text:
LDA and an aldehyde

Compounds A and B both react with LDA (lithium diisopropylamide, a strong base) to form the same anion C with the formula $\ce{C4H5O^-Li^+}$.

I am unfamiliar with LDA. My first thought was that the base would attack the H of the carbonyl C, I presumed that this H would have the lowest $\mathrm{p}K_\mathrm{a}$ because it is an $\mathrm{sp^2}$ C and the C is slightly positive from the carbonyl.
However, the structures this gave were confusing and didn't give the same product for both A and B. I tried to google LDA which showed that it was a big bulky base so it can't attack hindered H, but this shouldn't be a problem as the carbonyl C H isn't hindered. However, I found one mechanism for a different aldehyde that had the H taken from the $\mathrm{sp^3}$ C next to the carbonyl.
Where would LDA preferentially attack an aldehyde and why?
Additional note: I have now played around with the structures to find that the only way LDA could react with A and B to form the same anion C if it in both cases it removes and $\ce{H+}$ from the $\mathrm{sp^3}$ carbon in the structure, these seems counter-intuitive! But perhaps LDA only attacks $\mathrm{sp^3}$ C-H for some reason?

LDA is a very strong base and not fussy about whether it takes an alpha proton or a gamma proton because you will end up with the same extended enolate i.e. C, which is a stable species.

While this question has a good goal, I think it is a terrible question, if this is a homework or exam question, or just an incomplete question if you're asking it yourself. Which enolate should be produced under the circumstances? (E)- or (Z)- ? They are different, non-interconverting, and probably formed in a mixture whose composition is dependent on whether you start from A or B.

@Zhe The initial part of the question is an undergraduate past-paper question. However no one set me the past paper and I asked the question out of curiosity (and frustration at my own lack of understanding); so it's not a homework question in the strict sense.

Sorry, let me be clear: I am talking specifically about the quoted question, which, based on your comment, is an assigned question, and I find that specific question to be quite poor: C is a mixture of enolates! Your own questions are quite reasonable, and I have no problem with them.

If electronegativity alone were the only factor in determining the acidity of a carbon-hydrogen bond, the base would in fact attack the aldehyde hydrogen.  But that is not so, for delocalization of negative charge can also stabilize the anion.  If the base attacks at the initially saturated carbon in either compound, you get a delocalized negative charge distributed throughout the carbon chain and onto the oxygen.
Such delocalization stabilizes the anion so strongly that it overrides the original electronegativity difference between $\mathrm{sp^2}$ and $\mathrm{sp^3}$ carbon atoms.  So, the base will attack along the carbon chain to produce this delocalized charge anion, instead of at the carbonyl carbon which offers no such delocalization.

Is there a way to predict this other than by trying all options?


text:
Using Wilkinson's Catalyst to reduce esters?
I was solving some questions on alcohols when I came upon a question in which it was asked that whether esters can be reduced  using hydrogen in the prescence of wilkinson's catalyst.
Now I know that this reaction can take place using hydrogen in the prescence of nickel or platinum to give two alcohols.So is the same true for using wilkinson's catalyst also?
And if it cannot be so then on which functional groups can this be used on?
Also what would happen differently if wilkinson's catalyst is used?

https://aocs.onlinelibrary.wiley.com/doi/abs/10.1007/BF02678792

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text:
Scientific Reason for Salt Solution Gaining Volume
A little background to this question: my mom placed a glass with salt and enough water to just cover the salt in a room to "absorb negative energy" out of the room at her office. It obviously "worked". So when I arrive home from university she asked me if salt water draws moisture out of the air — and explains to me what happened.
The scientist part of me is curious, because you cannot "absorb negative energy" using salt water. So I placed salt in a glass, covered it with enough water and left it overnight. To my surprise, the volume of water increased by around 20-25%, the glass is significantly more fill.
My question is this: why?
I have 2 thoughts:

Somehow the ions in the solution as the salt dissolved have charges that repel other molecules making more space in between molecules and therefore increasing the apparent volume.
Because it is a salt solution some form of osmosis occurs to actually absorb moisture out of the air.

My next idea is to redo this later today but weigh the glass and solution first to see if it gains mass. That would support 2. If no mass gain, then I assume something similar to 1 is happening.
Thoughts?
P. S. I am an applied mathematician so I have no background in the specifics of chemistry and how molecules and ions etc. etc. interact.
P. S. S.: I also couldn't find an answer to this on the internet. IN FACT salt added to water appears to reduce the volume in a lot of information I read.

"1) Somehow the ions in the solution as the salt dissolved have charges that repel other molecules making more space in between molecules and therefore increasing the apparent volume. " The salt solution as a whole is neutral so this won't hold exactly as written.

Next time cover the glass with a lid and see what happens.

@IvanNeretin So if it is indeed pulling moisture out the air, then it's very effective... well at least until it reaches equilibrium.

Looks like so. Go on, try it with the lid. Chemistry is an experimental science, after all.

Ken it is lowering of the water vapour pressure. You can indeed interpret it as a kind of osmosis when coupled to the water equilibrium between the salt solution and the atmosphere around.  In  order for it to be very effective, it much depends on the relative humidity of the air that day in your room/area. Search for colligative properties, but I think you know them already and it is just this specific aspect of them that surprised you.

@Alchimista What surprised me, is how much the volume of water rised by. The properties are indeed interesting, I didn't know there were so many as well. Per Ivan's suggestion, I covered another glass with salt and water and as of now 7 hours later, no change.. and this is why I LOVE science :)

Two phenomena where water is removed from the atmosphere are deliquescence and hygroscopy. // Osmosis is a phenomena associated with two liquid phases.

@MaxW did not intended to make a new classification. The interpretation might be the same.

The vapour tension over the saturated table salt solution { the best  with the solid salt excess) is equivalent to the relative humidity about 75 %.
If the rel. humidity is lower than 75%, the solution evaporates and the salt crystallizes.
If the rel. humidity is higher than 75%, the solution absorbs the air moisture and the salt is dissolving.
Such systems may be informally called "relative humidity buffers". They are useful for hygrometer calibration or for making a constant humidity aerial baths for long term weathering of devices or materials.
See the articles with tables of relative humidity above saturated solutions of various salts:

Researchgate.net
PDF on NIST
PDF on Omega.com

Screenshots of scanned table from NIST PDF link, part 1:

and part 2:

You are right, when salt (NaCl) is added to water, there is a small reduction in volume. This is true for a closed system which is not exposed to water vapor in the atmosphere. An increase in volume by certain chemicals is not that unusual. Sulfuric acid will increase its volume because it is hygroscopic. It absorbs moisture from the air. Salt is no exception. This is from a research paper: J. Phys. Chem. B 2010, 114, 2435–2449
"NaCl has routinely been used as a model system for sea-salt aerosols. Previous measurements on bulk crystalline NaCl showed that above 70% relative humidity (RH), 3.5-4.0 monolayers (ML) of water are adsorbed onto the surface. As the RH above the NaCl surface increases, enough water adsorbs to the surface to initiate a deliquescence transition from the crystalline to aqueous phase at the deliquescence relative
humidity (DRH)."
Deliquescence of NaCl is keyword, which will help you in searching more about the phenomenon. Since you already have "wet" crystals, the equilibrium is not fully achieved. There will be a time, when the volume will no longer increase (and of course evaporation is also occuring). If you wish to speed up your observations, keep the negative energy absorbing in the bathroom, where RH is very high during shower.

related question: When it rains, it puddles. Spilled salt cycles between wet and dry with humidity. Is this akin to a phase change?

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text:
MO diagram of BeH₂
I was learning about the MO diagrams of triatomic molecules and was left slightly confused about $\ce{BeH2}$.
A picture of the MO is shown below:

When looking at the $\mathrm{1s}$ combinations, I saw that each of the hydrogen $\mathrm{1s}$ combinations only had $1$ electron, which confused me.
I thought each hydrogen atom in the $\ce{H-H}$ pair contributed one electron, so I expected two electrons on each line for the $\ce{H-H}$ LGO (ligand group pair).
Why is there only $1$ electron on each line?

It looks like that particular MO was created with the two possibilities of hydrogen combinations: Either the two 1s electrons are same phase or different phase. Combining those options, alongside the AOs for $\ce{Be}$ produces the MO for $\ce{BeH2}$
This picture would have been clearer if they had opted to use the MO of dihydrogen, (bonding AND antibonding) which should have ended up at the same result. But it has been way to long since I linearly combined anything so I'll leave it as an exercise to reader.
edit: further comment - I am also not 100% sure about the reason to put one electron in each - but from a purely statistical point of view: 1 in each is the average situation.

If they had used the MO of dihydrogen, then the lower energy orbital formed by the in-phase overlap of both the s-orbitals would contain 2 electrons; instead of one electron in in-phase orbital and one electron in out-of-phase orbital, wouldn't it?

@ApoorvPotnis Yes, it would have made more intuitive sense. I don't think the picture is wrong per se, but unecessarily complicated. But where the electrons are in dihydrogen, is not important, it is the energy levels of the orbitals (electron-containing or not) that contribute to the orbital diagram.

It is an odd rendering. I'm not sure what the actual mechanism is to create $\ce{BeH2}$ but the simplistic reaction would be
$$\ce{Be + H2 <=> BeH2}$$
The MO for $\ce{H2}$, which is shown in the figure below is taken from Wikipedia.

The right side of the diagram you showed neither represents a hydrogen molecule, nor two independent (and hence equivalent) hydrogen atoms.

so the correct diagram is what you represented above? is that correct?

The image in my answer if for the molecular bonding of the hydrogen molecule. Obviously for the two hydrogen atoms to be on opposite sides of the boron atom the hydrogen molecule must break apart. That is the sticky point on the image in your question.


text:
Can you lower the alcohol concentration of a drink using vegetable oils?
The log POW of ethanol is −0.18 according to Wikipedia, i.e. in a two-phase system ethanol would preferrably accumulate in the aqueous phase, but not by much.
Suppose I wanted to lower the alcohol concentration in a drink, how well would adding some vegetable oil, mixing and removing the top layer with a straw work? What other parts of the drink are likely to accumulate in the oil? Which (unwanted) side effects can I expect?
Edit: Definition of Drinks
I'm interested in the effects on different types of drinks/alcoholic beverages from low-alcohol content beers to wines, whiskeys, vodkas and Stroh.
PS: While this may sound like a question for the cooking subsite, I am only interested in the chemistry.

It's somewhat broad, depending on drink. In general, it's not particularly good idea. You sure could extract alcohol from, say, whisky, but a little of fat would  stay in water phase  and not only ethanol would get extracted... Taste would suffer.

Fats rather decrease the effect of alcohol in vivo. :-)

What kind of drinks are we talking about, strong liquor, longdrinks, wine? Depending on the alcohol content, you will have quite a large solubility of the oil in the aqueous phase. This is going to taste real funny.

Much simpler method: Freeze your drink, pour out the remaining high alcohol content liquid, and then thaw the rest. Probably more often used to rescue the pure spirits from nasty, colourful alcopops. ;-)

Liquid-liquid extraction, repeated one or more times, is a common isolation/purification method. However in a "drink" you have to consider both ways that this works. Some of the oil will dissolve into the water phase and compounds other than ethanol would go into the oil phase. The result would probably be a very funny tasting drink.

@Karl I included the information you requested.

To separate a mixture of alcohol and water you normally would use distillation, not extraction. The taste will be spoiled anyway, so better not to add other substances

Liquid-liquid extraction is a widely used method, but it's normally used the other way round: You add e.g. ether to a solution of your valueable substance X in water, separate the phases, and evaporate the ether to get pure X. Add some more ether to the aqueous phase, etc, two-three times.
Now you can be quite sure that you have recovered all of your substance X, but also your aqueous phase will be saturated with ether. No problem if you'd discard it anyway, not so good for drinking. :-/
It's also bad if X, like ethanol, is well soluble in both phases. For hard liquor, the whole mixture is likely totally miscible, and for beer, you'll need a lot of oil to get any significant effect. And then what do you do with the ethanolic oil afterwards?


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How can one distinguish an IR spectroscopy alcohol band on an organic molecule versus the alcohol bands from water in the sample?
I'm reading this IR spectra and trying to determine if the large broad band at 3388 is alcohol from an intermediate or from water in the sample. The details of the reaction, conditions, and mechanism can be found below if it's helpful.

I attempted to run a double aldol condensation of p-anisaldehyde and acetone using 2 mL of 3 M NaOH, 0.3 mL of p-anisaldehyde, and 1.6 mL of 0.625 M acetone in ethanol. I let the reaction run for 30 minutes at room temperature and recrystallized the product.
I drew all of the steps I think the reaction mechanism could have stopped that have an alcohol here:

Thanks for any advice.

The main working rule of a good FTIR spectrum is that everything must be bone dry! The absorbance in IR is additive just like Beer's law, i.e. your spectra is the sum of water traces in your sample as well as your suspected alcohol. There is no simple way to deconvolute it. Dry your product thoroughly in a vacuum dessicator for at least a day and re-run the IR. If your big -OH band vanishes, it means your product was not dry. If it persists, your product likely has an hydroxide group.
KBr must be oven dried well before use as well.

Thanks for the advice! I really appreciate it.

Also check the melting point before getting puzzled by spectra of all kind.

@Alchimista Thanks for the tip. The melting point is about 1.5 deg C less than the final product of the double aldol condensation [Bis(4-methoxybenzylidene)acetone], but my guess is that's due to minor impurities or my partner wasn't paying careful attention. I suppose there's a chance some of the intermediates with alcohol were in the sample, but I'm not sure the likelihood of that.


text:
Adding apple cider vinegar to a solid DIY deodorant
If I wanted to add a bit of apple cider vinegar to my oil based deodorant, to increase antimicrobial properties, would this cause me to need a preservative? My deodorant is not water based, so I don't need a preservative in my formulation currently. I also already use lemon in my recipe.

Vinegar has a strong odor plus tons of water. Your DIY may smell like Pringles with a vinegar flavor.

I feel like smelling like strong vinegar is not better than slightly body odor.

Hope your users don't work in a biology lab. Fruit flies are drawn to vinegar which is characteristic of rotting fruit.

How we shall know if something antimicrobial than ask for preservatives? It seems a contradictions, though preservatives is whatever preserves. I would buy a commercial stick and cook something nice instead ;-)

Hahaha! Guys, I know how vinegar smells and it was just a thought to add a small amount into my formulation. I have a successful formulation now that people love, but my other formulation seems a bit less effective since I removed the baking soda from it due to some being very sensitive to the baking soda. I'm looking for something else to add in. Apple cider vinegar does smell pretty bad by itself. So I was going to play around. I don't want to have to start adding in preservatives though. Thank you for your answers so far though.


text:
Carbonic acid, bicarbonate in water
When you add CO2 into water, carbonic acid is created. From carbonic acid (H2CO3), HCO3 + H is formed. When CO2, and H2CO3 leave the water, does HCO3 leave as well, often measure as kH, carbonate hardness?

The chemistry is a bit more complicated than you have described. When $\ce{CO2}$ dissolves in water, most of the carbon species are dissolved $\ce{CO2}$. Carbonic acid, the bicarbonate anion and the carbonate anion are minor carbon species. (See the Wikipedia article - Carbonic Acid.)
You can remove the $\ce{CO2}$ by boiling the water, or by sparging (blowing bubbles of a gas containing no $\ce{CO2}$ in the water). But it is only $\ce{CO2}$ that will leave the solution. The equilibriums with carbonic acid, bicarbonate, and carbonate all shift of course as $\ce{CO2}$ is removed.

Thanks for the answer. Do you know how the equilibriums with carbonic acid, bicarbonate, and carbonate all shift of course as CO2 is removed?

Yes, see the Wikipedia article on Carbonic acid.


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Which of the following compounds is NOT an isomer?

Which of the following compunds is not an isomer? I cant figure out how the answer is d

Labas.  Not sure if this is the answer, yet one might argue a double bond in the chain (a, c, e) or a ring closure (b) were equivalent as one unsaturation, which is absent in d.  (In this perspective, benzene would formally yield four DBE, one by cyclicity and three by the three double bond drawn.)

The simple way would be to count to the number of each type of atom in each molecule. Alternatively, you could also note that only d has 1 degree of unsaturation, while all the rest have 2.

thanks, but what do you mean by count to the number? like determening S and R configurations?

To quote "For a compound with formula $\ce{C_aH_bN_cO_dX_e}$ where X is F, Cl, Br or I, the degree of unsaturation is given by: degree of unsaturation = 1/2 (2 + 2a - b + c - e).  Hence 2 for $\ce{C5H9NO}$.  source: https://scilearn.sydney.edu.au/fychemistry/calculators/degree_of_unsaturation.shtml

In this case, the simplest way is to find the molecular formula of each structure. Then, you'd see why (d) is the correct answer (isomers must have same molecular formula).


text:
Antisymmetrized Product of Strongly Orthogonal Geminal (APSG) method
To get a deeper understanding of electronic structure methods, like Hartree-Fock, MP2, and Coupled Cluster method, there are many small programs (codes) available in Python, C++ or Fortran, for example, Daniel Crawford's website, Joshua website
Does any one know any implementation of a simple Antisymmetrized Product of Strongly-orthogonal Geminals (APSG) method (Geminal Model Chemistry) for, say, two geminals (LiH, or H4, etc.) in Python or another programing language?
I will be grateful if someone can share this type of code (implementation) to get a better understanding of geminal based methods.

I am not sure of the difference between strongly orthogonal geminals (SOG) and APSG, but Q-Chem has an implementation of the former. If you have access to the former, then you could at least run some calculations, but I do not think you will be able to access the actual source code. See this page of the manual.

Possibly related, here is an implementation of the Antisymmetric Product of 1 reference-orbital Geminals (AP1roG).


text:
What is the least dense liquid under normal conditions?
What is the least dense liquid under normal conditions, room temperature, one atmosphere of pressure, doesn't combust upon contact with air, also wouldn't kill a human just for being in the same room as it?

Isopentane $\ce{C5H12}$ has the density of $0.6201~\mathrm{g\,cm^{-3}}$ at $20~\mathrm{^\circ C}$ [1, p. 3-330].
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.

That is... surprisingly dense. Water is known to be 'heavy' in common speech, but there aren't really any liquids which are a lot lighter apparently.

@orlp That's due to the constraint imposed by OP: the compound must be liquid at NTP conditions. Once the constraint removed, there are more interesting things such as solution of lithium metal in liquid ammonia or good old liquid hydrogen with the density  of $0.0709~\mathrm{g,cm^{-3}}$, but that's not what the question is about.

@andselisk The famous blue solution is significantly lighter than plain liquid ammonia?

According to

Hans Jaffe: Eine metallische Verbindung von Lithium mit Ammoniak. Elektrische Leitfähigkeit und galvanomagnetische Effekte. Z. Physik 93, 741–761 (1935) https://doi.org/10.1007/BF01337859

the saturated solution of lithium in liquid ammonia (approx. $\ce{4NH3 \cdot Li}$) has a boiling point above RT, and is significantly less dense (=0.48 g/ml) than isopentane (0.62), or plain ammonia (0.77):

... Mit einer Dichte von 0,48 ist die gesättigte Lösung von Li in
NH3 die leichteste bei Zimmertemperatur existierende Flüssigkeit. Die
feste Verbindung ...

I wasn´t able to get the full text, unfortunately. And yes, there is a possibility for it to ignite if not handled under an inert atmosphere. How dangerous I don´t know.

You could allow non-german speakers who read your answer look for a translation elsewhere (not that the more clever readers wont intuit the meaning), or .... "... With a density of 0.48, the saturated solution of Li in NH3 is the lightest liquid that exists at room temperature."

@BuckThorn Well, whoever doesn´t get it now finds the translation right underneath. :-)

What a wonderfully unexpected answer!

@NicolauSakerNeto All thanks to anselisks intriguing comment below!


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Molarity Numerical on Mixing Solutions

If 3 L of 1 M NaOH solution is prepared by mixing properties of stock solutions of $x$ L of 2.5 M NaOH and $y$ L of 0.4 M NaOH and no water is to be used, find $x : y$.

My attempt:
$$3 = 2.5x + 0.4y$$
Now I am stuck how to calculate $x:y$.

In addition to the equation for the preservation of the amount of substance you've written, there is one more equation you can use for the system, namely the one reflecting preservation of volumes: $x + y = 3$:
$$
\left\{
\begin{align}
x + y &= 3 \\
2.5x + 0.4y &= 3
\end{align}
\right.
$$
$$x + y = 2.5x + 0.4y$$
$$1.5x = 0.6y$$
$$\frac{x}{y} = \frac{6\cdot 2}{10\cdot 3} = \frac{2}{5}$$
E.g. the answer is $x : y = 1 : 2.5$.

As a quicker way, you could also use the Pearson's square to solve this:
1.0 M - 0.4 M = 0.6 M
2.5 M - 1.0 M = 1.5 M
The ratio is 0.6 to 1.5 (which corresponds to 1 to 2.5). Notice that we did not use the final volume in this calculation, the answer does not depend on it.


text:
What am I doing incorrectly with this IR/ Mass spec data?

Hello, I am trying to determine the chemical formula for the following compound.  From the IR spec, I see there is a carbonyl present.  Moving on to the mass spec, I see an M+ peak of 142, with what appears to be no Cl/Br present.
This brings me to a chemical formula of C10 H6 O (12 x 10 + 6 + 16 = 142), with a degree of unsaturation of 8, which seems off.
I feel I am doing something wrong and any help would be greatly appreciated, thank you!

Related cross post homework

Look at your proton NMR integrals.This restricts the possible formulae with respect to the number of hydrogen atoms.

@JakeL The molecular mass which you have calculated is correct, but the formula instead of $\ce{C10H6O}$, it should be $\ce{C9H18O}$ with a degree of unsaturation $1$ , which corresponds to your ketone. Now your proton integrals add up to $9$, but here there are $18$, so the integrals now can be scaled up as $4:8:6$. Now it has to be a straight chain compound, and symmetric also... little thinking can give you the answer....it's $\text{Nonan-5-one}$.

Ahh gotcha, makes sense now!  It's been awhile since I've had to interpret spectra and I completely forgot about integral scaling in the hnmr.  Thanks so much!

What I was trying to say was that firstly the molecular formula has to be determined correctly. From the mass spectrum of the given molecule, the molecular ion peak ($\ce{M+}$) was determined as $\frac{m}{z}=142$, which corresponds to the molecular formula $\ce{C9H18O}$. Note that, it can also correspond to $\ce{C10H6O}$, but in your $^1\mathrm H$ NMR spectra, a total integration of at least $9$ is there, thus the latter consideration is anyway not possible to be the correct formula.
After determining the chemical formula, it's good to look for index of hydrogen deficiency(i.e. degree of unsaturation) of the molecule. If the formula is $\ce{C9H18O}$, the degree of unsaturation is $1$. From the IR spectra also, we can see a sharp distinct $\ce{C=O}$ stretch near $1700\ \mathrm{cm^{-1}}$, and also no aldehydic proton peak near $9.5{-}10\ \mathrm{ppm}$, therefore we can conclude there is a keto group present in the molecule.
Now, we come to analyse the $^1\mathrm H$ NMR spectrum, which shows only total relative integral of $9$, but remember the integrals in proton NMR are always relative, they are never absolute. So, you can scale them up or down relative to your choice. Here to match the total protons in our system, we have to scale it up by a factor of $2$. Therefore the integrals now correspond to $4,8$ and $6$ protons respectively, and also by this scaling we can also say that the molecule is symmetric in nature.
Now peak at $\delta_\mathrm H=2.5\ \mathrm{ppm}$ is splitted into a triplet and these correspond to the protons attached to your carbonyl, resulting in the most downfield shift, near $\delta_\mathrm H =1.5\ \mathrm{ppm}$, two peaks merge and therefore they are actually the methylene protons, and lastly near $\delta_\mathrm H\approx0.9\ \mathrm{ppm}$, the triplets correspond to methyl proton peaks and thus we arrive at the final most likely possible structure as,

Analysis of the splitting pattern to find out the $^3J$ values, I leave it to you...

I disagree vehemently with "that firstly the molecular formula has to be determined correctly." You don't jump to conclusions with sort of a problem you need to gather evidence, then decide how that evidence fits into the puzzle.

@MaxW Without the molecular formula in your hand, how do you proceed further ? How can you decide then whether there are one keto, or two keto or one ester etc. ?

What I am trying to say is that you don't just look at the mass spec and try to work out the formula first before doing anything else. The hydrogen ratios in the NRM are one of the key clues. Also as I said it is an iterative process. There isn't any nice flowchart for this sort of problem.

I think we're really having a violent agreement. I totally agree that figuring out the empirical/molecular formula is a very important step. I just disagree that it is the first thing to do.

A more by the numbers approach is this. Gather some evidence before speculating about the molecular formula.
Do a first pass to see what "jumps out" of each of the spectra.
From the IR spectra:

The bands below 3000 are C-H indicting that this is a totally unsaturated hydrocarbon.
The strong peak at 1700 indicates that there is a carbonyl group in the compound. So this is either an aldehyde or ketone.

From the Mass spectra:

The $m/z$ peak is at 142 and pretty clean, no other major peaks around it. So this is a "small" molecule.

From the NMR spectra:

There is no aldehyde proton at 9-10, so the molecule is a ketone, not an aldehyde.
There are three sets of peaks with integrals 2:4:3. So there must be 9, 18, 27, ... hydrogen atoms.
The fact that all the peaks are at low values indicates that there is no hydrogen bonded to a double or triple bonded carbon.

Now guess about a molecular formula
Now there is enough evidence to speculate an alkane with a ketone group.
Guess a formula, see how the pieces fit with a further analysis of IR, NMR and mass spec data.
It's an iterative process...


text:
Does dilithium exist?
Lithium has two electrons in $\mathrm{1s}$, filling the first energy level; and one valence electron in $\mathrm{2s}$. Could two $\ce{Li}$ atoms form a bond to create $\ce{Li2}$; like $\ce{O2}$ or $\ce{C2}$?

In an article about relativistic quantum chemistry successfully predicting low mercury melting point, there is mentioned that mercury is one of few elements not forming diatomic molecules in gaseous phase, calling it pseudo noble gas.

The dilithium dimer exists- but it isn’t particularly common, and we usually only see it in the gas phase.
Essentially, the 2s electrons interact and form a bonding orbital. (s-p mixing is particularly pronounced here, but I presume you don’t know MO Theory.)
Due to the diffuse nature of lithium’s atomic orbitals, the bonding is pretty weak (Wikipedia gives a binding energy of $102\text{ kJ/mol}$, about one fourth that for a hydrogen molecule); and the inter nuclear distance between the 2 lithium atoms is actually quite large.


text:
Why does chewing gum get hardened after drinking water?
Why does chewing gum get hardened after drinking water? This is a basic question, but please don't say temperature change, because I've tried that, and it never gets back to same elasticity. There must be some other chemistry at work. Even with lukewarm water or room temperature water it gets substantially harder.

Related: Why does gum get harder to chew after a while? Personally, I never noticed the gum altering its hardness even when I was a kid and consumed all kinds of garbage drinks while chewing the gum. I suspect it might be a subjective matter, or an effect from a particular brand only. It would be more helpful if you could back up your observations by a statistically significant number of similar precedents described in a literature.

Just for statistic, I have noticed the same. Consistent hardening of the chewing gum after drinking.  Just not sure it is brand independent, and even flavour independent :) but I was mostly chewing mint ones :)

@andselisk that's too much to ask from an software engineer, but yeah, I'll try to collect some data on it from colleagues, but that won't be random sampling, so have to deal with it. And what would be the benefit of such exercise BTW? Also it seems not related as the affect discussed in question in about hardening after chewing a while, i am specifically seeking for hardening after drinking water along gum in mouth.

Chewing gum can be made from quite a number of base materials, PIB, PVA, polyisoprene, mastic (arabic gum), Manilkara chicle, and many others (https://en.wikipedia.org/wiki/Chewing_gum has a list). I'll guess the effect (which I have also observed i think) will depend a lot on what you actually have.

@BhanuChhabra I'm not asking you to social poll. I suggested that since it might be a subjective matter, adding some data from third-party researchers would be helpful.

As you are doing experiments, try drinking high ionic strength liquids (e.g. salt water) and non-ionic high osmolarity liquids (sugar water). No need to swallow (in fact, no need to keep the gum in your mouth).

Well, i thought of doing so over weekend, but it seems impractical for me to gather un-biased data and opinion as the immediate access-group I've, are not so scientifically (and statistically) understanding.
and going by the data points I need(sample size, type of gums, type of liquids), it won't be feasible for me to gather the amount of data to draw a conclusion.

@andselisk let me search something of that nature, some research etc.


text:
Does the amount of solute affect rate of solvation?
I know that in order to increase the rate of solvation, you need to increase how often solute and solvent particles come into contact.
You can do it by:

agitating the mixture
increasing the surface area of the solute
heating up the solvent

But my question is: Does adding more solute (like adding more salt to water) increase the rate of solvation since the contact surface area is larger?
Edit: The original question in the book describes dissolving a certain amount of $\ce{NaCl}$ into water and asks, “What can be done to increase the rate of dissolution without affecting solubility?” I assume the rate is measured by how fast the mass of $\ce{NaCl}$ is consumed, which is $-\frac{dW}{dt}$.

Could you write down the chemical equation? Are we going from pure solid substance to completely dissolved substance? If so, point 2 is increasing the surface area of the solid substance, right? And how do you measure rate? Is it how fast the concentration of solute particles increases? I think technically, the process is called dissolution, not solvation. Solvation is more general, like when an acid dissociates and the products are now surrounded by more solvent molecules, or when a cation was bound to a protein and now is surrounded by bulk water.

Just made an edit.

If the answers are to be exclusive, then obvious choice is agitating the mixture.
If the answers are not to be exclusive, then agitating the mixture + increased surface by adding more solute increases $\frac{\mathrm{d}c}{\mathrm{d}t}$
Adding more solute does not increase rate of solvation,  but as the absolute surface increases, it increases $\frac{\mathrm{d}c}{\mathrm{d}t}$.
Note also without affecting solubility from your comment, so increasing temperature is out.

Thanks! I was curious because the question in my homework was like “What can you do to increase rate of solvation and not affect the solubility?”, and the option “Adding more solute” wasn’t one of the answers, so I went asking around.

@Alex I think Poutnik didn't get the point of the Q. Increasing amount of solute does not speed up the process.

@Alchimista Increasing of amount does not speed up diisolution. Increasing of surface does. Even if achieved by increasing the amount.Best if combined with agitating of mixture. I agree if solute sits passively on the bottom, the effective surface may stay about the same.

See also the answer update.

The original answer was wrong. Op ask if adding more solute speed up dissolution and answer is " it does. ...do you have any reason why it may not". The correct answer is no, from whatever point of views. The answer you gave is that to an hypothetical multiple choice exercise. Still the opening of your A is wrong. I do no bother, but there are major misunderstanding (while in your side it is likely a misunderstandng of the Q) that can convince the OP of something wrong (as OP accepted too quickly your A).

Now is somehow correct. If one focus on the concentration in solution it will go up faster. But it is not the solubilisation that is faster!  For me is fine as far as OP read my answers too.

But my question is: Does adding more solute (like adding more salt to water) increase the rate of solvation since the contact surface area is larger?

Not really.  If I get the point, you would like to know if x + y grams of a substance dissolve faster (as dn/dt) than an amount x. No. Ignoring activity of the solute, or in sufficiently diluted solution, the speed of solubilisation per surface stays the same, or decrease as activity increase if you look for an overall picture.
Thus, as per amount of substance the kinetic of solubilisation just profit of an effective augmented surface, ie the ratio between surface and mass of the solute (point 2 in OP list).   Smaller grains dissolve faster, and this is the basis for precipitate digestion to get bigger crystals.
This answer assume that the meaning of solvation is that of solubilisation, and point to a relative speed rather than the absolute time required to dissolve a substance. The latter goes up for both principle and, I would say so,  practical reasons.
Edit after reading an OP comment. The answer to the book Q "how to increase the rate of solvation without affecting solubility" are right stirring or finely grinding the substance to be dissolved, for instance.


text:
When does Hartree-Fock fundamentally fail?
The Hartree-Fock approximation (HFA) works by assuming each electron sees the effective electric field from all the other electrons as some self-consistent field.
The HFA is known to give pretty decent results, but often gets the last quantitative bit of accuracy wrong. This last bit is attributed to mysterious many-body or "strongly-correlated" effects.
There are many examples of HFA quantitatively giving wrong answers (energies are wrong, incorrect numerical results,  etc.). But, are there general phenomena that HFA gets totally wrong? I'm thinking of something where the chemistry/physics is fundamentally different from what HFA would tell you. Said differently, when do "strong correlations" actually matter and fundamentally change things in a chemical context?
I've tried reading up on this topic in the physics literature, and it is incomprehensible despite being an ex-physicist myself. Most of the discussion is about high temperature superconductors and complicated transition metal oxides. I can't understand the basic phenomena without all the specific details.

I suppose there is meant strong correlation of wave functions of some  electrons, while about independence may be expected..But I am a dumb chemist with settled dust on lectures of quantum chemistry.

Can't find a reference at the moment but in principle you can never have a metal at the Hartree-Fock level of theory - the unscreened exchange always results in a band gap, though  admittedly sometimes so small sometimes as to not be detectable in realistic calculations. Also please be careful to distinguish the inadequacies of HF with problems due to the incompleteness of the basis set being used.

Dispersion is another aspect

Funny you should ask that the same time my old question got some activity (Why do post-Hartree-Fock methods fail to predict the direction of the dipole moment of carbon monoxide?) Takeaway from this is the dipole moment of CO is completely wrong. In general I would not consider HF giving decent results except for molecular structures (for molecules). The neglect of most of the correlation energy in HF produces errors way beyond chemical accuracy (5 kJ/mol). (Quantitative accuracy is very hard to achieve in any case.)

Your wording seems weird. HF does not consider any electron correlation at all, by definition of the term "electron correlation energy". In your question, I would remove the "strong" and just talk about correlation in general.

@feodoran HF accounts for exchange correlation exactly. Admittedly the correlation energy is defined differently, but saying that it doesn't account for correlation at all is not quite correct. I agree that strong correlation is a different thing.

@Martin-マーチン If you consider exchange to be correlation ... yes. But then the terminology of having separate exchange and correlation functionals in DFT does not seem very consistent.

A related, previous answer of mine: https://chemistry.stackexchange.com/questions/108061/correlation-energy-in-quantum-chemistry/108911#108911

Restricted HF (RHF) fails anytime you have partial occupancy due to (near) degeneracy of the HOMO. For example, $\ce{H2}$ dissociates to a mix of $\ce{2H}$ and $\ce{H- + H+}$ and the dissociation energy is way off. Similarly, singlet $\ce{O2}$ is not well described by RHF if you can get it to converge at all.
The solution is to use more than one MO wavefunction (determinant) and this is sometimes referred to as "non-dynamical" (rather than "strong") correlation.
Unrestricted HF (UHF) can usually treat these systems qualitatively correct.

If I understand correctly, are you saying that RHF fails because the basis it uses is too limited? What about gross qualitative failures of unrestricted HF?

If by basis you mean basis set, then no. RHF will fail no matter how big the basis set is. The problem is that RHF requires double occupancy of all MOs. That's no good when the correct answer is single or partial occupancy. You need more than one determinant.  I don't know of any gross qualitative failures for UHF

So does UHF treat correlations exactly? Or are there still approximations being made there?

The correlation energy is defined as the difference between the exact solution and the HF solution in the complete basis set limit. So there is no correlation in HF methods. They are mean field methods. Some refer to the exchange energy as "spin correlation" but then you are using the term "correlation" differently from the original definition.

OK, so it sounds like my question has yet to be answered then.

There are also examples of qualitative failures in unrestricted HF (UHF). For example, UHF produces an unbound potential for the dissociation of $\mathrm{F}_2$ and predicts a square geometry for cyclobutadiene.

HFA really struggles when the correlation energy is too great. I dont know every example where this is true but I do know when you have a heavy metal it is. The increased mass of the metal imposes a relativistic effect as the inner electrons have much more energy (because they move much faster) and thus the correlation energy is also increased dramatically. This in turn makes HFA fundamentally wrong for a distincy physical reason.


text:
Regarding Standard Electrode Potentials
If the standard hydrogen electrode (SHE) is set arbitrarily to have a potential that is equal to 0 V, and has an absolute real potential. Then, in electrochemistry, how are the electrode potentials of other electrodes are determined using it?
Our high school textbook states that we form a galvanic cell from the SHE and another half cell of the electrode that we want to determine its potential. Then using the voltmeter, the voltage of the cell should resemble the standard electrode potential of the electrode.
I really don’t get it, because the voltage should resemble the difference in potential between the electrodes.

True, the voltage is the difference; what's unclear about that?

Then the voltage that the voltmeter will read is the difference between the potential of the electrode and the “real” potential of (SHE) .

That's right.$;$

So how are the electrode potentials determined?

Exactly as you described.

We are not determining the "absolute real" potentials of the other half cells, either. It's just like elevation about sea level. Mount Everest (सगरमाथा or ཇོ་མོ་གླང་མ) is not 8,848 m tall in absolute terms (not sure what that would be, from the center of the earth or how high you have to climb from the closest bus stop), it is 8,848 m higher than the sea level. As sea levels are rising, Mount Everest will shrink.

I think I wrote a relevant answer for a slightly different question: https://chemistry.stackexchange.com/questions/112316/reduction-potentials/112339#112339

You can only measure the voltage of a cell. $\ce{V_{cell} = V_{cathode} − V_{anode}}$. The SHE arbitrarily has  $\ce{V_{anode} = 0}$. So all other half cells are measured relative to it. // LOL - Remember Einstein "everything is relative..."


text:
Use a solid or liquid stationary phase in gas chromatography
In my textbook, they mention that the stationary phase used in gas chromatography is a solid, such as silica or an adsorbed high-boiling-point liquid.
My question is that what is the advantage of each one, in what situation one will be better than the other, or if it matters at all.
Also, how does the stationary phase not get carried away with the mobile phase? since a high-pressure gas is used doesn't the stationary phase just get carried along with the mobile phase, especially if it is liquid?

Solid adsorbents in gas chromatography were used about 30-40 years ago. They are obsolete now. The efficiency of solids (or packed column GC) was much lower than coated columns. Most modern GC stationary phases are coated or sometimes chemically bonded  to the silica surface inside the capillary.

how does the stationary phase not get carried away with the mobile
  phase? since a high-pressure gas is used doesn't the stationary phase
  just get carried along with the mobile phase, especially if it is
  liquid?

The viscosity of the coating is so high that it does not get carried away. Imagine a very viscous wax like substance stuck on a glassy surface. Smear it very thinly. The coating is now a few microns thick. Try blowing away thick waxy substance from a glass surface, can you ever get rid of it?
You may think that the pressure is high, but recall that Pressure= Force/Area. If the capillary area (i.e. cross section) is so small, the force is very small too. Capillaries diameters are on the order of microns. That is why you can use a silica capillary in GC. If we had used a wide bore capillary say 4 mm i.d., and used the same pressure with the same wall thickness, it would have shattered.

Solid phase may be useful in preparative chromatography.


text:
Temperature dependency of enthalpy H and entropy S
In textbooks, I often read that the temperature dependency of the enthalpy of reaction $\Delta_\mathrm{r}H$ and the entropy of reaction $\Delta_\mathrm{r}S$ can be neglected (in a limited $T$-range).
But why is this valid? What is the intention behind this?

Can you please provide more context?  Is this in connection with a chemical reaction or a phase change, for example?

In connection with a chemical reaction. Sorry, I am very new to this field.

The standard entropy change and the standard heat of reaction are functions of temperature, and this dependence can be expressed in terms of the heat capacities of the reactants and products.  However, over limited temperature ranges, these effects are typically small compared to the enthalpy and entropy changes at a specific temperature.

And what is about non-standard values?

Standard refers to reaction at 1 bar pressure.

Maybe you could add a specific reaction where the temperature dependence of the change in Gibbs free energy (and of the equilibrium constant) is discussed to make the question more specific. The ammonia synthesis might be a good one (I just calculated the equilibrium constant between 100 K and 500 K using those assumptions - not sure how much difference it would have made including the temperature dependence of enthalpy and entropy).

Three step process to evaluate temperature dependence
There is a change in enthalpy and in entropy when changing the temperature of a sample. It follows that the enthalpy of reaction and the entropy of reaction can also change when changing the temperature. The easiest way to get values for a temperature $T_2$ if you have values for a temperature $T_1$ is to do a thought experiment with three steps, starting with reactants at the temperature $T_2$:

Change temperature of reaction mixture from $T_2$ to $T_1$
Let the reaction proceed at a temperature of $T_1$
Change temperature of product mixture from $T_1$ to $T_2$

The net process is the reaction at the temperature $T_2$, where you don't know the entropy and enthalpy change yet.
Changes depend on differences in heat capacity
If we are working at constant pressure, the enthalpy change is (with $C_P$ the heat capacity of the reaction mixture):
$$\Delta H = C_P \Delta T$$ and the entropy change is:
$$\Delta S = C_P \ln \frac{T_\text{final}}{T_\text{initial}}.$$
Because steps 1. and 3. change the temperature in opposite directions, a lot cancels out. If we define  $\Delta C_P$ as the difference in heat capacity between reactant and product mixture, we get:
$$\Delta(\Delta_r H) = \Delta C_P \Delta T$$
$$\Delta(\Delta_r S) = \Delta C_P \ln \frac{T_2}{T_1}$$
Compared to other changes with temperature (e.g. Gibbs energy or equilibrium constant), these changes tend to be modest. To give an idea of the magnitude, the molar heat capacity of a diatomic ideal gas is $c_P = 3.5 R$, or about 30 J / (mol K). Given that values for $\Delta_r H$ are typically given in kJ/mol, the change in enthalpy when changing the temperature is relatively small. For the entropy, the "correction term" is not proportional to $\Delta T$ but to the logarithm of the ratio of temperatures. If you start at room temperature and increase the temperature by ten Kelvin, the value of the logarithm will be about 0.03, so the effect is also relatively small.


text:
Can neutralization ever be complete?
Suppose we have 1 eq of acetic acid and we want neutralize it using sodium hydroxide. After adding 1 eq of sodium hydroxide 1 eq of sodium acetate should form and neutralization should be complete but because of hydrolysis little bit of acetic acid will form again so we would have to add more sodium hydroxide and this process would go on forever so how will neutralization ever be complete?
My guess is that after adding 1 eq of NaOH the conc. of acetic acid becomes so low that it is neglected and the neutralization is assumed to be over.

So, is the solution of sodium acetate incompletely neutralised sodium hydroxide or incompletely neutralised acetic acid ? Neutralisation involving weak acids/basis does not involve completeness but it does involve equilibrium.

@Poutnik ,@Sanom Dane  I think he trying to tell , when dissociated h+ is consumed , more h+ ions are produces according to me chatliers principle.

I wanted to say, neutralisation is over, if the solution is equivalent to solution of the equivalent amount of sodium acetate.

@Poutnik What I wanted to say is that as equivalent amount of sodium acetate will never be formed(exactly) neutralization will never be complete.

I have said solution equivalent to sodium acetate solution, and that is achieved at a single point during neutralisation. At that point, there is equilibrium between acetate, acetic acid and hydroxide ion, no matter, if you achieved that by neutralisation of the acetic acid, or by dissolving of sodium acetate.

Ohk, so you are saying that neutralisation is complete when the conc. acid at equilibrium is same as that if only sodium acetate was dissolved.

Yes. Acidobasic reactions involving weak acids or bases ( their salts included) are always about equilibrium.

Also, the concentration of hydroxide ion at equivalence is the same as if sodium acetate were dissolved. So temptation to use extra hydroxide should be the same as to use the extra acid to get rid of the hydroxide. $$\ce{A- +H2O <=> HA + OH-}$$

Nice point !!!! So that makes neutralization a unique point atleast.

The $\ce{K_a}$ expression for acetic acid, in aqueous solution, is given by:
$\ce{K_a} =\dfrac{\ce{[CH3COO-][H+]}}{\ce{[CH3COOH]}}$
The expression can be rearranged to give
$\ce{[CH3COOH]} = \dfrac{\ce{[CH3COO-][H+]}}{\ce{K_a}}$
so no matter what the pH, all the acetic acid will never be reacted.
So in a titration let's assume that we start out with 0.1000 moles of whatever acid $\ce{[HA]_{initial}}$.

After base has added and $0.1000 \gg \ce{[HA]_{final}}$ and thus $0.0001 \gt \ce{[HA]_{final}}$, then the reaction is  quantitatively complete.
This is strictly a significant figures argument. You can't do a titration to 10 significant figures. 3 is decent, if you get 4 you're a magician.
When 0.1000 moles of base have been added to the solution the equivalent point has been reached.
The key point is that the moles of base is equal to moles of acid, regardless of the pH.
When ever you get the "signal" that the titration is over you have reached the end point.
The signal could be, for example, a color change of an indicator, the swing of the needle on a pH meter, or the formation of a precipitate.

Thanks, will there ever be a case where [Salt]>>[Acid] is unachievable (maybe when initial conc. of acid is comparable to $K_h$)

I don't understand what you mean by $\ce{[salt] \gg [acid]}$. Could you explain? example?

In your example if 0.1 mol of acid was taken then neutralization would be complete when [HA]<<0.1, also [BA(salt)]≈0.1? please correct me if I'm wrong, Thanks.

Not just 0.1 (1 significant figure), 0.1000 (4 significant figures). Let me clarify my answer about HA.


text:
α-Halo ketone and strong base: Favorskii vs SN2
Let us consider an α-halo ketone, with at least one α-H on the α carbon not containing the halogen. If I add a strong base, what is likely to happen: $\mathrm{S_N2}$ or Favorskii rearrangement?
I have learnt that α-halo carbonyls undergo excellent $\mathrm{S_N2}$ due to the lowered energy of its LUMO. And the Favorskii is a characteristic of the α-halo carbonyls. How to decide between the two?

By SN2 do you mean intermolecular attack of the enolate of the halo ketone on another molecule of halo ketone?

I meant the added "strong base" ,attacking the halogen atom.

What @jbussing said - are you talking about cyclic ketones or linear ones? Could you edit your question to include a specific example, and show the products of the two competing reactions you envision? Otherwise, the question might be closed because it is too broad and you weren't specific enough about your thoughts concerning the question.

I would comment however I am new and have yet to unlock that feature.
Do you have an exact substrate of interest? I have studied the Favorskii rearrangement a tiny bit in a graduate mechanisms course and if your seeking a very general answer to a general question; The favorskii seems to be favorable when your substrate is containing rings that are conformationally rigid where our alpha-halo-ketone moiety is found.
Check out Eaton's 1964 synthesis of Cubane to find two really elegant Favorskii reactions in the synthesis of Cubane. Notice for almost all of our examples of favorskii reactivity we notice a common pattern in the substrate scope?
Rings! My theory is conformational rigidity of a ring is partially responsible and we can confidently point our fingers to Favorskii where cyclopropanone intermediates are possible AND energetically favorable (conformation argument).
If our substrate scope is including non-ring alpha-halo-ketones.. We may want to consider alternatives like SN2 to predict reactivity.

Yes!! I have heard of the cubane synthesis via favorskii!

AS @Karsten Theis pointed out that this question is too broad. Yet, if you are willing to study, here is a review article (Ref.1), which included more than 250 references to go with:

Abstract: The skeletal rearrangement of α-halogeno-ketones, which is known as the Favorskii rearrangement, is met most frequently in aliphatic monocyclic, and polycyclic halogenated ketones. This molecular rearrangement is used in the synthesis of branched carboxylic acids and cis-αβ-unsaturated acids and for obtaining smaller rings in alicyclic and to a less extent heterocyclic compounds. The sterochemistry and the mechanism of the Favorskii rearrangement are also considered, these being of great interest for theoretical organic chemistry. A list of 261 references is included.

It is clear that the reaction reaction undergoes Favorskii rearrangement preferably when the substrate is cyclic. The review also discuss on the influences of $\alpha$-$\ce{C}$ on the rearrangement. Recent publication (Ref.2) has shown use of halocycloalkenones as Diels−Alder dienophiles and resulting adduct undergoes  Favorskii rearrangement (if you intereted to see). The graphical abstract is shown below:

References:

A. A. Akhrem, T. K. Ustynyuk, Yu A. Titov, “The Favorskii Rearrangement,” Russian Chemical Reviews 1970, 39(9), 732–746.
A. G. Ross, S. D. Townsend, S. J. Danishefsky, “Halocycloalkenones as Diels–Alder Dienophiles. Applications to Generating Useful Structural Patterns,” J. Org. Chem. 2013, 78(1), 204–210 (DOI: 10.1021/jo302230m).


text:
Cleaning a stubborn Electrospray residue
I have an electrospray ionization capillary tip that sprays into a chamber that contains some circuit boards. An aqueous CsI sample is apparently causing a white residue to build up on these boards after some time. I thought the residue was the salt itself, but I can't dissolve it with water. I tried methanol and that didn't work either. Does anyone know what this could be and how to clean it? I'm worried about excess build up over time.
I considered that it might be CsOH, but that should also rinse away with water because its so soluble.

Most likely impurities in the water, although some soft glasses can dissolve a bit in water (cactus,maize and some sponges reinforce their structure from dissolved silica). How pure is the water you use? Are any cleaning material residues left in the apparatus?

@DrMoishePippik - The water used is millipore water. However, the syringe has contained R6G before. it was rinsed and sonicated with MeOH thoroughly.

Millipore have various devices. What is the level of dissolved solids??

@DrMoishePippik - This is 18 Mohm millipore and we are using mM levels of CsI. When we use something other than CsI we don't get this build-up, so I'm pretty sure it's related to the CsI somehow. Also some relevant information that I omitted before by mistake: these boards are gold-plated, and as I understand it there may be an alloy forming? As for cleaning materials, I have only used water and MeOH, with some rubbing with a cotton cleaning tip.

Then I'd look to possible contamination in the CsI. Put a few drops of the plain conductivity water and a few drops of CsI solution on clean microscope slides and evaporate in a desiccator, then wash the slides with water and MeOH as you normally do. There should be no resdue.


text:
How can normalities be interchangeably used?
So I was doing the following question on stoichiometry . The part $\pu{1.80N}\ \ce{HCl}$ can also be written as $\pu{0.9 N}\ \ce{   H_2SO_4}$  but I don't understand why .
$\pu{10 g}$ sample of 'gas liquor' ($\ce{NH}^+_4$ salt) is boiled with $\ce{NaOH}$ and the resulting $\ce{NH3}$ is passed into $\pu{60 ml}$        of $\pu{1.8 N}\ \ce{HCl}$ . Excess $\ce{H_2SO4}$ required $\pu{10cm^3}$ of $\pu{0.40 N}\ \ce{NaOH}$. What is the % of $\ce{NH3}$ in gas liquor?
I got the answer but just can't understand why the two can be used interchangeably.

Lookup the chemical definition of normality.

No. Of gram equivalents/ volume of solution in litres

Dah... I have read the question poorly again. // You are right, there is a mistake. There is half as much acid in 0.90 N $\ce{H2SO4}$ as there is in 1.80 N $\ce{HCl}$. // There is as much acid in 0.9 molar $\ce{H2SO4}$ as there is in 1.80 molar $\ce{HCl}$.

Yeah that's what i was thinking. Its N=M$\times$ n factore ==> 1.8 =0.9 $\times 2$.

@gucci" "Its N=M× n factor ==> 1.8 =0.9 ×2"- You will never learn this gram equivalent concept if you memorize this relation. It does not work all the time, and fails for most redox reagents.

The fundamental idea behind gram equivalents and normality is that one gram equivalent of an acid reacts with one gram equivalent of a base. This idea eliminates the need of using molarity and thinking about mole ratios for any acid base titration.
Equal volume 1 N HCl will react with 1 N NaOH. Similarly, 1 liter of 1 N $\ce{H2SO4}$  or 1 N $\ce{H3PO4}$ will consume 1 L of 1 N NaOH.
The key calculations is that you should know how to calculate normality from a given molecular weight and a balanced equation:
For example:
$\ce{HCl}$ -> $\ce{H+}$ + $\ce{Cl-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 36/1 =36
$\ce{H2SO4}$ -> $\ce{2H+}$ + $\ce{SO4^2-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 98/2 = 49
Can you now make the connection that why "The part 1.80 N HCl can also be written as 0.9 N H2SO4" is quite wrong? because
1 N HCl = 1 N $\ce{H2SO4}$
and both solutions contain 1 gram equivalents of protons.
Yes if your teacher were comparing normality and molarity, then it is different.
Molarity calculations are different and far more easier. By definition, M of X= moles of A/ Total volume in which X is present.
A 1 M HCl just means there is one mole HCl (gas) dissolved in 1 L of solution
Similarly,
1 M $\ce{H2SO4}$ is 1 mole $\ce{H2SO4}$ dissolved in 1 L of solution.
If you were to titrate  1 L of 1 M NaOH with 1 M HCl you would require 1 L of HCl. However, since  sulfuric acid furnishes two moles of proton $\ce{H+}$ for each mole of sulfuric acid, only 500 mL would be required.
In short, 1 M HCl is equivalent to 0.5 M $\ce{H2SO4}$ in terms of titer for $\ce{OH-}$ ion.
Contrast this molarity unit with normality which is almost obsolete except in some South Asian colleges.

Ok.....but i am eager to know as to what would happen if we compared molarity and normality @M.Farooq

H2SO4: 1 M = 2 N // HCl: 1 M = 1 N

The student needs to learn the concept of normalities, otherwise they will have no clue why 1 M H2SO4 = 2 N HCl. They may apply the same theme to H3BO3 as well.


text:
Why is it aqueous and ethanolic conditions for nucleophillc substitution of cyanide ion with benzyl chloride?
I was doing an exam question where benzyl chloride was undergoing a nucleophilic substitution with cyanide ion to form benzyl cyanide.
Please can someone enlighten me as I have never seen both conditions before just one or the other. I have Googled it but no sites talk about the conditions of nucleophilic substitution with a cyanide ion in a benzene context. Does the benzene alter the conditions?

@orthocresol: Was benzene lost in the editing? Unclear where it comes into play.

@user55119 I’m pretty sure “benzene” here refers to the electrophile being benzylic.

Since $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is 9.21 (Wikipedia), the cyanide ion in water is pretty alkaline and contains significant amounts of hydroxide ions, according to the following equilibrium:
$$\ce{NC- + H2O <=> OH- + HCN}$$
That's why you smell strong $\ce{HCN}$ in aqueous  $\ce{KCN}$ or $\ce{NaCN}$ solution.  Thus, when you perform the reaction of $\ce{KCN}$ or $\ce{NaCN}$ with the haloalkane (including allyl and benzyl) in water you tend to get the substitution by $\ce{^-OH}$  instead of $\ce{^-CN}$. To get major product as cyanoalkane, you may need to do the reaction in absolute alcohol.
Not relevant, but note worthy that $\ce{^-CN}$ is an ambident nucleophile, meaning a nucleophile which can attack acitve site of a molecule through more than one sites (here either by $\ce{N}$ or $\ce{C}$). Thus, a reaction of alkyl halide ($\ce{R-X}$) with $\ce{KCN}$ gives alkyl nitrile ($\ce{R-CN}$) while a reaction of alkyl halide with $\ce{AgCN}$ gives an alkyl isonitrile ($\ce{R-NC}$).

Not that I am willing to find out, but does KCN in ethanol smell of HCN? Certainly, the pKa's of water and ethanol are similar. Although the solubility of KCN in ethanol is less than water, I suspect that the greater solubility of benzyl chloride in ethanol is the determining factor. If benzyl alcohol is formed in aqueous KCN, might a solvolysis (SN1) be competing? Food for thought.

Followup: Organic Synthesis: Benzyl chloride in ethanol added to aqueous NaCN. 80-90% yield distilled of phenylacetonitrile on a kilogram scale. http://www.orgsyn.org/demo.aspx?prep=CV1P0107

@user55119: I have seen this reference before I answered. Yet, I believe my argument has a sense. During the large scale synthesis, they have used about 35% water in ethanol and concentration of the solute is about $\pu{5 M}$. Thus,  $\ce{^-CN}$ ion predominate over hydroxide and ethoxide. That's why in these syntheses nowadays, phase transfer catalyst is popular. One more point about water Vs ethanol: If you use 50% aq. ethanol for solvolysis of 1-chloro-1-phenylethane, you always get 1-phenylethanol (no ethoxy product).

The cyanide ion is a very good nucleophile and reacts readily with benzyl chloride.
The problem is that the cyanide salts are soluble in water, while the benzyl chloride is not.
Therefore, ethanol is added so that both reagents are slightly soluble in the medium and the reaction proceeds.
An alternative to this system is to use phase transfer catalysts like tetrabutylammonium salts.
This cation can then transport the cyanide to  the organic phase, for instance toluene, in which the benzyl cyanide is dissolved, and the reaction occurs there.

More broadly speaking do reactions involving haloalkanes  or halogen functional groups require alcoohlic conditions as the h-x bond is polar?

No, it is not required. Apolar solvents are fine. In this case it is just a point of solubility. Note that the most polar reagent is the cyanide anion (and it's counter ion). The reaction itself is an Sn2, so not much polar stabilization is required. What is needed is to join the two reagents, either by choosing a solvent in which both are soluble or getting the cyanide in the solvent with a phase transfer catalyst


text:
Attack on cyclic intermediate during NGP
I have studied that when an epoxy ring opening takes place in acidic medium, the ring opens in a way that the most stable carbocation forms. This is due to the positive charge on the oxygen, which gives partial positive character to the carbon atoms.
A very similar intermediate is formed in NGP (neighboring group participation) reactions with sulphur as the neighboring atom (source). How is the attack carried out then? Is it decided sterically as a regular SN2, or does the logic with oxygen extend to sulphur?

Inductive reasoning in organic chemistry can be somewhat hard at times, given all of the exceptions that can pop up, for example the SN2-like mechanism of opening an epoxide in basic conditions versus the acidic conditions.
For the epoxide opening in base; as you know sterics wins, and in acid the stability of the reactive transition state is our reasoning for the product observed. I believe for the NGP reactivity that stabilization of transition states is an acceptable platform to reason the important question as to exactly how this reaction happens in the first place.
lets look at the NGP example you've pulled from Wikipedia
(Awesome btw! reading organic chem on wiki for fun?!)
I have a hard time fully following your question but here's the best answer I could come up with:
I see a symmetrical transition state where the reactive-intermediate charged cyclopropyl-S-heterocycle can probably just be attacked from either side in a SN2 like fashion. If you prefer charge polarization to be our main reasoning in the S-hetero cycle, given that it is a symmetrical substrate I believe the same product will form either way due to symmetry.
Did you read the part of that wiki article below that depicts NGP reactivity by an Alkene? The driving factor there appears to be pi-stabilization of the carbocation in the transition state.
This is a physical organic chemistry question that is not trivial. I applaud your curiosity.

Thanks for the answer. The picture I used was just to show the similarity between epoxide and the intermediate formed here. I'm mainly interested to know which way the attack happens predominantly (to the least hindered site or the site which stabalizes the positive charge the best) because the products will vary in case of an unsymmetrical substrate. I'm not able to find much regarding this

Thank you for clarification! I wasn't sure if that exact substrate was what was throwing you off or what..
So I am not a published peer reviewed expert on NGP reactions  - but I would cautiously advise that I believe that the charge stabilized product would be logically supported by inductive reasoning by comparing to epoxides. Sulfur and Oxygen both are chalcogens so they are similar electronic environments  - I would imagine that their reactivity wouldn't be too far off either.
What makes S so different than just being "fat oxygen"?
Hope it helps - this is a great question.

@himanshu if the question is adequate, I would encourage you to accept it (click the check mark)


text:
Is the change in enthalpy (ΔH) for dissolution of urea in water positive or negative?

To test the properties of a fertilizer, $\pu{15.0g}$ of urea, $\ce{NH2CONH2_{(s)}}$, is dissolved in $\pu{150 mL}$ of water in a simple calorimeter. A temperature change from $\pu{20.6^\circ C}$ to $\pu{17.8^\circ C}$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $Q = mc\Delta T$, and then dividing $Q$ by the moles of urea present. I can tell the process is endothermic because $\Delta T$ is negative, however my answer for $\Delta H$ comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
$$
\begin{align}
\Delta H &= \frac{(\pu{150ml}) \times (\pu{1g mL^{-1}}) \times (\pu{4.18J g^{-1} K ^{-1}}) \times (\pu{-2.8 K})} {(\pu{15g}/\pu{60.07g})}\\
&= \pu{-7030.59J/mol}\\
&= \pu{-7.03kJ/mol}
\end{align}
$$
TL;DR - question asks for $\Delta H$ of an endothermic process, not sure if my answer should be positive or negative

The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $\Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.


text:
What is meant by electrons moving to lower potential energy?
I am currently learning about chemical bonds in chemistry, and came across somebody who wrote this (source):
"Chemical bonds certainly "contain" potential energy, and the atoms want to move to a lower potential energy (become more stable).
When methane,
CH
4
, forms, the valence electrons end up in more stable (lower energy) C-H bonds.
These bonds are fairly strong, so methane is relatively inert.
However, if you add energy to the methane in the form of a flame or a spark in the presence of oxygen, some of the molecules will have enough energy to overcome an activation energy barrier.
Some of the
C-H
 bonds will break.
The electrons can then enter an even lower energy state by forming
C=O
 and
O-H
bonds rather than staying as
C-H
 and
O=O
 bonds."
What is meant by this? What does it mean when electrons enter a lower energy state, do they move to an even lower ground state?

Potential energy of the electrons has not much to do with the potential energy of the molecule in a reaction. Think then, Krypton or Xenon must be very unstable!

So when methane enters an even lower energy state (forming C=O and O=H), it has less potential energy, right? Do the electrons also enter a lower potential energy state?

What is meant by electrons moving to lower potential energy?

In a chemical reaction, the electronic states in the reactants are different from the electronic states of the products because the atoms are arranged differently into molecules. Lower potential energy of the electrons is associated with stronger bonds. When the products have stronger bonds than the reactants, energy is released in the reaction (exothermic). Conversely, when the bonds of the reactants are stronger than the bonds of the products, energy is absorbed in the reaction (endothermic).

do they move to an even lower ground state?

Yes. For example, the ground state in the hydrogen molecule $\ce{H2}$ is lower than in the hydrogen atom $\ce{H.}$ (because there is more room for the electrons as they are interacting with two nuclei instead of one). If you want to break the molecule into atoms, you have to provide energy to move the atoms apart (i.e. break the bond) because separating them will result in a higher ground state. For molecules reacting to other molecules, the energy balance depends on the relative strength of the bonds (and the number of bonds).
Follow-up questions

[OP's comment]: Could you just elaborate on the part where you say "because there is more room for the electrons as they are interacting with two nuclei instead of one"? Why does the ground state get lower because of that?

This might be jumping ahead a bit too much, but electrons don't get closer and closer to the nuclei (even when that would increase interactions) because "they don't like to be pinned down" (Heisenberg's uncertainty principle). If an electron can keep the same average distance to the nuclei but occupy a large space, it will lower its energy. This is reflected in the color of molecules with conjugated double bonds (carotene in carrots, for example), and also in a model called particle in a box. One analogy is strings of different length in music, where longer strings produce a lower pitch (lower energy) and short strings produce a higher pitch (higher energy).

[OP's comment]: I read this online. It talks about how the bonds formed in the products after exothermic reactions have lower potential energy. Is it the electrons which have lower potential energy or the new molecule? This answer states that it's the electrons which have lower potential energy, which one is more correct?

Typically, adjectives for bonds are strong or weak. If you are actually measuring an energy, it is the energy of the entire molecule. The reason energies are different is because the electrons have different energies (nothing happens in the nuclei). So saying electrons have lower potential energy makes the most sense to me. Distinct bonds don't really exist anyway, we just use them for making sense of the molecular properties.

[OP's comment]: So are bond strength and energy of the entire molecule both dependent on the electron energy?

Yes, much of chemistry is about the electrons. Chemical bonds and intermolecular interactions both are a reflection of "what the electrons are doing", and chemical reactions go along with "electrons doing something new", so knowing the electronic states of the molecules carries enormous predictive power.

Thanks for the very useful answer. Could you just elaborate on the part where you say "because there is more room for the electrons as they are interacting with two nuclei instead of one"? Why does the ground state get lower because of that?

This might be jumping ahead a bit too much, but electrons don't get closer and closer to the nuclei (even when that would increase interactions) because "they don't like to be pinned down" (Heisenberg's uncertainty principle). If an electron can keep the same average distance to the nuclei but occupy a large space, it will lower its energy. This is reflected in the color of molecules with conjugated double bonds (carotene in carrots, for example), and also in a model called particle in a box. The analogy is strings of different length in music

Thanks a lot for that. Just one more thing I want to clarify. I read this online https://socratic.org/questions/how-do-chemical-bonds-store-energy . It talks about how the bonds formed in the products after exothermic reactions have lower potential energy. This question may sound stupid, but is it the electrons which have lower potential energy or the new molecule? This answer states that it's the electrons which have lower potential energy https://socratic.org/questions/what-conditions-must-be-met-in-order-for-a-reaction-to-be-considered-exothermic, which one is more correct? Thanks so much

Typically, adjectives for bonds are strong or weak. If you are actually measuring an energy, it is the energy of the entire molecule. The reason energies are different is because the electrons have different energies (nothing happens in the nuclei). So saying electrons have lower potential energy makes the most sense to me. Distinct bonds don't really exist anyway, we just use them for making sense of the molecular properties.

okay, so are bond strength and energy of the entire molecule both dependent on the electron energy?

Thanks a lot. Just one more thing: When the new molecule is formed from the exothermic reaction, do all the electrons go to the ground energy level?

Yes, excited states are very short-lived in general.

But some electrons would obviously go in the second and third orbitals right? When we say it's in the ground energy level, there would still be some in the second, third oribtals, right? Would They would go to their lowest possible energy level, and not necessarily the ground energy level?


text:
Replacing Salt Bridge with Platinum Sheet
What would happen, if in a galvanic cell, the salt bridge is replaced by an inert platinum sheet?
Would charges accumulate on the sheet? Would the cell still work, and produce less emf than earlier, or stop working at all?
P.S.
I know why a salt bridge is necessary for the functioning of a galvanic cell - my question is rather specific to what would happen if I were to use a platinum sheet instead, and why we can't use it.
(Hence, not a duplicate.)

Possible duplicate of Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

You would have 2 cells in a serie, instead of one cell, each having Pt as one of electrodes.

So, according to you, the cell would operate?

You will not have that cell any more. You would have 2 cells different from the original one, with behaviour implicitly undefined.

In continuation of the discussion above, I have another opinion:

What would happen, if in a galvanic cell, the salt bridge is replaced
  by an inert platinum sheet?
The cell would stop working in most cases. Opposite charges would accumulate in the two > half cells, opposing the reduction and oxidation half reaction.

I am afraid the statement is not correct. The set up in which Cu and Zn are partitioned by a platinum sheet, is equivalent to two galvanic cells connected in series.
The two cells
(a) Cu-Pt cell
(b) Pt-Zn cell
which are connected via conductor. In this case, the conductor is also platinum.
I just was reading LeBlanc's historical textbook on electrochemistry (lovely book from German translation) for another purpose. He shows a nice picture of two cells connected in series.

Consider a similar situation in case of Pt. However we may not be able to calculate the potentials because Pt is not dipping in Pt ions in either beaker. Instead, one end is in copper ions and the other end in the zinc ions. It still may show a positive potential (Ecell) just like how a lemon battery does (=Zn and Copper rods immersed in a citric acid solution).
This has been discussed before in physics. https://www.physicsforums.com/threads/copper-wire-as-salt-bridge.146868/ The cell still works! Charges can never accumulate like static electricity in a galvanic cell in the presence of a conductor.

But the both Compartments would have just single solution, I=CuSO4, II=ZnSO4, and there would be no separation. There would be, from left to right: Cu/CuSO4, Pt/CuSO4, Pt/ZnSO4, Zn/ZnSO4. The Pt in CuSO4 would have to oxidize water to oxygen, right to reduce either Zn2+ either H+, depending on their potentials. But Cu/Zn Cell IMHO does not have enough EMF to electrolyze water.

"But Cu/Zn Cell IMHO does not have enough EMF to electrolyze water", yes a single cell. LeBlanc gives an example that the setup shown above can electrolyze water. This book is more than 100 years old and still a classic. Excellent German work.

I have thought the needed voltage is bigger, due hydrogen overvoltage. It may also depend on electrolyte.

This Daniel cell was a standard workhorse of electrochemical studies in the 19th and 20th century. Hydrogen overpotential is not that great otherwise Faraday would not have been able to electrolyze water with voltaic cells.

@Poutnik, your electrochemistry is very good. You must have had excellent teachers!

Thanks. It is from 80s, so it gets little rusty. I know the theoretical potential difference for water electrolysis is 1.23V, as as far as I know, the Daniel cell has EMF 1.1V.

according to https://en.m.wikipedia.org/wiki/Overpotential on the shiny platinum, hydrogen has overpotential -0.16V oxygen +0.95V. For platinezed(black) Pt,  it is -0.07V and +0.77V.


text:
Does Raney nickel deprotect acetals?
I know that Raney nickel can deprotect thioacetals. Can it deprotect acetals as well?
I found this image which shows that it deprotects and removes many protecting groups. But there was no reference to deprotection of acetals.

In my experience it does not,

Raney Ni cleaves C-S bonds.


text:
Degree of Hardness Numerical

To remove the temporary hardness from $5\ \mathrm L$ of $\ce{H2O}$ $5.6\ \mathrm{mg}$ of lime is required. Calculate the degree of hardness due to $\ce{Ca(HCO3)2}$ present in water.

My attempt:
$$
\ce{Ca(HCO3)2 + CaO -> 2CaCO3 + H2O}
$$
Since $5.6\ \mathrm{mg}$ of lime is required, amount of calcium bicarbonate present is $16.2 \times 10^{-3} $.
Molarity = ${16.2 \times 10^{-3}}/(162 \times 5)$
$= 0.002\ \mathrm M$.
This is equivalent to the concentration of calcium carbonate.
ppm of calcium carbonate =
 $ 0.002\times10^{-2}\times100\times10^6\times10^{-3} = 200 $
Therefore degree of hardness is $ 200\ \mathrm{ppm}$.
Is the solution correct?
Also I am looking for an alternative method for solving the problem.

HINT - ppm in this case is done on a weight basis.

Hardness is usually expressed in  mmol/L (1=equivalent of 56 mg CaO/L) or in dGH ( 1 dGH=56 mg CaO/L)

1 dGH is most likely a German unit. Double check you answer with the following approach as well. The traditional unit is ppm CaCO3 calculated in terms of mg/L. Find out moles of CaO, from the equation see the 1:1 mol ratio between CaO and CaCO3. If x moles of lime were used, x moles of CaCO3 is present in 5 L of water. Convert moles of CaCO3 into mg CaCO3 and express it per liter.


text:
What is meant by excess energy in exothermic reaction?
I was reading this article on exothermic reactions online: Exothermic reactions
It states that "In exothermic reactions, the products have less enthalpy than the reactants, and as a result, an exothermic reaction is said to have a negative enthalpy of reaction. This means that the energy required to break the bonds in the reactants is less than the energy released when new bonds form in the products. Excess energy from the reaction is released as heat and light."
I'm just curious, is the excess energy from the bonds being formed, or is it because of the fact that products have more potential energy than the reactants, and because of law of convervation of energy, is released as heat energy?

There is no "or". Both alternatives are correct, except that you have the second one upside down. Products have lower potential energy than the reactants, hence the extra energy.

The energy release is because of the total energy of valence electrons in the product bonds is lower  than in reagents.
In fact, the change of the mean total  potential energy of valence electrons is twice as much as the change of their total energy, as the half of that change is compensated by their increased mean total kinetic energy.
As a consequence, the energy released by forming chemical bonds of products is bigger than energy spent on breaking bonds of reagents.

But I thought the energy is released from bonds being formed? Is that incorrect?

Yes, That is right, it is the consequence of "low level quantum mechanics". See the answer update.


text:
Role of NaI in Formation of Silyl Enol Ethers using Chloro(trimethyl)silane
I am forming silyl enol ethers using Chloro(trimethyl)silane (TMSCl), and below is an experimental protocol I am following

Silyl Enol Ether Synthesis

In an oven-dried 100 mL three-necked flask, Et3N (37.2 mmol) and carbonyls (30.0 mmol) were added. The mixture was stirred and heated to 30~35 ºC, and TMSCl (37.2 mmol) was added dropwise over 15 minutes. The mixture was stirred at the same temperature for 0.5 hour.
A solution of NaI (37.2 mmol) in acetonitrile (35 mL) was added dropwise to the mixture over 0.5 hour and stirred for 1 hour. The mixture was then heated to 40-45 ºC and stirred at this temperature for 2 hours.
The mixture was then cooled to room temperature and 30 mL of cold pentane was added. The mixture was poured into 30 mL of ice-cold water and extracted with pentane (15 mL × 2). The combined organic phases were dried over anhydrous Na2SO4, filtered and concentrated in vacuo. After distillation under reduced pressure, the products were obtained as colorless oils and stored at 0 ºC under an atmosphere of nitrogen for use.

I don't understand the mechanistic background of Step 2. What does NaI do?
Side question: when adding TMSCl in Step 1, I saw some white fume. I am guessing TMSCl was reacting with moisture in the air but what would be the product of H2O+ TMSCl? Is it HCl and TMS-OH?

The use of $\ce{NaI}$ in $\ce{MeCN}$ (a solvent in which $\ce{NaI}$ is quite soluble) is forming $\ce{Me3SiI}$ from the $\ce{Me3SiCl}$. The reason for doing this is $\ce{Me3SiI}$ is significantly more reactive with the triethylammonium enolate and will increase the yield of the TMS-enolate.
You are correct in thinking that the white fume you see when adding the $\ce{Me3SiCl}$ is $\ce{HCl}$. The product of aqueous hydrolysis of $\ce{Me3SiCl}$ is $\ce{HCl}$ plus hexamethyldisiloxane (Wikipedia):
$$\ce{2Me3SiCl + 2H2O -> 2HCl + Me3SiOSiMe3}$$

OP has used oven-dried glassware and I assumed OP did the reaction under $\ce{N2}$ atmosphere. Hence, I think $\ce{HCl}$ was released by the reaction between $\ce{Me3SiCl}$ and $\ce{Si-OH}$ residue on glassware surface. Would you agree?

OP did not specify the reaction run under N2. I'm more inclined to think that the addition or possibly the weighing out of the TMSCl may have been done in air. OP comment?

@Waylander I did it under N2. However, I still could see some white fume. Maybe because it's not possible to completely eliminate water in the air?

possibly you are getting the TMSiCl reacting with Et3N vapour over the reaction mixture as you have excess Et3N

I carefully conducted every step under nitrogen but the reaction ultimately failed (twice already) so I am now doing with LiHMDS...

I would advise doing an anhydrous workup

@Waylander Is TMSiCl normally stored under N2? I believe I have done every step correctly but my reaction keeps failing. One possible explanation that I can think of is that TMSiCl was already degraded- the bottle I have used wasn't capped in N2.

Under nitrogen, in a fridge. If this is a bottle that is a lab resource someone may not have been as careful as needed. For certainty either use a new bottle or distill the material from the old one.


text:
Would Z and E isomers show different optical activity?

Consider this compound, which is in E-configuration and that red asterisk denotes the chiral carbon. If I were to somehow cause the E-configuration of the alkene part to Z-configuration keeping the rest of the molecule same, would it have the same optical rotation?

Not necessarily. Although the change you suggest would not alter the absolute stereochemistry (S in both cases) of the asymmetric center starred, the two molecules are diastereomers and therefore will have different physical and optical properties. Even if they did both rotate light the same way, they probably won’t have the same degree of rotation for the same concentration (specific rotation).
The problem boils down to whether we can predict the direction (or degree) of rotation from a drawn absolute stereochemical configuration. The answer to this is still “no”. We don’t currently have a theory that will allow us to say that, for a certain molecule, R gives us d or l, etc.
Some studies have been able to come up with some rules of thumb for small molecules, but there is no comprehensive theory to make this prediction in general. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2704476/

So there is no clear way of predicting the effects of such a change?

Not consistently. The rotation of plane polarized light it thought to originate from the interaction of the four groups off the asymmetric center with the two directions of circularly polarized light that are coincident in the plane polarized light. If one direction is slowed more than the other (circular dichroism: https://ja01.chem.buffalo.edu/~jochena/research/opticalactivity.html) then the plane of the polarized light is seen to rotate. We can't yet accurately predict how changing one of the four groups (like your E to Z change) will affect this interaction.

There is a great post on Physics Stack Exchange that deals with more of the details of how this effect could, in principle, be predicted from a molecule's structure. But I don't believe that studies in this field are comprehensive yet: https://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light


text:
A formula for the refractive index of a gas dependent on temperature and pressure
I am in search for a formula that gives me the refractive index of a gas at a given temperature and pressure. Does such a formula exist? Ideally the solution should have this form:
$n_g(t, p) = \ldots$,
where $n_g$ is the refractive index of the gas $g$, $t$ is the temperature and $p$ is the pressure.

n(g)-1=k* density, the k factor is called Gladstone-Dale coefficient. You can always look up gas density in the literature tables. https://books.google.com/books?id=HWtB2R0gWFgC&pg=PA26&dq=refractive+index+of+a+gas&hl=en&sa=X&ved=0ahUKEwiPrerS4-3hAhUPnKwKHSyhC6MQ6AEILjAB#v=onepage&q=refractive%20index%20of%20a%20gas&f=false


text:
Elements other than carbon that can form many different compounds by bonding to themselves?
My textbook says the following:

Unique among the elements, carbon can bond to itself to form extremely strong two-dimensional sheets, as it does in graphite, as well as buckyballs and nanotubes.

Is carbon the only element that can do this?
If not, then what are the other elements can also do this? Is there a term to describe such elements?
What is the chemical characteristic that allows this to occur?
I would greatly appreciate it if people could please take the time to clarify this.

Another case is black phosphorus allotrope https://en.wikipedia.org/wiki/Allotropes_of_phosphorus

Catenation : self linking property of an element (atom).

The right question isn't whether an element can bond to itself. Plenty do that. The issue is whether an element can form a wide variety of stable structures when bonded to itself.

I was also aware that boron sheets were recently synthesized, but did not realize that someone has even predicted nitrogen sheets that could be stable at room temperature!

Sulfur can also form rings and chains

@jeffB: Wow! That's really fascinating, but also kind of scary. They may say "stable at room temperature," I say "explosion waiting to happen." There's no way all those nitrogen atoms wouldn't much rather just pair up as $\ce{N2}$ and fly off to entropic heaven, as somebody once eloquently put it.

What is the actual question here? Your citation does not say that bonding between identical atoms is something unique - every single element could do this even if transiently. The point is that carbon carbon allotropes are somewhat exceptional.

@Mithoron Yes, I see now that I misinterpreted the author, but the questions are still valid and insightful.

Related, with some quantitative data in the answers: https://chemistry.stackexchange.com/questions/18843/what-makes-carbon-special-and-versatile

Is carbon the only element that can do this?

No, carbon is not the only element with such characteristics.

If not, then what are the other elements can also do this?

There is a whole number of elements such as silicon, arsenic, germanium.

Is there a term to describe such elements?

At least I'm unaware of such a term, which might be furnished by our far wiser community.

What is the chemical characteristic that allows this to occur?

Catenation.
More information:
According to the Molecular Orbital Theory, the condition for a compound to exist is that it should have more electrons in the bonding orbitals than in the anti-bonding orbitals. So, as long as you have the bonding orbitals filled more, you can have pretty anything, more than just chains of atoms.
Thus, the existence of a compound also depends on the precise conditions in which the compound is kept, for example sodium forms different types of chlorides under different conditions and that as pointed out by Poutnik in the comments, $\ce{He2^1+}$ and a ton of others are discovered and still more awaiting discovery.

No, carbon is not the only one that can bond to itself. It's a unique property of some elements mainly the group 14 elements like silicon, germanium, arsenic etc. This phenomenon is called catenation. It might be mainly due to presence of four valence electrons in their outermost shell. A large number of carbon atoms are linked with each other with sigma and pi bonds. However catenation gets limited as we move down the group in group 14.

You wanted to know if any elements were able to match carbon's ability to form 2D sheets, nanotubes, and buckyballs.
According to https://www.azonano.com/article.aspx?ArticleID=4863 and other sources, graphene-like 2D sheet formation has (thus far) been demonstrated in the following elements:
boron, silicon, germanium, phosphorus, antimony, bismuth, arsenic, and tin
In addition, nanotube formation has (thus far) been demonstrated for all of the above except tin.
Since any element that can form a 2D sheet should be able to form a nanotube, it is expected that tin nanotubes will be developed eventually. In the meantime, computational studies have been done to predict the properties of tin nanotubes.
By contrast, the subset of the above list that has been shown to form buckyballs is, thus far, limited to boron and silicon.
The formation of buckyballs (60 atoms) involves bond length and angle constraints, which may rule out some of the elements that can form 2D sheets. Though it is possible that elements that cannot form buckyballs could form fullerenes of other sizes.
I've repeatedly added the "thus far" qualifier to indicate that these properties may eventually be demonstrated for other elements as well.

So most of this has been answered, but the main elements that can bond to themselves are called diatomic atoms. This includes hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine, and bromine. There are other elements that can do this as well, however these main elements occur naturally in their diatomic state as gases.

"Diatomic atom" is a weird term, diatomic molecule makes way more sense. Besides, the way the answer is formulated, it appears to me that all elements that are capable of self-bonding are diatomic molecules, whereas in reality diatomic molecules is a tiny subset of the former.

It's pretty clear that the question is asking about self-bonding beyond simple diatomic molecules.

The ability to make a stable diatomic molecule might be the reason for not observing chains or 2D structures for these elements.

@Karsten Theis very much agree. Extreme Peierl distortion.