text:
Unusual hydrolysis of methyl 2,4,6‐trimethylbenzoate
Problem 223 from Conceptual Problems in Organic Chemsitry by Singh [1, p. 367]:

The given answer is b and I don't get why.
Usual ester hydrolysis would've given a.
I somehow managed to know that the mechanism is SN2, but why?
Also, if someone could provide the reaction mechanism, it would be wonderful.
Reference

Singh, D. K. Conceptual Problems in Organic Chemistry: For Engineering and Medical Entrance Examinations, 3rd ed.; Pearson India, 2013. ISBN 978-93-325-8207-1.

This is a known reaction mechanism for methyl esters with very good leaving groups (more commonly sulfonates, but benzoates as well). Here it's also favored by the steric hindrance of the carbonyl carbon. I'll try to dig up some refs when I have a few more minutes.

Here is a link to a paper that describes the reaction and various examples: http://folk.ntnu.no/audunfor/7.%20semester/Fysikalsk%20organisk%20kjemi/Artikler%20brukt%20i%20eksamensoppgaver/SN2%20Reactions%20with%20Carboxylic%20Esters..pdf

@Andrew you should make this an answer as you are undoubtly correct

The problem with this question is that you cannot be sure of the answer a priori. It would be a much better question to present the actual product and ask one to rationalize why that's the case.

In the typical reaction of hydroxide ion with a methyl ester, the hydroxide attacks the carbonyl carbon of the ester, resulting ultimately in elimination of methoxide. Given the basic conditions, the carboxylic acid formed immediately deprotonates to a carboxylate, and the methoxide is protonated to methanol. In older texts particularly, this reaction mechanism is referred to as BAc2, indicating "bimolecular base-catalyzed acyl-oxygen cleavage" (*) because the bond that is broken is between the acyl group and the oxygen of the methoxyl group [see for example Ingold, C.K. (1953) Structure and Mechanism in Organic Chemistry, Cornell Univ. Press].
An alternative mechanism is the BAl2 mechanism, designating "bimolecular base-catalyzed alkyl-oxygen cleavage". As the name suggests, in this mechanism, the bond between the alkyl group and the oxygen is broken. To do this, the hydroxide attacks the methyl carbon in an SN2 reaction, directly yielding the carboxylate and methanol products. This mechanism is also described in detail in Ingold's book and many other sources.
In the years since this mechanism was proposed, there has been some debate about how often it really occurs. Certainly for most esters, BAc2 is the dominant mechanism, and even when reaction by that mechanism is very slow, only a fraction of molecules, if any, react by the BAl2 option.
There seem to be two requirements for the BAl2 type reaction. First, the alkyl group must be methyl or a very ring-strained lactone. Other alkyl groups react much more slowly, as is observed in any SN2 reaction. Second, the carbonyl carbon must be sterically hindered to the point that the BAc2 mechanism is essentially impossible.
I am not certain that the tri-methyl benzoic acid in your question is hindered enough to react exclusively by the BAl2 mechanism, but it is certainly hindered enough that BAc2 would be substantially diminished.
You can find a more thorough discussion of these mechanisms in many textbooks, including the Ingold reference above. Another paper that may be useful is Mueller, P. and Siegfried, B. (1974) SN2 Reactions with Carboxylic Esters. Selective Cleavage of Methyl Esters. Helv Chim Acta 57:987.
*Note that since the net reaction in my example is $\ce{R-CO2Me + HO- -> R-CO2- + MeOH}$, the base is a stoichiometric reactant rather than a catalyst, so "base-catalyzed" is a misnomer. That terminology is used because the mechanism is the same if a different nucelophile is used, and the product is a non-acidic acyl derivative rather than a carboxylic acid. In that case, the base is indeed a catalyst.


text:
How does C−Br bond obtain energy through solvation?
I came across a question related to SN1 mechanism. In an example of SN1 reaction, i.e. hydrolysis of tert-butyl bromide to form tert-butyl alcohol, the following was given in NCERT Chemistry: Textbook for class XII, Part II [1, p. 304]:

Step I is the slowest and reversible. It involves the C—Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent. Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.
Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo SN1 reaction very fast because of the high stability of 3° carbocations. We can sum up the order of reactivity of alkyl halides towards SN1 and SN2 reactions as follows:
$$
\def\CL{{\large\text{Tertiary halide; Secondary halide; Primary halide;}~\ce{CH3X}}}
\ce{->[\text{For}~\mathrm{S_N2}~\text{reaction}][\CL]} \\
\ce{<-[\hphantom{[\CL}][\text{For}~\mathrm{S_N1}~\text{reaction}]}
$$

How does C-Br bond obtain energy through Solvation of bromide ion ?
References

NCERT. Chemistry: Textbook for Class XII. Part II, Reprint: August 2019 Bhadrapada 1941; M. Siraj Anwar, Shveta Uppal, Arun Chitkara, Bibash Kumar Das, R.N. Bhardwaj, Mukesh Gaur, Series Eds.; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-716-7. (PDF)

@andselisk Thanks for completing reference. I mentioned it because you have mention ACS style so I thought that we need to mention abstract of ACS research paper

No prob. You can find a brief summary with examples regarding using ACS citation style here or in the PDF from the corresponding chapter in the earlier edition of ACS Style Guide textbook. A CSL style file for your reference manager can be found on GitHub.

These release of energy is due to the neutralization between the oppositely charged halide ions and protons which get attached during solvation and hence they get stabilized by releasing energy.

I think the terminology is a little bit misleading in the book. They probably meant to say that the carbocation and bromide ion pair is stabilized by solvation by water. This means that the C-Br bond can break in a rate limiting step. Imagine that there was no water, the reaction was in vacuum, then you would never get a stable ion pair, because the electrostatic attraction would always push them towards each other.

@ShoubhikRMaiti. Yeah It sounds reasonable.

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text:
Disproportionation of silver(I) in aqueous solution

Use the following equations to predict whether or not $\ce{Ag+}$ ions will disproportionate in solution:
$$
\begin{align}
\ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\
\ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V}
\end{align}
$$

I used the method $E_\mathrm{cell} = E_\mathrm{red} - E_\mathrm{ox}$ and thought that $\pu{0.80 V}$ would be more negative. Therefore, it is oxidised and the reaction would be thermodynamically feasible.
However, my textbook and another webpage has a different answer. I am a bit stuck on how to get the right answer. Please someone explain me.

It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.
A generic Latimer diagram
$$\ce{A ->[$E_1$] B ->[$E_2$] C}$$
posesses the following properties:

If $E_2 > E_1,$ then $\ce{B}$ is thermodynamically unstable and disproportionates to $\ce{A}$ and $\ce{C}.$
If $E_2 < E_1,$ then the mixture of $\ce{A}$ and $\ce{C}$ is thermodynamically unstable and synproportionates to $\ce{B}.$

Now we can visualize your problem using a Latimer diagram
$$\ce{Ag^2+ ->[\pu{+2.00 V}] Ag+ ->[\pu{+0.80 V}] Ag}$$
and the condition of disproportionation resulting from the application of Nernst equation:

if the potential to the right of the species is higher than the potential on the left, it will disproportionate.

Since $E^\circ(\ce{Ag+(aq)/Ag}) < E^\circ(\ce{Ag^2+(aq)/Ag+(aq)}),$ disproportionation is thermodynamically unfavorable, and silver(I) can be considered stable in solution.
You can get the same result from a linear combination of both equations written for the disproportionation:
$$
\begin{align}
\ce{Ag+(aq) + e- &-> Ag} &\quad E^\circ_1 = \pu{+0.80 V} & \tag{1}\\
\ce{Ag^2+(aq) + e- &-> Ag+(aq)} &\quad E^\circ_2 = \pu{+2.00 V} &\quad|\cdot (-1)\tag{2}\\
\hline
\ce{2 Ag+(aq) &-> Ag + Ag^2+(aq)} &\quad E^\circ = \pu{-1.20 V}
\end{align}
$$
Since resulting $E^\circ = \pu{-1.20 V} < 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ > 0,$ and the disproportionation of silver(I) in solution can be considered a thermodynamically unfavorable process.

In this question, the equations given are:
$$
\begin{align}
\tag{1} \ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\
\tag{2} \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V}
\end{align}
$$
Now, since we know that $\Delta G$ is additive, we can use this property to proceed. (This is also the proof for why $E_\mathrm{cell} = E_\mathrm{red} - E_\mathrm{ox}$)
As the first step, we first find the value of $\Delta G$ using the formula
$$\Delta G = -nFE$$
So, for the first reaction, we see that $\Delta G = -0.8 \times 96500 = \pu{-77,200 J}$
For the second reaction, similarly we get $\Delta G = \pu{-193,000 J} $
Now, rewriting the two equations using $\Delta G$ instead of  $E$ in ($1$) and ($2$), we get:
$$
\begin{align}
\tag{3} \ce{Ag+(aq) + e- &-> Ag} &\qquad \Delta G = \pu{-77,200 J}\\
\tag{4} \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad \Delta G = \pu{-193,000 J}
\end{align}
$$
Now, the final reaction that we need is
$$ \tag{5} \ce{2Ag+ -> Ag + Ag^{2+}}$$
This can be achieved by subtracting ($2$) from ($1$). Now when we subtract the two, due to the additive property of $\Delta G$ we can simply subtract the $\Delta G$ of ($4$) from the $\Delta G$ of ($3$)
Doing so, we get:
$$
\begin{align}
\ce{2Ag+ -> Ag + Ag^{2+}} &\qquad \Delta G = \pu{115,800 J}
\end{align}
$$
So, we could end the question here, since we can see that the value of $\Delta G$ is positive and so the reaction is not spontaneous. However, since the question was asked in terms of $E_\mathrm{cell}$, we can convert this into $E_\mathrm{cell}$ using the given formula relating $\Delta G$ and $E_\mathrm{cell}$. We get:
$$
\begin{align}
\ce{2Ag+ -> Ag + Ag^{2+}} &\qquad E_\mathrm{cell}= \pu{-1.20 V}
\end{align}
$$
As you can see the cell potential is also negative. So, there was a mistake made in finding the right manipulation of the chemical reactions given.

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text:
What is "protein redox conformation"?
From an article by Frye et al. [1]:

Reduced glutathione (GSH) is the major intracellular redox buffer and is essential in free radical scavenging, redox homeostasis, maintenance of protein redox conformation and regulation of redox sensitive enzyme activity.

What is "maintenance of protein redox conformation"? Is it just fancy phrasing for "keeping proteins properly reduced, but not too much reduced, and properly oxidized, but not too much oxidized?
Reference

Frye, R. E.; Rossignol, D.; Scahill, L.; McDougle, C. J.; Huberman, H.; Quadros, E. V. Treatment of Folate Metabolism Abnormalities in Autism Spectrum Disorder. Seminars in Pediatric Neurology 2020, 100835. DOI: 10.1016/j.spen.2020.100835.

Ha, ASD seems to be quite a popular topic these days:) The term also appears in a 20 year-old publication https://doi.org/10.1046/j.1432-1327.2000.01601.x.

It seems like poor wording. It probably means "maintenance of equilibrium between protein redox conformations".


text:
Reaction to form amino acid

$$\ce{CH2=CH-CH=O ->[CH3SH][(1,4 addition)] Product (P_1) ->[(i) NaCN][(ii) NH3\\(iii) H3O+] Q_1}$$
Correct statement regarding $\ce{Q_1}?$

$\ce{Q_1}$ is essential amino acid
$\ce{Q_1}$ is acidic amino acid
$\ce{Q_1}$ is nonessential amino acid
$\ce{Q_1}$ on Lassign test gives prussian blue colour with $\ce{FeSO4}$ and $\ce{FeCl3}$ solution

By 1,4 addition (as mentioned in question) carbon-3 (taking carbon as carbon-1 of aldehyde group) will have $\ce{SCH3}$ added to it and oxygen must have $\ce{H}$ added to it.
After $\text{P}_1$ is formed, it can again tautomerize from an alcohol to an aldehyde.
The $\ce{^-CN}$ can now attack the aldehyde. Looking at the answer we can ay that the product must be an amino acid and so the $\ce{NH3}$ group and $\ce{^-CN}$ must be present on same carbon because then the $\ce{^-CN}$ will hydrolyze to give a $\ce{-COOH}$ group.
But how is $\ce{NH3}$ supposed to attack a carbon having $\ce{^-CN}$ and a hydroxyl group. Can we make the alcohol a leaving group somehow?
Or is the number of carbons in chain 1 less than what it should be? What should be the correct answer and approach for this problem?

Did you consider tautomerisation that could take place?

@safdar yes considered it after you mentioned and edited the question with a bit more attempt.

The starting compound is an organic compound known as acrolein. $\ce{CH2=CH-CH=O}$
Now your first reaction would be:
$$\ce{CH2=CH-CH=O ->[CH3SH][1,4 addition]CH3-S-CH2-CH=CH-OH} $$
Now, this compound tautomerizes to form methional
$$\ce{CH3-S-CH2-CH=CH-OH <-> CH3-S-CH2-CH2-CH=O}$$
Now, the second step in your reaction is a named reaction known as Strecker Amino Acid Synthesis. This converts an methional (in this case) into methionine.

Now, onto the question at hand, it asks about certain properties of methionine. Methionine is an essential amino acid. So A is correct. Since methionine doesn't have any extra acid or base groups, it is neutral.
As for the Lassaigne Test, if the compound had only nitrogen, it would have shown a prussian blue color. However since there is sulphur as well, it shows a violet colouration. So, the answer is A.

However in strecker amino acid synthesis reaction they have added NH3 in first step and -CN in the next. And its given vice versa in the question. So will we still get the same product? Btw thank you for the explanation

Cyanide adds to the imine of the aldehyde. Ammonia doesn't displace hydroxyl in a cyanohydrin but ammonia will reverse the formation of the cyanohydrin.

@user55119 So can I sum this up as after we add cyanide Methional adds up the cyanide to it's aldehyde group , however upon addition of NH3 , some percentage of reaction goes back and then Streaker Amino Synthesis is followed.

No. The ammonia condenses with the thioaldehyde with loss of water to form an imine (Schiff's base). Then the cyanide adds to the imine.


text:
Cyclic voltammetry for irreversible reactions
I'm new to electrochemistry and I was wondering two things, sorry if these two problems are trivial but I just can't find anything close to this in the literature.
First, the following reaction is irreversible, right? an intermediate Q is formed:
$B + e^{-} \rightarrow Q$
and another reaction follows where it is converted to a different product X:
$Q \rightarrow X$.
Then I'm supposed to sketch a cyclic voltammogram (CV) of this process, but I was taught that for irreversible processes a CV is not useful.
I looked pretty much all day for a similar reaction and the closest I could find is in "A Practical Beginner’s Guide to Cyclic Voltammetry" where the following reactions are presented:
$\text{Ox} + e^{-} \rightleftharpoons \text{Red}$
$\text{Red} \rightarrow Z$
These are defined as EC coupled reaction: reversible electron transfer Er following by an irreversible homogeneous chemical reaction Ci.
The ones I am given are not the same, I think. But I have no idea how to sketch a CV for that process. Am I interpreting the given reactions wrongly?

Look at https://en.wikipedia.org/wiki/Cyclic_voltammetry. In the cathodic scan, the scan voltage goes from more positive  to less positive w.r.t. the reference potential. Then you get a reduction of B to Q. Then Q irreversibly goes to X, which is not electroactive. So the reverse scan shows no oxidation peak current because Q is gone and X is not electroactive, i.e., it cannot be significantly oxidized in the CV situation.

Thank you for your reaction. What you describe seems like an irreversible process, which should look something like this, I think (see figure in question, don't know how to add it here).

Yes, that is correct.

Alright. I'm glad I asked this, otherwise I would have assumed that the first reaction was irreversible. So it is an entirely irreversible reaction then. Thanks!

Yes, I found a similar explanation in this source (see below). At slower scan rates the cyclic voltammogram would look similar to what I posted in my question. Thanks for the clarification!
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/JASDL/Courseware/Analytical_Electrochemistry%3A_The_Basic_Concepts/04_Voltammetric_Methods/A._Basics_of_Voltammetry/02_Potential_Sweep_Methods/b)_Cyclic_Voltammetry/iii)_Chemical__reactions__coupled__to__electron__transfer

Corrected version of previous comment: Just to be clear: the first scenario has two irreversible steps, while the second has a reversible redox followed by Red going irreversibly to Z. If that last step is slow relative to the reversible redox, then you would get anodic peaks. But repeated cycling would show all peaks gradually getting smaller as Z accumulates and does nothing in the CV situation.


text:
What temperature is really required to make fused quartz?
Glass transition temperature of fused quartz is 1200 °C. Does this mean it can be made by heating sand to 1200 °C, or does the melting point of $\ce{SiO2}$ has to be reached (~1700 °C) for that to happen?

See here for glass transition temperatures.

Most "glass" is not pure quartz (which is silicon dioxide). Usually many other ingredients are added to lower the melting temperature and/or to make the properties different.

Fused silica is not the same as fused quartz. Fused silica fluorescence cuvettes are an example. Even high purity natural quartz crystals have significant levels of impurities, e.g., aluminum.

Glass is not necessarily made of pure silica $\ce{SiO2}.$ Ordinary glass is a mixture of at least $50\,\%$ $\ce{SiO2},$ plus smaller proportions of $\ce{CaO},$ $\ce{Na2O},$ $\ce{K2O},$ and $\ce{Al2O3},$ with impurities like $\ce{Fe2O3}.$ Pyrex has also some $\ce{B2O3}.$
Glass has no defined melting point. It becomes like a very viscous paste at around 800 °C. Mixtures have melting points that are always much lower than their components.

Is explicitly stated im talking about quartz glass, which is 100% silica. I'll change it to fused quartz to avoid such misunderstandings


text:
Conductometric titrations: dilution effect
Since the conductivity of a solution changes linearly with the electrolyte concentration, during a conductometric titration the volume increases with each addition of the titrant aliquot. For this reason, each $\chi$ value should be corrected for the dilution effect.
Which of these two is the correct formula?
$$\begin{equation}\tag{1} \chi_{\text{eff}} = \chi_{\text{read}} \cdot \dfrac{V_i + V}{V_i} \end{equation}$$
or
$$\begin{equation}\tag{2} \chi_{\text{eff}} = \chi_{\text{read}} \cdot \dfrac{V_i}{V_i + V} \end{equation}$$
where

$\chi_{\text{eff}}$ = the effective value of $\chi$
$\chi_{\text{read}}$ = the value of $\chi$ read by the device
$V_i$ = the initial volume of the solution
$V$ = the volume of titran reagent added

"$\chi_{\text{eff}}$ = the effective value of $χ$" — for some reason this way of defining a variable reminds me of Lem's sepulki:) (since $χ$ is not defined anywhere else in the post)

Assuming we started from a initial volume $V_i$ of about 100 ml, some considerations must be made: at the end of the titration, the final volume, following the addition of the titrant reagent, will be more than 100 ml. Remembering that conductivity decreases linearly with dilution, the value provided by the instrument is influenced by the effect of dilution. Specific conductivity and concentration are directly proportional to each other
$$\chi = c \dfrac{n}{V}$$
with $c$ a proportionality constant and $n$ the total number of moles of free electrolytes in solution. The conductivity read by the instrument by adding a $V_a$ amount of titrant reagent will be
$$ \chi = c \dfrac{n}{V_i + V_a} $$
while the correct conductivity in absence of dilution is given by the formula
$$ \chi_{\text{eff}} = c \dfrac{n}{V_i} $$
so
$$ c \cdot n = \chi_{\text{eff}} \cdot V_i $$
replacing
$$ \chi = \dfrac{\chi_{\text{eff}} \cdot V_i}{V_i + V_a} $$
$$ \chi \cdot (V_i + V_a) = \chi_{\text{eff}} \cdot V_i $$
we get
$$ \begin{equation} \chi_{\text{eff}} = \chi \cdot \dfrac{V_i + V_a}{V_i} \end{equation} $$


text:
Vinegar reaction
please could someone help me. I poured white vinegar over a small hole where ants where coming out in my patio steps. There was some sand on the step and it started to bubble and fizz. We have been laying MOT one sub base and grano dust which may have gone on to the step also. I just wanted to know if it is safe and whether it is ok to leave.

What is "MOT one sub base" and "grano dust"? Do you have amazon links or an msds or something?

MOT Type 1 sub base appears to be made from limestone (which, chemically speaking, is mostly calcium carbonate).
Assuming limestone and vinegar are the only materials present, the relevant reaction is that of calcium carbonate with an acid (vinegar):
$\ce{CaCO3 + 2H+ -> Ca^{2+} + H2O + CO2}$
i.e. Assuming that nothing else is present, the gas is carbon dioxide. You will, however, slightly dissolve the sub base, so pouring large quantities of vinegar down there might not be desirable. And if your surface paving is limestone too, you’ll mark the paving.

On a non-chemistry note, from a vegetarian who used to have an ant farm as a kid: be nice to your ants. ;-)

Ants are not spared while making formic acid ;-)

Heavy CO2 gas likely suffocates ants. So, it was possibly also the CO2 gas that encouraged the ants to escape.

@NilayGhosh Historically, true! Not these days though.


text:
Comparing explosive properties of mercury(II) cyanate and mercury(II) fulminate
I found this question online on Jiskha Homework Help:

Of the compounds mercury(II) cyanate, $\ce{Hg(OCN)2},$ and mercury(II) fulminate, $\ce{Hg(CNO)2},$ one is highly explosive, the other is not. Explain.

My teachers says that most explosives are made of nitro compounds as abnormal expansion in $\ce{N-O}$ bond has been observed in nitro compounds above a certain temperature. So, fulminate will be more explosive as it contains $\ce{N-O}$ bond.
According to Wikipedia, explosiveness of nitro and other nitrogen containing compounds (like $\ce{NI3})$ is because of possibility of forming highly inert and stable $\ce{N2}$ molecules. Thus slight amount of energy causes breaking of relatively weaker $\ce{N-O}$ bonds and highly exothermic formation of $\ce{N2}$ triple bond making it act like an explosive.
Is my teacher correct? The reasoning given on Wikipedia also points to fulminate but some other concept may be at play here. So, which is correct? Also what would be the reaction mechanism if any?

"Obligatory" Breaking Bad reference: Fulminated mercury Wiki. As for the question, I bet there is an answer in one of Klapötke's published works:)

@andselisk That Klapötke is alive to write with all fingers and faculties intact is a majestic testament to either walking the razors edge with skill or armed with an unseemly supply of sheer luck. Or both.

See also Azidoazide azide: Chemistry in its element
12/11/20 by Chemistry World
Web player: https://podcastaddict.com/episode/116426632
Episode: https://traffic.libsyn.com/secure/chemistryinitselement/Ciie_Azidoazide_azide.mp3?dest-id=405840
A compound so explosively unstable that nobody has been able to measure how sensitive it is without it, well, exploding.

This can be attributed to the formal charges of fulminate ion and cyanate ion. Structural stability of a compound sometimes depend on formal charge.

We observe that formal charges of fulminate ion is higher than cyanate ion for which fulminate ions are said to have a less favorable molecular configurations which leads to facile decomposition of the fulminate ion.
Fulminate vs cyanate stability has been demonstrated by using their corresponding silver compounds. Here is a video demonstration.

In fulminate different minor structures then the ones you included are necessary. Also need for formal charge means virtually nothing on itself. .

I understand that you are saying that fulminate is less stable and thus more explosive, but what the abnormal expansion of N-O bond that my teacher said, have you ever heard of it?

One possible explanation is that compounds with formal postive charges on certain elements, such as nitrogen or chlorine, that are directly adjacent to formal negative charges AND with these charges not being able to be intramolecularly "neutralised" (were it not for formally "hypervalent" resonance contributors) are often explosive. This is evident from the two-year-old answer by Nilay Ghosh- cyanate has no such issues, while fulminate does, the latter being an explicit example in the paper.
*Before you comment things like "hypervalency has been disproven a long time ago" please read this paper.


text:
Hyperconjugation and inductive effect on carbocations
(I know there are many related questions, so please tell me which posts can solve my doubts before you close this question, thank you.)

$1$. $\ce{^+CH2F}$ is very unstable. This is because $\ce{F}$ is very electronegative, pulling the electron density from the already electrophilic carbon (empty $\text{p}$ orbital), thus increasing the positive charge.

For $\ce{CH3CH2+}$: E.N. of $\ce{H}$ = $2.2$, $\ce{C}$ = $2.5$, so the carbon in the methyl group is partially negative - like the $\ce{F}$ in $\ce{^+CH2F}$. To my understanding, the methyl group should be electron-withdrawing instead of donating, thus decreasing its stability, but this is false, why?
Some say that the partially negative carbon can donate some electron to the carbocation to increase stability, if it is true, why can't $\ce{^+CH2F}$ do the same thing?
$2.$ I know hyper-conjugation can form partial orbital overlap ($sp^3$ - $s$($\sigma$) for $\ce{H}$ & empty $p$ orbital in carbocation), thus delocalizing the electron density giving stability. However, the $\ce{C-C}$ $\sigma$ bond is always rotating, so from my view, hyper-conjugation is a bond that is continuously forming and breaking. Therefore, how come this is the predominant reason for carbocation stability?

Yes in hyperconjugation, the sigma bond of carbon with the carbon having positive charge rotate continuously and there is continuous bond forming and bond breaking. But as a result, the positive charge is not just on the one carbon, but on  alpha hydrogens also. So there is more dispersion of positive charge and this stabilize the carbocation.

@Manu You mean in as there are more alpha hydrogens present to overlap the empty p orbital (positive), thus better stabilise the charge? But can this 'override' the seemingly de-stabilization effect from my first doubt? Please correct me if I am incorrect, thanks!

In CH2F+, F is more EN than C, hence it attracts e-density, while in ethyl C+, both C have same EN, so they share excess charge to each other.

Inductive effect is weak as compared to hyperconjugation, so the instability caused by inductive effect is less than the stability imparted by the hyperconjugation, due to more dispersion of positive charge.

Can I clarify like this, the stability is achieved by 1. hyperconjugation 2. partial negative carbon on methyl group, donating electron density to positive charge? Thank you all :)


text:
Reactivity comparison between benzaldehyde and acetaldehyde
One of my teacher says that $\ce{Ph-CHO}$ is more reactive than $\ce{CH3-CHO}$ in performing nucleophillic addition reaction because $\ce{—Ph}$ is planar and thus offers less steric hindrance as compared to $\ce{—CH3}$ which is tetrahedral.
My other teacher gives the reason that the partial positive charge developed on the carbonyl carbon is resonance stabilised in $\ce{Ph-CHO}$ which makes the carbonyl carbon more prone to nucleophillic attacks.
What I think is that $\ce{—Ph}$ has its −I effect which deprives the carbonyl carbon of some amount of its electron density and thus makes it more prone to nucleophillic attacks.
I want to know which reason is more correct in explaining the reactivity of  benzaldehyde over acetaldehyde or if there is some other reason please specify.
The main thing I want to know by this is that whether planar substitutents offer less steric resistance than tetrahedral or not, and whether a partial charge on carbon can be stabilised by resonance or not.

The first reasoning reminded me of the prank often performed on first-graders. Shall one get a stuck glass joint during a lab practicum, the person is often told to soak the glassware in benzene, and here comes the sanity check. A first-grader ask why, and receives an explanation that planar benzene rings can easier penetrate the space between glass surfaces and wedge a joint at once. Student accepting such explanation usually don't do well during the first exam session.

$\ce{-CH3}$ has an A-value of 1.7 whereas $\ce{-Ph}$ has an A-value of 3. I'm also inclined to mark this as a duplicate of This answer

Be sure to check for resonances before inductives. Also, I'm not able to get the sense of "partial carbocation"?

@RahulVerma ; by partial carbocation I am trying to indicate the partial +ve charge on the carbonyl carbon

I'm confused in your initial lines where you write, "substitution nucleophillic addition elimination". I can't see which one are you asking?

@RahulVerma; actually I’m talking about the attack of Nucleophile on the carbonyl carbon.

If you're having 12th NCERT Chemistry part 2, then go to pg 358. Read it carefully, and you'll see why your teachers are wrong (really sorry for saying so)

I suspect that your second teacher has the right idea. Resonance stabilization renders benzaldehyde less susceptible to nucleophilic attack at the carbonyl group than acetaldehyde. Which brings up an additional question. How would you prove it?

@andselisk Do you mean first-year college students or first grade elementary students (6 year olds)? ;-)

@KarstenTheis Oops, I mean university students in their first year. Sorry for the lousy (plain wrong) choice of terminology, I still sometimes fail to figure out a proper term from the mishmash of education systems.


text:
Reaction of Silicon Dioxide with Hydrofluoric Acid
I learnt that $\ce{SiO2}$(Silicon dioxide) doesn't react with any acid except $\ce{HF}$. So what is special about $\ce{HF}$?
Why does only $\ce{HF}$ reacts with $\ce{SiO2}$ even though $\ce{HF}$ has a very high bond energy compared to other acids in that group?

Also: https://chemistry.stackexchange.com/questions/51223/can-hydrofluoric-acid-etch-glass-at-a-ph-of-7-0

If you are using comparative bond strengths to determine if a reaction is favorable, you need to consider the bonds in the products as well as those in the reactants. Although H-F has a stronger bond than H-Cl and other halogen acids, the Si-F bond is one of the strongest single bonds known, so forming four of them in the product makes the reaction thermodynamically favorable. Si-Cl bonds are not nearly as strong, and the difference is greater than 4x the difference between HF and HCl.

It is surprising that Silicon halides behave differently with water according to the choice of the halogen fluor or chlorine. The following reactions are known to happen : $$\ce{SiO_2 + 4 HF -> SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -> SiO_2 + 4 HCl}........ (2)$$ These equations are exactly the opposite from one another. Why does reaction ($1$) not proceed in the other way like ($2$) ? I don't know, but I would like present a personal interpretation. Maybe it is due to the fact that reaction ($1$) proceeds in two steps, with the intermediate formation of $\ce{H_2SiF_6}$. Maybe ($1$) proceeds like that $$\ce{SiO_2 + 6 HF -> H_2SiF_6 + 2 H_2O}$$ $$\ce{H_2SiF_6 <=> SiF_4(g) + 2 HF } $$$\ce{H_2SiF_6}$ is well known to be formed if $\ce{HF}$ is in excess. And of course HCl cannot form $\ce{H_2SiCl_6}$, because the chlorine atom is too big to act as a ligand this octahedral structure. $\ce{H_2SiCl_6}$ does not exist.
This is a personal idea. I would be pleased to hear about any comment.
Sorry for presenting these lines as an Answer. Alas the text is much too long for a comment...

But HF is a weak acid. How did it react in the first place? I'm confused

@Maurice I also think this is more like a comment, but I wanted to point out a formatting tip. Instead of using multiple dots like ........ for spacing out reaction numbers, please use \tag{<number>} macro. This way the numbers are perfectly aligned to the right margin and there is no distraction. Example: $$\ce{SiO2 + 4 HF -> SiF4 + 2 H2O}\tag{1}$$ $$\ce{SiO2 + 4 HF -> SiF4 + 2 H2O}\tag{1}$$

And if you add \label{<unique id>}, you can refer to the equation from the text with \eqref{<unique id>}. LaTeX writers usually use the following prefixes for <unique id> eqn: for equation, rxn: for reactions, tbl: for tables. Example: $$\ce{SiO2 + 4 HF -> SiF4 + 2 H2O}\label{rxn:sio2hf}\tag{1}$$ and later in text \eqref{rxn:sio2hf} for the dynamic reference.

@R.Anusha The reactions of HF are not about acid strength. They are more about the specific ability of fluoride to attack silicon dioxide. Acid strength has nothing to do with this reaction.

@andselisk what would be the right place to start learning about LaTeX and markup?

@Safdar There is MathJax basic tutorial and quick reference and TeX.SE. Just as pretty much always, one learns by doing. If you want to go further down the rabbit hole, find blogs and sites of interest that are using MathJax, and read books about LaTeX and literate programming. Just remember that MathJax ≠ LaTeX, they just share some markup commands, macro names and default fonts. Under the hood they are vastly different entities.

@maurice Your reasoning can be summed up as there being a kinetic barrier to formation of SiCl4, but a quick check of thermodynamic data shows that SiF4 is thermodynamically favored over SiO2, while SiCl4 is not. The key difference is the strength of the Si-F bonds compared to Si=O and Si-Cl. Much greater than the difference in bond strength of HF vs HCl

@ andselisk Thank you for your help in writing LaTeX symbols. You are an angel.

@ Andrew. Thank you for the thermodynamic data


text:
Steric effect of t-butyl group on SN1 on a cyclohexane ring
This is a question from GRB Kota Question Bank Organic Chemistry, Chapter 3, Reasoning type, Q. 4:

Assertion: trans-1-t-butyl-4-chlorocyclohexane is less reactive than cis-1-t-butyl-4-chlorocyclohexane towards an SN1 reaction.
Reason: More the steric factor near the leaving group, the higher the leaving group tendency.

The answer given is:

 Both the assertion and reason is correct and the reason is the correct explanation for the assertion.

The answer given feels wrong.
My reasoning for saying so is because the rate determining step in an SN1 is the formation of a carbocation. Due to this, the steric effect seems irrelevant. However, when the intermediate is same for both compounds undergoing substitution, I assume that the compound having greater stability would be less reactive since the threshold energy required would be greater.
Using the above statement, we can say that trans-1-t-butyl-4-chlorocyclohexane would be less reactive since both the groups are in equatorial position making it more stable whereas in its cis-isomer the t-butyl group would be equatorial whereas the chloride group would be axial.
Due to this, I presume that the assertion is correct. However, the reason seems vague as it does not mention whether the reaction taking place is SN1 or SN2 since this could change with respect to the reaction involved.
Is there anything wrong with my above reasoning? Can the validity of the reason be proved/disproved beyond doubt?

Two cyclohexane rings with no leaving groups are not reactive towards substitution.

@Zhe, I have rectified the issue.. It was chloride and not methyl.

An SN1 reaction will occur faster when the solvent is able to solvate the cation formed,thus compensating for the energy required in breaking the carbon-leaving group bond. Steric hindrance will inhibit this solvation. Does this hint help in answering your question?

@YusufHasan So, the reason is correct.. Thanks for that. However how does this make it the correct explanation?

Also, I am pretty sure that the "steric factor" as stated in the question is a textbook erratum: that term generally refers to the ratio between the experimental value of the rate constant and the one predicted by collision theory as stated here. I guess they meant to say steric hindrance

Also, the t-butyl group is in the farthest possible position, how does it affect it in this case? Or is the size so large that it is affected?

In the cis isomer, the solvation of the cation will be somewhat sterically inhibited as compared to the trans isomer. Therefore,theoretically, the trans isomer should be more susceptible to the rate-determining step(RDS) of the SN1 reaction(i.e. the formation of carbocation) than the cis isomer, and hence show a greater percentage of the SN1 reaction than the cis one

The t butyl group might be big enough to disrupt the solvation sheath that could have been formed over the cation even when it is that far away. Anyway,since this is a relative comparison,we are only interested in seeing how each factor plays out with respect to each other. The absolute percentage of the SN1 reaction doesn't matter as such here

Let us continue this discussion in chat.

@YusufHasan: I doubt that solvation would matter, by seeing the structure of cis and trans forms of 1-t-butyl-4-methylcyclohexane. As, the t-butyl group is on same side, so I think intermediate would be identical. And, therefore I prefer to look steric factors.

@RahulVerma I am talking about the solvation which would be present even before the C-Cl bond is broken. The breaking of the C-Cl bond doesn't happen on it's own; the Cl of the C-Cl bond is solvated by the positive end of the polar protic solvent,and these weak bonds which are formed b/w the solvent and the leaving group will provide the energy to ultimately break the C-Cl bond itself. This solvation will definitely be affected by the fact that whether the butyl is on the same or opposite side as the C-Cl. Anyway, the steric factor argument presented by the book doesn't make sense on it's own

...The presence of a sterically larger group can maybe increase the rotation barriers,and therefore reduce the number of conformers available,but I don't think it can have such a drastic effect so as to somehow weaken the C-Cl bond and allow the leaving group to leave on it's own accord

Hmm... I get it now. I too think that the reason in wrong.

@YusufHasan Can you post a definitive answer on this.. seems as though the question is pretty confusing.

@Safdar I would say that Zhe's answer provides the complete reasoning comprehensively. I myself had forgotten about Hammond's postulate and so was trying to answer this using indirect methods. As a final note, I feel that the book has got the correct answer with a somewhat incomplete explanation, as it is not immediately obvious why steric hindrance might lead to a quicker reaction(which is why I made a mistake in understanding it ). I would label the assertion as "correct", and the reasoning as "partially correct, but more elaboration is required, especially a reference to Hammon's postulate"

I would argue that the reasoning makes sense.
There are some important things to discuss here.
The cis isomer ($C$) and the trans isomer ($T$) will both react via the same carbocation intermediate. This means the energies of the intermediate are the same. Chloride has an A-value of 0.43, so $C$ is higher in energy than $T$. Since the formation of cation is endothermic, we may invoke the Hammond Postulate: the first step has a late transition state that structurally looks like the intermediate and is similar to it in energy. Finally, we conclude that the transition state for the reaction from $C$ has a lower barrier and thus a faster reaction.

Now, there is no way to "prove" that this or any other rationale is correct. There is no way to prove any model is correct. That's not how science works. You can only have consistent models that are good enough for now. They may even be "right," but you will never know. On the other hand, you could a reject the model based on inconsistency with experimental evidence, but there's no contradictory experimental evidence here.


text:
Calculation of pH in acid base reaction
This problem(from ionic equiibrium) states:

We are given $\pu{40 mL}$ of $\pu{0.1 M}$ $\ce{HCl}$ and $\pu{10 mL}$ of $\pu{0.45 M}$ $\ce{NaOH}$ in two separate beakers. These two solutions are mixed. Find the $\mathrm {pH}$ of the resultant solution formed.
Consider the concentration of the ions formed by self dissociation of water to be negligible.

My method
Number of moles of $\ce{H+}$ ion in the first solution is $\pu{0.004 mol}$ and the number of moles of $\ce{OH-}$ in the second solution is $\pu{0.0045 mol}$.
Now, after mixing the two solutions, the resulting concentration of $\ce{H+}$ ion becomes $\pu{0.08 M}$ and that of the $\ce{OH-}$ will become $\pu{0.09 M}$. Now these $\ce{H+}$ and $\ce{OH-}$ ions will react with each other such that the equilibrium constant of their reaction becomes $10^{-14}$.
So I formed the equation
$$(0.08-x)(0.09-x) = 10^{-14}$$
From here we can calculate $x$ and subsequently we can get the concentration of $\ce{H+}$ ion in the final solution.
I want to know is this the correct way to solve the problem?
The answer to this problem is given as $12$, which I doubt can be obtained from my way of doing the problem.

While this may be in the section on "ionic equilibrium," this is simply a limiting/excess reagents problem.

@Zhe can you please help me with an answer on how to solve this problem and why is my method incorrect.

Instead of solving it using Kw try using the principle of neutralisation $\ce{H+ + OH- -> H2O}$

@Safdar Ok let me try, but why is it wrong if I solve using Kw?

It isn't wrong. However solving this equation would lead to an approximation since you have to keep in mind the difference of orders of magnitude and is overcomplicating the question.

i think there is no need of ionic equilibrium , you have the concentration $\alpha$ would be 100% so just need to calculate which one is in excess and that will decide ph.

Hint: There are 0.5 millimoles of $\ce{OH-}$ in the final solution. Can you calculate the $\mathrm{pH}$ now?

No, you should not do the problem your way.
Given
$$(0.08−x)(0.09−x)=10^{−14}$$
You can solve for $x$ here, but its value will be quite close to $0.08$. This is the only way to make the product a small number, by making the one of the multiplicands very small.
However, $x$ will be so close to 0.08 that will have trouble figuring out the value of $0.08 - x$, which is what you want. Therefore, the right answer is obscured by your method of solving the problem.
In this problem, you can work around that by calculating the pH value from the hydroxide ion concentration, which, by the argument above, will be roughly $0.01\ \mathrm{M}$, consistent with a pH of 12. However, in the most general case, you may not be able to do that to determine what the final equilibrium concentrations are.

(+1) Very nice job explaining to the OP why their method is not optimal! ;)

As stated by Zhe, it is difficult to solve the equation as the value of x would be very close to $0.08$. However there is a workaround, if you wish to solve it this way using $K_\mathrm{w}$ from the side of the $\ce{H+}$ ion. The equation that you have is:
$$\tag{1} (0.08-x)(0.09-x)=10^{-14}$$
Instead of solving for $x$ here, we rather say $Y = 0.08-x$. Here $Y$ acts as the final concentration of $\ce{H+}$ in the solution (not counting dissociation of water). Using this, we get equation ($1$) to become:
$$Y(0.01+Y)=10^{-14}$$
Now, we assume that since the product is a very small number,  $Y + 0.01  \approx 0.01$. Now equation ($1$) becomes:
$$0.01Y=10^{-14}$$
Therefore, $Y = 10^{-12}$, pH is equal to $12$.


text:
Role of intermediate in determining the exothermic or endothermic nature of reaction
I was given a question to answer:

Which of the following reactions is “not” exothermic?

$\ce{CH3-Cl + CH3-CH3 -> CH4 + CH3-CH2-Cl}$
$\ce{CH3-Cl + (CH3)3-CH -> CH4 + (CH3)3C-Cl}$
$\ce{CH3-Cl + CH2=CH-CH3 -> CH4 + CH2=CH-CH2-Cl}$
$\ce{CH3-Cl + CH2=CH2 -> CH4 + CH2=CH-Cl}$

Now, what my teacher told was that; the 4th reaction proceeds through a very unstable free radical as compared to the other three reactions. Therefore the 4th reaction must not be exothermic.
But I have a doubt that stability of an intermediate in any reaction can only give us an idea of the activation energy required to reach the intermediate from the reactant via a transition state. Whereas the stability of product (with respect to the reactant) determines whether a reaction is endothermic or exothermic.
So how do we check here product stability to determine exothermic or endothermic nature of the reactions?
Or if something is wrong in what I think then please correct it.

Well, it may be true that #4 is the answer but not necessarily for the reason you state. These rxns. are, for the most part, hypothetical and are known as desmotic reactions. Heats of formation of the components are important. See: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/homodesmotic/homodesmotic.html

The difference of altitudes of 2 valleys (exo/endothermic reaction) is independent on the altitude of the mountain range between them ( energy of intermediates ).

You can get heats of formation here for free: https://webbook.nist.gov/chemistry/

You should accept the answer you find most helpful, it shows your appreciation for the effort people put into helping you out. If you are still unsure or unstatisfied, say so, so someone can try to clarify.

A typical method for determining whether a reaction is endo or exothermic is to compare the heats of formation ($\Delta_f H$) of the products to those of the reactants. If we do not have measured values for the enthalpies of formation, we can get a fairly good estimate by summing up typical values for the bond enthalpies of the bonds in the compounds.
When using bond enthalpies, we can simplify the calculation by considering only those bonds that are broken or formed during the reaction, since the other bonds appear in both the products and reactants and so cancel out when we look at the difference.
Thus,
$$\textrm{Enthalpy of reaction = }\Delta_r H = \textrm{[sum of enthalpies of bonds broken]} - \textrm{[sum of enthalpies of bonds formed]}$$
[Here I'm using the standard convention that bond enthalpies are reported for the breaking of bonds and so are positive values.]
Comparing the four reactions given, we can simplify even further by noting that all four reactions involve breaking of a $\ce{H3C-Cl}$ bond and formation of a $\ce{H3C-H}$ bond. Since those occur in all four reactions, we can ignore them for the purpose of comparison.
We now have reduced the problem to a simple comparison of the change of an $\ce{R-H}$ bond to an $\ce{R-Cl}$ bond.
You are correct that it does not matter what intermediates might occur on the reaction path. The radical stability is relevent, however, because the bond enthalpy is defined as the change in enthalpy needed for homolytic bond cleavage, ie
$\ce{A-B -> A. + B.}$
Since the products are radicals, the stability of the radicals gives us information about the bond enthalpy. It is for this reason that comparing the stability of radical "intermediates" is a way to tell which of your given reactions is not exothermic.

Your answer is very good, but I have still got one problem, i.e. , we could easily compare the enthalpy of bonds broken based on intermediate stability but how do we compare the enthalpies of bond formed for which we would have to think of product stability. So how do we do it ?

Now that you know that you are comparing $\ce{R-H -> R. + H.}$ to $\ce{R-Cl -> R. + Cl.}$, consider what factors determine how much energy it takes to break a bond homolytically. More polar bonds generally take more energy, which is why changing C-H (less polar) to C-Cl (more polar) is typically exothermic. What do you know about the relative polarity of the four possible C-H and C-Cl bonds in the question?

@Andrew has provided a lucid answer to the OP's question. I decided to calculate the heats of formation of the four isodesmic reactions by two methods ---namely--- via heats of formation (ΔHfo) and bond dissociation energy (BDE) to see if one of the reactions is actually "not" exothermic. The four reactions are shown in Chart 1 below with their respective heats of formation readily available at the NIST site. All four reactions turn out to be exothermic by computing ΔHfo(products) - ΔHfo(reactants) or (C + D) - (A + B). (continued)
Chart 1

These data are reproduced in the first four entries of Chart 2. The NIST site lists other values for the heat of formation of vinyl chloride (Entries 5 & 6). The three values are in red in column ΔHfo D.
Another approach to the solution of the thermicity of these reactions employs BDE's. BDE's compiled by Ellison1 are all available except for the carbon-chlorine bond of allyl chloride. The reactants involve breaking bonds. The are assigned a positive value. The products have bonds formed and are given negative values. Summing up the rows gives the heat of reaction. Both methods give good agreement for reactions 1 (in yellow) and 2 (green). The BDE method for reaction 4 gives a value of -1.7 kcal/mol which is in good agreement with heats of formation values when the ΔHfo of vinyl chloride is taken as 9.1 kcal/mol rather than 7.0 kcal/mol.
If one is willing to accept that the results by the two methods are compatible, then the BDE for the carbon-chlorine bond in allyl chloride may be computed. The heat of reaction for reaction 3 from the former method is -4.5 kcal/mol. Using this value (in red) can be taken as the heat of reaction for reaction 3 by the BDE method. Therefore, the BDE for the carbon-chlorine is estimated as 67.6 kcal/mol. This value should be less that for methyl and ethyl chloride given the stability of the allyl radical over that of the methyl and ethyl radicals.
Chart 2

S. J. Blanksby and G. B. Ellison, Acct. Chem. Res., 2003, 36, 255.


text:
What is the mechanism of AChE inhibition by Onchidal?
A naturally-occuring neurotoxin, called Onchidal, produced by a species of sea slug acts as an irreversible inhibitor of acetylcholinesterase. The structure of Onchidal is presented below:

How and why does Onchidal act as an AChE inhibitor? What does the adduct of Onchidal with AChE look like? Why is the inhibition of AChE by Onchidal irreversible?

Q.1: How and why does Onchidal act as an AChE inhibitor?
Onchidal is an antipredatory, feeding deterrent product secreted by marine organisms such as Onchidella binneyi and Elysia cripata, not having the protection of an external shell and thus relying on the production of a defensive secretion. [Gavignin et al., 2000, 1996, 1997; Ireland et al. 1978] To mammals, it is a neurotoxin that acts as an irreversible inhibitor of AChE, with a Kd of 300 μM, where it forms a covalent species with the enzyme.[Abramson, 1989]
Q.2: What does the adduct of Onchidal with AChE look like?
This molecule was isolated from the mucus obtained by squeezing the mollusc O.binneyi. [Ireland et al. 1978] It is a sesquiterpene acetate containing an α/β-unsaturated aldehyde and an acetate ester as shown above. It has also been speculated that onchidal’s active form is believed to be ancistrodial.

Abramson (1989) proposed multiple modes of inhibition:
(i) Schiff base formation with a lysine;
(ii) attack on the beta-unsaturated carbon in conjugation with the aldehydes by a Cys, Lys, His, or Tyr; and
(iii) the formation of 1,4-dialdehyde, which reacts with a Lys to form a pyrrole covalent adduct.
A look at the AChE co-crystal structure with an inhibitor (paraoxon in this case to locate the active site) as deposited with RCSB as 5HF5.pdb (DOI: 10.2210/pdb5HF5/pdb) , indicates that the closest reactive ε-amino group of a Lys residue is ~18  Å away. This distance is presumed too much for a covalent attachment of onchidal to AChE active site and thereby affecting AChE’s catalytic function. Likewise, there is no free sulhydryl available in the active site either, so a different residue may be involved as the point of attachment. The following figure below shows AChE with paraoxon shown with speres; Lys in green; and Cys in orange. The distance is in Å on the following figure I prepared:

Stoddard (2014) proposed three docking poses for onchidal. Here is the figure S2 from their paper:

Then, Stoddard (2014) further noted:

Only one His, the active site His-440, is located in the gorge. Several Tyr are located in the gorge; Tyr 442, 334, 130, 121, and 70. Based on the crystal structure and Abramson’s findings, the most probable mechanism is attack of the beta-unsaturated carbon in conjugation with the aldehyde by a Tyr; however a candidate Tyr for this mechanism was not identified from our docking analysis.

I’m wondering whether non-specific cellular esterases may convert Onchidal to an alcohol? If true, my speculation may change the docking analyses and ensuing inference. In any case, I’m not sure we have the full picture of Onchidal molecular mode of action, just yet.
Q.3: Why is the inhibition of AChE by Onchidal irreversible?
Onchidal contains an aldehyde and vinyls. Such reactive functionalities have literature precedents describing irreversible adduct formation with specific protein residues.

References
Abramson S.N., Radic Z., Manker D., Faulkner D.J., Taylor P. Onchidal (1989) “A naturally occurring irreversible inhibitor of acetylcholinesterase with a novel mechanism of action.” Mol. Pharmacol. 36:349–354.
Gavagnin M., Mollo E., Castelluccio F., Montanaro D., Ortea J., Cimino G. (1997) “A novel dietary sesquiterpene from the marine sacoglossan Tridachia crispate. Nat. Prod. Lett. 10:151–156. doi: https://10.1080/10575639708043731
Gavagnin M., Mollo E., Cimino G., Ortea J. (1996) “A new γ-dihydropyrone-propionate from the Caribbean sascoglossan Tridachia crispate” Tetrahedron Lett. 37:4259–4262. doi: 10.1016/0040-4039(96)00811-8
Gavagnin M., Mollo E., Montanaro D., Ortea J., Cimino G. (2000) “Chemical studies of Caribbean sacoglossans: Dietary relationships with green algae and ecological implications.” J. Chem. Ecol. 26:1563–1578. doi: 10.1023/A:1005526526884
Ireland C., Faulkner D.J. (1978) “The defensive secretion of the opisthobranch mollusc Onchidella binneyi. Bioorg. Chem. 7:125–131. doi: 10.1016/0045-2068(78)90043-3
Stoddard, S.V., Hamann, M.T., and Wadkins, R.M. (2014) “Insights and Ideas Garnered from Marine Metabolites for Development of Dual-Function Acetylcholinesterase and Amyloid-β Aggregation Inhibitors.”Mar Drugs 12:2114-2131. doi: 10.3390/md12042114

There is only scarce literature about Onchidal:(
The mechanism of the AChE blockade by Onchidal was suggested to be a novel one; it was not like the Sarin-like inactivation of the serine residue in the activation site (1).
Note that alpha, beta-unsaturated aldehyde in Onchidal is potentially reactive.  Onchidal has been shown to give some adducts to lysine residues in an enzyme (2).
(1) Abramson SN, Radic Z, Manker D, Faulkner DJ, Taylor P. Onchidal: a naturally occurring irreversible inhibitor of acetylcholinesterase with a novel mechanism of action. Mol Pharmacol. 1989;36(3):349-354.
(2) Cadelis MM, Copp BR. Investigation of the electrophilic reactivity of the biologically active marine sesquiterpenoid onchidal and model compounds. Beilstein J Org Chem. 2018;14:2229-2235.


text:
Solvent shell collapse
In "advanced organic chemistry part A" book page:306&307 ,  the author uses the term 'solvent shell collapse' when talking about the reason for retention of conformation in tertiary benzyl systems (under hydrolysis reactions) where he states that some of the reason for this retention might be the collapse of solvent shell in the front side of ion pair, the difficulty in solvation from rear side due to bulky tertiary system and hydrogen bonding between anion and water molecule which facilitates the capture of water molecule from the front side of the ion pair.
Could you clarify more about the reasons? what is solvent shell collapse? and why does the bulky tertiary system affect the direction of nucleophilic attack on the planar carbocation as the nucleophile attacks from above or below the electrophile?


text:
Converting fractional coordinates into cartesian coordinates for crystallography
So let's say for a given molecule I have the volume of a unit cell, the shape of the unit cell (monoclinic, cubic, etc), the cell angles alpha, beta, and gamma, the cell lengths a,b, and c, and a list of atoms with their corresponding fractional units. How do I convert these fractional coordinates of the atoms into cartesian coordinates for any given cell shape? Orthogonal or not?

https://www.ruppweb.org/Xray/tutorial/Coordinate%20system%20transformation.htm The mathematics behind it is explained briefly here. I wrote an excel sheet some years ago to perform that transformation, if you need it I can see if I can find it

A bit late but I wrote this to perform the calculations: http://ic50.org/fractorth/ Feel free to check it, thank you!

You have to use a matrix to convert. This is derived in most textbooks on crystallography, such as McKie & McKie 'Essentials of Crystallography'.
The matrix is
$$ M=\begin{bmatrix} a & 0 & \\b\cos(\gamma)& b\sin(\gamma) & 0\\c\cos(\beta) & cn_2 & c\sqrt{\sin^2(\beta)-n_2^2} \end{bmatrix}$$
where $$n_2=\frac{\cos(\alpha)-\cos(\gamma)\cos(\beta)}{\sin(\gamma)}$$
and $a,b,c $ are the unit cell dimensions and $\alpha,\beta,\gamma$ the angles in radians. The determinant of $M$ is the cell volume.
If $V_1$ is the vector of fractional cell coordinates $x/a,y/b,z/c$ etc. the matrix multiplication calculation $\hat d_{12}=(V_1-V_2)\cdot M $ and then $L_{12}=\sqrt{ \hat d_{12}\cdot \hat d_{12}}$ gives the bond length atom 1 to 2. The angle $\theta$ between two bonds 12 and 23 is given by a dot product $\cos(\theta)=\hat d_{12}\cdot \hat d_{23}/(L_{12}L_{23})$
An example calculation in python is shown below for a triclinic crystal. The @ is matrix multiplication.
import numpy as np
a = 7.55                # cell parameters
b = 4.99
c = 12.50
alpha = 122.5*np.pi/180
beta  = (95+18/60)*np.pi/180
gama  = (118+54/60)*np.pi/180

V1 = np.array( [-0.2812 , -0.0628 , 0.1928 ] ) # x/a, y/a, z/a
V2 = np.array( [-0.2308 , -0.0972 , 0.2931 ] )
V3 = np.array( [-0.3639 , -0.1913 , 0.3521 ] )

n2 = (np.cos(alpha)-np.cos(gama)*np.cos(beta))/np.sin(gama)
M  = np.array([[a,0,0],[b*np.cos(gama),b*np.sin(gama),0],
     [c*np.cos(beta),c*n2,c*np.sqrt(np.sin(beta)**2-n2**2)]])

dcm1 = (V1-V2) @ M             # row x matrix
L12  =  np.sqrt(dcm1 @ dcm1)   # sqrt(dot product)
L12

Is the matrix set up to multiply from the right or from the left? I'm more familiar with vector times matrix rather than matrix times vector.

It's allowed if the 1D matrix (row vector) is of length n and the matrix n x n so the columns of the vector (n) is the same length as the rows of the matrix (n).

I've added some code so you can see how it works.

It is allowed, but it gives a different answer unless you switch to the transpose when you switch the order, https://math.stackexchange.com/a/1623117/657106

yes, this is what I meant by using a row vector.

this saved me from having to interact with pymatgen.

The answer by porphyrin is correct and arguably the proper way to perform the transformation. But there is another / simpler way to convert between fractional and cartesian coordinates that requires a little less programming.
Provided you have the lattice vectors of the cell $\vec{a}_1$, $\vec{a}_2$, and $\vec{a}_3$ (in cartesian coordinates), you can organize these into a matrix with the lattice vectors stored column-wise:
$$A = \begin{bmatrix}\vec{a}_1 & \vec{a}_2 & \vec{a}_3\end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{bmatrix}$$
If we denote an atomic position in cartesian space as $\vec{x}$ and the corresponding point in fractional (or crystal) coordinates as $\vec{y}$, we can convert between the two using the following equation:
$$\vec{x} = A\vec{y}$$
To convert from cartesian coordinates to fractional coordinates, you can just trivially invert this equation to get:
$$\vec{y} = A^{-1}\vec{x}$$
So, now if you have a set of $m$ atomic positions in fractional coordinates, you can organize them into an $m\times 3$ matrix denoted as $Y$. To get the corresponding set of atomic positions in cartesian coordinates denoted as $X$ (also an $m\times 3$ matrix), you can use the following equation:
$$X = (AY^T)^T$$
and similarly to convert a set of cartesian coordinates to fractional coordinates you can use the inverse relation:
$$Y = (A^{-1}X^T)^T$$
In Python you can do the following:
import numpy as np

# Pick your favorite lattice vectors
a1 = np.array([...])
a2 = np.array([...])
a3 = np.array([...])

# Build the matrix of lattice vectors stored column-wise
# and get its inverse
A = np.vstack([a1, a2, a3]).T
A_inv = np.linalg.inv(A)

# A random set of fractional coordinates stored row-wise
Y = np.array([[0.25, 0.75, 0.1],
              [0.75, 0.25, 0.3],
              [0.25, 0.75, 0.5]]
             )

# Compute the cartesian coordinates
X = np.matmul(A, Y.T).T

# Perform the inverse operation to get fractional coordinates
Y_ = np.matmul(A_inv, X.T).T

You should be able to confirm that the arrays Y and Y_ are the same.

The lattice vectors should be given in cartesian coordinates, otherwise it does not work. The inverse of matrix A is the matrix M in the answer by @porphyrin.

Yes, the lattice vectors must be in cartesian coordinates. And yes, the inverse of matrix A is the metric matrix M. In my opinion this approach is a bit easier to remember, but it is entirely equivalent to the method described by porphyrin.

Seconded - life just becomes so much easier if you just work with the lattice vectors rather than mucking about with angles.

Without doubt, the previously provided answers will work well if the crystallographic model is a description in space group triclinic $P1$.  However on occasion, the symmetry of the unit cell is higher and the molecule of interest is  on a special position, for example a centre of inversion.  As a result, the underlying .cif file need not to contain as many explicit coordinate tuples as atoms in the molecule. On the other hand, on first glance by visual inspection, the model might appear incomplete.  This answer aims to complement the previous answers from the perspective not wanting to write new programs.
Entry COD 2102215 of the COD about 2,4,6-trimethoxy-1,3,5-triazine serves as an example. The initial display (of the asymmetric unit) by the COD and the complete molecular structure are provided below:

The .cif describes only coordinate tuples for seven atom as such, which is exactly a third of the molecule.  This is because the threefold symmetry axis of the molecule aligns with a threefold axis of the unit cell.  A script like codcif2sdf, part of the cod-tools collection, however is able to recognize the described symmetries of the unit cell acting on the molecule, and to join the molecular fragments into a .sdf:
codcif2sdf 2102215.cif > 2102215.sdf

The later format is one many other programs accept as input.  Equally, its content may be rewritten in other formats like the most elementary structure format .xyz, or to yield visualizations e.g., by  OpenBabel
obabel -isdf 2102215.sdf -oxyz -O 2102215.xyz
obabel -isdf 2102215.sdf -opng -O 2102215.png

Further details about OpenBabel are available here.


text:
Why should the scalar product of the components along two axes remain unchanged after transformation?
A couple of proofs from Chapter 2 of "Space Groups for Solid State Scientists" are giving me a hard time (see attached image).

So what I understand is:
"r" was the original lattice point. "r'" is the lattice point r after the mirror operation, which should also be a point of the original lattice. I also have no trouble with how the transformation r->r' works, and how dot products being zero imply a right angle between the vectors.
So why should the two dot products be equal? And how should one interpret this dot product intuitively (component of one axis onto another?) ? (I can kind of visualize how the mirror image of a triclinic lattice wouldn't coincide with the original if one of the axes is not perpendicular to the other two, but how should one prove it rigorously, and how does the aforementioned proof work)

Welcome on the Chemistry SE! We don't like screenshots, please use text copy-paste instead.

Think of the dot (or inner) product as a purely conceptual idea. It has some value associated with vectors, which are vectors regardless of the coordinate system. Naming the vectors differently under a, say, rotated coordinate system and with different coordinates does not alter the underlying geometric meaning or the value of the inner product.

Another way to say it is: if I draw two vectors at right angles on a page, and then I rotate the page, then two vectors are still at right angles. It's quite obvious, but the mathematical statement is more general.

@peterh-ReinstateMonica Sorry about that. Will take care from now on.

@Zhe I get that rotating the coordinate system would not alter the inner product. Wouldn't that be true for any two vectors with any angle between them, irrespective of whether were they are lattice vectors or not. In this proof, when we're rotating about the z axis/reflecting about the x-y plane, the angle between c axis and a-b plane gets restricted to 90°.

@NimanyuJoshi There are a lot of questions in your post. Can you maybe narrow it down to something more specific? I'm not really sure what we should be addressing here.

@NimanyuJoshi Click "edit", and simply copy-paste the content. It will remove the formatting of your beautiful formulas, but that we will be able to take care on.

Your question
The scalar product depends on the length of the vectors and the angle between them. Rotations, translations and mirror operations do not change distances or angles (i.e. don't distort objects), so the scalar product is unchanged as well. There are other operations that do (stretching, warping, etc).
The proof
Without knowing the context, the proof seems wrong. Equation [2.18] is not general and only works for orthogonal coordinate systems. To give a simple example: if the b axis is not perpendicular to the c axis, and we rotate on a twofold along c, the b axis will not be transformed into -b, but rather into a linear combination of b and c. What this proof shows is that if axes are orthogonal, they are orthogonal (tautology).
What is true is that if there is a two-fold axis along c, you can always choose the two other axes such that they are orthogonal to c. This is a choice, however, and other choices are possible.

In the figure, b and c are not perpendicular. Points at the origin, at the tip of b and the tip of c are equivalent by translation. If there is a crystallographic twofold rotation along c, there is another equivalent point at the tip of the rotated b. The rotated b is not the same as negative b, as the figure illustrates by counter-example. However, if you want perpendicular axes (and that is the choice recommended by the international tables), you can switch from the axis b to the thin line (shown as dot-dash-dot-dash), giving you a new axis perpendicular to c. In the case shown, the lattice is now face centered.


text:
The relation between q and T when the volume is constant

$$
\begin{align}
q_V &= ΔU \tag{1}\\
q_p &= ΔH \tag{2}\\
C_V &= \frac{\mathrm dq_V}{\mathrm dT} = \left(\frac{\partial U}{\partial T}\right)_V \tag{3}\\
C_p &= \frac{\mathrm dq_p}{\mathrm dT} = \left(\frac{\partial H}{\partial T}\right)_p \tag{4}\\
\end{align}
$$

Question 1: what is the relation between $q$ and internal energy and enthalpy? Why did I assume that when the volume is constant heat would relate to external energy and when the pressure is constant it would relate to the enthalpy change?
Question 2: why is in the equation of the heat capacity it’s equal to the change of heat over the change of temperature? Why is the change of temperature is in the denominator?

Could you explain what you meant in the two questions? Why you assumed something is not the right phrasing of your question, I assume.

Q2: How do you think the heat capacity is defined ?

These equations are all based on the definitions of internal energy, enthalpy, and heat capacity.

Internal Energy

The change in internal energy of a system is equal to the sum of the heat transferred $q$ and the work done $W$.
$$\Delta U = q + W$$
In simple systems, most work is pressure-volume work, $W = P\Delta V$. Thus, we can write
$$\Delta U = q - P\Delta V$$
At constant volume, $\Delta V = 0$, so no work is done. Thus we can write
$$\Delta U = q_\mathrm{v}$$
The v subscript in $q_\mathrm{v}$ indicates constant volume.

Enthalpy

Constant pressure is a much more common situation. Think of an open reaction vessel. This is constant pressure. The change in internal energy is difficult to calculate here, because we would need to measure the change in volume (might be minuscule).
Enthalpy is a correction to internal energy that takes into account the work needed to get the system to its current pressure and volume.
$$H = U + PV$$
The differential of enthalpy is
$$dH = dU + d(PV) = dq + dW + PdV + VdP$$
Substituting in $dW = -PdV$, we get
$$ dH = dq - PdV + PdV + VdP = dq + VdP$$
At the macro scale, we write
$$\Delta H = q + D\Delta P$$
A constant pressure, $\Delta P = 0$, so
$$\Delta H = q_\mathrm{p}$$
The subscript p in $q_\mathrm{p}$ indicated constant pressure.

Heat capacities

The heat capacity, $C$, of a system is the amount of heat needed to raise the temperature of the system 1 kelvin. Thus, heat capacity is the ratio of the heat exchanged to the change in temperature So:
$$ C = \dfrac{q}{\Delta T}$$
We can rewrite $C$ at the infinitesimal level:
$$C = \dfrac{dq}{dT}$$
Your two equations for heat capacity are just what happens at constant volume and constant pressure.

As I said in my answer, in thermodynamics, where heat capacity is a physical thermodynamic equilibrium property of the material (rather than a parameter related to the specific process), heat capacity is defined in terms of the internal energy U and the enthalpy H.  I stand by this statement.

In thermodynamics, the physical property known as heat capacity is not defined in terms of heat.  It is defined in terms of internal energy and enthalpy.  The change in internal energy and enthalpy between two thermodynamic equilibrium states of a material is a unique property of the material itself, and not of any specific process that brought about the change.  On the other hand, there are an infinite number of heat flow and work flow paths from the surroundings to the system that can take the system between these same two thermodynamic equilibrium end states.  The only constraint on these, according to the first law of thermodynamics, is that the difference between the heat flow and the work flow must be the same for all the paths (and equal to the change in internal energy).  So the reason that the heat flow Q can't be unique (without some constraint on the work) is that the work W can't be unique either.


text:
Is Na$^+$ more stable than Na?
I had a question in my textbook:-
Na -> Na$^+$ + e$^-$
Cl + e$^-$ -> Cl$^-$
But Na$^+$ and Cl$^-$ have a stable electronic configuration. So, why do they react to form NaCl?
So, this is my question. Is Na$^+$ really more stable than Na?
Na$^+$ doesn't react with atoms, only with ions, like Cl$^-$. Moreover, Na$^+$ has a noble gas electronic configuration, and Na is highly reactive BECAUSE of the extra electron in its valence shell. So, in this respect, Na$^+$ should be more stable, right?
Edit: I looked this up on StackExchange, and got varying answers. That's why I'm confused.

The question makes no sense. You can't compare these two in terms of stability.

Why not? I mean, na cation does have a stable electronic configuration?

Na+ does not react with Cl-. NaCl(s) is collection of alternating Na+ and Cl- ions, each surreounded by 6 opposite ions.

So, is the question given in my book wrong? I suppose they were trying to ask why does NaCl form, when Na+ and Cl- already have stable electronic config

As @Poutnik mentioned, I would say that they're basically held together by electrostatic forces, which can break on dissolving it in water.

@stonecraft bros If you had 2 sets of balls charged positively and negatively, what would they do if you mixed them together ? NaCl as a molecule does not exist. Even in a gaseous or plasmatic phase, it is a ion pair. If dissolved in water, you get separated independent hydrated ions. If melted, then again independent "naked" ions.

$E=mc^{2}$, so no you cannot compare two things that have different amounts of matter. One of those things is missing an electron, which is a lot of energy.

I think you are mixing things up in your question and it would be helpful to specify what the question in your textbook was.
NaCl is a solid made up from Na$^+$ and Cl$^-$ ions, not from uncharged atoms as stated in your question,  arranged in a periodic crystal lattice. So the redox equations for oxidation and reduction in your question are not relevant for the formation of NaCl from its ions e.g. from a supersaturated solution, where the formation of the solid releases lattice energy. Although the redox equations are relevant, when looking at the formation of NaCl$_{(s)}$ from its elements Na and Cl, which you rightly pointed out yield the favorable noble gas configuration for both. (See Born-Harber cycle for more information on the formation of salts from its elements and the related energies)
As stated in Nilay Ghosh's comment, a question (and answer) comparing the stability of Na vs Na$^+$ can be found here.
Also: Sodium ions do in fact react with other (uncharged) atoms, it's just a question of a strong enough reducing agent under the right environment. Commercially, sodium was produced in the Deville process at 1100 °C with carbon as the reducing agent:
$\ce{Na2CO3 + 2 C -> 2 Na + 3 CO}$

I think I understand it better now. Thanks! Just have to read up on Born-Haber cycle and stuff, and will update you


text:
What did Rutherford expect in his alpha scattering experiments?
At the time when Rutherford's gold foil experiment was performed, Thomson's plum pudding model was believed to be true (at least by Rutherford himself and his students).
With this model in mind Rutherford predicted that most of the alpha particles will be deflected by at most a fraction of a degree (sourced by this Wikipedia page), but why?
In my opinion, since according to the plum pudding model the mass of an atom was assumed to be uniformly distributed and the atomic mass of gold is nearly 50 times larger than the mass of an alpha particle, and gold is solid, therefore much less intermolecular space will be present, so, most of the alpha particles should rebound or get deflected by a large angle.

Perhaps see this answer to get started: https://chemistry.stackexchange.com/a/10392/79678. Search for “Rutherford” here for more information.

Strongly related (almost duplicate): https://chemistry.stackexchange.com/questions/106819/thought-process-behind-rutherfords-gold-foil-experiment?r=SearchResults

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

What is the alpha particle bouncing off of? What are the kinematics of that situation? Without using backspin, you don't get backscattering in pool/snooker because the balls are all the same mass.

Further, given that Thompson's model was proposed after $\alpha$ backscattering was observed, it is clear that the assumption in the first sentence is not valid.

The nucleus has a radius roughly 10⁴ times smaller than the size of the atom itself (imagine a sports ball in a stadium). That would mean that its volume were 10¹² smaller than the volume of an atom. Sure the gold nucleus is ~30 times as charged and is ~50 times heavier. But dilute that charge and mass by a factor of a trillion, and suddenly those don't seem so significant anymore.
That's the worst thing about those schematic diagrams that you see for the Rutherford experiment. They blow up the size of the nucleus to prove a point, but those diagrams are definitely "Not to Scale™".

I know that but I asked what led Rutherford think that Alpha particle should pass through the gold foil keeping in mind plum pudding model

he expectesdthis i.e. before conducting the experiment

That's what I'm saying to you. Rutherford is able to divide, so he can already estimate the density of matter and charge within the gold foil, and it's not that dense.

Why is a question with 3 upvotes closed?  Surely there was a known difference between an alpha particle and gold and this had to be in his mind. Ostensibly one conducts experiments to answer a question, What was the question? The 16in projectile analogy is foolish; shooting a pingpong ball at a pillow filled with pingpong balls and having it bounce back is more appropriate and more exciting.

@jimchmst I would expect that most ping pong ball collisions result in its bouncing back. That seems less than exciting.


text:
Rate of dehydration for butan-2-ol and butane-2,3-diol
A question taken from a test series (Vibrant, India):

Amongst butan-2-ol and butane-2,3-diol, which has higher rate of dehydration?

The carbocation formed in butane-2,3-diol is more stable as it will rearrange to get +M stability from alcohol group. However, the product will be aldehyde (due to tautomerism) for butane-2,3-diol and alkene for butan-2-ol.
As rate is determined by stability of carbocation, butane-2,3-diol should have faster rate. Also, the products don't give any extra hint. Right? But the answer given for higher rate is butan-2-ol. Is there any point that I have missed out?

I would have refused to answer this question on the grounds that "rate of dehydration" is ambiguous when comparing a mono-ol with a diol.

Not to say about lack of any info about circumstances, like temperature, solvent, catalyst, or even phase!

Sorry, but I was given this question in a test and devoid of any such details, hence I didn't any info over here.

For this question, given the lack of initial information, I'm making the following assumptions:

Butane-2,3-diol dehydrates to give butan-2-one
2-butanol dehydrates to give but-2-ene

Butane-2,3-diol dehydration is nicely described in this paper from Emerson1:

Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of sulfuric acid to form methyl ethyl ketone. The reaction could be described by a pinacol rearrangement mechanism and had an activation energy of 36 kcal/mol.

Finding the data for dehydration of 2-butanol proved more elusive. One study2 conducted dehydration rate experiments for secondary alcohols at 250 °C and 40 bar pressure and obtained a rate constant for 3-hexanol to be $k = \pu{(0.067 \pm 0.021) h^{-1}}$. Assuming this rate would be similar to 2-butanol, we can find the activation energy via the arrhenius equation, which comes out to be around 48 kJ/mol, which is around 11.52 kcal/mol. Another source3 conducted the same experiment over a copper-chromite catalyst, and obtained an activation energy of 13 kcal/mol. This means that 2-butanol will dehydrate faster than butane-2,3-diol.
From a conceptual standpoint, this makes sense: The pinacol rearrangement mechanism would require a hydride shift, which would be the RDS and would take more time and energy than a simple carbocation formation. However, if you're comparing absolute reaction rates, a large number of factors come into play, and given the right conditions, butane-2,3-diol can dehydrate faster than butan-2-ol.
References:

Emerson, Richard R., et al. “Kinetics of Dehydration of Aqueous 2,3-Butanediol to Methyl Ethyl Ketone.” Industrial & Engineering Chemistry Product Research and Development, vol. 21, no. 3, 1982, pp. 473–77. doi:10.1021/i300007a025
Bockisch, Christiana, et al. “Kinetics and Mechanisms of Dehydration of Secondary Alcohols Under Hydrothermal Conditions.” ACS Earth and Space Chemistry, vol. 2, no. 8, 2018, pp. 821–32. doi:10.1021/acsearthspacechem.8b00030.
Joseph Sires Cantrell, "Rate and activation energy of the first order dehydration of 2-butanol over a copper chromite catalyst", 1954, source

Why do you think hydride shift is RDS? All literature I have seen says it’s the cation formation that is rate-determining. It would be slower than 2-butanol because of electron withdrawing effect of second OH which destabilizes cation, apparently enough to overcome the statistical advantage of two OH groups.

I just looked up the Emerson reference that you cited, and it also says the departure of H2O is generally believed to be rate-limiting, although it may be concerted with and aided by the hydride shift. Certainly, the hydride shift alone is not the RDS and would not "take more time and energy than a simple carbocation formation". In fact, concerted hydride shift would lower the reaction barrier by assisting the H2O elimination and leading directly to a more stable carbocation.

@Andrew: The mechanism is similar to neopentyl alcohol elimination. Primary carbocation formation is assisted by concomitant shift of methide group. Nevertheless, the reaction is slow.


text:
Role of alcoholic potassium hydroxide in an E2 reaction

Which of the following compound will give least substituted alkene as major product with alcoholic $\ce{KOH}$.
The options are:
1.
2.

3.
4.

I am confused between options 1 and 3. Both seem to give more substituted alkene as the major product. Does alcoholic $\ce{KOH}$ always mean it is an E2 elimination?
What is the mechanism here?

Comments are not for extended discussion; this conversation has been moved to chat.

Haloalkanes react with alc. $\ce{KOH}$ mainly via $E_2$ mechanism. But, in some cases, where $\beta-\ce{H}$ isn't present, it proceeds via $E_1$, as in case of $(4)$ one. Let's analyze each option,
Option $(1)$:
Firstly, the $\ce{-Cl}$ departs, and then the base picks up a $\beta-\ce{H}$ from the terminal methyl group (i.e., $\ce{C-5}$), resulting in mono- substituted alkene (Hofmann product as major). Steric factors are significant in selectivity of $\beta-\ce{H}$. Therefore, Zaitsev product will be formed in very minor concentrations, bcoz $\ce{-C(CH3)2Ph}$ is more bulkier substituent than $\ce{^tBu}$.
Option $(3)$:
This one is similar to $(1)$ case, except for the fact that due to less steric hindrance, it will give di- substituted alkene (Zaitsev product).
Option $(2)$: It's clear that this molecule will form a tri- substituted alkene.
Option $(4)$: As hinted above, it will proceed through $E_1$, resulting in a tri- substituted alkene.

Different levels of substitution are shown below, (source: www.chemistrysteps.com)

Related: https://chemistry.stackexchange.com/questions/135700/which-will-be-major-and-minor-alkene-formed-if-e1-reaction-occurs-on-2-chloro-2


text:
Hydrolysis of alkali and alkaline earth metals

Which of the following cations are not hydrolysed in aqueous solution?
1) $\ce{Mg^2+}$ 2) $\ce{Ca^2+}$ 3) $\ce{Na+}$ 4) $\ce{K+}$

I thought nothing would undergo hydrolysis, so all the options would be correct.
But my answer key says only $\ce{Na+}$ and $\ce{K+}$ wouldn't hydrolyse.
Why is the case?

Think about it this way, If it forms a hydroxide in water which is precipitated, can't we say it is hydrolyzed?

Neither calcium nor magnesium comes close to precipitating a hydroxide as a result of hydrolysis.  It takes an accurate pH measurement to detect hydrolysis of either of these ions in dilute solution, especially calcium.  I would have used iron(II) and zinc for my two hydrolyzing choices, that would better correlate with the real chemistry.  Too bad we can't give -1 to the book.

Hydrolysis is a reaction with water, producing two new ions. Doubly charged ions, like $\ce{Ca^{2+}}$ or $\ce{Mg^{2+}}$ may react with water and produce complex ions like :$$\ce{Ca^{2+} + H2O -> [Ca(OH)]^+ + H+}$$ $$\ce{Mg^{2+} + H2O -> [Mg(OH)]^+ + H+}$$ These complex ions are well known in chemistry. Their existences explain why poorly soluble calcium salts like $\ce{CaSO_4}$ are much more soluble in water than what could be calculated from their solubility products. The solubility product gives only the value $\ce{[Ca^{2+}]}$ and does not give the concentration of the hydrated complex $\ce{[Ca(OH)]^+}$.
This sort of reaction is impossible with singly charged ions like $\ce{Na^{+}}$ or $\ce{K^{+}}$, because these alcaline ions cannot fix a $\ce{OH}$ group without being immediately dissociated.

While aqueous calcium does form $\ce{CaOH^+}$ as a minor species, the "excess" solubility of  calcium sulfate could be more related to the $\ce{CaSO4}$ ion pair.

@ Oscar Lanzi. Right. Thank you for your information.


text:
When you melt an ionic compound, do you "break" its electrostatic force of attraction, or its lattice energy?
I know that the melting point and the boiling point of ionic compounds are very high. However, when I was trying to find the reason for this, I found that this is because of the high electrostatic forces of attraction.
Does this mean that when we melt an ionic compound, we break its electrostatic forces? But then that means that in the liquid state, ionic compounds have only ions? That somehow doesn't seem right to me. I tried to find out if its the lattice of the ions that we are breaking instead but found no clear answer.
So, that is my question. When melting an ionic compound, do we break its lattice or its electrostatic forces? what about boiling?
Edit: I got this from Wikipedia:

Even when the local structure and bonding of an ionic solid is disrupted sufficiently to melt it, there are still strong long-range electrostatic forces of attraction holding the liquid together and preventing ions boiling to form a gas phase.[65] This means that even room temperature ionic liquids have low vapour pressures, and require substantially higher temperatures to boil.[65] Boiling points exhibit similar trends to melting points in terms of the size of ions and strength of other interactions.[65] When vapourized, the ions are still not freed of one another. For example, in the vapour phase sodium chloride exists as diatomic "molecules".[66]

Here, the last point. "Sodium chloride, in the gas phase, exists as diatomic molecules." This means, the electrostatic forces of attraction between the ions don't break ( The ions don't get separated). Like, for NaCl, you still have "molecules" of NaCl, you don't have ions of Na$^+$ and Cl$^-$ in the gas phase.
So, in a way, the only the lattice of NaCl broke, NaCl itself ( with the electrostatic forces) didn't break.

And the difference between them is?

Edited with more info

The term "lattice of NaCl" is misleading, possibly making false impression the lattice is formed by NaCl molecules.

No, your interpretation is still wrong. Covalent component of NaCl bonds is always present, as long as there's a bond.

Related https://chemistry.stackexchange.com/questions/136299/what-happens-to-boiled-bmimpf6 https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state https://chemistry.stackexchange.com/questions/46/are-there-any-ionic-amorphous-solids

What should "break its electrostatic forces" mean? As long as you have ions, there are electrostatic forces between them. There is nothing you can do about that, just as you cannot disable gravity between any two bodies.

Related: https://chemistry.stackexchange.com/q/59462

When you melt an ionic compound, do you "break" its electrostatic force of attraction, or its lattice energy?

When you melt an ionic compound, the ions are still there, and the electrostatic force of attraction is still there.
There is no longer a lattice. In the solid, there is a long-range order of the positions of the ions that you can describe by the translational symmetry of a repeating unit cell. This is gone in a liquid.
Lattice energy is not something on top of electrostatic potential energy. Instead, it is a way to describe these for a crystal structure.

Here, the last point. "Sodium chloride, in the gas phase, exists as diatomic molecules." This means, the electrostatic forces of attraction between the ions don't break ( The ions don't get separated). Like, for NaCl, you still have "molecules" of NaCl, you don't have ions of Na+ and Cl− in the gas phase. So, in a way, the only the lattice of NaCl broke, NaCl itself ( with the electrostatic forces) didn't break.

It is just the opposite: In a solid or liquid, there is no reason to pair up a cation with a specific anion. Each anion has several cations as closest neighbors, and each cation has several anions as closest neighbors (to be specific, 8 neighbors of the opposite charge in the LiCl crystal structure).
The difference between solids and liquid ionic substances is the number of neighbors while the distances don't change a lot. In lithium chloride, for example, the Li:Cl distance is 2.57 Å (and the Cl:Cl distance is 3.64 Å). In a simulation of molten lithium chloride, these distances are similar:

Source: https://www.jstor.org/stable/79992
You can tell that the liquid LiCl is less densely packed by comparing the density of solid LiCl to that of liquid (or molten) LiCl, which is about 25% less (2.07 g/cm³ vs 1.502 g/cm³, source: http://moltensalt.org/references/static/downloads/pdf/element-salt-densities.pdf).

As Karl mentioned in his comment, the electrostatic forces are not being "broken" per se. The kinetic energy gained by the particles upon heating allows them to work against the electrostatic forces that are pulling them together, which results in a change of state (solid -> liquid -> gas). In simpler terms, the forces are present between the ions, but their effect is overcome. So, to answer your question:
Melting an ionic compound would break its lattice.


text:
Confused Reading of Pressure on Vacuum Gauge
I have a gauge used to measure vacuum but I'm a bit confused as the measurements state "Hg" (I would have supposed inches of mercury except it goes to -30) and "Bar" going to minus 1.
I need the pressure in Torr, how can I properly convert my readings?
I made a google conversion but it gives me a negative result in Torr which cannot be right.

A picture of the set up would be helpful.

Vacuum is negative with respect to 1 atmosphere.

See Pressure measurement. The table there will be helpful.

This already in Torr - millimetres of Hg.

OP's question: I need the pressure in $\pu{Torr}$, how can I properly convert my readings?

$\pu{1 Torr} = \pu{1 mmHg}$ and $\pu{1 inch} = \pu{25.4 mm}$.
Thus, any reading in $\pu{inchHg}$ can be multiplied by $25.4$ to convert it to $\pu{Torr}$. For example, if your reading is $\pu{-20.4 inchHg}$:
$$\pu{-20.4 inchHg} = \pu{-20.4 inchHg} \times \frac{\pu{25.4 mmHg}}{\pu{1 inchHg}} \approx \pu{-518 mmHg} = \pu{-518 Torr}$$
If the barometric pressure is $\pu{760 Torr}$ at where you live, then the reduced pressure in your container is $\pu{(760-518) Torr} = \pu{242 Torr}$.
Similarly, $\pu{1 Torr} = \pu{0.001333 Bar}$. If your reading is $\pu{-0.46 Bar}$:
$$\pu{-0.46 Bar} = \pu{-0.46 Bar} \times \frac{\pu{1 Torr}}{\pu{0.001333 Bar}} \approx \pu{-345 Torr}$$

Pressure gauges typically measure pressure relative to ambient, so "0" on a gauge means equal to ambient (~ 1 bar absolute pressure depending on where you are). In some cases, the units have a "g" at the end, for example "psig" to indicate that the measure is "gauge pressure" rather than absolute, but that isn't always the case.

Thanks for your answer Andrew! How do I make sense of it? How do I convert the reading into Torr?

@Andrew: This is useful information, but did not give answer to OP's question.

Sorry. Thought it was obvious how to convert once you know absolute pressure is gauge + ambient.

I'd think so too, but OP's question is specific.


text:
Correlation between redox pseudocapacitance, surface area, and cell potential with metal oxide electrodes
I'm having difficulty intuitively reconciling two different ideas concerning metal oxide thin films in contact with aqueous solutions:
1) that charge storage by both redox pseudocapacitance and electrical double layer formation depend on the thin film's surface area (i.e. more surface area = more active sites and interstitials = more redox reactions at surface groups & more counterion adsorption at interstitials = more adsorbed charge),
2) but that interfacial potential due to the presence of the same protons / hydroxyls / H2O does not depend on surface area (i.e. hydrolysis of H2O and adsorption of, or reaction with, H+ and OH- just changes the electrochemical potential of the surface groups, leading to an interfacial potential, measured as a cell voltage between two electrodes, that doesn't depend on surface area).
For the sake of clarity, the specific relation I'm describing in 2) is the pH-dependent voltage of certain metal oxide films. There is a lot of literature for, for example, iridium oxide showing that pH sensitivity of these electrodes is independent of surface area (from 10 square microns up to several square millimeters).
I can clearly see that, for example, the Nernst equation for electrodes at equilibrium features no term for amount of charge present; in this case, cell potential differences are altered only by the number of electrons transferred in the dominant redox couple(s) and their relative activities. This makes sense from a potential energy standpoint. However, from an electromagnetism perspective, additional point charges (assuming for a second that orbital electrons not used in bonds or that free protons can be described only as point charges) generate a linear, additive electrostatic potential; of course, solution counterions and/or free charge carrier redistribution in the metal oxide balance to negate an electric field outside of this localized area, but this doesn't change the fact that the potential energy in this regions is higher.
Of course, voltage is an integral over a path (in this case, between the ohmic contacts of a working electrode and reference electrode), and the electric field should be symmetric about the solution interface, but if all charges are balanced at equilibrium then how do we measure a cell potential at all? Is there just some scaling law at play here, where the charge magnitude inherent to each point charge multiplies, when redox reactions take place, at a faster rate than the number of point charges (active sites) per surface area?
Perhaps someone with a good intuitive understanding of cell potentials and redox potentials could shed some light on where my thinking is incorrect here.
Side note: I have viewed this, this, this, and this but they didn't have the answer I was looking for. I also thought about writing this question a little more generically than what I have here but it's long enough as it is.