text:
Single crystal x-ray diffraction experiment backwards: from CIF to genuine set of raw data without performing actual experiment
Single crystal x-ray diffraction stands among the most complex yet trustworthy analytical techniques when it comes to structure determination of solids. But how hard it would be to simulate(fake) the entire raw dataset, including all reflections, possible artifacts, experiment strategy etc. in order to fit a certain given structure? Is it even theoretically possible? Wouldn't it happen at some day that with all the computational power and trained neural networks XRD suddenly looses its good reputation?
Here
is a brief report from the editorial section of Acta Cryst. E which overviews fraudulent content, which is still relatively easily detectable with the help of tools like CheckCIF even without looking at raw data. But can one commit a "perfect crime" by faking raw data entirely?

I have a vague apprehension this is going to be moved to Academia.SE or elsewhere. Long story short, yes, XRD is trustworthy, but not because it is hard to fake (which, arguably, is not all that hard).

I'm completely OK with that, though I was thinking that maybe anyone could point to an existing precedent, or state whether it is doable at all, and if yes, then how. Maybe I should've asked this on chemport.ru instead:)

@andselisk  On terminology: In the experiment you refer to, the CCDs (and other sensors) on a diffractometer record the diffraction intensities, rather than the reflection intensities.  Regarding your theoretical question ... a short answer is "yes".

@Buttonwood From my knowledge it's actually exactly the reflections intensities. I specifically just checked CrysAlisPro and PLATON manuals for that. I've heard people using diffraction intensity when it comes to powder XRD, but in single-crystal XRD it's reflections what you are working with,  indexing, applying weight scheme etc. Answering yes, could you probably elaborate more in a form of answer in details?

Simulation of XRD patterns are straightforward (both powder and single crystal). Powder diffractions can be easily obtained even with e,g. Mercury. Would you elaborate the question: why do you think it is a challenge?

@Greg I'm not asking about powder XRD. The title has "Single crystal x-ray diffraction" in it, I thought it was explicitly enough.

@andselisk Ok, ShelX lists the data under "reflection", yet the title of the question joins with "diffraction" ... why not "intensity data".  Well, perhaps "diffraction" is more general (http://pd.chem.ucl.ac.uk/pdnn/diff2/structf.htm) here?  While the original question is about single crystal diffraction, I intentionally included powder diffraction as the two share the principle to minimize the difference of experimentally accessible electron density map and calculated electron density map; the former recorded by CCD or point detector.

I guess I never thought that x-ray diffraction was particularly complex, and should be trusted no further than any of a number of other experimental techniques. Bottom line, if somebody sets out to consciously do wrong, they can for quite some time, depending on how important what they do is to other people. On the flip side, there are a variety of alloy phases where the exact crystal structure is still being debated, a 100 years after they were first discovered (some are really hard to pin down).

@andselisk I said BOTH powder and single crystal... It is still not clear what is your asking / why do think calculating reflection from crystal structure would be any more challenging than the inverse problem?

I do recall a case of someone faking quite a few submissions to Acta Cryst E, so faking them is quite possible. I don't have a citation for that though, so can't give you a posted answer.

Analytical methods are not "trustworthy", this adjective applies to humans and collections based on human decisions. And of course can you fabricate data of any analytical method for a fake substance. Any exact (artificial) 3D electron density distribution has a clearly defined diffraction pattern. It's the opposite direction that is not as simple.

@JonCuster alloys are rarely the subject of SC-XRD, at least in the industrial scale, I agree. Powder XRD and  x-ray dynamic defectoscopy are used way more often instead. But I don't just want to focus on particular case, I would like to find a unified answer for both small molecules and proteins single crystal diffraction.

@Greg it's not just about just calculating the reflections, but faking the entire dataset, including adsorption correction, reverse-engineer experiment strategy for goniometer, broadening of diffraction beams, cryojet parameters, CCD frames etc.

@Karl True, but partially. Commercially available SC-diffractometers where you can just mount a crystal and in a span of few hours you get the structure submitted and registered exist for decade or so. It doesn't work in all cases, more often than not human intervention is required, but not necessarily required anymore in simple cases. And yes, it is the opposite direction I'm curios about.

@Canageek This is a good example, if I find a decent link to that case I'll add it to the question. But these fakes were quite obvious and they tended to serve more like an advertisement for checkCIF utility as they primarily contained improper atom assignments, ridiculous ORTEP plots and one didn't even need to have access to their raw data to prove these persons wrong. I'm asking about a "perfect crime" if I'm allowed to say so, where it's going to be impossible to find evidence even on a data collection level (which is 100% simulated).

Faking XRD data is not a perfect crime, but stupid, expecially if it is submitted to some database, where the fake will definitely get detected sooner or later.  Or pointless, how would you know the structure is indeed 100% correct if you don't have an xray pattern?

@Karl Submitting the dataset (which is several Gb in size) to database is not required as of now. You submit the result of your refinement - usually, a CIF file. Again, I'm speaking about single-crystal x-ray diffraction, not the powder one. You used the term "xray pattern", that arouse my suspicion that you don't fully understand the question. And let me remind you that a century ago Edison also was also labeling the idea of using AC as a mean of electricity delivery as stupid.

@andselisk I thought you wanted to fake the complete dataset? But nevermind, you can only do that perfectly if you have the correct crystal structure, which you can only have if you have an actual crystal and measured it. But then why fake it?

Structure determination by X-ray diffraction, regardless if powder diffraction, or single crystal diffraction analysis, is a spatial mapping of electron density.  After solving a crystallographic structure (phase problem), the subsequent structure refinement includes steps to complete the crystallographic model, and to minimise the differences in the Fourier maps mapping observed electron density ($F_O$), with electron density predicted by your current model ($F_C$).  This minimisation is done until this said difference is below a threshold value, and if the experimenter is satisfied with the result. For the later, this includes the complete and reasonable attribution of atoms (does this electron density belong to an atom of Carbon, or Nitrogen?), checking if bond distances and angles included in the model are consistent with the body of data determined in other experiments.  Some of the critera applied are interdependent to others.
If the crystallographic model is completed (in layman's terms, if there is a *.cif file), than it is easy to predict a theoretical powder diffraction pattern.  Programs like CCDC Mercury just ask about the wavelength to consider and offer graphical and numerical output like

Such a theoretrical diffractogram is then compared with an experimentally recorded one; for example to check the newly prepared sample's phase identity (polymorphism) and purity.
In PXRD, spatial arrangement and conformations of subunits in the crystallographic model are optimised (refined) until yielding a model that chemically and mathematically is reasonable. This is the reason you see in publications a difference trace below the the superposition of experimental and predicted diffraction pattern:

(source)
For single crystal diffraction, the generation of a 3D diffraction model computationally might be more costly than for PXRD, on the other hand, as said, programs like ShelX routinely compare experimentally recorded $F_O$ with predicted / calculated diffraction intensities $F_C$ derived from the current set of atoms (their coordinates, occupancy, etc.) of the model. It is done internally. ShelX' *.lst listings for example include a section like:

Hence I answer your question, if diffraction pattern may be simulated / faked with a "yes it is possible, provided there is a good data set about how the atoms / molecules are spatially arranged".  Initiatives like the recurrent CSP blind tests to predict crystal structures and databases like PCOD may provide a foundation for this.
Then how to spot an artifical and wrong diffraction model?  Checkcif / Platon are tools to check for consistency and integrity of such models against the raw data (hkl-files), structure factors. The recent installation of checksums into *.cif files (currently under watch as Platon errors PLAT012, 013, and 014), may be seen as one element of protection if few entries were changed after completing the work described above.  These bells probably remain silent if data were compromised very early on.

I'm sorry, but this not an answer. These are the steps you routinely perform to obtain CIF-file. You also conveniently jumped over the phase problem and proceeded with fancy outcome. Sure, having larger dataset (SC-XRD experiment) you can always derive a subset (simulated powder diffractogram), this is trivial. My question is, would it be possible to produce initial/raw dataset just as if it would be collected by single-crystal diffractometer, only having desired crystal structure - all that info that you find in CIF files -- coordinates, occupancies, thermal parameters, etc.

@andselisk As the listing + the next-to-last paragraph aim to show, yes it is possible.  Quantum chemistry may give you reasonable distances of atoms within a molecule, and may optimise the packing (hence the contests mentioned).  Thankfully not too often the atoms of a skeletton of organic molecules may be described with a SOF equal to 1 (or 11, in ShelX' notion), easing the situation.  And with more than 500k entries in the CCDC, there are plenty examples of anisotropic displacement parameters, for many atoms even in different bonding environment and experimental temperature, available, too.

@Buttonwood Anisotropic displacement parameters is not something you may just copy from another file. Making the ellipsoids "look right" would probably be the hardest part of the job. Then again, it is not terribly hard.

@IvanNeretin I agree that ADPs are not just a simple scalar, and split into components (A in ADP).  A speculative criterion were C(sp3), chain teriminal != near a rigid aryl. Yet for my computational colleagues around, interested in homo/lumo calculations, just to know the centre of the atom in question often is already enough (or, if facing disordered Me / $\ce{CF3}$ / tBu groups to drop one of the two populations at all). If not too distorted (under their visual inspection), they don't mind if the refinement included a set of EADP's, some RIGU, or the simple kept ANIS. Sigh.

@Buttonwood That's right, but we are talking about the reverse problem: to produce some realistically looking ADPs without any actual data.

@Ivan Neretin Thank you, that's exactly what I meant.

@Buttonwood There is no need to use quantum chemistry for establishing the geometry. My assumption is that that ALL info you find in CIF file already given. The assignment is to generate a plausible "fake" dataset, which upon solving and refinement routines will again give our CIF.

@Buttonwood And I'm not going to downvote your answer, it's 100% correct and beautifully crafted, but it answers another question -- "How do I perform a single crystal diffraction experiment and use this data to prove the polycrystalline sample I synthesized is what I think it is?"


text:
Selenium oxydichloride bond angle
According to me , the $\ce{Cl—Se—Cl}$ bond angle should be greater than $\ce{Cl —Se—O}$
bond angle because since oxygen is more electronegative than $\ce{Cl}$ , the orbital directed towards $\ce{O}$ will have more p character hence, the bond angle should be less than $\ce{Cl—Se—Cl}$ bond angle.
But in my book its written that the $\ce{Cl—Se—O}$ bond angle is greater. Why?

Yes  they coexist within a single molecule [SeOCl2]

My suspicion is that the info in the book relies on single-crystal x-ray diffraction experiment (1), where geometry of selenium oxydichloride cannot be determined solely from MO principles. So, here is the structure of dioxane-seleninyl-dichloride:

As far as you can see, Se1 atom has three more neighbors: O1 from another $\ce{[SeOCl2]}$ unit, and O2 + O3 from dioxanes. The bond angles correspond to those you discovered in the book, $\angle \ce{Cl1-Se-Cl2} = 95.6(3)$, and $\angle \ce{Cl1-Se-O} = 101.6(7)$:

The reason for that is given in the paper as well:

APPLICATION of the Gillespie-Nyholm theory to the molecules of
  seleninyl dichloride, $\ce{SeOCl2}$, ... leads to the same prediction
  for each:  pyramidal geometry with the fourth position of a
  tetrahedron occupied by a lone pair. However, more detailed study of
  many non-metal compounds containing lone  pairs has shown that the
  Gillespie-Nyholm approach gives an incomplete description of the
  bonding. The central atoms very frequently form additional bonds, of
  greater length than the primary covalent links. In the case of
  pyramidal molecules, it has been found that there are often three of
  these secondary bonds, completing a distorted octahedron around the
  central atom. Striking conformation of this has come from complexes
  (or solvates) of $\ce{SeOCl2}$. All show $\ce{Se-O}$ and $\ce{Se-Cl}$
  secondary bonds, with distorted octahedral geometry, although in some
  cases the third secondary bond opposite the $\ce{Se-O}$ bond may be
  significantly longer than the other two secondary bonds. Furthermore,
  there are rare examples where a secondary bond has been replaced by
  two longer contacts, and it may be inferred from these results that
  the steric importance of the lone pair of electrons on $\ce{SeOCl2}$,
  varies. In those cases where an octahedron can be observed, the angles
  $\ce{X-Se-Y}$ ($\ce{X, Y}$ = $\ce{O}$ or $\ce{Cl}$) have been found to
  lie in the range 142 -- 173, and there appears to be no clear way of
  deciding which atoms, chlorine or oxygen, are going to form the
  secondary bonds. The present structure of $\ce{SeOCl2 * C4H8O}$, is,
  however, the first in which $\ce{SeOCl2}$, is involved in three
  secondary bonds to oxygen with the $\ce{SeOCl2}$, being classed as
  amphoteric.

Original
Long story short, your assumption would probably be correct for the "idealized" molecule in gas phase, whereas in reality there always will be a distortion as $\ce{As}$ strives to complete unsaturated coordination environment.
(1) Alcock, N. W.; Sawyer, J. F. J. Chem. Soc., Dalton Trans. 1980, No. 1, 115–120. DOI 10.1039/DT9800000115

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What did actually happen when Staudinger tried to synthesise diamond?
It is a famous reaction that used to be a teaching lab experiment, but is now banned in Germany, because it is too dangerous. To quote-translate quite liberally from the German www.seilnacht.com:

In a stone quarry near Zurich, Hermann Staudinger attempted to artificially produce a diamond by reacting the potassium-sodium melt with tetrachlorocarbon in a bomb tube. The attempt failed as the mixture tore the bomb tube apart.
In einem Steinbruch bei Zürich versuchte Hermann Staudinger einen Diamanten künstlich herzustellen, indem er die Kalium-Natrium-Schmelze mit Tetrachlorkohlenstoff in einem Bombenrohr reagieren ließ. Der Versuch misslang, da die Mischung das Bombenrohr zerfetzte.

I was probably one of the last students who were still able to see this experiment, and while considerably down scaling the actual set-up, it was still one of the most powerful explosions I ever witnessed.
To me it was actually surprising that I found pages, while researching this reaction, that asked whether $\ce{CCl4}$ can act as a solvent in the Wurtz Reaction, or that in the Wurtz reaction of carbon tetrachloride and sodium no product would be formed, because $\ce{CCl4}$ is too stable to react.
However, due to this I found the following passage in Inorganic Chemistry:

Tetrahaloalmethanes can be reduced by strong reducing agents, such as alkali metals. For example, the reaction of carbon tetrachloride with sodium is highly exoergic:
  \begin{align}
\ce{CCl4 (l) + 4 Na (s) &-> 4 NaCl (s) + C (s)}&
\Delta_\mathrm{r}G^\circ &=\pu{-249 kJ mol^-1}
\end{align}
  This reaction can occur with explosive violence with $\ce{CCl4}$ and other polyhalocarbons, so alkali metals such as sodium should never be used to dry them.

Unfortunately I reached a dead end with finding more literature about the reaction. I am especially interested in the mechanism, and the final products. I would assume that the above equation only represents a theoretical model, and the products will likely be carbon dioxide and other combustion related products. I'd be interested to know if there are any practical and/ or theoretical studies to the kinetics and thermodynamics of the reaction.
If the source does not cover tetra-, but tri-, or bihalogenated methane, I assume it is fine, too, but it should go beyond the Wurtz reaction ($\ce{2 R-Cl + 2 Na -> R-R + 2 NaCl}$).

We need a tag for this kind of question... like... [tag:reaction]?

@orthocresol I tried that, but it wouldn't let me. They really hate fun at stack exchange :(

@Martin-マーチン So how was this reaction initiated? And are you interested in the products of a controlled reaction with dilute carbon tetrachloride in an inert solvent, or in Staudingers bomb?

I know very little about chemistry, so I don't know whether there's any connection, but.... https://en.m.wikipedia.org/wiki/Detonation_nanodiamond

Isn't  it possible that the $\ce{NaCl}$ gets incorporated into the diamond structure. In such case, the diamond wouldn't be pure though.

@Karl I am generally interested in the reaction mechanism, and at what point it would explode, or what is the reason for the explosion, and why it is so terrifying. If there are studies that do the reaction in a controlled manner, that would be interesting, but I guess those studies wouldn't stop only at this level.

Studies of the controlled reaction would stop quite a way from the point of reaching explosiveness. Chemists value their fingers. And groups that do scientific work on explosives are few and far between.

@Karl Without encouraging further comment, I am not interested in those. There are perfectly solid studies on the hydrogen-oxygen system, without the researchers blowing up the lab. If you cannot extrapolate your controlled reactions towards the catastrophe, how good are they then anyway.

This is not really an answer, but more like a historical and literature review. Anyway, I hope it can help to further understand the reaction the the ideas behind.
First of all, thaditional Wurtz reaction suggests formation of 2D bonds such as $\ce{R^1 - R^2}$, whereas the idea in Staudinger's experiment is to obtain hree-dimensional network, ideally the the $\mathrm{sp^3}$ diamond hybrid state. Reaction of alkali metals with organics with high halogen content most likely will always lead to an explosion. For this reason qualitative Lassaigne's test should always be performed carefully, and one must not use fire extinguishers with tetrachloromethane for burning sodium.
According to the great review in [1, p. 8], the ﬁrst documented record of a combustion reaction between a metal and halocarbon compound refers to the ways of obtaining aluminium carbide for technical methane production. In 1907 Camille Matignon observed a very exothermic reaction between hexachlorobenzene and aluminium [2]:
$$\ce{10 Al + C6Cl6 ->[\pu{225^\circ C}] 2 Al4C3 + 2 AlCl3}$$
In 1913 H. Staudinger published his attempts to synthesize ethylenedione $\ce{O=C=C=O}$ [3]. He suggested  abstracting the halogen atoms ($\ce{X}$) from either oxalyl chloride and -bromide $\ce{(COX)2}$ by a reaction with potassium or its sodium alloy ($\ce{NaK}$). This is one of the experiments performed by Staudinger (doing science was much more fun back then):

In einer starken Bombe wurde aus 5 g Kalium und 5 g Natrium die flüssige Kalium-Natrium-Legierung hergestellt, dazu ein Röhrchen mit 8 g Oxalylbromid gegeben, und die Bombe nach dem vollständigen Evakuieren zugeschmolzen. Wenige Sekunden nach dem Zertrümmern des Röhrchens trat äußerst heftige Explosion unter gewaltiger Detonation ein, die nur durch glücklichen Zufall ohne erhebliche Beschädigung des einen von uns verlief.
  Liquid potassium-sodium alloy was prepared from 5 g of potassium and 5 g of sodium in a strong bomb, then a tube of 8 g of oxalyl bromide was added, and the bomb was melted after complete evacuation. A few seconds after the tube had been trimmed, and an extremely violent explosion ensued, with a great detonation, which proceeded only by chance without considerable damage to one of us.

Formation of an unknown highly reactive intermediate such as Dioxyacetylenkalium was suggested, but it was not confirmed experimentally.
A decade later, in 1922, Staudinger reported explosive reactions of alkali and alkaline earth metals with partially and perhalogenated solvents ($\ce{CH2Cl2}$, $\ce{CCl4}$). He again assumed the formation of different very instable species on contact that would be very sensitive to mechanical impact and thus trigger an explosive reaction. [4, part "5. Explosion mit Alkalimetallen"].
At the same time he also proposed usage of reaction between various alkali and alkaline earth metals ($\ce{Na, K, Ba}$) with halogenated organics with short $\ce{R}$ ("Rest des Moleküls") ($\ce{CH2Cl2}$, $\ce{CHCl3}$, $\ce{CCl4}$, $\ce{CH2Cl-CH2Cl}$, $\ce{CHCl2-CHCl2}$, $\ce{CHCl2-CCl3}$, $\ce{C2Cl6}$, $\ce{CHCl=CHCl}$, $\ce{CHCl=CCl2}$, $\ce{CCl2=CCl2}$, also corresponding $\ce{Br, I}$-derivatives) as an ammo detonating charges:

DE Patent 396,209 1922 "Verfahren zur Darstellung von Sprengmitteln" (DPMA Publikationsnummer CH 100 199 A);
DE Patent 391,346 1922 "Verfahren zur Initialzündung von Sprengstoffen" (DPMA Publikationsnummer CH 100 200 A).

He also underlined that the more halogenated the molecule is and the less other groups are present, the more vigorous the explosion will be. In addition, potassium shows better overall performance due to electropositivity ("Vergleicht man die Reaktionsfähigkeit der ver
schiedenen Alkalimetalle, so ist das elektropositivste, das Kalium, das reaktionsfähigste, Lithium das reaktionsträgste"), but various amalgams or alkali metal mixtures will outperform potassium (flux).
Based on this, later its been proposed that the explosive reaction between sodium and tetrachloromethane to make diamond. His idea was implemented in late 1980s for obtaining diamonds in TNT/RDX detonation soot [5].
Finally, an autoclave synthesis based on original Staudinger work (funny enough, they didn't cite it) implementing sodium and tetrachloromethane [4] resulted in successful nanodiamond synthesis (in a 2% yield), and, subsequently, a well-recieved report in Nature in 1998 [6]:
$$\ce{CCl4 + 4Na ->[\pu{700^\circ C, 48 h};  Ni-Co (cat)] C(diamond) + 4 NaCl}$$

We call it the reduction-pyrolysis-catalysis (RPC) route. According to the free energy calculations $\Delta G^\circ (\text{diamond}) = \pu{-416.7 kcal mol^{-1}}$ and $\Delta G^\circ (\text{graphite}) = \pu{-417.4 kcal mol^{-1}}$,
  $$\ce{CCl4 + 4 Na -> C(diamond) + 4 NaCl}$$
  and
  $$\ce{CCl4 + 4 Na -> C(graphite) + 4 NaCl}$$
  are thermodynamically spontaneous. Formation of graphite (amorphous carbon) and diamond is possible, and the yield of graphite (amorphous carbon) and diamond may be determined by kinetics.

The role of $\ce{Ni-Co}$ catalyst is not clear yet, and athors suggested than experimentation with other transition metals can potentially improve the process.
Review [1] points out that idea of diamond synthesing is mentioned in Staudinger's monography [7], though I failed to find this book, so unfortunately I cannot provide any substantial details from there.
Bibliography

Koch, E.-C. Metal-fluorocarbon based energetic materials; Wiley-VCH: Weinheim, 2012. ISBN 978-3-527-32920-5.
Matignon, C. Comptes rendus hebdomadaires des séances de l’Académie des sciences 1907, 145, 676–679.
Staudinger, H.; Anthes, E. Ber. Dtsch. Chem. Ges. 1913, 46 (2), 1426–1437. DOI 10.1002/cber.19130460222.
Staudinger, H. Angewandte Chemie 1922, 35 (93), 657–659. DOI 10.1002/ange.19220359302.
Greiner, N. R.; Phillips, D. S.; Johnson, J. D.; Volk, F. Nature 1988, 333 (6172), 440–442. DOI 10.1038/333440a0.
Li, Y.; Qian, Y.; Liao, H.; Ding, Y.; Yang, L.; Xu, C.; Li, F.; Zhou, G. Science 1998, 281 (5374), 246–247. DOI 10.1126/science.281.5374.246.
Staudinger, H. Arbeitserinnerungen; Hüthig Verlag, 1961.

Thanks for the extensive review, it's a good read, and much more thorough than mine. I assume you also couldn't find any mentions of the kinetics or proposed reaction mechanisms? I also was looking for [7] without any luck, maybe somewhere in Germany there still is a physical copy...

It looks like UB Gießen, UB Frankfurt, ULB Darmstadt, UB Mainz, Stadtbibliothek Worms, and, what's more relevant for you, UB Marburg (Signatur = O 61/11) have physical copies. But again, it looks like an interesting reading, but I don't think there is anything mechanism-related (see the book's TOC)

Indeed, it is unlikely there is any of that in there. But most certainly it will be an interesting read. I mean the scales they were operating with at the time, incredible...

A great answer, but I'd like to add that explosive reactions like this don't go to equilibrium. The full reaction products would include everything from unreacted alkali metal to unreacted carbon tet, and a wide range of bizarre (and not necessarily stable) carbon-chlorine compounds with a Cl:C ratio less than 4. These will be very hot, and nearly all gases, so they will also burn when they hit the air, further complicating things. But the mixture will be dominated by the energetically favoured materials you mention.

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text:
Sulfonyl azides: diazo or azide transfer?
Sulfonyl azides, $\ce{RSO2N3}$, are capable of acting as sources of electrophilic azide ("$\ce{N+3}$"), but are also capable of effecting diazo transfer, as shown in the scheme below, taken from ref 1:

In 1990, Evans and coworkers1 showed that the aryl group on the reagent was important in controlling the outcome of the reaction:
$$\begin{array}{ccc}
\ce{Ar} & \text{yield of }\textbf{3a-S}\text{ / %} & \text{yield of }\textbf{5a}\text{ / %} \\ \hline
\ce{(4-NO2)C6H4} & 15 & 70 \\
\ce{(4-Me)C6H4} & 51 & 26 \\
\ce{(2,4,6-^iPr3)C6H2} & 91 & 0 \\
\end{array}$$
which suggests that $\ce{TrisN3}$ (the third entry) is the best azide transfer reagent.
However, confusingly, it has also been reported that under phase transfer conditions, $\ce{TrisN3}$ outperforms $\ce{TsN3}$ (the second entry) in diazo transfer to ketones.2
With this in mind, Evans concluded that

Collectively, these independent observations underscore the lack of mechanistic sophistication which currently surrounds these reactions.

Since then, have there been any efforts to rationalise the effect of the aryl group on the product distribution? And if so, how can the results above be explained?

References

Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc. 1990, 112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.
Lombardo, L.; Mander, L. N. A One-Step Synthesis of Cyclic α-Diazoketones. Synthesis 1980, Vol. 5, 368–369. DOI: 10.1055/s-1980-29022.

I'm kind of surprised that the diazo compound 5a survives in acid. Are putative mechanisms proposed for both pathways?

@Zhe partially yes for the azide transfer - not for the diazo transfer. Educated guess: https://i.stack.imgur.com/hVs8J.jpg

0r other unaccounted for factors, like sensitivity to lab light, a source UV.

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text:
How does P4O10 make sense?
I'm trying to understand the gains phosphorus has from forming $\ce{P4O10}$. The molecular structure is such that every oxygen atom forms two bonds (either two with one phosphorus atom, or one with two), while every phosphorus atom forms five bonds.
I'm clear about the gains the oxygen has from this -- it fills its outer shell to the octet rule, mimicking neon effectively. But what about phosphorus? We're talking about a covalent bond, so forming five bonds would leave it with five extra electrons, which is two more than it needs to reach an optimal configuration.
I remember reading somewhere (possibly on this site) that ionic bonds can be regarded as an extreme form of covalent bonds, where the difference in electronegativity is so great, that the metal gives its extra electrons to the non-metal. Can we assume that the difference in this case is also large enough, such that phosphorus effectively 'gives' away five electrons, therefore reaching neon too? Or am I oversimplifying it?

The octet rule isn't universal, atoms don't have to satisfy it all the time.

sigh These are single dative bonds between P and O, it's just dumb archaic way of drawing compounds.

Related: https://physics.stackexchange.com/questions/308494/how-does-covalent-bonding-actually-work

"ionic bonds can be regarded as an extreme form of covalent bonds." They "can be regarded" as magic, too. What's your point? In what specific context is such regard useful? (I'm not questioning that there may be some specific context that you could regard them that way, rather I AM saying that it is not generally useful.)
 The octet rule is used to predict valence.  Obviously H doesn't obey it. It's quite useful for the chemistry of C,N, O, and F.  But consider ClF5. The Rule is not going to get you there. Nor is it useful, imho, for P4O10 --- but there are ways to bend the meme to "explain" the deviancy. see for example the discussion of hypervalent compounds in https://en.wikipedia.org/wiki/Octet_rule#Hypervalent_molecules. Once you get away from the 2nd & 3rd Period and the 13th - 17th Groups, it shouldn't be relied upon. So, its applicability is limited, but fortunately it's applicability is for a lot of elements which are interesting for us in our environment here on Earth. Since d-orbital (and f-) contributions to hybridizations is generally small, the octet rule is generally useful. What sort of confidence should you have in using it to predict valence? Well, there's two ways to answer this: 1. as applicable to a student learning basic chemistry or 2. as applicable to experts who have in addition to the rule a vast literature on the subject. It is useful for the first and nearly useless for the 2nd.


text:
Relative orbital energies of Mn and Ni
In this MO diagram, why are the atomic Mn $3d$ orbitals higher in energy than the Ni $3d$ orbitals?

Those elements are in the same period, but nickel has a higher atomic number and therefore greater effective nuclear charge.

Higher effective nuclear charge means increased stabilization of electrons.


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Reaction of PLA plastic with methane or carbon dioxide
PLA (polylactic acid) plastic is commonly used for 3D printing, and I would like to create a 3D printed part that can be used in an experiment with methane ($\ce{CH4}$) and carbon dioxide ($\ce{CO2}$) gas sensors.  Does $\ce{CH4}$ or $\ce{CO2}$ gas react with PLA plastic?  What are the long-term effects of these gases in contact with PLA plastic?
Anaerobic biodegradation can break down PLA, producing $\ce{CH4}$ under some environmental conditions (DOI: 10.3390/ijms10093824).  But is PLA resilient to $\ce{CH4}$, or is there a chemical reaction?  I can't seem to find any reference to such an effect in the literature.
The experiment that I am conducting involves the measurement of environmental sources of $\ce{CH4}$ and $\ce{CO2}$ gas.  A hose needs to be passed into a 3D printed chamber that encapsulates the sensors.  A molecular sieve material blocks water and sources of moisture before the gas enters the chamber.  The chamber will not be heated in any fashion (room temperature or colder), and detection of the gas by the sensors is by diffusion.

Do you intend just to place PLA altogether with $\ce{CH4}$ or/and $\ce{CO2}$ alone in a dry, closed container, at room temperature?  Then the occurrence of a degradation / decomposition reaction is unlikely. Or do you add some other chemicals?  Or do you intend to warm / heat PLA and $\ce{CH4}$, or even ignite the methane in presence of PLA? Please add some details of the intended experiment.

@Buttonwood: Thanks, I've added some text to the description above.


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How do I get a value of the Gibbs energy of formation at 0 K and 0 bar?
The standard Gibbs free energy of formation for copper(II) oxide, $\ce{CuO}$, is $\Delta_\mathrm fG (\pu{298.15 K},\pu{1 bar}) = \pu{-129.7 kJ mol-1}$.
How can I get $\Delta_\mathrm fG (\pu{0 K},\pu{0 bar})$, the limiting value of this?
I want to compare it with the approximate value obtained by density functional theory calculations:
$$\Delta_\mathrm fG (\pu{0 K},\pu{0 bar}) \approx E^\text{Total}_\ce{CuO} - E^\text{Total}_\ce{Cu} - \frac{1}{2}E^\text{Total}_\ce{O2}$$

Long story short, you can't. That is, unless you find the heat capacity data for the whole range 0 - 298K and do some integration.

This doesn't make much sense. If you go down to 0K then deltaG is deltaH, so why not compare those? And why not do the calculations for reasonable temperature and pressure and cwork with that? For example if you run a frequency calculation in Gaussian without any keywords you get the values for 298.15 K and 1 atm.

VASP takes a different approach to Gaussian (planewaves). So you cannot add temperature and pressure unless you are doing molecular dynamics, which is not the case.


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Determining geometry of trisubstituted alkene
The way I know to tell whether an alkene is E or Z -is to look at the coupling constant of the protons across the doble bond. Large value (16 Hz) indicates an (E)-alkene where as smaller values (11 Hz) usually indicate a (Z)-alkene.
In my project I've been doing a cross metathesis between two alkenes which gives me a tri-substituted alkene, and I cannot observe the coupling constant across this double bond. The reaction gives a mixture of products which I think are the (E)- and (Z)-isomers. I'm able to separate them on HPLC but they do not crystallise.
Is there another method of determining this? A lab member suggested I could derivatise the alkene but I don't think this would provide any more useful information.

I think you are more likely to get a helpful answer if you show us what the structure of the alkene looks like.

Also, what about NOESY to check if which hydrogens are spatially close on the mono-substituted side to others on the di-substituted side.

Zhe I don't know whether I'm permitted to share my chemistry on this forum but the alkene is trisubstituted. On one side there is a ch2 and in the other side a CHOH

Moot point now, but you can obscure the functional groups that are further out, but any handles close to the alkene will help if you're looking for suggestions with derivatization. For example, in the answer, you might show that you have methylene and hydrogen on one side and TMS and methylene on the other.

As @Zhe points out, its not possibly to definitively answer your questions without knowing the structure of the olefin, as its important what is around the olefin, as well as just how many protons are attacked to the olefin.
If you read many of the original Grubbs' papers (and indeed any papers using a Grubbs' metathesis to make tri-substituted alkenes), the common method to determine the geometry is to use the nuclear Overhauser effect, or nOe for short.
The nOe identifies interactions through space (as opposed to standard NMR where we're looking at the interactions through bonds. By recording the nOe, one can essentially 'measure' how close two protons are, which in this case would allow you to distinguish E and Z.

Practically speaking, there are several different NMR experiments that one can run to observe the nOe.

The 1D nOe (1D-NOESY) difference experiment involves irradiation of a particular signal (for instance the one alkene proton), to allow measurement of what other protons are proximal in space. This experiment has the advantage of being quick to run, but in order to gain useful information, you have to run several of them (irradiate the alkene proton, then run a different 1D-NOESY in order to make sure that the proton that the alkene proton is 'seeing' also 'sees' the alkene proton
The 2D nOe (2D-NOESY) experiment shows all nOe enhancements in a molecule, this is incredibly useful and often used in structure elucidation, but can be slow to run, especially if the amount of sample available is limited.

How can you tell me to run this using a Bruker NMR machine ? I don't see the option on the drop down list there is only 1H 13C 19F COSY HSQC-Dept and something called dqfcosy

I suspect you're talking about an open-access NMR machine where you just leave your sample in the carousel and wait for an email. The nOe experiments require a bit more setup, so usually need running manually. I have no idea what the situation is at your lab, but either the lab members or the NMR department can likely sort this out (the NOESY experiments are present on all Bruker machines by default so there should be no problems)

As an example, FYI, see the MIT guideline on running a 1D nOe here

@JamesDealon if you use IconNMR, you're only getting a restricted list of preconfigured experiments. There are something like 20 different NOESY sequences included with Bruker Topspin, which ones you can run is the decision of your NMR facility. This is something you just have to ask your NMR facility manager, if there's no NOESY in the list in IconNMR, you can't run one without help of an administrator.


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Can Cellulose Molecules be Synthesized from Glucose?
I am reading that cellulose may be broken down into glucose. This is usually done with some sort of solvent.
Is there any way to 'rebuild' these cellulose molecules from glucose?

Are you interested in a lab synthesis (in a flask) or the biosynthesis?  Cellulose is biosynthetically derived from glucose derivatives

Cellulose may be broken into smaller pieces (though not necessarily into separate glucose molecules) and regenerated into cellophane with the viscose process.
For the home lab, rather than using the poisonous and extremely flammable $\ce{CS2}$ in the viscose process, the somewhat safer cuprammonium process is a better choice. Do be careful -- don't wind up with blue stains on skin or clothing, it shows poor technique. See this video for a demonstration.


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Why BH3 its classified as a "Molecular Hydride" and AlH3 its a "Intermediary Hydride"
First, I understand that, on $\ce{BH3}$, they eletronegativity difference between $\ce{B}$ and $\ce{H}$ is very small. Which mean that the $\ce{B-H}$ bond is less polarized.
But, I couldn't find a relation of this, to the fact that he is a Molecular hydride.
And, secondly, I couldn't understand why $\ce{AlH3}$ its "intermediary", and not molecular, as well...

(What do you mean by "relation"?) BH3 exists as the gaseous dimer molecule diborane. AlH3 is a solid that decomposes at 100°C. It's not molecular.

I see. But, i still don't understand why this happen, you know?

"Intermediary" is meant in comparison to BH3 on one side and "real" metall hydrides on the other. The differences are explained in any advanced inorganic chemistry textbook.

Alane is not a molecule: https://en.wikipedia.org/wiki/Aluminium_hydride You can't isolate molecules easily because they polymerize.
Borane on the other hand is known to exist as a dimer: https://en.wikipedia.org/wiki/Diborane
The difference in bonding is likely due to "hypervalency" (or similar concept with different terminology) in aluminum.

Article you linked shows that alane monomer was isolated and how its polymers  bonding looks, so I encourage you to expand answer.


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Concentration of a substance in a solution
Will the Concentration of any dissolvable substance be uniformly distributed in any type of liquid?
I am willing to read more about this topic , so if you have good suggestions for helpful books , please share it with us

That depends on (1) how much time you give the system and (2) what exactly you mean by "uniform."

For instance if you let sugar dissolve in water without stirring, then the concentration of sugar at the bottom of the liquid will be more sugar concentrated than the liquid at the top. The solution will eventually equilibrate, but it will take a "long" time.

Uniform: Every unit volume has the same amount of particles of that particular substance

I was thinking about gravity factors actually. That the concentration must be higher the lower we go due to gravity force acting on the particles .

Not if it's a solution. In suspensions gravity matters, but in molecular scale it's not important.

Also I have no idea why you thought that solutions could be always isotropic. On contrary, they cannot be perfectly isotropic.

Using a FULLY MIXED sugar solution (in a closed container to prevent evaporation) as an example, if you could measure out successive 1-ml samples with perfect accuracy, and could count the sugar molecules in each sample with perfect accuracy, you would expect to find that the number of sugar molecules varies and follows a normal distribution with a mean value corresponding to the "nominal" concentration of the solution.
You can observe this phenomenon experimentally in the case of standard solutions containing a given DNA sequence. When you create a series of increasingly diluted samples and then analyze each dilution multiple times using a PCR assay designed to detect that sequence (modern PCR assays can literally detect one molecule of DNA!), you begin to observe the expected proportion of negative sample results as you approach a dilution factor corresponding to one copy of DNA per sample. That is because when you repeatedly sample your solution containing a nominal concentration of one DNA copy per sample, some physical samples will have zero copies, some will have one (highest probability), some will have two, etc. with decreasing probability as you go up.


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How the electronic configuration of Hydrogen can explain the strength of the Hydrogen bonding?
So, i understand that, in this case, the partial positive charge its on the Hydrogen. And that his electronic configuration is H = 1s^1
How this fact will explain the strength of the Hydrogen bonding?
Is it because we are talking about an 1s orbital, that is much smaller, and closer to the nucleus? Or is it because the 1s orbital is electron-defficient, and, the fact that the hydrogen has a partial positive charge means that he is even more defficient on electrons. Consequently, he will "hold" this bond in a stronger way(I had a hard time on making a conclusion here).
Help on this...

There's no simple way to tell what exactly are properties of element knowing only its configuration, and bonding depends also on its partners in bond. Much like its very difficult to tell how protein behaves only from aminoacid sequence. Complexity arises in nature from simpler things.


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What are the plastic connectors called (for glassware)?
I can't remember for the life of me, and google isn't making it easy. e.g. the things labeled 24/40 etc., as in this picture:

Those are called Keck Clips. 24/40 is the size of the joint.

in mm, I guess?

@khaverim very likely, as it is a Swiss idea (https://worldwide.espacenet.com/publicationDetails/originalDocument?CC=US&NR=4442572A&KC=A&FT=D&ND=&date=19840417&DB=&locale=#)

Yes more or less, it's not the exact value. https://en.wikipedia.org/wiki/Ground_glass_joint describes the different types and the german wikipedia page actually shows the biggest diameter mm values. It's all in ISO 383

They are called Keck clamps.
Here's a link to the page for ChemGlass: http://www.chemglass.com/product_view.asp?pnr=CG-145

There are versions in metal, too; like https://www.vaporbrothers.com/Metal-Keck-Clip-19mm_ProductID_8766

They are clips not clamps

@JamesDealon Please direct your comment to ChemGlass. I don't have strong feelings about the name. The important part is "Keck."

The clips are named after Hermann Keck - Chemistry World's Classic Kit columnist, Andrea Sella, spoke to him to get the story of the invention before he died a few years ago. https://www.chemistryworld.com/opinion/classic-kit-keck-clip/3004921.article


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What determines "polarizability" of an element?
In this question, the answer given discusses that fluorine has a good orbital overlap with hydrogen and also that fluorine is not very polarizable. However, is it always true that strong bonding is inversely related to polarizability?

Fluorine is not very polarizable because it is small. Its electrons, therefore, are all close together. A polarized atom has shoved all its electrons to one side. Since they are close together in Fluorine, the negative-negative repulsion is too large for that. In another halide such as Iodine, whose valence electrons are much farther from each other, much better polarization can be achieved.
More polarizability does NOT always mean better or weak bonding. Generally, polarizable atoms like to bond to each other, and non-polarizable atoms like to bond to other non-polarizable atoms. For more info, see:
http://www.adichemistry.com/inorganic/cochem/hsab/hard-soft-acid-base-theory.html

The polarisability is a measure of how an atom's electrons can be displaced from equilibrium by a nearby electric field.
The polarisability is approximately inversely proportional to the ionisation energy (taken across periods in the periodic table); the easier it is to ionise the easier it is to alter the average electron's positions because they are less energetically bound to the nucleus.
Forming a chemical bond is an all together different thing and depends on the balance of the electron's kinetic and potential energies and these are determined by electron-electron repulsion, electron-proton attraction and proton-proton repulsion energies. These energies are calculated using a quantum mechanical approach that involves Coulomb and related integrals.

It is not true that the strength of a bond is inversely proportional to the polarisability of the atoms involved in the bond.
Bond strength involves various factors, including the electronegativity difference between the atoms, the orbitals involved in bonding and electrostatic interactions. The list is non-exhaustive and I don't think all the factors can be easily summarised in a list even.
In the post you have linked in your question, "low polarisability" of the fluorine atom was discussed as a reason for the strong hydrogen-fluorine bond. Now, let us make sense of that...
When an atom is not very polarisable, we are saying that the electron cloud of the atom cannot be easily distorted to give an uneven electron distribution, such that there are partial charges induced on different sides of the atom. Such atoms usually have high electron density. A good example often mentioned is the trend in polarisability in the halogens: Fluorine is the least polarisable while iodine is the most polarisable. This is due to the different sizes of the atom. Iodine having a larger and more diffuse electron cloud, allows for ease of electron movement within the electron cloud.
The effect of this is significant in bonding with hydrogen. When a halogen atom is bonded to hydrogen, there is evidently a partial positive charge on the hydrogen atom and a partial negative charge on the halogen atom. The polarisation of the iodine atom in hydrogen iodide is particularly significant due to its polarisable nature. The very uneven electron distribution of the iodine atom causes the hydrogen-iodine bond to be slightly unstable. On the contrary, since there is not so much of polarisation within the fluorine atom and thus, the hydrogen-fluorine bond is much more stable.
However, that is not the most important explanation for the bond strength to be as such. In fact, the "polarisability reason" as far less important, compared to the "orbital reason", which is that the fluorine atom has more defined orbitals used in bonding with hydrogen, compared to iodine. The interaction of more defined and electron-dense orbitals used in bonding gives rise to a greater accumulation of electron density in the bonding region. The attractive forces of the nuclei acting on the bonding electrons is thus stronger. Thus, there is a stronger bond.


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Action of conc H2SO4 on pinacol
Question: Please suggest the correct answer along with suitable mechanism

I'm voting to close this question as off-topic because there is no effort in the question.

I guess dehydration can occur if heated though, so  can be (B).

Answer is A. But dont know why. B and C also seem correct

I would also say A.{PINNACOL PINNACOLON REARRANGEMENT)
First H+ can create carbocation at any one of the carbon and then methyl shift(CH3-)   from other carbon can occur and we would get positive charge on carbon attached directly with OH ,  and then OH can share its lone pair and form double  bond O.   So ketone on left and four degree carbon on right
What's the answer given?
And pls also post the efforts you had made for the question in problem description :))


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Dehydration: Methylcycloexanol + H2SO4 + THF
I came across this exercise:

Why do we need THF?
How the THF influence the mechanism?

I couldn't find any link or external explanation, if you do please share.
Thanks in advance.

It's a solvent. You need to keep the reagents in an environment that's good for reacting,

That's it?
So we would still have methylene cyclohexane as a product like we would find in a normal dehydration?

You won't get methylene cyclohexane, you get 1-methylcyclohexene as mentioned in the picture. But THF doesn't do anything other than that. The solvent was mentioned just so that the reaction is more "descriptive".

related https://chemistry.stackexchange.com/questions/41836/reactions-of-alcohol-with-sulfuric-acid


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Use of concentrated sulfuric acid in Fischer esterification
Concentrated sulfuric acid is often used as a catalyst in Fischer esterification reactions. To my knowledge, the role it plays in the reaction is not as a reducing agent or dehydrating agent but only serves the role of creating an acidic environment, as seen in the mechanism.
Thus, dilute sulfuric acid would be sufficient. In fact, it would be better since it ionises to give an even more acidic solution. However, chemists choose to use the concentrated one instead, suggesting that the role of sulfuric acid is more of a reducing agent. Is there a reason for this?

That is absolutly not true what you say about H2SO4 acidity vs concentration. Conc. H2SO4 can be milion times stronger then diluted. This seems to be common misconception. In reality addition of water only weakens any strong acid.

@Mithoron What do you mean by "strong". If you mean shows acidic properties well or to ionise completely in solution , then I don't think what you said is true. My teacher once mentioned the example of 20M nitric acid being much less acidic than 2M nitric acid because the acid fails to get ionised in the much more concentrated solution

Well I guess you both didn't hear about Hammett acidity function. Nothing needs to disassociate  to be acidic, water's levelling effect weakens every acid which is stronger by itself (like H2SO4 molecule) then H3O+ Also acids are often used in other solvents then water, much less basic.

This brings back old times, I was telling a new user about it more then 2 years ago, being relatively new here myself: https://chemistry.stackexchange.com/questions/27905/ Also this is related: https://chemistry.stackexchange.com/questions/53580/

In fact, sulfuric acid is not a must. Many other H-acids, as well as Lewis acids and their combinations would be sufficient.
The trick here is not to get concentrated acid, but to reduce amount of water in the system. Consider the simpliest transformation, which is typically performed with a large excess of methanol and trace amounts of acid:
$$\ce{R-C(O)OH + CH3OH_{(excess)} <=>>[HCl] R-C(O)OCH3 + H2O}$$
This is an equilibrium reaction. The catalyst allows us significantly faster achieve the equilibrium state, but it does not shift it. Catalyst is equally promoting both direct and reverse reactions. A large excess of alcohol, in accordance with the law of mass action, shifts the equilibrium toward the formation of the ether and thus increases the degree of conversion of the acid into ether.
But if water is not eliminated from the system, or is initially present, hydrolysis occurs:
$$\ce{R-C(O)OCH3 + H2O_{(excess)} <=>>[HCl] R-C(O)OH + CH3OH }$$
Hydrolysis of ether will happen even faster, if a strong base is present, as in this case reaction equilibrium shifts further to the right due to the salt formation.
The bottom line is, in order for the reaction to go to the end, you either add an excess of acid (or alcohol), or remove water as it is produced. Starting with concentrated or water-free acid you are not only simplifying your life, but you are also getting better yield.

But isn't water concentration constant? Also why we need to add more alcohol (Le Chatelier principle) if also alcohol as a liquid have a constant concentration? Is because the volume of the products is reduced therefore and their concentration whereas for reactants only the volume of the acid is reduced (alcohol as solvent is assumed to have constant concentration?) ?

In fact, it would be better since it ionises to give an even more acidic solution.

Yes, a dilute acid does dissociate more, but that doesn't mean it furnishes more $\ce{H3O+}$ ions to the solution.
As a simpler example, consider hydrochloric acid. It dissociates as:
$$\ce{HCl <=>[K] H+ + Cl-}$$
It's equilibrium equation would be:
$$K = \frac{\ce{[H+][Cl-]}}{\ce{[HCl]}}$$
Let's say the initial concentration of $\ce{HCl}$ was $x$, and the final concentration of $\ce{H+}$ is $y$. Plugging these values into the equilibrium equation, you get:
$$K = \frac{y^2}{x-y}$$
Rearranging the terms,
$$y^2 + Ky -Kx=0$$
Using the quadratic formula:
$$y = \frac{\sqrt{K^2+4Kx}-K}{2}$$
Plotting this on a graph, we get:

You can clearly see that the amount of $\ce{H+}$ you get increases as you increase the amount of $\ce{HCl}$ you add.

In short, while low concentrations increase the extent of dissociation, high concentrations increase the total amount of $\ce{H+}$ ions furnished.

I get your point. However, I was thinking along the lines of self-ionisation of sulfuric acid, which occurs to a large extent. In fact, according to Wikipedia, the self-ionisation constant of sulfuric acid is on the magnitude of 0.0001. Compare that to that of water to give you a better picture of the large extent of ionisation.

So I was thinking that it was due to this large extent of self-dissociation which gives the highly acidic solution.


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Band structure in half-metals
In a typical band diagram of a half-metal, the band structures for electrons of up and down spin are different. Why is this so?

My issue here is with the word "why." It's a half-metal so they have to be different. Can you clarify what justification you want to have provided?

You got it reversed: it is not a half-metal properties that make band structure different but it is the band structure that gives rise to half-metal properties.
That said, the proper question would be: why band structure may be different for electrons with different spin orientation?
The answer is that electrons do not exist in vacuum and unpaired electrons with different spin interact with each other through exchange which lead to having unpaired electrons having lower energy when their spins are parallel (please, note: what 'direction' we use for 'up' and 'down' when performing this type of calculation is completely arbitrary). So, when we have a half-filled band, we may expect electrons on it to have parallel spins. In reality it is usually not the case, because exchange interactions are fairly weak.
For more details you probably should repost on Physics.SE or go for specialized literature. You need at least basic understanding of quantum mechanics to understand the details. Some insight may be obtained here.


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Simple Ethyl Acetate synthesis in absence of catalyst
To smoothen 3D printed PLA parts I need ethyl acetate. This is a pretty cheap solvent to synthesize but still to expensive to buy, so I would like to make it myself.
I have seen that ethanol and acetic acid are combined in a Fischer esterification with sulfuric acid as catalyst, then the solution gets distilled afterwards to yield ethyl acetate. I don't have sulfuric acid but I have:

Ethanol 95% (with impurities to make it non consumable)
Acetic acid 8%

Also, I don't really have to distillate the solution since the other chemicals don't harm the PLA. Will simply mixing these two solutions produce a little bit of ethyl acetate? And how can I maximize the yield?

It's actually kinda funny because I answered a very similar question about an hour ago. I wouldn't use acetic acid, even $\ce{HCl}$ will be significantly better.

In the absence of other acids you can give acetic acid a shot, though make sure to distill the water out with a $\ce{CaCl2}$ drying tube.

And why don't you just use acetone for finishing your parts? It's dirt cheap and readily available.

Acetone works on ABS, I have some CaO will that also work to get water out?

My bad, I didn't know that. You can use calcium oxide, but keep in mind that it reacts with water exothermically, and don't let the forming $\ce{Ca(OH)2}$ contact with your mixture.

so I have to minimize the amount of water in the staring acetic acid solution before mixing it with ethanol (and maybe HCL)? And need to slowly do the CaO mixing because its an exothermal reaction?

Pretty much, unless somebody else suggests a different approach.

My suggestion is to never attempt to make a commodity solvent when you can buy it. It's a waste if time and money. You'll also need to purify it...

@Zhe understandable suggestion, but as a hobby its a nice project, also for experience/learning

You'd need glacial acetic acid, there's no sense in starting from diluted. Also catalyst is a must have.

If you are willing to do this purchasing it, then you will rapidly notice the close boiling points of ethanol ($\pu{78 ^\circ{}C}$), freely water soluble and able to solute the product ethyl acetate ($\pu{77 ^\circ{}C}$, equally English wikipedia). Of course, for the pure compounds. I wonder why solid state acids like some zeolithe and acidic ion exchange resins like Amberlyst 15 were not yet mentioned here. The methyl ester prepared at room temperature in  www.arkat-usa.org/get-file/44731/ is interesting, yet isolation and purification there is much easier than here (reactive distillation?).


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Metals Vapor Pressure
I know metals like copper, silver and gold for example have vapor pressures at high temperature but do their vapor pressures still exist at standard temperatures (290 kelvin) and pressures (100 kilopascals)?

Of course they exist, it is just that they are low enough to be considered 0 for all practical purposes.

Also, importantly, the vapor pressure is independent of external pressure.

@Zhe What's it dependent on?

Temperature only.

This question has a couple of answers. But first we need to distinguish between a quantity or value existing and a quantity having a value of zero. I won't get into the weeds on that, but if you think about it, there is a clear difference between existence and non-existence on one hand and between zero and non-zero on the other. So, on to the question do these metals have a VP at STP? Yes. Is it greater than zero? Probably. Is it significantly greater than zero? Depends on what the purpose is for asking, that is:it depends on context. For most things, their STP VP is effectively zero because other sources of Cu, Ag, and/or Au contamination become dominant. Related factors to consider are: What is meant by a clean surface? Is VP really independent of P? (No) What is the lowest, most sensitive range over which we can measure (vapor) pressure? How good are our best man-made vacuums (Ultra-high vacuum)? In the real world, how well isolated are our most isolated physical systems (from leaks, neutrinos, cosmic rays, etc.)? To me, I consider a metal surface is almost never clean. It is quite difficult to clean - I mean really difficult. For a VP of zero, we require that an atom (of the metal) on the surface of the crystal face never ever obtains sufficient energy to break all of the bonds to its neighbors and has enough kinetic energy to escape the surface. I'd not expect that to happen very often at temperatures that are ~700 K less than the metals MP (and > 1700 K less than its BP). It is much more likely that migration across surfaces will dominate any signal detected, at least that type of problem would have to be ruled out. So, I'd sum up to say that while I'd expect almost all measurements to return a zero value, that I'd expect to see a stray atom of the metal from time to time, as well. The kinetic (statistical) theory of gasses is an infinite distribution of energies. That is, at temperatures above 0 K, the chances of an atom having near infinite energy is not zero (although it is near zero). Meaning the chances of an atom of the metal having enough energy to leave the metal surface is not zero. I'd suggest that that means theoretically the metals have a VP, while practically we may or may not be able to measure it. (I've not done a literature search, sorry.)


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Canonical vs localized orbitals - stationary towards changes?
So regarding Koopman's theorem I found the the explanation that you should not use it to find the ionization potential of localized orbitals because, much unlike canonical orbitals they will not be stationary towards small orbital changes. So any minor perturbation will cause a big difference in energy.
Usually I would simply say that the canonical orbitals are the eigenvalues of the diagonalized fock matrix while localized orbitals are not direct solutions of the fock matrix.
But isn't localization simply another transformation of the matrix that keeps the energy but sort of contracts the density more around bonds or atoms to resemble the bonds or lone pairs we know from organic chemistry? How does this effect the way they behave towards small changes to the orbitals?

Does this answer your question?

Thank you for the link but I pretty much based my explanation given above from said discussion. The Koopman's theorem here is just an example where it can be used but I was rather referring to the question what this 'stationary towards small orbitals changes' of the energy means for the canonical and localized orbitals.

Hm. I don't think I understand your question then. It's not that you shouldn't use localized orbitals to calculate the IE via Koopmans', it's that you can't: localized orbitals have undefined energies. Separately, the electron density field calculated from localized orbitals should be identical to that calculated from canonical orbitals. Can you clarify further what you mean by "the way they behave" and "small changes to the orbitals", e.g., in your last sentence?

I'll just quote the sentence directly from the original source:
'this expression should not be used for non-canonical orbitals since these values are not stationary with respect to small orbital changes'. This was the explanation of why you cannot use loc. orb. in Koopman's theorem. I just don't understand what this stationary here means and how it differs from canonical to localized. (The text is from my university script)

Stationary means the orbitals won't react to removing one electron. So this basically refers to the assumption in Koopmans' theorem.
For localized orbitals you cannot make such assumptions, they are not even uniquely defined.

Ah. Okay thank you. I had the assumption that stationary refers to the energy. Because I remember there were comparisons to the potential energy surface at some point. So I assumed that perhaps the position on the PES or the PES itself might change with small perturbations for only one of the two. So this refers to the orbital relaxation term in Koopman's theorem ?

Well I guess it refers to the orbital energy not being changed, thus being stationary.
PES is a whole different story, it is about the nuclear problem, where nuclear coordinates vary. The orbitals belong to the electronic problem, where nuclear coordinates are fixed. Both problems are treated separately due to the Born-Oppenheimer-Approximation.

If I understand you correctly, @Feodoran, effectively the problem arises because any given localized orbital can be thought of as being composed of "pieces" of a number of the canonical orbitals. Thus, if you "remove" an electron from any of them, you're guaranteed to be "removing" a portion of the wavefunction that is of lower energy than the HOMO. Thus, such a removal would result in the wavefunction 'relaxing' appreciably afterward.  (Is this close?)

@hBy2Py I guess yes. But maybe one could even find (or construct) a set of non-canonical orbitals for which Koopmans' Theorem will work even better, then for the canonical ones. But this would only be by chance, not for any physical reason.


text:
Choice of equipment for distilling
I need to distill some reagents, and the book Purification of laboratory chemicals suggests first refluxing over $\ce{P2O5}$ for several hours before distilling across.
The obvious choice is a solvent still head like the ones used to distill THF, but I don't need very much of this so it seems wasteful.
The way suggested that I do it is to set up a normal reflux, let it cool, and then quickly change the condenser for a side arm. This will obviously work but seems like a lot of effort. I think there should be some other piece of glassware to allow this process. Is this the case?

A Perkin triangle springs to mind, though I don't think I've ever actually came across one in my lab

http://www.sigmaaldrich.com/catalog/product/aldrich/z516333?lang=en&region=US
This condenses, and depending on what you have the knob turned too will either send it back to it's origin or to a collector. (I think, I have something similar in my lab but the link doesnt have a great drawing)

Never came across one of those before. Very nice!

There are distillation heads below the reflux condenser with a central bulb / reservoir of about $\pu{250 mL}$, obviously not filled this much during operation:

(source)

Alternatively, if you need not to be super dry, if you may distil under a blanket of Argon and perhaps a colleague lends you a hand (or two), you may mount an intentionally empty Soxhlet between still pot and reflux condenser instead. (The small ones at Aldrich claim an extractor capacity of $\pu{50 mL}$.)   After $\pu{x h}$ of refluxing, lower the heating intensity so that the condensed liquid is siphoned less frequently.  After some stabilisation, to harvest your solvent, let the level of the condensed liquid in the Soxhlet almost reach the level of the syphon. Then 1) dismount the ensemble of (Soxhlet and condenser) from the still pot and 2) mount instead a dry / Argon flushed two necked flask below the Soxhlet.  3) Now tilt the Soxhlet just enough to let the siphon transfer all the dried solvent in the lower dry receiver flask. After this rapid fill, 4) mount Soxlet and condenser on the distil pot, and 5) syringe your liquid from the two-necked flask refilled with Argon.  Well, this second ballet-like version requests some choreography, and definitely is not something for volatile solvents you get from the normal solvent still / bulb (vide supra) or a Grubbs system. To render the performance save, smooth (elegant), and quick, rehearse once (or twice) so that every hand does know what to do when and where.


text:
Shouldn't the Hofmann product be preferred in this elimination?

Shouldn't the exocyclic alkene (Hofmann) be the major product? The base is $\ce{t-BuO-}$ (tertiary butoxide ion), which is pretty bulky. I think the Saytzeff product should be the minor product.
Also, the solution says "E2, some E1". It's a 3° alkyl halide (more sterically hindered than 1° and 2°), so shouldn't it undergo E1 elimination?

You have reasoning already in your picture. Also don't put "thanks" in posts. You can do it in comments.

I don't understand the reason given, according to that E1 Hoffman elimination should be favoured. Please explain.

"good base" for mechanism, high temp - isomer

The base is sterically hindered.

Maybe it would matter matter in low temp.

Yup, $\ce{t-BuO-}$ is a bulky base.
But is it bulky enough that it can't even abstract a hydrogen from the ring? Nope.
Take a look at this picture of a $\ce{t-BuO-}$ molecules approaching the hydrogen of 1-bromo-1-methyl-cyclopentane.

Does that $\ce{t-BuO-}$ look like it's struggling to pull that hydrogen out? Definitely not. It's doing what it does with ease.

The picture given in the question states some conditions. What do you think about their relevance to the Zaitsev Product?


text:
Corrosion issue at large DC current
I'm no expert in chemistry, but I know that corrosion can cause great problems in systems. I'm working on a prototype, and trying to figure out if corrosion will cause issues.
The setup
Imagine the following system. Two aluminium rods make contact with thin aluminium plates (3 mm thickness). Each aluminium plate is connected to the battery with copper wires:

The surface of contact between the aluminium rods and plates is subject to moist and dust infiltration (see figure above). Also the connections from the copper wires to the aluminium plates are in that "dirty zone".
When contact is made, a DC current of 10 Amp flows:

The contact surface between aluminium rods and plates is about 3 to 4 cm^2. The contact force is about 2 to 3 kg. Contact is not made in a very smooth way. Some gliding or even scratching can occur.

Possible issues
I'm afraid corrosion will appear on the contact surface between the aluminium rods and plates. I'm also worried about the copper to aluminium connections.

Possible solutions

Perhaps I should replace all aluminium in the system by copper. Copper is a more "noble" metal compared to aluminium. But copper lacks the same protective natural oxide coating like aluminium has.
Perhaps I should insert a "sacrificial metal" somewhere in the system. But I have no idea where to put it in this specific case.
Perhaps I have to put some sort of coating on the metallic parts. But I have no idea what coating would be best, and if it can withstand scratching and gliding when the rods make contact to the plates - in a rough way.
Perhaps I have to switch from DC to AC current. This would be a last-resort option. For reasons I won't explain here (because it would lead us too far off-subject), this solution would cause an economical issue.

Unless all air and water is excluded, I'd suggest using another metal than aluminum, not only because of likely electrolytic corrosion, but because a build-up of aluminum oxide on the surface would likely make current flow erratic, or even stop it entirely.

Thank you @DrMoishePippik . Which material would you suggest?

Your suggested use of copper sounds good. depending on the actual application. Though bronze or brass is stronger, it suffers corrosion at the grain boundaries. Another choice might be stainless steel or monel metal (https://en.wikipedia.org/wiki/Monel), again depending on the environment.

Thank you @DrMoishePippik , Monel seems a very interesting option. Would the copper wire connections to the Monel be a spot where corrosion occurs?

Yes, if not thoroughly protected, e.g. by a tightly-adhering varnish.


text:
Can one turn a nickel blue?
I've heard of artificial coin toning, but haven't been able to find a recipe for creating a blue (or blue-ish) nickel coin.
While not a metallurgist, I work with one. She tried nitric acid, which looked more like a coating and it rinsed off -- hoping for a permanent color change.

I've read that nickel is not a reactive metal, so not sure if this is possible.

Paint it blue ;) On a more serious note: Why do you want a blue nickel coin? Does the surface necessarily have to be nickel, or can you treat it with something else?

@paracentamal - I'm in a band called "Blue Nickel". Was thinking of creating some jewelry or similar. So no, it doesn't have to be a nickel surface, but it should look like a nickel.

watch out - Gov't rules against defacing / altering / deforming currency. Not sure it's worth their effort to go after you over $0.05, but just saying...

"Look like a Nickel"Ah... all this time I thought you were talking about the element :D.

@paracetamol Good point on the distinction. The US Nickel coin is actually made of an alloy of 25% nickel metal and 75% copper. Some sort of reaction with the copper to produce a coating of something like azurite or copper sulfate might work.

Copper sulfate hardly can be made stable. My bet is on heat tinting.

Austria Mint (Münze Österreich AG) since early 2000s issues bimetallic coins of various colors. Coloring is achieved by anodizing various metals and alloys. Basically, anodizing is metal surface oxidizing via a controllable electrochemical deposition of oxide layer of negligible thickness (about $\pu{1e-6 mm}$), which is physically very strong and can have many shades.
For example, anodized niobium among other colors (depending on current applied) can have an intense sky blue color. A 2003 commemorative "700 Jahre Stadt Hall In Tirol", as well as latter 2010 "Erneuerbare Energie" coins have an anodized niobium core in a silver ring:

The problem with US nickel coin assuming it's made of 25% nickel and 75% copper is that copper forms two stable oxides instead of one, which happens, for example,  in case of well-suited metals for anodizing, such as $\ce{Al}$, $\ce{Ti}$, $\ce{Nb}$, $\ce{Ta}$ and certain steels. This adversely affects the adhesion and significantly increases the risk of cracking the oxide film.
Attributing blue tint to copper and its alloys is possible via patination, which, I believe, happened in a case of Nelle Findlay (great observation, by the way!). The downside is that the coin will look like it has seen better days and also this tinting can be easily worn off (layer made of copper and nickel carbonates and oxides is rather loose), which will make the coin look less attractive (or more, from the collectors' point of view).
From what I see, this leaves us with these options:

Use pigment coloring as suggested numerous times in comments;
Apply patina and subsequently protect the surface with varnish;
Create a coin replica (a distinctive one from the original to avoid problems with law) from aluminium and anodize it with virtually any color possible.

Of course, anodized niobium will also do, but I'd rather leave it to an album that goes platinum, or as an attractive factor for a pre-order campaign:)

I did it accidentally, I had coins in the bottom of my school bag and a packet or paracetamol and ibuprofen leaked without me knowing, the tablets got crushed into a powder and were left in the bottom of my school bag with the coins for about 6 months. They're now green, and it doesn't wash off with water and it doesn't look like a coating.

As noted : An old nickle or new one ? Old nickles are nickle , new ones are cupronickel ( I think more like 30 % Ni , bal Cu).Quarters are clad with monel ( 70% Ni, 30 Cu). Cu has several blue compounds but they would not adhere well so would wear off . I vote for paint.

Or a blue sharpie marker

Maybe try cleaning it and pass a water break  test, then try following up with a nickel strike solution while still wet ( not too much as it will dull )to help with adhesion. Then carry on with your over plate( just enough to prepare it for coloring.) Then color (but not too much as it strip away the preparation) Finally rinse with deionized water and dry with dry compressed air or better yet dry with nitrogen. Regards

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


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How is work converted to heat?
My book says work can be converted into heat but I want to know how work affects kinetic energy of molecules which, when transferred, is called heat?

What is work? Roughly speaking, it's when you move things around. Now, those things move smaller things, which in turn move something smaller yet, all the way down to molecules, and when you move those, you increase their kinetic energy, which is heat.

Work is not always converted to heat. For example, if you expended 1000 calories moving a one liter thermos bottle of water up an incline the water would not change temperature. However, if you expended 1000 cal stirring the water, it would warm up by one Celsius. Also you could expend energy a certain way and end up cooling the water (less efficient!)

The disorderly motion of molecules is called thermal motion. Work is
identified as energy transfer making use of the organized motion of atoms in the surroundings, and heat is identified as energy transfer making use of thermal motion in the surroundings.

Work can be understood as a variation of energy during which the transporters (may they be molecules) are spatially aligned or bonded, meanwhile heat is a variation of energy in which the transporters aren't aligned (therefore heat dissipates energy).
With this in mind, any machine that uses an ordered energy transfer (like your hand waving) can induce heat in a fluidic means, as particles in fluids are initially moving randomly and are not likely to move far on the orientation of your hand.
In this example, the induced work is little and restricted to the analysis on the surface of the moving hand, but if the analysis is made from further on a wider system, we can assume you're doing work to generate heat alone.

Work is the change in kinetic energy, and in some cases, say you do some work to move a bunch of molecules; if it results in them colliding, thereby releasing heat in the process, due to their kinetic energy during collisions converting into heat.


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Cannot optimize transition state
During a Gaussian scan calculation, I got a high negative frequency for a transition state candidate, but it doesn't matter what I do (Berry algorithm, qst methods) I cannot get a stable TS, I always get a really low frequency that's clear a rotation in a methyl group.

By 'scan calculation' I am assuming that you haven't done a fully relaxed optimization.  Frequencies are meaningless unless you've fully optimized a geometry to reach a minima, TS, or higher-order saddle point.  What method/basis are you using?

DFT? In that case turn the methyl group a bit and try opt=(ts,calcfc,noeigen,Verytight) and int=ultrafine.

@LordStryk, I did method 4 here https://joaquinbarroso.com/tag/qst3/ but I also ran a opt with modredundant as a keyword. With the former, I got a much negative frequency for TS candidate ("top of the scan"). I'm using  DFTB. I got a stable TS with PM6 and PBE for this structure.

@ziulfer Try DSVA's suggestion.  Be warned that DFT approaches can give rise to spurious imaginary modes of vibration (small imaginary magnitudes on the order of a few cm-1 or less).  I don't know what your trying to characterize so its hard to tell you how to get around it.  In general, try a set of functionals, increase your basis set, and use better convergence criteria.

@DSVA  I tried your suggestion, but unfortunately, it didn't work. Do you  think that a stable TS doesn't exist for my reaction doesn't exist for the DFTB Hamiltonian? I'm studying the H-transfer in this  reaction:
C(CH3)2(O2.)(CH2)2CH(CH3)2 --> C(CH3)2(O2H)(CH2)2C.(CH3)2


text:
In a single cell battery, is voltage exactly equal to cell potential?
I've been researching electrochemical cells. In chemistry-focused sources, you usually see the potential difference between the electrodes referred to as cell potential, but in engineering/industrial sources(where you talk about "batteries") you usually see plain "voltage." Are the two always the same, or are there some situations where they're different(excluding multiple cells)? I want to know whether you can use the cell potential formula $$E^o_{cell} = E^o_{cathode} - E^o_{anode}$$ to calculate an accurate standard conditions voltage for a battery. If not, what gets in the way?

I did not alter my answer in response to your edit as the answer basically remains the same. The formula for the cell potential is correct as you have written it and can be used to calculate standard cell potentials so long as you know how to use it.

It is not uncommon for two identical phenomena to be described by different terminology in different contexts or scientific disciplines. In this case, voltage and electric potential difference refer to the same thing. According to this Wikipedia article:

Voltage, electric potential difference, electric pressure or electric tension (formally denoted ∆V or ∆U, but more often simply as V or U, for instance in the context of Ohm's or Kirchhoff's circuit laws) is the difference in electric potential energy between two points per unit electric charge. The voltage between two points is equal to the work done per unit of charge against a static electric field to move the test charge between two points. This is measured in units of volts (a joule per coulomb).

Electrochemical potential is referenced (mostly arbitrarily) towards the standard hydrogen electrode, while "electrical potential" has "ground" as reference (which usually means "zero excess charge"). "Voltage" is the difference in potential, measured in Volts, for both.
The standard hydrogen electrode has 4.44 V relative to an ideal ground (absolute potential, at 25°C), but thats's hard to reproduce, so for all practical purposes, it ($E^0$, standard electrode potential) is defined (by IUPAC) to be zero at all temperatures. It's all relative.


text:
Chemical oscillators that behave like a double pendulum
Oscillating chemical reactions such as BZ-reaction or chemical clock typically result in sequential systematic changes, most notable of which is arguably coloring. Though underlying mechanism can be extremely complicated, we always expect the same sequenced cycling set of properties. For iodine clock it's always will be

... dark-blue > colorless > dark-blue > colorless ...

colors that we observe. Just like ordinary pendulum. But are there oscillating chemical reactions with the non-linear outcome and "chaotic" sequences for each cycle, e.g. behaving like a double pendulum?
And if such systems do exist, can we also just use similar mathematical apparatus like a system of differential equations used for a double pendulum to describe such behavior?

Differential equations with chaotic behavior certainly do exist. Whether they can be materialized on the basis of chemical reactions is another question.

Then I guess you'll be surprised that even BZ reaction shows chaotic regimes.

@Mithoron Nicely noticed, I agree. But I'm in doubt whether it is correct to compare these two systems (double pendulum and BZ-reaction) -- if you've seen a reference where it is done explicitly, or you can prove the similarity yourself, then please feel free to post an answer.

BZ seems to be more complicated then wackiest pendulum on Earth ;)

related https://chemistry.stackexchange.com/questions/79106/how-does-an-oscillating-reaction-work

The ODEs that model the double pendulum are nonlinear but also non-polynomial (due to the sines and cosines). The ODEs that model a chemical reaction should be nonlinear but polynomial, due to mass action kinetics. Hence, it should be impossible to simulate the double pendulum using a chemical analog computer.

@RodrigodeAzevedo Great point, thank you! I guess this might be the answer – please feel free to elaborate a bit more and post one :)

Before considering a double pendulum, let us consider a simple pendulum, which is modeled by
$$\ddot \theta + \frac{g}{\ell} \sin (\theta) = 0$$
Choosing units that make $\frac{g}{\ell} =1$, we have
$$\ddot \theta + \sin (\theta) = 0$$
Let $x_1 := \theta$ and $x_2 := \dot \theta$. Hence, the 2nd order ODE above can be rewritten as follows
$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - \sin(x_1)\end{array}$$
The 2nd ODE is not merely nonlinear, it is non-polynomial (due to the sine).  However, the ODEs that model elementary chemical reactions should be polynomial due to the law of mass action. Hence, it should be impossible to simulate the simple pendulum using a chemical analog computer. Simulating the double pendulum should also be impossible.
What may be possible is to simulate an approximation. For example, if $x_1$ is sufficiently "close" to zero, then we could use the approximation $\sin(x_1) \approx x_1$, which produces a system of coupled linear (and, thus, polynomial) ODEs
$$\begin{array}{rl} \dot x_1 &= x_2\\ \dot x_2 &= - x_1\end{array}$$
which approximate the original pendulum provided that $x_1$ remains "small" (in absolute value).
Lastly, since concentrations cannot be negative, we have non-negativity constraints $x_1, x_2 \geq 0$.

I follow most of your answer (both pendulums are impossible in perfect form but we can come close to simulating the single pendulum by following a certain constraint). What I fail to see is an answer to the question ‘can the double pendulum be approximated chemically?’

@Jan This answer is a work in progress. I know virtually no chemistry. I am still trying to come up with a mechanism that can simulate the harmonic oscillator. I will let you know when I have progress.

I would like to supplement this excellent answer by a different argument. Chemical reactions are typically done in bulk, involving numbers of particles on the order of $10^{17}$ - typically greater. In statistical thermodynamics, this is considered an ensemble, and while a single pair of molecules (that can react which each other) cannot be predicted in their behavior (when will they react? what will be the product?), the behavior of the ensemble can usually be predicted. I think that the same may be true for an ensemble of double pendulums. Thus, the analogy would hold, but not in the way the question intended; rather, it would be on a molecular level.

This is about kinetics or reactions not behavior of single molecules. Also kinetics can be chaotic as I pointed out earlier, just not in the same way as in pendulum.


text:
Are there any examples of activating meta directors?
I have heard about activating ortho-para directors. For example: methoxy group, hydroxy group, alkyl group, et cetera.
I have heard about deactivating ortho-para directors. For example: halogen substituents
I have heard about deactivating meta directors. For example: nitro, trifluromethyl group, carboxylic acid group, et cetera.

So, naturally, I have embarked on a quest to find Activating Meta directors.
I have found one potential candidate, which is dimethylphenylborane:

Boron possesses a vacant $\mathrm{p}_z$ orbital, which can interact with the $\pi$-cloud of the benzene ring.

Considering the fact that boron is treated as a metalloid, it will definitely be less electronegative than carbon, so the dimethylborane group should offer a good +I effect, which could possibly outweigh the -M effects.
Can anyone shed some light on this issue? Are there any other examples of activating meta directors?

Related: https://chemistry.stackexchange.com/q/15973/4325

You're talking about for EAS?

@Zhe That's a positive.

OK. Got it. So what makes you think this is a +I effect? For example. how is having a borane on the ring any different from having a carbonyl attached to the ring?

@Zhe First of all, I never confidently said that borane will give a +I effect, I only assume it to be so, since boron is quite electropositive compared to carbon, so it might tend to push the $\sigma$-electrons towards carbon.

Screwed up the first response I was trying to write...  I think this is an interesting idea. You might even consider other substituents instead of methyl on the borane.

@Zhe I first considered having hydrogens instead of methyl, but I decided not to do that since $\ce{B-H}$ bonds impart reducing properties to the molecule.

This is very theoretical, but what if you added a subtituent such as orthonitrate  so that the formula is Na2O3NPh, basically a nitrogen attached to a phenyl with 3 oxygens single bonded to it. The meta directing would be done similar to the nitro group, due to recent evidence pointing towards the pentavalent nitrogen structure being a part of resonance, and then the carbon gains electron density because of the pentavalent structure being in resonance with the traditional structure (following octet rule).

@AS_1000 I just looked up what orthonitrates were, and they don't seem to activating groups to me. Also, can you elaborate because of the pentavalent structure being in resonance with the traditional structure?

The activation is because it's like a phsophonium ylide (ammonium ylide) and the traditional structure is that of which follows the octet rules, I'll probably end up posting an answer to clarify, structures are required to explain.

Note: This requires an extremely high energy scenario to occur.
This is another example that looks promising, it is essentially an ylide that has enough EWing power to meta direct while giving some density to the ring in another resonance structure. []
The phosphonium ylide derivative (phenylphosphonic acid) may provide the same effect if the hypervalent nature of nitrogen is challenged, or if this substance is unobtainable under any circumstances.

Can you explain whether there's an +I effect or not?

Also, can you comment on dimethylphenylborane, the example that I suggested for an activating meta director?

Product distribution is around 1:1:1 for the boranes, so a +l effect is definitely there, there is probably going to be a +l effect for this group, just as a phsophonium ylide used in the Wittig reaction gives that effect.

This is highly dubious and speculative.


text:
Enantiotopic hydrogens in cyclopropanone

Are the hydrogens that on the same carbon and those that are cis enantiotopic? And are they isogamous or anisogamous?
And will they anisochronous in a chiral environment NMR?
Isochronous - hydrogens that are in the same chemical environment leading to the same shift and splitting in NMR. If the hydrogens are enantiotopic they will be isochronous in an achiral nmr environment but anisochronous in a chiral environment.
I can see there is there is a C2 and two planes of symmetry but the TA said they will be isochronous in a chiral NMR resulting in a single singlet but as i did the replacement test I got two enantiomers which means that they will be anisochronous in chiral NMR, but I'm not sure.

They should be different in NMR in a chiral environment because they would appear to be diastereotopic in the presence of an external chiral reference.

Cyclopropanone is a reactive species, as such there aren't many experimental reports in which it is characterised, however some (possibly questionable data) does exist:

Cyclopropanone is a highly reactive compound. It is stable for a few days at liquid nitrogen temperature but rapid polymerisation takes place above o or even below if traces of water are present. The reaction is highly exothermic.
Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.19660851113

.

The NMR spectrum  (neat) shows a singlet  at 1.72 p.p.m.  relative  to  tetramethylsilane.
Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.196608511

It appears that in the absence of a chiral environment, all of the protons are in a single environment at 1.72 ppm (Hans Reich quotes a single peak at 1.65 ppm) which is what we'd expect (the molecule is essentially planar, and all of the protons are equivalent).
As you point out in your question, replacing any one of the protons (hypothetically) would yield two enantiomers, which, in the presence of a chiral environment, should be distinguishable. It can therefore be concluded that in a chiral environment, the protons should appear to be enantiotopic.*
* there are two slight caveats to this. Firstly, the compound would still be unstable and prone to polymerising, therefore there is no way of actually testing this. Secondly, the chiral environment must be able to interact with the molecule, for things like chiral alcohols, lanthanide shift reagents are (/used to be) commonly used, but I'm less convinced how you put a ketone into a chiral environment.

A chiral solvent should perfectly present a chiral environment, should it not?

I can't think of any chiral solvents suitable for NMR (?). My point was that it should work, but I'm not sure if it does work.

Every solvent is suitable for NMR afaik. You just want your hydrogens to be deuterium if you’re using proton NMR.

aha. One of us is being pedantic, I just can't quite tell whom. But yes, fine, I can't think of any chiral deuterated solvents. I mean actually, it doesn't even have to be deuterated (I often run in non-D THF/dioxan for reaction monitoring). I wonder if the viscosity becomes the issue, as a chiral alcohol is necessarily 'bigger' than methanol/chloroform etc. I think we need long (or Loong).


text:
What are stereospecific and stereoselective reaction?
I have read the defination but couldn't grasp the concept, what is the key factor that says this is stereospecific or this is stereoselective and what are the main intuitions i can get if i am given that a reaction is stereospecific or stereoselective.

If you have a question, ask it. You say you've read the defination[sic]. What it is that you fail to understand?
IUPAC Gold Book
stereospecificity (stereospecific)

A reaction is termed stereospecific if starting materials differing only in their configuration are converted into stereoisomeric products. According to this definition, a stereospecific process is necessarily stereoselective but not all stereoselective processes are stereospecific. Stereospecificity may be total (100%) or partial. The term is also applied to situations where reaction can be performed with only one stereoisomer. For example, the exclusive formation of trans-1,2-dibromocyclohexane upon bromination of cyclohexene is a stereospecific process, although the analogous reaction with (E)-cyclohexene has not been performed.
The term has also been applied to describe a reaction of very high stereoselectivity, but this usage is unnecessary and is discouraged.

stereoselectivity
(Also contains definition of: diastereoselectivity)
The preferential formation in a chemical reaction of one stereoisomer over another. When the stereoisomers are enantiomers, the phenomenon is called enantioselectivity and is quantitatively expressed by the enantiomer excess; when they are diastereoisomers, it is called diastereoselectivity and is quantitatively expressed by the diastereoisomer excess.


text:
Complexation of hydroxy chalcone oxime with metal
Does the complexation of 2-hydroxy chalcone oxime (below) with a metal require any catalyst? Is the complexation pH sensitive?

Like what catalyst?

Here Hydroxy is ortho to oxime so that it can form six membered chelated ring..and catalyst like any acid to deprotonate hydroxy proton ?

@user46820 maybe if the hydroxyl was on the other ring, but not in the structure shown

I'm pretty sure that the use of acid/alkali to change the protonation state of a ligand to allow it to complex is generally not considered catalysis.


text:
What category do Gd3Fe2Fe3O12 belong?
After polishing my $\ce{GdFe2}$ sample with silicon carbide emery paper it developed trace amount of $\ce{Gd3Fe2Fe3O12}$ on the surface. Why is it being formed, are these type of compounds called anything?

Is this something you've actually done? If so, how do you know that this is the exact compound being formed?

@chipbuster Yes I am currently doing this in my summer internship. I know that this exact compound is being formed because I performed grazing incidence X-ray diffraction and while indexing the diffractogram I came to know this.

Formula isn't correct and why wouldn't it be simple corrosion.

@Mithoron I did the analysis (indexing) using computer program and this formula is the correct one. Can you please explain further how can corrosion form a complex compound like this. My doubt is that why the presence of this compound is not detected before polishing in either powder or GI XRD and what exactly happened during polishing that led to the formation of it.

You'd need to mark Fe as II and III to be correct. It's just some mixed oxide.


text:
Existence of polyaminomethyl compounds
Amines such as triaminomethane, tetraaminomethane, and many polysubstituted amines do not exist because of their degradation into ammonia.  Is it possible for derivatives of these amines to exist at all? Take for example tetra(t-butylamino)methane. Will there be enhanced stability due to the tert-butyl groups or will degradation to t-butylamine still occur?  Is there a point at which this degradation will stop, if so where?

I wouldn't say that they don't exist, but there's very little to keep them from collapsing to an amidine, which makes isolation quite difficult.
This is completely analogous to orthoformic acid, which, as a tetrahedral intermediate, rapidly collapses.
Substitution helps to stabilize the tetrahedral intermediates a lot. For example, both trimethyl orthoformate and tris(dimethylamino)methane are commercially available.
EDIT:
Forgot to address this in my answer, but having t-butyl groups probably makes things worse due to steric hindrance. Neopentyl is bad enough...