text:
Mechanism for Diels–Alder of alkene with substituted thiophene sulfone with loss of sulfur dioxide
I came across a synthesis of centrohexaindane that has the following steps:

This is a Diels–Alder reaction that gives an aromatic ring. I know that alkynes will give this reaction as follows:

But in the scheme, after the initial Diels-Alder to form the product with bridging $\ce{SO2},$ it cannot just lose $\ce{SO2}$ to become aromatic. Does the molecule with the bridging $\ce{SO2}$ get oxidized to have two double bonds like in the second image and then lose $\ce{SO2}?$
The sulfone is the only possible oxidizing agent here (toluene solvent doesn't seem like it would oxidize) and Wikipedia says on its page on sulfones that:

Sulfone is a relatively inert functional group, being weakly basic(compared to sulfoxides). They are non-oxidizing.

A few references into the rabbit hole and we have an answer. The major reference was given in the link mentioned by OP which was Chem. Ber. 1994, 127 (2), 409–425[1], in which the reaction mechanism of the synthesis is given as follows:

[...] Reaction of 33 with tetrathiophene-S,S-dioxide is performed by heating the reactants in highly concentrated toluene solution to $\pu{120 °C}$ for a total of
24 h. The cycloadduct 34 is obtained in high yield.

Now the question that has arisen in the step from 33 to 34 is the following, under normal conditions of Diels-Alder, there would only be one double bond in the final product. Here there are two. This can be answered going one down into the rabbit hole and using J. Org. Chem. 1980, 45 (5), 856–867.[2]
This paper deals with the specific reaction that causes confusion - the annulation using tetrachlorothiophene 1,1-dioxide. According to this paper:

The compound has a combination of stability and reactivity that makes it a useful synthetic reagent. It undergoes Diels-Alder addition with double bonds to form an adduct which spontaneously loses sulfur dioxide to leave the double bond annelated with a tetrachlorobutadienediyl group.

Furthermore, it talks about the reactivity to olefinic bonds and states the following:

Example 18, from l-hexen-5-yne, reveals the greater reactivity of alkene vs. alkyne. Reaction is exclusively with the alkene bond as far as NMR reveals. Thus, the kinetically favored product is formed rather than the thermodynamically favored one.

There is a minor error in the reaction mechanism as given in the link of OP as the first paper shows that the tetra-chloro derivative needs to be oxidised in order to be converted into cyclohexaindane
References:

Kuck, D. Benzoanellated Centropolyquinanes, 15. Benzoanellated Fenestranes with [5.5.5], [5.5.5.6], and [5.5.5.5] Frameworks: The Route from 1,3-Indandione to Fenestrindan. Chem. Ber. 1994, 127 (2), 409–425. DOI: 10.1002/cber.19941270218.
Raasch, M. S. Annulations with tetrachlorothiophene 1,1-dioxide. J. Org. Chem. 1980, 45 (5), 856–867. DOI: 10.1021/jo01293a019.


text:
Procedure for measuring EC
Most of the procedures on the internet are of a university research lab level. Is there a simpler method I could use to determine the concentration of EGCG

Seems like this one might be helpful: https://chemistry.stackexchange.com/a/128553/79678. So I think you can do it.

I have not done the red wine titration, but the linked protocol seems doable. I would just substitute the green tea for the red wine, keeping everything else the same. The indigo carmine indicator might be replaceable by another indicator, but I would make no changes in the protocol, at least at first. Frankly, it would be best to have an ECGC standard so you could get the stoichiometry nailed down: as it is, you are not really going to oxidize ECGC all the way to carbon dioxide etc. That is all I can say, so best of success if you go for it!

Ok. Thanks for your help @EdV - BTW, what do you mean when you say: "Frankly, it would be best to have an ECGC standard so you could get the stoichiometry nailed down"?

Well, then you would know the stoichiometry, i.e., how many permanganate ions per ECGC? You know that the green tea to be titrated has some unknown ECGC concentration. So titrating to the endpoint, and knowing the stoichiometry, then allows you to deduce the ECGC concentration. Without knowing the stoichiometry, you just get within scale factor of correct.

Do you have access to a UV-vis spectrophotometer?

I am conducting a school-based research project into the antioxidative properties of green tea. Part of the procedure is to determine the concentration of antioxidant in green tea. This antioxidant is EGCG (a type of catechin).

First of all, you cannot determine the concentration of just EGCG (epigallocatechin gallate) alone, because any extract of green tea contains polyphenols other than EGCG, which include catechin, and/or other flavonoids such as flavanols, flavandiols, and phenolic acids. Most of the green tea polyphenols are flavonols, commonly known as catechins, which are flavan-3-ols (Ref.1):

As shown in the diagram, There are four kinds of catechins mainly find in green tea: epicatechin (EC), epigallocatechin (EGC), epicatechin-3-gallate (ECG), and epigallocatechin-3-gallate (EGCG). The amount of catechins varies in the original tea leaves due to differences in variety, origin, harvesting time, and growing conditions.  Also, the preparation of fresh green tea cannot totally extract catechins from the leaves, therefore, the amount determine by you may not reflect the absolute values determined through the complete extraction of leaves. Thus, your findings may differs from the accurate benefit of green tea.
Nonetheless, there are methods to determine polyphenol content in particular extract as the amount equivalent to a reference. Easily, your reference could be gallic acid. One easy method I found in literature is the estimation of total phenolic content and other oxidation substrates using Folin–Ciocalteu reagent* (Ref.2; PDF is attached for convenience):

Determination of Total Polyphenols by Folin-Ciocalteu Method: One milliliter of the tea extract was diluted with water to $\pu{100 mL}$. The diluted extract solution $(\pu{1.0 mL})$ was then mixed with $\pu{5.0 mL}$ of 50% Folin-Ciocalteu reagent. After $\pu{5 min}$, $\pu{4.0 mL}$ of 7.5% $\ce{Na2CO3}$ solution was added and mixed. Then the mixture was incubated at room temperature for $\pu{60 min}$. The absorbance of the solution was measured at $\pu{765 nm}$ with a Shimadzu UV-2550 UV-Vis spectrophotometer against a reagent blank. The determination was three times. The total polyphenol content was expressed as gallic acid equivalents (GAE, units: $\pu{g}/\pu{100 g}$ sample).

Preparation of Tea Extract: Each tea sample was ground to a fine powder and $\pu{1.0 g}$ of the powder was extracted with $\pu{25 mL}$ of methanol/water (7 : 3, v/v) in a water bath at $\pu{70 ^\circ C}$ for $\pu{10 min}$. The extraction was repeated twice and the extracts were combined. After filtration through a filter paper and a $\pu{0.45 \mu m}$ membrane filter (Millipore), the extract volume was adjusted to $\pu{100 mL}$.

According to the procedure, you need just a spectrophotometer in visible range (do not need an UV-Vis spectrophotometer, which is an expensive one). If you do not have a membrane filtration setup, you may extract leaves without powdering them, but need extra extracting time (may be pre-prepare extract overnight at $\pu{70 ^\circ C}$ using the same solvent system). If you want to check the green tea as we prepared them to drink, use that way but you may have to adjust reagent concentrations accordingly to test.
If you have a 96-well plate-reader and some micro-pipettors, there is a reliable high-input method is listed in Ref.3 for the same method (PDF of the paper has been attached to compare).

*Note: Folin–Ciocalteu reagent is a specific redox reagent containing phosphotungstic-phosphomolybdenum. The reaction forms a blue chromophore constituted by a phosphotungstic-phosphomolybdenum complex with phenols that can be quantified by visible-light spectrophotometry (Ref.4,5).
Gallic acid could be used as calibration standard for this determination and the total phenols could be reported as gallic acid equivalent (GAE). Other alternative is pyrogallol. Choose the concentration range enough to give absorbance 0.4-1.0 (at $\pu{760 nm}$) so that the assay is linear (Ref.5). Following is the published calibration curve for gallic acid using Folin-Ciocalteu Assay (Ref.6):

The procedure used: In brief, aliquots of a standard solution of gallic acid $(20, 40, 60, 80, \pu{100 mg L-1})$ were mixed with $\pu{500 \mu L}$ of Folin-Ciocalteu reagent (previously diluted 10-fold with distilled water) and $\pu{400 \mu L}$ of $\ce{Na2CO3}$ (7%). After $\pu{40 min}$ of incubation at room temperature $(\pu{23 \pm 2 ^\circ C})$, the absorbance was measured at $\pu{760 nm}$ using a Spectrophotometer against a blank sample.
Note that the temperature of $\pu{23 ^\circ C}$, time of $\pu{40-60 min}$, and  $\lambda = \pu{760 nm}$ are optimized condition for Folin-Ciocalteu assay (Ref.5). Also make sure that you add $\ce{Na2CO3}$ solution at least after $\pu{10 min}$ of the addition of Folin-Ciocalteu reagent (not before). Folin-Ciocalteu reagent is unstable under alkaline conditions.

References:

Sarah C. Forester and Joshua D. Lambert, "Antioxidant effects of green tea," Mol. Nutr. Food Res. 2011, 55(6), 844–854 (doi:10.1002/mnfr.201000641)(PDF).
Chunjian Zhao, Chunying Li, Shuaihua Liu, and Lei Yang, "The Galloyl Catechins Contributing to Main Antioxidant Capacity of Tea Made from Camellia sinensis in China," The Scientific World Journal 2014, Article ID 863984 (11 pages) (doi: https://doi.org/10.1155/2014/863984)(PDF).
Elizabeth A Ainsworth1 and Kelly M Gillespie, "Estimation of total phenolic content and other oxidation substrates in plant tissues using Folin–Ciocalteu reagent," Nature Protocols 2007, 2(4), 875-877 (doi:10.1038/nprot.2007.102)(PDF).
P. Schofield, D. M. Mbugua, and A. N. Pell, "Analysis of condensed tannins: a review," Animal Feed Science and Technology 2001, 91(1–2), 21-40 (doi: https://doi.org/10.1016/S0377-8401(01)00228-0).
Andressa Blainski, Gisely Cristiny Lopes, and João Carlos Palazzo de Mello, "Application and Analysis of the Folin Ciocalteu Method for the Determination of the Total Phenolic Content from Limonium Brasiliense L.," Molecules 2013, 18, 6852-6865 ( doi:10.3390/molecules18066852).
A. Bouyahya1, N. Dakka, A. Talbaoui, N. El Moussaoui, J. Abrini, and Y. Bakri, "Phenolic contents and antiradical capacity of vegetable oil from Pistacia lentiscus (L)," Journal of Materials and Environmental Sciences 2018, 9(5), 1518-1524 (doi: https://doi.org/10.26872/jmes.2018.9.5.167).

I already upvoted this comprehensive and useful answer! Well deserves the bounty and green checkmark, IMHO!

@andselisk♦: I don't know a private conversations with users over phone or email are allowed by CSE. However, I preferred conversation in the chat room. I have one created and check for "The antioxidant properties of EGCG."

@MathewMahindaratne I am puzzled by some changes in the question and why you only got 25 bounty, even though you have the only answer and it is nicely upvoted and accepted. Feel free to ignore this, but it seems weird.

@Ed V: Thank you for your concern. I realized they have deduced 25 points from not only this question, but also from other question, which is also accepted as the answer. I felt I got the step-mom treatment from here whoever (moderators?) responsible for awards. But, I just keep working on my love for chemistry and to help others who need assistance. Thanks again for your kindness.

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Comparing van der Waals constant for gases
How can I compare the van der Waals constant $a$ (the liquefication constant) for different compounds based on my knowledge of bonding in the compound? Let's pick the following sample set:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Compound} & &\hspace{-1em}a \\
\hline
\text{Helium} & \d{0}{0346} \\
\text{Argon} & \d{1}{355} \\
\text{Xenon} & \d{4}{25} \\
\text{Water} & \d{5}{526} \\
\text{Ethanol} & \d{12}{18} \\
\text{Acetone} & \d{16}{02} \\
\text{Acetic acid} & \d{17}{71} \\
\text{Benzene} & \d{18}{24} \\
\text{Toluene} & \d{24}{38} \\
\hline
\end{array}
$$
A general trend that I can see is $a$ is proportional to molecular mass. Am I wrong to compare on the basis of intermolecular forces like hydrogen bonding, which is clearly very good in the water but the $a$ value isn't so promising?
Is it my misconception that $a$ depends on intermolecular forces? As far as I know, the stronger the forces, the more easily it will be liquefied, and greater should be $a.$

See the related post about correlation of a and b parameters: how-are-the-van-der-waals-constants-a-and-b-related-to-each-other

The common factor could be polarisability which is proportional to the volume of an atom or molecule and is one  property responsible for intermolecular interaction. So mass is misleading here.

@Random Quark ”Can I compare the Van der waals constant a  for different compounds on the basis of bonding?” — Random Quark, Yes but I think theoretically there are too.. many things for considerations if one wants to compare between any two(or any number of) random gases but it is still possible to do so, but if it is to be compared in a certain trend, then it may be very easy to do so.(maybe I may expand upon this and write an answer)

How can I compare the van der Waals constant $a$ (the liquefication constant) for different compounds based on my knowledge of bonding in the compound?

Actually, as you indicated in the question, this comparison is not realistic since $a$ is unique to a particular compound/atom. As Chet Miller pointed out $a$ is depend on the each molecule/atom's critical pressure $(P_c)$ and temperature $(T_c)$ as:
$$a = \frac{27R^2T_c}{64P_C}$$
It is also depends on other factors such as the size, shape, mass, intermolecular forces, etc.

A general trend that I can see is a is proportional to molecular mass. Am I wrong to compare on the basis of intermolecular forces like hydrogen bonding, which is clearly very good in the water but the a value isn't so promising?

I'd say, yes! It is definitely proportional to molecular/atomic mass of the molecule/atom of interest. But it is true that it is also depends on specific types of molecules/atoms:

As shown in the plot of the noble gases, the slope of the plot is $0.0304$ ($1.74^\circ$ to the $x$-axis), the smallest of all compounds. That is expected since all noble gasses are believed to have no intermolecular (interatomic here) forces other than London forces (also they are mono-atomic and even those forces are minimal). I also included hydrogen gas here (which is the only bi-atomic molecule in the set) because it fits in without significant deviation. The biggest deviation is shown by neon, which is always like to be behave alone. Even with this hiccup, the relationship is almost perfect $(R^2 = 0.9907)$.
On the other hand, to compare with noble gases, I chose a second set of compounds, which included water and ammonia (considered to be having strong intermolecular interactions). I also included hydrogen gas hoping to have a cross-section between two lines. Even with vastly different compounds having mixed properties, $a$ and molar masses of this set have a good relationship $(R^2 = 0.9773)$. The most significant deviations are from benzene (probably due to $\pi$-$\pi$ interaction or $\pi$-stacking) and acetic acid (probably stay as a dimer in vapor state). The slope of the plot is $0.2564$ ($14.38^\circ$ to the $x$-axis), a significant increment from noble gasses as expected, but it is not the largest of all compounds I have so far studied in this regard (Nonetheless, the angle between two straight lines is $12.64^\circ$, a significant deviation):

The expectation was the slope of the plot with polar compounds to be the largest since this particular set of compounds believed to be having the strongest intermolecular (or interatomic) forces in addition to London forces.
Surprisingly,  That credit goes to the group of compounds, mostly of which are hydrocarbons:

It is noteworthy to mention that 17 of 20 compounds included in the plot are hydrocarbons. The rest is alkyl primary amines, which are also capable of having strong inter-molecular interactions (e.g., interactions due to polarizability; see porphyrin's comment elsewhere). The slope of the plot of this specific set is $0.3106$ ($17.25^\circ$ to the $x$-axis). The data are in a good agreement $(R^2 = 0.9757)$. Only outliers are hexane, cyclopentene, and methylamine, but they are not off by much. Except for three amines, all other 17 compounds are alkanes, alkenes, and a alkyne, all of which are unable to make hydrogen-bonding. Thus, it is safe to suggest that the better influence on $a$ values are from mostly non-polar interactions among molecules, which are having higher polarizability  would show the highest influence. I disagree with Ritam_Dasgupta's idea of comparison of water with molecules with similar molar masses. The plot with polar molecules including water $(MW = \pu{18.02 g mol-1})$ and ammonia $(MW = \pu{17.03 g mol-1})$ would disagree with that idea. For instance, water is much polar than ammonia and make stronger hydrogen-bonding (c.f., boiling points, $\pu{100 ^\circ C}$ vs $\pu{-33.3 ^\circ C}$), yet $a$ value of water and ammonia are $5.53$ and $4.23$, respectively with close proximity.

Oh, the work you put into that! Thanks, the trend is much clearer now.

The reason that in your sample set water seems to have a low value of $a$ despite having hydrogen bonding is because most of the molecules that you chose were of much higher molecular mass. You should compare water with some gas of approximately equal molecular mass, then you'll see that value of $a$ in water is quite high. Your other observations are generally correct.

That's what I've been thinking that London dispersion forces depend greatly on molecular weight.

Yes, that is true. It is easier to liquefy polyphonic gases for this reason

In terms of the critical temperature and critical pressure, $$a=\frac{27R^2T_c^2}{64P_c}$$

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text:
Is a hydrocarbon bio-battery feasible?
Hydrocarbons have some of the highest energy densities and specific energies of any chemical fuel.  Unfortunately, our primary use of them is in thermal engines, whose efficiency is limited by Carnot's Theorem.  Batteries, on the other hand, do not have this limitation and can be engineered to arbitrary efficiency.
Sony has designed a biobattery which runs on glucose.  While this is an impressive achievement, liquid propane and butane have a much higher specific energy (> 3x) than glucose, as well as a higher energy density (not surprising, since glucose contains oxygen atoms which are basically dead weight for a battery).
Is there any reason that the enzyme catalyst approach could not be applied to simple short-chain hydrocarbons like propane through octane?  I assume that the battery's performance (current discharge) is proportional to the amount of enzyme present, and thus, the weight overhead for the battery depends on the amount of desired current.  Are the enzymes expected to be a significant fraction of the mass/volume of the battery?
It seems that an electric airliner running on Jet-A fed to bio-batteries could theoretically have more than twice the range of best current turbofans.  Is this an unreasonable conclusion?

There are millions of people thinking like you. But nobody has found a way of having an alcane react to produce electrons in a cell, as it theoretically should. Oxygenated derivates of alcanes are able to produce electricity, but not alcanes. Maybe somebody will find a trick or a catalyzer in the future. Who knows ?

They were called, at least the highest terms, paraffins, from parum  affinis, scarce reactivity.

Hydrocarbons could be used in a fuel cell at high temperature. Alkanes can be cracked with heat (and catalyst, to make the process faster and more efficient), yielding hydrogen suitable for use in the cell, and a residue containing more carbon. In a similar way, ammonia can be used in fuel cells (though NH3 is far more reactive than CnH2n+2).

@DrMoishePippik I suppose fuel cells are the general form of "turn combustion into a battery"... it boils down to finding the right catalysts and electrolytes, then?

Worth noting that typical bio battery designs do not even capture all of the energy in glucose. They only convert to gluconolactone not CO2. Just doing that would greatly increase energy yield before you even consider a switch to alkane


text:
Reason for protonation of amino group in amino acids at near neutral pH
What happens to the amino group when amino acid is added to a near neutral solution? How does it come about gaining an extra proton?
I understand what happen to the carboxyl group, since it is a strong acid it will dissociates quickly to donate a proton.

Prerequisite knowledge

A proton is deprotonated when the $\mathrm{pH}$ exceeds the $\mathrm pK_\mathrm a$ of the proton.

Analysis
There are 3 different forms that an amino acid can exist in. For the sake of simplicity of analysis, I am taking the simplest amino acid - glycine.
$$\underset{\mathrm{glycine}}{\ce{HOOC-CH2-NH2}}$$
Continuing to state the conditions of analysis, let's take a highly acidic medium to begin with and observe the changes that take place on increasing $\ce{pH}$.
Case 1: pH of solution < 2.34
In such a scenario, the $\mathrm{pH}$ is lower than the $\mathrm p K_\mathrm a$ of the acid group and lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that both the amino group and acid groups are protonated. The amino acid is positively charged at this $\mathrm {pH}$.
$$\underset{\mathrm{glycine\,(pH \,<\,2.34)}}{\ce{HOOC-CH2-NH3+}}$$
Now, we increase the $\ce{pH}$ to a value greater than $2.34$ but less than 9.6  Specialty of these numbers? We'll see at the end.
Case 2: pH of solution greater than 2.34 but less than 9.6
In such a scenario, the $\mathrm{pH}$ is higher than the $\mathrm p K_\mathrm a$ of the acid group but lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that the amino group remains protonated and the acid group is now deprotonated.
$$\underset{\mathrm{glycine\,(2.34\,<\,pH \,<\,9.6)}}{\ce{^-OOC-CH2-NH3+}}$$
Case 3: pH of solution greater than 9.6
In this last case, the $\mathrm{pH}$ is higher than both the $\mathrm pK_\mathrm a$s of the amino group and the acid group. This means that both groups are now deprotonated. The amino acid is now negatively charged.
$$\underset{\mathrm{glycine\,(pH \,>\,9.6)}}{\ce{^-OOC-CH2-NH2}}$$

Observations and Conclusion
As you noticed, there are some specific numbers that are so important they are mentioned. They are the $\mathrm p K_\mathrm a$ values of the acid group ($\mathrm pK_{\mathrm a_1}$) and the amino group ($\mathrm pK_\mathrm {a_2}$). There is one more important number and that is the exact point of net average neutrality. It is known as the isoelectric point.
$$\mathrm {pI} = \frac{\mathrm pK_\mathrm {a_1} + \mathrm pK_\mathrm {a_2}}{2}$$


text:
Possible carbocation rearrangement in substitution reaction to give two different products
In the following substitution reaction, is compound I formed by $\mathrm{S_N}$2 and compound II formed by $\mathrm{S_N}$1 and carbocation rearrangement?

Or is there some rearrangement I'm not aware of?

I would say yes. Your reasoning is OK.

$\mathrm{S_N2'}$ is also a possibility

Is compound I formed by $\mathrm{S_N}$2 and compound II formed via $\mathrm{S_N}$1 after a carbocation rearrangement? Or is there some rearrangement I'm not aware of?

For the first question, the answer is probably not because the mechanism of a given reaction is mainly based on the condition used in the reaction (at least at the undergraduate level of teaching). The the answer for the second question is yes, you are unaware of the mechanisms called $\mathrm{S_N}$1' and $\mathrm{S_N}$2' (as Shoubhik R Maiti mentioned in the comments section). This type of reactions are common when the allylic compounds are the substrate, which is appropriately called allylic rearrangement or allylic shift. Allylic shift is possible under either $\mathrm{S_N}$1 or $\mathrm{S_N}$2 conditions, and commonly known with $\mathrm{S_N}$1' and $\mathrm{S_N}$2' notations, respectively.

Strictly speaking, if you have used condition favoring $\mathrm{S_N}$1 reaction, the intermediate is a resonance stabilized allylic carbocation. Thus nucleophile would attack at two positions (either carbon 1 or 3) to give the resulting products, I and II. For instance, see your reaction in a protic solvent ($\mathrm{S_N}$1), which would given the same type of product mixture (for convenience, I have used the labelling to show the difference):

Molecular modeling calculations would shows you the allylic secondary carbon position is more stable than the allylic primary carbon (recall your ab initio calculations for kinetic versus thermodynamic control in Org.Chem.II lab course; a simple and effective lecture demonstration of thermodynamic versus kinetic control, see: Ref.1) in intermediate carbocation. Accordingly, compound II would be the major product.

Similarly, if you are using the condition favoring $\mathrm{S_N}$2 reaction (strong nucleophile in a polar aprotic solvent) in this type of reaction, it is also possible to get two product due to $\mathrm{S_N}$2' mechanism. Usually, a bulky leaving group in $\mathrm{S_N}$2 conditions or bulky non-leaving substituent on the framework which give rise to significant steric hindrance would thereby increase the conjugate substitution ($\mathrm{S_N}$2'). Following scheme show the mechanism for two products:

Late Edit:
Since it seems OP is still cannot see how compound II is formed, I included the arrow-pushing mechanism for $\mathrm{S_N}$2 versus $\mathrm{S_N}$2' products:

References:

I. J. McNaught, "Thermodynamic versus kinetic control: A lecture demonstration," J. Chem. Educ. 1978, 55(11), 722 (https://doi.org/10.1021/ed055p722.1).

ahh I see. Yeah, I'm not aware of SN1' and SN2', I'm not from the US so maybe that wasn't in my course, please share some links ;-; ? I did think of allylic rearrangement, I went though the wiki page but I want to know more about it. I don't really know how compound II is formed.

I do not have access to the experiment at home, but it is based on molecular modeling using SPARTAN-8 program, which allows to do Ab initio calculations. Nice lab to show allyl shift in theoretical manner.

@sam: I did some editing to my answer to clear some issues you still have in this reaction.

Assuming an $\ce{S_N}1$ mechanism, there really is no rearrangement of atoms within the carbocation.  When the chloride ion leaves, the resulting carbocation is allylic (which itself favors $\ce{S_N}1$), thus potentially electrophilic at both ends of a conjugated three-carbon chain.  The mixture of products then corresponds to the cyanide ion adding at either end of the conjugated carbons.

You are right. You may also note that in polar protic solvent, such as water or alcohol, compound II would form the major product because of the stability of allylic carbocation, whereas in polar aprotic solvent compound I would form major product, because SN2 is preferred in such solvents due to its  relatively poor ability to hydrate and stabilize ions.


text:
Absorption spectrum of common plastic polymers
I am looking for data regarding absorption spectrum of common used plastic polymers, in the range of 200- 1.000 nm
Moreover, I am looking for inorganic or organic compounds, which can be mixed together with common plastics (PC, ABS, PMMA, POM) and have an absorption peak at 450 nm

Do you want only around (or about) 450 nm or would something like this be acceptable?

It seems both question can be entries for a search. Ex: absorption spectra of pmma.

What @EdV suggests would be fine. We need different polymers, not only PMMA

@Francesco my meaning was that first you search and than eventually come here to ask.

@Alchimista, I can assure I did extensive search. As I wrote I am not interested only in the absorption spectra of the different polymers, but also in knowing if there is any special compound with a strong absorbance of light around 450 nm.


text:
What exactly is the idea of gaining stability by the various chemical species?
When we consider the alkaline metals for example, we say that an atom of a metal like sodium gains stability by losing an electron, however this doesn't seem to be a very meaningful statement. Either it should be like it gains stability and finds a lone electron to be a hindrance in being stable and thus loses it. Even then, on losing the electron, it gains a positive charge and thus become unstable in just another way. So how does the mechanism of attaining stability exactly work? How does a species know that some act or process is going to give it stability? Since it doesn't know this how does stability reach it?
Further, what exactly does it mean to be in chemical stability by losing/gaining/sharing electrons, forming ions/bonds[of various forms] or by changing chemical partners etc.? What is their purpose exactly? Is it to have the least energy possible? And why is it that some things are more stable than the others from the start[not the very start possibly]?
I wanted a notion of stability + the mechanism to get it and retain it. [in simpler words that one would likely use for addressing high school science students, please]

The first point is that Na loosing an electron does not become more stable. It can be if another species is involved. It happens many times that sentences take their status in a truncated form. (Similar to E = mc2 :))

Stability is much like patriotism. Whoever talks about it is most probably trying to fool you.

The enthalpy of formation of $\ce{Na}$ is $\pu{0 kJ/mol}$. The enthalpy of formation of the ion $\ce{Na+}$ is $\pu{-240 kJ/mol}$. So the ion $\ce{Na+}$ is much more stable than the neutral atom.

@Maurice Formation enthalpies of ions are placed on an entirely different scale with $\Delta H_\mathrm{f}(\ce{H+}) = 0$ by definition, see e.g. Atkins et al. Physical Chemistry. It is not right to compare the two numbers you have compared.

@orthicresol Hf of Na as equal to zero is even that of a piece of metal. I don't know how that comment came up.

Dear Lumbini A Tambat:
Thank you for pointing out one of the simple concepts that get overlooked in teaching. Teaching chemistry especially! Stability. With reference to what? It's easy to imagine so many ways to get un-stable, but how can you get more stable, if you are an atom?
Well, darn it, a sodium atom is quite stable, all by itself. You can shake it, heat it, freeze it, bang it (within limits), and it remains quite unchanged. Sodium metal at 0 $^o$C and 1 atmosphere pressure is in its ground state, as stable as it can get because there is no state more stable. This is Standard Temperature and Pressure. (https://www.thoughtco.com/difference-between-standard-conditions-state-607534) Of course, we have to complicate it a bit, so we also have Normal Temperature and Pressure (NIST uses 20$^o$C), and there are 21 different modifications of the exact temperature and pressure that constitute a standard or normal temperature and pressure. https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure
But the point is that there is some generally well-defined set of conditions under which we classify materials as being in the lowest energy state possible, or most stable, or in a ground state, from which we do all our calculations and reactions.
It is possible to take two materials, both in their respective (lowest energy) ground states and allow them to react to form a new material which is now more stable because it has released some amount of energy - if you consider the two original materials, e.g., sodium and chlorine, to be in their ground states, imagine that the energy released digs a hole to an even lower ground state. That lower energy state exists because Na$^+$ ions are strongly attracted to Cl$^-$ ions in a sodium chloride lattice, and that energy of stabilization was enough to knock an electron off a sodium atom and force it onto a chlorine molecule, splitting it into pieces, one of which become a chlorine anion.
Now we explain the final situation as having a stable sodium cation, so it must have stabilized (because it  can't get changed any more under ordinary chemical reactions). So now it's really stable. Unchangeable. Well, you could dissolve it in water and hydrate it, and it could become a little more stable, but not much. And the chlorine atom has an extra electron and seems quite stable: it doesn't explode or anything. But were you looking when the sodium and chlorine were reacting? They didn't look at all like they were getting stable - it would have been quite exciting!

http://derekcarrsavvy-chemist.blogspot.com/2016/12/lattice-energy-1-calculating-lattice.html
In the picture, NaCl is the lowest energy material. The sodium is "stabilized" by the chlorine. This level is the ground state for NaCl. The next line above is a ground state for two elements, sodium and chlorine. They are stable in themselves, if kept separate, but they have reactivity, which also conveys the idea of instability. So which is it? Stable or unstable? This is where the teaching function often breaks down: one word means two entirely different things, yet we manage to communicate successfully - if we have been forewarned. We still sometimes trip ourselves up.
In the picture, we start with stable Na (solid), put in energy to vaporize it to Na (gas), put in more energy to ionize it to Na$^+$ plus an electron, then shift to the chlorine, dissociate it to atoms by putting in energy, then let a chlorine grab the electron (giving off some energy), then allowing the sodium ion to condense with the chlorine ion to form a lattice of NaCl and give off the lattice energy, which makes the final product (NaCl) so very stable the we have tons of it in the sea and in salt mines all over the earth.
Now if you have a supernova available, you could turn the sodium into iron, which is really the lowest energy you can get to, but in most labs, the diagram illustrates several levels of energy, and conditions where the materials are stable, but could react to form something more stable. Sodium is reactive and stable at the same time, depending on the conditions.

This might turn to be useful many times.

https://hal.archives-ouvertes.fr/hal-00516229/document is an interesting reading. The misconceptions are so widespread that it is difficult not to blame school and teachers.

@Alchemista: Very interesting link (43 pp!). I suspect that is why lab work is so important. And demonstrations of effects that are not suitable for individuals in the lab so as to illustrate some examples with real, visible material, not just a chart or diagram. I was captivated in freshman college chem course by a demonstration of acetone vapor pressure in a mercury barometer: more acetone did not give more depression of the Hg column! How could that be??? Even more memorable than the H2 +O2 soap bubbles exploding best at a 2:1 ratio.


text:
What is the density of nitrogen at standard conditions?
I find diverging answers to what the density of nitrogen at standard conditions is.
By standard conditions I mean T=273.15 K and P=1 atm (=1.01325 bar).
Different sources state:

Wikipedia : 1.2506 g/L
Fluidat : 1.250 g/L
NIST : 1.2504 g/L
When I use ideal gas law : 1.2498 g/L (using molar mass as 28.0134 g/mol and gas constant as 8.31446261815324*10^3 J/(K mol)

Which one is more correct and why ?

The first 3 values probably rely either on measurements either on values provided by some of  real gas state equations. Try to calculate density from the van der Waals equation, probably by numerical finding the  root of the cubic equation.

Nothing is "more" correct. Something is correct or isn't.

Looks identical to me.

@NisargBhavsar As this is science, not math, with errors, assumptions, conditions, fuzziness and uncertainties, it is not so much black and white. Gramatically, you are right though, like there is nothing more optimal than optimal.

@NisargBhavsar not true in most sciences, and definitely not true in chemistry.

The deviation is in order of one-ten thousandths. So, I'd say, the values are pretty much equal.

One possibility for differences in measured density of an element is isotopic variation. For example, sulfur isotope abundance may vary, depending if the element is derived from soil, from sulfide minerals or as the native element from a volcanic vent. Nitrogen has two common stable isotopes: 14N and 15N. It is possible that the measurement of density varied because of natural variation.

However, it is possible the density varied because the definition of "Standard Temperature and Pressure" (STP) has changed over time (e.g., 101.325 kPa pre-1982, 100 kPa afterwards).

It is possible the density measurements varied because the definition of mass has changed over time.

It is possible the density measurements varied because of experimental error.

So if the discrepancies are a cause for concern in your work, check the source given for each measurement to help explain the differences.


text:
How can I save an EEM file on a Fluoromax4?
I'm trying to run excitation-emission matrices on some water samples, but I'm running into some software issues that the manuals aren't helpful for. In order to use R studio to extract my data, I need to save my output as a .eem file, but I don't see how to do that.
When I try to save my matrix, the only option I get is an .spc, which it doesn't even let me do that - I get an error saying I need to change my matrix into a worksheet. I can't even figure out how to do that.
Is it saving these eems automatically somewhere I can't find? All I've been able to do is save an ASCII version of my data, which isn't a format the eemR package lets me import.
Hardware: Horiba Fluoromax 4
Software: Horiba FluorEscence, R studio eemR, Matlab drEEM
Thanks

I have never used the software, but most spectroscopy softwares have an option to extract data as .csv files. See if you can do that and then convert the csv to eem, or read the csv with R studio. As .csv is almost an universal data format, most softwares should be able to read it too.

@ShoubhikRMaiti thank you; I have the .csv, but the specific matlab/R packages I'm using to analyze the data work best/only with the proprietary .eem files from different fluorometers - you specify the model you have and give it the file and it knows what to do from there - trying to go around this with a csv would make the process more time consuming. Starting to think it'd be faster than waiting on Horiba support, though.

Is this useful? https://www.rdocumentation.org/packages/hyperSpec/versions/0.98-20140523/topics/read.spc


text:
what is the use of finding out the radius of an atom
I happen to come across this, but what is the radius of an atom, velocity, KE, PE, etc.. actually used for?
like the formula r=0.529n^2/z and so on

Firstly, the atomic radius is a bit arbitary, you could say that it is the typical distance from the center of a nucleus to the boundary of surrounding electrons. Therefore more widely used definitions are the Van Der Waals radius, ionic radius, covalent radius etc. The notion of atomic radi as a 'length' value came about when it became possible to measure lengths using X-ray crystallography. They are normally measured in the bonded state since it is difficult to measure an isolated atom. The Radi therefore also sometimes depend on what they were bonded to during measurement too.
The atomic radius is a crude approximation, but it can give explanations and predictions of different phenomena eg. density of liquids and solids, diffusion of liquids through molecular sieves, the arrangement of atoms in crystal structures and the approximate size and shape of molecules.
The formula you refer to is the radius of the n'th Bohr orbit. Remember that the Bohr model only holds for one-electron atoms so it's applications are fairly limited. However using the Bohr model, rough energy estimations can be acheived for these atoms.
You can use calculated energy values of atoms and groups of atoms to realte back to the feasability of certain reactions occuring, products forming etc.  (from a thermodynamic approach). The Bohr model is what lead on to more compelx quantum chemical ideas and methods such as Hartree-Fock methods which allow for more accurate energy estimation of both atoms and molecules. Using this, the energies of specific orientations of molecules and atoms can be determined which can be use to estimate how likely a reaction is to proceed or what the energy of a trasnition state is!


text:
Does a reaction proceed through SN1 if rearrangement results in a stable carbocation?
While going through my book I came across the following problem:

I thought it was safe to assume that the secondary alcohol would give SN1(although secondary alcohols give significant amount of both SN1 and SN2) but I was sure that the 2nd reaction would proceed through SN2 since it was a primary alcohol.
But it turned out to be the opposite as my book said that both would proceed through SN1. Why does this happen?

The products shown were also rearranged. But isn't rearrangement a thing that happens after the carbocation has already formed?
And if the carbocation is 1 degree, then why would it form at the first place?
How does the compound know that it can rearrange once it forms the less stable carbocation, so it compromises some stability to gain more?

The acidic conditions make the SN1 pathway more favorable.

You are considering every step as distinct with sharp boundaries. However, the real molecules can exist in a range of possible states. For example, after the OH group is protonated the C-OH bond is weakened, and it sort of starts to look like a carbocation. Then CH3 can migrate, pushing out the OH2. Even tertiary carbocations are probably coordinated by solvent molecules in actual solution.

According to Advanced Organic Chemistry[1],

Neopentyl systems are typically resistant to nucleophilic substitution reactions. They are primary, so do not form stable carbonium ions, and the tert-butyl substituent effectively hinders back-side attack. [...]  Under conditions that
favor ionization, rearrangement usually occurs, and the products are derived from tert-amyl cation. Substitution reactions of neopentyl tosylate without skeletal rearrangement can be effected, however, by using good nucleophiles in hexamethylphosphoramide as solvent.

(emphasis mine)
The stereochemistry of the reaction was found as follows,

The use of optically active neopentyl-l-d tosylate allows
the stereochemistry to be established. Complete inversion of configuration was observed, again consistent with a direct displacement of the leaving group

This shows that the reaction takes place via an internal rearrangement which is similar to the mechanism seen in SN2 except that the $\ce{-CH3}$ group acts similar to the nucleophile in this internal rearrangement. After this, the reaction proceeds similar to a normal SN1 reaction. Its energetics diagram looks similar to an SN1 reaction as well.
Why not SN2? It is a primary alcohol after all. This is explained in J. Am. Chem. Soc. 1942, 64 (3), 543–546.

From the fact that 1-bromo-4,4-dimethylpentyne-2 is not hindered in comparison to 1-bromo-heptyne-2 it is concluded that the "neopentyl effect" is not capable of transmission through an unsaturated linkage and is hence not a chemical but a steric effect

This mentioned "neopentyl effect" refers to the lack of SN2 reaction seen even though the halide/alcohol is primary.
References:

Carey F.A., Sundberg R.J. (1977) Nucleophilic Substitution. In: Advanced Organic Chemistry. Springer, Boston, MA. https://doi.org/10.1007/978-1-4615-8882-5_5


text:
Comparing nucelophlicity order of acetate anion with different nucelophiles
We have hydroxide anion, methoxide anion, bisulfide anion , triphenyl phosphine,  azide anion , 1-pyrrolidine cyclohexene.
We want to tell which of the above species are better nucleophiles than acetate  anion.
We know that good nucleophile  is a weak base, (I know there might be exceptions like cyanide anion), so hydroxide anions, bisulfide anions, methoxide anions are much more basic than acetate ion, so acetate ion should be better nucelophiles than these. However, my question is that whether the rule of  good base   implies a bad nucleophile  will imply here? Also, I'm not sure which criterion do we use to decide for the other three species.
Thanks.

Is it all of them except the last?

@Ritam_Dasgupta, it is all of them, how can we compare the last one?

Dunno what made you think such rule even exists.


text:
Calculate Specific Heat given initial temp, final temp and equilibrium temp
I would appreciate if someone could enlighten me on this question.
Question:
A metallic object with a mass of 2 kg at 50ºC is placed in 10 L of water at 20ºC. When the
thermal equilibrium the temperature is 20.64 ºC What is the specific heat of the metallic object?

How I tried to solve it:

Correct answer: 455.6 J/kg ºC
Why would it be this? Where did I go wrong?
Thanks.

Using units would be a good start.

Hello, welcome, please check out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?.

You're using the wrong value for the specific heat of water, because we're dealing with kilograms and joules here, and not calories and grams. It should be:
$$4184\ {\rm J\over kg\cdot\ ^{\circ}C}$$
and the minus sign goes away because the heat that is lost by the metal is gained by the water, hence $\Delta Q_\text{water} = -\Delta Q_\text{metal}$.
Thus, we have:
$$c_\text{metal} = 4184\ {\rm J\over kg\cdot\ ^{\circ}C}\times 10\ {\rm kg}\times 0.64^{\circ}{\rm C}\times {1 \over 2\ {\rm kg}}\times{1\over 29.36^{\circ}{\rm C}} = 456.02\ {\rm J\over kg\cdot\ ^{\circ}C}$$

The principle of calorimetry states that in an isolated system such as a calorimeter, the total heat gained by the colder bodies is equal to the total heat lost by the hotter bodies. So since we are calculating heat lost by metal, you should take
$$\Delta T = \pu{50 ^\circ C} - \pu{20.64 ^\circ C}$$
What's more, the specific heat of water is $\pu{1 cal g-1 ^\circ C-1}$, so you should convert to SI units if you wish to obtain the answer in SI.

I've upvoted this because in part you have explained some of the answer but not all of it. Your expression for $\Delta T$ is not correct: you have $T_{initial} - T_{final}$. Please see my answer as an example of what constitutes a reasonable, full answer that addresses the OP's query.

@Todd you have done exactly the same calculation though. I have explained why I took $\Delta T$ as such, from the principle of calorimetry. You have done the same thing, but it is you who hasn't given any explanation for the -ve sign.

As far as I'm aware this 'principle of calorimetry' is just known as conservation of energy in most places of the world.


text:
Where do the electrons for which RTG's use for the Seeback effect come from?
I know that RTG's use Radioactive Isotope that emit Alpha particles generally (ie. NASA uses Pu 238), I also know that these Alpha particles are consiting of 2 Protons and 2 Neutrons, since the RTG has two thermocouple plates that force the protons and the electrons to the cooler side. I was wondering where the electrons come from? Also I was wondering how exactly does the Seeback effect utilize these positive and negative charges with the junction formed to create the electrical charge.

Did you look at the Wikipedia article on plutonium-238? That isotope has atomic number 94, so it has 94 electrons. When one of those atoms emits an alpha particle, the atom becomes U-234. Since uranium has atomic number 92, it has 92 electrons. And that is where the two electrons come from: 2 = 94 - 92.

The Pu is contained so the alphas don’t make it to the thermoelectric material. It just gets hot to drive the thermal gradient. No net charge is produced.

@EdV I was also wondering, in an RTG the electrons come from the decay and not just the free-flowing electrons from the metal of each Semiconductor right (as NASA uses Skutterudite to build the semiconductors)? I was also wondering for the two semiconductors, would one let in protons and the other electrons to create the volage diferential or just electrons for both semiconductors. Please refer to the photo of the diagram above and see if it is accurate. Thank you very much.

@JonCuster Same question as to EdV above but I was also wondering what material does NASA use in their MMRTG to create the semiconductors, I know they have to be out of two dissimilar metals and read a post on Wikipedia which wrote "The MMRTG design incorporates PbTe/TAGS thermoelectric couples (from Teledyne Energy Systems), where TAGS is an acronym designating a material incorporating tellurium (Te), silver (Ag), germanium (Ge) and antimony (Sb). But that they are trying to use skutterudite, since it needs to be made out of dissimilar metals how would that work?

Please look at this. The decaying Pu-238 yields U-234 plus helium gas. There are no electrons left over: the alpha particle gets two electrons that U-234 cannot hold onto. Alpha particles are powerful oxidizing agents: they can easily oxidize fluorine, for example. They cannot be allowed to oxidize the RTG components. So it is only the heat released by the radioactive decay that powers the RTG. As per the linked article, other non-radioactive heat sources can also be used. As for your diagram, I have no idea what it tries to show.

@EdV Thank you for the clarification, but I was still wondering that to create the voltage differential we have the hotter side ie. the heat from the RTG and the cooler side (space), the atoms vibrate less and scatter towards the colder end, I was wondering in the RTG's is that just electrons, just protons, or one of both in each Thermocouple for each semiconductor, and do they just come from the material of the semiconductor?

The semiconductors in typical examples are doped N type (so you have donor impurities resulting in electrons are majority charge carriers) and P type (so you have acceptor impurities resulting in positively charged  holes as majority charge carriers). No protons are involved. You should do some deeper digging on the Seebeck effect : lots of semiconductor physics books discuss its fundamental principles, and that is more than I can do so long after my last courses on that.

I just looked up skutterudites and learned they are promising thermoelectric semiconductors. So P and N doped skutterudites are the active elements. Also, a quick search for “seebeck effect” shows some nice cartoons illustrating how the temperature gradient drives both the electrons (in the N type semiconductor) and the holes (in the P type semiconductor) away from the heat source and toward the heat sink. So a potential difference is created and this gives the usable electrical energy output. So heat energy in, electrical energy out. I have nothing further to add.


text:
Why is osmium(VIII) oxide so toxic?
Osmium(VIII) oxide or osmium tetroxide is a volatile chemical compound with the formula of $\ce{OsO4}$. It is highly toxic for supposedly irritating mucous membranes, but I'm not too sure if it is that or because it is a really strong oxidizer.
PubChem rates it as corrosive and highly toxic and according to Wikipedia, the toxic effects of $\ce{OsO4}$ are as follows:

$\ce{OsO4}$ is highly poisonous. In particular, inhalation at concentrations well below those at which a smell can be perceived can lead to pulmonary edema and subsequent death. Noticeable symptoms can take hours to appear after exposure.
$\ce{OsO4}$ will irreversibly stain the human cornea, which can lead to blindness. The permissible exposure limit for osmium(VIII) oxide (8 hour time-weighted average) is $\pu{2 µg/m3}$

How is $\ce{OsO4}$ toxic in nature? What makes it so?

You seem highly interested in understanding what makes compounds toxic or reactive. The following few links might help with the first question: ccohs.ca/oshanswers/chemicals/oxidizing/oxiziding_hazards.html nature.berkeley.edu/~dnomura/pdf/Lecture6Mechanisms3.pdf http://www.organic.lu.se/education/Kemiteknik&Bioteknik/EnvirChem/7_ToxCompounds.pdf

Osmium tetroxide might be particularly damaging because it acts catalytically.

Metallic osmium is harmless but it reacts with oxygen at room temperature, forming volatile osmium tetroxide. Any other osmium compounds are also converted to the tetroxide if oxygen is present. This makes osmium tetroxide the main source of contact with the environment and human body. Exposure to osmium tetroxide cause harm to the skin, eyes, and respiratory system.

So, why makes osmium tetroxide so harmful? (TL;Dr: Because to its oxidization ability)

There is a paper1 which studied the effect of osmium tetroxide in human body and how it can be a lethal bioweapon. The following is the text from the paper:

Osmium tetroxide is highly volatile and penetrates skin readily, and
is very toxic by inhalation, ingestion, and skin contact. Airborne low
concentrations of osmium tetroxide vapor can cause lung congestion and
skin or eye damage.
[...]
Osmium tetroxide is a rapid, indiscriminative oxidizer that does not distinguish between organic tissue and inorganic materials. An
inhalational toxicity study with rabbits proved futile, because of the
rapid reduction of $\ce{OsO4}$ by the skin, hair, mucous membranes,
etc., as well as by the chamber walls. Inhalation, ingestion,
contact with skin and with mucous membranes may all result in severe
consequences. Due to its high vapor pressure (7 mmHg at 20 °C), most
exposures are to vapor. These can cause severe chemical burns to the
eyes, skin and respiratory tract. Very short-term contact with the
vapor may cause lacrimation, accompanied by cough, headaches and
dizziness. $\ce{OsO4}$ may cause irreversible blindness by turning the
cornea black. Symptoms may not be noticeable until several hours after
the exposures. Another severe delayed effect following inhalational
exposure is acute lung injury, which may be followed by
non-cardiogenic pulmonary edema Direct contact with osmium tetroxide
solution will turn the skin black (severe chemical burns due to strong
oxidizing properties). Painful burns or contact dermatitis may
result, depending on the concentration.

The paper compares the LD50 value of osmium tetroxide with other chemical warfare agent and thus proves how lethal it can be as a bioweapon:

The paper also notes that osmium tetroxide can rapidly reduced to relatively inert compounds by e.g. ascorbic acid or polyunsaturated vegetable oils (such as corn oil which contains double/triple bonds. $\ce{OsO4}$ can attack those bonds and can convert to non-toxic vicinal diols).
Reference

Makarovsky, Igor & Markel, Gal & Hoffman, Azik & Schein, Ophir & Finkelstien, Arseny & Brosh-Nissimov, Tal & Tashma, Zeev & Dushnitsky, Tsvika & Eisenkraft, Arik. (2007). Osmium Tetroxide: A New Kind of Weapon. The Israel Medical Association journal : IMAJ. 9. 750-2.
McLaughlin AIG, Milton R, Perry KMA. (1946). Toxic manifestations of osmium tetroxide. Br J Ind Med 3:183-186.

I upvoted, of course! Maybe ruthenium tetroxide will pop up next! ;-)

http://www.chm.bris.ac.uk/motm/oso4/oso4h.htm
https://www.epa.gov/sites/production/files/2014-09/documents/osmiumtetroxide_interim_ornl_jun_2008c.pdf
https://www.michigan.gov/documents/mdch-osmium_tetroxide_fs_109244_7.pdf
Osmium tetroxide is an incredibly strong oxidant: it is stronger as an oxidant than the halogens and also oxidizes halides back to their halogen form.  When it comes into contact with skin, it will readily reduce into osmium dioxide, causing irritation and leaving a noticeable black residue.
It is also notable for its volatility.  Osmium Tetroxide will sublime, which makes it incredibly easy for vapors to be inhaled through the nose or come in contact with the eyes.  This allows it to potentially oxidize and irritate vital organs the heart and lungs, as well as stain the cornea of the eyeball.  Those actions, in turn, will lead to the symptoms you describe.

I am new to StackExchange, so please do not hesitate to give feedback/criticism to my answer! :D

You should turn urls into links, in text or bibliography.


text:
Rearrangement in dehydration of alcohol
I came across the following problem in my book:

Now my confusion is how will cyclohexene be obtained? I read that primary alcohols dehydrate by E2 mechanism because of formation of a primary carbocation. Further I read in Solomons organic chemistry that rearrangement in Primary alcohols occur due to the reversibility of the reaction:

So I tried to figure out the mechanism:

So I am unable to find how cyclohexene will be obtained. For ring expansion to take place there must be positive charge on the alpha carbon but in E2 mechanism there is no possibility of carbocation formation

Concentrated sulphuric acid is a highly acidic medium and thus it very exclusively prefers the $E_1$ mechanism over $E_2$ mechanism and thus a carbocation is surely formed in this conditions.
The example you provide proceeds through the following mechanism:

Now the mechanism you provide, other than following the wrong pathway(very minor pathway) also has a major mistake. During the protonation of 1-methylenecyclopentane the tertiary carbocation is not more stable than the primary carbocation. The tertiary carbocation is $\mathrm{sp^2}$ hybridized and thus prefers a bond angle of 120° which cannot be provided by the cyclopentane ring as it is not flexible enough.
It prefers the very less angle strain in case of $\mathrm{sp^3}$ hybridized orbitlas but with less positive inductive effect over higher angle strain but more positive inductive effect. That's the reason the below equilibrium is highly backward favoured.

Can you cite a reference or evidence for the final point? I would have always said the tertiary cation was more stable, but perhaps I've been wrong in this assumption?

I agree. I would like to see a reference for this

(-1) for the last step. A reference would be appreciated since this seems very unlikely.


text:
In benzene and 1,3-cyclohexadiene, which one has greater C=C bond length?
In benzene and 1,3-cyclohexadiene, resonance will occur, so both will have a partial pi character. Technically, all bonds in benzene have partial pi character, whereas in 1,3-cyclohexadiene, some have partial pi character and the rest are pure sigma. So considering this, 1,3-cyclohexadiene must have a greater bond length. But the answer in the book says benzene.

Would be nice to point at the bonds you are looking at. Your text assume equal bond length which is true only in benzene (and is even the underlying answer to your question).

Sorry, I meant the C=C bonds. Have changed the question.

In benzene, the double bonds are not localized. There are no real double bonds. Each C-C bond is of an order situated between $1$ and $2$.

It would be interesting to know the sources cited in your question (both the book, as well as the primary literature) as reference because one complementary approach to the (assumed) spectroscopic determination is the analysis of the crystalline state.  While there are multiple records about pure benzene, it seems there is no public model of the crystalline structure of pure 1,3-cyclohexadiene ; on the other hand, there are multiple records where the two motifs are present simultaneously.
From those with little geometric distortion by organometallic bonding, a framework of 1,4-benzenedicarboxylates with incorporated (guest molecules) of 1,3-cyclohexadiene nevertheless appears instructive:

(CCDC entry IXODUV, based on Wang et al. in 2011InorgChem2028)
For the moderately distorted 1,3-cyclohexadiene, the intermolecular distances are $\pu{1.35 A}$, $\pu{1.45 A}$, and $\pu{1.33 A}$ along the sequence of double, single, double bond $\ce{C=C-C=C}$. In the p-terephthalate, the model describes all aromatic $\ce{C\bond{~-}C}$ bonds with $\pu{1.38 A}$.  Despite the constraints of the structure model, it is an example for the difference of aromatic  $\ce{C\bond{~-}C}$ bonds from conjugated double bonds.
In the context of your question, the paper is instructive because the framework of the host, the p-terephthalate, chemically stays the same and with only minor changes of the lattice constants if it is loaded with guest molecules of benzene, 1,3-cyclohexadiene, 1,4-cyclohexadiene, cyclohexene, or cyclohexane.

I'll guess: The question comes from an exercise book which is designed to "teach" rules of thumbs and educated guessing concepts. As such the target of this was probably to count some resonance structures with charge separation and seeing that the others they are equal in benzene and that is much better and whatnot and then conclude that there is more delocalisation in benzene than there is in the other one and hence the double bond in benzene is less of a double bond and must be weaker and therefore longer. Anyway, I really like your very scientific answer to such a terrible exercise.


text:
Should cyaphide create similar chemical compounds as cyanide?
The cyaphide anion $\ce{(CP)-}$ should be chemically similar to its nitrogen analogue cyanide. $\ce{(CN-})$ But unlike it's counterpart it readily polymerizes  so it has only been found in outer space naturally. However, it should form compounds that should be the same  like cyanides.For example sodium cyanide ($\ce{NaCN}$),would be called "sodium cyaphide" ($\ce{NaCP}$) and hydrogen cyanide ($\ce{HCN}$) would be called hydrogen cyaphide or as it's called methylidynephosphane ($\ce{HCP}$) and the anions: phosphaethynolate (cyaphate) ($\ce{PCO-}$) and thiophosphaethyonlate (thiocyaphate), ($\ce{PCS-}$) and also potentially the cyaphogen ($\ce{P2C2}$) have been observed but as far as I know the only reliable info I found on cyaphides were on Wikipedia and they were very short as well. Which source on information do I go with?
More information: https://en.wikipedia.org/wiki/Methylidynephosphane

Also I'm basing the nomenclature from the PCO-/PCS- anions from the cyanate and thiocyanate anions

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

Methylidynephosphane or phosphaethyne is known and it has been achieved in 1961 by Thurman Gier. He passed phosphine gas at low pressure over an electric arc produced between two carbon electrodes. Condensation of the gaseous products in a –196 °C (–321 °F) trap revealed that the reaction had produced phosphaethyne along with acetylene and ethylene.
$$\ce{PH3 ->[electric arc][carbon electrodes] HCP + H2C=CH2 + HC#CH}$$
The ions $\ce{PCO-}$ and $\ce{PCS-}$ are also known. From the abstract of this paper1:

A range of λ3-phosphaalkynes $\ce{P⋮C−R}$ $\ce{(R = F, Cl, O-, OLi, OH,}$
$\ce{OSiMe3, S-}$, $\ce{SLi}$, $\ce{SH, Me,}$ $\ce{SiMe3, H)}$ has been geometry-optimized at the HF, MP2, and CCSD(T) level of theory. Vibrational frequencies as well as 31P and 13C NMR chemical shifts were calculated and compared to experimental values of the large variety of λ3-phosphaalkynes prepared over several years via differing synthetic routes. With a few obvious exceptions, a good agreement of calculated and experimental parameters was found. This also opens up the possibility to make predictions on
spectroscopic data of compounds not yet synthesized or fully
characterized, e.g., $\ce{PCOSiMe3}$ and $\ce{PCSH}$. The bonding situation
especially of the negatively charged species $\ce{P⋮C−O-}$ and $\ce{P⋮C−S-}$ was
studied via NBO analyses to determine the degree of phosphaalkyne and
heterocumulene contributions to the mesomeric anions.
$$\ce{^{-}X-C#P <-> X=C=P- (X=O,S)}$$

Cyaphogen ($\ce{(CP)2}$) has not been synthesized yet. It is considered a pseudohalogen. Structurally, it is formulated as $\ce{P#C-C#P}$.
Currently, no binary cyaphides are known but numerous organometallic cyaphide complexes are known. Some of them are listed in the Wikipedia article.
Reference

Theoretical Studies of NMR Chemical Shifts and Vibrational Frequencies in λ3-Phosphaalkynes P⋮C−R, Klaus Hübler and Peter Schwerdtfeger, Inorganic Chemistry 1999 38 (1), 157-164, DOI: 10.1021/ic9811291


text:
change in bond Length due to resonance
for example, if we consider benzene ring ;

all the pi bonds will show resonance and therefore a partial pi bond character will be observed over all the bonds in the ring

my question is

is the c1-c2 bond length equal to c2-c3 bond length???
according to me since they all have equal pi bond character , the bond lengths should be equal
am I right?????

All the bonds in benzene are equivalent and have bond order $=1.5$ due to equivalent resonating structures. Hence, you are correct, all bond lengths will be equal.

In this specific case you are right. But don't generalise too much. In general resonance would tend to equalise the relevant bond lengths, but geometry and symmetry considerations shall be taken in account. But the underlying principle is this one.


text:
Preferred radical attack on a polyester
My question concerns the relative stability of a radical next to an ester oxygen compared to a radical next to an ester carbonyl carbon.

The stabilizing effect of the electron pairs of ether oxygen is known, but the situation may be different for an ester oxygen, which donates fewer electron density.
On the other hand, can the pi electrons of the carbonyl contribute to the stabilization of the radical next to it?
What is the preferred point for a radical attack on such a (poly)ester structure and, in your opinion, is there a clear preference? References are welcome.


text:
Bond formations in doping of semiconductors
I was reading about semiconductors and found out that semiconductors are doped with either 13th group or 15th group elements which are named p-type and n-type semiconductors, respectively. In n-type semiconductors I found out about the following:

Why does antimony form four bonds with silicon when it could just bond with three silicon atoms and have a lone pair instead of having an electron excited and have a free electron?


text:
Mechanism of combustion of hydrocarbons
While this question comes close, it doesn't quite answer my question.
Adding heat to oxygen & a hydrocarbon breaks the atoms apart and they reform into carbon dioxide and water. I get that.
I want to know specifically at what point the water and carbon dioxide are formed?
As I understand it the heat vibrates the atoms, breaking bonds.  Those atoms reform with oxygen forming stronger molecules ($\ce{H2O}$ and $\ce{CO2}$) which can withstand the combustion heat.  I reached this far with help from a Reddit thread.
Where that thread seemed to disagree is in my main question, on where/when the formation actually happens, it also raised another question of how long atoms are broken apart for.
In this article, it would suggest that free atoms are created which find oxygen:

$\ce{CH4 + 2O2 → CO2 + 2H2O + energy}$
[...] the arrow represents the moment the reaction happens. But there are a lot of interesting things happening that are hidden behind that arrow. A more detailed equation would look something like this:
$\ce{CH4 + 2O2 + a little energy →C + 4H + 4O →CO2 + 2H2O + lots of energy}$
The first arrow represents the breaking of the bonds, which requires energy. On the middle line are the atoms, now broken out of molecules and free to react. [...]

Which at least one of the Reddit thread contributors disagreed with.
So! (TL;DR)

When are the new molecules formed (cylindar, afterwards or both)?
How long are atoms "free", how long do they take to form new molecules, is it instantaneous or ... what?

Atoms are pretty much never free.

That's what the Reddit user said, but the article linked above disagrees.  You can see my confusion? :P

Ross beware that the mechanism(s) of combustion will be dependent on fuel type and engines. The correct word for you to search for is "mechanism of combustion of... in...". I am not a specialist at all but I can tell you that you might find perhaps hundred reactions going on. The formation of water and CO2 should be at the very end. Still, CO2 is prone of equilibrium with CO.

related https://chemistry.stackexchange.com/questions/14704/reaction-mechanism-of-combustion-of-hydrogen

I went away like I do with all SE questions, expecting trickled responses.  These are great, thanks.  I'll go through them and the suggestions...

Although some of the detail of the combustion reaction is provided in the duplicate, those details don't quite address the question as posed. While the proposed dupe is relevant, it is not an exact duplicate.

I agree, that's why I accepted an answer as well, it addressed my question more directly.

Combustion reactions are complicated even when they result in simple products
Part of the problem with answering the question "at what point are water and carbon dioxide formed" is that the reactions occuring in combustion are very complicated and not at all well represented by the simple versions shown in the question.
We can say for sure that the picture where all the components are atomised and then recombine is not the way the reaction works. This simplistic idea is part of the reason for disagreement on the reddit thread.
It is an incorrect view for many reasons but mostly because the reaction doesn't all happen at once. Take petrol (that's gasoline for americans) engines. The initiation of combustion is started by a spark and then (relatively slowly in chemical terms) progresses through the rest of the fuel/air mixture (if it were very fast the engine would explode rather than smoothly driving the cylinder, if it were only a little too fast the engine would "knock", damaging the cylinder). So we need to think–even before considering the detailed chemistry–about the (slow) steps involved in the reaction.
The major steps involved in most combustion reactions involve radical chains. Though the full detail is not understood (though this answer gives some more detail), it is thought that the initiating step involves a hydrocarbon reacting with an oxygen molecule to give a hydroperoxide radical and a hydrocarbon radical. Very simplistically this looks like:
$\ce{O2 + CH3(CH2)nCH2-H -> HO2. +CH3(CH2)nCH2.} $
But neither of those radicals is stable or long-lived. In an environment flooded with more hydrocarbons and oxygen, they can both react further to give more products and more radicals. For example, the hydroperoxide may also fall apart to hydroxide radicals which can react with parts of other hydrocarbon molecules to give water and yet more radicals. In the early stages, most reactions create more radicals; later in the process some radicals can combine into stable products (like water and carbon dioxide). Once the reaction is started it can be self-sustaining as the formation of those stable products release a lot of heat, enough to keep creating new radicals and new chains until the source hydrocarbons or the oxygen is depleted.
So, even thinking about the detailed, step by step, chemical reactions suggests that the process is messy. Even early on in the process some final products are being created, though this continues until the fuel/air is mostly used up. if you plotted the amount of final product over time you would see some very early and a rising amount until the reaction had finished. It is pretty meaningless to ask "at which point" the products are formed.
In short, the initiation does not "atomise" the components: it creates radicals (this doesn't need as much energy input as "atomisation"). They can continue to react creating chains of yet more radicals and some chains can stop when stable products are formed, even early in the reaction. The net reaction releases enough heat to sustain combustion while fuel remains. Final products start to form early in the process and continue to be formed until the air and fuel have gone.

Thank you for explaining how stupid I am in a very respectful way :)

@RossDrew there is nothing stupid in being used to just the "net" result of a chemical reaction, especially as we normally present reaction like combustion, some fermentation, etc. Spectacular cases are our respiration in detail or photosynthesis!

The hydrocarbon and the oxygen begin to combine when the spark initiates and they continue to combine in a burning, and not homogeneously (not an explosion - except if the combustion begins too early because of knocking, which occurs because the hydrocarbon is too sensitive to the heat and pressure build-up from the compression in the cylinder). Typically, essentially all of the hydrocarbon is burned before the exhaust valves open up, unless the rpm is very high.
That's the beginning. Then, because some of the hydrocarbon escaped burning, or was transformed into other unburned material, and undesirable CO and NO$_x$ was formed, further reactions occur over the catalytic converter, and finally, CO$_2$ and H$_2$O come out the exhaust pipe.
"Specifically at what point" is a question that does not have a short answer. The actual combustion in one cycle is a time line, a chain reaction over which many different types of reaction occur. A common thought is that explosions occur in the cylinders of an internal combustion engine, but the goal is to produce a rapid, even burning over several milliseconds so as to develop even pressure gradients to push the pistons. Knocking is really an explosion; not a terribly fast one, but one that could eventually overstress the engine. Even that explosion isn't a point in time.
In a car engine, we would be talking about femtosecond reaction times (10$^{-15}$ seconds) for individual molecules or perhaps less for specific reactions like ionization. The time and distance scale of reaction for an individual molecule could be plotted down to less than a micron and a microsecond, but in any real reaction, there is so much material involved that the "point" is totally spread out in space and time. The overall combustion reaction from beginning to end is a bit more than the few milliseconds in the cylinder because of the catalytic converter, but for most purposes, you could consider the reaction to take the whole time between the spark and the exhaust.
There is an extensive discussion of combustion on Wikipedia: https://en.wikipedia.org/wiki/Combustion

https://corels.ibs.re.kr/html/corels_en/research/research_0305.html


text:
Effect of short lived chemicals on global warming exceeding effect of carbon dioxide
From Wikipedia:  Global Warming Potential

GWP values and lifetimes:
Nitrous oxide lasts about 121 years in the atmosphere, causes about 280 times as much warming over 20 years, but about 300 times as much over 100 years.
HFC-23 lasts about 222 years, causes about 11,400 times the warming over twenty years, but about 13,600 times over one hundred years.

Carbon dioxide is widely believed to last at least 300 years, if not longer, in the atmosphere.
How can certain short-lived chemicals cause more global warming in the long term, relative to carbon dioxide, than they do in the short term?

CO2 is ~constant in the atmosphere (yes, increasing but…). Now release a burst of something that decays away - what is the impact of it over time?

Note that estimated halftime of atmospheric $\ce{CO2}$ is dozens of years. It is not that obvious, as it is constantly supplied by natural and human-triggered processes, so its level is a steady state with long term drift. So it should not be surprise those components are evern more effective in long-term, beincg compared to $\ce{CO2.}$


text:
Why do selenium , tellurium and polonium have more negative electron gain enthalpy than oxygen?
In my book(NCERT Chemistry Part I, Textbook for Class XI[1]) a list of values of electron gain enthalpies of various elements is given. According to that list, oxygen has less negative electron gain enthalpy than selenium, tellurium and polonium. Shouldn't it be the opposite because electron gain enthalpy becomes less negative down a group?

Reference:
[1]: Rajput, Anup, Kumar. Classification of Elements and Periodicity in Properties. NCERT Chemistry Part I, Textbook for Class XI, First Edition(October 2019 Reprint); Uppal, Shveta, Banerjee, Binoy, Eds.; National Council of Educational Research and Training: New Delhi, India, 2006; pp 90.

It's due to extremely small size of oxygen as compared to others in the group.

Electron Gain Enthalpy becomes less negative down the group.

This is true, as a general trend. But the 2nd period is an exception to this trend.
In p block, it is observed that the EGE of some 2nd period elements is actually less negative then their 3rd period counterparts.
In the case of oxygen, it’s EGE is even less than selenium, tellurium and polonium. It’s because oxygen has very small radius and thus have high electron density. Due to this, the incoming electron faces a lot of repulsion from the already present electrons.
So, it takes more energy to add an electron to an oxygen atom i.e. oxygen has less negative EGE. On the other hand, selenium and tellurium have much larger radius and thus the electron density isn’t very high. Hence, their EGE is more negative as its easier to add an electron to them.

Please take into account that my answer applies to Po only because your data supports it. I myself have not checked if it is true or not.


text:
What reaction happens between glucose and ammonium chloride when drying?
I have the following observations:

When I put a solution of glucose (80 g/ℓ) and ammonium chloride (20 g/ℓ) in a drying cabinet at 37°C (maybe a bit warmer) a brown substance forms when all (or most) of the water has evaporated.
When I add water to the reaction product, the substance dissolves within minutes (under shaking), but the resulting solution is brownish.
From this I conclude that a reaction happened and the brown colour isn’t just the colour of some salt crystal.
If I perform the same procedure on glucose solution or ammonium chloride solution alone, nothing similar happens, which indicates that it’s not a reaction with the air or the container. PS: If I redissolve the product of drying ammonium chloride alone, the result is acidic (pH of 5 or lower for 1 g/ℓ).
No browning happens if I replace ammonium chloride with sodium chloride, which indicates that chloride is not responsible for the reaction.
The same thing doesn’t happen in the presence of water.
Even when the solution has already been dried to a highly viscous sludge, I still don’t observe any browning.
It only happens in final stage of drying.

My questions are:
What reaction is this, why does it only happen when drying, and is there some way to prevent this, e.g., with a lower temperature?
I ideally want to get a dried residue containing glucose, ammonium, and chloride in my container.
Just filling in the powders doesn’t help, because I cannot do this efficiently and precisely.
What I found so far is this:

Typical browning reactions of sugar (Maillard reaction, caramelisation) require much higher temperatures or additional ingredients.
According to the paper Reactions of Free Sugars with Aqueous Ammonia, some reactions between sugar and ammonia happen at this temperature, but with a low yield and in the presence of water.

When a sugar solution turns brown (or black), it's generally a sign of polymerization of the sugar molecules triggered by dehydration reactions that create more reactive molecules. A classic example of this is caramelization, in which sucrose is dehydrated by heating and forms many types of furans and pyrans and the oligomers caramelan, caramelen and caramelin. The latter three all contribute to the brown color.
Although caramelization typically requires heat, a similar effect can also be achieved by use of an acid catalyst. A simple example is treatment of sucrose with concentrated sulfuric acid, which very quickly results in a solid black mess.
In your case, you've started with a nearly neutral solution of glucose in ammonium chloride. The solution is nearly neutral because it is a 1:1 mix of a moderate base (ammonia) with a strong acid (HCl). As you evaporate off the water, however, some ammonia evaporates as well, resulting in a slight excess of HCl and a lowering of pH. The combination of mild heat and acid is likely causing a caramelization-type reaction at a lower temp than is usually required. The ammonia might also enhance the color via a Maillard-type reaction.
A good review of carmelization is: Sengar, G.; Sharma, H. K. Food caramels: a review. J. Food Sci. Technol. 2014, 51 (9), 1686–1696. DOI: 10.1007/s13197-012-0633-z

The pH of 0.1 M NH4Cl is given as 4.6 in  https://scrippslabs.com/ph-of-common-reagents-at-room-temperature/. Altho this is lower than most values given on the internet, all agree that NH4Cl is acidic.

@JamesGaidis - I also would expect a solution of NH4Cl to be somewhat acidic, but OP said in a now deleted comment that it was pH 7 and went down to 5 during evaporation. I assume he adjusted the pH with a bit of base when he made the original solution or something like that. I wrote "nearly neutral" as a hedge to just keep it simple and because I wasn't sure exactly how he made the solution. I should have mentioned that  the relative weakness of ammonia as a base compared to the acidity of HCl means net acidic.

OP said in a now deleted comment that it was pH 7 and went down to 5 during evaporation – It turns out that my pH measurement did not work on NH₄Cl for some reason. I now repeated it with another method and it is indeed acidic (pH of 5). Presumably my dried and redissolved NH₄Cl solution has a much lower pH than 5 as the same problems should apply. However, I cannot easily measure that one with the other method. CC@JamesGaidis


text:
Carbocation formation and stability from bromo compounds

When heated,the following bromo-species all form carbocations via $\sigma$ bond cleavage. For each structure draw the corresponding carbocation and briefly describe how it is stabilised.

This is my attempt at answering the question:

I would like to know whether I have drawn the correct carbocation forms of each of these bromo-compounds and if my reasoning is sound.

Yes, both your answers and reasoning are correct. For the second compound a slight ambiguity might arise, because the $\mathrm{sp^2}$-hybridized carbons show $\text{-I}$ effect which would destabilize the carbocation, however considering the fact that there are not only  $\ce{2 Me}$— groups attached which show $\text{+I}$, they are also closer than one of the $\mathrm{sp^2}$- carbons, it is quite natural that the net inductive effect would stabilize it.


text:
Acid titration of sodium acetate in the presence of sodium ethoxide, is it possible?
I have an aqueous solution of ethanol of $30 - 40\text{ V%}$ that contains a few (unknown) $\text{%}$ ethyl acetate (ethyl ethanoate - $\ce{EtOAc}$).
I wonder if I can quantitatively determine the $\ce{EtOAc}$ content by back titration of a small excess of strong $\ce{NaOH}$.
The idea is to add a stoichiometric excess of $\ce{NaOH}$ to a known volume of sample solution, then reflux this mixture to achieve saponification of the $\ce{EtAc}$, acc.:
$$\ce{EtOAc}+\ce{NaOH} \to \ce{NaOAc}+\ce{EtOH}$$
The excess $\ce{NaOH}$ would form sodium ethoxide with the ethanol, acc.:
$$\ce{NaOH}+\ce{EtOH} \to \ce{NaOEt}+\ce{H2O}$$
Since as $\ce{NaOEt}$ is a strong base and $\ce{NaOAc}$ a weak one, I think it should be possible to titrate them sequentially.
Any thoughts would be appreciated.

As ethanol is much weaker acid than water, the second reaction would occur significantly only in NaOH solution in (almost) pure ethanol. Otherwise, equilibrium is pushed leftwards.

I though that too, later on.

The pH of 0.1 M NaAc is calculated to be 8.9. https://pubs.acs.org/doi/pdf/10.1021/ed079p29.1
The pH of NaOH will be much higher. The 30-40% of ethanol may have some effects, but in general, you should be able to add about twice as much NaOH as you think will react with all the EtAc, reflux and back titrate the NaOH. As Poutnik comments, there won't be significant NaOEt.
The reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH
If you can get some thymolphthalein, you could use that as an indicator.

https://en.wikipedia.org/wiki/PH_indicator
Or, you can use pH paper strips if you are satisfied with a cruder analysis.
Possibly a more intricate back titration might work: add about twice as much NaOH as you think will react with all the EtAc in your ethanol, reflux as above, then add the same number of moles of acetic acid as you added moles of NaOH. Then the reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH;   reflux;   + 2 HAc -->
NaAc + EtOH + NaAc + H$_2$O + HAc
Then back titrate the excess acetic acid with phenolphthalein (common enough - you must have some). If there were no EtAc in your ethanol, the pH of the solution after adding the 2 HAc will be about 8.9; if there is any EtAc in your ethanol, you will have excess acetic acid due to the saponification of the EtAc. And you back titrate that with NaOH. So this is sort of a back to back titration, or a back and forth.

Could you please use MathJax for the chemical equations? You have been on this site for long enough. The equations are really hard to understand as they are now.

In addition to the rightful suggestion by @ShoubhikRMaiti regarding mhchem, note that tables are essentially textual information and should be ideally presented as such, not as images (as long as it's technically viable, which is clearly your case). Also, note that "Ac" is a symbol for actinium or acetyl, whereas it looks like you are talking about acetate with standard symbol "OAc".

Thanks for your answer. Good news really on the non-formation of $\ce{NaOEt}$. I'll be titrating against a $\mathrm{pH}$ meter, I think.

Have you any suggestions for 'optimal' reflux conditions, to ensure complete saponification?

@Gert: You mean how long? Check the pH. Bring it to a boil, then cool it right down and check the pH at room temp. If there's a significant change, try it again to see if there is another change (maybe boil it for a few minutes). If no further change, you're done! Or, check it first with an artificial solution made up with reagent EtOH, EtAc and water, as a rehearsal. Maybe it doesn't even need to be boiled!

Maybe it doesn't even need to be boiled! There's actually some evidence for that.


text:
Do helium and neon form "stable" chemical compounds?
Helium and neon are the two most unreactive elements in the Periodic Table, but could they form  compounds with an element such as fluorine that won't spontaneously explode and decompose outside of the lab? Radon, xenon, krypton, and even argon already do form fluorides(well argon forms argon fluorohydride). It said in my Wikipedia source that helium supposedly made a compound with fluorine,but I'm confused with that. Is there any known evidence that this is true?
https://en.wikipedia.org/wiki/Noble_gas_compound
https://en.wikipedia.org/wiki/Helium_compounds

Yes, sure helium and neon are the most unreactive elements but it doesn't mean it can't form compounds. Given the condition, they are known to make clathrates, fullerenes etc. Have a good read: https://en.wikipedia.org/wiki/Helium_compounds https://en.wikipedia.org/wiki/Neon_compounds

I think the OP might be looking for the mechanism behind this reaction. Maybe he should include in the title


text:
Why are tri-organotin chloride compounds so dangerous?
Organotin compounds are compounds with the tin-carbon bond and some of them(specifically the tri-organo chloride ones) are as toxic as hydrogen cyanide. Why is this though as I don't really know some of the information that this link gave me. Can someone please help?
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1475111/#:~:text=The%20organotin%20compounds%20can%20be,of%20the%20central%20nervous%20system.

I Googled the title of your question - omitting the part after the question mark - and got 241,000 hits. Is that of interest to you?

Still getting downvotes even though this question is closed.You really have to put salt on the wound?

To get the gravity of how toxic organotin compounds can be, let us compare it with inorganic tin compounds. Most of the inorganotin compounds are non-toxic because of their low solubility and low absorption in human body. Workers mining tin oxide suffer from "stannosis", a benign form of pneumoconiosis without any tissue reaction or harmful reaction inside human body. But don't get me wrong! They are still harmful: High intakes of it may cause abdominal pain and anaemia. Inhaling tin hydride can cause nerve damage. Tin(II) chloride are found to induce vomit and diarrhea in cats. But they are nothing when compared to organotin compounds.
Organotin compounds are more harmful due to the fact that it can create small chemical and biochemical reactions in environment to convert to more toxic form. In human body, they are far more dangerous. The toxicity of the compounds depends on the structure of the compounds and its physicochemical action on human body. Small chain alkyltin compounds are by far the most toxic especially trimethyl and triethyl tin compounds because they can  absorb in gastrointestinal tract. Triethyltin particularly produces status spongiosus of the white matter of the central nervous system and causes encephalopathy and cerebral edema. Workers exposed to tributyltin reported to have suffered from dermatitis. Long chains alkyl and aryl tin are poorly absorbed and are less dangerous but still they are classified as neurotoxin.

So, why are organotin compounds toxic?

Organotin compounds are known to penetrate cell membranes due to its lipophilicity and causes damage to cell membrane, interrupt oxidative phosphylation and damage mitochondria. They can inhibit synthesis of heme oxygenase and can be immunotoxic and genotoxic. It is also reported that they are known to have carcinogenic or tetratogenic effect although not proved. You can read more on the adverse effect of organotin compounds on human body by reading paper 3 and 4.
References

CHAPTER 42 - Tin, ELENA A.OSTRAKHOVITCHM. GEORGECHERIAN, Handbook on the Toxicology of Metals (Third Edition) 2007, Pages 839-859, DOI: 10.1016/B978-012369413-3/50097-5.
Organotin Compounds Toxicity: Focus on Kidney, Carolina Monteiro de Lemos Barbosa, Fernanda Magalhães Ferrão and imageJones B. Graceli, Front. Endocrinol., 22 May 2018 | DOI: 10.3389/fendo.2018.00256
Toxicity of organotin compounds: Shared and unshared biochemical targets and mechanisms in animal cells, Alessandra Pagliarani, Salvatore Nesci, VittoriaVentrella, Toxicology in Vitro, Volume 27, Issue 2, March 2013, Pages 978-990, DOI: 10.1016/j.tiv.2012.12.002


text:
Can acetaldehyde be condensed with methyl vinyl ketone to form 2-Cyclohexen-1-one as per the Robinson annulation
I have seen that cyclohexanone can react with methyl vinyl ketone to form a fused bicyclic cyclohexenone, I am wondering if the same reaction can be used to produce cyclohexenone the single ring product.
I have since seen that the process can be acid catalyzed in Tetrahedron Lett. 1971, 12 (52), 4995–4996, might there be any obstruction to the formation of the cyclohexenone product?

There's just too much that can go wrong here. Acetaldehyde is very good at self-condensing when exposed to strong base.

What if the acetaldehyde was diluted and added drop wise to vinyl ketone slowly.

Acetaldehyde also polymerises under acid catalysis. I can find no references to a succesful reaction between acetaldehyde and MVK, though I do not have access to Scifinder.


text:
Deciding locants of double bonds vs. other substituents
After having decided that the keto-group is the highest priority, this is a ketone. However, does the double bond become C2, or does the amino become C3?
In locant placement, do we prioritize double bonds over substituents, or other substituents over double bonds? (Or is it just a matter of having the smallest starting number?)

5-aminocyclohex-2-enone
3-aminocyclohex-5-enone

Double bonds are preferred over substituents according to IUPAC nomenclature, so the first name should be correct.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) while numbering the parent chain, an unsaturation is preferred over a substitutent.

P-14.4 Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned  to them in the following decreasing order of seniority.
.. (a) fixed numbering of a ring or ring system;
.. (b) heteroatoms in chains and in rings and ring systems;
.. (c) indicated hydrogen;
.. (d) principal characteristic groups and free valences (suffixes);
.. (e) added hydrogen;
.. (f) saturation/unsaturation (hydro/dehydro prefixes and ‘ene’ and ‘yne’ endings);
.. (g) detachable alphabetized prefixes, all considered together in a series of increasing numerical  order;
.. (h) lowest locants for the substituent cited first as a prefix in the name;
..

As per the preference order the ketone will get the priority first as it is the principal characteristic group. Thus the name will end in an -one.
Now as the unsaturation is preferred over a substituent the numbering will be done in such a way that the double bond gets the lower number.
Thus the preferred IUPAC name of this molecule would be (5S)-5-aminocyclohex-2-en-1-one.


text:
Practices while using a boiling tube
When using a boiling tube to boil a liquid, I am frequently advised to move the boiling tube in and out of the Bunsen flame and when the liquid starts to boil, remove the tube from the flame for a few seconds before resuming. Why is this so?

That piece of glassware is called test-tube. "Why it is so" seems pretty obvious.

You should also keep the open end pointed away from you (and everyone else!) and, of course, wear safety glasses.

Usually, if people have experience of boiling water at home and think a while about it, they are able to answer it themselves. These "boiling tubes" are used primarily for ambient temperature testing, so they are called test tubes, or test-tubes.
Taking a test-tube in and out is for decreasing the heat flow to prevent violent overheating and the burst of boiling liquid shooting out of the tube. Similarly, removing the boiling test-tube from the flame serves the same purpose, avoiding violent outbursts. Bear in mind there is high ratio of provided heat versus the test-tube heat capacity, so warming up is fast and subsequent boiling can be violent.
Additionally, as a welcome side effect, moving of a test-tube keeps mechanical disturbances and traps air bubbles, creating boiling centers. This in large extent prevents overheating above the boiling point and then sudden boiling outbursts. This is especially important when heating liquid with strong tendency to be overheated, like strongly alkalic solutions. A burst of boiling hydroxide solution is nothing I would recommend to experience.
Another precaution is warming up rather the upper part of the liquid in a tilted test-tube. If an outburst happens in spite of being careful, less liquid is shot outside.


text:
Is Sodium Laurate (SL) the same as Sodium Lauryl Sulfate (SLS) and Sodium Laureth Sulfate (SLES)?
Is Sodium Laurate (SL) the same as Sodium Lauryl Sulfate (SLS) and Sodium Laureth Sulfate (SLES)?
There's controversy surrounding using SLS and SLES ingredients due to possible carcinogens. However, some soaps doesn't have SLS or SLES but has SL. Does SL have the same controversy surrounding it as SLS and SLES have? Is SL a skin irritant?

Short answer: no. SL, SLS and SLES are three different compounds. You can find their structures quite easily on Wikipedia and other general information sites.

Though health is not a topic https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4651417/#:~:text=There%20is%20no%20scientific%20evidence%20supporting%20that%20SLS%20is%20a%20carcinogen.&text=SLS%20is%20not%20listed%20as,Agency%3B%20and%20the%20European%20Union. and stuff like that...

No they are not the same thing.
Sodium laurate is the sodium salt of lauric acid $\ce{CH3(CH2)10CO2Na}.$
Sodium lauryl sulfate is the same compound as sodium dodecyl sulfate (SDS) and is the sodium salt of a sulfonic acid $\ce{CH3(CH2)11SO4Na}.$
Sodium laureth sulfate is a sodium salt of a sulfonate ester. There may be multiple ether linkages in the long chain giving several variants e.g. laureth-2 sulfate. The generic structure is $\ce{CH3(CH2)11(OCH2CH2)_nOSO3Na}.$