text:
Why does heating transform the aquacomples of slightly acidified cobalt(II) chloride solution to the chlorido complex?
There is a famous demonstration where, if you mix a $\ce{CoCl2}$-solution with the right amount of $\ce{HCl}$ you will end up with the pink $\ce{[Co(H2O)6]^2+}$ complex but when heating it to boiling temperatures it will change to blue as the $\ce{[CoCl4]^2-}$ complex is formed. This reaction is reversible as the mixture is cooled again. Now literature always describes that this works because that shift towards $\ce{[CoCl4]^2-}$ is endothermic, but why is this the case?
What causes one to be more stable here than the other one but being substituted when heated?
I have some ideas but I can't really compare them:

First of all there should actually be an entropic advantage towards the chlorido-side, as 6 water ligands leave, meaning that the exothermic step here is against the natural direction of the entropy(?).
Second, perhaps the chlorido ligand is the better leaving group. I know usually $\ce{H2O}$ is a good leaving group in organic molecules but then its usually a $\ce{H2O+}$. So perhaps chloride is the better leaving group here.
Maybe there is a kinetic effect, that it is harder to displace 6 ligands than just four ligands
Or, as much more water is present it just shifts towards the water side because of the huge excess of water?
Chloride is a weaker ligand than water. But on the other hand the iron-fluorido complex is quite stable in water and towards other ligands.
So could it be due to the different ligand fields? The splitting in the tetrahedron is much smaller than in the octahedron. Although I don't know how this actually affects the stability.

So those were the ideas I had but I can't really find a common thing among them besides the excess of water which often causes an equlibrium to shift.
By the way $\ce{[Cu(H2O)6]^2+}$ and $\ce{[CuCl4]^2-}$ do the exact same thing.
Does anyone have an idea why this reaction is endothermic?

Most of your ideas don’t matter at all. The thing about ‘leaving groups’ (in coordination chemistry one does not typically call them that) is not helpful. It is typically a property of the entire complex, most importantly of the LFSE, whether the complex tends to lose ligands easily or not. Most ligand exchanges also happen via a dissociative mechanism (one ligand leaves, another takes the free space) rather than the associative mechanism (one ligand joines in, this leads to another being kicked out) seen in organic nucleophilic substitutions. Therefore, I didn’t talk about those in the answer

Anhydrous cobalt(II) chloride is blue in colour but on dissolving in water it changes to pink in colour  with the change in coordination number to six. $\ce{Co(II)Cl2}$ has a blue colour in tetrahedral geometry.
$$\ce{CoCl2·6H2O-> [Co(H2O)6]Cl2}$$
The anhydrous salt $\ce{Co(II)Cl2}$ has a blue colour in tetrahedral geometry and the hexahydrate $\ce{[Co(H2O)6]Cl2}$ is octahedrally coordinated complex.

You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as:
$$\Delta G = \Delta H - T\Delta S\tag{1}$$
In this case, we would be observing $\Delta G$ for equilibrium $(2)$.
$$\ce{[Co(H2O)6]^2+ + 4 Cl- <=> [CoCl4]^2- + 6 H2O}\tag{2}$$
Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.
You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.
One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.


text:
How to dissolve PLA (polylactic acid)?
I'm a maker and I've been interested in 3d printing for years.
I have found a technique for dissolve the PLA supports in a model printed in 3D in HIPS

I have printed the model in HIPS and the support in PLA
I have put the 3D model in a ultrasonic cleaner in the solution with water and sodium hydroxide

(1 liter H2O + 200 grams NaOH)
The PLA has dissolved after 3 hours with temperature 65° and 1 hour of ultrasonic sound.
I attach a youtube link with a similar experiment
https://www.youtube.com/watch?v=uWYn5wJ72zY
Is it possibile use another substance / solution less dangerous and less toxic?
In theory, the PLA is biodegradable. Is it correct?
Ps:
I want use this combination of materials HIPS for model and PLA for Support.
Thank you in advance for the answers.

Ethanol? (And what is toxic here?)

Hello, I know that sodium hydroxide is dangerous for vapor.

I have tried with denatured alcohol in temperature 23° but didn't work. Do you mean pure alcohol for drink? What temperature?

by dangerous do you mean hazardous?

Yes, I mean hazardous. I have tried a lot substance: Petrol, Acetone, trichlorethylene, denatured alcohol. But unfortunately no one worked. :(  After I discovered that the sodium hydroxide can solve the PLA because has Ph basic (14 Ph). Do you think there are another substance?

Trichlorethylene? That stuff is really unhealty.

In theory, PLA is biodegradable, yes. In real life, it is a hard nut to crack, as you already found out.

Ethyl acetate disolves PLA - which is typically the ingredient in non-acetone nail polish removers.
https://en.wikipedia.org/wiki/Ethyl_acetate
According to Wikipedia:

"If the filament is soaked in a small amount of ethyl acetate, it will dissolve and can be used to clean 3D printing extruder heads or remove PLA supports. The boiling point of ethyl acetate is low enough to also smooth PLA in a vapor chamber, similar to ABS and acetone."

I'm not sure this information is reliable. I've heard people call it into question, failing to achieve any vapor smoothing with it, and in my own trials soaking pure-ish PLA in (non-pure; just non-acetone nail polish remover with EA as the active ingredient) ethyl acetate, it had almost no effect on the PLA.

I realise this post is over 2 years old but since nobody had mentioned this I'm gonna add it here for any future readers.
DCM (DichloroMethane) is a solvent for PLA and PLA+. I realise not everyone has access to it, but for those who do, it is great for solvent-welding of PLA parts. it dissolves PLA almost instantly and produces a seamless and clean weld once it evaporates away. (works on acrylic and Polycarbonate as well)
Just a safety note, DCM is fairly nasty stuff, you don't want it on your skin, it dissolves and damages many paints and coating as well as epoxies and plastics. the fumes are also fairly toxic. do your homework before getting or using it.
DCM is not really suitable for removing support materials. it's hard to get, it is fairly expensive and it's a toxic, also the waste is expensive to get rid of properly.
As for where to get DCM, it is the main ingredient in some brands of paint stripper. there are instructions online as to how to distill it for use.
Cheers

When I started 3D printing I did some trials to dissolve PLA. I tried acetone, caustic and some other solvent. The only one that worked very well at room temperature was methylamine in water.
Unfortunately, it is not a product to use at home. Caustic is corrosive, but the vapours of methylamine can irritate the eyes and the respiratory system. It should be used in a fume cupboard.
It is very effective because instead of a saponification, the amine reacts with the ester forming an amide, which is much more efficient.

It is true that a mist of aqueous sodium hydroxide can be hazardous to eyes or lungs (or other mucous membranes, even skin when present in large amounts for extended periods of time.) (It is not true that NaOH vapor poses any significant risk at or near room temperature, including at 65°C.) My first suggestion would be to stop the use of the ultrasonics, and try gentle stirring instead. If you must use an ultrasonic bath, then it should be sealed to prevent leakage of the mist into unprotected areas. There are many other chemicals which can provide a high pH.
I doubt your explanation that it is simply the pH which "dissolves" the PLA is correct. Na+ will react with PLA to form the salt in a de-esterfication reaction. It is possible that KOH may be more effective, either on a mass or a cost basis - you'd have to try it. The most obvious high pH chemicals are NaOH, KOH, and Ammonia (NH4OH/NH3(aq)), but you could also try lye, dry lime, CaO and Ca(OH)2. You could try mixed salts such as Sodium bicarbonate or sodium biphosphate to increase the metal ion concentration. There are also known catalysts for the desterification reaction, one well known one is iron (but many catalysts may discolor the plastic).
De-esterification also occurs in acidic conditions. You should consider the use of sulfuric acid or acetic acid (perhaps even hydrochloric acid) as well as the bases I've mentioned. In either case (acid or base), any residual contamination (due to poor rinsing or even absorption by the remaining part) can cause various structural and aesthetic problems over time. Be warned.
To answer your question in general (I do not have any specific experience with PLA), NaOH will be quite effective, it is unlikely that other possible bases or even KOH will improve your situation. CaO has obvious advantages in terms of both waste disposal and possibly for effects of aging, but it is also less soluble than K/Na hydroxides, so might be more difficult to rinse away, and it may not result in equally effective decomposition. (And note that your principle waste problem is the PLA waste.)

Would you mind structuring your answers a bit? You always write terrible text-worms.

I'll bet the calcium salt of lactic acid precipitates and gives a horrible slime. Don't.

Ok thanks, I think will try also CaO. But maybe NaOH make less residues. You have wrote that I can put together NaOH the Salt or sodium bicarbonate is it correct? In my solution,  what quantity of Salt or sodium bicarbonate I can mix in (1 liter H2O + 200 grams NaOH) Thanks everybody for the advice.

alphonse is right: What you get with NaOH (or KOH) is not dissolved PLA, but a solution of saponified lactic acid. The good news is you can just pour that down the drain as long as we don't speak of many liters per day and your sewage gets treatment and doesn't just flow into the next river.
You can calculate the amount of NaOH needed from the mass of the PLA printed. I'll leave that little stochiometric exercise to you, just use a reasonably small excess of NaOH.
With any organic solvent (if it works), you will soon sit on liters of waste for which you have no legal and affordable way of disposal.

Simple Solution to Dissolve PLA is THF (Tetrahydrofuran) whose preferred IUPAC name was changed in 2013 to oxolane. It is used in the Pool industry, as well as in 3D Printing, and vinyl printing (like Vehicle wrapping)
See https://en.wikipedia.org/wiki/Tetrahydrofuran

If you consider IUPAC nomenclature important here, why inconsistently capitalize one of the names?

PLA is biodegradable in the sense that it takes less time to biodegrade than other plastics. It normally takes around 80 years to biodegrade if I remember correctly, compared to a few hundred years for most other plastics. Acetone is able to partially melt PLA, but it won't dissolve; instead it will make the PLA sticky and soft, so I would normally use acetone for smoothing print layers.

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text:
Most contributing resonance structure of phosphoric acid
I am uncertain about the structure of $\ce {H_3PO_4}$ because different sources show different ones.
The structure shown here is:

while Wikipedia gives this one:

Arguments in favour of the 2nd structure:

Structure having filled octet for second row elements (C, N, O, F) are stable.
Structure having minimum number of formal charges and maximum number of bonds.
Structure in which there is minimal charge separation while keeping the formal charges closer together.

Arguments supporting the first structure:

Structure in which negative charge appears on the most electronegative atom should be chosen.
I feel that this could be more contributing because in the first structure sulphur has an expanded octet while $d$ orbitals are of higher energy so it's more favourable to form bonds without using them.
Structure having filled octet a for second row elements ($\ce {C, N, O, F}$) are stable. (followed by both structures)

$\ce {H_3PO_4}$ is different from phosphate ion.None of the answers in the linked question directly state what would be the more contributing structure. They focus on the hybridisation of P in phosphate.

Why the downvote? Please let me know.

There is no sulphur here and no such thing as expanded octet.

Why can't phosphorus have an expanded octet @Mithoron?

"Expanded octet" exists mainly in outdated textbooks.

SO is my rationalisation for the first structure correct @Mithoron? Also, does sulphur too not show expanded octet?

Well, the first structure isn't correct either, formal charges are missing. Please don't use MathJax in titles.

Moderators cannot tell you why a question is down-voted because thats every users own decision. I agree that your question itself is different, but the answer will still be extremely similar. The core remains the phosphate ion, only that three of the four oxygen bond to protons now. The so called 'octet expansion' utilises d orbitals which has been disproved long time ago. The guideline for fewer formal charges being better is also incorrect as it often does not reflect the actual bonding situation. The first structure would be correct, if formal charges were added.

You are asking a question about phosphoric acid, there is no mentioning of sulfur whatsoever. I guess that sparked the comment. The whole concept of extended octet is deprecated, obsolete, wrong. Better forget it completely.

Both structures are not the main contributing structures. In fact, the first structure is outright wrong and should be marked as an error as the electrons of phosphorus do not add up and the non-protonated oxygen is implied to be a radical.
However, the correct most contributing structure can be derived from the first structure simply by adding formal charges: a formal positive charge on phosphorus and a formal negative charge on the non-protonated oxygen. This structure satisfies the stability conditions well:

All main group elements have octets (dublets for hydrogen)
The second proposal violates this most important principle

The number of formal charges is minimalised
We cannot go any lower without violating the more important first point

The type of formal charges matches the electronegativity difference
And also we cannot distribute them differently

The distance between formal charges is minimal
But again we have no choice.

(+1) Can you add a diagram of the correct structure too?

"Most important first point". Well, today itself I read an answer on Chem.SE that octet isn't essentially important for all atoms. Let me see if I can find it.

@Abcd I wrote more important. And I also added the caveat for hydrogen. Most other candidates for non-octet structures are either not main-group elements or are rarely found in molecules (although lithium, magnesium and aluminium may be counterexamples, I agree)

Apologies. Yes you said more. By main group elements are you referring to representative elements? How can we ignore the transition elements when they form the bulk of the periodic table?

@Abcd When a rule is only valid for the elements in groups 1, 2, 13, 14, 15, 16, 17 and 18 (the main group elements) then of course we can ignore the transition metals in this rule. They follow entirely different principles, more subtle, harder to explain. Other rules are made for even less elements, for example the rule that only the second period prefers forming double bonds rather than maximising coordination numbers.

Is there any answer on SE that deals with the rules that the transition elements follow?

@Abcd The key word is 18 electron rule, but as I said, this is much less important and many compounds are stable despite violating the 18 electron rule significantly.

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text:
How exactly lithium metal is produced?
Wikipedia and other sites say that lithium metal is produced by making the electrolysis of a mixture of half lithium chloride and half potassium chloride "at about 450°C"... What is the chemical reaction of this electro-chemical reaction?

The electrolysis is done as a molten salt.
$\ce{LiCl(liquid) ->[LiCl/KCl @ 400-460 C] Li(liquid) + 1/2Cl2(g)}$


text:
Which compound or compounds are responsible for color change in blue butterfly pea flower tea?
Blue Butterfly Pea flower tea is blue by default. Add citrus juice and it turns purple. What compound or compounds are responsible for this reaction to change in pH?
Wikipedia doesn't say, nor can I find any source on the internet that does.

Many pH indicators (all, actually) do similar colour changes. https://en.wikipedia.org/wiki/Red_cabbage#pH_indicator is a well known example. What did you do during chemistry lessons in school? ;-)

@Karl - Yes. Which pH indicator(s) is in the tea?

I don't know what's the blue stuff in Clitoria Ternatea. Some biologist might be able to find out. Is it important for you? The colour change surely comes from one of the typical reactions that occur in pH indicators.

@Karl. I'm simply curious. I hope that's a good enough reason to post the question.

OK, that was simple. This
https://www.google.de/search?q=Clitoria+Ternatea+blue+color
gave this
https://www.ncbi.nlm.nih.gov/pubmed/14568080
which lead me to this
https://en.wikipedia.org/wiki/Anthocyanidin
and I guess that's the culprit. I judge from that abstract linked above that the actual plant can contain a wide mixture of those anthocyanidines.

I guess my internet research chops need work :-/  Thanks.

Another interesting article that also mentions this class of compounds in the context of potential anti-inflammatory uses: https://pubs.acs.org/doi/abs/10.1021/acs.jafc.5b00928?src=recsys&journalCode=jafcau

So it is the same group of chemicals as in red cabbage. Interestingly if you get red cabbage juice on eggs they go blue, so I guess that's the inverse reaction to lemon in blue pea.


text:
What is K2O-nSiO2-xH2O?
In need of $\ce{K2SiO3}$ for my cucumber plants, I purchased a bag of potassium silicate. The molecular formula is $\ce{K2O-nSiO2-xH2O}$. Normally, my bags of fertilizer are easy to figure out. For example, $\ce{Ca(NO3)2}$, or calcium nitrate, has a molecular weight of 164.0878 g/mol. The purity is also 95%. Knowing the purity and the molecular weight, I normally can find out how many ppm of Ca or N I want to use in my hydroponic nutrient tank. With $\ce{K2O-nSiO2-xH2O}$, I don't understand $n$ and $x$. I've googled this only to be stumped. I just don't get it.
The bag of fertilizer ($\ce{K2O-nSiO2-xH2O}$) has the following information listed on it:

Wt% $\ce{K2O}$ 26.5~29.5,
Wt% $\ce{SiO2}$ 55~60,
Weight ratio ($\ce{SiO2/K2O}$) 2.0~2.2,
Mole ratio ($\ce{SiO2/K2O}$) 3.2~3.4,
Bulk Density g/L 400~600,
solid % 82~89,
particle size 270+/-50 and
purity 77%

I need to know the molecular weight, so I can figure out how many ppm of $\ce{Si}$ and $\ce{K}$ to add to the solution.

You're dealing with a mixture of similar hydrated salts instead of a single salt (as in the example of $\ce{Ca(NO3)2}$). The notation $\ce{K2O-nSiO2-xH2O}$ means that there are different crystals with varying proportions of silicon dioxide and water (hydration). i.e. $n$ could be $1$, or $2$, $3$, etc. and $x$ likewise, for any given molecular "speck". A hydrated salt can be de-hydrated with heat (thus changing relative weights of elements), so be careful where you keep the fertilizer, because it will change the math.
This means, to measure a reasonable concentration, you need to consider the reported bulk average quantity of each component. Based on the percent by weight you reported, you can approximate the material as being (by mass -- that is wt%)

$28\%~\ce{K2O}$
$57.5\%~\ce{SiO2}$
(Thus) $14.5\%~\text{water}$

Say you want to know the $\pu{ppm}$ of $\ce{K+}$ for a solution of $\pu{1 g}$ of this fertilizer in a gallon ($\pu{3.78 L}$) of $\ce{H2O}$. Using the good approximation $\pu{1 mg/L} = \pu{1 ppm}$, We have
$$\begin{multline}
\pu{1 g}
\times \frac{\pu{0.28 g}~\ce{K2O}}{\pu{1 g fertilizer}}
\times \frac{\pu{1 mol}~\ce{K2O}}{\pu{94.2 g}~\ce{K2O}}\\
\times \frac{\pu{2 mol}~\ce{K+}}{\pu{1 mol}~\ce{K2O}}
\times \frac{\pu{39.0983 g}~\ce{K+}}{\pu{1 mol}~\ce{K+}}
\times \frac{\pu{1000 mg}}{\pu{1 g}}
\times \frac{\pu{1 gal}}{\pu{3.78 L}}\\
 = \pu{61.5 ppm}~\ce{K+}
\end{multline}$$
Here I guessed a concentration that may be relevant for you, but you can see the way to move forward. I would note, though, that since purity is reported pretty low ($77\%$), you should multiply the above result by $0.77$. Hard to say what the impurities are though, if they are mostly other potassium salts, the concentration could be higher. At any rate it's safe to say the above example is $\approx \pu{45-65 ppm}~\ce{K+}$.
You can also think of the compound like this, based on component masses:
$$\text{Molar mass} = 0.28 \times M(\ce{K2O}) + 0.575 \times M(\ce{SiO2}) + 0.145 \times M(\ce{H2O}) = \pu{63.532 g/mol},$$ and the average molar formula is $\ce{K2O - 3.2SiO2 - 2.7H2O}$.
Finally, I would recall that dissolution of $\ce{K2O}$ in general produces a basic solution, so if cucumbers are fond of acid (I don't know), you may consider adding a little acid ($\ce{HNO3}$, perhaps) to the solution to avoid perturbing the greens. But, you're the farmer!

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text:
Does the distance between an electron and the nucleus change for excitation and relaxation?
According to the Bohr model, when an electron moves from a ground state to the excited state, it will absorb energy, but what about the distance between electrons and nucleus, will it be changed? Why ? How?
Also, if there is the distance change when from a ground state to the excited state, will there be also any change in distance when electrons move from excited state to ground state ?

What do you want to achieve? The Bohr model is not useful to describe chemical reality. You can use it to calculate numbers, but those cannot be reproduced with other methods. And I don't understand you last sentence. You really ask is the ground state before and after excitation is the same?

Yes.
The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons.

Electrons undergo exothermic or endothermic reactions during ground state or excited state. When the electrons move from ground state to excited state, it means that they gain more energy, resulting in higher energy levels (higher energy shells), and this causes the electrons to orbit further from the nucleus. The same explanation is applicable when electrons move from excited state to ground stage- the electrons lose energy and this results in lower energy levels (lower energy shells), causing the electrons to orbit nearer to the nucleus.

I am not sure whether the higher excited state always increases distance between electron and nucleus, especially when considering orbitals of higher angular momentum ($d, f \dots$). There might be some exceptions.

@Feodoran Bohr knew nothing of orbitals, at least not when he devised the model named after him. And in Schrödingers model of the hydrogen atom, the different orbitals with same quantum number $n$ are degenerate. For anything else than hydrogen atoms, the original Schrödinger is quantitativly nearly as bad as Bohr (although the concept is much better of course).

For Hydrogen yes, but I was thinking about atoms and molecules in general.
Why do you say "Schrödinger is quantitatively nearly as bad as Bohr"? The Schrödinger equation is in principle exact (at least in terms of chemical accuracy). Whether you need to make certain approximations to make solving it feasible, is another story.

I may be wrong here as this was long ago, but if I remember correctly absorption of light and the excitation process of an electron (for example) are faster than the electronic movement and therefor appear as a straight vertical line on the energy to distance plot (Jablonski-diagram). So you start from an optimized geometry for the ground state which is not necessarily the optimized geometry for the excited that and it will relax in the time while it is being excited before it falls back to the ground state. I remember we used this electronic change to calculate pKs values of acids for example. So yes, if I remember correctly the x-axis on this diagram was the distance of the electron to the nucleus (but I could be wrong there).
There was a rule for the excitation process, I think by Fermi (?). Historically seen, when they first analyzed the emission spectrum of for example hydrogen, they found that not every possible transition is equally likely, so there has to be quantum-mechanical process which descides what transition is more likely. And as the excitation process is much faster it needs to be almost straight up, so without any change in the geometry or any distances. And essentially you will need a good overlap of those amplitudes.
If you look here for example:
http://www.geoset.info/wp-content/uploads/2012/05/Electronic-spec.png
You can see that both of these curves represent the different vibrational states and those have amplitudes, which tend to be bigger and the sides and not in the middle. I think in classical mechanics you'd expect that to be somewhere around the point where the oscillating spring is relaxed. And you can show, that the absorption process is better if the overlap of those amplitudes is at it's biggest, so you start from the ground state and then you go straight upwards till you find a higher vibrational mode of the excited state where you end up in a high amplitude. This is why you end up in v'=2 in the picture. Then you still have the same geometry but it is a vibrationally exicted state and will first relax to the excited vibrational ground state v'= 0. But that curve is shifted to the right and this is how the geometry changes and I believe in you example this would be the orientation of the electron to the nucleus.
At least I hope so. If not I'm truly sorry but this was the only example I came up with right now.

In principle you are right, but you are answering a different thing. The question was about change of electron density and you are talking about change in the nuclear geometry.

And Fermi and Jablonski are post-Bohr.


text:
Friedel–Crafts alkylation of benzene with propyl chloride
This is a question from my textbook:

Why does the alkylation of benzene with 1-chloropropane give (propan-2-yl)benzene and not (propan-1-yl)benzene?

The book shows the reaction with chloromethane and chloroethane, and there is no such ambiguity in the two cases as the obvious products are methylbenzene and ethylbenzene respectively.
Is there a rule which I need to know in order to figure out which substituted benzene of the two would be more stable? Up till now I've only encountered Markovnikov's rule, Kharash/peroxide effect, Zaitsev's rule and Popoff's rule in the book; none of these seem to be applicable in this case.

Related question without answer

The gist is that 1,2-hydride shift (Wagner-Meerwein rearrangement) can create a more stable secondary carbocation from the primary pseudocation. This process is much more rapid than addition to the aromatic system. As it is half one in the morning, I’ll leave the full answer to someone else.

related https://chemistry.stackexchange.com/questions/38454/anomalous-friedel-crafts-acylation-with-tertiary-acid-chlorides


text:
Can fat make rubber gloves stiff?
Sometimes I use dishwashing gloves when carving cooked duck, which is very fatty. I have noticed that over time my dishwashing gloves seem to be shrinking a little bit and getting stiff. Could this be because of the fat?

Maybe, if it was "washing off" plasticiser from surface.

Fat could diffuse into the glove rubber and let it swell a bit. That would make the material become stiffer, and because it is originally stretched, shrink a bit in the direction it was stretched.

Plasticiser in gloves? Come on!

@Karl PVC glove (have seen thin one time use ones) need plasticisers, because pure PVC is just hard and brittle.

For a good answer, we need to know the material of the gloves - just roughly is enough, but details help, of course. The product name may be helpful.


text:
How to separate two diastereomeric amines?
I am developing a method to synthesize amines using organometallic catalysts and my reaction yield these two diastereomeric amines ($\ce{R2N}$ is, in this case, a morpholine group).

With both being formed in approximate equal amounts, making it almost impossible to identify their structures by NMR without separating them. All the NMR spectra look like a mess since some signals get duplicated and some not. I have a hint that they are actually these two amines because they show the same molecular ion in the MS spectrum and the methodology used for synthesis offers little space for doubt, since it always forms the predicted structure.
I would like to know, if there is a practical way of separating those two compounds without using something super fancy like chiral column chromatography decorated with gold and antibodies kind of stuff. I have tried column chromatography but couldn't find a solvent system that could separate them. I also tried recrystallization (though I don't know exactly why I tried it), it also didn't work.

When you mentioned column chromatography I had to assume you meant flash chromatography or biotage, which by implication is normal phase. When people mention HPLC they are normally defaulting to reverse phase. Now, if you can get reverse phase TLC plates (C18 are available) , you could make some advances.  The issue though is then getting the silica and you also mention large quantities.
Large lab scale to me is say, 50g plus. For this, I would advocate crystallisation,  but to guarantee some differentiation,  I would make salts with organic acids. Nitrobenzenesulfonate salts are renowned for giving well defined crystals.
If you needed something to better differentiate again the two amines, I would try a camphorsulfonate salt (pick an enantiomer).  These tend not to be too expensive.  The salts then have different solubilities and the right solvent becomes a screening exercise. Otherwise, tartaric acid is very cheap and easy to salt break after crystallisation.

Thanks for the advice @Beerhunter , by large amounts I mean about 5g of the mixture... At least for the compounds I work with this is a HUGE amount haha. I was trying to avoid crystallization at all costs since it is the most boring procedure ever to be done at a laboratory, but if I'm not successful with chromatography and preparative HPLC then I will have no better option than making crystals.

Raul, for me there was nothing more satisfying than having a nice swirl of crystals after a crude solution had been made, and ending up with clean solid on a filter. While more is better, a few grams still plenty to play with to develop the crystallisation. It's all good scale up fodder for your new technique.

Separating the two compounds should not be the issue. They are diastereomers, meaning they show different chemical properties which should be different enough to allow separation. The more difficult question is how to locate the separated substances.
I assume very strongly that standard, average HPLC should do the separation trick. It is commonly used to separate very similar compounds. There is no need for a chiral stationary phase as you already have diastereomers. The issue remains how to identify the products since they likely won’t show any significant absorption at any accessible wavelength.
You did mention that you performed MS measurements. Most MS measurements I have seen were actually HPLC-MS or GC-MS. If you have access to a HPLC-MS, you can check whether the standard gradient gives two peaks or a double peak in the graph that shows the total mass caught by the mass analyser. (Careful: these peaks are not quantitative!) If you already have two peaks, you should be able to directly port the separation method to preparative or semi-preparative HPLC. Be sure to collect a fraction after every ten or so millilitres (assuming a column designed for up to $\pu{60mg}$ and a flow rate of about $\pu{7ml/min}$). Then, analyse all these fractions by HPLC-MS again to identify which contain the first eluting and which the second eluting peak.
If you observe a double peak in the HPLC-MS, you will first need to improve the gradient to allow for separation. For close peaks it is often a good idea to choose an isocratic method over a gradient method. The rest remains the same.
If you have access to other, more rapid methods of identification (i.e. you can tell the two compounds apart on TLC) you can of course use these rather than running tens of HPLC-MS experiments and thereby take strain off your colleagues’ nerves. (However, if you have an autosampling mechanism you might run the experiments overnight instead.)

Thanks Jan, I mentioned chiral column decorated with gold and antibodies as a joke, but I know I don't need any of these to separate such compounds. TLC was the hell for me, I tried every possible combination of solvents and proportions but nothing worked. HPLC is probably the best way out, but I don't have access now to a preparative HPLC, and I need to separate big amounts of the two amines.

@RaulLuciano Did you try re-evolving the TLC, i.e. use only apolar solvent (e.g. pentane, hexane) and do the TLC several times. Often this does the trick for me if the compounds are very similar.

@RaulLuciano In those tricky cases where I had to separate isomers, it often helped to do a normal-phase gradient column, e.g. start with pure pentane, and then slowly mix in EtOAc. Alternatively, you could try to do a reversed-phase TLC. That would be a nice test to see if you get separation in RP.

@deusexmachina yes, I tried all types of gradients... starting with pure hexane, mixing EtOAc, then a gradient of dicholoromethane and finally incresing methanol up to 100%. I can separate the amines from everything else, but they are still eluting together... I tried to run preparative TLCs multiple times with hexane, haxane/EtOAc, pure EtOAc, CHCL2, none worked. I feel terrible because I had no problem separating the other amines I obtained using the reaction, this pair is killing me though.

There is also a terrible feature to these two amines, they are not visible under UV light. I can only see them using a GC or spraying a small TLC plate with ninhydrin. However, I can't use ninhydrin on a preparative TLC to see where the amines band are since it would destroy my product. That's why TLC didn't work for me even though I have separated all the other amines using only TLC.

@RaulLuciano Would creating a Toluolsulfonate salt help? But then they are probably to polar to even elute on the column.

@RaulLuciano Did you try adding NEt3 to the eluent? I have heard this works 'like magic' in some cases...

@deusexmachina a funny thing about this reaction is that the major byproduct is NEt3 , so I'm not sure if adding it to the eluent would be helpful, but I will try, thanks :D

@RaulLuciano Should be around ~1%, if I remember it correctly (better look it up :D). Amines are typically very sticky, so NEt3 should be particularly effective. You could also think about ammonia (see https://orgprepdaily.wordpress.com/2006/10/05/puryfying-amines-on-silica/), both NEt3 and ammonia should be pretty effective. Hope that helps : )

@RaulLuciano For a preparative TLC (assuming they separate), you can spray just one side of the plate and collect all the remaining stuff. Or, if worst comes to the worst publish ‘the diastereomers could not be separated.’


text:
Is zero dipole moment possible in a square pyramidal complex?
Say, the complex is [ML5] and is repelled in magnetic field. Given, its dipole moment is zero. Can its shape be predicted as trigonal pyramidal by the given information or square pyramidal structure may also fullfill them?

These are two different unrelated questions that are only held together by the fact that both concern pentacoordinated species.

It appears that you have created two accounts, you can find more information in the help centre about merging them.

A square pyridimal structure can have a dipole moment of zero, e.g. a squarepyramidal cluster of atoms of the same element would have no dipole moment. However for most cases where the bonds are polarized, the dipole moments will not cancel as they would for other symmetries (e.g. tetrahedron). So basically, most square pyramidal complexes with polarized bonds will have a net dipole.
If you have a complex in mind just calculate the dipole computationally or look at the ir spectrum.


text:
Electrolysis of brine with filter paper bridge

What would be the problem with connecting the two dishes up with a piece of metal wire?

You would have electrolysis running on both ends of that wire. Other than that, no problem.

That's a bit of an understatement, @IvanNeretin. Depending on the wire material, one end would be gone within short time, OR you get electrolysis also there, and why then split the thing into two dishes in the first place? You also need twice the voltage (or a bit less, if the wire dissolves). Question to OP: What is that thing supposed to do? Why two dishes?

If  the purpose is to electrolytically seperate a salt, then you need move ions from one cell to the other. The purpose of the bridge is to transport those ions. Solid metal will conduct electrons but not ions. In the past mercury (liquid) was used industrially to transport sodium ions from one cell to the other, but its is too toxic and contaminates the fluids in the sodium hydroxide, and brine cells.


text:
Does temperature affect the %oxygen in air by volume? (even though oxygen density varies inversely with temperature according to P/RT=n/V)
I want to investigate how temperature affect the concentration of oxygen in air as it will affect how much oxygen we inhale per breathe. Oxygen by percent composition in air is 20.95%. However, the oxygen concentration/density in air varies due to the expansion of gas particles and varies inversely wth temperature according to P/RT=n/V (where n/V is the density of gas).
I did an experiment that involves the rusting of steel wool to determine the percent oxygen in air by volume (determined by volume of water rose in a test tube divided by the total volume of air in the test tube). However, I'm not sure if the % oxygen I determined is a measure of the oxygen density (if so I can convert the percentage to ppm).
My initial hypothesis is that as there are less oxygen molecules in hot air, water level will rise less since there is less oxygen reacting with the steel wool (compared to cold air). Hence lower % oxygen concentration. Vice versa for cold air.
Then later, I'm thinking that as oxygen molecules (or any gas particles) are heated, they move quicker so each oxygen molecule will occupy more volume. Vice versa if the gas particles are colder. However, since there are more oxygen molecules in cold air than warm air, the increase in number of oxygen molecules will occupy more volume and make up for the smaller volume each oxygen molecule occupy by slower movement.
Therefore, the % oxygen in air by volume I determined by the volume of rising water level over the volume of air, will theoretically be the same and hence it can't be used to determine the oxygen density in air.
But then I also saw a picture of a vernier O2 probe with percentage O2 reading (which im not sure if it's percent by volume or percent composition) that can be converted to concentration in ppm by a click of a button. I then did some research and found out percentage concentration (by mass or by volume) can be converted to ppm (which will be a measure of oxygen density).

No, the concentration does not depend on density in a well-mixed ideal gas.
The key principle here is that the atmosphere is close to being an ideal gas. In an ideal gas the components are well mixed and the composition will not change unless there is a chemical change that acts selectively on one of the components.
The ideal gas equation applies equally to each component of the mixture of gases that is air. If you warm a vessel of fixed size, the pressure goes up. If you warm the mixture in a variable volume, the volume increases. But nothing in either process causes the composition of the gas to change, so the concentrations of the components will remain the same.
You seem to be applying the gas equation to just the oxygen and ignoring the other components (Nitrogen, argon etc.). Yes the warmer air will have a lower density, but the composition will be unaltered. So the relative amount of oxygen in the air remains the same. Warm air has an overall lower density (which is why hot air balloons work) and the volume density of all the components will decrease. So the percentage of oxygen per litre will be lower but so will the percentage of Nitrogen. Mostly when measuring gas composition chemists talk about the concentration relative to all the components in the mixture not to the volume (and in an ideal gas, the volume/pressure/temperature relationship is the same for all components). So, trivially, the volume density of all components is lower when the gas is warmer at constant pressure. There is no selective change for just one component.

Below 90.19 K there is definitely an effect of temperature on the % oxygen in air, as it condensates out.
So as general as the question is stated in the header, the answer is 'yes'.

If we observe a closed air parcel so that no gas can go in or out, the amount-of-substance fraction $x$ of each component is given as
$$x_{\ce{N2}}=\frac{n_{\ce{N2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.7812=78.12\ \%$$
$$x_{\ce{O2}}=\frac{n_{\ce{O2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.2096=20.96\ \%$$
$$x_{\ce{Ar}}=\frac{n_{\ce{Ar}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=0.0092=0.92\ \%$$
Since molar mass is defined as
$$M=\frac mn$$
we can directly obtain the corresponding mass fraction $w$.
$$w_{\ce{N2}}=\frac{m_{\ce{N2}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{N2}}\cdot M_{\ce{N2}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.7557=75.57\ \%$$
$$w_{\ce{O2}}=\frac{m_{\ce{O2}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{O2}}\cdot M_{\ce{O2}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.2316=23.16\ \%$$
$$w_{\ce{Ar}}=\frac{m_{\ce{Ar}}}{m_{\ce{N2}}+m_{\ce{O2}}+m_{\ce{Ar}}}=\frac{n_{\ce{Ar}}\cdot M_{\ce{Ar}}}{n_{\ce{N2}}\cdot M_{\ce{N2}}+n_{\ce{O2}}\cdot M_{\ce{O2}}+n_{\ce{Ar}}\cdot M_{\ce{Ar}}}=0.0127=1.27\ \%$$
Assuming ideal gas behaviour, i.e.
$$pV=nRT$$
we can also calculate the corresponding volume fraction $\varphi$.
$$\begin{align}\varphi_{\ce{N2}}=\frac{V_{\ce{N2}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{N2}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt]
&=\frac{n_{\ce{N2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{N2}}=0.7812=78.12\ \%\end{align}$$
$$\begin{align}\varphi_{\ce{O2}}=\frac{V_{\ce{O2}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{O2}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt]
&=\frac{n_{\ce{O2}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{O2}}=0.2096=20.96\ \%\end{align}$$
$$\begin{align}\varphi_{\ce{Ar}}=\frac{V_{\ce{Ar}}}{V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}}&=\frac{n_{\ce{Ar}}\cdot\frac{RT}p}{n_{\ce{N2}}\cdot\frac{RT}p+n_{\ce{O2}}\cdot\frac{RT}p+n_{\ce{Ar}}\cdot\frac{RT}p}\\[6pt]
&=\frac{n_{\ce{Ar}}}{n_{\ce{N2}}+n_{\ce{O2}}+n_{\ce{Ar}}}=x_{\ce{N2}}=0.0092=0.92\ \%\end{align}$$
Note how $R$, $T$, and $p$ cancel since they are they same for all involved gases. Thus, volume fraction $\varphi$ is equal to amount-of-substance fraction $x$, but only for ideal gas.
Temperature $T$ and pressure $p$ of the air parcel may change, and the volumes $V_{\ce{N2}}$, $V_{\ce{O2}}$, $V_{\ce{Ar}}$, and $V_{\ce{N2}}+V_{\ce{O2}}+V_{\ce{Ar}}$ may change accordingly; however, the volume fraction $\varphi$ remains constant since the volumes are proportional to each other.

Example
The following values were obtained from the GERG-2008 mixture model, i.e. without using the ideal gas equation.
Assuming dry air with the following initial conditions:
$T_1=10.000\ \mathrm{^\circ C}$
$p_1=1.0000\ \mathrm{bar}$
$V_{\text{air},1}=1.0000\ \mathrm{m^3}$
$\rho_{\text{air},1}=1.2306\ \mathrm{kg\ m^{-3}}$
$w_{\ce{O2},1}=0.23160$
Thus, the mass of $\ce{O2}$ is
$$m_{\ce{O2}}=V_1\cdot\rho_{\text{air},1}\cdot w_{\ce{O2},1}=1.0000\ \mathrm{m^3}\times1.2306\ \mathrm{kg\ m^{-3}}\times0.23160=0.28501\ \mathrm{kg}$$
The density of $\ce{O2}$ at the initial temperature and pressure is
$\rho_{\ce{O2},1}=1.3603\ \mathrm{kg\ m^{-3}}$
Thus, the initial volume of the $\ce{O2}$ alone is
$$V_{\ce{O2},1}=\frac{m_{\ce{O2}}}{\rho_{\ce{O2},1}}=\frac{0.28501\ \mathrm{kg}}{1.3603\ \mathrm{kg\ m^{-3}}}=0.20952\ \mathrm{m^3}$$
Therefore, the initial volume fraction of $\ce{O2}$ is
$$\varphi_{\ce{O2},1}=\frac{V_{\ce{O2},1}}{V_{\text{air},1}}=\frac{0.20952\ \mathrm{m^3}}{1.0000\ \mathrm{m^3}}=0.20952=20.952\ \%$$
If we now heat the air to $30\ \mathrm{^\circ C}$ at a constant pressure, the volume expands, so that the new parameter values are as follows.
$T_2=30.000\ \mathrm{^\circ C}$
$p_2=1.0000\ \mathrm{bar}$
$V_{\text{air},2}=1.0708\ \mathrm{m^3}$
$\rho_{\text{air},2}=1.1492\ \mathrm{kg\ m^{-3}}$
The density of $\ce{O2}$ at the new temperature and pressure is
$\rho_{\ce{O2},2}=1.2703\ \mathrm{kg\ m^{-3}}$
Thus, the new volume of the $\ce{O2}$ alone is
$$V_{\ce{O2},2}=\frac{m_{\ce{O2}}}{\rho_{\ce{O2},2}}=\frac{0.28501\ \mathrm{kg}}{1.2703\ \mathrm{kg\ m^{-3}}}=0.22436\ \mathrm{m^3}$$
Therefore, the new volume fraction of $\ce{O2}$ is
$$\varphi_{\ce{O2},2}=\frac{V_{\ce{O2},2}}{V_{\text{air},2}}=\frac{0.22436\ \mathrm{m^3}}{1.0708\ \mathrm{m^3}}=0.20953=20.953\ \%$$
We can see that, although the volume increased by about $7\ \%$, the volume fraction of $\ce{O2}$ did not significantly change.


text:
Would cerium(III) in solution be oxidized to cerium(IV) in ESI mass spectroscopy?
During ESI (assuming positive ion mode for now), we apply a strong voltage ($1.0{-}\pu{5.0 kV}$) to induce ionization.
Therefore, it is common to see oxidation of metal ions (e.g. $\ce{Cu^{+I} -> Cu^{+II}}$).
I wonder if a $\ce{Ce^{+III}}$ solution would be observed as $\ce{Ce^{+III}}$ or as $\ce{Ce^{+IV}}$ due to in-source oxidation.
I feel like there should be a systematic study on this kind of effect but could not find any references for it. Any hints would be appreciated.

I remember I submitted a few $\ce{Ce^3+}$ complexes with oxoanions in DMSO and DMF for ESI-MS (4.5 kV) a while ago, as from what I remember there were no shards attributed to $\ce{Ce^4+}$ species. But I suspect it may depend on solvent and compound itself.

Systematic study maybe not. But you could get lucky in checking a few papers that report such solutions and the species therein by MS.

@Jan I tried that (google scholar), but couldn't really find anything.

It would be hard to have a systematic study as it strongly depends on the nature of the ligands / solvent / counteranions. Although not systematic, some studies have looked at several cations (http://doi.org/10.1016/S1387-3806(98)14044-7) and even shown that some reduction might occur due to the flow of electrons (http://doi.org/10.1016/S1044-0305(98)00100-7).

@PLD Nice references, thanks : ) Yeah, I figured that the coordinative surrounds can be very different. On the other hand, if the bare ions (e.g. weakly coordinating anion salts, let's say Ce(OTf)3) don't get oxidized, it will certainly not happen for stronger coordinating ligands. So a study of ESI oxidation in "bare metal ions" is certainly very useful for a broad range of applications.

I found one study that examined cerium(III) polyoxometalate salts and concluded that at least under the conditions of their study, $\ce{Ce(III)}$ did not oxidize under electrospray conditions.  The study is Bray et al. 2011 in the International Journal of Mass Spectrometry.
They say:

The presence of the redox active $\ce{Ce}$ metal ion introduces additional complicating effects as it would be difficult to differentiate such ions as $\ce{HCe(IV)P2Mo22O75^3−}$ from $\ce{H2Ce(III)P2Mo22O75^3−}$. However, successive degradation of $\ce{H_{g}Li_{2−g}Ce(III)P2Mo22O75^3−}$ to eventually form $\ce{HCe(III)PMo7O26−}$ yields an ion that can be directly compared to $\ce{LiCe(III)PMo7O26−}$. In particular, the two ions $\ce{HCe(III)PMo7O26−}$ and $\ce{LiCe(III)PMo7O26−}$, are separated by ∼6 m/z (Fig. 6), supporting the proposition of substitution of H+ by Li+ and, therefore, the identity of $\ce{LiCe(III)PMo7O26−}$. These results also aid in assigning the proton count on $\ce{H2Ce(III)P2Mo22O75^3−}$, $\ce{HCe(III)P2Mo18O62^2−}$, and $\ce{HCe(III)PMo13O44−}$, thereby indicating the trivalent oxidation state of cerium throughout. Although Ce(III) would not necessarily expected to be oxidized to Ce(IV) under the conditions of these experiments, confirming that cerium remains in the Ce(III) oxidation state from solution throughout the electrospray and CID processes is a useful methodology for oxidation state determination of redox active metal-complexes with f-block elements such as Ce(IV) and Pu(III/IV).

So they didn't expect Ce to be oxidized by electrospray in general, although the basis for this expectation wasn't clear.  They also report hard data showing that at least for polyoxometalate salts, no oxidation was apparent under electrospray conditions.
In general, the thousands of volts applied during the electrospray process does not mean that individual ions are exposed to fields of thousands of volts, because electrosprayed solutions are conductive and field strengths in solution are lowered because of Debye shielding.


text:
Oxide content description
From time to time, I come by to see what people do with my mhchem package. Here, I came across a notation that I did not know yet: $\ce{K2O-nSiO2-xH2O}$ and $\ce{K2O - 3.2SiO2 - 2.7H2O}$
What is this? Is this really a bond? Is this a short form for writing $\ce{K2O-Si_nO_{2n}-H_{2x}O_x}$ and $\ce{K2O - Si_{3.2}O_{6.4} - H_{5.4}O_{2.7}}$? Is this an established notation that mhchem should support? (It doesn't yet with the numbers.) Do you have authoritative References?

My gut feeling is that they should be hyphens.

Dots, if anything. $\ce{K2O\cdot nSiO2}$...

I don't think this is standard notation, and the linked question suggests that it is rather a recipe or composition of a mixture of salts rather than an actual compound formula.

They should be dots (i.e. \cdot, $\cdot$) rather than hyphens- they may or may not be bonded. See here for an explanation https://chemistry.stackexchange.com/questions/26699/what-is-the-meaning-of-the-dot-notation-in-chemical-formulas
See the formulas here https://en.wikipedia.org/wiki/Water_of_crystallization

@IvanNeretin Dots make even more sense. I had thought the manufacturer had at least gotten the type of symbols right so my brain was stuck in the distinction between different types of horizontal lines ^^'

The notation, upon replacing potassium with sodium, can refer to a water glass composition. The unique notation, in my opinion, is warranted based on the following comment from a paper1:

Despite of years of investigations, there are still questions about
the molecular species configuration of these solutions that need to be
answered to gain clarity over the best conditions for the various
applications
When adding sodium hydroxide ($\ce{NaOH}$) to a water glass, the ratio (molar
or weight) $\ce{SiO2/Na2O}$ is decreased. This ratio is called water glass
modulus (n) and determines various physical and chemical properties
such as the pH and the viscosity of the solution.
[...]
the viscosity is extremely dependent on the modulus and increases as
the solution becomes either more siliceous or more alkaline, i.e. at
both higher and lower $\ce{SiO2/Na2O}$ molar ratios (n). Since the viscosity
of such disperse systems like this is given by the silicate
conformation i.e. extended chain conformation and the degree of
polymerization
[...]
the degree of polymerization of the predominant silicate species in
the alkali-solution [...]

So we have apparently a weakly specified structure characterized by extended chain conformation and degrees of polymerization with its rather unique notation (seems right to me).
Reference

Helén Jansson, Diana Bernin, and Kerstin Ramser , "Silicate species of water glass and insights for alkali-activated green cement", AIP Advances 5, 067167 (2015), DOI: 10.1063/1.4923371


text:
Relationship of CO2 in water
We have an evaporative cooling system here at the power plant that uses city water.  The water is pumped up and cascades down across a filter media cooling the intake air to the gas turbine.  This is a recirculating system but a certain amount of the water is blown down continuously to a floor drain.  The water goes through the floor drain system to an oil-water separator then to a collection sump.  The water is then pumped out to the city sewer system.  There are pH monitors in the discharge to the city sewer.
When ever the evaporative cooling system is in operation we have a problem with the discharge pH meters going low on pH causing alarms.  It only happens when the pumps are off and the water is sitting stagnant in the wet loop where the pH meter probes are located.  Over time the pH will drop from around 7.5 down to a reading of around 6 which brings in the alarm.  If we recirculate the sump water through the line the pH will go back up to match the sump pH but once the pumps are turned off the pH will start to slowly drop again till the alarm comes back in.  Again, this only happens when the evaporative cooling systems are in service.
My question is could the drop of pH be caused by CO2 being dissolved into the water as the air is being draw across it in the filter media then coming out of solution and turning to carbonic acid as the water sits stagnant in the pipe where the pH probes are located?  It is definitely not an instrumentation error but we are at a loss to explain why this happens.
Thank you for your help.

You shouldn't get that much dissolved carbon dioxide if the water is stagnant... Could be dissolved ions where you're measuring.

The use of the word stagnant probably was not the best choice.  The pH probes sit in a drop loop that makes sure that the probes are always submerged in water.  When the pumps that discharge to the city sewer system are off the water sits in this pipe until the level in the sump/tank gets high enough for the pumps to turn on again.  It is during the periods of no flow when the pH drops.  I was thinking that the water might be picking up the CO2 as it is cascading down over the filter media in the evaporative cooling system as the air is drawn across it.


text:
HCH bond angle in cyclopropane and cyclobutane
I understand why the $\ce{CCC}$ bond angles in cyclobutane and cyclopropane are less than ideal $109.5^\circ$ due to the geometric restrictions they encounter in a cyclic structure. However, it is not clear to me how the $\ce{HCH}$ angles (which turn out to be greater than ideal) are affected by this constraint.

This earlier answer may be helpful.

Computational investigations of geometry, aromaticity and bond energies at HF/6-31G** theory level [1] revealed the following deformed  geometrical ($\alpha$) and interpath angles ($\beta$ -- the angle between two bond paths) for $\ce{H-C-H}$ bond, which are slightly deviating from the strain-free $109.5^\circ$ due to varying Baeyer strains:
\begin{array}{lr}
\hline
\text{molecule} & \alpha,~^\circ & \beta,~^\circ \\
\hline
\text{cyclopropane} & 114.1 & 117.0 \\
\text{planar cyclobutane} & 107.9 & 109.7 \\
\text{puckered cyclobutane} & 108.3 & 110.0 \\
\hline
\end{array}

P. S. Unfortunately, this data cannot be taken from single-crystal x-ray experiment as this method doesn't allow precise determination of H-positions, and I didn't find a neutron diffraction data on solid cycloalkanes mentioned.
Reference

Cremer, D.; Gauss, J. J. Am. Chem. Soc. 1986, 108 (24), 7467–7477 DOI: 10.1021/ja00284a004.

Thanks for your help. I'm wondering if there was a way to explain this in an introductory organic chemistry course level

@Diracc Well, for the introductory level the best would be to take off with Karl's answer; if you want to illustrate with the numbers, maybe you want to use $\beta$ values, as the paper claims they reflect the real geometry better. Also, maybe introduce Baeyer and Pitzer strains at some point.

I believe NMR together with qc modelling should allow to determine the bond angles quite accurately.

If one angle is smaller than the ideal, the others have more space to spread out. They won't spread out much, because the hydrogen atoms are small (not much repulsion) and rather happy at 109.5, which is also the inherent angle between bonds at an sp3 hybridised carbon atom.
The latter point changes a bit when the constrained bond angle is really far off, so the whole thing is no longer properly sp3 hybridised.

One may approach it from a hybridisation standpoint. Take cyclopropane for example: the C–C–C bond angle is 60°, as is required by an equilateral triangle.
If the AOs at each carbon were simply four sp3 hybrids, the ideal bond angle between them would be 109.5° - which creates a great deal of angle strain. Consequently, there is some redistribution of s- and p-orbital character between the C–C and C–H bonds. Like hybridisation, this is not a physical process whereby the carbon atom "moves" p- and s-orbitals around. All it means is that the hybridisation deviates from the ideal 4 × sp3 hybrid case, where each sp3 hybrid has exactly 75% p-character.
When the s-character is decreased and the p-character increased, then the ideal bond angle will go down (pure p orbitals would have an ideal bond angle of 90°). So, it makes sense that there is more p-character in the C–C bonds. However, it's also a zero-sum game: since we only have three p orbitals to use, if we put more p-character into the C–C bonds, then the C–H bonds have to have less. And since there is less p-character, one can predict that the H–C–H bond angle will be increased.
According to ron's earlier answer, it turns out that the C–H bonds in cyclopropane can be described using a sp2.46 hybrid (i.e. 71% p-character), and the C–C bonds with a sp3.74 hybrid (i.e. 79% p-character).
Now, cyclobutane is exactly the same as above, except that the deviation from ideality is smaller.
As an addendum: the hybridisation of the carbon orbital used to form the C–H bond can be related to the one-bond C–H coupling constant in NMR. The larger the s-character, the larger the coupling constant; and Gunther's NMR Spectroscopy (3rd ed., p 424) states that for hydrocarbons an empirical correlation has been found:
$$^{1}\!J_\ce{C-H} \approx 500\cdot s$$
where $s$ is the fractional s-character. In methane, the coupling constant is 125 Hz, which corresponds to exactly 25% s-character, as expected for an ideal sp3 hybrid. On the other hand, in cyclopropane, the coupling constant is 161 Hz. The correlation with the s-character calculated above isn't perfect, but the fact that it is larger is already indicative of the increased s-character. For cyclobutane, it is 136 Hz. (NMR data taken from Hans Reich's tables.)


text:
How deep does a counter electrode need to penetrate the solution in electrochemical experiments?
I ordered platinum-wire counter electrodes that are shorter than I was expecting. How much should it be submerged in my solution (right now it is only the tip that is submerged)? My experiment is hydrogen permeation using an H-cell with counter and reference electrodes in each side and my working electrode in the middle. My solution is 1 M NaOH. Should I order longer electrodes? How do I know if it is submerged enough?

TL;DR As long as you are not getting overloads on your instrument, it is submerged enough.
Longer answer:
Your potentiostat will increase the potential on your counter electrode until it achieves the desired potential across the working electrode(WE) reference electrode(RE) circuit. The exact potential applied on the counter will depend on a lot of factors like solution resistance, amount of current flowing on the working at the potential desired, material your counter is made of and finally, the area of the counter electrode.
If you end up with a situation where your potentiostat hits its electronic limits before the potential difference WE-RE is the desired value, you will start seeing overloads.
If you are curious, you can always put a voltmeter between your WE and your counter electrode to see how much of a potential your instrument applies.


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Replacement for Centrifuge in Calcium Content Determination
Basically, I'm performing a lab investigation to determine calcium content in different milk sorts. Since my school didn't have any EDTA, I'm using Sodium Oxalate to form a precipitate with Calcium Oxalate. However, the source I have says that the best way to make these two things react is by centrifuging, a tool which my school lacks. How could I do this without centrifuging for 5 minutes on 1500 RPM? Can I just wait or heat up increase the rate of reaction? Any tips?

Centrifuging won't make the reaction go, but rather isolate the precipitate at the bottom of the centrifuge tube. If you heat the milk then you are going to denature milk proteins  which will precipitate.

You can probably get pretty good settling by just waiting. But if you're trying to measure the mass of precipitate, it won't matter. Just filter, collect, and weigh.

Use a washing machine in dry mode 1200 RPM should be avaible :p

Why reopen a five years old question ? The user was in school in 2017. Does he or she still need our advice ?

If your "source" claimed that a centrifuge increased the reaction rate, then you need to find a better source. (I admit that it is possible that the compressive work a centrifuge does causes heating which for high speed centrifuges can be significant, but it isn't likely to be a factor here with small analytical amounts at relatively low speeds, <300 g's, imho) Without some calibration, you really need to find a method you can follow closely. If you have some knowns, say some calcium phosphate and some dietary calcium supplements, then you may be able to validate any deviations from the given method by showing that both highly insoluble salts and highly bioabsorbable calcium (usually calcium carbonate, acidified with HCl to the Ca++ ion) both can be analyzed with the modified method. (I also would guess that the centrifuge wouldn't "push down" other suspended (colloidal) material in the milk (which might either also contain calcium or improve the precipitation rate or amount of the calcium oxalate).)


text:
Propylene Glycol as a water main anti-freeze
We are a young student science group looking for solutions to real world problems. We are currently working on ways to keep our city water mains from freezing in the winter. Our team is researching adding propylene glycol to the city water supply in the winter when the weather forecast is really cold and could freeze older water mains (approximately -20 degrees F). Our older water mains are not deep enough to provide adequate protection against freezing and it is too costly to replace them all.

How much and what kind of food grade propylene glycol we need to add per gallon to keep the mains from freezing (or how to calculate it)? The city uses about 10,000 gallons of water per day.
When would we need to add it to keep the mains from freezing? We don't want to add it to the water if it's not needed.
How easy it is to dissolve and will it stay dissolved for at least 24 hours?
Will it interact with what the city already adds to the water supply, chlorine, polyphosphate and fluoride?

You'd have to add too much propylene glycol (50-50 mix with water) to protect to -20 F for this method to be practical. Also the public would go nuts with so much of a "contaminate" in the water supply. It would be a mess to flush out too. ref: http://www.ttequip.com/knowledgelibrary/Proper%20Use%20Of%20Inhibited%20Propylene%20Glycol.pdf

I think you might want to look at the pumping system, if you can keep the velocity high the water won't be able to freeze. You'll have lower water pressure though.

Picture this: Bob decides to boil some noodles. He fills a pan and places it on his stove top with the heat on high (since he is in a hurry). Then his phone rings and its the neighbor calling to say his dog has gotten loose. He runs out looking for his dog. While he is searching, the water has boiled away, leaving the propylene glycol heating. It bursts into flame, the kitchen and then the entire house is engulfed, his two year old daughter sleeping upstairs is burned to death.
Your "solution" is worse than the problem. The most obvious other problem with your idea is that for some odd reason (I can't explain it, because it is contrary to what your group must observe every day!!) you think water temperature and air temperature are directly related (or the same??). They are not. A simple search for "freezing point depression of propylene glycol" would give you plenty of information. Water use per day (in the USA) is about 80 to 100 gallons per person. That means your "city" has a population of 100. (????!!!!).
Other problems with your idea include washing your car - leaving a sticky residue which obscures your vision and leads to an increase in fatal traffic accidents. Killing your lawn (do you really think spraying PG onto your lawn won't have serious consequences.). And that doesn't even get to the various industrial uses of water which couldn't tolerate such adulteration.
Finally, PG is a food for bacteria. Do you really want to grow bacteria in your "water"? Or how about this, you wash your car, the water runs into the street. A car driving by attempts to stop for a stop sign, unfortunately you've lubricated the street with PG and a fatal traffic accident ensues. Below the Frost Line ground doesn't drop below 0°C. The deepest Frost Line in the Continental USA is 8 ft. It would be difficult to understand why your city's water pipes weren't placed well below this depth to begin with.

Thank you so much.  You must be so proud of yourself for insulting a bunch of children literally in grade school and their parents (ie their coaches who are not chemists (I have an electrical engineering degree not chemical engineering)) trying to find solutions for real world problems while teaching the kids about science. So, we posted the kids' ideas and what questions we were trying to answer for our project only to be belittled by you.  Pat yourself on the back.

Not to mention you clearly didn't read the entire post.  They were only looking to use PG DURING THE WINTER, not only that but only when it would be horribly cold!  So, no, we wouldn't be putting PG on the lawn or washing our vehicles when the outside temperature is below -10F.  Unless your pranking someone I guess in which case they wouldn't be able to use their vehicle for probably the rest of the winter.

We obtained our daily city usage information from the city workers who actually work for the city water works.  Not sure why you felt compelled to insult the children for that one!  And I'm not sure where you live, but our city is about 200 years old and just 3 years ago they pulled up some of the first water mains that were put in (not below the frost line mind you because 200 years they didn't get the memo) and those mains were made of .......wait for it.......wood.

So, I'm gonna go with they really didn't understand water distribution and what that all entails 200 years ago as we know it today.  But hey, you just had to insult our city's founders while you were insulting children.
Lastly, PG is in everything from drugs to shampoo.  Maybe you should run around and tell all the companies that use it that they are horrible for putting something that breeds bacteria in our medication.  I'm sure you will get award for that one.

You know, you could have just nicely explained what the errors were with the idea and maybe, i don't know, have helped us think it through instead of being a jerk.


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How do chemists measure purity in high purity metals and other reagents?
I have noticed that gold bars are often marked "999.9" fine:

Whenever I have metallurgical assays done, they always seem to do them with ISP or DSP or whatever, some spectrometer-based system. These systems have errors of 1% to 2%. When I complain about the high error, they say "it's good enough" and "nobody does the old fashioned stuff anymore". So, if ± 2% is the new standard for assayists, how are these banks determining that the gold is pure to a 9999 out of 10,000 standard?

I think XRF is the "gold" standard for this analysis, and can measure to better than 4-nines fine (999.9 = "4-nines fine").

related https://chemistry.stackexchange.com/questions/65180/obtaining-100-pure-metal

How about ICP-MS or ICP-AES?

@deusexmachina Why use destructive methods which require probe preparation when there is XRF? :)

@andselisk Right, XRF. Just didn't know it is also quantitative : )

I did XRF for years. For such a high purity (99.99%) you'd have to analyze for the impurities not the gold itself.

I think light spec is better for traces than XRF , although I could be obsolete.

As the comments alluded, to analyze purity at high purity, you analyze the impurities. I'll use XRF for this example, but ICP, MS, HPLC, or other methods could work in the same way.
With XRF's $1-2\%$ precision directly measuring the gold would give you a maximum supportable purity of $98-99 \%$. To verify a metal as $99.99\%$ the impurities are measured which are usually in the parts per million range, added up and subtracted from $\pu{100\%}$.
For example, let's say you have a gold bar with $\pu{35ppm}$ silver, $\pu{20ppm}$ copper, $\pu{10ppm}$ lead, and $\pu{5ppm}$ mercury with all other elements below a , $\pu{0.5ppm}$ detection limit (also called a threshold limit). The total impurity would be the total of the measured impurities and the detection limit of the unmeasured impurities. Note: most analysis is by metal basis which excludes the other elements that may be present.
$$\mathrm{Impurity = 35 + 20. + 10. + 5 + [0.5 \times (59-5)] = \pu{97 ppm} }$$
The $59$ is for the $59$ naturally occurring metal elements and the $5$ is for the five elements we measured ($\ce{Au, Ag, Cu, Pb, Hg}$). $\pu 2 \pu\%$ of $\pu{97ppm}$ is $\pu{1.94ppm}$ and the measurement plus the maximum error is $\pu{99\!.ppm}$. Taking $\pu{99\!.ppm} = \pu{0.0099\%}$ and subtracting from $\pu{100.00\%}$, we get an inferred purity of $\pu{99.991\%}$ Gold ($\mathrm{100\% - 0.01\% = 99.991\%}$). The gold would still be sold at $\pu{99.99\%}$ regardless of it being purer than needed as selling the gold by lots would be unnecessarily complicated and time intensive.


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How do reaction mechanisms go from proposed to approved?
Pretty much what it says in the title, I've just been wondering lately, what makes a mechanism be recognized by the scientific community as the most likely thing that happens on the molecular/atomic level when the corresponding chemical reaction takes place in real life?


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How to replicate compositions of discontinued paints?
I am a house painter, over the last 20 years I've noticed that discontinued lines of paint are much better. These paints were vinyl acrylic. Now the vinyl has been removed. I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula? I think the old ingredient was polyvinyl acetate.

I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula?

No.
Paints are a specific mixture of a number of components designed to fulfill their role well in this specific combination. If you go and add a random additional ingredient, which in itself is not a pure substance but a nondisclosed mixture, you cannot predict what will happen. At best, the additional ingredient does no good and no harm. At worst, you can throw away the entire can of paint.
Paint companies also do not disclose key data about their mixtures for the obvious reason that they don’t want to be copied. So even though it may say X, Y and Z on the back of the can, that doesn’t mean you can infer much about the mixture. Even if they give percentages, you don’t know what all the rest is. Therefore, you cannot duplicate the old formula unless you have access to the full blow of analytical methods and a sample of the old paint.
Finally, I’m not even sure if changing vinyl acrylic on the packaging to just acrylic has any implication on the content of the can. It could just be a new choice of description.

Paint formulations continue to evolve to humor the EPA. Basically is is against EPA rules to make older more effective paints. Long ago I had some chlorinated rubber paint , it was great for rusty steel ; been illegal for decades.


text:
Why does soapy dishwater lose soapiness over time?
I wash dishes by making a concentrated solution of dishsoap (eg dawn) and water.  Sometimes I'm lazy and forget to wash the container containing the soap solution, and it sits on the counter overnight.  In the morning, the solution always loses its "soapiness" - agitation no longer induces bubbles, and there is no characteristic slippery feel to the solution.  Why?  This happens regardless of the container material - glass, plastic, ceramic, or stainless steel.  Tagging oxidation because it's the only potential cause I can think of.  If so, how though?

Probably you deal with hard water, so that soap gets "neutralized" over time (M = Ca, Mg):$$\ce{2 C17H35COONa + M^2+ -> (C17H35COO)2M v + 2 Na+}$$ Maybe try to prepare the solution with boiled water instead.

Huh - if ion exchange was the mechanism, I would have expected it a lot quicker - Na+ in the Dawn would solve pretty fast (arguably just as fast as you dissolve the concentrate).  The resulting solution would be a mix of Ca,Mg, Na, and detergents - are you saying that aybe Mg/Ca might slowly bond more tightly and denature one of the detergents?
I will try both and get back to you.  I suppose I could also make a solution in an airtight container and see if it still happens - that would be an argument for the ion exchange.  thanks!


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Anthracine blue composition
In 1873 a German chemist named Springmühl announced a derivative of alizarin which acted as a blue dye and he sold this dye for a high price. The dye was sometimes called "anthracine blue" and Springmuhl kept the process for producing it a secret.
Has the composition of this dye ever been revealed?
From Chemical News, April 25, 1873:

Anthracene Blue.— A few years since aniline was the great source of new and beautiful colours. Now that every possible shade of colour, surpassing in number and beauty the hues of the rainbow, have been produced from aniline, the chemist has taken up the study of anthracene and alizarine, also coal-tar products. While preparing artificial alizarine from anthracene, Springmühl has obtained a by-product, from which he has made a beautiful blue colour, superior in some respects to any of the aniline blues. The process by which it was prepared he keeps a secret. Dried in vacuo, it is a blue powder with a few little crystals. In this it differs from the aniline dyes, which are one colour when dry, another when in solution. When pure hot water is poured over anthracene blue it mostly dissolves, but leaves a little insoluble residue. The addition of an alkali destroys its colour, which is restored, however, by an acid. The strongest mineral acids are unable to destroy its colour, but rather heightens its tone. Unlike aniline dyes, it is insoluble in alcohol and ether. Experiments show that it withstands the action of light better than aniline blue. Unfortunately, it is at present very expensive, for Springmühl obtained but 2'5 grains of anthracene blue from 25,000 grains of anthracene, which makes it cost about 3000 dols, per pound at present. A cheaper method of making it is certainly desirable.—American Artizan.

Springmühl indeed reported Anthracene Blue (orig. "Anthracenblau" (German)) from an intermediate with brutto-formula $\ce{C14H7O3}$. It seems like this method only worked well at a lab scale.
According to Springmühl's Lexicon der Farbwaaren und Chemikalienkunde [1, p. 111-112], Anthracene Blue is obtained as a byproduct of oxidation of anthracene to anthraquinone by fuming nitric acid. When gas (hydrogen in original text) evolution stops, resulting solution is neutralized with potash and heated under pressure in a sealed glass tube for 8 hours. Dark-green precipitation is then filtered off, and deep-blue solution containing Anthracene Blue is then treated with sodium chloride to precipitate the product in a form of blue water-soluble powder. But, treated with tannic acid, Anthracene Blue forms dark-blue precipitate and cannot be dissolved neither in water, nor in alcohol; but slightly soluble in acidified solutions.
Original German text:

Springmühl, der Entdecker des Anthracenblau, stellte dasselbe zuerst dar, indem er eine Verbindung von der Zusammensetzung $\ce{C14H7O3}$, welche er bei der Oxydation des Anthracens neben Anthrachinon erhielt, mit rauchender Salpetersäure behandelte und auf das erhaltene Product Wasserstoff in statu nascendi einwirken liess, die resultirende Lösung mit Kali versetzte und unter Druck erhitzte. Es bildete sich eine tiefblaue Färbung in der zugeschmolzenen Glasröhre und nach 8stündiger Erhitzung fand sich ein dunkelgrüner fast schwarzer Niederschlag in derselben, während die Lösung tiefblau gefärbt war und das Anthrancenblau enthielt, es liess der Farbstoff aus filtrirter Lösung sich mit Chlornatrium niederschlagen und mit Gerbsäure konnte der in Wasser wieder gelöste Farbstoff völlig als dunkelblaues Pulver gefällt werden, welches weder in Alkohol noch Wasser, hingegen leicht in Säuren sich löste.

Later synthetic process was changed to adapt it for industrial production. Beilstein catalogue from 1925 lists Anthracenblau-WR as 1,2,4,5,6,8-hexahydroxy-9,10-anthraquinone $\ce{C14H8O8}$ [2, p. 569], which can be obtained from 3-hydroxybenzoic acid with concentrated $\ce{H2SO4}$ in the presence of $\ce{H3BO3}$ at $\pu{260 .. 280^\circ C}$. The product is then poured into water and boiled. Alternatively, 3,5-dihydroxybenzoic acid can be used. There are also plenty other synthetic techniques slightly deviated from the mentioned above.
These days Anthracenblau is also known as commercially available dye Anthracene Blue (S)WR.

Reference

Springmühl, F. Lexicon der Farbwaaren und Chemikalienkunde: practishes Handbuch für Kenntiniss und Prüfung aller in der Industrie und im Gewerbebetrieb verwendeten Farbstoffe und Chemikalien; Leipzig: Gustav Weigel, 1877. (in German, Archive.org)
Beilsteins Handbuch der Organischen Chemie, 4th Ed.; Prager, B., Jacobson, P., Schmidt, P., Stern, D., Eds.; Springer: Berlin, 1925; Vol. 8. (in German, Archive.org)

Translated from your citation: Springmühl […] produced it first by subjecting a compound of the consitution $\ce{C14H7O3}$ […] with fuming nitric acid and subsequently letting nascent hydrogen act on the obtained product. Potash was added to the resulting solution and it was heated under pressure.

I find it highly unlikely that the sum formula remains the same after those transformations and the source also says that further details about it are not known.

The compound $\ce{C14H7O3}$ was obtained as a byproduct of the oxidation of anthracene to anthraquinone.

@Jan Oh, I see. Thank you very much, I screwed up with translation.

None of this seems to be telling us what the structure is. Does anyone know? wasn't that the actual question?

@matt_black Thanks for pointing this out, I added some info that I think should clarify the situation.


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Can a used-up buffer solution be used again in the other direction?
Suppose a strong base is added into an acidic buffer (such as acetic acid buffer) up to the endpoint. This means that the acetic acid molecules neutralize the hydroxide ions to the point that the solution contains mostly acetate ions (with their counter ions, of course), a negligible amount of acetic acid molecules due to Le Chatelier's principle and water.
My question is: can this resulting solution be immediately used as a basic buffer, since it contains a base with its conjugate acid?
Can the same buffer solution be used back and forth in the lab?

I am confused by the way you describe the use of buffer solutions. I usually know them being used to guard against excessive pH changes in a medium in which something else is being performed. Obviously, that something else would mean that you can’t reuse the buffer. But maybe I am grossly misunderstanding.

Yeah, I'm actually revisiting the basics of the subject of buffers (where you are just playing with buffers and acids\bases to see what happens...), so I thought about this situation. I guess it doesn't really come up in the lab, as you said. But perhaps as an exercise in a buffer teaching lab? Would the question be valid there?

A buffer is an equimolar mixture of a weak acid and it's conjugate weak base. Right? What you have after adding a lot of strong base is a spent buffer solution, which has practically no conjugate acid.

In principle, your suggestion works though you should recognize that the buffer is not active until you have added enough of the other component to bring it back to the buffer range. There's no reason in practice that anyone would want to do this, since it's much easier just to create a new buffer solution.


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Why is an energy cost associated with splitting the HOMO spins in singlet oxygen?
The molecular orbital schemes for both forms of singlet oxygen ($\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.

Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.
Hund’s rule sufficiently predicts that the triplet $\mathrm{^3\Sigma_g^-}$ state is the most stable, i.e. lowest in energy. When one speaks of singlet oxygen, one typically has the $\mathrm{^1\Delta_g}$ form in mind which is $\pu{0.98eV}$ higher in energy than triplet oxygen according to a group member’s recent literature seminar. This is often explained on this site by the fact that spin pairing puts another electron into the space that is already populated by one electron, causing electron-electron repulsion that must be overcome by adding energy.
Extending this explanation, I am inclined to think that the $\mathrm{^1\Sigma_g^+}$ form of singlet oxygen, whose MO scheme is depicted in the centre of figure 1, should have a lower energy than the $\mathrm{^1\Delta_g}$ form because the spins are occupying spacially different (and orthogonal) orbitals. However, the same seminar talk also included the energy difference of $\pu{0.65eV}$ between $\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$ with the $\mathrm{^1\Sigma_g^+}$ being higher in energy.
Why is this so?

I used to think of it in terms of Hund's second rule; higher angular momentum = lower energy. But I'm not sure about the validity of (1) the extension to molecules; and (2) the extension to comparing excited states. Technically, Hund's rules are only used to find the ground state. On a side note, I'd suggest actually putting the term symbols into the title; this is one rare case where it's really helpful to have MathJax in the title.

I need to clarify if I understood this correctly, You are asking: Why is $\mathrm{^1\Delta_g}$ lower in energy than $\mathrm{^1\Sigma_g^+}$?

@Martin-マーチン Yes.

@orthocresol you are correct. I think the extension to molecules and excited states is valid (with caveats, but that is just being technical)

@getafix - both the Mulliken and Herzberg references in my answer give examples where the ordering of excited states predicted through Hund's rules can be wrong. They're a good rule of thumb, but that implies exceptions can and do exist.

@Geoff Yeah that's what I meant when I said "with caveats"; I guess the latter half of that sentence sounds a bit flippant. But yeah, long story short, ortho was right in this case, and well your answer is excellent; I made the comment before reading through it.

Short story: The comment by orthocresol is correct, based on Hund's rules. Given the same spin multiplicity (i.e., the two singlet states), the state with higher orbital angular momentum ($\mathrm{^1\Delta_g}$) wins.
Loosely speaking, one can imagine that comparing the two singlet states, the higher angular momentum implies that the electrons are a bit further out, and suffer less electron-electron repulsion.
Longer story...
The "unified atom" model can apply to simple diatomics - that is, many of the same considerations for the electronic structure can be used for multi-electron atoms. There's discussion in both Mulliken (Rev. Mod. Phys. 1932, 4, 1. and Herzberg's Spectra of Diatomic Molecules).
The vibrational-electronic spectra of $\ce{O2}$ was determined fairly readily:
 (Figure from Chem. Rev., 2003, 103 (5), pp 1685–1758, adapted from Herzberg).
We can assign the lines as discussed by Mulliken in 1932:

The $\mathrm{^1\Sigma_g^+}$, is 1.62 volts above the $\mathrm{^3\Sigma_g^-}$ according to theoretical considerations,[118] the $\mathrm{^1\Delta_g}$, is probably about halfway between $\mathrm{^1\Sigma_g^+}$, and $\mathrm{^3\Sigma_g^-}$ (cf. Fig 48).

Reference 118 is to Hund and Hückel:

F. Hund, Zeits. f. Physik 63, 726 (1930).
E. Huckel, Zeits. f. Physik 60, 442-3 (1930).

Both Mulliken and Herzberg discuss the Hund's rules in regards to many diatomics - that the triplet will lie lower than the singlet because of higher S, but then the two singlet states will differ based on angular momentum (here L).

Incidentally, the $\mathrm{^1\Delta_g}$ state is doubly-degenerate, but that doesn't affect the ordering of the states.


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H-NMR spectroscopy of [18]annulene
Why does [18]annulene show only one peak in its proton NMR at $\pu{100^\circ C}$? Does any structural change happen such that all the hydrogens are alike?

I’m guessing the key point to solving this is the temperature.

Ya. My I'm stuck with my attempt because I'm not sure of what happens to the compound at 100°C. And I meant H not H'

Actually, this question is more interesting than it seems. I would prefer if this were not closed. Some references: J. Chem. Theory Comput. 2015, 11, 5083; also J. Am. Chem. Soc. 2000, 122, 722; also Pure Appl. Chem. 1971, 25, 573.

The answer obviously has to be yes since drawing the simple structure yields two sets of equivalent protons....

Chem. Commun. (London), 1966,0, 904-905 refers to an earlier work (which I haven't found) describing the temperature dependence of the NMR spectrum of [18]annulene due to rapid proton interchange at high temperature.

A good first exercise is to consider what the 3D structure of [18]annulene looks like.

@zhe Which is apparently not as trivial as it seems. DOI: 10.1039/C4OB01018C (Paper) Org. Biomol. Chem., 2014, 12, 6145-6150

@Martin-マーチン Yup, and I think that's why this problem is interesting like orthocresol says. :)

And I too would oppose closure.  The question seems to invite discussion of what aromaticity is, vs. what it not always is.

The aromaticity of [18]-annulene1 was confirmed by room temperature 1H NMR (60 MHz) in 1964.2 [τ values have been converted to δ values; δ = 10 - τ] When the spectrum was recorded at room temperature the 12 deshielded, exterior hydrogens appeared at δ 8.9 while the 6-interior hydrogens were highly shielded (δ -1.8). Evidence for flexiblity in the ring at room temperature was confirmed by observing a further downfield shift of the 12 hydrogens (δ 9.28) and an upfield shift of the 6 hydrogen signal (δ -2.99) when the spectrum was recorded at -60oC. Recording the spectrum at 110oC caused the two signals to coalesce to a singlet peak at δ 5.45. This value is close to the weighted average of the resolved chemical shifts. Ostensibly, the coalesence is due to the interchange of external and internal hydrogens. When the sample is cooled, the resolved spectra are restored.

1) Sondheimer, F., Wolovsky, R., Amiel, Y. J. Am. Chem. Soc., 1962, 84, 274.
2) Gaoni, Y., Melera, A., Sondheimer, F., Wolovsky, R. Proc. Chem. Soc., 1964, 397.

Very roughly, we would expect aromaticity to do two relevant things:

It tends to make the ring rigidly planar, but...
Its energetic effect decreases with increasing ring size.

With $18$ atoms in the ring the energetic effect is decreased to the point where the ring isn't really all that rigid. It becomes fluxional on warming.
This is one of many examples of aromaticity not always having all the effects we're accustomed to seeing in benzene rings.

So as temperature increases, the molecule becomes more destabilized?

In a sense all molecules become less stable when heated, they vibrate more from their most stable confirmations.  In [18]-annular one such vibration mode  scrambles the hydrogens.


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Why does a Diels–Alder reaction not occur here?
I have  recently come across this organic chemistry problem.
I thought it would undergo simple Diels–Alder reaction
But it didn't turn out to be. The product to me looks like a nucleophilic attack of (I) on (II).
Can anybody say me why exactly this abnormality seen and what is the mechanism of this reaction?

You're not going to achieve a s-cis conformation for a [4+2] cycloaddition. The dithiide protecting group is way too bulky to be in plane with the vinyl ether. This is a conjugate addition into the enone, followed by aromatization of the ring, and addition of the phenol moiety into the oxocarbenium.

I thought about the bulkiness of the dithiane group but I couldn't get any further ideas. Can you please put up the solution.

You should have enough hints from my first comment. Please think about the requirements for a [4+2] cycloaddition and the implications thereof. Build models if necessary.

I suggest editing the title. As it stands now, it is misleading. See, there is a thing called inverse electronic demand D.-A., and that's an abnormal Diels-Alder reaction for sure. Yours is not a Diels-Alder at all.

@IvanNeretin an inverse electron demand DA is not abnormal. Diene + dienophile leading to a cyclohexene derivative is a DA reaction, doesn't matter what's the mechanism.

Have you thought about the HOMO-LUMO of either of these two species? It may be that the HOMO and LUMO are not compatible since the signs are opposite.

I guess because they are not coplanar

Firstly, Why doesn't Diels-Alder reaction happen here?
On first sight the reaction seem completely feasible as both the diene and the dienophile are good species that could undergo Diels-Alder, but no. It could be proven by just analyzing the products (the actual one and the expected one).

Resonance. The property that literally defines organic stability. The DA product has not got even a trace of the stabilization that the actual product receives. The actual product gets to experience extended resonance due to the benzene ring that is formed.
The second factor being that the actual product has got TWO (yes...two) aromatic rings fused together, each having 6 electron resonance.
Steric hindrance. The DA product experiences Steric hindrance while the reaction happens due to the bulky dithiane group. While the other path is free from this.
Strain. Consider the carbon that is holding the dithiane group in both the products. In the DA product the carbon is $sp_3$ while the other product has $sp_2$ (perfect 120 degrees required, and 120 degrees received). In fact, all carbons in the actual product are $sp_2$, other than the ones in the dithiane group obviously.

All of these factors encourage the reaction to proceed in the second path which leads to an amazing yield of 62%.
P.S. I kinda feel stupid after typing "actual product" like 10 times.

Now, the MECHANISM (ohh yeahh!)
Let's start with para-benzoquinone. It is an $\alpha-\beta ketone$. Ergo, the $\beta$ hydrogen is acidic. So, lets prepare it for the incoming nucleophile.

umm...something's missing. Yeah!

OK. Looks better.
Now that we are done with this, we will get to the other reactant. Observe that the methoxy group on the $sp_2$ makes it a good nucleophile. But there is a problem, the dithiane group is too bulky to be drawn again and again. Therefore,

Properties of 'G' :
*bulky, must be kept away
*observing the product, where it is intact, we can say that it is useless for the reaction.
Coming back to the reaction, this nucleophile attacks our first molecule.

Here the (*) marked carbon is activated due to the methoxy group, hence the attack is at that site.
Now the carbocation can stabilized using proton transfer onto the oxygen.

The encouraging factor for this is the achievement of an aromatic ring in the molecule.
Observe the (*) marked carbon in the above image. We see that it is strained due to the thiane group present on + the methoxy group. Therefore the methoxy could abstract the proton from the adjacent carbon and exit as methanol.

Yet again, another encouraging factor for this exit is the formation of another aromatic ring.
Replacing the 'G" with it's true form we have...(drum roll)

Resonance is only an argument if all potential products are in full thermodynamic equilibrium and can convert back and forth which should not be the case here. 2) I am really not sure whether you should count the benzofurane system as two aromatic rings; it’s not like there are two benzene systems being generated. 3) Angle strain shouldn’t play any significant role in any of the products or intermediates. 4) I believe your mechanistic steps to be … let’s say suboptimal.

Resonance stabilisation could be used as an arguement between two reaction paths (right?). As the product will be more thermodynamically stable hence easier to be formed. 2) Benzofurane system can be considered as two separate aromatic rings due to the Sacrificial Aromatic Character displayed by them (as seen between Napthalene and Benzene). 3) I strongly believe that angle strain should play a role. If the system is too strained there would be a higher chance of reversion of the product into the parent molecules. 4) Could you please be more specific about the steps you find "suboptimal".

Only if there is indeed a conflict of pathways. In OP’s example, the DA pathway fails long before the question of resonance or not can even be considered, it simply does not happen so it is not useful to compare product stabilities. 2) Eh, more like 1.5 aromatic systems but that doesn’t have much bearing on the answer. 3) There is no intermediate in this pathway (or the impossible DA pathway) where any carbon atom is susceptible to any angle strain so it is irrelevant to this question. 4) The cyclisation is most certainly not concerted, …

The proton transfer will most likely happen before cyclisation, and methanol elimination will probably follow a different order of steps to name just the most striking.

All of your comments make so much sense!! Anyways, I think there is a serious defect in the way things are happening in the mechanism I proposed. Thanks for pointing it out (still_a_kid...got to learn a lot :) ).