text:
Fractal patterns on water
I stored water in a bucket (of aluminium probably), and some random fractal-like patterns are formed on the water:

See here for some more pictures.
Why this happened?
I'm unable to reproduce it. How can I reproduce it ?

Cross posted from physics SE (https://physics.stackexchange.com/questions/352377/fractal-patterns-on-water)

The carbonates precipitate that appears after boiling due to water hardness would often form on the surface (instead of bottom) and look just like that. Others sources seem less probable, though not impossible. As to its shape, it is explained in the answer on Phys.SE.


text:
pH of dilute acids in hexane
Consider a strong acid $\ce{HA}$. In a $\pu{1000 mL}$ of hexane (assume it is unable to accept a proton) we "dissolve" (this can mean forming a suspension with vigorous stirring - otherwise we can use a polar solvent which cannot accept a proton) $\pu{1e-30 mol}$ of the acid along with enough water to remove all the protons.
Now we have extremely dilute solution of $\ce{H+}$ where autoprotolysis is insignificant compared to the volume of solvent, so according to $\mathrm{pH} = -\log[\ce{H+}]$, pH should be extremely high. However, we are still using an acid, so pH should not be too high. What accounts for the difference?

Interesting. pKa values are given for alkanes (generally pKa >60) and other non-water compounds. pKa is the pH where half of it exists protonated/unprotonated /deprotonated, so this would suggest pH can be measured in non-aqueous solutions.

https://chemistry.stackexchange.com/questions/72/determining-the-ph-of-an-organic-solvent

pH is defined for water solutions; using it elsewhere is either impossible or leads to contradictory results like yours. Who said pH should not be too high? Well, you might argue that high pH implies alkaline solution, does it not? No, it doesn't. Alkaline solutions are (roughly speaking) those with high concentration of $\ce{OH-}$. In water this means the same thing, because $\ce{[H+]\cdot[OH-]=10^{-14}}$, so low $\ce{[H+]}$ means high $\ce{[OH-]}$ and vice versa. In hexane, not really.
To determine pKa values which are way beyond the physically possible range of pH, we extend it with things like Hammet's $H_0$, and thus build a ladder of measures suitable for a much wider range. You may call it "effective pH" or like that, but remember: it has little to do with literal concentration of $\ce{H+}$, and your hypothetical solution of $\ce{10^{-30}M\; HCl}$ in hexane is in fact neither acidic nor basic.

Well, I think the pH (activity) can be theoretically applied to non-aqueous solutions (http://onlinelibrary.wiley.com/doi/10.1002/chem.201003164/abstract;jsessionid=6237E1A09F29EE5D07AA17B2658C677B.f03t03), but it isn't very convenient.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Can 4-aminophenol be used to produce 2,5-cyclohexadien-1-imine?
If the hydroxl group on 4-aminophenol was protonated by a strong acid in aqueous solution, and the compound further to that dehydrated, could a rearrangement of the subsequent carbocation lead to formation of 2,5-cyclohexadien-1-imine and thus regenertion of the strong acid before SN1 was possible?
Is there a more common name for this sort of reaction (assuming it is possible)?

I'm afraid you didn't notice that there would be reduction needed there.

And it wouldn't work because hydride donor can't reduce in highly acidic solution.

TL;DR
No
Explanation
As far as I understand, the reaction being suggested is as follows:

The first thing one should notice is that 2,5-Cyclohexadien-1-imine is not going to stay as it is. It will tautomerise to the more stable and aromatic aniline.

Let us go back to the starting material - 4-aminophenol. Both the oxygen and the nitrogen are directly attached to the aromatic $\pi$-system. Both can donate a lone pair and cause a +R effect. Their electronegativities would result in a -I effect, which is very small, compared to the +R effects. The point is, the starting material is electron-rich, i.e. nucleophilic. It would not leave a carbocation (if it does form) alone.
Also, the thing about the lack of a hydride donor...

Thanks for your answer and comments! But what was that about the lack of a hydride donor??

See the difference in molecular formulae, for instance. The difference is not "$\ce{H_2O}$", but is "$\ce{OH}$". There needs to be something that provides that $\ce{H}$

Ah yes I see I may have overlooked that, my bad. Thanks anyway, an interesting discussion nonetheless.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Dilution of whisky – the molecular perspective
Not a chemist. I'm trying to understand this abstract. In particular:

We found that guaiacol is preferentially associated with ethanol, and, therefore, primarily found at the liquid-air interface in mixtures that contain up to 45 vol-% of ethanol.

So guaiacol is amphipathic which means that is a largish molecule in which one end is hydrophilic and the other hydrophobic (or lilophilic). At least that's my understanding from Wikipedia. So, how did they arrive at the "therefore primarily found at the liquid-air interface" part? At the surface the molecule attaches one end to the alcohol and one end to the air?
Sorry for any naive interpretation of chemistry on my part.
Edit:
Just run across this article discussing the matter.

Their meaning is that guaiacol is somewhat " extracted" from the bulk of whisky by the diluted, water richer layer formed upon dilution. Guaiacol moves towards the interface not because of air but because of the water. As you first sip and taste that layer, this is beneficial to the taste. Until here, it sounds nice. They also suggest that a prior dilution would affect the entire bottle, but I do not get the reasoning for this latter statement

Is the assumption that water is distributed preferentially on the surface, not throughout? Isn't water heavier than alcohol such that it would sink, even if added to the surface?

Yes, but not if you consider a glass situation. Especially with ice cube floating :) A water rich layer is surely atop. Again, I do not see the reasoning for the other conclusion...

It occurs to me that given a finite amount whisky in a glass and thus a finite amount of guaiacol, that the first half of a drink will have more guaiacol (and hence more flavor) and the second half of the drink will have less guaiacol (and hence less flavor) than if the guaiacol were evenly distributed.  So I guess you just pour the second half down the drain?

I agree with your point. But surely the tasting happens at the first sip. If we have to take this seriously, the first sip might be anvenhanced gustative experience, but nevertheless the rest of the whisky is not totally depleted on guaiacol. Less seriously, after three on the rocks nobody cares :)

When you put ice cubes in whisky, any discussion about flavours is rather pointless.  Otherwise: The density is irrelevant, molecules like guaiacol, which are not well soluble, do tend to move to the phase boundary, because that minimises the interfacial tension.

@Karl. I was expecting a purist not getting the point. The density itself is irrelevant unfortunately without gradient of it, the guaiacol concentration at the interface would be the same as in the ou- of- the- bottle whiskey or actually less. With this I quit this discussion as it is getting too much. Authors should be happy to have passed a Nature out of it, for what my thinking is concerned.

@Alchimista, if you like, you could post an answer based on your comments

What the paper says is that in that range of concentrations, that the excess of alcohol at the air-liquid interface is enough to extract some of the guiacol from the bulk solution. At higher bulk concentrations of ethanol, there's no particular benefit for the guiacol to separate to the interface (there's enough ethanol in the bulk to keep the guiacol satisfied, so to speak). I forget whether the paper determined if there was a concentration (proof) below which the guiacol did not segregate, but if there was, it would be because the slight increase in alcohol concentration at the interface wasn't attractive enough. In other words the guiacol only is preferentially concentrated at the interface when the difference between the concentrations of alcohol at the interface to the concentration in the bulk makes it worth-while for it. (Worthwhile here means the difference in energy is large enough). Here's an analogy (My family would now roll their eyes, my analogies are lousy). A little kid is deciding whether to get up from his video game and go trick-or-treating. There's no candy in his house, and maybe he'll get a half dozen or a dozen pieces if he hits 20 or 30 houses/doors. Nah, he says, not worth it. But what if he's in a rich neighborhood? If he hits 10 doors, he's sure to get at least 2 dozen, maybe 3 or 4 dozen pieces of candy. Well, if his home hasn't much candy then he'll probably go, but if his home is chock-full of candy (rich neighborhood, remember?) then he may decide it's not worth it either. The choice he makes depends not just on the difference in the absolute amount of candy he will end up with, but also depends on how much he values an extra couple of pieces. It's non-linear, or as Liberals might say "context sensitive". (added: just to be clear, going trick-or-treating is analogous to concentrating at the air-liquid interface, not going is analogous to staying in the bulk phase.)

Yes we knew it. Could you help me to understand why dilution (a bit more water than what is currently added) before bottling should be beneficial? It is a suggestion of the paper that I do not understand


text:
Inductive vs hyperconjuative effects in electrophilic aromatic substitution
$\ce{C6H5C(CH3)3}$ is attacked faster than $\ce{C6H5CH3}$  in nitration, presumably because of the greater +I effect. However, for chlorination $\ce{C6H5CH3}$ is attacked faster i.e hyperconjugation takes over. Why?
One explanation given in my textbook is that chlorination is a less polar reaction. But I don't understand how that affects the relative importance of hyperconjuative/inductive effect.


text:
Why is fluorine deactivating while OH is not?
Both have 2p-2p overlap and oxygen and fluorine have similar electronegativities but OH is strongly activating while fluorine is deactivating

What does the oxygen have attached that the fluorine doesn't, start from there.

How could only the presence of a hydrogen bond, which slightly decreases OH's inductive effect completely reverse the trend from deactivating to strongly activating?

Think of another interaction that the hydrogen might be involved with, look at the resonance structures as well.

F is the most electronegative element of the periodic table.Due to its -I effect, it shifts the electron density of any alkyl group towards itself, resulting in less density of electron cloud over the alkyl or phenyl group.This deactivates the ring for electrophilic substitution reactions.However , oxygen also being an electronegative element , has capacity to donate its lone pair electrons forming a back bond in case their is a deficiency in alkyl group.In case of cyclic rings that is in aromatic structures , the lone of oxygen is in  resonance with the ring during attack of the reagent , that is during formation of sigma complex.However this is not shown by fluorine due to its high electronegativity.Do not get confused with other halogens.Cl ,Br and I are ring activating groups as they also can form the same back bond as oxygen.


text:
How to calculate the number of skeletal electrons
The number of skeletal electrons in CH is 3 and the number of skeletal electrons in BH is 2. How

Why not? Carbon has one more electron than boron. I assume you're asking about this in the context of Wade's rules?


text:
When two containers at different T and P are kept together, will pressure become same or temperature?
Two containers, kept at different temperatures are connected through a stopcock and two different ideal gases at different pressures are filled inside them. Initially the stopcock is closed. When the stopcock is opened, will the pressures become same or will thermal equilibrium be established first? Why?
Here's an extract from an online source about thermal equilibrium-"As we have seen in the zeroth law of thermodynamics, when two objects are placed in contact heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer."
We are taught in class that after the stopcock is opened, gaseous molecules travel from container at higher pressure to the container with lower pressure. There wasn't any discussion on thermal equilibrium. If the temperature of two gases kept at different temperatures don't become equal, isn't that a violation of zeroth law?
When the stopcock is opened, the gas molecules at different temperatures do come in contact. So, why thermal equilibrium is not established and instead there is a greater driving force to make pressure in both the containers same?

Of course thermal equilibrium will be established, otherwise that would be a violation of the said law.

In your class, have they now explained to your satisfaction why the final temperatures of the two chambers are not equal?

The situation that is being covered in your class is where the two containers are thermally insulated from one another.  In this case, the only place where heat can be transferred from one container to the other is through the gas in the valve (by conduction).  But the assumption is being made in your class problem that heat conduction through the gas in the valve is negligible.  In any event, even if the valve were left open until the pressures effectively equilibrated, the temperatures in the two containers would require a much larger time to equilibrate.  So, if the assumption is made that, once the pressures effectively equilibrate, the valve is closed again (and does not allow heat transfer between the containers), then at final thermodynamic equilibrium, the temperatures of the gas in the two containers will be different.

Do you have a problem with my answer?

Thermodynamics is only concerned with equilibrium situations and tell us nothing about how quickly things happen, so all that thermodynamics can say is that the pressure will be equalised and thermal equilibrium will be established.
We know from experience that the pressure will be equalised very rapidly as one gas flows into the other (think of gas expelled on opening a cola bottle) and with it some energy will be transferred. At the far end of one container we can imagine that there will still be some hot molecules and some cold ones at the end of the other, thermal diffusion, which is a slow process, will then eventually equalise the temperature.
(Assuming that the whole assembly is thermally isolated from the surroundings and that heat transfer is only through the gas).


text:
Predicting sigma bond overlap strengths of s-s, p-p, s-p, sp-sp etc
How can we qualitatively predict $\sigma-$bond strengths of overlap between:

$\ce{s-s}$
$\ce{s-p}$
$\ce{p-p}$
$\ce{sp-s}$
$\ce{sp-p}$
$\ce{sp-sp}$

etc.?
My school-book says $\ce{s-s}$ overlap bond strength is greater than $\ce{s-p}$ overlap which is greater than $\ce{p-p}$
Another book Inorganic Chemistry by JD Lee states this:

Are the facts stated in the above books correct or justified? Is there any order of orbital overlap strengths? If so can we predict it qualitatively?

I think the issue might be what orbitals are being referred to. Both books could be correct if they are referring to different $\mathrm{p-p}$ overlap. For example, suppose two atoms are bonding along the z-axis. $\mathrm{p_z}-\mathrm{p_z}$ overlap would be greater than $\mathrm{s-s}$ because the lobes of the p orbitals extend out to meet each other between the two molecules. If however we looked at $\mathrm{p_x}-\mathrm{p_x}$ overlap, the $\mathrm{s-s}$ overlap would be greater because the p lobes don't extend toward each other. Increasing hybridization increases overlap because it combines the directionality of p orbitals (that they extend out away from the atom) and the electronic density of s orbitals (larger space for overlap to occur).

The schoolbook asked about p orbitals on internuclear axis only! I feel schoolbook may be wrong because as you said p orbitals are more directional and should form stronger bonds..

@samjoe Well then for now, I would say trust the Inorganic chemistry book and that hybridization increases overlap. I'm not certain what the other textbook is referring to.

@Tyberius According to this reasoning, shouldn't a p-p overlap be stronger than all of the overlap of hybridised orbitals?

@Taufeeque I believe these measures are for the orbitals overlapping head on. The OP mentions the reasoning that the hybrids have larger directed lobes and so should have stronger bonding.

@Tyberius I am unable to visualize how $\ce{pz}$ orbitals extend their lobes for bonding while $\ce{px}$ does not. I believe that the z-axis has been taken as the molecular axis by convention, but I can't really see how that helps. Also, if we are increasing the p- character in our hybridization(as an example, let it be $\ce{sp3}$), then isn't it possible that the overlap becomes weaker along any given axis(say y-axis) as compared to a bond formed by a pure $\ce{py}$ orbital, as percent y character in the former is surely less than a pure $\ce{py}$ orbital?

@YUSUFHASAN what I was basically trying to convey is that internuclear bonding of p-orbitals (a sigma bond) should be stronger than noninternuclear bonding (like a pi bond). For the hybrids, we can always reorient the atoms such that the largest lobes are facing each other. So it two atoms went from sp to sp2 hybridization, we could always reorient them so their lobes are along their internuclear axis.

@Tyberius Again, didn't get you on the second part there. The sigma overlap of two sp2 hybridized atoms will always take place on the molecular axis itself, so where would the reorientation occur? Also,(unless you are somehow involving them in some kind of chemical reaction) how can atoms simply "go" from one hybridization to another?

@YUSUFHASAN so is your confusion why a hybridized orbital overlaps better than a pure p? It's really just a matter of  hybridized orbitals having their density focused in one lobe, rather than equal lobes in the bonding and opposite direction. You could think of a chemical reaction changing the hybridization, but we can be more abstract and just think of the hybridization changing. Hybridization is just a mathematical construct and we can always choose an arbitrary mixing of the orbitals; We just happen to know that certain mixings due a better job at describing bonding.

Although the general trend $\mathrm{s} < \mathrm{p} < \mathrm{sp}^n$ makes sense, these magic numbers $1.73$, $1.93$, $1.99$, and $2.00$ seem to have just been pulled out of a hat. If these are really "approximate strengths of bonds", a good book would justify these by showing which bonds they use to come up with these numbers.
Just to show one counterexample, it is commonly known that the strength of C–H bonds increase going from alkanes to alkenes to alkynes. That is to say, the bond strengths increase in the order $\ce{C_{sp^3}-H} < \ce{C_{sp^2}-H} < \ce{C_{sp}-H}$. This clearly contradicts the supposed trend given in the book of $\mathrm{sp}$ orbitals forming slightly weaker bonds than $\mathrm{sp^3}$.
From the second table in this Wikipedia section:

Compound
Carbon hybridisation
C–H bond strength (kcal/mol)

Ethane
sp3
101

Ethylene
sp2
111

Benzene
sp2
113

Acetylene
sp
133

As far as I am aware, this trend is mainly rationalised using bond lengths.

Are the facts stated in the above books correct or justified?

Some parts of it (e.g. the general trend of $\mathrm{s} < \mathrm{p} < \mathrm{sp}^n$) might be correct, as already well addressed by Tyberius' answer, but other parts (especially these magic numbers) are highly debatable.

Also, I think it will depend on principal quantum number $n$ of inter-mixing orbitals.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How does activated carbon trap impurities?
I've tried and I've tried to find an answer to this all over the web, but I can't seem to find an in depth explanation.
To the best of my knowledge, activated carbon has a very microporous structure and uses a process called adsorbption to "trap" particles. This isn't very in depth though and I'd like to better understand the process of what's going from a scientific standpoint. How does the carbon/charcoal keep the adsorbed material attached or trapped within the carbon?

I don't have time right now for a proper answer, but the keywords you're looking for are chemisorption and physisorption.

@Todd Minehardt cheers for this, I'd be grateful though for a proper answer at your nearest convenience.

Never forget to interpret in terms of physical/chemical forces, by order of energy: van der Waals, hydrophobic (pseudo forces), hydrogen bonds, disulfide bridges, covalent, electrostatic (for the most frequent).
Indeed, in charcoal, there is a large number of $\pi$-electrons that can make bonds, in particular. And yes, the porosity is important.
But perhaps could you have a look at Boehm H. P. Some aspects of the surface of carbon blacks and other carbons, Carbon, 1994, 32(5), 759-769. DOI 10.1016/0008-6223(94)90031-0.

any chance you could transcribe a full answer?

@Charlie. In addition to the phisisorption and chemisorption, it might help you if you consider that there is not in general selectivity. Activated charcoal is very good to capture impurities by virtue of the fact that impurities are in little/trace amounts.

Then very quick, consider the surface of your charcoal. Consider all the possible interactions listed by user50565. The impurities you want to remove have a high chance to get immobilised at the huge surface. Your impurities basically stain the charcoal, to give you an idea.

A spoon of activated charcoal might have an effective surface of few soccer fields !

So do the impurities react chemically with the charcoal or are they physically trapped within the pores?

Let us continue this discussion in chat.

Now I am in app.... Well say interact rather than absolutely react. In this sense user59565 gave you a list of possible "bonds". I wanted to stress that charcoal doesn't necessarily block impurities more selectively than the substance(s) you want to purify. Being the impurities already in very small amount (but enough to  impart colour, or prevents crystallisation, for instance), they are still effectively removed (along with some negligible amount of the target substance(s).

@Alchimista I see. The reason I am asking this question is I'm particularly interested in charcoal as a teeth whitener at the moment, and wanted to know more about how it may work. So based on what you said, is it likely that using activated carbon as a teeth whitener actually removes enamel (hydroxylapatite) as well as the stains?

@andselisk nice edit!

Not really a dentist here. Therefore I cannot say if it is good or not for your teeth, and body. I do not think that charcoal is a good whitener as for the stains you want to remove are not diving around but already sticking to the enamel... It is as replacing baking soda ( with something presumably even less hard than the salty grains). You basically relies on abrasive properties and not to the - sorption processes mentioned before. I would rather stay with sodium bicarbonate, perhaps not daily.


text:
Why aren't chaperones considered catalysts?
I'm reading about protein folding on Wikipedia and I stumbled on a bit about a class of proteins called chaperones that aid in the folding of proteins by:

...reducing possible unwanted aggregations of the polypeptide chain that might otherwise slow down the search for the proper intermediate...

However, they aren't considered to be catalysts because they do not actually speed up the folding process. My thought is that without these chaperones, the polypeptides chain would go through a long process of finding a proper intermediate and therefore should be considered catalysts.
Where am I going wrong in my thinking?

Without these chaperones, the polypeptide chain would end up folded in a wrong way or not folded at all. Then again, who said they are not considered to be catalysts?

The usual definition of catalyst is that it has to take part in the reaction. The reaction is usually of zeroth order kinetic, regarding the catalyst. Here it sounds like those chaperones are inhibiting an unwanted side reaction. That could possibly lead to catalyst-like kinetics. Any way, catalyst is just a name.

I don’t think you are wrong. Actually molecular chaperones have been named enzymes in some texts. The molecular chaperones comprise several unrelated classes of proteins that have rather different functions.

Most molecular chaperones are ATPases (enzymes that catalyze ATP
  hydrolysis), which bind to unfolded polypeptides and apply the free
  energy of ATP hydrolysis to effect their release in a favourable
  manner.

It is also made evident in the way some of them function (they possess some ATPase activity), it suggests they’re kind of  catalysts  enzymes:

They function in an ATP-driven process to reverse the denaturation and
  aggregation of proteins (processes that are accelerated at elevated
  temperatures), to facilitate the proper folding of newly synthesized
  polypeptides as they emerge from the ribosome, to unfold proteins in
  preparation for their transport through membranes.

Two classes of molecular chaperones have been well studied. Both are found in organisms ranging from bacteria to humans.

The heat shock proteins 70 (Hsp70)

The first class is a family of proteins called Hsp70. They function in an ATP-driven process to reverse the denaturation and aggregation of proteins (processes that are accelerated at elevated temperatures), to facilitate the proper folding of newly synthesized polypeptides as they emerge from the ribosome, to unfold proteins in preparation for their transport through membranes and to subsequently help them refold.

Chaperonins

Are also heat shock proteins that form large, multisubunit, cagelike assemblies that are universal components of prokaryotes and eukaryotes. They bind improperly folded globular proteins via their exposed hydrophobic surfaces and then, in an ATP-driven process, induce the protein to fold while enveloping it in an internal cavity, thereby protecting the folding protein from nonspecific aggregation with other unfolded proteins.
Other chaperones e.g. Hsp90 proteins (have a regulatory role in that they induce conformational changes in nativelike substrate proteins that result in their activation or stabilization) and Trigger factor (prevents the intra- and intermolecular aggregation of newly synthesized polypeptides as they emerge from the ribosome by shielding their hydrophobic segments) However, unlike most other chaperones, trigger factor does not bind ATP.
Enzymes are more or less catalysts in biological systems, so yes its right to synonimise them with catalysts.
References

Biochemistry (Voet and Voet) Section 9.2 Folding Accessory Proteins, Molecular Chaperones: The GroEL/ES System.
Lehninger Principle of Biochemistry Chapter 4, Protein Denaturation and Folding.
Biochemistry R. Garret, C. Grisham. Chapter 31 Completing the Protein Life Cycle: Folding, Processing, and Degradation

There are quite a few reasons why one would say that chaperones are not catalysts. For one, they only assist by shielding away unwanted influences. They don’t really actively assist in folding a protein — if you have an image of a chaperone perfoming some kind of origami, that is inaccurate to the best of my knowledge.
Therefore also, folding is not necessarily faster, it only has a greater success rate. According to the currently accepted thermodynamic model (again to the best of my knowledge), the folded protein is a pretty significant local minimum of the potential energy field so that if no extremely wrong paths are taken the correct structure should result.
The most important reason, however, is that folding is simply not a chemical reaction that could be catalysed. It is merely a physical process like rotation, no bonds are formed or broken in the process. (Hydrogen bonds don’t count.) If there is no reaction, then whatever assists or speeds up cannot be considered a catalyst.

Your last paragraph isn't true. https://en.wikipedia.org/wiki/Topoisomerase

Rest pretty much too. https://en.wikipedia.org/wiki/Chaperone_(protein) You should have checked sources :( Afinsen made rather a "guideline" ;)

@Mithoron What does topoisomerase have to do with it? An enzyme that breaks a phosphate backbone bond in DNA to assist in unwinding and then reforms it? Not only is that clearly a catalytic process but it also does not concern proteins.

You are right about the rest, though. I didn’t check sources, I relied on my memory from the biochem classes back at uni some eight years or so ago O:-)

Topoisomerases aren't best example but difference in product and substrate does lie only in intermolecular interactions likewise in proteins. Second link covers it though, stating that some chaperones do catalyse, even use up ATP for it. Also h-bonds in proteins definitely do count ;)

@Mithoron I did write no bonds are formed or broken in the process wanting to imply that if a bond is broken and reformed we have a total of two bonds broken and/or formed ;) And hydrogen bonds don’t count because the net effect of a formed or broken intramolecular hydrogen bond in aquaeous solution is zero.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is there an electronic structure method which accounts for 3-, 4-, ..., n-electron exchange integrals?
Something I learned recently which really helped me in understanding exchange is that the case of exchanging two electrons is really a special case of enforcing symmetry (or anti-symmetry) in which the permutations of labels in cycles out to length $N$ are used. $N$ is the number of particles.
Just to write it out, the two-electron exchange integral is defined as,
$$K_{ab}=\int d\textbf{r}_1d\textbf{r}_2\psi^*_a(\textbf{r}_1)\psi_b(\textbf{r}_1)r_{12}^{-1}\psi_b^*(\textbf{r}_2)\psi_a(\textbf{r}_2)$$
where everything means what it normally means and I took this integral and notation straight out of Szabo and Ostlund.
Now, this is clearly only a permutation of two electrons, and introduces the only correlation present in Hartree-Fock by allowing for exchange of spin-paired electrons.
Shifting gears, I've been reading Quantum Mechanics and Path Integrals by Richard Feynman and Albert Hibbs. In the chapter on Statistical Mechanics, they talk about liquid helium, which has a peculiar transition around $2K$ in which the heat capacity begins to increase. Feynman was the first to explain that this is due to the exchange interaction, and at this temperature is when permutations of the atom labels larger than just one (i.e. two particle exchange) become important. The following statements from the book are relevant to electrons (as opposed to helium because $\ce{^4He}$ is a boson!).
In a different context, the calculation of partition functions, the authors state:

If we were dealing with Fermi particles, e.g., the isotope of helium which has three nucleons, we would have to include an extra factor of $\pm1$, positive for even permutations and negative for odd permutations. There would also be some extra features which depend upon the spin of the atom in our result.

And more directly about electrons,

Consider the behavior of electrons in a solid metal. The mass of the electron is so much smaller than that of a molecule that the critical temperature is much higher. At room temperature, electrons in a metal are described accurately only by equations which include the exchange effects of these cyclic permutations. From this point of view, room temperature is very cold for electrons.

So, hopefully we have now reached a place where it is clear why I am a bit confused. Namely, to my knowledge, we only ever consider the exchange of two electrons of the same spin in our electronic structure methods. First, is this actually true? That is, are there methods which implement (in principle) $N$-electron integrals to describe all permutations of the labels?
Second, is the comparison I am making even strictly valid? That is, in the case of a partition function, which is what Feynman is talking about, permutations of even length tend to cause the electrons to stay further away from each other. This is the behavior we are used to with exchange. In contrast, for permutations of odd numbers, the contribution to the partition function picks up a minus sign. So, the total partition function for a system of fermions is an alternating sum of positive and negative terms in the path integral formalism. Can I simply substitute partition function for wavefunction and sum for integral and find what would happen in the electronic structure case? That is, will the 3-electron exchange integral (assuming this is real and I'm not mistaken) be a negative number, so that we can picture the wavefunction relaxing a little bit from the so-called "heaps" which form due to exchange?
Finally, if I'm not just delusional about this whole thing, is there evidence that permutations beyond the two labels are important for describing any known chemical processes?

References:
[1]: Szabo, A., & Ostlund, N. S. (2012). Modern quantum chemistry: introduction to advanced electronic structure theory. Courier Corporation.
[2]: Feynman, R. P., Hibbs, A. R., & Styer, D. F. (2010). Quantum mechanics and path integrals. Courier Corporation.

The electron-electron interaction is a 2 particle operator ($\frac{1}{r_{ij}}$), which means only terms with 2 electrons ($i$ and $j$) show up.  There us no need to consider 3-electron integrals (or higher).

Right. But what I'm saying is in principle there should be a 3-particle exchange and 4-particle exchange, etc.  Or at least there is when considering fermionic systems in calculating a partition function. It could be that all higher order permutations are zero but I'd have to be shown why this should be the case.

You might be confusing partition and wave function. The partition function is about the distribution of states, which in turn are defined by wave functions. Those are two different layers of abstraction. Considering permutations about which particle occupies which state is a different thing than the Exchange Integral in HF.

No. I was just drawing an analogy. In looking a bit deeper I think that explicitly correlated (F12) methods actually deal with precisely these integrals. There are"cyclic" operators of the $\frac{1}{r_{12}}\frac{1}{r_{13}}\frac{1}{r_{23}}$. I think this is what I'm asking about but these papers are very heavy on math and very light in physical interpretation. Haha. I'll add some references tomorrow.

I would be careful with an analogy between permutations of particles and the HF exchange integrals. This exchange integrals is only named like this because in its mathematical equation you exchange two electronic coordinates with respect to the coulomb integral. Since there is no classic analog to this, it is difficult to interpret this physically. So yes, analogies due to the mathematical form of the equations may show up. But with analogies in the physical interpretations I would be careful.

About the $\frac{1}{r_{12}}\frac{1}{r_{13}}\frac{1}{r_{23}}$ those are still 2-body. A 3-body term would be something like $\frac{1}{r_{13}r_{12}}$.
But yes, there are some cases/methods which deal with 3 electron integrals, maybe even higher. But I am not familiar with those.

The exchanges I'm talking about are still between two electronic coordinates, but simply accounting more than one exchange at a time. So 1 with 2, 2 with 3, and 3 with 1. I am pretty sure this would still be interpreted as the same exchange effect but simply a higher order term.

Since the exchange integrals show up as a sum in the Hamiltonian $\sum_{i, j>i} K_{ij}$ those three terms ($i=1,j=2$; $i=1,j=3$ and $i=2,j=3$) do show up  in HF.

I am posting this answer not as the definitively correct answer, but as an example of the type of thing I am talking about.
azago's answer is correct in pointing out that the Slater determinant contains within it the correct permutational symmetry of the wavefunction required by electrons. Namely, it is antisymmetric upon interchange of two orbitals. Additionally, it symmetric upon interchange of three (or any odd number of) orbitals. This is the symmetry which must be present in higher cycles of exchanges.
What I have found upon further reading is that I could have been more precise in the wording of my question. Namely, these higher-order exchange interactions are not surprising if I had made it clear that these are necessarily three-center, four-center,... integrals. Then it is more clear that HF, while having the correct symmetry to adhere to Pauli exclusion, does not include these integrals as it only deals with 2-electron integrals at the most.
So, I will discuss three contexts I have found in which the types of integrals I am thinking of show up.
Hamiltonian for a Three-Atom System:
Following along with ref. [1], the Hamiltonian for the three atom system, $\ce{H3}$ is,
$$
H=\sum_i\left(-\frac{1}{2}\nabla^2_i-\sum_{\alpha}\frac{1}{r_{\alpha i}}+\sum_{j\gt i}\frac{1}{r_{ij}} \right)+\sum_k\frac{1}{R_k}
$$
where $i$ and $j$ are labels for the three electrons, and $\alpha$ is the label for each Hydrogen atom with locations $R_1$, $R_2$, and $R_3$.
It is clear from this Hamiltonian that we have the usual coulomb terms and the usual two-electron integrals. Something weird is going to happen with the exchange terms though. There are actually three possible configurations of the atoms, but the only point I'm making here is that the matrix elements for one of the atomic configurations can be written as,
$$
\langle\psi_1|H|\psi_1\rangle=4J+4K_1-2K_2-2K_3-4K_{123}
$$
I have deviated from their notation here for consistency with current convention. These integrals are written explicitly as:
$$
\begin{align}
J&=\langle abc|H|abc\rangle \\
K_1&=\langle abc|H|bac\rangle \\
K_2&=\langle abc|H|acb\rangle \\
K_3&=\langle abc|H|cba\rangle \\
K_{123}&=\langle abc|H|cab\rangle \\
\end{align}
$$
where now it is quite clear that $K_{123}$ is a double exchange. That is, two electron labels are changed, and it is clear this is distinct from the other exchange integrals which are the usual exchanges we are used to.
I do not think this has anything to do with the fact there are three atoms present, however. Only that there are three electrons. Thus, I believe that even a complete description of Lithium would include these three-electron exchanges. Thus, the reason we do not see these integrals in HF is that they are left out of the Hamiltonian.
Explicitly-Correlated Methods:
I'll be brief here because explicitly-correlated methods are pretty complicated and I haven't spent enough time studying them to really know what's going on, but one thing which does happen is that you get matrix elements of the following form[2]:
$$
Z_{\nu w,p}^{xy,q}=\langle w_{\nu w,p}|(1-\hat{p}_{13})r_{13}^{-1}+(1-\hat{p}_{23})r_{23}^{-1}|w_{xy,q}\rangle
$$
The only important in this horrible integral is that the exchange operator $\hat{p}_{ij}$ appears with the labels of three-distinct electrons. Now, I have no idea if it's valid to interpret this as the type of integral I'm talking about, but these kinds of integrals appear all over the place in explicitly correlated methods.
Three-Body Symmetry-Adapted Perturbation Theory
Now, two-body symmetry-adapted perturbation theory (SAPT) is quite a common method, but there is a 3-body version of SAPT which can be quite a useful tool at times.
In Ref. [3], the authors present the relevant equations for 3-body SAPT, but they note that they approximate the exchange term out to the fourth power of orbital overlap. That is, for the exchange operator they do the following:
$$
P\approx 1+P_2+P_3+P_4
$$
They describe this as follows:

In this expression, $P_2$ represents the operator which interchanges
  one electron from $X$ with one electron from $Y$ and
  therefore produces terms proportional to $S^2$. The operator
  $P_3$ includes double exchanges between three monomers involving
  three electrons. This operator, absent for a dimer,
  generates terms proportional to $S^3$. The operator $P_4$ is a
  double exchange operator which interchanges four electrons
  among the monomers and generates terms proportional to
  $S^4$. The double exchanges produced by $P_4$ are of the
  $(XY)(XY)$ or $(XY)(XZ)$ types and can be thought of as two
  exchanges of the type produced by $P_2$ occurring simultaneously.
  In contrast, double exchanges of the $(XY)(XYZ)$
  type are included in $P_5$ and double exchanges of the
  $(XYZ)(XYZ)$ type are included in $P_6$, leading to terms proportional
  to $S^5$ and $S^6$, respectively.

The whole purpose of three-body SAPT is to be able to describe the 3-body non-additivities which arise from intramolecular interactions. It has already been pointed out that the electrostatic interaction is additive up to 2-body because Coulomb's law follows the superposition principle. This then makes it clear that these higher-order exchanges involving more than two electrons generate some of the non-additive forms of coulomb correlation.
Finally, the authors of ref. [3] provide the form of these three-electron cycles I was trying to describe which can be written as:
$$
P_{111}=\sum{i\in A}\sum{i\in B}\sum{i\in C}[P_{ij}P_{jk}+P_{ik}P_{jk}]
$$
which interchanges electrons in a cycle on atoms $A$, $B$, and $C$. This operator generates all possible three-electron exchanges between three atoms. There are no intra-monomer exchanges however.

[1] Porter, R. N., & Karplus, M. (1964). Potential energy surface for H3. The Journal of Chemical Physics, 40(4), 1105-1115.
[2] Hättig, C., Klopper, W., Köhn, A., & Tew, D. P. (2011). Explicitly correlated electrons in molecules. Chemical reviews, 112(1), 4-74.
[3] Lotrich, V. F., & Szalewicz, K. (1997). Symmetry-adapted perturbation theory of three-body nonadditivity of intermolecular interaction energy. The Journal of chemical physics, 106(23), 9668-9687.

The Hartree–Fock wave function in fact has the correct permutational symmetry for any $n$-particle exchange, because the determinant has it. This follows directly from the definition of a determinant as a sum over all permutations of rows (columns). But as @Feodoran says in the comment, the Coulomb potential is two-body, and it's expectation value, $\langle\psi|\hat V|\psi\rangle$, picks only the two-electron exchange.
So, to directly answer your question: First, we actually do consider all $n$-particle exchanges, but they simply do not show up in the relevant integrals. Second, I'm no expert on statistical physics, but I don't see why there should be any strong analogy between the wave function and the partition function in classical statistical physics, which is a thermodynamic quantity and does not consider the microscopic dynamics of a system, that involves correlations and state transitions. (There is also a partition function in QFT, and there perhaps may be some analogies between that and the wave function, but I know even less about QFT.)

Maybe an analogy to CI, where you consider all possible permutations of $N$ electrons in $M$ orbitals, will work better here?


text:
Why aromatic compounds have pi-electrons clouds above and below the plane
I was reading a text and it asked "aromatic compounds have pi-electron clouds above and below the plane due to?". Someone, please help.

Hmm, do you know what aromatic compounds are in the first place? If not, check the wikipedia article for Aromaticity.

That is just how Aromaticity is defined. Simply put, it means you have delocalized $\pi$ electrons in some circular structure. Being above and below the molecular plane is simply a property of $\pi$ orbitals/electrons.

The orbitals are akin to normal modes of vibration. Sigma bonds have no nodes, like the fundamental vibration of a guitar string. Pi bonds have one node (in the same plane as the atomic nuclei), and that is like the first harmonic of a guitar string (pluck it while holding your finger gently against the midpoint of the string). The note is one octave higher than the fundamental--higher frequency means higher energy.


text:
Formation of hydrogen bromide: Does the concept of "order" apply to this reaction?
Started reading Levine's Physical Chemistry [1] yesterday, and I hit upon (another) wall.
On page 517, Chapter 15, she includes a reaction resulting in the formation of $\ce{HBr}$ from $\ce{H2}$ and $\ce{Br2}$ (it's a homogeneous, gas-phase reaction) along with the corresponding rate equation:
$$\ce{H2 + Br2 -> 2HBr} \qquad r= \frac{k[\ce{H2}][\ce{Br2}]^{1/2}}{1 + j[\ce{HBr}]/[\ce{Br2}]},$$
where $k$ is the rate constant and $j$ is...um...just a(nother) constant O:)
Levine goes on to mention:

[For this reaction] the concept of order does not apply.

Now I'm still new to the whole Chemical Kinetics thing... But what's this?!
How come the "concept of order does not apply" to this reaction? I've never come across this situation in any book so far. Well, the reaction seems trivial (though the rate equation says otherwise...); I fail to understand why an order can't be assigned to this (overall) reaction.
(Note: Until I know otherwise, I'm assuming "concept of order doesn't apply", is the same as saying "this reaction has no order". Do correct me if I'm wrong in doing so...I tend to miss out on subtleties...)
Even having an order of zero still counts as having an "order". But what could Levine possibly be talking about when she says that the idea of an "order" doesn't apply to this reaction? What are the (physical) implications of a reaction lacking an "order"?

My questions, explicitly put:

Why can't an order be assigned to the above-mentioned reaction? Or can it?
What are the physical implications of a reaction having "no order" (as compared to reactions with an order)?

References

Levine, I. N. Physical chemistry, 6th ed.; McGraw-Hill: Boston, 2009. ISBN 978-0-07-253862-5.

there is some more information in replies to this question https://chemistry.stackexchange.com/questions/79080/is-there-any-empirical-upper-limit-for-order-of-reaction/79084#79084

@porphyrin Gee thanks, that helped! Your answer there would work here...maybe you could slightly alter it and re-post it here? It'd get my upvote. Your call though O:)

@paracetamol No prob, great question! An edit is the least I can do.

We don't assign orders. An order either simply is there, or it isn't. In particular, this expression doesn't look like $\ce{\underbrace{(something)}_{\substack{\text{may include }\ce{[H2]},\\ \text{but not }\ce{[Br2]}}}\cdot[Br2]^a}$ for any $a$, hence there is no order.
Alternatively, if we look at the reaction at its initial moments when $\ce{[HBr]}\approx0$, the denominator simplifies to just $1$, and we may say that the reaction has order $1\over2$ with respect to bromine.
Having or not having an order doesn't have any particularly far-reaching physical consequences. Moreover, it is not even an intrinsic property of a reaction; it may depend on the approximation used, as you saw above.
The scary rate equation comes from the so-called steady state approximation and the detailed reaction mechanism with intermediates and stuff, which is about two times longer than my answer, so I'd better refrain from including it. Well, here is the first line, just to give you the taste of the whole thing:
$$\ce{Br2}\underset{k_{-1}}{\stackrel{k_1}{\longleftrightarrows}}\ce{2Br^.}$$
To tell the truth, most real reactions are like that. The well-known $r=k\ce{[A]^a[B]^b}$ expression with stoichiometric coefficients for exponents is but a great oversimplification. It applies to reactions with no intermediates, which are quite rare.


text:
Is there a highly exothermic reaction that the reactants can be easily regenerated from the products?
Is there a highly exothermic reaction that the reactants can be easily regenerated from the products? Let's say you put in contact two reactants and they react and produce large amount of heat. Then can you take the products and regenerate the reactants easily?
If we put acid and alkali together, an exothermic reaction will take place. How can we regenerate the acid and alkali?

What does "easily" mean for you? Burning hydrogen is exothermic, producing hydrogen from water is relatively easy (though not necessarily cheap)

I was thinking something like pouring a liquid on a metal surface and then taking the liquid and dissolve it, electrolysise it or heat it or somehow regenerate it.

There is no free lunch, and also no perpetual motion machine. If that is not what you want, please elaborate a bit more.

I’m voting to reopen. It might not be a good question, but it’s sufficiently clear and can be answered using the fundamental laws of thermodynamics.

Why, many reactions are like that, but there's a catch. To regenerate the reactants from the products, you'll have to put in at least as much energy as was gained from the reaction in the first place. This is the energy conservation law; there is no going around it.
Other than that, the examples are plenty. You may go with burning of hydrogen and electrolysis of water.


text:
Cp-Cv For a gas following Van Der Waal equation of state
How do you derive the relation
$$C_p-C_v= R\left(1 + \frac{2a}{RTV}\right)$$
for a gas obeying van der Waals equation of state? Any leads?

How do you know you can obtain an equation of this form by making certain assumptions if you don't know what those assumptions are? Wherever you found this formula should say what assumption they had to make for it to be true.

It is a question in a competitive exam, hence the doubt.

Does the test question include any extra context? If so, you should include the text of the actual question.

No sir this is the only text.

Problem 2 here seems to answer your question after a little algebra and I believe with the assumption that V>>b.

Also related, p.2 and 3 here

You could start with $C_V=(\partial U/\partial T)_V=T(\partial S/\partial T)_V$  and H and p instead of U and V for $C_p$ as appropriate. Then generate an expansion for S as $dS=(\partial S/\partial V)_T dV+(\partial S/\partial T)_VdT$ and differentiate wrt T. You should then get an expression in $C_V$ and $C_p$ plus other terms that you can find using the vdw equation.

Yes but since the vdw equation is not explicit in V, it does not boil down to the above expression. The above expression can be obtained using a specific assumption which is what I am after.

You should find that the difference in heat capacity is $T(\partial P/\partial T)_V(\partial V/\partial T)_p$ so calculate the derivatives and keep going. It is messy. However, since a is in the answer given in the question a likely solution is that $b=0$. Making this assumption does simplify the answer a bit although its best usually to apply specific limits at the end of a calculation if possible.


text:
IR spectroscopy in CCl4 Solution v/s Liquid film
Why there are differences in these spectra ? Which one is more useful and accurate ? What is the significance of using these two different methods ?

Both are useful as for with one of the spectra and some usual extra work you can identify the molecule, for instance. In the couple of spectra you show, the main difference is seen in the O-H str region. Differences are due to hydrogen bonds in neat and the lack of in CCl4. The "hole" at low wavenumber is due to the latter ( abs). .

A real use of this is the addition of traces of water to a sample in CCl4 solution. Like this you can further confirm an alcohol, for example.


text:
Anhydrous ammonia vapor-liquid equilibrium
I have been given a anhydrous ammonia vapor liquid system in an air-tight container where the pressure is 30 bar and the temperature is 30 degrees. They claim the system is in equilibrium, which I have two doubts about.

The vapor pressure for anhydrous ammonia at 30 degrees is less than 30 bar, how is it possible to have this pressure in the given system?
Will the volume ratio of vapor phase with regards to liquid phase affect equilibrium? If so, then how?

What's so special about 30 bar? You can have this pressure in a closed system just as well as you can have any other pressure.

@IvanNeretin the problem is not with pressure, but with the state being in equilibrium. I assumed if you have 30 bar pressure, system's temperature will change to make '30' as the saturated pressure

Well, yeah, something will change, that's for sure. If the vapor pressure of liquid ammonia is less than 30 bar, then some of it will condense until the pressure lowers down.

So how do i deal with the question (it is for assignment purposes), because it is said system is in equilibrium, i am rather confused regarding the same. Is the question wrong?


text:
When heat is increased, does water move upward in a straw due to increased capillary action or increased pressure?
I left a water bottle with an internal straw in my car throughout a long, hot day. When I opened the bottle at the end of the day, water immediately splashed out from the straw. This made me wonder if the increase in temperature caused the liquid to move with greater capillary action upwards in the straw due to higher movement of $\ce{H2O}$ atoms OR whether it was an increase in pressure that led the water molecules to automatically expel outwards from the straw once the bottle was opened.
I know this may seem like a silly question but what explanation would you give for the expulsion of the water from the straw? (I find this recurring event strange because the shape of the bottle never shrinks at all from the heat yet something forces the water outwards but I cannot discern what force causes this. By the way, this never happens to me with regular water bottles - only with ones containing straws/tubes on the inside.)

The effect of heat would be negligible on the water itself (pressure, volume, capillary action, etc.). The cause just has to be air pressure. I'm not sure I have a very good image in my mind of your water bottle, but presumably there is air inside your bottle along with the water, it is under pressure from being warm, and when you remove the cap the pressure is allowed to escape. This happens to squirt bottles of  water and other liquids in laboratories all the time when there is poor temperature control. On hot days the headspace air in the bottles heats, expands, and the liquid comes out.

I ran out of space above and I wanted to keep it in a comment rather than answer because I'm not sure I understand the design of your bottle yet. Does what I said make sense based on what you've seen happen?

Capillary action, if anything, decreases with temperature.

@airhuff Yes. I understand what you mean. A large volume of water was expelled so I assumed that the water had undergone more chemical change than the air. But, in reality, the air had experienced even greater thermal expansion than the water.

It almost certainly was due to thermal expansion of water and the air above it. Consider that the thermal expansion coefficient of water near room temperature is ~$0.000214$ K$^{-1}$. [Though it is not linear over a broad range, assume for the small temperature rise in your car that it is.]
If the water was cold (e.g. ~$277$ K, maximum density) and it warmed to ~$300$ K ($80$o F), a temperature change of ~$23$ K, it would have expanded by a factor $(300 \ce{K} - 277 \ce{K}) \cdot 0.000214 \ce{K^{-1}} = 1.005$.
Assume you have a one-liter bottle filled completely with water when you start; by the time you return it's $1.005$ L. In the whole bottle, the change would hardly be noticeable, but the narrow straw greatly magnifies the rise of liquid. If the straw is $6$ mm in diameter, the volume of liquid is given as: $$V=\pi\cdot r^2\cdot h$$
$$5 \mathrm{mL} = 5 \ce{cm}^3$$
$$r = 0.3 \ce{cm}$$
$$h = \frac{5 \ce{cm}^3}{\pi \cdot (0.3 \ce{cm})^2} = 18 \ce{cm}$$
So the water would have risen about $18$ cm in the straw!
This exercise only took into account the water's expansion... for any air above the water in the bottle, and for other (approximately) ideal gases, $PV=nRT$ and thermal expansion would be far greater.
To some extent, the bottle itself expands, decreasing the rise of the liquid, so these calculations are not very accurate. Most likely, air expansion contributed more to the "mass ejection" than did that of water... it depends on whether you view your container as half-full or half-empty.

Thank you. That was a very thorough answer. I have a much better understanding of this phenomenon now. :)


text:
Simulating H1 NMR spectra using chemical shift/multiplicity data?
I've encountered a number of free programs/internet plugins that can output IR spectra and the corresponding chemical shift/multiplicity data if you input a chemical structure, but are there any where you can input chemical shift/multiplicity data and get an output of at least the H1 NMR spectra?
Note: by "chemical shift/multiplicity data", I am not referring to chemical shift tensors (σ). I mean simple data like "1.92 ppm (2 H, m)".

I am not aware of any software that can transform a spectrum into a molecule. But if you are asking for look up database have a look at Spectral Database for
Organic Compounds SDBS
Here you can input simple data as you asked for like ppm, and then get suggested moleulces as output.

If you know the R language, there is a package SpecHelpers which draws proton and carbon NMR spectra given input similar to what you describe.  For proton spectra, it is limited to the basic multiplets, there are no doublet of doublets possible for instance.  Disclaimer: I am the author of SpecHelpers.


text:
Determining the purity of commercial NaCl
I am trying to figure out how to determine the percentage of NaCl in salt. My purpose is to be able to test the purity of salt that I received before washing so that I can determine the quality of the salt and how much I should pay for it. I also want to be able to determine how good a salt washing machine works by comparing the before and after purity of the salt.
My basic understanding is that there is a standard minimum percentage of NaCl in after wash salt (usually above 99% depending on where you are), the percentage is lower for a pre-wash salt.
Is there any test to know the purity of Salt?
Thank you..


text:
Synthesis of methylenecyclohexane from cyclohexylmethanol
I want to form methylenecyclohexane from cyclohexylmethanol

I'm thinking about an $\mathrm{E1cB}$ mechanism since the β-hydrogens will be quite acidic (similar electronegativity to fluorine). So I can use a strong and hindered base (to reduce $\mathrm{S_N2}$) to form methylenecyclohexane.
Another possibility is a weak hindered base for an $\mathrm{E1}$ Hofmann elimination (since the carbocation will rearrange). But not only will this lead to a some amount of Zaitsev product, $\mathrm{S_N1}$ will dominate because of better thermodynamics (net 1 σ formed vs net 1 π formed that is not very stabilized either).
I feel like $\mathrm{E1cB}$ would be a better idea.
Are there any better methods?

Tosylate and E2?

I can't pick any compound. I gotta do it through the one I've attached in the picture. OH- is a poor leaving group. It won't undergo E2.

Tosylate the alcohol, and E2

As @orthocresol has suggested I would form the mesylate or tosylate and treat with DBU

Off the top of my head no direct methods will work as $\ce{OH-}$ won’t leave no matter what you throw at it and strong bases will deprotonate the alcohol before any $\ce{C-H}$ bond of your compound. So you have to modify and use a two-step (or two-stage) procedure.

@Jan why wouldn't an E1CB mechanism work? E1CB requires a poor leaving group like OH, or F

The CB in there means conjugate base. You deprotonate the beta-hydrogen before elimination. But there is no way in hell that hydrogen (tertiary unactivated $\ce{C-H}$ bond) would deprotonate even slightly while the hydroxy proton would leave very quickly.

@Jan Hmm so the beta hydrogen isn't acidic enough.  Seems like the only way is through tosylation.

One could also use phosphoryl chloride coupled with a non nucleophillic base to generate the Hoffman product.

Whichever method of elimination you end up preferring, there is no way to directly eliminate the hydroxy group without derivatising it first in some way. The hydroxide ion is a very bad leaving group (but a rather good nucleophile) and transforming a hydroxy group into a better leaving group by protonation ($\ce{R-OH2+}$) requires acidic conditions that do not really allow the (basic) abstraction of a β hydrogen. For each of the simple options:

A hydroxide ion will never leave as part of an $\mathrm{S_N2}$ or $\mathrm{E2}$ reaction; it is too weak a leaving group.
An $\mathrm{E1}$ reaction would require strongly acidic conditions to protonate the hydroxy group; under these conditions the β hydrogen cannot be abstracted. $\mathrm{S_N1}$ would dominate.
An $\mathrm{E1cb}$ mechanism is absolutely impossible. The $\mathrm pK_\mathrm{a}$ value of the β hydrogen is around $50$ while that of the hydroxy group is around $15$. The only conjugate base being formed will be the alcoholate anion.

This means we have to resort to two-step procedures. A number of them have been published in literature in various different contexts. Some bear a name, others don’t. Some follow one of the paradigms mentioned above, others follow a completely different mechanism. Since there is only one elimination product possible, you need not worry about selectivity. I shall present a few possibilities grouped by the mechanistic type where possible but note that this list is not exhaustive and does not even begin to scratch the surface.
$\mathrm{E2}$ type mechanisms
The simplest is the one Orthocresol mentioned in the first comment: transform the alcohol into the corresponding tosylate, e.g. by adding $\ce{TsCl}$ and pyridine, then eliminate with additional potentially stronger base.
A variant of this would be to use mesyl chloride/DBU instead. Depending on how activated your proton is, a simple addition of more DBU to the reaction mixture may suffice.
Considering that your starting material is a primary alcohol, $\mathrm{E1}$-based reactions are very unlikely. Considering again the acidity of the β hydrogen, so are $\mathrm{E1cB}$; no matter which the reaction protocol.
syn-selective eliminations based on cyclic transition states
One example is the ester pyrolysis: use an acid chloride (e.g. acetyl chloride $\ce{AcCl}$) to synthesise an ester, then heat up to initiate thermal rearrangement to the corresponding acid and alkene. The transition state is cyclic, so this reaction is syn-selective.
A similar reaction is the Chugaev elimination. React the alcohol with a base and $\ce{CS2}$, then add an alkyl halide (typically iodomethane) to generate a xantogenate. This also fragmentates thermally to give an alkene and a dithiocarbonate while syn-selectively removing the β hydrogen.
And finally my favourite one: the selenoxide pyrolysis. First synthesise a selenoether with $\ce{NCSe-C6H4-NO2}$ (ortho-nitrophenyl selenocyanate) and $\ce{PMe3}$ in a Mitsunobu reaction, then oxidise the selenoether e.g. by using $\ce{H2O2}$ at room temperature to smoothly give the corresponding alkene — again via a cyclic transition state practically identical to the former two.


text:
How can I get more information about the Voronoi integral in Gaussian09?
I am working on metal based organic complexes to find out catalytic characters. I am using the B3LYP density functional level of theory in Gaussian09.
I need to know the possible keywords for getting effects of Voronoi integral where I use the keyword EmpiricalDispersion = GD3. This is not a cluster, but the central metal atom is $\ce{Fe}$. How can I obtain more output about it?

cross-post: http://www.chemicalforums.com/index.php?topic=92398.0

I know quite a bit about the D3 correction, but have never heard of a Voronoi integral being involved in that. Note that the Grimme corrections do not affect the electron density/KS-determinant (beyond the change in geometry if they were used in obtaining coordinates).

Thank You Martin and TAR86 for giving concentration on my topic. During study I obtained some distorted results. Someone suggested to apply Voronoi Integral, but I couldn't find any information about it in Gaussian Reference book and in available network search. But one thing I know that for more accurate transition states or structure conversion we can include 'int=fine' or 'int=ultrafine' in ipput line. But not sure that it is the same as that of Voronoi integral. That is why I just expect suggestion from the experts. Is there any other keywords that can convert in more accurate structure?

The only thing I was able to find is the VDD: https://en.wikipedia.org/wiki/Voronoi_deformation_density - which either is or is not implemented in the QC code of your choice. The int= choices you mention are most likely unrelated to the VDD, this is a different integration altogether. As for accuracy of structures and methods for that, I have to refer you to the literature. We cannot reasonably be expected to tell you which is the best method without more details (and even then, it will be nigh impossible).

These things are completely unrelated to each other. Voronoi population analysis is a real-space partitioning of the electron density, where the boundary lines are given by Voronoi cells. It has nothing to do with empirical dispersion corrections. It is also only available in ADF.

By introducing the following keywords in inputline produces the relative better results. "EmpiricalDispersion=GD3 Int=ultrafine"


text:
The "Ostwald Isolation Method": How does this work?
Recently started reading up on Chemical Kinetics, and I came across a certain "Ostwald Isolation Method" that figures in my school-issued workbook. Apparently, it's used to determine (approximately) the order of a given reaction.
However, the painfully short passage that constitutes that section of the book, coupled with my teacher's reluctance to take up the said section in class, means I still don't know how this "method" works. Heck, I don't even know if the said "method" even exists (at least not under that name anyways...the book is notorious for its numerous errors that have gone unchecked over the years. We fondly view it as the Mater Omnium Errorum)

My book, on the "Ostwald Isolation Method":

This method is particularly useful in case of reactions where two or more reactants are involved.
Consider the reaction between $a$ moles of species $A$ and $b$ moles of species $B$ to give product(s) $P$: $$\ce{aA + bB -> P}$$
The reaction is first carried out with $B$ present in great excess. So the reaction must only depend on the concentration of $A$: $$\ce{r = k[A]^{x}}$$
Similarly, when the reaction is now carried out in a large excess of $A$, the rate of reaction must only depend on the concentration of $B$: $$\ce{r = j[B]^{y}}$$
The order, $n$, of the overall reaction is therefore: $$\ce{n = x + y}$$

Okay, I understand how the first two bits work (i.e- Concentrations of whatever species present in excess can safely be considered constant/invariant).
But I don't get how directly adding $x$ and $y$, both obtained from different rate equations, having different rate constants , gives us the overall order of the original (without excess) reaction.
Could someone explain the logic behind this to me? That is, if there is any logic behind it...

Order of reaction has nothing to do with rates.

The powers $x$ and $y$ are intrinsic to the reaction (at least if the reaction pathway remains the same) and will not vary with reaction parameters, so that adding them then poses no difficulty.
More specifically, it is presumed that the reaction rate is given by an equation of the form $$r = c\ce{[A]}^x\ce{[B]}^y;$$ the Ostwald isolation method finds the powers $x$ and $y$ by flooding the system with $\ce{B}$ and $\ce{A}$ respectively in excess in two separate experiments, so that in the former case we have $$r = (c\ce{[B]}^y)\ce{[A]}^x \approx k\ce{[A]}^x$$ and in the latter $$r = (c\ce{[A]}^x)\ce{[B]}^y \approx j\ce{[B]}^y.$$
Adding the powers $x$ and $y$ gives the order of reaction by definition.

Incidentally, Wikipedia lists this method as the 'method of flooding', here.


text:
Heat of reaction for a reversible reaction
If we have a reversible reaction
$$\ce{A <=> B}$$
and we have the activation energies of the forward and reverse reaction can we use
$$E_a^\mathrm{rev} = E_a^\mathrm{fwd} - \Delta H$$
to find the heat of reaction? $E_a^\mathrm{rev}$ and $E_a^\mathrm{fwd}$ are the activation energies of the reverse and forward reactions, respectively; $\Delta H$ is the heat of reaction.
I have tried using this but am unsure as to the layout of the equation.

I don't see a question here. Maybe provide a specific example?

Suppose that the forward reaction is exothermic so that the product is $\Delta H$ below the reactant then $E_a^{fwd} +\Delta H= E_a^{rev}$. Draw a sketch.

As you can see from energy diagram of the reaction that 'Heat of Reaction' is
ΔH = Efwd- Erev.
So your expression is correct.

No your equation is not correct;  $\Delta H + E^{fwd} =E^{rev}$:  you should not assume that $\Delta H $ is negative.

It is ...in that case Efwd > Erev and ΔH > 0 .

still no; in your diagram $\Delta H $ has the same sign as the $E_a$.

oh.. I just took the diagram from net..overall ΔH < 0 as Erev> Efwd. But expression will always hold as Efwd, Erev will always > 0 . their magnitude decides whether ΔH is positive or not.


text:
Correlation between 1H and 13C shifts - coincidence or not?
A quick glance at any table of typical chemical shifts will reveal that the chemical shifts of protons $(\delta_\mathrm H)$ tend to correlate pretty well with the chemical shifts of the carbons to which they are attached $(\delta_\mathrm C)$. This correlation is frequently taught in introductory organic chemistry and the rule of thumb
$$\delta_\mathrm C \approx 20 \cdot \delta_\mathrm H,$$
although simplistic, actually holds up fairly well. To illustrate the point further, here is a diagram taken from the textbook by Silverstein et al. on organic spectroscopy:

Image taken from: Silverstein, R. M.; Webster, F. X.; Kiemle, D. J. Spectrometric Identification of Organic Compounds, 7th ed.; Wiley: Hoboken, NJ, 2005, p 207.
To the beginner such a correlation might seem intuitive. However, delving deeper into NMR theory, one learns that the $\ce{^{13}C}$ chemical shifts are generally dictated by paramagnetic shielding, while $\ce{^{1}H}$ chemical shifts are generally dictated by diamagnetic shielding and effects due to neighbouring groups. See, for example, this question:  NMR chemical shift range of different elements.
In light of this, how much should we make out of the general correlation seen? i.e., is it best understood to be a pure coincidence, or is there more to it than that?

I've never actually been taught that rule of thumb regarding dH shifts and dC shifts. Learnt something new today. THANKS!

I don't have the understanding to actually answer, but the conclusion has been drawn before and was definitely pointed out during my undergrad tutorials. Seems to be a coincidence that holds up okay for simple things. See  J. Chem. Educ. 1991, 68,  284.   An old article, with no explanation, but does offer an empirical conversion based on a training set of 53 (complex-ish) molecules.

Apart from other specific effects, chemical shift is a function of electron density around the nucleus, so it is reasonable that proton shifts roughly correlate with those of the attached carbon.

Very interesting, beautiful old-school diagram.  I will just add that an HSQC gives a very visual insight that increasing shift in 1H corresponds to a commensurate shift in 13C--but you can also see many peaks off of this correlation diagonal to get an idea where this doesn't hold up

This is not my best answer, but because I'm at work and because I'm the person who asked the question, I'll forgive myself.
Quoting from Hans Reich's website:

All nuclei heavier than $\ce{^1H}$ (with the possible exception of $\ce{^6Li}$ and $\ce{^7Li}$) have $\sigma_\mathrm{p}$ [the paramagnetic shift] as the principal chemical shift effect.
It is fortunate that $\sigma_\mathrm{p}$ often responds in the same way to electron density as does $\sigma_\mathrm{d}$ [the diamagnetic shift], since this results in parallel $\ce{^1H}$ and $\ce{^{13}C}$ chemical shift features. However, a qualitative description of the relationship between $\delta$ and molecular features must consider two factors which affect $\sigma_\mathrm{p}$ and $\sigma_\mathrm{d}$ quite differently and may cause very counterintuitive behavior: the $1/ΔE$ dependence, and orbital symmetry effects.

Reich then goes on to provide specific examples where this dependence falls apart, such as anti-aromatic compounds, the pyridine/pyridinium pair, and phenyllithium. Mathew Mahindaratne's post also provides some counterexamples. (More broadly speaking, this analysis of paramagnetic and diamagnetic shifts also tends to neglect things such as alkynes, where anisotropic effects lead to unusual chemical shifts.)
I'd say that yes, the correlation is coincidental; and Reich's word choice of 'fortunate' supports this.
PS: There are a number of arguments, even in this very post, about how chemical shift is dictated by 'electron density'. These, however, fail to acknowledge the different mechanisms by which electron density affects proton and carbon chemical shifts (which I pointed out in the original question). I think it's very unfortunate that this correlation exists, because it's clearly led to a lot of misconceptions amongst non-NMR specialists.

It is true that a quick glance at any given table of typical chemical shifts will reveal that the chemical shifts of protons $(\delta_\ce H)$ tend to correlate pretty well with the chemical shifts of the carbons to which they are attached $(\delta_\ce C)$. The rule of thumb of this correlation is as shown in the question:
$$\delta_\ce C \approx 20 \cdot \delta_\ce H,$$
This simplistic rule actually holds up fairly well. I wonder is it because of $\ce{^{13}C}$-NMR spectroscopy, characterized by a large range of variation in $\ce{^{13}C}$ nuclear shielding (about thirty times that of protons) and by a simpler analysis of the proton decoupled spectra (Ref.1). My aim at this answer is to provide some contradictory case against it norm.
I think cyclopropane and cyclobutane would provide evidence against the rule of thumb in this case (Ref.2):
$$
\begin{array}{c|lcr}
\text{Compound} & \ce{^{1}H}\ (\pu{ppm}) & \ce{^{13}C}\ (\pu{ppm})  & 20 \cdot \delta_\ce H \\
\hline
\text{Cyclopropane} & 0.22 & -0.28 & 4.4 \\
\text{Cyclobutane}  & 1.94 & 22.4 & 38.8 \\
\text{Cyclopentane} & 1.51 & 25.8 & 30.2\\
\text{Cyclohexane} & 1.44 & 27.0 & 28.8\\
\text{Cycloheptane} & 1.54 & 28.7 & 30.8\\
\hline
\end{array}
$$
References:

O. A. Subbotin, A. S. Kozmin, Yu K. Grishin, N. M. Sergeyev, and I. G. Bolesov, "$\ce{^{13}C}$ NMR spectra of cyclopropane derivatives. Stereoisomeric substituted 2-phenylcyclopropanes," Organic Magnetic Resonance 1972, 4(1), 53-62 (ODI: https://doi.org/10.1002/mrc.1270040106).
Horst Friebolin, In Basic One- and Two-dimensional NMR spectroscopy; 3rd Edition, WILEY-VCH: Weinheim, Germany, 1998 (ISBN: 978-3-527-32782-9).

Maybe it would help to comprehend the coincidental changes in $\ce{^1H}$ and $\ce{^13C}$ chemical shift by understanding the origin of chemical shift. In both cases, very crudely speaking, the influence of ring currents and electronegative elements/groups cause a downfield (away from zero) shift. The factor of 20 quoted is a function of the number of electrons (usually outer ones) involved in forming the bond. So the trend will also be true for other elements e.g. $\ce{^15N}$ where aromatic N compounds i.e. pyridine have a larger downfield shift.


text:
Chemistry behind lip balm coloring based on skin pH
I stumbled across an ad for a clear, colorless lip balm that purportedly tints your lips a shade of pink determined by your skin's pH.

This balm stain applies on clear then reacts with your skin's pH levels, uniquely creating the perfect shade of pink lips just for you.

Out of curiosity (and the fact that I'm an easy target for marketing), I read further. The product contains the following ingredients:

Hydrogenated Polyisobutene
C12-15 Alkyl Benzoate
Diisostearyl Malate
Octyldodecanol
Dibutyl Iauroylflutamide
Dibutyl Ethylhexanoyl
Thylhexanoyl Glutamide
Isononyl Isononanoate
CI 45410
Chrysanthemum (a whole flower added to the bottom of the tube, presumably just to make it look nice.)

Other than CI 45410, I can't find anything on the above list that mentions anything about color -- and there's nothing listed that I think could change color based on pH.
Note: Some of the ingredients are very hard to find data for, but most seem to be thickening agents.
Does this clear lip gloss actually have the ability to change color based on pH or does the color come from the CI 45410? If it's from the CI 45410, what causes it to go from clear to pink?

I will hazard a guess here. It would be easy for you to confirm or falsify this guess. Looking at the ingredients (some of which have had their names/spelling mutated beyond recognition) I suspect this question is equivalent to the question: "If water is transparent, why is snow white?" That is, the pigment (or dye, not sure) Red 27 Lake isn't going to become not red at any remotely reasonable skin pH. So, the only explanation I can think of is that the red is obscured by the presence of liquid crystals. Those could be (mildly) sensitive to pH but are [con't below]

https://en.wikipedia.org/wiki/Phloxine

[con't] far, far, far more likely to be temperature sensitive. This is just speculation on my part, but I suspect the lc phase is simply acting as a light scatterer rather than changing the absorption wavelengths and that the amount of lc is highly sensitive to temperature near the temperatures of skin (lips).

CI 45410 (A.K.A. Phloxine B or acid red 92) is a pH dependent dye which is clear below about pH=4.1 and pink at pH>4. The formulation is mildly acidic and is neutralised by saliva upon application to become pink.
(Thereby "uniquely creating the perfect shade of pink lips just for you.")


text:
Crystallization in a solution of 2 or more solutes
This is a very basic question I found myself wondering. For simplicity's sake, we'll say it is a super saturated aqueous solution of an equivalent measure of sugar and salt. As the solution cools and crystals form, do you get sugar crystals, salt crystals, cocrystals or discrete crystals of both?

When the solution contains several substances, it is very often possible to achieve their separate sequential crystallization due to the different nature of the solubility of the substances as a function of temperature. In a certain temperature range, the solution will always be saturated with respect to one substance and unsaturated with respect to the other. This leads to the isolation of only the first substance from the solution, while the other remains in the mother liquor.
The difference in the compositions of the liquid (vapor) and solid phases formed during the partial crystallization of the solution, melt, and gas phase always leads to the fractions enriched with one component, and sometimes even to practically pure components. Possible degree of separation during crystallization is determined by the type of the phase diagram (composition vs temperature) in the solid-liquid(gas) system.
Obviously, salt and sugar do not form solid solutions, therefore depending on the concentration and temperature one component will crystallize first, and, when water is fully evaporated, a mechanical mixture of crystals is left.
This is a basis of various crystallization methods of mixtures (binary or multicomponent) separation, such as:

Fractional crystallization;
Crystallization in the presence of auxiliary substances;
Fractional melting and sublimation;
Multiple recrystallization.


text:
Rubber composition A13 B13 C20 F17 K11 L14
I am looking at a rubber compound with the following description:
A13 B13 C20 F17 K11 L14
I cannot find the meaning for "A" and "L".
I am not a chemist.

I don't think either of those are elements.

Whatever it is it does not make sense as a chemical formula.

@M.A.R. These are not chemical elements. It looks like they are ASTM D2000 specs.

I'm voting to close this question as off-topic because it has nothing to do with chemistry. Just because something looks like a chemical formula to some laymen does not mean it is one. The answer confirms the suspicion.

TL;DR I think you might need two more things to interpret this line: the first code part of the standard, and the standard itself - ASTM D2000 (needs to be purchased).

These descriptors are not chemical elements, they seem to be the part of ASTM D2000 "Standard Classiﬁcation System for Rubber Products in Automotive Applications", so what you are looking at is the suffix part of the full spec, which is usually written as:

The part after ASTM D2000 includes information about units of measure, grade durometer hardness and tensile strength. This is a rather essential sequence and if you don't have it it is impossible to define the exact meaning of the given suffix, which are:
A: Heat resistance;
B: Compression set, standard test specimen cut from a slab;
C: Ozone or weather resistance;
F: Low-temperature resistance;
K: Adhesion;
L: is not explicitly defined in D2000−12, probably falls into Z: Any special requirement, which shall be specified in detail.

Each suffix letter should preferably be followed by two suffix numbers [...] The first suffix number always indicates the test method; time of test is part of the method [...]. The second suffix number, if used, always indicates the temperature of test [...]. Where three-digit numbers are required, they are separated by a dash.

Exact characteristics of the rubber can be found in numerous tables from ASTM D2000-12 datasheet, but without having the first part of the descriptor there is not much to be done.
The exact set of suffixes you provided can be found in this report, the rubber they refer to is ASTM D2000 5AA425 A13 B13 C20 F17 K11 L14 used for shock absorbers.


text:
MOLCAS CASPT2 energy
I have calculated all the roots for a specific symmetry and I need to calculate the CASPT2 energy for all the roots. For example: here I have 4 roots and I need to calculate the CASPT2 energy for all 4 roots. However If I only use the CASPT2 keyword, only the energy for first root wil be calculated. How can I calculate the energy for all of the roots?
Natural orbitals and occupation numbers for root  1
    sym 1:   0.999414   2.000000
    sym 2:   2.000000
    sym 3:   2.000000
    sym 4:   1.000587   2.000000
    Natural orbitals and occupation numbers for root  2
    sym 1:   1.501037   1.504109
    sym 2:   2.000000
    sym 3:   2.000000
    sym 4:   1.496113   1.498741
    Natural orbitals and occupation numbers for root  3
    sym 1:   2.000000   1.999999
    sym 2:   0.999213
    sym 3:   1.000789
    sym 4:   2.000000   1.999999
    Natural orbitals and occupation numbers for root  4
    sym 1:   2.000000   1.011097
    sym 2:   1.999999
    sym 3:   2.000000
    sym 4:   2.000000   0.988905

Have a look at the manual of Molcas. Especially keyword 'MULTistate' for CASPT2 input. http://www.molcas.org/documentation/manual/node70.html#SECTION05233100000000000000


text:
Looking for a two step first-order reaction with a useful intermediate
I am working on a mathematical model for my homework, and would like to do some tests to see if it is accurate in real life as well. The model involves a two step chemical reaction in the form of the following:
\begin{align}
\ce{A + B &-> C + D} \tag{1}\\
\ce{C + D &-> products} \tag{2}
\end{align}
Is anyone aware of such processes where either $\ce{C}$ and $\ce{D}$ are:

measurable (not necessarily continuously, could be done by say, quenching the reaction at intervals);
are also useful in real life in some way or another;
where the order of reaction with respect to the chemicals are first order?

Any ideas are welcome!


text:
IUPAC Naming of CH3C(p-ClC6H4)2CH(Br)CH3

According to the locant rule, the numbering should be done from left end, as from left end locant positions are 2,2,3 and from right it is 2,3,3.
So, the IUPAC name should be 3-bromo-2,2-bis(4-chlorophenyl)butane. But according to most textbooks, its name is 2-bromo-3,3-bis(4-chlorophenyl)butane. Is alphabetical prefrence given to bromine, and if so, why?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.
For the compound that is given in the question, the preferred IUPAC name is actually generated using multiplicative nomenclature (see also What is the rationale of the IUPAC name for DDT?), which yields the name 1,1'-(3-bromobutane-2,2-diyl)bis(4-chlorobenzene).
In general nomenclature, nevertheless, you may choose the butane chain as the senior parent structure and generate the complete name using substitutive nomenclature.
The most important criteria for the numbering in substitutive nomenclature according to the current version of the IUPAC Recommendations are:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Since the compound that is given in the question has no principal characteristic group and the butane chain, which is taken as the parent structure, has no multiple bonds, the rules (c) and (e) are not relevant in this case.
Thus, all substituents are considered together according to rule (f). Therefore, the correct name is 3-bromo-2,2-bis(4-chlorophenyl)butane and not 2-bromo-3,3-bis(4-chlorophenyl)butane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.


text:
Confusion about Sodium Nitroprusside
In Sodium Nitroprusside and related compounds (eg. brown ring complex) why is the oxidation state of NO +1 and not zero? It seems pretty unusual without any explanation.
Could anyone please explain this, and maybe provide a satisfactory theoretical explanation for the exhibition of 'unusual oxidation state'?

I have checked these. There is no proper theoretical explanation, though.

I'd like to know when NO exists in 0 and when it exists +1, with some sort of theoretical explanation.

@NilayGhosh Did you delete those links? Even vaguely related stuff should stay linked. Also why you didn't CV duplicate?

Those links do not answer my confusion. Some say that Fe is in +3 and NO is in -1, while some say that both are in +1 oxidation state. Which one is right?

@user28968 read the answers in by F'x and DavePhD, they agree that Fe is 3+ and NO is -1, point to literature that supports this, and that both being +1 was a misconception.


text:
E2 Reactions in Cyclic Structures
In my textbook, the percentages of anti-eliminations are given which are quite contrary to expectations. The reason is also not given quite clearly. Why does the preference for anti-periplanarity change and what are the reasons for the disparity between the different cyclic structures?
\begin{array}{lr}
\text{Ring Size} & \%\text{ SYN Elimination} \\
\hline
\text{Cyclobutyl} & 90 \\
\text{Cyclopentyl} & 46 \\
\text{Cyclohexyl} & 4 \\
\text{Cycloheptyl} & 37
\end{array}
The values are for $\ce{(CH2)_nC-HY}$.

You need to build some models or look at some computed examples. It's really hard to achieve a conformation where the hydrogen atom and the leaving group are anti-periplanar in some cases.

To begin with, I want to present you approximate representations of the four cycloalkenes in question. Unfortunately though, these representations cannot, in this case, be used to fully replace building the molecules; I suggest you do so with the molecular modelling kit of your choice.

I’ll start off with the six-membered ring. This — as is often correctly pointed out — is essentially strainless and the chair conformation displayed is the most stable one. While frequent interconversion between different chair forms can occur, switching axial and equatorial substituents. We can thus expect that a reaction will, in principle, always take place in this chair form.
If we examine the dihedral angles, we notice that an axial substituent is anti-periplanar to the adjacent two carbons’ hydrogens. This means, it can easily anti-eliminate in both directions. No substituent whether equatorial or axial has any ecliptic hydrogen in either direction. This means, that a syn-specific elimination, which requires an ecliptic configuration, cannot occur out of the chair conformation. The $4~\%$ syn-elimination the table references must occur from other, minor configurations such as the boat.
Cyclopentane is also hardly strained. However, the configuration of the ring is much more volatile. Furthermore, even in the shown configuration there are possibilities of syn-eliminations; for example see the two carbons opposite to the upwards-pointing one. Still, anti-eliminations are also well possible, especially from the other three carbons that form the upwards arch, and account for most of the elimination.
A similar case can be made for cycloheptane, which is even more strained. There is not much for me to say here that I haven’t said for cyclopentane other than that you should build it.
And finally cyclobutane. This is the most strained of the systems considered and probably the one with the most fluctuating configuration. Here, it looks like an anti-elimination is possible; but this requires the substituent to occupy an axial position. This configuration is strongly sterically hindered. Syn-eliminations can occur from any position by only moderate shifts from the geometry shown. This may explain why anti-eliminations are strongly disfavoured here. I would, however, once again stress how helpful it is (especially in this case) to build the model.


text:
How many bonds can nitrogen form?
As known, nitrogen could form 3 bonds based on octet rule, because it has 5 valence electrons. That means it needs 3 bonds.
On the other hand, why sometimes nitrogen forms 4 bonds?

Because lone pairs can make dipolar bonds...

The title and the question are somewhat incoherent. The question is rather trivial, whereas the title is quite interesting. If this is about the maximum number of  bonds, I would answer I saw a structure with 7.

@andselisk True, but the original title (prior to the edit) was even more interesting ;)

@paracetamol Ah, the mighty sodium. Probably OP should clarify what exactly is expected from the answer.

I want to know in which conditions could nitrogen or other elements form more than it is supposed to?

Well, for nitrogen the richest coordination environment I know is a capped trigonal prism, C.N. 7 (Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288). But this is a rather non-standard case as the $\ce{N}$ atom is trapped inside a metal framework.

Insert Professor Klapötke who is certain that $\ce{NF5}$ exists but hasn’t been synthesised yet.

I'd like to try answer the question from the title regarding the maximum number of atoms nitrogen is capable of bonding with, and also slightly expand my comment.
Metal nitrido complexes are commonly known to have up to 6 metal centers bound with a single bridging $\ce{N3−}$ ion, located in an octahedral cavity. An interstitial nitrogen can contribute 5 electrons, and the rest is provided by the group 9 and 10 metals which are electron-rich (typically, $\ce{Rh}$, $\ce{Ir}$).
There are few more exceptions where formal C.N. for nitrogen is 7: complexes of lithium amides based on $\ce{\{Li14N10\}^{6-}}$ cluster framework [1,2] and an inclusion nitrido-cluster $\ce{\{Co2RhN2\}^{3-}}$ [3]. Unfortunately in both crystal structures [1,2] with 6- and 7-fold coordinated nitrogens of $\ce{N-Ar}$ groups those are heavily disordered. Structure [3] is more suitable for the representation.
tris(Tetramethylammonium)($\mu_7$-nitrido)-($\mu_6$-nitrido)-decakis($\mu_2$-carbonyl)-undecacarbonyl-deca-cobalt-rhodium(I) $\ce{[Co10RhN2(CO)21]^3-}$ [3] contains two non-equivalent 6- and 7-fold coordinated nitrogen atoms ($\mathrm{N2}$ and $\mathrm{N1}$, respectively), sharing a triangular face:
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$;
$\color{#0A7D8C}{\Large\bullet}~\ce{Rh}$;

Cluster core wireframe model without carbonyl ligands:

Atom $\mathrm{N1}$ with C.N. 7 is coordinated with 6 cobalts and 1 rhodium, forming a capped trigonal prism. Interestingly enough, $\mathrm{Co1}$ is a capping atom, not rhodium:

     N1   SYMM Co5   Co4   Co6   Co3   Co2  Rh1  Co1
Co5  1.90 I    -     -     -     -     -    -    -
Co4  1.91 I    135.0 -     -     -     -    -    -
Co6  1.92 I    79.5  80.2  -     -     -    -    -
Co3  1.98 I    129.7 85.1  80.2  -     -    -    -
Co2  2.00 I    82.3  140.2 128.4 75.4  -    -    -
Rh1  2.18 I    80.7  81.1  128.3 144.9 95.3 -    -
Co1  2.43 I    143.5 70.3  136.5 66.4  70.1 78.6 -

Both interstitial nitrogens play the role of internal ligands, which provide
cluster valence electrons (CVE), but don't contribute to steric hindrance between external ligands such as carbonyls, making the cluster more stable [4, ch. 1.18]
Bibliography

Armstrong, D. R.; Barr, D.; Clegg, W.; Drake, S. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wright, D. S. Angew. Chem. Int. Ed. Engl. 1991, 30 (12), 1707–1709. DOI 10.1002/anie.199117071.
Armstrong, D. R.; Ball, S. C.; Barr, D.; Clegg, W.; Linton, D. J.; Kerr, L. C.; Moncrieff, D.; Raithby, P. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wheatley, A. E. H.; Wright, D. S. J. Chem. Soc., Dalton Trans. 2002, 0 (12), 2505–2511. DOI 10.1039/B107970K.
Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288.
Metal clusters in chemistry; Oro, L. A., Braunstein, P., Raithby, P. R., Eds.; Wiley-VCH: Weinheim; New York, 1999. ISBN 978-3-527-29549-4.

while esoteric examples with high numbers of "bonds" are fascinating, the OP was asking about the far more basic case of why nitrogen can form 4 covalent bonds. For that question, this answer probably just confuses (though it would be a good one to a different question).

How can nitrogen form this compound then? It cannot donate 2 electrons (coordinate covalent bond) to the R-group as the R-group only needs 1 electron for an octet.

@tryingtobeastoic I'm not sure how this is related to the answer or why post a simple formula as an image, but it's just an isocyanide which exists as resonant hybrid of two canonical forms: bipolar with a negative charge on the carbon atom and carbene with divalent carbon.

Nitrogen has three electrons in its 2p orbital. Therefore, it can form three bonds by sharing its three electrons.
It cannot accept any more electrons but here's how it forms the fourth bond.
Nitrogen has one lone pair of electrons in its 2s orbital. It can donate this electron pair to form a coordinate bond. This coordinate bond that nitrogen forms by donating its electron pair to the vacant orbital of other atom is how it can form 4 bonds.

How can nitrogen form this compound then? It cannot donate 2 electrons (coordinate covalent bond) to the R-group as it only needs 1 electron for an octet.