\ ΔG$ is always zero for a reversible process in a closed system.
$3. \ ΔG°$ for an ideal gas is a function of pressure.
$4.$ For an irreversible, isothermal expansion process in a closed system with PV-work only, $ΔG$ must be negative.
The answer: only statment $4$ is correct.
I tried to use Gibbs-Helmholtz equation for $3.$
$$H = -T^2 \left( \frac{\partial (G/T)}{\partial T} \right)$$
and that for a process at equilibrium
$$ΔG° = -RT\ln(K_{\text{eq}})$$
I am still unable to conclude. How to comment about all the statements?

If this is the only question you get wrong on a test you are fine. I don't know what an isothermal expansion doing P-V work irreversibly is. G is a state function most useful for reactions that reach equilibrium and therefore are reversible. The Second Law removes the deficiency in the First Law that reactions do not go to completion but to equilibrium, the free energy functions.

The standard change is for a pressure of 1 bar.

For 1/ and 2/, consider when the Helmholtz energy is used instead of Gibbs energy. For 3/ Consider equation for dependence of Delta G on pressure. What is the constant term at the given temperature?

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Difference in reactivity of sodium ethoxide and NaOH in ethanol
What is the difference in the reaction that proceeds when we use the reagent $\ce{C2H5O- Na+}$ and $\ce{NaOH}$ in $\ce{C2H5OH}$ in with, say ethyl chloride. Is the latter considered to be "alcoholic NaOH"?
$$\ce{CH3-CH2-Cl}$$
As per my understanding, $\ce{OH-}$ in $\ce{C2H5OH}$ exist in an equilibrium with $\ce{C2H5O-}$ and $\ce{H2O}$, with the formation being favoured due to a basic medium provided by $\ce{NaOH}$
$$\ce{C_2H_5OH + OH- <=> C_2H_5O- + H2O}$$
$$\ce{C_2H_5O- + CH_3-CH_2-Cl -> C_2H_5OC_2H_5}$$
$$\ce{C_2H_5O- + CH_3-CH_2-Cl -> CH_2=CH_2}$$
Given that the substrate is a primary chloride, nucleophilic substitution by $\ce{C2H5O-}$ seems to be the most favourable pathway, as compared to E2-elimination. (Williamson synthesis)
However, I have seen in multiple books that both alkenes/ethers are given as products in different instances, with the reagents I mentioned being used interchangeably. What is the difference between them and what is the major product formed?

With sodium ethoxide, there are no traces of water.

@Karsten what are the implications of that?

Depends on the specifics of the reaction.

@Karsten Could you give some examples of the variables that would decide the major product in the two cases?

Temperature, solvent, nucleophilicity/basicity of reagents, steric hindrance of the base are some examples.

What about this particular example?considering ethyl chloride

If the nucleophile/base is not a variable, you still can choose different temperature and solvent.

NaOH dissolved in ethanol gives a mix of hydroxide and ethoxide depending on how much water is present. To run a reaction with a pure alkoxide the procedure is to dry the alcohol and to generate the alkoxide by careful reaction of the alcohol with the appropriate metal preferably under an N2 atmosphere. Alkoxide solutions can be purchased.
The balance of E2-SN2, E1-SN1 can be complicated and is best investigated on an individual basis. If interested in the Williamson synthesis check here:
https://en.wikipedia.org/wiki/Williamson_ether_synthesis

And how does that affect their reactivity towards ethyl chloride?

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Question about chemical potential of a multicomponent ideal gas
Consider a pure ideal gas, i.e. only having one component. The chemical potential and free energy of this gas are as follows:
$$\mu_1=\mu^\circ(T)+RT\ln\left(P\right)$$
$$G_1=n\mu_1$$
Where n is the total number of gas moles.
Now, just assume that we divide the molecules in gas by two, and treat each of these tow parts as a separate component. This is just an imaginary division, and none of P, T, composition, number of molecules are changed. So, we expect the free energy to remain the same as well.
Now, the chemical potential for each of these two component is identical and reads as
$$\mu_2=\mu^\circ(T)+RT\ln\left(P\frac{1}{2}\right)$$
and the free energy is
$$G_2=\mu_2\left(n\frac{1}{2}+n\frac{1}{2}\right)=n\mu_2$$
Now my question is that why is not G_1 equal to G_2?
Edit: To illustrate this I have envisaged two cases for making the gas divided in two parts, please see that attached image. In case 1, the gas is divided into to compartments with half of the volume of the initial compartment. In case 2, the gas is divided into to compartments with the same volume as the initial compartment.
I got two different Delta G values for these two cases. In the first case Delta G is zero and chemical potential does not change, but for case 2 chemical potential changes as described above.

Why did you write P/2 when pressure is supposed to remain same?

The partial pressure of each part will be p/2. See Eq. 12.24 here https://www.bcp.fu-berlin.de/en/chemie/chemie/forschung/PhysTheoChem/agkeller/_Docs/StatTD19/12_MixturesOfIdealGases.pdf

Thanks for your comment! No, there is no change in the gas, no dimerization or change in molecular interactions.
It is just an imaginary division of the gas molecules to two identical parts.

@bloodybutunbowed0 are the volumes of our imaginary compartments equal or not?

@NatashaJ yes it is equal along with all other properties such as pressure and temperature.

@Poutnik I think it should be possible to do such separation. Consider that the gas free energy is the sum of chemical potentials of all its molecules, so it should be fine to assume that its free energy can be the sum of two imaginary parts assumed by us.

@Poutnik could you explain more? Why is wrong to assume that the gas is mixture of two separate molecules? I’m sure that I’m doing sth wrong, but cannot just find it.

To clarify, is the scenario like this? : Case 1: given p,T,V, nA of ideal gas A with mi0,A. Case 2: p,V, T as Case 1, but I. gases B and C with mi0,A=mi0,B=mi0, C, also nB = nC=nA / 2? // If yes, then calculate miB and miC before and after mixing them up at constant pressure, with the the difference of  G as the Gibbs energy change of mixing. Before mixing Case2 G = Case1 G. After mixing, it is lower.

But if you mean just formal dividing molecules of an ideal gas to two substances, while keeping the molecules intact, then your calculation  is off. It must be based on exchange of volume work, heat and entropy from standard to particular state. You need to formally derive contributions of the formal components to these parameters.

@Poutnik Thanks a lot! Based on your points, I have considered two scenarios for separating the gas into two identical parts (I have edited the question, have a look please).

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
// You may find useful formatting guides for texts and formulas/equations/expressions.


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Difference in 13C NMR spectrum
The tasks proposed in Spectroscopy challenge, I found two structures that are mostly similar in allylic moiety, the contradiction is presented below:

Considering two carbons of the alkene bond, it could be easily seen that while the mesomeric effect contributes to the deshielding of beta-carbon on the right structure (143,0 ppm) which is much bigger than alpha-carbon (125,2), the opposite is predicted for the other structure, why is this?

Presumably the lower degree of substitution on the beta carbon compared to the alpha carbon in the open structure happens to cancel out the extra deshielding from the mesomeric withdrawal? Note that the methoxy CH3 in the open structure is also ~15 ppm more shielded than the CH2 in the closed structure.


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Catalytic hydrogenation of Benzene
Why is Benzene not hydrogenated when a side group is catalytically hydrogenated? Some books say that the Benzene ring is also hydrogenated, others say the opposite.

It depends on the conditions - catalyst used, hydrogen pressure etc. The aromatic system makes the double bonds particularly resistant to reduction, but it can be done with the correct choice of conditions  see https://www.chemengonline.com/cyclohexane-production-benzene-hydrogen/


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Acidity of Benzoic acid
Please help me understand the electronic effects used to explain the acidity of benzoic acid. I was taught that there are two major effects namely: a) -I (negative inductive effect) b) +M (positive mesomeric effect). So my doubt is:

COOH shows -I AND -M effect so, here are we talking with respect to -COOH or the phenyl group. Please explain why and also how to identify with respect to which group i'm supposed to look for other molecules as well.

-I is shown by both COOH and Phenyl group. Now,for the above question if the -I effect (as my instructed by my teacher) shown by the molecule is w.r.t phenyl then how can we explain it w.r.t the inductive effect order i.e. -I shown by COOH is greater than that shown by phenyl so then the inductive effect of -COOH group should dominate and it will be like the phenyl group is donating or showing +I effect.

This should help: https://chemistry.stackexchange.com/questions/9334/understanding-mesomeric-effect/9342#9342

Maybe you can choose a more descriptive title? Acidity of benzoic acid is only an indirect consequence of what you discuss in the question.

Thank you!Will try!

You'll find a table here: https://en.wikipedia.org/wiki/Electrophilic_aromatic_directing_groups

There will be a mutual effect but here you are interested in the effect that the substituent on the aromatic ring has on substitution reactions (or other reactions) on the ring, so it is the M,I effect of the substituent on the ring that is of interest. Of course the substituent can be another ring.


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Reaction between copper rod and carbon rod in hydrochloric acid
The situation is this: a copper rod and a carbon rod, touching each other, are submerged in HCl(aq). We're asked to describe the reaction that occurs, but as far as I can see, there's no reaction? The copper wouldn't react with the acid and I don't see why the carbon rod would add anything to the equation. Is there something I'm missing?

Consider electrolytic corrosion, i.e., a short-circuited galvanic cell.

The carbon as an inert electrode can serve as the conductor for an oxygen electrode due dissolved aerial oxygen.
But the reaction
$$\ce{2  Cu(s) + 4 H+(aq) + O2(aq) -> 2 Cu^2+(aq) + 2 H2O(l)}$$
would be very slow.
The respective redox half reaction on the cathode:
$$\ce{O2(aq) + 4 H+(aq) + 4 e- ->  2 H2O(l)}$$
and anode:
$$\ce{2  Cu(s) -> 2 Cu^2+(aq) + 4 e-}$$

As in any other galvanic cell. Oxidation of copper on the anode, reduction of aerial dissolved oxygen on the cathode. Note that the mentioned reaction happens even just with the copper in acid. But in both cases it very slow reaction.
But it may happen the task author does not consider this, so in such a case, the expected answer may be there is no ongoing reaction.

Carbon as electrochemically inert material acts similarly a platinum merely just as a conductor, providing/ collecting electrons for adjacent redox system, if any present. In our case, it is possibly(not necessarily) $\ce{O2/H2O}$ system.
In a way, it is similar as if there were zinc instead of copper. Without a connected second electrode, zirc would reduce hydrogen ions to hydrogen on its surfof coace. With the electrode, the hydrogen would be formed on carbon, reduced by electrons coming via a wire from the zing electrode.
For a copper, it would be very slow oxidation and dissolving, supported by oxygen.
Theoretically, it is $\ce{Cu/O2}$ galvanic cell,but practically, it would be very soft power source with very low available current, possibly hard to measure by less sensitive devices.

BG LAU does not speak of oxygen in his experiment.

@Maurice He does not speak about water either.

Sorry, how does this work exactly? I've never seen something similar talked about in class.

So could I treat this as some sort of chemical cell?

If so, then is the role of the carbon rod simply to make the oxygen some sort of electrode?

Answer was updated.

If no air and no oxygen gas is involved in this system, there is no reaction :  copper should not react. If oxygen $\ce{O2}$ is present (for example in solution), the reaction proposed by Poutnik occurs, and it is : $$\ce{2 Cu + O2 + 4 H+ ->  2 Cu^{2+} + 2 H2O}$$ It can be considered as a redox reaction, with two half-equations like : $$\ce{Cu -> Cu^{2+} + 2 e-}$$ $$\ce{O2  + 4 H+ + 4 e- ->  2 H2O}$$ These two half-equations can occur in a galvanic cell : the two electrodes being, one copper dipped in water for the first electrode (anode), and a platinum plate (or a carbon rod) in contact with oxygen and water for the second electrode (cathode).


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Optical activity in tritiated cyclobutane
How would we determine the optical activity in a compound as such:

The biggest difficulty I'm having is determining the R-S configuration as that will determine whether the pair are enantiomers or diastereomers.
My attempt:

We know that the carbon having the tritium atom will always have a higher priority than the carbon having a regular hydrogen (as per CIP rule).
Since $C_1$ has the lowest priority group (hydrogen) at the line and not the dash, and the rotation is clockwise, it has S configuration.
Similarly for $C_2$, since it has hydrogen at the dash, and the rotation is anti-clockwise, it has S configuration.
For it's mirror image,

$C_3$ has hydrogen at the dash and the rotation is clockwise, so it has R configuration.
As for $C_4$, it has hydrogen at the line and not the dash, and rotation is anti-clockwise, so it has R configuration.
Since configuration of both carbon atoms changes in the mirror image, they should be enantiomers, however they are supposedly diastereomers.
Why are the pair of compounds diastereomers and could someone point out where I'm going wrong?

Where you're going wrong is to trust the source that tells you that these are diastereomers instead of trusting your own work. Just rotate the right structure by 180-degrees, and the mirror plane is pretty clear.

@Mithoron Is that a justification for the downvote?

Edited post can be downvoted. https://chemistry.meta.stackexchange.com/questions/57/should-we-standardize-capitalization-of-elements-and-compounds

@Mithoron If you feel it hinders the search features of Chem.SE or the "discoverability" of the question, feel free to make the suitable edits, I'll gladly approve them :)

The down vote was probably because the first illustration simply switches the configurations not showing the mirror image. the diastereomer would be the cis isomer RS becoming SR

You are absolutely right! The compounds are enantiomers. The second compound is just rotated by a certain angle(Rotating a compound doesn't change its configuration). Most of the questions do this in order to confuse you.


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Why do halobenzenes not give Birch reduction?
I was studying Birch reduction and my book says halobenzene and phenol do not give Birch reduction. I want to know why these two type of compounds do not give Birch reduction.
I also want to know if using reagents of Birch reduction on halobenzene would give halocyclohexane.

Phenol is acidic and reacts with the Na metal to give sodium phenoxide which is not going to reduce under these conditions

What about chlorobenzene?

I don't think it would work like that.

Ok. Maybe book is wrong. Sometimes this happen.

https://doi.org/10.1021%2Fja01407a038

Birch reduction of chlorobenzene has been achieved using gas-phase hydrated electrons $\ce{(H2O)_n^−}$, n = 30–65 leading to formation of a chloro-cyclohexadienyl radical $\ce{C6H6Cl.}$1.
Birch reduction of phenolic species has been achieved with β-napthol and some of its derivatives. They have further said that such systems are nearly impossible to undergo reduction is due to high potential energy barrier to electron addition to the phenolate anion to form the required dianion radical intermediate2. But still theoretical studies are still made on this reaction on such species along with system containing other substituents like $\ce{CH3, CN, COOH, NO2, OCH3, NH2}$ to understand the stability of cyclohexadienyl and homocyclopentadienyl anion3.
References

Yali Cao, Robert F. Höckendorf, Martin K. Beyer,
Birch reduction of chlorobenzene in gas-phase hydrated electrons,
International Journal of Mass Spectrometry,
Volume 277, Issues 1–3, 2008, DOI: 10.1016/j.ijms.2008.05.025
Birch Reduction of Phenols, Josef. Fried, N. A. Abraham, and T. S. Santhanakrishnan, Journal of the American Chemical Society 1967 89 (4), 1044-1045, DOI: 10.1021/ja00980a063
A theoretical approach to the Birch reduction. Structures and stabilities of cyclohexadienyl anions, Arthur J. Birch, Alan L. Hinde, and Leo Radom, Journal of the American Chemical Society 1980 102 (21), 6430-6437, DOI: 10.1021/ja00541a009


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How do upper layers of urethane adhere or bond to lower layers after curing?
I have been painting fine art subject matter, mostly on wooden panels, with urethane (not polyurethane) for a number of years and it has turned out to be a useful medium, particularly because it's not gummy like acrylic and can be occasionaly lightly sanded. I understand the urethane I use may cure ("dry") after 24 hours and perhaps longer, depending on the humidity. In a semi-wet state, the layers meld, which seems obvious and intuitive. The urethane can dry pretty flat if brushed on relatively evenly, which gives rise to the question of whether the adhesion is a physical bond - as a layman I picture a bumpy molecular upper surface settling into a bumpy lower layer molecular surface -- but that notion would not necessarily explain why one layer wouldn't fall off if one turns the painting upside down. So perhaps there are chemical electrical bonds being formed -- but how would they bond if the layer is no longer wet -- are the bonds physical after all? Thanks for any info.

Urethane is a liquid. in hardening it polymerizes, becoming polyurethane.

If a layer is allowed to completely polymerize, the next layer does not adhere as strongly to the first, unless it sanded before applying, to provide that bumpy layer you describe.

Adhesion is a powerful force. Even without chemically combining, the two layers are still attracted electrostatically. Certainly, the second layer will not simply fall off, just as decal placed on glass adheres.
A classic experiment to show the strength of adhesion is to take two clean glass microscope slides, put a drop of water between, and quickly pull them apart. The adhesion between glass and water is sufficient to snap one of the slides.

That said, after sufficient weathering or abrasion, the second layer is likely to start to peel off.


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Different definitions for work in physics and thermodynamics
In physics, work is defined as the force applied on an object multiplied by the distance traveled by that object. However, in thermodynamics, work is defined as a function of the opposing force. For example, if a gas expands inside a cylinder with a piston, work is defined as the opposing force (exerted as atmospheric pressure for example) times the displacement of the piston. Why is it not defined as a function of the pressure exerted by the gas? The gas exerts pressure on the piston and the piston moves. As defined in physics, the work would be defined as a function of the gas pressure. Is this some kind of misconception?

Is the force exerted by the gas on the piston equal in magnitude to the force exerted by the piston on the gas?

Note thermodynamics is also a sub-discipline of physics, so physics is internally inconsistent according to this question.


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Resonance Structures for mCPBA
Is this a valid resonance structure?

I.e. can oxygen donate to positive carbon (i.e. are all oxygens sp2 hybridized)? If not, then why?

Oxygen can't donate to oxygen, are you even serious with this last bit?

Yes, an oxygen atom adjacent to a carbocation «can lend» electrons.  Similar to the mesomeric formulae drawn by you, one can equally formulate formulae for instance about ethyl acetate:

The charged structure to the right contributes less to the present state, than the first one to the left because the form with separated charges is higher in energy, than the uncharged one.
Regarding the hybridization and a difference of the oxygen atoms in an ester vs. for instance the oxygen in diethyl ether

count the number of non-bonding electron pairs of the atom in question which are not moved drawing mesomeric formulae.  Then add the number of $\sigma$ bonds between this particular atom, and adjacent atoms bound;

if the sum equals to two, the atom formally is $sp$ hybridized
if the sum equals to three, the atom formally is $sp^2$ hybridized
if the sum equals to four, the atom is formally is $sp^3$ hybridzed

Hence: in diethyl ether, consider the oxygen as $sp^3$ (similar: water, alcohols); but in esters both oxygen atoms as $sp^2$.
The last structure depicted by you (lower row to the right) however is wrong.  Atoms up and including neon can not extend their valence shell beyond an octet (a bit different to e.g., S/P in the next period).


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Why does Cs2SO4 have more lattice energy that Na2SO4?
My chemistry teacher told us that $\ce{NaF}$ has a higher lattice energy than $\ce{CsF}$. He explained it by telling lattice energy is inversely proportional to size of ions added. He then mentioned that $\ce{Cs2SO4}$ has a higher lattice energy compared to $\ce{Na2SO4}$ (which contradicts the above relation of lattice energy and size of ions). Is this true or is he wrong? I'm not able to find an answer anywhere.

You absolutely don't write element names in lowercase.

Lattice energies are directly comparable, regarding ion sizes , only for the same lattice structures. If ions are arranged differently, ion sizes are not the only parameters to evaluate. E.g. for 1:1 stoichiometry with spherical ions, the radius ratio <0.414 leads to coordination number 4,  <0.732 to 6, >0.732 to 8

You will get in trouble with non-standard element symbols; stick to the convention! Because e.g., Sn in Hill formulae is about one atom of tin, SN  would be about an atom sulfur and an other about nitrogen etc.  In SMILES convention, uppercase C1CCCCC1 would be about cyclohexane, lowercase c1ccccc1 about benzene -- two different compounds.  You surely want to be called by your name proper, too -- don't you?

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and report numerical values (\pu{}) including a non-breakable space.

Petrusevski, Vladimir & Stojanovska, Marina & Šoptrajanov, Bojan. (2007). DEMONSTRATIONS AS A TOOL FOR IRONING-OUT PRECONCEPTIONS: 1. ON THE REACTIONS OF ALKALI METAL SULFATES WITH CONCENTRATED SULFURIC ACID. The Chemical Educator. 12. 71-74. contradicts the above assertion where it gives the lattice energy for Na2SO4 as 1827 kJmol-1 and for Cs2SO4 as 1596 kJmol-1 . This also briefly discusses the different structure of the two compounds.

What may be happening (and I stress this is a complete guess) is that the lattice energies of the primitive cells are being compared. This causes problems as the primitive cell for the Na salt contains half the number of formula units of the Cs salt. What you should compare is the conventional cells - the Na salt crystallizes in a higher symmetry, cubic, system and the conventional cell has twice the number of atoms of the primitive, so the same number as the Cs salt. If this mistake is being made you can half the above number for Na, and you get the claimed ordering.

Also @Poutnik is right, it is difficult to compare lattice energies for different crystal structures. In this case, however, one can view the Cs structure as a distortion to the Na case, and while not perfect a comparison is possible. That said if I weren't about to get on a bus I would like to do some proper calculations ...

Generally speaking, large cations and large anions ($\ce{Cs2SO4}$) form more stable lattices as compared to small cations and large anions($\ce{Na2SO4}$). One reason could be that the ionic character decreases as the size of cation decreases and thus, the factors that influence the stability of a lattice are diminished with covalent character of the bond increasing.
Otherwise, your teacher is correct regarding the factors that influence the stability of a lattice. It's just that in this case those factors are diminished in their effect, due to the characteristics of bond.

This is more applicable on spherical big anions like I-, less so on multiatomic oxo-anions, that are much less polarizable.


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Make the plastic cat flap clear again?
I used something on the clear plastic cat flap to clean something sticky and lasting off and the clear plastic is now cloudy.  Is there anything I can use on the clear plastic to restore it?

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.


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Variable in Thermodynamics
While studying Physical Chemistry, I came up with this question.
As I derive equation C_p-C_v, I found the equation (dQ/dT)p = T(dS/dT)p (d-> round)
As I know, dQ=TdS so dQ/dT=(Tds/dT). and I'm still curious that why (Tds/dT)= dS+T(dS/dT).
Can we consider T as a constant value although the differentiating variable is T?
Thank you!

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 How to use MathJaX.


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Behavior of polyanionic silicate chains in silico
I am trying to understand the stability of charged linear molecules, more specifically these, which contain periodically repeated high electron affinity parts, such as single-bonded oxygen (radicals). The example given in the title is chosen because it is clearly very common in solid state (natural minerals) and hopefully well studied by chemists.
As pointed out in the comments, long chain anions almost never form in vacuum, far away from cations. To be clear, the theoretical situation is similar to a silicate chain in the beginning of the simulation of the process of decaying/discharging/changing shape of the molecule.
What happens to a silicate chain without cations, in vacuum?
I guess, first it repels some of its extra electrons, but I can't imagine the changes in its shape nor the quantity of remaining charge per unit length.
Is it correct that atoms stay in the chain so that it is still $\ce{SiO3}$ (stoichiometrically)?

Separated from cations in vacuum? What makes you think anything like that happens... looks like another XY problem.

Why do you think anionic species would float in vacuum?

@Mithoron The tag "theoretical-chemistry" tells us, anything like that is actually too improbable to ever happen, that's why I can't just do the experiment and find my answer. However, it is still physically plausible: say, a long sequence of lucky collisions with high energy particles. That is not chemistry, but the resulting ion's stability and structure certainly is.

@NilayGhosh To help avoid interactions with irrelevant species. I am talking about theoretical situation.


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Is there a symbol or representation for tolyl groups as there is for the phenyl group?
Is there a symbol or representation for tolyl groups as there is for the phenyl group?
Phenyl group is often represented by "Ph" when for a matter of space you can not draw the entire benzene group. Is there any accepted or extended representation for a tolyl group?

Yes there is Tol if I presume you refer to p-tolyl.  See for instance page 16 of the author guidelines (revision December 2014) of the Encyclopedia of Organic Reagents initiated by Leo Paquette.
On occasion o-Tol about the ortho isomer, for instance $\ce{P(o-tol)3}$ for tris(o-tolyl)phosphine  ligand (or as $\ce{P(o-Tol)3}$), too) though this arguably appears more often in figure / table captions to shorten line formulae, than in structure formulae.  And $\ce{P(m-Tol)3}$ as an example with the meta isomer.  However outside catalysis I wouldn't consider them as usual as (Ph, Bn, Bz; Tf, Ts, TsO, etc.)

(Clayden, Greeves, Warren Organic Chemistry, Oxford University press, 2nd edition 2012, ISBN 978-0-19-927029-3, p. 1080).

Use tol as abbreviation for tolyl group.
Ref.: https://www.axeleratio.com/ip/ligands/abbrev/IndexGrpLigAbbrev.htm


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Is ammonium a coordination compound?
According to IUPAC Gold Book, the definition of "complex" is preferably called "coordination entity" which is stated that:

An assembly consisting of a central atom (usually metallic) to which is attached a surrounding array of other groups of atoms (ligands).

So, it is not forbidden for a non-metal to be called "coordination entity" which in fact, lots of them exhibit this term like [PCl6]-, [SiF6]2-,... The problem that I have is that of ammonium cation, it is an adduct.
So, is ammonium a coordination compound?

Nobody calls it as such - that's the important thing.

You mean nobody calls ammonium a coordination compound?

There is a key point to coordination compound. Here the metal atom accepts electrons from ligands (neutral species like H2O, NH3), forming coordination bond, hence the name. The whole species can be charged and that can be considered a complex ion, for e.g. $\ce{[Co(NH3)6]^3+}$. Ammonium ion on the other hand is two atoms covalently bonded and the species is charged, hence it is a polyatomic ion.

I want to be a little specific:
a) There is a coordination bond in NH4+ which is H3N-->H. Is it compulsory for a coordination compound to be fully coordination bond and no covalent bond involved?
b) According to coordination complex of wikipedia: https://en.wikipedia.org/wiki/Coordination_complex. I could say [PCl6]- a coordination compound. Is it wrong?

There's no "coordination bond", it's dative, or coordinate, bond. This is an arbitrary distinction, as all bonds in ammonium are exactly the same and dative bonds are no different from "non-dative" ones.


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Trends in pKb for both acetic acid and water
I am trying to rationalise a result that I have recently measured in the lab. I'm measuring the pKb of different compounds in anhydrous acetic acid, using the procedure outlined in this paper (https://pubs.acs.org/doi/abs/10.1021/ic50084a055). I'm able to re-produce their calibration plot for my pH probe set-up, so no issues there.
I notice that in their table however, that the pKb of diphenylamine is greater than that of triphenylamine (9.72 and 9.20, respectively). This is confusing to me, because I thought that when considering the pKb scale, a low pKb means that it is a stronger base. So the data recorded states triphenylamine is a stronger base than triphenylamine. But in triphenylamine, the nitrogen lone pair is less accessible for protonation, and therefore it should be a weaker base than diphenylamine (i.e: triphenylamine should have a higher pKb than diphenylamine).
I also see in anhydrous acetonitrile, you get the expected trend as you increase the number of phenyl rings attached to a nitrogen (shown below with pKa values). As more phenyl rings are introduced, the pKa decreases, which means that the pKb increases in that solvent. The rational in the literature agrees with what I describe above.

Why does glacial acetic acid act in the opposite way to other solvents when it comes to the pKb (and therefore the pKa)?


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Ionic Character of Metallic (Alkali Metals) Hydrides , and Reactivity
In my book , the following line has been written:-

All the alkali metal hydrides are ionic solids with high melting points
The ionic character , however, decreases from Li to Cs The reason being that as the size of the cation increases , the anion hydride ion can polarise the cation more easily. As a result , the covalent character increases and hence the ionic character decreases.

Now my doubt is that , is Fajan's rule violated here? Because Fajan's rule states that for bigger sized cations, the corresponding covalent character should be less , and therefore the ionic character should be more.. but why is this not the case here as well? Why does the anion hydride ion polarise the cation more easily?
Moreover the book also says that Lithium will be the most reactive towards Hydrogen though it does mention that the reaction takes place at considerably higher temperature (1073K) compared to other alkali metals (673K).. Could anyone explain why lithium should be the most reactive metal (if it is so) despite the above given fact?

First and foremost, hydrides of alkali metals are ionic in nature. So, you cannot use Fajan's rule to determine ionic character as it is only used for covalent compounds. For ionic compounds we use something called cation-anion radius ratio RC/RA (C = cation, A=anion). For small cations like $\ce{Li+}$, small anions like $\ce{H-}$ are preferred. This is because if the cation is small and the anion is large, the electron clouds of both don't overlap enough for a strong electrostatic attraction to form. This can lead to unstable lattice energy. Hence, as we go from top to bottom in the first group of the periodic table, the size of cations increases thereby decreasing the stability(ionic character) of their respective halides

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Thanks a lot! Just one more question... For a stable lattice energy , or in other words , for more ionic character , the sizes of cation and anion must be somewhat similar?

And... you also mentioned that halides of alkali metals are ionic in nature but isnt LiCl and lower halides have a predominant covalent character?

Typo! I meant hydrides not halides. Really sorry

Yes, we majorly use the relative size of anions and cations

basically fajan's rule is for determining covalent character and relative size for ionic character


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Would it be safe to melt plastic bottle caps and remold them into a form?
Ever since I got a dog who loves to sniff and chew most things that come across her path, I have been hyper-aware of plastic bottle caps thrown onto the street/sidewalks/grass/etc. Most caps are HDPE 2 and HDPE 5. I have been collecting for a year now and wonder if I could safely melt the caps and remold them into something useful. My questions:

Is it safe to melt HDPE2 and HDPE5 to a pourable form? I know that burning plastic emits harmful fumes, but what about melting it?
Would melting and re-shaping HDPE 2 and HDPE 5 compromise the integrity of the plastic or does it cool and set to its original integrity?
Do I need to melt HDPE2 and HDPE5 separately? Can they be mixed?

I do not want to do this in my home. There is a local glass blowing studio I would consider talking to about using equipment. Any other suggestions?
Please consider saving your plastic bottle caps and donating them to a responsible recycler. Dog and animal owners thank you!

Why would it mater to the dog? Are you afraid it will eat them? I mean, of all the things dogs come across every day, bottle caps seem like the last thing to worry about. I've had dogs all my life and while they'll happily gnaw at plastic, I've never had one actually eat it. What you are suggesting is still a wonderful idea and I absolutely hope you'll do it (if it turns out to be feasible), I just don't understand why it would make any difference to your dog one way or another.

Consider https://www.preciousplastic.com/ has details and plans for small batch plastic recycling.

Terdon: Dogs will eat anything We should be more concerned about not littering the environment.

Home extrusion of LDPE or HDPE is impractical. Products made from these plastics are extruded under pressure (as much as 500 to 1,000 atmospheres!) at lower temperature, since raising the plastic to a temperature at which it could be poured  leads to decomposition.
As for safety, even commercial equipment operated by professionals can fail catastrophically.

With care and proper equipment plastics especially the thermoplastics such as polyethylene, polypropylene and the terephthalates melt and do not decompose at the melting points. They will ignite if finely divided and the foamed plastics might be dangerous. Plastics will decompose at a specific, rather narrow, temperature range usually above the autoignition T of the vapors, so high temperatures in air can result in a BLEVE [look it up] and must be avoided. I recommend that you do not do it. Recycle the caps and work towards laws that demand more and more complete plastic recycling.
I, possibly We, applaud your efforts. Give your Dog an extra pet from all of us.

Thank you for mentioning BLEVE.

Note that polystyrene plastics are known for releasing styrene fumes even in the proper temperature range. This is an issue for 3D printing with ABS or ASA filament. Some plastics can also absorb water and foam, spatter, or potentially cause steam explosions if confined and heated.


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Lone pairs are more spread out, but how does this increase repulsion?
Both lone pairs and bonding pairs consist of 2 electrons.
From my understanding, bonding pairs are localised between two atoms and thus lie further from the nucleus, and have a longer sausage shape. Lone pairs are attracted to a single nucleus, and thus can spread out more, creating a rounder shape.
Now repulsion, is really determined by couloumbs law. Both lone pairs and bonding pairs have the same charge, so it must come down to distance. Does being more spread out mean, that at a particular instance, the distance is less, thus the repulsion larger? After all the shapes of bond, only create a probablistic area where electrons may be found at an instance.
My question then comes down to, how can we use the fact that the electrons are more spread out and closer to the nucleus in a bonding pair to explain why the repulsion is greater? Just because the former is true, I struggle to immediately conclude that the repulsion is larger. So can we use the former and coloumbs law to clarify this. Repulsion must be determined by coloumbs law right?

You struggle right and it isn't easy. It is a postrationalisation of vsepr shoehorned into valence bond theory. There are attempts to use the Pauli Principle for this, but that is really hard to explain. And I'm not sure it is the truth either.

https://chemistry.stackexchange.com/questions/32211/why-are-lp-lp-repulsions-greater-than-bp-lp-and-bp-bp https://chemistry.stackexchange.com/questions/157867/quantifying-location-and-volume-of-lone-pairs-for-vsepr

"Repulsion must be determined by coloumbs law right?" Not really, it's more likely related to Pauli exclusion, mentioned earlier.

Bonds are of Lower energy that implies that the electrons are Closer to the 2 nuclei than the nonbonded electrons were to their individual nuclei. Take your thoughts one at a time and think them thru. There is so much volume about an atom or in a molecule if the bonds take up less space the nonbonded are free to take up more space and of course the inner electrons are repelling everyone. Things happen the charge explains it but use some common sense. The diffuse unshared pair pushes the bonds or the more compact bonds make space for the unshared pairs or both acting in concert.

I don't believe it is clear what repulsion you're talking about. The two electrons repelling each other? The net repulsion?

You make contradictory statements: "how can we use the fact that the electrons are more spread out and closer to the nucleus in a bonding pair to explain why the repulsion is greater?" and "bonding pairs are localised between two atoms and thus lie further from the nucleus, and have a longer sausage shape"

"Does being more spread out mean, that at a particular instance, the distance is less, thus the repulsion larger" This is very confusing. Less spread out would bring the electrons more close together and thus they would repel more.

Please check your statements, as it stands this risks being closed as unclear.


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What is the difference in reactivity between the various borates (trimethyl, triethyl, triisopropyl borate) to form boronic acids?
For the transformation of 1,3-dibromobenzene to bromobenzene boronic acid, different groups use different boration reagents namely trimethylborate, triethyborate and triisopropylborate).
The reaction is straightforward and involves doing a lithium-halogen exchange using nBuLi then quenching with the borate species. The addition of hydrochloric acid will then yield the boronic acid through the removal of the aliphatic chains.
Using triethylborate (reagent = 1,3-dibromobenzene), the yield is 85%. Using trimethylborate (reagent = 3,5-dibromotoluene) the yield is 65%. Using process chemistry for 1,3-dibromotoluene using triisopropylborate (same reaction with optimized parameters such as lithiation reaction time), the yield for 1,3-dibromotoluene is 91%.
Assuming the reactivity of dibromotoluene and dibromobenzene are equivalent, how exactly does the boration reagent affect the reaction?
Carbon is more electronegative than Boron and a methyl group is more electrodonating than hydrogen according to this response:
Why do larger alkyl groups show a greater inductive (+I) effect?
Hence the trimethylborate carbon (CH3) is more positively charged than the triethyl and triisopropylborate.
This higher positive charge will lead to a stronger inductive effect B->C (electron density will go from the boron to the carbon due to the difference in electronegativity).
Hence the boron will be "more positively charged/ more electron deficient" for trimethyl borate than for longer alkyl chains. Since the reaction simply involves the attack of the benzene anion onto the electron-deficient boron, one would expect a higher yield with a more electro deficient boron, which doesn't seem to be the case.
Theoretically speaking, none of this should really matter since the benzene anion is extremely reactive and that the difference in reactivity between the trimethyl borate and triethyl borate should be extremely small. Thus, yields would be expected to be equivalent for all borating reagents. Steric effects also do not seem to play a role.
Hence my question. How does one chose which borate to use for these kinds of conditions and why?
Triethylborate:
Kim, H. J., Kim, M. J., Park, H. D., Lee, J. H., Noh, S. T., Lee, Y. C., & Kim, J. J. (2010). Near-IR electromer emission from new ambipolar carbazole containing phosphorescent dendrimer based organic light emitting diode. Synthetic Metals, 160(17–18), 1994–1999. https://doi.org/10.1016/J.SYNTHMET.2010.07.024
Triisopropylborate:
Usutani, H., & Cork, D. G. (2018). Effective Utilization of Flow Chemistry: Use of Unstable Intermediates, Inhibition of Side Reactions, and Scale-Up for Boronic Acid Synthesis. Organic Process Research and Development, 22(6), 741–746. https://doi.org/10.1021/ACS.OPRD.8B00118/ASSET/IMAGES/LARGE/OP-2018-00118Q_0004.JPEG
Trimethylborate:
Group 4 Post-Metallocenes Supported by [OCH2N,C(σ-aryl)] Auxiliaries Bearing a Seven-Membered Metallacycle: Synthesis, Characterization, and Catalysts for Olefin Polymerization
Cham-Chuen Liu, Qian Liu, Shek-Man Yiu, and Michael C. W. Chan
Organometallics 2019 38 (15), 2963-2971
DOI: 10.1021/acs.organomet.9b00307

A potential reason suggested to me by a fellow chemists is a s follows. Trimethylborate is more reactive (as explained in the question) and thus is more prone to hyrdolysis over time (degradation of starting material). Introduction of even slight amounts of boronic acid instead of pure trimethylborate would quench the anion (in this case the bromo benzene anion), forming the dehalogenated specie sideproduct thus inducing a loss in yield.


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Hyperconjugation in o-Cresolate and p-Cresolate anions
I have read that hyperconjugation as an effect works only with an empty orbital available to share sigma-electrons with. Based on that, the 'destabilizing' effect of hyperconjugation on alkoxides is not taken into account when judging their relative basic strengths.
$\ce{MeO- > EtO- > iPrO- > tBuO-}$
based on only inductive effects.
In judging the stability of cresolates, the 'destabilizing' effect of hyperconjugation is taken into account for the ortho and para positions.
My question is why is hyperconjugation being taken into account even when there is no vacant orbital to overlap the electrons with in the first place?