text:
Proton NMR Spectrum for Nicotinic Acid
I have been able to slightly unpack the NMR spectrum for the nicotinic acid to understand the order of signals and what ones correspond to what protons but I am having a problem with why some of the signals produced a different number of peaks I can not explain. I understand how simple and complex spin-spin splitting works to an extent when producing a number of different types of groups such as, s, d, t, m, dd, dt and many others.
The spectrum below is one for nicotinic acid, and the peaks are in order of most downfield to least, signals for H on carbons 2, 6, 4 then 5.
Looking at the environment proton two is in it has no equivalent neighboring pairs so why is it producing a doublet?
Same goes for the proton at carbon six.  Why is it producing a double of doublets when it only has one neighboring pair?
As for the signal for proton five that is simple to understand and it should produce a double of doublets because the neighboring protons are in-equivalent, but yet again this is still not the case for the splitting of the signal. The same for proton 4 which produces a double triplets.
Would some one be able to explain why this is occurring?

In an aromatic system, you can usually see couplings higher than 3J.

First, a great source of information on this kind of problem is the website of Hans Reich at UW Madison, see e.g. here. Couplings can occur with protons at ring positions ortho, meta and para, with typical values of $\pu{7-9}$, $\pu{2-3}$ and $\pu{<1 Hz}$, respectively.
This  is (roughly) the information I extracted from your spectrum:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz}$
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz}$
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz }$
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz}$

You should report the field at which the spectrum was acquired since otherwise the magnitude of the J couplings is ambiguous. Assuming that the largest J is $\pu{8 Hz}$ leads to the guess that it is a 400 MHz ($\ce{^1H}$) magnet. You should also report the solvent.
Second, your spectrum is very nice, showing great resolution. Compare for instance to this spectrum (also at 400 MHz but presumably much higher concentration -  $\pu{100 mM}$). Either the concentration of the solvent and temperature may be reasons you obtain higher resolution, resolve very small couplings. The BMRB spectrum was acquired in D2O, pH 7.4, 298 K. Your spectrum was acquired perhaps in $\ce{H2O}$, and perhaps at low pH, resulting in protonation of the pyridine N, but probably not.

Based on the structure you expect 2 to have the weakest couplings to other H. It displays a small coupling, not easily quantifiable because you clipped the spectrum in the figure. 5 Is expected to be a doublet of doublets, with both couplings similar and large. It also displays a smaller third coupling. Therefore lets start by tentatively assigning 2 to the resonance at 9.15 ppm, and 5 to 7.6 ppm. The resonance at 8.3 ppm is an apparent doublet of triplets, the triplet splitting being small, 2 Hz, the larger 8 Hz matching up with that from 5, so this H is vicinal to 5. One of the smaller couplings is presumably to 2, the other to 6, explaining the apparent triplet multiplicity. The last resonance, at 8.8 ppm, has a coupling matching up to 5, and another smaller one to 4, so we assign this to 6. So, to summarize, here are the assignments:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz, \space }$ 2
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz, \space }$ 6
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz , \space }$ 4
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz, \space}$ 5

This matches your assignments and those in the BMRB. So this analysis explains the small coupling for 2. The simplest explanation: that's a meta coupling to 4. It does not explain the small coupling for 5, though. That might be a trans coupling to 2.
The magnetization transfer pathways between the assigned H can be verified by inspecting the $\ce{^1H-^1H}$ TOCSY available at the BMRB. In particular, there is no pathway between signals at 9.1 and 8.8 ppm.


text:
What will happen to bottle of cola after 50 years?
I'm going to make a time capsule for 25 or 50 years, and I don't know wether it's safe to put food, e.g. bottle of cola there. Will it explode and ruin the capsule? Will it be ok to drink it, if it won't explode?
Also, what about Snickers or a hamburger in a ziploc package with air sucked out of it with vacuum cleaner?
So: will Snickers, Hamburger in a ziploc, Bottle (or can) of Cola, all not opened, go crazy in 25 years? In 50 years? If so, can they go out of their packages and ruin the contents of the time capsule? If not, will it be safe to consume one of them?

My money is on the bottle of cola making it, the others not so much. I'd stay away from the Snickers and hamburger in the ziploc bag with the air sucked out by using a vacuum cleaner.

I like your idea, even if none of us will still be around to witness the outcome. Now, hamburgers are known to remain visually unchanged for a few years at least (not that I recommend consuming them afterwards). Cola bottle, on the other hand, is made of plastic, and it may grow brittle over decades.

Glass bottle of Cola will do fine, I guess. Also, you get some spare money if the future won't turn out to be so bright.

Yeah, if you want liquids, use glass bottles.

Hummmm... glass bottles have a metal cap (steel if I remember correctly). A  time capsule would be in the dark. It is light that degrades plastic the most not time.

You might want to include some vapor phase corrosion inhibitors in a sealed bag containing any bottle with metal top. BTW, why not freeze-dried food, such as freeze-dried ice cream?

Maybe you should purposefully put in some things you think might not make it, along with a list of items inside, and preferably photos of how they looked prior to burial. Not often are everyday foods and drinks left forgotten for 25 or 50 years. I imagine in most cases we don't actually know what would happen to them. Could be an experiment of sorts! But definitely don't consume them.

I'm going to make a time capsule for 25 or 50 years, and I don't know whether it's safe to put food, e.g. bottle of cola there.

Depends on the food and the way it's packed, but food is not something I'd suggest you pack for such a long period..

Will it explode and ruin the capsule ?

Food or drink are very unlikely to explode (maybe alcohol would in the wrong circumstances).
But food can be unsafe to eat in quite a short time, depending on the food, preparation and packing and how you store it.  Bacteria could be a problem.
Most food will go bad over a period as long as you're describing without very specific storage (like refrigeration, and even then I'd say most things won't work).
The safest ones would be :

popcorn (kernels)
honey (jar of)
sugar (most forms)
alcohol (spirits like vodka, whiskey)
dried beans, dried lentils

I would not be planning to eat any of these stored for 25 years myself.  And in general I'd suggest testing the items before trying them after 25 years or more (if you feel you must).
I would not expect cans or glass or plastic bottles of soda to be in good shape after anything like 25 years.  The plastic might not survive without degrading.  The can and plastic might react with the liquid over that timescale and the glass would survive but I'd be less optimistic about a sugar laced chemical soup like soda or cola not undergoing some changes.  Hard to say.
If you want more info on this try this website.

Will it be ok to drink it, if it won't explode?

I would not try it.  At best it soda would be flat and possibly not taste the same (chemical changes over that timescale ?) and at worst it could actually be harmful.
Exploding seems very unlikely.

Also, what about Snickers or a hamburger in a ziploc package with air sucked out of it with vacuum cleaner?

Air isn't the issue.  There are bacteria that will happily live (and increase in numbers) on what's in the food.  Well, it is food, after all.  There are bacteria that will survive refrigeration as well.  Over the timescale you're talking about I'd say it's all bets are off territory.

So: will Snickers, Hamburger in a ziploc, Bottle (or can) of Cola, all not opened, go crazy in 25 years? In 50 years?

All of those could be dangerous over such a long time period, IMO.  At the very least they'd taste bad and at worst they'd kill you if you consumed them.

If so, can they go out of their packages and ruin the contents of the time capsule? If not, will it be safe to consume one of them?

Depends on the packaging.  Glass would last indefinitely baring physical force or extreme of hot and cold (which might possibly cause fatigue cracking).  The other wrappers would last pretty well (structurally), but 25 years is way past their design intentions.  It would be a dice throw.

I know! I'll just make two capsules: one with stuff and one with food, and bury them together!

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text:
Proper common notation for a ligand
I've observed several notations for the ligand in the literature in general, say, for the typical complexes like $\ce{MX_n}$ ($\ce{M}$ - metal) where ligand $\ce{X}$ is denoted as (according to my subjective frequency of occurrence):

$\ce{L}$;
$\ce{Lig}$;
$\ce{Lg}$;
$\ce{Lgnd}$.

I can imagine $\ce{Lg}$ being a poor choice as it can be easily confused with leaving group, and $\ce{Lgnd}$ which looks more like a company's NASDAQ code, but what about the first two options? And what normative document governs the proper notation anyway?
Regarding the $\mathrm{L}$ use for leucine (see comments section here and a thread on Biology.SE What is the preferred way to abbreviate amino acids?), I encountered more sources using $\mathrm{Leu}$ for that matter rather than $\mathrm{L}$. I guess 1-letter code for 20 amino acids is quite suitable and eliminates confusion between leucine ($\mathrm{L}$) and lysine ($\mathrm{K}$), but for me handling $\mathrm{Leu}$ and $\mathrm{Lys}$ is mnemonically easier.

(1) is out due to ambiguity - it is the one-letter code for leucine. I'd go with (2), as that is what I've seen in the literature.

I often use Lig. for ligand.

@ToddMinehardt I think that is a useful distinction, but I would say it is context dependent. If you working with biological systems, it would be wise to use Lig to avoid ambiguity, but for purely inorganic systems, I think L should be used for the sake of brevity and the limited likelihood that it would cause confusion in that context.

Honestly I don't see why a specific convention is needed. It might as well be implied from the context. You could just say Let $\ce{A}$ be a ligand and go on with it.

@PrittBalagopal Following this logic one can say Let Tg be element tungsten, but people don't do that because there is a general agreement in a form of standardized periodic table. Same goes for classes such as lanthanides (Ln, not Ld or anything else), or coordination number (C.N. to avoid confusion with cyanide). I'm trying to find the same standard way for another entity, which this time happen to be ligand.

@andselisk There's a little flaw in your reasoning. You can't assume an acronym for tungsten since it's an element with it's own unique properties. Lanthanides are a class of elements showing a specific property. A non-ambiguous acronym is needed in such cases so that people all over the world can understand precisely what element we are talking about. However ligands are a bit different. They are a very broad category of chemical species. They come in several types (bidentate, unidentate, tridentate, etc.), so they aren't really that organized to need a specific notation imo.

@PrittBalagopal If a term ligand would have been that broad, it wouldn't exist in the first place as an unifying class. Morphologically there is no difference between isotopes of elements and denticity/hapticity of ligands. Also, any science is by definition implies search of generalization. If there is something that seems too vast for you, it doesn't mean it shouldn't/cann't be generalized. I also don't see the message you are trying to deliver. There are certain notations already established, and the only question is, which one fits better.

Going off Linear Christmas's comment, I will note that the notation from the referenced article has become relatively common for inorganic chemists. I was introduced to this notation in Chapter 2 of
The Organometallic Chemistry of the Transition Metals
by Robert Crabtree, most of which is provided in the linked Google Book. You may also want to look at
Application of the Covalent Bond Classification Method for the Teaching of Inorganic Chemistry (DOI 10.1021/ed400504f)
by M.L.H Green et al, who developed the Covalent Bond Classification scheme.
The notation largely boils down to describing metal complexes using three classes of ligands: $\ce{L}$ type ligands, which are neutral ligands that give $2\ce{e^-}$ to the complex (Lewis Base) and $\ce{X}$ type ligands which are radicals that contribute $1\ce{e^-}$ to a neutral metal (or alternatively are negative ligands interacting with a cationic metal). There are also $\ce{Z}$ type ligands, which aren't named as such in Crabtree, but are mentioned in regards to being $0\ce{e^-}$ ligands. An example used in Crabtree and also Green's paper is $\ce{BF3}$, which like the other $\ce{Z}$ types is identified as a Lewis acid.
To more directly reference how Green's paper describes this notation, I include the passage below:

Adopting the view that the bonding in many covalent molecules can be represented in terms of 2-center-2-electron bonding interactions, there are three possible scenarios that describe the construction of these bonds in a molecular orbital sense, as illustrated in Figure 4. Thus, with respect to the central element of interest (M), the neutral ligand can contribute either two (L), one (X), or zero (Z) electrons to the bonding orbitals. The classification of ligands as L-, X-, or Z-type, as featured in a variety of textbooks,4 is now well established, and some simple examples of these ligands are provided in Table 1.

Some examples of classifications you can make are given in the table below or in the linked preview:

We can note some of the more complicated ligands we can simplify with this notation. $\eta^5$ cyclopentadienyl can be thought of as a
$\ce{L_2X}$ type ligand based on the number of electrons contributed and it's charge. Again from Green:

Although many ligands coordinate to a metal center by a single covalent bond, a variety of multidentate ligands coordinate via more than one covalent bond. Such ligands are classified as $\ce{[L_lX_xZ_z]}$, where $l, x,$ and $z$ are the respective number of $\ce{L, X,}$ and $\ce{Z}$ functionalities that are associated with the frontier orbitals of the ligand in the geometry that corresponds to its binding mode.

This notation provides an easy means of electron counting, which is useful for determing whether a compound follows the 18 electron rule (a useful heuristic for determine the stability of metal complexes).
In regards to your main question, as far as I can gather from this, the reason $\ce{L}$ has become widely used as a generic ligand (or in this case, a generic $2\ce{e^-}$ donating ligand) is that, at least within the realm of inorganic chemistry, it is short and unambiguous when used to write a formula.
In addition, along with your answer that the ACS uses $\ce{L}$ for ligand, Green mentions that a number of textbooks have adopted the convention since their original paper in 1995, which means that the current generation of teachers and students is thinking of the material with this notation. So in addition to researchers that simply liked the scheme and chose to use it in their writing, students who learned this scheme from the beginning are starting to publish and make use of it because its what they know.
Some reasons why I expect there isn't a lot of confusion with leucine are:

Even now, authors will still specify what they mean when they throw in an extra L.
Leucine is uncommon outside biochemistry, but the term ligand is used in essentially every sub-discipline of chemistry.
I would imagine that the three letter amino acid codes are used more commonly anyway because they are clearer in their meaning (besides situations where space is at a premium and the context makes it clear, as is the case when writing out amino acid sequences).

The following is not a directive, more an advice. I guess it's better either to refer to the original paper from Green, or it's more recent adaptation (Green, M. L. H.; Parkin, G. Journal of Chemical Education 2014, 91 (6), 807–816. DOI 10.1021/ed400504f). Also, according to the CBC method Crabtree is citing, there are 3 types of ligands based on how they contribute electrons to a bonding orbital: 2 (L), 1 (X), or 0 (Z). It also would be great if you could draw some sort of conclusion why $\mathrm{L}$ is preferred over other abbreviations (also, situation with leucine).

My main reason for referring to the book was the Google preview allowed anyone to see at least a brief source of my information (I also didn't have access to the paper myself over the weekend). I will try to add in some additional information.

GoogleBooks also don't guarantee 100% accessibility to the given pages over time, and there is a queue for pages being visible to certain geolocations at certain periods of time. I think quoting the primary source is the most robust way to go.

@andselisk anything else you are looking for out of answer?

Great edit, I guess it addresses all the issues now. Very well done!:)

In my limited experience, three-letter-codes are used for all applications except for amino acid sequences which tend to use one-letter-codes to save space.

The ACS Style Guide suggests using $\ce{L}$ [1, p. 186] in APPENDIX 10-2 Abbreviations, Acronyms, and Symbols:

\begin{array}{ll}
[\ldots] & \\
\small{L} & \text{absolute conﬁguration} \\
\mathrm{L} & \text{leucine} \\
 & \text{ligand} \\
 & \text{liter} \\
\mathrm{L_I} & \text {spectral line} \\
[\ldots] &
\end{array}

IUPAC "Green Book" Quantities, units, and symbols in physical chemistry also defines $\ce{L}$ [2, p. 160] in chapter 9 ABBREVIATIONS AND ACRONYMS:

\begin{array}{ll}
[\ldots] & \\
\mathrm{KS} & \text{Kohn-Sham} \\ \\
\mathrm{L} & \text{ligand} \\
\mathrm{L2TOFMS} & \text {laser desorption laser photoionization time-of-flight mass spectroscopy} \\
[\ldots] &
\end{array}

Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995) uses $\ce{L}$ [3, p. 1349]. For example, see this entry:

metal-carbyne complexes:
  Metal complexes of the type $\ce{RCML_n}$, ($\ce{M}$ = metal, $\ce{L}$ = ligand) in which formally a carbyne is coordinated to a metal. E.g.

$\mathrm{Lig}$ and other notations are not mentioned in the above sources as suitable ligand notations.
Bibliography

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI 10.1021/bk-2006-STYG.
IUPAC “Green Book” Quantities, units, and symbols in physical chemistry, 3rd ed.; Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Moss, G. P.; Smith, P. A. S.; Tavernier, D. Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375 DOI: 10.1351/pac199567081307.

A paper by M. L. H. Green, 'A new approach to the formal classification of covalent compounds of the elements' (J. Organomet. Chem, $1995$, vol. 500, issues 1–2, pp 127–148, DOI: 10.1016/0022-328X(95)00508-N), might be of interest. It denotes (or classifies) general ligands mostly by letters $\mathrm{L,\ X,\ Z}$. This is based on the sometimes ambiguous number of electrons donated by the ligand. Green himself calls it the covalent bond classification.

@LinearChristmas Wow, that's a great find, and quite a lengthy reading. If you have some time, please do feel free to post this comment as an answer (maybe you can add a few crucial details/examples). Even though it seems like the paper doesn't touch biochemistry (the leucine case), it is still an interesting concept (and a new one to me).

Thanks for the encouragement to turn the comment into an answer! I'll hold off for now since I have not read the whole thing. I might have time to do so in the second week of September; if it's still relevant by then... :) Sorry!

@LinearChristmas Aside from the bounty expiring in 6 days there is no hurry, I already started to read this thing. Maybe you can just post it now as is, and expand later if necessary. If nobody else answers this question in 6 days, I'm going to loose 100 xp for nothing (for the second time, first time I lost 50:)).

Haha, I didn't notice there's a bounty (honestly). Unfortunately, I wasn't lying about not having time, in that I'm not just being lazy ;). If I don't make it before the bounty expires, and there's a good chance of that happening, it's my fault. I wouldn't be worthy...

@LinearChristmas Anyway, thanks for the paper, I would really like to see another answer. But feel free to get it done when it's done:)

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Hyperconjugation in cresol
In cresol both $+I$ effect as well as hyperconjugation work together to determine its acidity. When the methyl group is present in the ortho- and para-position both of them can act upon phenol.
Hyperconjugation as a rule dominates which increases the electron density of the aryl. This makes it less acidic because the conjugate base is now less stable. But when its meta-cresol, hyperconjugation no longer works.
Why is it so? It happens even though an empty $\mathrm{p}$-orbital is present near the $\alpha$-hydrogen. What else does it need?

Above are the resonance forms of m-cresol after deprotonation. As you can see, the negative charge resonates around the aryl ring, but never resides on the carbon directly bonded to the methyl group at the meta position.
Thus, the hyperconjugative effect from the meta methyl group does not affect the acidity of cresol in this isomer. Notice how the destabilising hyperconjugation effect would destabilise the resonating negative charge, if the methyl group was at the ortho- or para- position.

Would you please elaborate on the destabilising effect of hyperconjugation. How does it decrease the stability when the methyl group is at the ortho/para position? Thanks.

@ParthChauhan Methyl groups are electron donating (from the electron density of the C-H sigma bonds donating into adjacent parallel(ish) orbitals i.e. hyperconjugation). When the adjacent parallel(ish) orbital is a lone pair of electrons (which it would be in this case), the interaction between the lone pair and C-H sigma electrons is destabilising as one would expect through simple unfavourable Coulombic interactions.

It is known that cresol containing methyl group destabilises the conjugate base ion by releasing electrons which results in decreasing of acidic strength.
But in meta-cresol, this destabilisation is only due to $+I$ effect. While in ortho- and para-cresols, the destabilisation is also due to hyperconjugation with $+I$ effect as the methyl group is attached in the para- and ortho-positions (since the hyperconjugation is only shown in para- and ortho-substituted).
Means the factors which are trying to decrease the acidic strength is 2 in
para- and ortho-cresols, while in meta-isomer, only one $+I$ is decreasing the acidic strength.


text:
Closed shell structure & intrinsic molecular (electric) dipole moment
Does the closed shell structure of a molecule imply that it has an intrinsic (electric) molecular dipole moment? If so, why is that?
I came across this in paper [1], which seems to suggest that this is the case on page 6 (top left-hand corner):

While both inorganic thin ﬁlms and organic monolayers alter the charge rearrangements between the two contacting bulks, organic molecules add their own, “intrinsic” dipole caused by the closed shell nature of molecules. Thus, electron-withdrawing (e.g., halogens, $\ce{CN}$, and $\ce{NO2}$) or -donating groups (e.g., $\ce{OMe}$ and $\ce{NH2}$) can induce huge electric ﬁelds within a
  single molecule. Adsorbing molecules so that they will be at solid interfaces can in principle aﬀect, and thus modify, any of the interfacial mechanisms: interface traps ..., molecular dipoles ..., and charge rearrangement ...

Bibiliography

Vilan, A.; Cahen, D. Chemical Reviews 2017, 117 (5), 4624–4666.
DOI 10.1021/acs.chemrev.6b00746.

I added the text from the paper I think you meant for a quick lookup, also for those who is having hard times getting the paper itself. If this is not the text you've intended to quote, please let me know or just feel free to roll back.

Thanks! That was the text I intended. This makes it much clearer! :)

I don't think it can.  Methane is closed shell but has no electric dipole moment by virtue of tetrahedral symmetry.
Would be interested to know what they mean though. http://www.sciencedirect.com/science/article/pii/0166128088802665 page 304 ; http://www.users.csbsju.edu/~frioux/h2bond/MethaneMOBonding.pdf ; http://www.science.oregonstate.edu/~gablek/CH334/Chapter1/methane_MOs.htm


text:
Minimizing Catalyst Residue in Vegetable Oil Hydrogenation
We are using a nickel catalyst in our hydrogenation process to help convert vegetable oil to vegetable wax. The issue is ensuring all additives are successfully removed once hydrogenation is complete.
There are numerous catalysts that can be used to assist this process.
The question is: what catalyst is effective for hydrogenation but also extremely efficient to filter back out of the finished wax? Or, is there a cleaner way to hydrogenate oil such that almost no residue remains?

Nickel is usually a good catalyst for this sort of hydrogenation as it's cheap and effective. As to residues, I'm assuming that alludes to residual catalyst; there are a few ways to mitigate against this. Initially, ensure you have minimal fines in the catalyst. The easiest way is to ask your supplier to ensure this. If it's not that sort of process and you just use off the shelf material, then settle your catalyst in water, give it a mix and remove supernatant after a few seconds. Fines will suspend for longer and can be removed in this way. Large particle size material will sink faster. By using the larger particles, you make it easier to remove the catalyst by filtration later. The definitive check is residue on ignition of the final samples. The downside to nickel in batch (as opposed to continuous or flow)  is that it's very dense and needs the right agitator type and speed to get it suspended effectively. However,  this is not normally an issue unless you have inappropriately sized the agitator and type e.g. anchor, to the vessel size and geometry. It may not be worth using precious metal (PM) catalysts for this sort of transformation, but they are easier to suspend when supported on carbon. The advantage of PM catalysts is lower loading.
In all references to catalysts above, they are of the heterogeneous type.

Excellent suggestion - thank you. Perhaps it would be easier/faster if we also used a solvent like hexane which we could boil off. We're just trying to minimize any possibility of contamination in the finished wax. I would upvote your response but not enough reputation yet. Thanks for the ideas!

Another advantage of Nickel is that it is magnetic. On a lab scale Ni residues can be removed by standing the suspension over a fridge magnet for a few minutes before decanting.

Excellent point Waylander. When I did screening, I would use a cross magnetic stirrer. At the end if the reaction, most of the nickel was stuck to the stirrer and caught most on decanting/filtering also. @Gr3go for minimal contamination, I would use a packed bed reactor i.e. A column packed with catalyst, like a hplc column. However, this technology is not always available. I would get nickel test strips to get a good idea of any nickel content. Also, use heptane instead of hexane. Less toxic, category 3 solvent and faster static discharge i.e. Much less build up of charge

Great ideas. I had hoped nickel's magnetic properties would help us and this is a good confirmation. Also, heptane and nickel strips are perfect suggestions to minimize and verify purity. I assume heptane has a low boiling point as well, but will confirm in a bit. Thanks guys!

Heptane actually has a higher boiling point than you'd anticipate but is easy to remove under vacuum. In fact, I've removed it to appropriate levels in a custom synthesis - confirmed by GC - simply by bubbling nitrogen through a solution in a fume cupboard, back in the day.


text:
What are the side effects of adding dry ice to a swimming pool?
My apartment complex pool was closed today due to a resident dumping a large quantity of dry ice into the pool. In the notification from my landlord, they stated:

Dry ice is frozen carbon dioxide and not only imbalances the pool, but can cause burns due to the below freezing temperatures.

My understanding of dry ice being added to water is that it would sublimate and carbonate the water, causing a minor temperature decrease, and creating a low-hanging cloud of carbon dioxide gas on top of the surface (since it's heavier than air).
However, it would seem that if the dry ice indeed does cause "below freezing temperatures", that would just freeze the water; so, unless you already in the water, you shouldn't get a freeze-burn.
I've done a bit of searching, but most material talks about burns from solid dry ice, not from adding it to water.

One thread on Reddit suggests that freeze-burns are more possible from dry ice gas if the skin is wet, but there are no citations.
CCOHS states,

Direct contact with the liquefied [carbon dioxide] gas can chill or freeze the skin (frostbite).

... but refers to liquefied gas, not gas in the gaseous state.
Wikipedia doesn't list any particular safety concerns aside from handling of liquid or solid carbon dioxide.

So, here's what I'm wondering: How does dry ice affect the pool's pH, and to what extent? How does it affect the temperature, and to what extent? And what other side effects (mostly harmful ones) would arise from the addition of dry ice to a swimming pool?

(1) Negligibly (if at all) given the small amount of dry ice relative to that of a buffered system in equilibrium with the atmosphere (small bump in the positive direction until equilibrium is reestablished); (2) Negligibly (if at all) given the small amount of dry ice relative to that of a buffered system in thermal equilibrium with the surroundings (small bump in the negative direction until thermal equilibrium is reestablished); and (3) None unless you happen to contact it with your bare skin.

@ToddMinehardt For (3), you're referring to gas-to-bare-skin contact, or solid-to-bare-skin contact?

Bare skin to solid contact. Sorry for not being clearer. That said, you wouldn't want to be exposed (bare or not) to something like a jet of carbon dioxide gas (really, any gas) blasting at you, but that's far from you scenario. What I'm seeing here is a larger version of a standard intro chemistry demonstration of dry ice in water that "boils" the water at room temperature, and is frequently seen in Halloween and similar mock-ups of "boiling" liquids in laboratory apparatus.

There are only two problems. First if someone comes into contact with the solid dry ice then they can suffer frostbite burns. There would be no "liquid" dry ice (carbon dioxide) at the pressure of the atmosphere. The second problem is that carbon dioxide is heavier than air. so if the pool were in a recessed area the carbon dioxide could "pool" displacing the air (and hence the oxygen in it). Thus swimmers could suffocate.

I would ask the land lord to walk on the ice formed on the below freezing surface of the pool.  Sounds like someone who wants to inconvenience people to try and force social pressure onto some pranksters because some other mother Grundies take it on themselves to complain about pranksters having fun.  Ask him to calculate how much pocket money it would cost to purchase sufficient dry ice to freeze the pool, then ask if that is likely to occur.

Global warming.

I’m reminded of the Russian guy who brought dry ice and a bottle of vodka on a field trip. At lunch, he threw chunks of dry ice into the vodka to cool it down. We then drank the vodka.

Solubility of $\ce{CO2}$ in water is $\pu{1.45 g/L}$ at SATP. Apparent dissociation constant for carbonic acid at the same conditions is $K_\text{a}(\ce{H2CO3}) = \pu{4.47e−7 mol L^{-1}}$.
Now lets speculate how much dry ice could be disposed and what the pool capacity might be, guessing that the water warmed back up. It's hard to tell which pool size your apartment complex (hopefully not this one) might have, so I'm going to propose some values (marked in $\mathrm{\color{red}{red}}$) which look plausible to me.
Assuming $m(\ce{CO2}) = \color{red}{\pu{10 kg}}$, and the pool volume of approx. $V = \color{red}{\pu{5e5 L}}$ (olympic pool is $\pu{2.5e6 L}$, I took $\frac{1}{50}$ of that), one notices that in order to completely dissolve $\pu{10 kg}$ of $\ce{CO2}$, $\pu{6.9e3 L}$ of water is required (based on solubility data), which means that we have more than enough water in the pool to completely dissolve all $\ce{CO2}$.
For the weak acid
$$[\ce{H+}] \approx \sqrt{K_\text{a} C} = \sqrt{\frac{K_\text{a} m}{M V}} = \sqrt{\frac{\pu{4.47e−7 mol L^{-1}} \cdot \pu{e4 g}}{\pu{44 g mol ^{-1} \cdot \pu{5e5 L} }}} = \pu{1.43e-5 mol L^{-1}}$$
so the pool water will indeed be acidified to the level of
$$\mathrm{pH} = -\log[\ce{H+}] = 4.85,$$
assuming the values I've willfully established, and that all carbon dioxide completely dissolved in water. You can put your values into the equation and argue with your landlord further.

There's no way you're hitting pH of 4.85 in system buffered by chemicals in the pool (basic salts) as well as atmospheric carbon dioxide levels. At worst, pH is 5.65. The scenario you're describing is on that does not include time: equilibrium will very quickly be reestablished.

@ToddMinehardt You are absolutely right, I took the idealized boundary simplest case because there is too many variables to take into account; it's more like a model for OP to get initial idea of what's causing the pH shift.

Another factor that will limit the amount of CO2 that is dissolved in the water is the loss of gas as it boils off and bubbles escape the water's surface. The constraint of complete dissolution was named in the calculation to establish it as a theoretical limit. The system is not closed and the large surface area will help to reestablish equilibrium.


text:
What is the difference between a glycosidic bond and peptide bond
What is the difference between a glycosidic bond and peptide bond because both of them involve the elimination of water but what exactly is different in between them?

Is Wikipedia broken?

Eaten and broken them everyday... what bonds make carbohydrates? what about protein?

The bond connecting the anomeric carbon to the acetal oxygen is termed a glycosidic bond.

In general speaking, with respect to sugars,  a bond between the anomeric carbon atom and the oxygen atom of the alcohol is called a glycosidic bond (this arrangement is termed an O-glycosidic bond).

Unlike peptide bond, the glycosidic bond can have several variations that it can be formed by through association of the anomeric carbon to other heteroatoms such as nitrogen as in nucleosides (anomeric carbon atom of a sugar linked to the nitrogen atom of an amine) and is specifically called an N-glycosidic bond.
e.g the glycosidic linkage is from the C-1’ carbon of the sugar to the N-1 nitrogen of pyrimidines or to the N-9 nitrogen of purines. Its not necessarily the bond between monosaccharides only
Another unique aspect of this bond is can exist in several structural forms depending on how anomeric form of the sugar is involved in the bond, and this is the basis of structural forms of polymers and physical properties (starch, cellulose, glycogen etc)
To illustrate this:

The notation for the glycosidic linkage between the two sugars
  specifies which anomeric form of the sugar is involved in the bond and
  also specifies which carbon atoms of the two sugars are linked
  together.
For example two ways in which two α-D-glucose molecules can be linked
  together are α(1>4) and α(1>6). Firstly, the α anomeric carbon (C-l)
  of the first glucose molecule is joined in a glycosidic bond to the
  fourth carbon atom (C-4) of the second glucose molecule;  or the C-1
  of the first glucose molecule is linked to the C-6 of the second
  glucose .

Another important difference with respect to resulting polymers formed from these bonds - polysaccharides, in contrast to proteins and nucleic acids, form branched as well as linear polymers
α-Amylose is a linear polymer of several thousand glucose residues linked by α(1 >4) bonds. Note that although α-amylose is an isomer of cellulose, it has very different structural properties. This is because cellulose’s β-glycosidic linkages cause each successive glucose residue to flip 180° with respect to the preceding residue, so that the polymer assumes an easily packed, fully extended conformation.
Peptide bond
The resulting linkage formed when α-amino acids polymerize, through the elimination of a water molecule is known as a peptide bond (sometimes called an amide bond):

Peptide bond (shown in red)
Glycosidic bonds between monosaccharide units are the basis for the formation of oligosaccharides and polysaccharides.
The glycosidic bond is therefore the carbohydrate analog of the peptide bond in proteins. (The bond in a nucleoside linking its ribose residue to its base is also a glycosidic bond)

Great answer! Molecular structures look so nice and crisp, did you use chemfig $\LaTeX$ package to craft them?

@andselisk Thanks no I didn't use that package, instead I used some illustrations in my biochem resources and texts :)

Sources of pics?

A glycosidic bond is one between monosaccharides, while a peptide bond is one between amino acids. That is essentially their only difference in terms of structure.


text:
C-O bond strength in metal carbonyls
The metal carbonyls (and similar organometallic compounds) involve a combination of a $\sigma$-bond, a $\pi$-bond and backbonding. The bond strengths under consideration are the metal-carbon bond and the carbon oxygen bond. Here is the original question:

Predict the order of $\ce{C−O}$ bond strength in the following:
I) $\ce{[Mn(CO)6]+}$
II) $\ce{[Cr(CO)6]}$
III) $\ce{[V(CO)6]-}$

All of these are isoelectronic, and hence whatever the difference of bond strengths, must arise from the difference in the resultant nuclear force of attraction. Since $\ce{Mn+}$ has the highest charge density, I would assume that the $\ce{M−C}$ bond strength of the manganese compound is the strongest (vanadium being the weakest) and therefore, the $\ce{C-O}$ bond strength must follow the opposite order, as the strengthening of the metal carbonyl bond should weaken the carbon oxygen bond.
But the answer given is $\mathrm{III<II<I}$, exactly the opposite of my prediction. Why is it so? A relevant factor which I didn't consider is the back bonding, but I am unsure of how to integrate that into my prediction. Also, how would we make a prediction if the given compounds were not isoelectronic?

You did learn about backbonding, right? With simple CO complexes like these, the M to C backbonding is likely to be more significant than the ordinary sigma bond from C to M. And backbonding increases in the order Mn+ < Cr < V-.


text:
Taste same but why not LiCl a substitute for NaCl (common salt)?
A clinical paper from 1949 [1] observes the toxicity of LiCl, at very low concentrations. That just went over my head... Is toxicity the only reason to avoid LiCl as a substitute for NaCl (specially for those who were on a salt restricted diet)? Why did they even consider it to be a substitute in the first place?
What physical and chemical property of Li behaves differently than Na (e.g. being toxic)?
P. S.: not sure what tag to put in.
Bibliography

Hanlon, L. W. Journal of the American Medical Association 1949, 139 (11), 688. DOI 10.1001/jama.1949.02900280004002.

All of the alkali metals will have different physiological properties since the ions are of different sizes. Exactly how is a subject for books, not a paragraph answer. // Lithium has been used medically for various brain problems. https://en.wikipedia.org/wiki/Lithium#Medicine

https://en.wikipedia.org/wiki/Lithium_chloride#Precautions - I guess you found it here.

Reviews of other alkali chloride salts (e.g. LiCl, NaCl, KCl, RbCl, CsCl)

The reason why anything is ever considered a substitute for normal salt (NaCl) is that some people take too much salt with their diet and this is widely thought to be bad for your blood pressure. This, however, is disputed by some mainly on the grounds that only a small proportion of people are sensitive to hight sodium in the diet and the rest of us can cope with widely differing amounts with no discernible harm.
However, this doesn't mean that anything that tastes like salt is necessarily a good substitute. Not having enough salt is also a problem as the Na/K balance in cells is important and a serious lack of sodium can be an acute danger. For most people with an adequate diet, using a salt substitute (for table salt) can reduce their sodium intake but, since there is already a lot of sodium in the diet, won't cause a problem. But not for all people.
The most common table salt substitute is KCl, which, again, won't be harmful to people with a normal diet. The body is good at processing potassium and sodium and can regulate their levels within reasonable bounds. So KCl is safe as the body is used to coping with it.
This isn't true for lithium. There is no major biological role for lithium in the body, and, while lithium is regarded as an essential trace element, nobody really knows why. Whatever the normal role, the body is not designed to cope with significant amounts of it (unlike K and Na which are major components of many biological systems). So this might suggest that using large amounts a salt substitute has an unknown risk for harm far greater than the use of, say, potassium. Moreover, lithium does has significant known effects on the nervous system, though the mechanisms are poorly understood. This is illustrated by the use of controlled amounts of lithium salts as a significant treatment for bipolar disorder. This fact alone (regardless of any other toxic effects in high doses) would suggest it isn't a safe alternative when used as a substitute for table salt in uncontrolled doses.
The use of lithium as a treatment for bipolar disorder is relatively recent (started in the early 1950s) so it might have looked like a harmless alternative to sodium chloride in earlier times.

But Li has an outright negative side effects on those who take it for even bipolar. (Another paper said it so or in wiki). I am going to leave this as it is. Sometimes it's really bugging me with these questions along the way. I just want to put it aside and move on (which doesn't feel right...oh well...)


text:
Unix/MobaXterm Project Assistence (Knowing the Error in my Output File)
I'm a student and I recieved a project in which I am supposed to simulate a transition state in an $\mathrm{S_N2}$ reaction.
The first instruction in this project is to use a benchmark reaction in order to establish the best level of theory. The reaction is $\mathrm{S_N2}$ with the following reactants: $\ce{MeCH2Cl + Cl-}$ and with the following level of theory: HF/6-31+G(d,p).
In summary this is what I did:

I used Gaussview to create the reactants (in the same file) and the transition state.
I saved the files and moved them to Mobaxterm for editing the text and setting the level of theory.
I condensed all the data into one input file and used the Opt = QST3 method for simulating the transition state.

I recieved the error:
Constants:
Undefined symbol, angle beta, card   9.
Error termination via Lnk1e in /private/chem/g09E01/g09/l101.exe at Sat Aug  5 09:07:08 2017.
Job cpu time:       0 days  0 hours  0 minutes  0.9 seconds.
File lengths (MBytes):  RWF=      5 Int=      0 D2E=      0 Chk=      3 Scr=      1
Error: segmentation violation

Here is my input file:
%chk=Reactants.chk
%mem=4000MB
%nproc=4
#P opt HF/6-31+G(d,p) Opt=QST3

MeCH2Cl (Chloroethane) + Cl^-

0 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

Product MeCH2Cl + Cl^-

0 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

TS MeCH2Cl2

0 2
 C
 H                  1            E1
 H                  1            E2    2            F1
 C                  1            E3    3            F2    2            G1    0
 H                  4            E4    1            F3    3            G2    0
 H                  4            E5    1            F4    3            G3    0
 H                  4            E6    1            F5    3            G4    0
 Cl                 1            E7    4            F6    7            G5    0
 Cl                 1            E8    4            F7    7            G6    0

   E1             1.06999987
   E2             1.06999978
   E3             1.54000000
   E4             1.07000000
   E5             1.07000000
   E6             1.07000000
   E7             2.20000000
   E8             2.20000000
   F1           120.00007951
   F2           119.99972580
   F3           109.47120255
   F4           109.47120255
   F5           109.47123134
   F6            90.00004829
   F7            90.00004855
   G1           180.00000000
   G2            59.99830388
   G3           179.99831867
   G4           -60.00168873
   G5            29.99845985
   G6          -150.00183223

I also don't know how to format the input file text in a way that it will keep its original form.
Thanks for the help. Any guidance would mean a lot to me.

For code I'd use \...`` (backquote symbol, typically located on a tilde (~) key), also check out How can I format math/chemistry expressions here?

"I saved the files and moved them to Mobaxterm for editing the text and setting the level of theory." What do you mean? How did you move them to MobaXTerm (do you know what MobaXterm is?) and why didn't you just set the correct input paramaters in the first place? This problem clearly doesn't have anything to do with MobaXterm or Unix...

The method I learned in class was to first build the molecule in GaussView, after that save the file in my desktop (as .jgf). Then I would drag it into MobaxTerm and edit the text there.

I already tested the TS search calculation of this reaction based on your proposed structure coordinates. The calculation has been terminated quickly at link9999.exe sub-routine program in which its error message being implies that your initial geometry coordinate is not good enough for TS search by using QST3 method.
When you deal with QST3 you have to carefully consider whether your initial guess structure for Reactant, Product, and TS are consistent or not. I did adjust the position of some atoms and their order. As you see the figure below, the Cl atoms (number 8 & 9) which are opposite side have the similar bond distance for TS structure. Also note that better guess structure, better optimized structure (garbage in, garbage out).

By the way, the number of electron and charge and spin multiplicity that you set are not correct. Since there is Cl anion they should be 52, -1, and 1, respectively.
Here is your fixed gaussian input
%chk=TS-OPT-QST3-HF.chk
%mem=1000MB
%nprocshared=2
#p opt=qst3 freq hf/6-31+g(d,p) gfinput

MeCH2Cl (Chloroethane) + Cl^-

-1 1
 C                  0.00000000    0.00000000    0.00000000
 H                  0.00000000    0.00000000    1.07000000
 H                  1.00880567    0.00000000   -0.35666667
 C                 -0.72596296    1.25740473   -0.51333333
 H                 -1.73476864    1.25740473   -0.15666667
 H                 -0.22156012    2.13105608   -0.15666667
 H                 -0.72596296    1.25740473   -1.58333333
 Cl                -0.82967196   -1.43703398   -0.58666667
 Cl                 1.87443357    0.38660782    1.34616685

Product MeCH2Cl + Cl^-

-1 1
 C                  0.05335154    0.01100392    0.03831562
 H                 -0.62960365   -0.59030929    0.60124957
 H                  0.25817241   -0.64302643   -0.78338549
 C                 -0.67261142    1.26840866   -0.47501772
 H                 -1.68141709    1.26840866   -0.11835105
 H                 -0.16820858    2.14206000   -0.11835105
 H                 -0.67261142    1.26840866   -1.54501772
 Cl                -1.06646990   -1.92858370   -0.75351772
 Cl                 1.44289489    0.29760161    1.03624759

TS MeCH2Cl2

-1 1
 C                  0.00002466    0.00041433   -0.00004215
 H                  0.37333169   -0.90857336    0.42337605
 H                  0.56365981    0.90508352    0.09370541
 C                 -1.33365185    0.00041433   -0.77004668
 H                 -1.39031750   -0.87322218   -1.38522816
 H                 -2.14693922    0.00038473   -0.07473118
 H                 -1.39034379    0.87408048   -1.38518364
 Cl                 1.01053970   -0.23517338   -1.75028838
 Cl                -1.01018689    0.24110207    1.74968530

I also strongly suggest you to add Freq keyword to invoke the program to print out you the vibrational mode and frequencies values for confirming later if the last optimized TS searched is the desired TS.

"Also note that better guess structure, better optimized structure (garbage in, garbage out)." not necessarily, once you are close enough the result will be the same. It takes longer if you are further away but that's it.

Thanks. How in the first place do I determine the best guess structure? It seems like I'm more likely to get it wrong.

I tried running your inpout file but I recieved this error message in the output:                                                                                Symbolic Z-matrix:
Charge = -1 Multiplicity = 1
(had to delete coordinates)
End of file in ZSymb.
Error termination via Lnk1e in /private/chem/g09E01/g09/l101.exe at Mon Aug  7 15:02:53 2017.
Job cpu time:       0 days  0 hours  0 minutes  0.3 seconds.
File lengths (MBytes):  RWF=      5 Int=      0 D2E=      0 Chk=      2 Scr=      1
Error: segmentation violation

@HB It's hard to fix the issue without seeing the complete input or output file.
Can you please upload to a public website (e.g. http://pastebin.com/). I would be happy to help you.

It worked like a charm. It made me realise that I wrongly used the same cordinates in the Product and reactant sections. I rebuilt with GaussView. I did use a different kind of coordinates (I think I saved without using "write Cartesians). My new input file terminated normally. Thanks a lot.

I very, very strongly discourage compound jobs with Gaussian (adding the freq keyword). If anything fails in the second part, it compromises your checkpoint file which might not be recoverable. You can always read in the results of a completed calculation, but if something fails you might carry on a mistake and don't even realise it.

I copied your input into a new file, added some empty lines at the end and changed the input line to be "#p opt=qst3 hf/6-31..." instead of redundant (and maybe erroneous(?) "opt ... opt=qst3"). I dropped the memory and processor line, as I defined some defaults in the Default.Route file. Using the following input, the calculation runs smoothly towards a transition state.
%chk=Reactants.chk
#P opt=qst3 HF/6-31+G(d,p)

MeCH2Cl (Chloroethane) + Cl^-

0 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

Product MeCH2Cl + Cl^-

0 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

TS MeCH2Cl2

0 2
 C
 H                  1            E1
 H                  1            E2    2            F1
 C                  1            E3    3            F2    2            G1    0
 H                  4            E4    1            F3    3            G2    0
 H                  4            E5    1            F4    3            G3    0
 H                  4            E6    1            F5    3            G4    0
 Cl                 1            E7    4            F6    7            G5    0
 Cl                 1            E8    4            F7    7            G6    0

   E1             1.06999987
   E2             1.06999978
   E3             1.54000000
   E4             1.07000000
   E5             1.07000000
   E6             1.07000000
   E7             2.20000000
   E8             2.20000000
   F1           120.00007951
   F2           119.99972580
   F3           109.47120255
   F4           109.47120255
   F5           109.47123134
   F6            90.00004829
   F7            90.00004855
   G1           180.00000000
   G2            59.99830388
   G3           179.99831867
   G4           -60.00168873
   G5            29.99845985
   G6          -150.00183223
<- blank line
<- blank line
<- blank line
<- blank line

The result is a transition state with one imaginary frequency at $-823.9116~\mathrm{cm^{-1}}$.
Then I realized the Charge and Multiplicity as was also criticized by rangsiman. So your slightly modified input works for me, but the result is a transition state for "charge=0 multiplicity=2" instead of "charge=-1 and multiplicity=1".
The optimization for the correct charge/multiplicity took roughly 4 minutes on my computer and the corresponding frequency calculation took 35 seconds. The result is a transition state with an imaginary frequency at $-363.5245~\mathrm{cm⁻1}$.
%chk=Reactants.chk
#P opt=qst3 HF/6-31+G(d,p)

MeCH2Cl (Chloroethane) + Cl^-

-1 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

Product MeCH2Cl + Cl^-

-1 1
 C
 H                  1            B1
 H                  1            B2    2            A1
 C                  1            B3    3            A2    2            D1    0
 H                  4            B4    1            A3    3            D2    0
 H                  4            B5    1            A4    3            D3    0
 H                  4            B6    1            A5    3            D4    0
 Cl                 1            B7    4            A6    7            D5    0
 Cl                 1            B8    4            A7    7            D6    0

   B1             1.07000000
   B2             1.07000000
   B3             1.54000000
   B4             1.07000000
   B5             1.07000000
   B6             1.07000000
   B7             1.76000000
   B8             2.00000000
   A1           109.47122063
   A2           109.47122063
   A3           109.47122063
   A4           109.47122063
   A5           109.47122063
   A6           109.47122063
   A7           115.75316236
   D1           120.00000000
   D2          -180.00000000
   D3           -60.00000000
   D4            60.00000000
   D5           -60.00000000
   D6           120.46043739

TS MeCH2Cl2

-1 1
 C
 H                  1            E1
 H                  1            E2    2            F1
 C                  1            E3    3            F2    2            G1    0
 H                  4            E4    1            F3    3            G2    0
 H                  4            E5    1            F4    3            G3    0
 H                  4            E6    1            F5    3            G4    0
 Cl                 1            E7    4            F6    7            G5    0
 Cl                 1            E8    4            F7    7            G6    0

   E1             1.06999987
   E2             1.06999978
   E3             1.54000000
   E4             1.07000000
   E5             1.07000000
   E6             1.07000000
   E7             2.20000000
   E8             2.20000000
   F1           120.00007951
   F2           119.99972580
   F3           109.47120255
   F4           109.47120255
   F5           109.47123134
   F6            90.00004829
   F7            90.00004855
   G1           180.00000000
   G2            59.99830388
   G3           179.99831867
   G4           -60.00168873
   G5            29.99845985
   G6          -150.00183223
<- blank line
<- blank line
<- blank line
<- blank line

Perfectly, Gaussian ends with: EXPERIENCE IS THE FRUIT OF THE TREE OF ERRORS.

Instead of rangsiman’s hint to add the freq keyword to your input line, I would suggest you to do it additionally and not as the combination opt freq as the latter can lead to problems by not copying your whole input into the frequency calculation.

Is there a reason you used 4 blank lines? Afaik two of them are interpreted by Gaussian as end of input file.

I just go straight to the end of any input that I write and add a random number of blank lines >= 3. So less could also work, yes.


text:
Hartree Fock and DFT - Exchange term
this might be a stupid question but I am really confused at the moment. I'm currently preparing for an exam in computational chemistry. In the lecture I noted always down that the exchange term (Pauli-Principle), which arises as part of the two-electron functions is not determined in Hartree-Fock and added to the cusp-problem, when two electrons come close together. Those two then form the correlation energy, which is determined in post-Hartree-Fock methods.
Then in DFT we have several attempts to solve the exchange part, and I took a note back then, that basically in DFT or in the Kohn-Sham approach, as we do not know the DFT-functional, you create something, that looks like a Fock-operator, with the difference, that it also contains the exchange part.
Now I looked at some tasks and one question was why you do not simply take the HF expression for the exchange part and insert it into the DFT equations. I was a bit shocked what this expression might be and it turns out, when I look into other scripts I cannot find any note that in modern HF the exchange part would be neglected in any way.
So for simple HF (not post-HF) calculations, is the exchange term calculated as well and the only thing you miss would the the correlation?

If I'm understanding your question correctly, yes, HF does include exchange (sometimes referred to as exchange correlation) interactions between electrons with the same spin. It does not include the overall correlation of electron movement, as HF uses the mean field for the potential.

Ah thank you, I found the problem, I thought xc was the abbreveation for exchange, not exchange correlation. So my Professor probably meant, that you seperate it into x + e and use post-HF methods from this point and I thought you also take the exchange part.

Most of the original question has been solved by discussion in the comments, but a pertinent question remains:

[W]hy you do not simply take the HF expression for the exchange part and insert it into the DFT equations [?]

The answer is that it has been tried and does not work well. In a sense, the DFT quantities "exchange energy" ($x$) and "correlation energy" ($c$) do not quite match the HF quantities "exchange energy" and "correlation energy".$^1$ In fact, even within DFT, the quantities do not match between functionals, as can be seen in some $x$- and $c$-functionals working together well and some not so well. In particular, those that have been fitted in combination with each other often only work in that combination. This suggests to me an illusion of a clear-cut way to differentiate the two in the DFT framework. It is also true that in some functionals, the $x$-term appears to account for some static correlation (which, broadly speaking, often becomes important when a single Lewis structure does not describe the molecule well: transition states, resonance structures). Entirely replacing it with Fock exchange forgoes this advantage.
Some programs, such as NWChem, and to some extent Gaussian, allow you to combine functionals and Fock exchange (often with arbitrary weights) to conduct your own experiment. A small benchmark set for main-group thermochemistry would probably give you an indication.
I wish I could you references for this, but this has been related to me verbally by my advisor back in the day and I no longer have access to most of the literature. If anybody has such references, I would be happy to see them included.

$^1$ The notation is a bit sloppy, but we're not striving for rigor in this answer.


text:
Relationship between surface area and rate of reaction
Say I add a whole piece of calcium carbonate to some $\ce{HCl}$ and then measure the reaction rate by measuring how quickly the mass changes (how quickly $\ce{CO2}$ is being evolved).
Then, I repeat the experiment for the same size calcium carbonate piece but broken into 2 pieces, then 4 pieces and finally, 6 pieces. If I plot the rate of reaction vs. the surface area of the calcium carbonate, would I get a linear relationship?

You mean initial rate? I don't think you could measure it quickly enough.

Yes, initial rate. So, let's say the grams evolved over 120s


text:
When are all cannabinoids infused into grape seed oil using non-decarboxylated, dried plant matter?
I am unsure which is the correct terminology, infusion or extraction? For all practical purposes I will continue using infusion here. (What is the difference between the two?)
When making the infusion1 I'm using a ratio of 2 oz of dried, but not decarboxylated plant matter to 1 quart of grape seed oil. Any brand will work. The plant matter mainly consists of the buds and colas. These have gone through the curing phase already and are dry.
I put the plant material in the oil inside of a small 1.5 qt crockpot. The last time I did this, I let the crockpot sit on high for 8 hours before switching to low for 3 days (72 hours).
A lot of tutorials and online videos will say leave it sit for 4, 8, 24, or all the way up to 72 hours and more.
How long will it actually take until most cannabinoids are infused into the oil, without infusing other, unwanted compounds like chlorophyll?
I'm assuming that there is a breaking point where the plant matter is done decarboxylating, and the oil has extracted all of the cannabinoids.
I've done this kind infusion using coconut oil and I didn't get the same potency as when I did the infusion with grape seed oil. What is the reason on a molecular level that allows grape seed oil to infuse cannabinoids more easily?

1 Legality is not an issue since I live in a state where I am allowed to grow up to 6 plants of varying maturity and also produce my own grape seed oil infusions.

J. Robinson - I think I can answer your question but you need to dissociate the decarboxylation from the extraction (you say infusion). Otherwise, your extraction selectivity is different for the carboxylic acid opposed to the decarboxylated material, which makes your extraction oil and time a moot point. Time, temperature and solvent play a factor. Do you want to extract, then decarboxylate or decarboxylate then extract? The extraction selectivity will guide you

I'm hoping that through the process of extraction the cannabis plant material decarboxylates itself and then infuse itself into the oil? I'm not a chemistry expert by any means, so sorry for the layman's terms. Is it an extraction and not an infusion then?

Infusion could be used as an every day term. The chemical term would be extraction.   To help answer your main question, I'm assuming it's a CBD or THC (or CBDA /THCA,  or both?) compound that's being extracted?

cross-post: https://cooking.stackexchange.com/q/83527

The question in its current state is considered by some as too broad to be treated scientifically and is likely to be closed. The reason for this is that there are simply too many unknown variables. For you personally the brand of grapeseed oil might not matter, but it is already a mixture, which makes the starting conditions quite unpredictable. Then obviously high and low are no temperatures. And we would need more information on which cannabinoids are being extracted, all that and more plays into extraction time/efficiency.

The time taken (before there are diminishing returns) of extracting the herb is totally dependent on the solvent used, or in your case, the oil. If you were to use water, the answer might be days. If you used heptane, a couple of hours.
The decarboxylation and the extraction - a catch all chemical term for removal of components or a component from a particular solid liquid or gas phase, as opposed to infusion which seems to be specific to botanical extractions (so both are correct in this instance) - should be decoupled. If you start with the cannabinoidal acid and heat in solvent, you'll start extracting components at different rates, while also decarboxylating and extracting the cannabinoid. This reduces the selectivity of the extraction.
Again, depending on your chosen oil, the rate at which your desired components and undesired ones are pulled into your oil will differ. You seem to have a handle on  which of your two oils is better. I would try two experiments:

Heat the dry herb to decarboxylate; less than two hours should be ample and may even be too much. Then extract/infuse your oil. Bear in mind, the more oil you use also leaves you open to extracting more rubbish you don't want, as does extending the time. You're better cutting your time down first and monitoring potency (as extracting more undesired components will lower your potency in the final extract, so potency should increase as you bring your time down) and then experimenting on reducing the oil content until your potency decreases. A decarboxylation should take about one hour at around 110C, but it depends on the amount of herb you have and if it's whole leaf,  shredded,  milled etc. I'm talking fine material at kilo scale. Extraction under ideal conditions is done in six hours, depending on solvent.
The second option is to extract at cooler temperatures (cool temp to 50-60C absolute maximum) first to get the cannabinoidal acids. If you have the facility, add base (diluted sodium hydroxide is ideal and would take minutes . Sodium carbonate dissolved in water may work but would be slow (days?) and you'd have to monitor carbon dioxide evolution). The acids you want will form salts and extract into the water.  The neutral stuff you don't want (terpenes, waxes - long chain alkanes) stay in the oil. Separate the oil and water. You have a choice here.

Add acid to your water layer - this may cause your cannabinoid acid to precipitate, so filter it off.
Otherwise add fresh oil to the water,  then acid and your cannabinoid acid will now be in the oil after mixing.

Whatever you choose, heat your cannabinoid acid to over 100 degrees Celsius and you should be done in an hour or two. Over 120 degrees Celsius may be too much. As I mention, some experimenting will be needed but try to ensure your herbs are consistent and not mixtures of different varieties each time.
As for your oils. One is more like a liquid "fat" - the coconut oil. The grapeseed oil is still a mixture, but is primarily "polyunsaturated" ( it's actually di-unsaturated) and about 50% bigger than the main coconut oil component. These oil mixtures are not usually used in chemical extractions as they're not pure. I assume you use digestible, non-toxic solvent oils. Otherwise I'd suggest limonene.

Please use paragraphs. This sort of wall of text is very tiring to read.

I am down-voting this simply because its terribly formatted.


text:
How to theoretically predict thermodynamic stability?
Let's say we have $\ce{CH3-NH2}$ undergoing Hoffman's exhaustive methylation with $\ce{CH3-Cl}$. In which case side reactions such as $\ce{CH3-NH2 + CH3-NH2 -> CH3NHCH3 + NH3}$ would also occur.
How do we theoretically determine the relative difference in thermodynamic stability between the products of the side reaction ($\ce{CH3NHCH3 + NH3}$) and the main reaction?

I imagine it wouldn't dominate because we take methyl iodide in excess.

Okay I'll make this question more specific

The side reaction products will be more stable but would be kinetically unfeasible(comparitively high activation energy)

What is your side product: dimethyl amine (as suggested by the hydrogen on nitrogen)? The bonds of the leftmost carbon don’t add up …

@jan yes, made a mistake while typing this question sorry


text:
Providing more than required amount of energy to an electron in an energy level
I know that an atom has energy levels i.e the energy of each electron in the atom falls in a discrete energy level.
I am also aware that, if the exact difference of energy between two levels is provided to an electron, then it jumps to the higher energy level, and that is a lower amount of energy is provided, the electron stays where it is.
My questions are:

What happens to the energy in the second case? Does it get radiated back as heat? If not, where does that energy go?
What happens if more than the requisite amount of energy is provided? Does the electron stay where it is, or does it go to the higher energy level?

https://chemistry.stackexchange.com/questions/10334/how-strict-is-the-to-excite-electrons-the-energy-must-equal-the-energy-state-di

If the energy exceeds the minimum needed, there are a few possibilities:

More than one electron may be affected.
The electron(s) can be moved to yet higher levels within the atom.
The electron can be expelled (raised to a very high level) with the electron(s)' kinetic energy making up the difference.
At sufficiently high energy, pair production can occur.
Also, rather than expending all its energy on the electron, the photon can be scattered as a lower frequency (less energetic) particle (since it interacted at a point, I'd consider it a particle for that duration).

So -- the excess energy, in effect, can be shared between the photon and electron in varying proportions.
If less than the minimum amount to raise the electron's level is supplied, then the electron stays put... except for quantum phenomena such as tunneling. The Esaki diode exhibits negative resistance, where electrons are more likely to pass through the p/n junction with lower bias.


text:
Why does lithium forms no ethynide on reacting with ethyne?
Sodium Metal when treated with ethyne forms mono and di sodiumethynide why doesn't  lithium ?

Related: Why are lithides not known?

@andselisk I'm not asking about lithides. In mono sodium ethynide sodium has a positive charge

It looks like it does form monolithium ethynide, which should also react to form dilithium ethynide. http://onlinelibrary.wiley.com/doi/10.1002/047084289X.rl035.pub2/abstract;jsessionid=B2FF10E486A8DAB6875934ED0FBB0D1D.f04t02?userIsAuthenticated=false&deniedAccessCustomisedMessage=

@Tyberius Yep! You're right https://en.m.wikipedia.org/wiki/Acetylide anyways I asked because my school textbook was saying lithium does not form Acetylide

@NilayGhosh Thanks for the explanation. But why lithium doesn't reacts directly with terminal alkynes ?

Sodium alkynide was formed by reacting sodium and acetylene but in case of lithium alkynide, it can be formed by reacting acetylene with organolithium reagents like BuLi, Lithium amide, LiHMDS. From wikipedia:

Lithium amide, LiHMDS, or organolithium reagents, such
  as butyllithium, are frequently used form lithium acetylides:
$\ce{H-C#C-H + BuLi ->[THF][-78℃] Li -#-H + BuH}$

The above process is well described in this paper which @Tyberius linked in the comment section.

Lithium actually can be reacted with acetylene to produce the ethynide:

In the laboratory samples may be prepared by treating acetylene with a solution of lithium in ammonia, on −40°C, with creation of addition compound of $\ce{Li2C2 • C2H2 • 2NH3}$ that decomposes in stream of hydrogen at room temperature giving white powder of $\ce{Li2C2}$.

$\ce{C2H2 + 2Li -> Li2C2 + H2}$

However,

Samples prepared in this manner generally are poorly crystalline.

Various alternatives are therefore often preferred, and (some of) these are also given in the Wikipedia article, and by Nilay Ghosh in his answer.


text:
Why does the presence of EWG's on the aryldiazonium ring increase the yield of a Meerwein arylation?
In my textbook, Strategic Applications of Named Reactions in Organic Synthesis, the author says the following regarding properties of the Meerwein arylation:

The general features of this reaction are: 1) the procedure is simple; no special laboratory equipment is needed;  2) the aryldiazonium halides are prepared by the diazotization of aromatic amines using sodium nitrite and aqueous hydrohalic acids and are not isolated, rather immediately reacted with the alkenes in the presence of an organic solvent (e.g., acetone, acetonitrile); 3) the presence of electron-withdrawing substituents on the aromatic ring tends to increase the yield, whereas electron-donating groups often give lower yields; 4) the alkene component usually has an electron-withdrawing substituent and mostly α,βunsaturated carbonyl compounds are used;

I don't really understand why that is. If anything, I would think that EDG's (in para position) would increase the yield because they stabilize the aryl cation that would form if the aryl diazonium decomposes and loses its diazo group as nitrogen gas (N2).

Could it be that an EDG would make the aryl diazonium compound too unstable and decompose too quickly? Allowing for side reactions to occur?


text:
What should I do about working in a laboratory that I believe is unsafe?
I currently work in a lab that frequently gives me anxiety about safety. We work with toxic fumes daily. The sliding sashes on the fume hoods are held up by cinder blocks. There is a large crack in the glass of one of the hoods. We have an abundance of work so everyone is always rushing through SOPs. I see people pull beakers out while toxic gases are still fuming. We've never even had a safety drill on what to do if there is a chemical spill. We had an audit a few months ago and somehow passed? How should I address this situation? Thank you.

I don't know that this deserves a downvote, but it is a very broad question.  Issues of such importance are probably best addressed through your institution, but the means by which to do so vary from country to country.  It sounds like your lab passed "inspection", but perhaps the examiners didn't look closely enough, or things were covered up.  Failing correction at the level of the institution, you have a right to go to OSHA (in the States) or the equivalent governing body.

Please don't deface your question.  You can flag it for deletion if you so choose.

You should be cautious and follow written down procedures because whistle-blowers are often treated badly.


text:
Is an imine considered a secondary amine?
I am looking at the difference between primary, secondary, and tertiary amines.

What about if the nitrogen is double bonded to a carbon?

Is this carbon counted twice, and is the molecule above a considered tertiary amine?

Imines can never be amines. Amines are amines. Imines are imines. Can an alkene be an alkane?

Imine is to amine as carbonyl is to ether. Do you think a carbonyl is an ether?

Also, the abbreviation using a degree sign is not common in chemistry.

No. Imines cannot be considered an example of secondary amines any more than pyridine can be considered a tertiary amine.
The easiest way to think about this is with oxidation state: imines are at a higher oxidation level than amines, just as aldehydes are at a higher oxidation level than alcohols.
Aside from this 'intuitive' explanation, the IUPAC gold book (a source of chemical nomenclature) defines both amines and imines, clearly differentiating the two:

amines. Compounds formally derived from ammonia by replacing one, two or three hydrogen atoms by hydrocarbyl groups, and having the general structures $\ce{RNH2}$ (primary amines), $\ce{R2NH}$ (secondary amines), $\ce{R3N}$ (tertiary amines).
imines. Compounds having the structure $\ce{RN=CR2}$ (R = H, hydrocarbyl). Thus analogues of aldehydes or ketones, having NR doubly bonded to carbon; aldimines have the structure RCH=NR, ketimines have the structure $\ce{R'2C=NR}$ (R' ≠ H). Imines include azomethines and Schiff bases. Imine is used as a suffix in systematic nomenclature to denote the C=NH group excluding the carbon atom.

Generally NO. Tertiary amine is an ammonia derivative with 3 hydrogen atoms substituted to 3 alkyl/aryl substituents. Whereas imines exhibit some properties of tertiary amines (like pyridine). It can be protonated or alkylated and form iminium salts.

Pyridine is not a tertiary anime...


text:
What kind of energies are present in an electron?
I learnt that an electron loses energy as it goes closer to the nucleus and gains energy as it goes further from it. But what type of energy is involved in this process? If it's kinetic energy, then the electron must also lose its speed, but does this happen? In addition, how do I calculate the energy gained or lost due to the electron's movement?

From your question I assume you are talking about an electron bound to a nucleus, probably hydrogen. Since the mass of the nucleus is ~1836 times larger than the mass of the electron, we may assume the nucleus to be stationary. In a classical model of the problem, we thus have an electron with a certain velocity $v_n$ and kinetic energy $\frac{1}{2}m_ev_n^2$ at a position $r_n$ from the nucleus where it is bound by a Coulombic potential $-Ze^2/(4\pi\epsilon_0r_n)$ (where for hydrogen $Z=1$). The total energy is given by the sum of the two
$$
E_n=\frac{1}{2}m_ev_n^2-\frac{Ze^2}{4\pi\epsilon_0r_n}
$$
In order for the electron to make a (classical) stable trajectory around the nucleus, the electrostatic force should be equal but opposite to the centripedal force, hence
$$
\frac{m_ev_n^2}{r_n}=\frac{Ze^2}{4\pi\epsilon_0r_n^2}
$$
from which it follows that $\frac{1}{2}m_ev_n^2=Ze^2/(8\pi\epsilon_0r_n)$ and the total energy is given by
$$
E_n=\frac{Ze^2}{8\pi\epsilon_0r_n}-\frac{Ze^2}{4\pi\epsilon_0r_n}=-\frac{Ze^2}{8\pi\epsilon_0r_n}
$$
This equation shows that the total (average) kinetic energy in the hydrogen atom is half the size and opposite to the (avarage) potential energy, i.e., $\langle E_\text{kin}\rangle=-\frac{1}{2}\langle E_\text{pot}\rangle$, a result that is also predicted by the virial theorem.
Using Bohr's postulate concerning the quantization of angular momentum, the last expression can be recast in the familiar expression for the hydrogen energies.


text:
What is the best way to illustrate the recovery of an extraction method?
I'm currently working on PFAS (perfluoroalkyl compounds) extraction from soils and their analysis by HPLC-MS/MS.
I wonder what kind of graph can be the best to illustrate the recovery of the method. I have tried with a boxplot on R and I think it's interesting because it shows the variability between all the soils that I used for the validation. The problem is that if their is different recovery between this matrixes (depending on the physico-parameter of the soils) we can't see it neither identify it.

What do you think about it?  According to you, what is the best way to illustrate recovery and how can I do it on R?
Thank you for your answers!

It would maybe help if you can state more clearly what it is that you are interested in. If I understand it correctly, you want to plot the recovery against a physochemical property?


text:
Why does a pickled gherkin glow only at one end when a current is passed through it?
I recently saw this Periodic Video.
The Professor (Martin Polyakoff) notes that,

The Gherkin normally glows only at one end due to excitation of $\ce{Na+}$ ions.
The end which glows can either be the neutral end or the positive end. It is impossible to predict which one will glow.
During the video, both ends glowed together only for a few seconds.

Even the Professor admitted that probably glowing at one end happens due to pure chance.

Why should the gherkin glow only at one end? Is there any reason? Normally one would expect it to glow symmetrically around the center, that is, at both ends.

The question has been quite well received. However no one has commented or answered. Does the question need to be re-worded or better explained?
Do let me know in the comments.

The title of this question competes with some of the best from Arqade.

I don't quite get your meaning...?

yeah but is the title good?

@DeltaScuti_Fomalhautb I personally like it. Sincerely. It meets the requirements of Chemistry.SE Help Center.

Oh, no, the title is great. Arqade is just known for some titles that look really  funny out of context. e.g.: https://gaming.stackexchange.com/q/303625/ and https://gaming.stackexchange.com/q/293615/ (@andselisk)

ohh, now i get it:)  cooking a fairy! those titles are weird

It looks like an electric arc to me, more precisely, a single-phase arc -- it seems like AC current was applied, as "ground" and "neutral" ends were used. Hence the light emission, high observed current density, and high temperature. As you noticed, the yellow color is most likely due to the presence of $\ce{Na+}$ ions.
Any conducting (or not so, if the voltage is high enough) object put in contact with a couple of separated electrodes and applied current (DC, single-/three-phase AC) can undergo arc discharge. For example, with the current from the standard household outlet one can do the same with lemons (it's going to smell better though than the gherkin, at least at the beginning due to evaporating essential oils); one can also light a tip of a cigarette moistened with saliva; or put a piece of paper with a pencil drawing on fire within seconds (starting the ignition from the graphite layer) and so on. Even a piece of wood can also  glow when arc discharge paves the way in a form of graphite path after dielectric breakdown  in between electrodes of HV-generator ($\pu{5 .. 10 kV}$ depending on wood and moisture).
It is not the particular end of the gherkin that glows, it is an area where arc discharge takes place. One can also observe arc(s) hopping along between two elongated parallel electrodes in the very same gherkin:

Video courtesy of YouTube channel "ПРОСТАЯ НАУКА" (Russian; English "SIMPLE SCIENCE")
Predicting the path of the arc is a rather non-trivial physics problem. To put it simply, it partially depends on the contact area, which constantly changes mainly due to vigorous boiling substance, as well as the local pressure and composition of formed gases, which results in equilibrium or non-equilibrium plasma formation in positive column (luminous region in the environment between the electrodes).
The process of forming an electric arc can be simplified down to three distinguished stages:

Arc ignition, when, as a result of shock ionization and emission of electrons from the cathode, an arc discharge begins and the ionization intensity is higher than deionization.  As the electrodes are diverged from each other, the contact pressure and the contact surface are decreased due to increased transient resistance in addition to the resistance of the media (mostly vinegar/salt solution in this case), overheating  starts. High temperature subsequently contributes to thermionic emission, when the electron velocity increases so they can be eliminated from the electrode surface. The voltage on the contact gap is quickly restored, and since the distance between the contacts is small, an electric field of high intensity arises, under the influence of which more electrons get accelerated and escaped from the electrode surface. When they struck a neutral atom or molecule, they give away their kinetic energy. If this energy is sufficient to remove at least one electron from the shell of a neutral atom, then the ionization process takes place.
Stable arc combustion maintained by thermal ionization in the arc itself, when the ionization and deionization intensities are approximately equal. Free electrons and ions form plasma arc content, that can be represented as an ionized channel, in which the burning act takes place and continuous motion of the particles is ensured. Negatively charged particles (primarily electrons) move toward the anode, and positively charged particles (deprived of one or several electrons atoms and molecules of gases) in the opposite direction (toward the cathode). This contineous flow makes the conductivity of plasma comparable with the conductivity of metals. So the voltage drop along the length of the arc is small due to the intensive thermal ionization.
Arc extinction, when the deionization intensity is higher than the ionization. In addition to ionization, there is also deionization of atoms and molecules. The latter occurs mainly through diffusion of charged particles to the environment, and the recombination of electrons and positively charged ions, which are reunited into neutral particles releasing energy in a form of heat.

Video bloggers all over the internet popularized the heck out of this effect, which is, strictly speaking, has more to do with physics, rather than with chemistry.

Your answer is good. But I still have not understood why? does it happen. Can you elaborate on the positive column part? Or should I ask it on Physics SE?

@DeltaScuti_Fomalhautb Thank you! I tried to comment further on the physics of the process, but for more hardcore stuff explaining plasma effects I would bet on another answer or guys at Physics.SE :)

The electric power released in a certain volume element of the gherkin is
$$P = U \cdot I = R \cdot I^2$$
where $U$ is the voltage, $R$ the electrical resistance, and $I$ the electric current through this volume element.
Due to the biological nature of the gherkin, the electrical resistance is not the same for all volume elements. Thus we can expect that in some regions more energy will be released than in others. If the total power available does not suffice to let the whole gherkin glow, we have an explanation for the observed asymmetry.
The effect could be amplified if the specific resistance would increase in the glowing region.
Note: The above explanation needs to backed by experiments.


text:
Best reagent to convert 4-chloro-4H-pyran into an aromatic compound
My question:

My attempt: (I am not really good at mechanisms, so a little help would be good. Thank-you)

I do not know what kind of reactions these will give.

In order to choose the best reagent to convert the compound to an aromatic one, let's ask the question of what is preventing this compound from being aromatic? At carbon-4, the pi-system is disrupted by an sp3-hybridized atom. If carbon-4 could be converted to sp2 by some reaction, then the compound could become aromatic. At C-4, we see a chloride. We know that chlorides are good leaving groups, so let's look at the product of that step.

After chloride leaves, there's a cyclic, fully conjugated pi-system containing six electrons (two for each double bond, two from the oxygen, and zero from the carbocation). Drawing a resonance structure shows that this is an oxygen analog of benzene (known as pyrylium ion).
What kind of reagent would promote this process, i.e., a halide leaving? A Lewis acid, such as $\ce{AlCl3}$ could promote this in two ways. First, it could interact with the chloride in the starting material, building up positive charge on the chloride, making it a better leaving group. Secondly, the Lewis acid can coordinate the chloride on the product side, preventing it from attacking the pyrylium ion.

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


text:
Determining the calcium content in cashews via titration?
I have been planning on an experiment for quite some time now (as part of my after-school chemistry club), but I still feel like I am missing something.
Due to my family's calcium deficiency, I decided to measure calcium content in various foods. It's something rather personal, therefore it will make the experiment more interesting overall.
At first I wanted to choose dietary supplements since they can dissolve easily, though then I realised how flawed of an idea that is, since the contents are usually given on the package in full detail anyway. In then moved onto fruit and other various foods, though it all seemed rather messy/ hard to achieve.
I then realised I could measure the calcium content in cashews. I was thinking about using EDTA and Eriochrome Black T, as well as methyl orange. Here is where I hit my first struggle - how do I dissolve the nuts in the first place? I thought about acids, such as hydrochloric acid, though that would as a result affect the pH of the solution (which, I am not sure, could have an effect on the way the titration procedure functions).
I'd use methyl orange together with Eriochrome as the transition is much clearer since it goes green then grey, whereas the transition from purple to blue is rather subtle - in my opinion this would increase the accuracy of my experiment.
Alternatively, I could use absorbance technique to determine the calcium content - would it be more appropriate for this scenario? I could use a stock solution to find for example the absorbance of 10ppm before calculating the absorbance of the cashew. Does this sound more appropriate?
Any input would be greatly appreciated!
These can be hints, I definitely want to work through the issue myself as much as possible. I just seem to be stuck and don't know where to go from there!

I would recommend you try any procedure with chemicals that you know the content of already to determine the accuracy of your experiment (blind test). If I remember correctly, calcium itself was not easy to determine, especially in presence of other elements. Also the overall concentration on cashews might not be enough to be suitable for titration.