text:
Why is sublimation considered a zero-order process?
In my Chemistry book, it states that sublimation is normally a zero-order reaction or process, as the concentration of molecules that have sufficient energy to sublime is limited by the solid's surface area. My book states that the concentration of these molecules does not change when the amount of subliming substance decreases.
However, I am struggling to understand one thing: if the amount of subliming substance decreases and the object gets smaller, wouldn't that have an effect on the surface area of the solid as well as a consequent effect on the rate of reaction?
Why exactly is sublimation zero-order when clearly the surface area of a solid is variable in different instances? According to me the only way for sublimation to be zero-order is if the solid is held in a sealed container and is under dynamic equilibrium with its vapor pressure. Am I correct about this? I would appreciate it greatly if someone could help me out about all this.

Well, there are about 1E15 at/cm2 atoms on a surface, and 6E22 at/cm3, so it takes quite a lot of atoms leaving a surface to change the volume or surface area much...

Thank you for replying. That makes sense and provides me with some helpful insight, but what if you considered a case where on a macroscopic level you did notice a large difference in volume or surface area, such as a heap of snow reducing in size by a few factors after a period of a few days? Surely the rate of sublimation can't be considered approximately constant in that case? There would be a disparity between the initial and final rates, correct? Thanks for your help.

I think that the rate of sublimation even on large time-scales can be taken as constant because even though it might change to some degree across that entire length of time, the rate of sublimation decreases at such a slow rate in general that we can simply consider it approximately constant.

Zero'th order only means the rate does not depend on the concentration of the substance in question in the gas phase (or in solution). It's up to the experimentalist to keep everything else (temperature, catalyst concentration, active surface area, ...) constant.

In the case of your book, you just have to put the substance in an open box, and voilà, the surface doesn't change any more during sublimation.

Thanks for the additional input, Karl. Correct me if I'm wrong on this please. Let's use an ice cube as an example. Increasing or decreasing the concentration of the ice means adding more or less of the substance in a container. From what you are saying, you can limit or control the amount of subliming surface of the ice by placing it in an open box. The only way to do that would be to place it in a box whose dimensions were close to the dimensions of the ice cube, so that only the top surface of the cube is exposed. Therefor sublimation occurs on one face of the cube and the rate Is constant.


text:
Which component in my tap water "scale" does not react with acid?
Where I live we have pretty hard tap water and so I'm used to scale build-up in kettles and on bathroom tiles, easily removed with acids (it's mostly calcium carbonate I guess). However it seems a part of the "scale" is resistant to acid. For example, I have a glass that I use in my office to drink tap water throughout the day, so the remaining drops of water dry often at the bottom. This results in a circle of "scale", but a significant part of that remains even after treatment with sulfamic acid. Similarly, it seems a part of the residue that forms on my bathroom tiles is not removed by acidic cleaners.
Looking at the water quality charts of my utility, the only part that I think could be responsible just by guessing at the amounts present is $\ce{SiO2}$, at 29 mg/l. Does this form a white, scale-like substance that's hard to remove, at least with acid?
(Tap water analysis here, fold down the Wasserwerk Schulensee part.)
Below an image of my glass after treatment with warm sulfamic acid. The limescale present bubbles off. The remaining white stuff can be scratched off with a knife, so it's actually a residue and not just "etched" glass, but it is pretty tough.

A likely candidate is calcium sulfate, which forms a relatively soft mineral (gypsum), hence removable with a knife, but it has limited solubility in water and does not react with common acids.
See here for more on water hardness.

Where I live, we get our water from deep wells in sandy soils. Silicates are the big issue, particularly since I don't like using HF at all (note: do not try this at home!).

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text:
Why does some coffee smell like urine?
I can smell a urine like odour from some coffees. I don't seem to be alone in this as there are multiple reports on the internet of people having the same experience.
I have read from a few posts that results of the Maillard reaction create ketoamines that could be causing it.
What's the science behind this?
References:

Reddit
Yahoo Answers

Actually I have similar experiences. Often, after drinking caffeinated beverages or coffee-flavoured pastries, I observe greenish tinge of colour in my urine and a familiar coffee-like smell. This question may also fit well on Biology StackExchange.

@TanYongBoon well, coffee-smelling urine or urine-smelling coffee is not exactly the same thing ...

@mannaia That is true. Actually, I have never drunk urine-smelling coffee before. It would be weird to think that coffee could smell like urine. Unless... One drinks a lot of coffee, such that one's urine smells like coffee. Then, one can assert that "my coffee smells like my urine"... And that would be correct :)

There are surely plenty of molecules in coffee and surely lots of them contain nitrogen. Ammonia and amines are most likely formed or released upon roasting. Coffee is good because it comes in a huge number of reach and complicated aroma. As stated in other places here urine doesn't smells particularly unless for specific states of the body or dietary reason.

This answer is purely speculative in nature fueled by high school science and a research paper
From medicalnewstoday

Urine does not usually have a strong odor to. However, occasionally, it will have a pungent smell of ammonia.

Googling coffee + ammonia led me to this little article Filter coffee nanoparticles to turn ammonia sensors.
The article  says that the researchers at IISC Bangalore were using coffee powder for detecting ammonia because coffee powder contains carbon and nitrogen in abundance and they've used that to make carbon nanotubes whatnot.
Then I hit gold when I found the research paper itself
- Understanding the ammonia sensing behavior of filter coffee powder derived N-doped carbon nanoparticles using the Freundlich-like isotherm.
The detecting technology is based upon adsorption which is a surface phenomenon basically the adsorbate(ammonia) gets concentrated only at the surface of the adsorbant(coffee powder).
Now, there are two types of adsorbtion.

Chemisorption
Physisorption

Now comes the speculation I'm assuming that the process over here is physisorption because physisorption is inversely proportional to temperature.
How it works?.
Filter coffee is essentially a powder this provides a large surface area for adsorbtion.
Intially when the coffee in your kitchen at room temperature some amount of ammonia gets adsorbed on it. Now, when you make the coffee it is exposed to high temperatures and as you can recall I just discussed that in physisorption the amount of gas adsorped on the surface is inversely proportional to temperature so the hot coffee can hold less amounts of ammonia on its surface as compared to the room temperature coffee.
So that's how ammonia is released on making coffee. Presumably.

Intriguing, but where does the ammonia come from in the first place?

The atmosphere?

If urine is stinking ammonia that's a sign someone is close to death. If in your atmosphere there's so much ammonia that coffee is stinking then better get out of there.

Interesting speculation though

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text:
Density of d-block elements
Something that confuses me slightly is the trends in density when comparing periods 4, 5, and 6 in the d-block. Looking at periods 5 and 6, the density peaks at group 8, with ruthenium and osmium, respectively. The density slightly drops at group 9 and falls further through group 10-12 and the p-block.
But in period 4, just from assumption you would expect iron, the first group 8 member to be the densest of period 4, but it's not at all. Sure, iron is the densest yet, but the trend increases up to copper and only starts dropping at group 12 and beyond. Why is there this discrepancy?
Furthermore, why does density seem to peak at group 8 anyway? What is it about the d-shells that create this general pattern?

Accept it; it is the d-block :)

Density varies depending mainly upon i)atomic mass, ii) atomic radius and iii) packing type.
i) Atomic masses increase across the period and this tend to increase densities across the period
ii) Across the first transition series (3d series)-as in any other d-series- electrons are entering the same set of orbitals and so atomic sizes tend to decrease due to enhanced nuclear electron attractions. Beyond manganese electrons start pairing up in d- orbitals which has a counter effect of electron-electron repulsion that tend to expand the electron cloud as a whole. So we atomic sizes decrease up to the middle and then remain nearly same or register  small increases as we see: Sc(162pm), Ti(147pm), V(134pm), Cr(128pm), Mn(127pm), Fe(126pm), Co(125pm), Ni(124pm), Cu(128pm), Zn(134pm)
iii) Packings: Sc(HCP), Ti(HCP), V(BCC), Cr(BCC), Mn(BCC),Fe(BCC),Co(HCP),Ni(FCC),Cu(FCC)),Zn(HCP); here BCC has 68% packing efficiency where as HCP as well as FCC(CCP) packings have 74%. So based on packing efficiencies an drop of densities are expected for V,Cr, Mn and Fe
Corresponding Densities:Sc(2.98), Ti(4.5), V(6.0), Cr(7.2), Mn(7.2), Fe(7.87),Co(8.9), Ni(8.9), Cu(8.96), Zn(7.14)
P.S. For the first transition series there are no inner d-electrons or f-electrons to influence the atomic sizes and so densities.
Just qualitative evaluation alone cannot help to predict an order of densities because the factors are not all acting in the same directions.Probably Mother Nature has no intention that things always must be in regular order so that man can study them easily ;)

Electron are added when you go to the right of a $d$ line, they have the same principal quantic number and so they are not making the atom bigger, but at the same time there is more protons and neutron resulting in more attractive force : the density increase because the volume is slightly the same but the masses of nucleus increase.
This phenomena start to be inverted at some point of the line because the "shell" (repulsive force due to other electron) of previous electron start to be important enough, the new added electron is not able to penetrate this cloud of electron and make the volume of the atom bigger, resulting in a tendency of decreasing density.
This phenomena of "shell" is explain by Slater and even have rule.
https://en.wikipedia.org/wiki/Slater%27s_rules?wprov=sfla1
EDIT : The explanation for the peak of density of the last group is explained by the lanthanide contraction. The $f$ orbitals have really low shield power and will cause a contraction of all the new added electron : it will be easier for them to penetrate the cloud of electron. This result in higher density
https://en.wikipedia.org/wiki/Lanthanide_contraction?wprov=sfla1


text:
Is 1-(1-Methyl cyclopropyl) carbocation [1] as stable as Cyclopropyl Methyl Carbocation[2]
Does the extra Methyl group interfere with the exceptional stability of Cyclopropyl Methyl Carbocation?

As you can clearly see in the canonical structures,
The extra methyl group stabilizes 1 by positive inductive effect. Hence 1 is more stable than 2


text:
Stability of tertiary carbocation
Our teacher told us that a carbocation connected to 3 cyclopropane rings is among the most stable carbocations. But I am unable to understand how, as resonance (the most important factor governing stability) is absent. How did other factors bring so much stability?

It is not called ‘three degree’, it is tertiary. Likewise, they are secondary and primary carbocations. The degree notation should have died out a long time ago.

Related: https://chemistry.stackexchange.com/q/48264/54655

https://chemistry.stackexchange.com/q/19090 Half-way dupe. If I vote to close it will be close-hammered because of my badge so I will leave it to others to decide if this is dupe-y enough.

Your teacher is correct, that

a carbocation connected to 3 cyclopropane rings is among the most stable carbocations.

This because of various electrical effects, such as:

The Inductive Effect
Resonance
The Mesomeric Effect
Electromeric Effect
Hyperconjugation

Here's a link for you, to study them: Click Here.
Please make sure that you learn this correctly and from standard resources, in order to avoid confusion.
There may be a little confusion between mesomeric effect and resonance effect, so here's another link just to avoid confusion. : refer to the answer by Maria Dar
Please, do ask your teacher to help you understand this topic better :-)

To state the blooming obvious, only one of those effects has anything to do with cyclopropylmethyl carbocations. Further, are you affiliated with that study link site because it may be considered spammy.

It is spammy as it is really bizarre to start speaking of electronic effects in chemical reactions than afterwards recalling what molecules are :) But the user invite to study standard resources, this is also true


text:
Can't decide between E1 and E2 mechanisms for this reaction
I can see that due to the substrate being bulky, an elimination reaction will take place and not substitution. The reagent is a strong base. E1 reactions are independent of the base chosen, and E2 requires strong base. There is no information about the geometry of the substrate (i.e., whether the beta hydrogens are anti to the halide). How do I decide between the two mechanisms?

I’m not sure that the E1/E2 question is as important for answering what the major product is. Just think about which hydrogen is most acidic.

Oh, I get it now. It is the beta hydrogen attached to the benzylic carbon. Since its removal would form a stable carbocation, that is where the elimination would take place, right?

And now, since the reaction does depend on the stability of carbocations, can I say it proceeds through E1? Or is there still no conclusion to be drawn about that?

What do you mean about the geometry of the substrate?

That was about the anti geometry required between the beta hydrogen and the halide, in order for E2 to take place.

Is the answer 3?

No, it's actually 2. But I marked 3 initially too.

The acidity reason makes sense.

But they're in methylene groups on both sides. ?

You have a tertiary halide. These dissociate rather easily into the relatively stable tertiary carbocation and the free anion. That in and by itself should be indication enough that the reaction proceeds via E1. Alcoholates are only moderately strong bases so the mechanism cannot be $\mathrm{E1_{cb}}$.
You could consider E2 because the substrate is non-cyclic which means that it can always orient itself in such a way that a hydrogen atom is anti with respect to the halide. The only thing to take care of is steric hindrance; just draw the appropriate Newman projections to determine the likelyhood.
Once you have the tertiary carbocation, any of the β hydrogens could dissociate. You would likely initially get some kind of mixture based on the corresponding $\mathrm pK_\mathrm a$ values weighted by the number of hydrogens. Looking at the three different carbons, one is obviously much more acidic because it is benzylic. This explains the major product.
Even if initially another hydrogen is deprotonated, you can assume that equilibration takes place towards the conjugated styrene system given catalytic excess base.


text:
How would I know that a compound and its isomers are enantiomers?
How would I know that a compound and its isomers are enantiomers? Considering the chiral carbons
Consider the compound is 2-bromo-2-chlorobutanedioic acid.

I am assuming that your question is this:
"How to check whether two given compounds are enantiomers of each other?"
First, you check for structural or geometrical isomerism.
If there isn't any, you check whether the compounds are diastereomers. (Check whether they are not the mirror images of each other.)
If they are not diastereomers, you have two possibilities:

Either the compounds are enantiomers of each other.
Or they represent the same compound.

You could try this approach:
First note the groups attached to the chirality centre. The Carbon is obviously an sp3 hybridized atom with four different groups attached to it.
Interchanging any two groups attached to the chirality centre produces the chiral compound's enantiomer.
So you could try interchanging the positions of the groups attached to the chirality centre until you get the other compound. If the number of swaps is even, the two compounds are identical. If odd, they are enantiomers of each other.

Nice first contribution! As you are undoubtedly aware not all enantiomers are based on sp3 carbons, i.e., allenes, helicenes and atropisomerism.


text:
Abnormal reaction of grignard reagent
My book states that:

When isopropyl magnesium bromide is added to diisopropyl ketone, the expected tertiary alcohol is not obtained; instead the secondary alcohol; diisopropyl carbinol is obtained resulting from reduction".

I know this has happened due to steric crowding but can anyone tell the mechanism for this?

https://www.organic-chemistry.org/namedreactions/grignard-reaction.shtm see "single electron transfer"

Compare this reduction with the Meerwein-Pondorff-Verley reduction. https://www.organic-chemistry.org/namedreactions/meerwein-ponndorf-verley-reduction.shtm

The reaction you are suggesting is:

ChemLibreTexts mention using $\ce{LiAlH_4}$ and $\ce{NaBH_4}$ for the reduction of carbonyls to alcohols:

Again, Grignard's reagent does not show.
The Wikipedia article on Grignard's reaction also shows conversion of secondary ketones to tertiary alcohols:

Organic-chemistry.org shows a similar reaction:

However, no alcohol is obtained as a product. Meerwein-Ponndorf-Verley reduction, mentioned in the comments, requires an aluminum catalyst. However, as pointed in the comments, it says "Grignard Reagents will sometimes yield the result of an MPV reduction if the carbonyl carbon is too hindered for nucleophilic addition."

This should answer the question.

Did you read the whole entry on MPV reduction at organic-chemistry.org that you linked to? Specifically this sentence "Grignard Reagents will sometimes yield the result of an MPV reduction if the carbonyl carbon is too hindered for nucleophilic addition."

The following mechanism results in the secondary alcohol:

This answer is just a snap shot of a diagram. Must described what you think in words please.

Technically this is correct, but I "don't know" why only a hydride ion is transferred. A complete answer gives this reason.

This proposed diagram is wrong. The grignard reagent is a very strong base and the alpha carbons of the ketone have relatively acidic hydrogens. I think a more likely diagram would be the grignard carbanion abstracting one of the alpha hydrogens, pushing its bonding electrons up to for a double bond with the carbonyl carbon and displacing the carbon-oxygen double bond electrons to a negatively charge oxygen whose charge is counter-balanced by the MgBr+. This is a tautomer of the original ketone and in a protic solvent could probably re-equilibrate to the original ketone form.

The comment at the end of the original post of " the secondary alcohol; diisopropyl carbinol is obtained resulting from reduction" to me likely means a reduction reaction (LiAlH4 or NaBH4) followed by acid work up. It is worth noting that these conditions would also yield the secondary alcohol from the starting ketone without the application of the Grignard. As a practical note, this would be a very expensive and roundabout way of making an alcohol lol.


text:
Software for the atom model of all elements
I want an atom model with shells but I haven't found one yet and the only software that created an atom was Cinema 4D. This software is not for beginners like me and I used a tutorial to make an atom but it was really hard until I knew how to create shells, electrons, protons and neutrons.
Does anyone know a replacement for Cinema 4D? I checked the list in Google and I found molecules which is not what I wanted to find.

Something like this in that picture or something in 2d if it's not possible in 3d?

For $\mathrm{\LaTeX}$ users there is bohr package by Clemens Niederberger which

provides means for the creation of simple Bohr models of atoms up to the atomic number 112.

Example:
\documentclass{article}
\usepackage{chemmacros}
    \chemsetup{modules = all}
\usepackage{bohr}
    \setbohr{
        shell-options-add = dashed,
        shell-dist = .5em,
        insert-missing,
        atom-style={\footnotesize\sffamily\ch}
    }

\begin{document}

\bohr{}{He}
\bohr{10}{O^2-}
\bohr{18}{Ca^2+}
\bohr{30}{As^3+}

\setbohr{
    distribution-method = quantum,
    shell-options-add = {solid,red!25!},
    electron-options-set = {red!75!},
    nucleus-radius = 1.5em
}
\bohr{55}{Ce^3+}
\setbohr{
    distribution-method=periodic,
    shell-options-add = {solid,green!25!},
    electron-options-set = {green!75!},
}
\bohr{55}{Ce^3+}

\end{document}

Ok but how do I use it?I tried to write in the Latex software and it showed me an error.

Do I install the package or something like that?

That's not very informative. What "Latex software" and what error? Are you sure you have necessary packages installed since they are not a part of "minimal" installation of neither TeXLive nor MiKTeX?

File `libertinehologopatch.sty' not found. }{}

I downloaded the package from the internet.Is that how the steps work?

Well, you are missing either one of or both packages since libertinehologopatch.sty is written by the same author. I suggest to use your package manager to fetch the packages from the nearby mirror (you still haven't mentioned what LaTeX distro you are using). As for the future, I would recommend to make a full install with all the packages. It's going to take up to 5 GB or so with all the docs, examples and so on, but IMO it is worth it.

I'm using TeXstudio and  MiKTex 2.9.Do I have to install the package in both of these programms?

No, TeXstudio is an IDE, you don't install packages there. You need to open MiKTeX Console and make sure both chemmacros and bohr packages are installed via package manager.

Ok now I have both packages.Is there a tutorial to use bohr?(I don't know how to write the code and I want to learn from somewhere)

Yes, there is: bohr's manual.


text:
Why Does 0.10 M Dihydrogen Phosphate Dissociate Rather than Hydrolyse in a Neutral Solution of Water?
Dihydrogen Phosphate has a Ka of 6.2x10^-8, yet it seems to form solutions with an acidic pH when it is in water.  Why is that?  Shouldn't it dissociate back to phosphoric acid and form a basic solution, since its Ka is less than 1.0x10^-7?
It seems as though Ka does play a role in determining whether an amphoretic substance will act like a base or acid, since 0.10 M Hydrogen Phosphate has a Ka of 4.8x10^-13 and creates a basic solution.  However, I feel like I am missing another mechanism that determines the behavior of an amphoretic substance in a neutral solution.

The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species $\ce{H2PO4-, HPO4^{-2} , PO4^{-3} ,H3PO4}$  in solution, that ($\ce{H2PO4-}$)   is amphoteric that behaves both:
as an acidreleasing $\ce{H+}$ in aqueous solution and( forming $\ce{HPO4^{-2}}$ ions which can produce very few of $\ce{PO4^{-3}}$ ions)
$$\text{
Ionization:}\ce{H2PO4- <=> HPO4^{-2} +\color{red}{H^+} \quad\left(K_\mathrm{a}=\color{red}{6.2\times10^{-8}}\right)}$$
OR
as a base(can undergo hydrolysis forming ($\ce{H3PO4}$) acid
$$\text{Hydrolysis:}\ce{H2PO4-  +  H2O <=> H3PO4 + \color{blue}{OH-} \quad\left(K_\mathrm{b}=\color{blue}{1.3\times10^{-12}}\right)}$$
So ,$\color{red}{K\mathrm{a}}>\color{blue}{K\mathrm{b}}$ ,thus the first reaction is the dominant,so ,$\pu{pH}<7 $ and the solution is$~\color{red}{\text{acidic}}$.

You don't really compare the $K_a$ value with the extent of autodissociation in pure solvent.  Rather, for an sacid compare the generation of solvated hydrogen ions by the solute dissociation versus solvent (water) dissociation.
Here, solvent dissociation would give up to $10^{-7}\text{ M}$ solvated hydrogen ion.
Compare this with the solvated hydrogen ion coming from the dihydrogen phosphate ion.  Letting $x$ be the amount of solvated hydrogen ion that would be formed this way you have:
$K_a=6.2×10^{-8}=\dfrac{[\ce{H^+(solv)}][\ce{HPO4^{2-}}]}{[\ce{H2PO4^-}]}=\dfrac{(x)(x)}{0.1-x}$
where you use the stoichiometry based on the reaction $\ce{H2PO4^-<=>H^+(solv) + HPO4^{2-}}$.  When you solve the quadratic equation for $x$ you will get $7.9×10^{-5}\text{ M}$, which overwhelms the maximum autodissociation by a factor of almost eight hundred.  In fact all this excess solvated hydrogen ions make a negative feedback, cutting down the autodissociation more via Le Chatelier's Principle. Your solvated hydrogen ion concentration is dominated by the solute, so the solute registers as an acid.

Solving the same equation for hydrogen phosphate also gives you a value for x that is over 1.0x10^-7, which is 2.12x10^-7, yet it tends to hydrolyze backwards.

Be sure you have the amount of monohydrifen phosphate ion actually formed in the first dissociation, use that for the second dissociation equilibrium.

Sorry I wasn't clear enough.  I meant that if you put in 0.1M sodium hydrogen phosphate in neutral water, it would still produce a greater H+ ion concentration than water would.

Next comparison -- use the Kb value to see how much hydroxide comes from the monohydrogen phosphate ion.  Which dominates now?

Ah, I see.  However, I'm still confused.  Plugging in the Kb value for dihydrogen phosphate gives me a greater concentration of OH- than H+, with [OH-] = 1.2x10^-4.  I used the equation X^2/(0.1-x) = 6.2x10^-8/10^-14


text:
Alkyl halide addition

For the above question....i did the usual addition of HF across double bonds and  joined both of them and.i thought the answer would be option(b) but the answer is something else please help!!

F- is a pretty bad nucleophile

So do you mean to say if it were a Cl/Br ....what i mentioned above would have happened?

yes, if this was an HBr reaction you would be correct.  The H-F bond is much stronger.  https://www.chemguide.co.uk/organicprops/alkenes/hhal.html

Thank you so much... they have given option c as an answer....could you explain the  way it has attained

Hi thanks i found the link...https://books.google.co.in/books?dq=Benzene+reacts+with+Cyclohexene+in+the+presence+of+HF&hl=en&id=0X4cQus2gz8C&pg=PA285&sa=X&ved=0ahUKEwicmpjh0NzVAhVCO48KHQo2AHMQ6AEIJzAA

HF is just there to release H+ to protonate the cyclohexene...and the carbocation so formed will be used for EAS on benzene...and so (c) will be the answer...your welcome


text:
What is the resulting shape all orbitals are added up?
In all scientific books and Wikipedia (e.g. this video), the shapes of s, p, d, and f orbitals are always drawn individually, but not the sum of them.
If we add up all the orbitals of an atom, for example for the atom of krypton, what will be the resulting shape, or what space will be occupied?

If you take a look at all of the electron density of an isolated atom, it must be spherically symmetric over a time average. Why? Because space is isotropic, that is, it looks identical in any direction. Specifically, our choice for $x$-, $y$-, $z$- axes is arbitrary.
If you say that the electron distribution is not symmetric, there must be some preferred direction, call it $z$. But our choice of $z$ is arbitrary in space, so some other $z$ will do just as well, and we expect that the atom will have the same preferred direction, which is a contradiction. The original assumption of non-spherical distribution is wrong.


text:
Calculating volume of gases in different conditions

Doctor Who has a secret laboratory on Venus where atmospheric pressure is $92$ times that of Earth and room temperature is $\pu{735 K}$. There, Doctor Who mixes $\pu{4.00 g}$ of $\ce{CaCO3}$ with $\pu{40.00 mL}$ of $\pu{1.00 M}$ $\ce{HCl}$. What volume of carbon dioxide gas at VSTP (Venusian Standard Temperature and Pressure) is produced?

I know that the chemical equation is
$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$
Since there are $\pu{0.04 mol}$ of $\ce{CaCO3}$ and $\pu{0.04 mol}$ of $\ce{HCl}$, $\ce{HCl}$ is the limiting reagent; thus, $\pu{0.02 mol}$ of $\ce{CO2}$ are produced.
This is equivalent to:
$$V_\ce{CO2} = \frac{nRT}{p} =\frac{0.02 \times 0.08205 \times 735}{92} = \pu{0.013 L}.$$
However, the answer key says $\pu{26.2 mL}$ of $\ce{CO2}$ is produced, which is double my answer. Where did I go wrong in my reasoning?

You should include units in your calculations, but otherwise your maths seems to be correct. Does it ask for how much CO2 gas is produced, or how much gas is produced? At 735 K water would be gaseous, and assuming ideality (of course) this would lead to twice the volume of gas.

@orthocresol thank you for checking the calculations. The question asked for the amount of carbon dioxide produced.

@DrPepper The CO2 and the H2O will occupy the same volume of container.  So it seems to me that orthocresol is correct.

I haven't found any flaws in your math either, and I can only join those who recommended to include units in your calculations. By googling around I found this problem appeared in 2012 Australian Science Olympiad Exam, and $\pu{13.1 mL}$ was one of the options there. Maybe your source of the answer key is incorrect?
Now to the fun part. I'm pretty sure you were expected to use ideal gas law, but come on, it's Venus and no way under the pressure of $\pu{92 atm}$ any gas would behave even remotely as the ideal one. In fact, $\ce{CO2}$ on Venus with its critical temperature of $\pu{304 K}$ and critical pressure of $\pu{7.39 MPa}$ is going to be in supercritical state!
You have to use Van der Waals' equation:
$$
pV_\mathrm{m} = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2},
$$
where $p$ – pressure; $V_\mathrm{m}$ – molar volume; T – temperature; R – molar gas constant; $a$ and $b$ – characteristic parameters related to the critical temperature $T_\mathrm{cr}$ and pressure $p_\mathrm{cr}$ (reduced form):
$$
a = \frac{27R^2T_\mathrm{cr}^2}{64p_\mathrm{cr}};
\qquad
b = \frac{RT_\mathrm{cr}}{8p_\mathrm{cr}}
$$
Using the above-mentioned values coefficients $a$ and $b$ can be found:
$$
a = \frac{27\cdot (\pu{8.314 J mol-1 K-1})^2\cdot (\pu{304 K})^2}{64\cdot\pu{7.39e6 Pa}} = \pu{3.65e-1 J2 mol-2 Pa-1}
$$
$$
b = \frac{\pu{8.314 J mol-1 K-1}\cdot \pu{304 K}}{8\cdot \pu{7.39e6 Pa}} = \pu{4.28e-5 J mol-1 Pa-1}
$$
Now you need to plug both values in the first equation and solve quadratic equation for $V_\mathrm{m}$. Wolfram Alpha claims $V_\mathrm{m} = \pu{2.56e-2 m3 mol-1}$, so
$$
V(\ce{CO2}) = \pu{2.56e-2 m3 mol-1}\cdot\pu{2e-2 mol} = \pu{5.12e-1 L},
$$
which, of course, deviates quite a lot from the value found using ideal gas law.


text:
Is it possible to unambiguously identify a substance with a rapid test?
I've seen a documentary about the narcotics control at the Jorge Chávez International Airport in Peru. The officers used a rapid test to identify cocaine. They put a cotton swab into the substance. The test was positive if it turned turquoise.
How does the test work? Is it really possible to unambiguously identify a substance by the color change? How can any known or even yet unknown substance be ruled out to react in the same way?

I don't know how advanced forensic science is. But to me, a more likely scenario is that this is just a quick test to determine whether somebody should be reeled in for further questioning/investigation. It would need to be combined with other evidence to be conclusive.

They were arrested and handcuffed on the spot (and questioned further). The alleged cocaine was weighed exactly to the gram. The potential prison sentence was based on the weight. I found that strange, too, since even if the test was right, the drug could have been diluted.

What documentary was that?

@RodrigodeAzevedo Don't know for sure (and didn't even know back then), but it might have been one of those: https://www.natgeotv.com/za/shows/natgeo/airport-security-peru


text:
Dependence of rates of neighbouring group participation on length of alkyl chain
On the topic on neigbouring group participation, it is mentioned in Carey & Sundberg (2007)[1] that the effectiveness of the participation is dependent on on the ease with which the molecular geometry required for the participation can be attained. The example of the cyclization of ω-hydroxyalkyl halides was then given:

The rate of cyclization of ω-hydroxyalkyl halides, for example, shows a strong dependence on the length of the chain separating the two groups.

To prove the point, they showed the data for the solvolysis rates for ω-chloro alcohols with various chain lengths. Clearly, when a 5-membered ring is form, the rate of reaction is fastest. This is reasoned with the fact that a 5-membered ring has relatively little ring strain and thus, the rate of forming it would be relatively higher. However, this analysis does not seem to be convincing. If this reasoning was indeed correct, then wouldn't the next member in the series, with the capability of forming the 6-membered ring, have a higher solvolysis rate?
\begin{array}{|c|c|c|c|} \hline
\text{Substrate} & k_\mathrm{rel}{}^\text{[2,3]} \\ \hline
\ce{Cl(CH2)2OH} & 2000 \\ \hline
\ce{Cl(CH2)3OH} & 1 \\ \hline
\ce{Cl(CH2)4OH} & 5700 \\ \hline
\ce{Cl(CH2)5OH} & 20 \\ \hline
\end{array}
In general, why doesn't the rate increase as the alkyl chain increases in length, based on the reasoning using ring strain in the intermediate? We see that 3-chloropropanol is also another outlier.
Additional information
It seems that the relative solvolysis rates of ω-methoxyalkyl p-bromobenzenesulfonates also follow the same trend, with the rate peaking with the formation of the 5-membered ring.
\begin{array}{|c|c|c|c|} \hline
\text{Substrate} & k_\mathrm{rel}{}^\text{[4]} \\ \hline
\ce{CH3(CH2)2OSO2Ar^*} & 1 \\ \hline
\ce{CH3O(CH2)2OSO2Ar} & 0.28 \\ \hline
\ce{CH3O(CH2)3OSO2Ar} & 0.63 \\ \hline
\ce{CH3O(CH2)4OSO2Ar} & 657 \\ \hline
\ce{CH3O(CH2)5OSO2Ar} & 123 \\ \hline
\ce{CH3O(CH2)6OSO2Ar} & 1.16 \\ \hline
\end{array}
*The first entry shows the rate of solvolysis without any neighbouring group participation from the methoxy group.

References

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.
Capon, B. Neighbouring group participation. Q. Rev., Chem. Soc. 1964, 18 (1), 45–111. DOI: 10.1039/QR9641800045.
Richardson, W. H.; Golino, C. M.; Wachs, R. H.; Yelvington, M. B. Neighboring oxide ion and fragmentation reactions of 1,3-chlorohydrins. J. Org. Chem. 1971, 36 (7), 943–948. DOI: 10.1021/jo00806a019.
Winstein, S.; Allred, E.; Heck, R.; Glick, R. Neighboring methoxyl participation in solvolytic nucleophilic substitution. Tetrahedron 1958, 3 (1), 1–13. DOI: 10.1016/S0040-4020(01)82605-3.

I think for 2-chloroethanol, the rate is high since only a single bond rotation is required for achieving the configuration favourable for NGP while for 4-chlorobutanol the 5mem ring formation increases the rate but when we go to 5-chloropentanol, the rate drops probably due to the low probability of achieving the right configuration required for NGP... however there might be other factors operating here too...

Longer the chain the two reacting sites can span over a bigger volume therefore have littler chance to meet and react. In other words is an entropy effect.

This is a standard case of balancing entropy of activation (as Alchimista mentioned) vs enthalpy of activation (as mentioned in your question body) for ring closure reactions. Clayden 2ed pp 805–807 covers it quite well. And for the second table, the authors explain that NGP is not competitive for the 3-membered ring, so the rate is about the same as in the first entry. (I don't immediately see why this is the case, admittedly..)

@orthocresol The gibbs free energy of activation has to do with kinetics. Thus, perhaps, the rate of formation of the 5-membered ring is the greatest. However, if we look at thermodynamic constants, would the equilibrium constant for the formation of the 6-membered ring always dominate? Under the thermodynamic conditions, what would be the major change?

@TanYongBoon, under thermodynamic control it's difficult to get anything apart from 5- and 6-rings, and all else being equal the 6-membered ring would be preferred, as you said. That said, it will probably also depend on the exact substitution pattern; if you have substituents which can stabilise or destabilise the preferred conformations of the 5- or 6-ring, then things can change. I can't think of a good example – maybe the hemiacetal forms of hexose sugars, where you only really see furanoses (5-ring) and pyranoses (6-ring).

@orthocresol Why is the entropic factor not significant under thermodynamic control yet rather significant considering kinetics?

Hmm, I'm actually not sure. We might have to look at actual numbers (which I don't have and don't know how to get).

Thanks to orthocresol's suggestion to look at Clayden's (2nd ed. p. 805-807), I now have a satisfactory answer which I can now present.
We can explain these trends using the Gibbs free energy equation. Except that now, we are now interested in the quantities: Gibbs free energy of activation, enthalpy of activation and the entropy of activation, since we are analysing rate.
$$\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger $$
Explaining the decreasing of $\ce {\Delta H^\ddagger}$ as the ring size increase
$\Delta H^\ddagger$ is associated with the ring strain of the resultant ring formed. The larger the ring strain, the larger the value of the enthalpy of activation. Clayden, Greeves, & Warren (2012) explains$\ce{^1}$:

Firstly, small rings form slowly because forming them introduces ring strain. This ring strain is there even at the transition state, raising its energy and slowing down the reaction. The activation energy for forming a three-membered ring is very high, due to strain, but decreases as the ring gets larger.

The data (from pg. 108 of ref. 2) presented here of the strain energy determined for cycloalkanes of varying sizes also supports this trend.
\begin{array}{|c|c|c|c|} \hline
\text{Ring size} & \text{Strain energy}^\text{2,*}/ \text{kJ/mol} \\ \hline
\ce{3} & 115 \\ \hline
\ce{4} & 110\\ \hline
\ce{5} & 26 \\ \hline
\ce{6} & 0 \\ \hline
\end{array}
*Reported to the nearest whole number.
Explaining the increasing of $\Delta S^\ddagger$ as the ring size increases
Think of it this way: a long chain has a lot of disorder, and to get its ends to meet up and react means it has to give up a lot of freedom$\ce {^1}$. However, for shorter chains, it is far easier for their ends to meet since these chains have fewer degrees of freedom and thus, the loss of entropy upon cyclisation is less negative.
Consolidating the trends
Clayden, Greeves, & Warren (2012) reasons that based on the above trends for $\Delta H^\ddagger$ and $\Delta S^\ddagger$, the activation energy for the formation of the 4-membered ring is the largest. This is because the formation of the 4-membered ring likely has similar $\Delta H^\ddagger$ to that for the 3-membered ring while its $\Delta S^\ddagger$ is relatively more negative.
This seems to agree with the solvolysis rates I have presented:
For the ω-chloro alcohols, we see that the relative solvolysis rate is lowest for $\ce {Cl(CH2)3OH}$. For the ω-methoxyalkyl p-bromobenzenesulfonates, the formation of the 4-membered ring also has the 2nd lowest rate ($\ce {k_{rel} = 0.68}$), with the rate for the 3-membered ring being the lowest ($\ce {k_{rel} = 0.28}$).
Why does the solvolysis rate not peak at the formation of the 6-membered ring and instead, peak for the formation of the 5-membered ring? This is probably a coincidence. There are 2 opposing trends here so it may be the case that the combined effects of the two trends give different results under different circumstances.
ron has provided a graphical representation of the data for strain energy and the entropy of activation here. We do see that in general, there is a peak in the rate of formation of the 5-membered ring. In other words, from the formation of 5-membered ring to formation of 6-membered ring, $\Delta S^\ddagger$ becomes significantly more negative.

References

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.
Anslyn, Eric V., and Dennis A. Dougherty. (2006). Modern Physical Organic Chemistry. Sausalito, CA: University Science.


text:
Explanation of phase changing/solid and liquid water existing in the same container at same temp
I was trying to explain to a friend the other day why when ice and water are in the same container, the water is 32 degrees Fahrenheit or 0 celcius, but not frozen, because the ambient temp is above freezing and the ice I'd going through a phase change. Can anybody explain or label this please?


text:
Total number of stereoisomers of 2,3,5,6-tetrachlorooctane

Total number of stereoisomers for:

I tried the following combinations:
RRRR, SSSS, RRRS, SSSR, RRSS, RSRS, SRRS, RSSR, RSRR, SRSS, RRSR, SSRS, SSRR, SRSR.
So, total number of stereoisomers I got is 12, but my teacher says it's 16. I don't know how to figure out what other four isomers are.

Just list all four-letter combinations of R and S, and that would be it.

You have just missed a few, for example you also need RSSS as well as SSSR and so forth.

If you have studied combinatorics, and if you know the fundamental principal of counting, then this is basically making a four letter word with two letters- R and S. Each place has two choices, and you have four spaces to fill, so you get 2×2×2×2= 16 optical isomers for this molecule. Since the terminal alkyl groups are different, so there is no chance of a meso-compound. And as Ivan rightly stated in the comment, to list the isomers, make all possible combinations of R and S(you have missed a few as pointed out by porphyrin in the comments)

In such a case (1), once you have made sure that not 2 configurations are the same (because of symmetry) for n chiral centers, then the number of stereoisomers is 2^n, because there are 2 possibilities for each center, then 2 possibilities for the next one and so on. Overall, you get 2 possibilities for 1 chiral center, 2 * 2 possibilities for 2 chiral centers, and then 2^3, 2^4 etc.
You gave 14 configurations, so you obviously missed 2 configurations to get to 16 (i.e. the closest power of 2).
(1) There are cases where the overall shape of the molecule adds asymmetry but it is not the case here.

If u apply simple counting for every chiral atoms present in the molecule there are 2 possible stereoisomers so by fundamental principle of counting you can directly say that total number of stereoisomers are 2^n where n represents the total number of chiral atoms it's just a convenient formula based on the method described but others in their answers above

There would be 16 stereoisomers as suggested  by other's answers.But there would be less no
of enantiomer as this molecule posses plane of symmetry in certain configurations

Aren't you talking about 2,3,5,6-tetrachloroheptane instead?


text:
When the present state of knowledge concerning central noradrenergic transmission was established?
Some time ago I began to search for descriptions of the mechanisms responsible for the regulation of noradrenergic transmission in the central nervous system of man and to my great surprise the most recent relevant publication I found in periodic literature was published in 2004 (by Eisenhofer, G. et al) and reviewed metabolism of noradrenaline and other catecholamines only.
A bibliography appeared in 2012 under the title "Catecholamine Research in the 21st Century" "provides a guide to the diverse catecholamine research efforts across key themes including, Synthesis and Storage, Release and Re-uptake, Metabolism, Catecholamine Receptors and Catecholaminergic Signaling", but is it a review of the findings latest for the present moment?
I tried to find out looking for books published since 2012 reviewing neurochemistry, neurophysiology, neurobiology, mathematical and computational modelling with application to neuroscience and examining their contents did not find anything like a comprehensive review of the chemistry of noradrenaline.
For comparison a number of such reviews were dedicated to the chemistry of dopamine. An article authored by Meiser, J. et al. is "looking at DA biosynthesis, sequestration, degradation and oxidation chemistry at the metabolic level, as well as at the transcriptional, translational and posttranslational regulation of all enzymes involved. This is followed by a short overview of cellular models currently used in Parkinson's disease research".
Alternative models were provided in the following publications:
Best, J. A., Nijhout, H. F., & Reed, M. C. (2009). Homeostatic mechanisms in dopamine synthesis and release: a mathematical model. Theoretical biology & medical modelling, 6, 21. doi:10.1186/1742-4682-6-21
Qi, Z., Miller, G. W., & Voit, E. O. (2008). Computational systems analysis of dopamine metabolism. PloS one, 3(6), e2444. doi:10.1371/journal.pone.0002444
A mathematical model of serotonin synthesis, release and reuptake was also created:
Best, J., Nijhout, H. F., & Reed, M. (2010). Serotonin synthesis, release and reuptake in terminals: a mathematical model. Theoretical biology & medical modelling, 7, 34. doi:10.1186/1742-4682-7-34
Can anyone provide an authoritative commentary to the picture outlined above?
References:
Eisenhofer, G., Kopin, I. J., & Goldstein, D. S. (2004). Catecholamine metabolism: a contemporary view with implications for physiology and medicine. Pharmacological reviews, 56(3), 331-349.
Eiden, L. E. (Ed.). (2013). Catecholamine Research in the 21st Century: Abstracts and Graphical Abstracts, 10th International Catecholamine Symposium, 2012. Elsevier.
Meiser, J., Weindl, D., & Hiller, K. (2013). Complexity of dopamine metabolism. Cell communication and signaling : CCS, 11(1), 34. doi:10.1186/1478-811X-11-34

Hello @Noir, welcome to Chem.SE! If you haven't already, check out the [tour] and [help]. We hope you enjoy your stay! Right now, it looks like your question is definitely within scope for the site, but it seems very specific and I want you to know that it is unlikely for this question to be answered in a timely manner (due to the very specific skill set that is necessary to answer this question).


text:
Photoelectric effect — Expected observations for waves
I understand why treating light as particle-like photon packets leads to the observed results.  I seek to understand why one would expect different results if light were treated as a wave (and thus why these results imply particle-like behaviour).
Experiment 1:  Increasing the intensity.
I imagine a light wave incident on the surface as Ψ = Asin(kx-ωt).  I imagine increasing the intensity of light as adding multiple parallel waves incident on the surface.  If each wave deposited enough energy to kick out 1 electron per second, wouldn’t adding multiple waves kick out more electrons per second without changing the energy per electron?
Is treating increased intensity as more parallel rays too much like particles?  If I instead treat increased intensity as increased amplitude squared, then I think of the wave as pouring energy into the metal surface at a faster rate, like a thicker stream of water from a spout. In this case, wouldn’t the light have deposited enough energy to free an electron more quickly, and thus still eject more electrons per second?
As a corollary here, is it right to equate A² ∝ number of photons?
Experiment 2:  Increased frequency.
I do understand why a continuous stream of energy from a wave is inconsistent with a threshold frequency to elicit a current.
Why would increasing the frequency (and therefore energy) of a wave not increase the kinetic energy of the most energetic ejected electrons?
Wouldn’t a single, particle-like wave-packet, also have energy proportional to its amplitude² as well as its frequency, and thus kick off more energetic electrons?

Would this not be better placed in Physics.SE?

I learned about the photoelectric effect in chemistry, and my chemistry lecturers have always given the photoelectric effect as the origin of quantised electron energy levels. They give the observed results as evidence that light travels as packets of discrete energy.

I mean I don't really care where it's put, I thought that here was appropriate, but I just want an answer.

Yes, to your 1st point based on my understanding. Higher intensity means to say electrons are receiving more photons per second, so rate of ejection is higher. See here: https://www.quora.com/If-we-want-to-increase-the-number-of-ejected-electrons-in-a-photoelectric-effect-of-light-what-should-we-do

Regarding your second point, recall that E_photon = Work function (W) + KE of ejected electron. Since E_photon = hf, and W is constant, by increasing frequency, KE of ejected electron necessarily increases. I wonder why you said that it doesn't...

Regarding your last para, clearly you have made an error when you stated energy is proportional to intensity. Based on my understanding, amplitude can be seen as the number of photons but not the energy of the photons.

Have you obtained a satisfactory answer?

I understand the reasons for all these observed results, such as increased kinetic energy with frequency, with the photon model.  However I have been led to believe that this experiment was used as evidence for the particle nature of light. I want to know why we would expect Different results with the wave model as opposed to the particle model, and thus why this is evidence of the particle model.

Regarding the last paragraph, perhaps it did not come through, but I had intended a superscript 2 after “amplitude” to indicate that energy was proportional to the square of the amplitude of an EM wave.  I do not see why it is an error to assume that intensity is proportional to energy. My understanding of intensity is power/area (J·s⁻¹m⁻²).  Is this wrong?


text:
Half-Reactions of Hydroxide
The following is the equation I have:
$$\ce{Ca(s) + 2H2O(l) -> Ca(OH)2 v  + H2 ^}$$
I am tasked to write the half-reaction equations for both the oxidation and/or reduction process. So, currently I have:
$$\ce{ Ca(s) -> Ca^{2+}(aq) + 2e-}\quad\text{(oxidation)}$$
but I'm not sure how do you write the reduction equation for the water and hydroxide.
Any help is appreciated :)
P.S.: I have no idea how to superscript or subscript in the forum.

Balance this for the reduction step: xH2O ---> H2 + y OH-

that would be: 2H2O --> H2+ 2OH-    Is that correct?

Right! since the oxidation is -2 electrons (Ca to Ca++) and the reduction (0 to -2) +2 electrons, just add the two reactions and simplify if needed.

Just so I'm getting this right: My Reduction equation reaction would be 2H2O --> H2 + 2OH-. Is there no e- required in this case?

Of course e- is required, otherwise there would be no reduction. Besides, your equation must be balanced in charge as strict as it must be balanced in all elements.

Grimlock: You may want to look at this link to balance redox equations in elements and electrons in base and acid: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/redox.htm

So, I've updated my answer to include the electron gained. Is this correct?

Alright, so I've worked out my answer (thanks to @user55119). The Reduction step was:

$$\ce{ 2H2O_{(l)} + 2e^- -> 2OH^-_{(aq)} + H_2_{(g)}}\quad\text{(Reduction)}$$


text:
Effect of solvent polarity on SN1 reactions of different charge types
On the topic of the ionisation mechanism (i.e. $\mathrm{S_N1}$), Carey & Sundberg (2007) discuss the effect of polarity of solvent on the rate of $\mathrm{S_N1}$ reactions of different charge types (p. 392). It is well-known that the rate of $\mathrm{S_N1}$ reactions are increased by the use of polar, protic solvents. Hence, the energy profile diagram shown in a is no surprise. This is type of $\mathrm{S_N1}$ involves the initial ionisation step $\ce{RX -> R^+ + X^-}$ and it is typically seen.
However, the energy profile diagram shown in b is less straightforward. The reaction that it describes has the first ionisation step as $\ce{RX^+ -> R^+ + X}$. In the caption, it is written:

Polar solvents decrease the rate because stabilisation of the [transition state] is less than for the more polar reactant.

Firstly, I would like to ask if "decrease the rate" here means that this type of $\mathrm{S_N1}$ reactions occur more slowly in polar solvents compared to in nonpolar solvents. If we look at the activation energy $E_\mathrm a$ and assuming they are relatively drawn to scale, then we can conclude that for such reaction type, the rate would be faster in nonpolar solvents since it has a comparatively smaller $E_\mathrm a$.
Actually, we do notice that the TS is still lower in energy for the polar solvent, which is logical since although it may be less charged overall, the TS is nevertheless charged and would obviously be stabilised by a polar solvent.
Finally, I would like to ask if the reason why this type of $\mathrm{S_N1}$ reactions proceed faster in nonpolar solvents because, the charged reactants $\ce{RX^+}$ are in fact stabilised in polar solvents to a much greater extent compared to in nonpolar solvents. Thus, the $E_\mathrm a$ in polar solvents is thus much larger. This point was not discussed in the text and it is purely my deduction. Could I confirm that it is a correct interpretation?

Reference
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.

Pretty much yes to everything you've written.


text:
Determining the more acidic proton in cyclopenta-1,3-diene
I am in the David Klein textbook in Chapter 3 going over rule 2 of finding the more acidic proton: if the electron pair in the conjugated base is delocalized, then the conjugate base is more stable, ergo, the proton at that location is more acidic. Here's the practice problem I am stuck on:
I generated two different conjugated bases. The book indicates the red proton is the more acidic proton. Is this the proper resonance structure for the red proton's conjugate base?

If so, then this explains why the red proton is more acidic, because it has a delocalized pair of electrons, while the blue proton does not have a delocalized pair of electrons.

Yes, your reasoning sounds right.

What is your question? You seem to have answered it yourself.

@orthocresol, I apologize if I have hidden the question with adding too much detail. I am trying to confirm with someone if I properly drew the resonance structure that would explain why the top proton is more acidic. Thanks Tan Yong Boon! I feel much better with you thinking my reasoning is correct. I suppose that will answer my question.

Just an additional information, as soon as you made the conjugate base, the compound became aromatic(Huckel's rule), so the red proton is definitely very acidic.  Had you got a conjugate base which was antiaromatic, then inspite of conjugation, the acidity of the proton would have been quite less.  So while checking acid base equilibrium, Gibbs free energy as a whole is to be considered

More is involved here than just resonance structure.  When you have pi electrons delocalized around a single ring there is the $4n+2$ rule where the most stable configuration of pi electrons in a ring has $4n+2$ delocalized electrons.  A planar, conjugated ring with the proper $4n+2$ number of pi electrons is aromatic and will be strongly stabilized.  If instead, you have $4n$ delocalized electrons it doesn't work, and especially for smaller rings you may end up with something highly unstable.
When you draw the resonance structures by removing the red proton, you count six delocalized pi electrons -- $4n+2$.  So you get that highly stabilized aromatic ring.  That's why the red proton is a relatively strong acid.
Bonus questions:  1) Would it work if you used cyclopropene instead of cyclopentadiene?  2)  What if you had 3-chlorocyclopropene instead and you used a Lewis acid to remove a chloride ion(see slides 30 and 31)?

Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.
If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.


text:
IUPAC nomenclature of aspartame
I'd like some help in constructing the IUPAC name of aspartame:

From what I understand there are carboxylic acid, amine, amide, ester, and phenyl groups - but I'm having trouble figuring out the order in which to name them.

Aspartame is (a methyl ester of) a dipeptide. Therefore, you could use the nomenclature of amino acids to find a descriptive name for aspartame.
However, since you are asking about the correct order of the individual groups, apparently, you want to generate the fully systematic name according to nomenclature for organic chemistry.
Since there are various groups, you have to identify the principal characteristic group in accordance with the seniority of classes and suffixes first. In this case, the characteristic group with the highest seniority is the carboxylic acid group.
The principal characteristic group is chosen for citation at the end of a name by means of a suffix or a class name (or implied by a trivial name). Since there is only one carboxylic acid group in aspartame, the principal characteristic group also decides the choice of a principal chain. Thus, the parent structure of aspartame is butanoic acid.
The remaining substituents are expressed as prefixes in the name in alphabetical order. In this case, the remaining substituents are one amino group, one oxo group, and the complex substituent group [(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino.
Therefore, the complete systematic name of aspartame is (3S)-3-amino-4-{[(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino}-4-oxobutanoic acid.


text:
Friedel–Crafts with unsaturated halides
Which of the following compounds will not undergo Friedel–Crafts reaction with benzene: vinyl chloride or allyl chloride?
According to me vinyl chloride will be a better electrophile than allyl chloride, hence allyl chloride should be the answer. Is this a correct way to see? As my answer does not match with the textbook

https://chemistry.stackexchange.com/questions/24404/friedel-crafts-alkylation-and-hybridization

Friedel-Craft reactions only occur with alkyl halides, vinyl or aromatic halides do not react  further information here so the answer is vinyl chloride.

So, acc. To you, allyl chloride do not react with benzene?

No! The opposite; vinyl chloride does not react with benzene under F-C conditions,


text:
Could Na+ ions in water used with a humidifier cause dust everywhere?
We have very dry air in our small family house.
So we bought a humidifier ( https://www.klarstein.com/All-products/Monaco-Digital-Ultrasound-Humidifier-White.html )
But here comes the issue: the tap water is very hard and this causes that dust is everywhere.. due to the humidifier + hard tap water.. like everywhere, even the notebooks fan are full of the white dust, overheating, brutal.
So we need to buy a "water-softener" ex.: "BlueSoft S30 SVR34" ( http://euro-clear.eu/uploads/content/catalog/EC_residential_product_catalog.pdf )
But: I can read that this "water-softener" device "replaces the Ca++ and Mg++ to Na+ ions"
The big question to chemistry: if the water will have many "Na+ ions" and not "Ca++ and Mg++" ions, would it solve the "dust" issue? So we will won't have any more white dust everywhere, if the humidifier is turned on?
We already asked the vendor, but they are not sure that this will solve the issue.

Sorry, but you'll just end up with sodium salts instead of calcium or magnesium salts. You need a different type of humidifier. Probably several for a whole house, or a large one added to the heating system.

Why not using filtered rain water?

Anyway you won't get dust sticking and adhering everywhere (insoluble carbonates). It could be an improvement.

The easiest solution would be to buy some distilled water and use that in the humidifier.

In the end, the solution was that instead of using an "ultrasound" based humidifier, we bought "evaporative humidifier".
delete this line if it cannot be said, but we bought a "trotec aw 10 s"


text:
Is the concept of frontier orbital interactions chiefly a kinetic concept?
Frontier molecular orbital theory has proved to be an immensely useful concept. The concept of HOMO-LUMO interactions are frequently employed in detailing the mechanistic descriptions of reactions in organic chemistry. It seems to me that there are numerous examples of HOMO-LUMO interactions seem to be responsible for kinetic phenomena.
For example, secondary orbital interactions allow the preference for the endo product in the Diels-Alder reaction under kinetic control.
Nucleophilicity, which is often linked to the energy of the HOMO of the nucleophile, is also often regarded as a kinetic phenomenon. In fact, Lewis basicity itself may even be seen as a kinetic concept. For example, we make the assertion that phosphine is a better nucleophile (and also better Lewis base) compared to ammonia because $\ce {P}$ is less electronegative than $\ce {N}$ and thus, the HOMO of $\ce {PH3}$ is thus higher in energy, compared to that of $\ce {NH3}$.
Should we thus classify frontier orbital interactions as a kinetic concept?

This apply in thermochemistry as well once you are under kinetic control.


text:
Determination of molecular formula of hydrocarbons by combustion analysis
So, I've this question in my textbook:

$\pu{20 mL}$ of unknown gaseous hydrocarbon reacts with excess oxygen gas. It decreases by $\pu{30 mL}$ upon cooling down. It decreases farther $\pu{40 mL}$ after it passes through an alkali solution. What is the formula of the hydrocarbon?

There is an answer provided, and I get the part where they calculate the "$x$" or how many carbon atoms one molecule of the hydrocarbon has. But, I cannot understand how they calculate the number of hydrogen atoms. There is a strange line,

$$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20x mL} = \pu{30mL}$$

The statement is:

combustion of hydrocarbon and the following cooling down of the resultants mean the decreased volume of water vapor is $\pu{30 mL}$.

I do not get this line. Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?
P.S. Sorry for poor translation. The question is in my native language.

The combustion can be described by the following reaction:
$$\ce{C_xH_y (g) + ($x + y/4$) O2 (g) ->[\Delta] x CO2(g) + $y/2$ H2O (g)}$$
All products are gaseous at first, but as the system cools down to the room temperature, water condenses attributing to the decrease in $\pu{30 mL}$. Further decrease in $\pu{40 mL}$ is due to reaction of alkali with the remaining gaseous product, $\ce{CO2}$:
$$\ce{CO2 (g) + 2 M^IOH (aq) -> M^I2CO3 (aq) + H2O}$$
From the ideal gas law it's obvious that amounts of gaseous substances are proportional to their volumes:
$$n = \frac{p}{RT}V \quad\to\quad n\propto V$$
Let's rewrite the first equation with assigned volumes:
$$\ce{\underset{\pu{20 mL}}{C_xH_y (g)} + ($x + y/4$) O2 (g) ->[\Delta] \underset{\pu{40 mL}}{x CO2(g)} + \underset{\pu{30 mL}}{$y/2$ H2O (g)}}$$
Since $n\propto V$, and for the balanced chemical equation amounts of reactants and products are equal, so are their volumes. Hence:

Q: Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?

A: Yes; no, this is correct for the given reaction.
That "strange line"

$$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20 mL} = \pu{30mL}$$

is a direct consequence of that as it can also be rewritten as
$$\underset{\ce{C_xH_y}}{\pu{20 mL}} + \underset{\ce{O2}}{20(x + y/4)~\pu{mL}} = \underset{\ce{CO2}}{\pu{20 mL}} + \underset{\ce{H2O}}{\pu{30mL}}.$$
However, as this is a single equation with two unknowns, it only illustrates the preservation of the total amounts of products and reactants. In order to find the formula of unknown hydrocarbon, you just have to equate corresponding coefficients with the ratio between $\ce{C_xH_y}$, $\ce{CO2}$ and $\ce{H2O}$ ($20:40:30 = 1:2:1.5$):
$$x = 2$$
$$y/2 = 1.5 \to y = 3$$
which results in empirical formula $\ce{C2H3}$. In order to find out the molecular formula you would also need auxiliary data such as molecular weight of the hydrocarbon.

This is a combustion reaction, so it is an exothermic reaction. Shouldn't the T increase and as P is proportional to T, it'll also increase? Does this affect the "n is proportional to V" statement? and 2. I did a mistake, the equation should be '20mL + 20(x + y/4)mL - 20xmL = 30mL" the "x" is getting multiplied with "20", so the answer should be C2H2 (provided in my textbook)

I edited the question and added the "x"


text:
Where will a H+ attack on 6-hydroxyhexan-2-one?
Where will a H+ attack on 6-hydroxyhexan-2-one? Protonation can take place either on the ketone or the alcohol. So where will it attack? In short, on which of the two oxygen atoms is electron density higher?

Well,what do you think?

As a hint, consider which lone pair is more "free" i.e. less tightly bound to the oxygen nucleus

You mean on alcohol?

Yep.. If you have understood why... Then you have your answer

Please note that the site of preferred protonation may not be the one that leads to reactivity. Since proton transfer between heteroatoms is very fast, you are operating under Curtin-Hammett kinetics, so the more reactive intermediate (regardless of its concentration) will be the one that reacts.

Well put @Zhe! This is not an either/or situation. I hope Franshu isn't under the misapprehension that protonation occurs at only one of the two oxygens. In CH3OH/H+ this hydroxyketone will lead to the cyclic, pyran mixed ketal (and/or 2-methyl dihydropyran) whose mechanism of formation requires protonation of the ketone oxygen.


text:
Suitable safe cleaning solution for aliphatic urethane
I am looking for a suitable cleaning solution for a sprayed on coating that is an acrylic aliphatic urethane. My end goal would ideally be something that can neutralize rust, whiten and brighten the surface, and also include a powerful enough solvent that can loosen stuck on dirt/soap scum but not harm the urethane.
The urethane is approximately 70% solids (don't know if that matters or not). I'm not a chemist so layman's terms would help greatly.

There may be specific solutions designed to work with that sort of surface, but a good solvent to try if you can't find any information on them would be isopropanol which tends not to destroy plastics and is good at removing grease and contamination from surfaces. As always, try on a small area first in case there is some interaction.

Oxalic acid is good at removing rust (iron oxides) and generally not agressive against polymers like polyurethanes.
Chemical resistance of polymers is however a complicated area, because polymers comes in a wide spectrum of formulations, additives etc. Some indications can be found on different manufacturers websides


text:
When is donation into an anti-bonding MO stabilising?
This has been puzzling me for quite a while and I'm not sure if it could be a misconception on my part. When is interaction of a filled orbital with a antibonding MO stabilising and when is the interaction destabilising? I will now detail a few examples of the "stabilising interactions" and the "destabilising interactions". I will use Clayden's Organic Chemistry (2nd ed.) as my primary source of examples.
Destabilising interactions

Nucleophilic attack at the carbonyl carbon (p. 127)

Here, interaction of the nucleophile's HOMO with the $\ce {C=O \pi ^*}$ LUMO destabilises the $\pi$ bond, resulting in its cleavage. This is thus a destabilising interaction.

Nucleophilic attack in the $\ce {S_N2}$ reaction (p. 340)

Here, again we see the interaction between a filled HOMO and an empty antibonding MO ($\ce {C-X \sigma ^*}$ MO) destabilising the $\ce {C-X}$ bond.
Stabilising interactions

Stabilisation of cis configuration in esters (p. 805, explained in detail here)

Here, we see that the interaction of a filled, nonbonding MO centred on the $\ce {O}$ interacts with the $\ce {C=O \pi ^*}$ LUMO to stabilise the molecule overall.

Unreactivity of dichloromethane (p. 804, see here also)

As can be seen in diagram, interaction of nonbonding electrons centred on $\ce {Cl}$ with the $\ce {C-Cl \sigma ^*}$ stabilises the molecule, as it decreases is reactivity, by raising the level of the new resultant LUMO of the molecule.

Explaining Zaitsev's rule

This is a classic explanation for the increasing stability of alkenes as substitution increases. Interaction of $\ce {C-H \sigma}$ MO with the $\ce {C=C \pi ^*}$ MO increases the stability of the alkene.
I am sure there are likely to be more examples (e.g. various other instances of hyperconjugation) but I believe these examples are sufficient to prove my point. Is this inconsistency in the use of frontier MO theory or am I merely misinterpreting its use?

Update
After pondering over this for a while, I was thinking, perhaps we should make use of an MO diagram to more clearly describe the interaction. Below is a simple, typical diagram (from p. 111, with edits). Based on my understanding, all of the above examples I have presented can in fact be described by this straightforward MO diagram. Of course, the energy gap between the two interacting MOs may be different and thus, the strength of interaction would also differ.

Now, may I alert readers to something:

In no way does the MO diagram indicate any sort of destabilisation of the molecule.

Such an interaction between a filled MO and an unfilled MO only produces stabilisation, as seen in the diagram. Not only is the filled orbital of a lower energy and hence less likely to interact with LUMOs of other molecules, the empty orbital's energy has also been raised, making it less energetically accessible from the point-of-view of a nucleophile. In other words, the molecule is both less reactive to an electrophile, due to a energetically-lowered filled orbital, and also less reactive towards a nucleophile, due to a energetically-raised empty orbital.
If this is the case, may I question further:

What is the basis of destabilisation in the first 2 examples I have presented above?

Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

The difference between the "stabilising" and "destabilising" examples that you have chosen are that the "stabilising" examples are static molecules, whereas the "destabilising" examples are chemical reactions, in which a lot is happening. We'll look at the "stabilising" cases first, as these are easier to understand.
As with any question involving "stability", it is important to consider what we are measuring stabilisation with respect to. In the example of the anomeric effect (e.g. in esters), we are comparing the true electronic state of the ester against a hypothetical scenario, where the lone pair and σ* do not overlap, but everything else is kept the same. In such cases, it is nearly always true that allowing overlap, or delocalisation, leads to a lower energy. This is captured quite well in the last picture in your question body. (See also: Physical reason for why delocalisation leads to stability?)

In a chemical reaction, as we move along the reaction coordinate (RC), we are creating overlap between two different orbitals, just as in the previous case. Naively we might think that this necessarily leads to stabilisation. But this is not physically sensible: if the reaction proceeds via a transition state (TS), this TS must be higher in energy than the reactants (at infinite separation).[1] The problem is that when we bring two molecules together in space, it is not just overlap that is affected: the energies of every orbital will change as nuclear positions are moved, electron-electron repulsions are different, etc. This means that all the orbital energies change in a way which depends not only on overlap with other orbitals (as was the case before), but also a number of other factors.
Instead of going from reactants at infinite separation to the TS directly along the RC, as would be physically realistic, it may be instructive to consider hypothetically doing it in two stages. In the first step, we bring the reactants together from infinite separation to the TS geometry while forbidding any orbital overlap; secondly, we stay at the TS geometry, but turn on the overlap. In real life, you cannot separate these two things: changing the geometry comes with formation of overlap. But thinking about it this way allows us to link it to the "stabilising" cases talked about before.
Let's use the addition of a nucleophile to a carbonyl group as an example to make this more concrete. In the reactants, there are three key frontier orbitals we need to consider: the nucleophile lone pair ($n$), and the π-type orbitals of the carbonyl group ($\pi$ and $\pi^*$).
In general, as we bring the nucleophile and carbonyl together, mixing of all three orbitals will occur continuously. As the reaction proceeds, we will therefore have three MOs $\{\psi^{(i)}\}$ $(i = 1,2,3)$ which are linear combinations of the three frontier orbitals listed earlier:
$$\psi^{(i)} = c^{(i)}_n\phi_n + c^{(i)}_\pi \phi_\pi + c^{(i)}_{\pi^*}\phi_{\pi^*}$$
For well-separated reactants, the relevant MOs will (for the most part) be the original three frontier orbitals. So, for example, we could have $\psi^{(1)} = \phi_\pi$, $\psi^{(2)} = \phi_n$, $\psi^{(3)} = \phi_{\pi^*}$. But at the end of the reaction, the MOs of the product will be a complicated mixture of the three orbitals. For example, the $\ce{C-Nu}$ $\sigma$ orbital might be made up of something that resembles this:
$$\psi^{(1,\text{new})} \approx \frac{1}{\sqrt{3}}(\phi_n + \phi_\pi + \phi_{\pi^*})$$

The reason why these linear combinations (or mixtures) can form is because there is non-zero overlap between all of these orbitals as the reaction proceeds. If there is no overlap between two orbitals, then they cannot mix. So, the way to forbid any overlap from occurring in the first step is to simply not change the coefficients of the MOs as we bring the reactants closer to each other (towards the TS geometry). Of course, this is not physically possible, which is why it is purely a hypothetical scenario.
Even though the MOs themselves do not change in this first step, their energies do change, because the Hamiltonian is changing (by virtue of the nuclear positions changing). So, the energies of the MOs will change. To throw an additional wrench into the works, it is not just the electronic energy that we must consider, but also the nuclear energy (when we bring nuclei closer to each other, they will repel each other, and this is not captured in MO diagrams which only illustrate electronic energy). As you can imagine, it is very hard to make any real analysis as to what happens in this step.
In any case, once we reach the TS geometry, the energy of the system will definitely have changed in some way. We can now carry out the second step by "turning on" the overlap, i.e. finding the appropriate linear combinations of ${n, \pi, \pi^*}$. Since nothing else here is changing (in particular, no nuclear movement), except for us "allowing" overlap to occur, it is exactly analogous to the "stabilising" case which was discussed earlier. So the second step always leads to a decrease in energy.
The problem is that, from the infinitely-separated starting materials, we have carried out two steps to get to the TS. In the second step, the energy has decreased – a "stabilisation" of sorts. But in the first step, we don't know what has happened. If a TS is to be higher in energy than the reactants, then it follows that there must have been an increase in energy in the first step.[2]
Simple LCAO-MO diagrams don't quite capture the complexity involved in the first step – they can only really properly describe the second step. A better way of drawing it might be to do something like this:

I haven't drawn the orbitals in the TS, but you can imagine that they are starting to morph into what they will be in the products. So $\psi^{(1)}$ will be somewhat intermediate between $\pi_\ce{C=O}$ and $\sigma_\ce{C-Nu}$. Of course, this is purely hypothetical - especially in the middle, the orbitals and their energies have no physical meaning at all. In a real reaction you would go straight from SM to TS, and not via an unphysical intermediate stage as is depicted here.
So, therein lies the difference between the two cases you have described. In the "stabilising" case, the only thing that is happening is a magical "switching on" of overlap. In the "destabilising" case of a chemical reaction, in addition to the overlap (which is stabilising), you also have a glut of other factors which can lower or (more frequently) raise the energy of the system as it move towards the TS. The problem is that we cannot separate the two from each other. Creating overlap comes with all the other strings attached, and so overall as we move along the RC towards the TS, there is (typically) an increase in energy.

Finally, how do we relate this back to the description of nucleophilic addition in Clayden? The wording is definitely somewhat loose (organic chemists frequently do this for ease of use), but it does have a grounding in quantum theory. I'll reproduce the image here and try to relate each phrase:

Electrons in HOMO begin to interact with LUMO

This refers to the developing overlap that we looked at in step 2, i.e. the formation of MOs which are linear combinations of the HOMO ($n$) and LUMO ($\pi^*$). On its own this seems to be a stabilising effect, but recall that this does not come on its own: we also have to consider the different nuclear positions and its implications on the energy.

Filling of $\pi^*$

I think the way to interpret it is that, at the start of the reaction the two filled orbitals $\psi^{(1)}$ and $\psi^{(2)}$ are simply $n$ and $\pi$, so the coefficients $c_{\pi^*}^{(1)}$ and $c_{\pi^*}^{(2)}$ are zero. As the reaction progresses, these coefficients steadily increase, which in a sense corresponds to a population of the $\pi^*$-orbital, even though the $\pi^*$-orbital ceases to exist in its own right once we bring the starting materials together.

causes $\pi$ bond to break

In a sense we could define a "$\pi$ bond order" to be some kind of difference between the coefficients of the $\pi$ orbital in the filled MOs, and the coefficients of the $\pi^*$ orbital in the filled MOs. At the start these are $1$ and $0$ respectively, for a bond order of $1$. As the reaction proceeds and the orbitals mix, we find that $c_\pi$ decreases in the filled MOs, $c_{\pi^*}$ increases, and at the end of the reaction we have a bond order of $0$. So loosely speaking, it is true that "population of the $\pi^*$ orbital" (i.e. increasing $c_{\pi^*}$ in the filled MOs) leads to the cleavage of the $\pi$-bond.

Electrons from $\pi$ bond end up as negative charge on oxygen

This is somewhat true again since the filled MOs morph from $\pi_\ce{C=O}$ plus a lone pair to $\sigma_\ce{C-Nu}$ and a lone pair, but it's important to remember that it is not a one-to-one correlation, i.e. the $\pi_\ce{C=O}$ MO does not become the oxygen lone pair all by itself.[3] More properly, the oxygen lone pair is a linear combination of multiple MOs in the starting material.
It's also important to not ascribe too much importance to exactly which electrons go where, as this violates quantum indistinguishability.

Notes
[1] There are barrierless reactions which do not proceed via a transition state, such as $\ce{H + H -> H2}$, and in such cases it is quite true that we always get stabilisation as we move along the RC towards products. However, these reactions don't fully capture the complexity involved in other reactions with TSes and it's better to think of these as being special cases of the other reactions.
[2] Barrierless reactions don't obey this line of thinking, of course. In fact, they don't even have a TS, so there isn't a well-defined geometry for us to distort towards in the first step. You can pick any geometry along the reaction coordinate between reactants and products, and it would probably work.
[3] Under some special circumstances it is possible to have a one-to-one relationship between reactant MOs and product MOs. See e.g. the first correlation diagram in Theoretical basis behind orbital correlation diagram for pericyclic reactions. But note that here there is only one orbital of each symmetry species, so there is no mixing going on and no linear combinations being formed. See also the last section in the answer on "Real pericyclic reactions".

Ok I think your answer definitely consolidated our discussion from yesterday. But I would like to ask further regarding the change in the nuclear coordinates (i.e. the first step): Firstly, do MOs only depict the electronic energies? Secondly, from the point of view of an electron, when it is brought near to another nuclei, there would be net fall in PE or is it net gain in PE due to the increased inter-electron repulsion?

(1) Yes, for a better understanding of this look up the Born–Oppenheimer approximation in a textbook. The bottom line is that we separate the Schrödinger equation for the whole molecule into a nuclear component and an electronic component, and the MOs are part of the (approximate) solution for the electronic bit, so they don't reflect nuclear repulsion energies (which fall out of the nuclear component). (2) Not 100% sure, but I don't think this can be qualitatively predicted. You would probably need to work out the actual integrals.

In that case, is it possible to pinpoint what is the exact cause of destabilisation in the "first step"? Can we even confirm that there is destabilisation in the "first step" to begin with?

I don't think there's much insight that can be gleaned there, and that's why I deliberately refused to say anything about it in the actual answer. There are several components in the Schrodinger equation: (1) nuclear-nuclear repulsions, (2) electron-electron repulsions, (3) electron-nuclear attractions, (4) kinetic energy. It seems intuitive to me that an increase in (1) and (2) causes destabilisation, but an increase in (3) would also be stabilising. For different reactions these factors would contribute to different extents. [The KE should be constant if the wavefunctions don't change.]

Ok I think we need to exactly quantify what is this so-called "overlap". "Overlap" means that the electrons are no longer localised on a single nucleus and now becomes shared with the nucleus of the atom being attacked (e.g. the C in C=O). In other words, during "overlap", we are concerned with terms (3) and (4) alone? While increase in (1) and (2) are what happens in "step 1"?

If you change the distribution of electron density, the electron-electron repulsions will also be changed, so (2) + (3) + (4) would be more accurate for the "second step". And it's not a mutually exclusive thing, so the "first step" doesn't just affect (1), but rather all of (1) + (2) + (3) as described in my previous comment.

Hang on... If "MOs are part of the (approximate) solution for the electronic bit, so they don't reflect nuclear repulsion energies" then why are we even including (1) in our current discussion?

Because if you are looking at the energy of a system overall as it undergoes a chemical reaction, then you need to consider all components of the system. You could ignore the nuclear repulsions, but that would be quite unphysical: there is no point solely monitoring the electronic energy as a reaction proceeds, as there is no useful information to be derived from it. For your "stabilising" cases there is no nuclear movement and hence nuclear energies stay constant throughout, so it's OK to just ignore them in order to focus on how delocalisation affects the electronic energy.

Let's talk about the decreasing value of the coefficients of the n and $\pi$ MOs and the increasing value of the coefficients of the $\pi ^$ MO. What does that physically correspond to? I can imagine that as they come closer to each other, the electrons interact possibly overall repulsively at first, raising their overall energies, such that they move into the $\pi ^$ gradually. Would that be a reasonable interpretation?

By the way, is your approach of segmenting the interactions into the so-called "step 1" and "step 2" an original, novel approach? Or has it been developed by others in the past?

I don't really want to continue this discussion much further. I'd be wary of physical interpretations... truthfully, I'd say I don't know. As far as I know it's my own approach (which I really only developed for this question because "step 2" allows us to draw a parallel to your "stabilising cases"). But many people have written stuff on how to actually interpret chemical bonding (it's a complicated subject), and everybody has a different take on it, see e.g. https://doi.org/10.1002/9783527664696.ch1 (it's a chapter in a book), so who knows, maybe somebody else has written about it better...

Thank you very much for spending so much of your precious time in having this discussion with me! I have definitely gained much from it. Merry Christmas and happy New Year!

No problem. Merry Christmas and happy New Year to you too! I think, you can now see why organic chemists don't bother thinking about the physical basis of their MO diagrams ;) I found that book in my last comment to be very insightful, but also very difficult to read. I wouldn't claim I understood much of it.

You are making the mistake of looking at part of the molecule instead of the whole.  Let's look at your example involving esters, or better yet carboxylate ions.  You have the carboxyl group and then a second oxygen atom.  When the latter donates its pi electrons to the molecular orbital, they go into an antibonding orbital of the carboxyl group.  So that part of the bonding is weaker.  But this interaction also makes a new pi bonding interaction involving the second oxygen atom.  You have to count that, too, to evaluate the bonding in the whole molecule.  When the extra bonding outweighs the extra antibonding in the whole molecule, as in pi electron radicalization in an water or carboxylare, the interaction is favored even though not every part of it is.

Ok I think we are off to a good start. I know exactly what you are saying. The donation to the antibonding MO destabilises the particular bond but brings about a stronger bonding interaction that can overall bring down the overall energy of the molecule.


text:
What is a probable mechanism for acid catalysed decarboxylation of alpha-hydroxy acid?
The following question was on one of the tests we had recently
Now i worked out the intermediates and the product as follows:
The first step is an intramolecular Cannizzaro reaction in which the aldehyde is oxidised to acid and the ketone reduced to base
In the next step decarboxylation takes place and we are left with benzyl alcohol (B)
Oxidation of (B) using Pot. Permanganate gives benzoic acid and decarboxylation of benzoic acid using soda lime gives us benzene as the final product which is the correct answer to the question.
Now i am quite satisfied with the validity of all the other steps excluding the acidic decarboxylation of the alpha-hydroxy acid
Can someone explain to me the probable mechanism for that reaction?
Or correct any other mistake i may have made in any of the other steps/conversions

You can turn PhCH(OH)CO2H to benzaldehyde (PhCHO) by direct heating (http://dx.doi.org/10.1039/CT9048500827), but I'm not sure if you can do an acid-catalysed decarboxylation...

@orthocresol Then by what way would the following set of steps give benzene as the final product? Or is that not the final product at all? (The answer key for our tests has been wrong before!)

I don't know, to be honest. I am always rather skeptical of some of these questions, because they are often not grounded in fact, but rather in some made-up synthetic sequence which nobody has actually carried out before. Not all of them are fake, of course. But I cannot tell which are and which aren't without actually looking up the reaction in a database. Suffice it to say that I couldn't find any transformations of PhCH(OH)CO2H to PhCH2OH (or PhCHO, which you can oxidise to PhCO2H as well) with just acid as the sole reagent.

@orthocresol ok fair enough! However if any alternative probable pathway comes to mind please do let me know

I prefer to call the Cannizzaro rearrangement as a benzylic acid rearrangement to give mandelic acid (A). Attack of hydroxide at the aldehyde with subsequent hydride transfer.  I agree with @orthocresol about made up conditions. I doubt that H3O+ will work to give benzaldehyde (B). But strong H+ sources (H2SO4, polyphosphoric acid, etc.) can initiate loss of water and CO (decarbonylation) to get B. Oxidation of B to C gives benzoic acid, which, when heated with CaO as Eilhard Mitscherlich did by dry distillation (1834). No NaOH needed.

@user55119 okaay thankyou that basically solves thr question


text:
The nomenclature behind “Diacetonamine”
Please explain how the name “Diacetonamine” can be found by looking at the formula:  $$\ce{(CH3)2C(NH2)CH2C(O)CH3}$$

Why is it called “Diaceto” if there only one $\ce{-C(O)CH3}$ group?

Your formula doesn't show diacetonamine; it shows 4-amino-4-methylpentan-2-one.

I’ve added a picture - did I write the formula down wrong?

Isn’t there only 1 “aceto-“ group then why is it called diacetoamine?

I think I understand what you mean. I edited your question a bit (added a reference to ChemSpider to backup your doubt; fixed a typo in "Diacetoneamine") and cast a vote to reopen. But Loong is right, diacetonamine is a different compound, I think it should have $\ce{NH(CH2C(O)CH3)2}$ formula.

This compound is obtained by reaction of two molecules of acetone and one molecule of ammonia , so it is named diacetone amine

So the name cant be obtained by just looking at the compound's formula as it is? (if we don't know how it's formed)

It's likely a trivial name constructed in a similar way like its alcohol analogue $\ce{(CH3)2C(OH)CH2COCH3}$ 4-hydroxy-4-methylpentan-2-one, better known as “diacetone alcohol”, which is prepared by dimerization condensation of acetone.