text:
What does "spin degeneracy" mean in this context?
I am currently studying Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Masanovic. In chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS, the authors say the following:

On the other hand, in a covalently bonded solid like the semiconductor materials we use to make diode lasers, the uppermost energy levels of individual constituent atoms each broaden into bands of levels as the bonds are formed to make the solid. This phenomenon is illustrated in Fig. 1.4. The reason for the splitting can be realized most easily by first considering a single covalent bond. When two atoms are in close proximity, the outer valence electron of one atom can arrange itself into a low-energy bonding (symmetric) charge distribution concentrated between the two nuclei, or into a high-energy antibonding (antisymmetric) distribution devoid of charge between the two nuclei. In other words, the isolated energy level of the electron is now split into two levels due to the two ways the electron can arrange itself around the two atoms.$^1$ In a covalent bond, the electrons of the two atoms both occupy the lower energy bonding level (provided they have opposite spin), whereas the higher energy antibonding level remains empty.
$^1$The energy level splitting is often incorrectly attributed to the Pauli exclusion principle, which forbids electrons from occupying the same energy state (and thus forces the split, as the argument goes). In actuality, the splitting is a fundamental phenomenon associated with solutions to the wave equation involving two coupled systems and applies equally to probability, electromagnetic, or any other kind of waves. It has nothing to do with the Pauli exclusion principle.

If another atom is brought in line with the first two, a new charge distribution becomes possible that is neither completely bonding nor antibonding. Hence, a third energy level is formed between the two extremes. When $N$ atoms are covalently bonded into a linear chain, $N$ energy levels distributed between the lowest-energy bonding state and the highest-energy antibonding state appear, forming a band of energies. In our linear chain of atoms, spin degeneracy allows all $N$ electrons to fall into the lower half of the energy band, leaving the upper half of the band empty.

What does "spin degeneracy" mean in this context?

You might want to look at calculation of the energy levels in polyenes as they get longer, familiar to chemists, and which show the same behaviour as in your figure.

I'm not a solid-state chemist, but I think the meaning seems reasonably clear from context to me. "Spin degeneracy" here means that each energy level is capable of holding a spin-up electron as well as  a spin-down electron, i.e. in each orbital there are two different spin states which are degenerate (some would call these spin orbitals). Thus, $N$ atoms form $N$ different energy levels which $N$ electrons need to be distributed between. Because of this degeneracy, only the lowest-energy $N/2$ energy levels are actually populated: each energy level has $2$ electrons.

Thanks for the answer. What does it mean for a spin state to be "degenerate", as you described?

It’s not one spin state that is degenerate, it’s two spin states. Degenerate means that they have the same energy, so it doesn’t make sense to say that one spin state is degenerate.

Hmm, I'm trying to understand this part: "... in each orbital there are two different spin states which are degenerate ...". What does this mean? The wording is confusing to me.

Each orbital corresponds to two spin states. You must be familiar with the standard atomic orbitals, 1s, 2s, ... The 1s orbital can hold two electrons: one spin-up and one spin-down. That is the same thing as saying that the 1s orbital has two different spin states. The shape of the orbital in a solid is different, but it doesn't change the fact that each orbital can hold one spin-up and one spin-down electron.

"... in each orbital there are two different spin states which are degenerate ..." sounds like it's saying that the spin states are degenerate, but I think the point you're making is that the orbitals themselves are degenerate because they are capable of holding two different spin states, right?

This is getting into the territory of semantics, and I'd rather not go too deep down that rabbit hole. My take on it is that when you say two things are degenerate, it means they have the same energy. You can't say that one thing (like an orbital) is degenerate, because effectively you are saying that "an orbital has the same energy". Same energy as what? It's an incomplete sentence, which doesn't really make sense. I'm saying that the two spin states are degenerate, i.e. the two states spin-up and spin-down have the same energy. Someone else might have a different take.

Ok, thanks for taking the time to clarify.


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Is it possible to measure the entropy of a chemical reaction through calorimetry?
My broad question is how do you measure entropy change? I was doing a bit of digging into how thermodynamic tables are developed at I got this from NIST. Basically it seems like you need to indirectly determine the entropy of a substance from other experimental parameters. What NIST says is

You use heat capacities and phase transition enthalpies to do a ground up calculation
Statistical mechanics methods once you have a partition function
Measure temperature dependence of equilibrium constants from van't hoft plot
Calculate it from previously determined enthalpies and free energies.

The thing that's bothering me is that all of these measurements depend on other experimental parameters and it just seems to be a raw calculation of entropy after determining these other quantities.
My primary question then is it possible to do a calorimetry experiment to calculate $\Delta S_{reaction}$? After all, how did people measure entropy experimentally in the olden days before we had copious information on heat capacities?
I tried to think about this for a constant pressure apparatus. A reaction happens and I have a thermometer to measure $\Delta T$. I can correlate this to $\Delta H$ using the heat capacity of the calorimeter. If I have a resistive circuit running through the calorimeter, I can supply heat externally to lower $\Delta T$ so now I can say that $\Delta H$ occurred at the same initial and final temperatures
However, this only tells me about the $\Delta S_{surroundings} = -\frac{\Delta H}{T}$. How do I calculate the change in the reaction system? I tried to imagine the reversible path but if I start and end at the same point, then isn't $\Delta S_{system} = 0$? Where is the flaw in my reasoning or is it just something we cannot end up measuring?

You have the right idea in devising a reversible path and measuring the reversible heat from the surroundings for that path.  But, it isn't that easy to devise a reversible path for a chemical reaction, starting with pure reactants in stoichiometric proportions at T and P and going to pure products in corresponding stoichiometric proportions at T and P.  Are you familiar with the concept of the van't Hoff equilibrium box?

(1) I don't understand what you mean by a "raw" calculation of entropy.  I would have just written "a calculation of entropy". Raw means "unprocessed", as in raw data. Given that the methods you list determine entropy by going through the process of a calculation, rather than by direct measurement, it's the opposite of raw.  (2)  It sounds like you are trying to measure entropy directly, rather that by calculating it from other measurable quantities and, in so doing, you are presenting it by contrast with $\Delta H$, which you seem to think can be directly measured...

....but no state function can be directly measured.  They all need to be calculated from measurable quantities.  E.g., we can't directly measure $\Delta H$.  Rather, we can only determine $\Delta H$ by measuring the heat flow (a path function) at constant pressure.  So I don't see a significant qualitative difference between calculating $\Delta H$ from $q_p$ at a single temp., vs. calculating  $\Delta S$ from the temp-dependence of $q_p$ via the van't Hoff eqn....

...I.e., we can determine $\Delta S$ calorimetrically the same way we determine $\Delta H$ calorimetrically—by measuring $q_p$; the only diff. is that, for $\Delta S$, we need to measure at multiple temps. (3) You wrote "in the olden days before we had copious information on heat capacities". You don't need copious info. on heat capacities to determine $\Delta H$. You just use a small quantity of reactants, such that the heat capacity of the rxn mixture is small (and can thus be ignored) relative to that of your calorimeter. Thus the only heat capacity you need to know is your calorimeter's.

You can do a titration monitored by isothermal calorimetry. Once the titration is done, you can figure out the enthalpy of the completed reaction. Then, you know much reacted at each point (from the enthalpy fraction at that point), so you can figure out the equilibrium constant.
Taking the equilibrium constant and the enthalpy, you can calculate the Gibbs energy and the entropy. This requires that the reaction is complete at the end of the titration, but at equilibrium in the first part of the titration.
Below is an illustration of the (perfect) results of such an experiment:

Source: https://upload.wikimedia.org/wikipedia/commons/5/59/ITC_THERMOGRAM.png
In general, measuring equilibrium constant and reaction enthalpy is a common path to calculate the entropy of reaction. The equilibrium constant can be measured in many different ways (or you can use electrochemistry to directly measure the Gibbs energy of reaction).

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How to (numerically) model a phosphoric acid titration curve
I've been learning how to simulate a 1M phosphoric acid titration curve using numerical methods in R.
So far has this has been the best curve:

Notice how it flattens out (artificially) towards the left.
I am not certain if this is a programming and / or a "chemistry" issue.

The system is defined by charge balance, mass balance, and equilibrium equations:
$$[\ce{H3A}] + [\ce{H2A-}] + [\ce{HA^{2-}}] + [\ce{A^{3-}}] = P_{CA}$$
$$[\ce{H+}] + [\ce{Na+}] = [\ce{H2A-}] + 2[\ce{HA^{2-}}] + 3[\ce{A^{3-}}] + Kw/[\ce{H+}]$$
$$K_{a, 1} = [\ce{H2A-}] [\ce{H+}] / [\ce{H3A}]$$
$$K_{a, 2} = [\ce{HA^{2-}}] [\ce{H+}] / [\ce{H2A^-}]$$
$$K_{a, 3} = [\ce{A^{3-}}] [\ce{H+}] / [\ce{HA^{2-}}]$$
I've posted a very similar question at SO and shared code for the simulation over there, using R's nleqslv::nleqslv (go have a look if you'd like to).
I thought this is also a good place for this issue. For instance, this answer (and the ones it links to) do seem to be useful, but I'm not sure how to solve that equation for $[\ce{H3O+}]$ (called $x$ over there).
I would like to know which is the best way to simulate a titration curve.
And also: are the equations I used correct? do you know what am I missing?
Thanks!

You may compare the visual result with other's solutions in the field, e.g. CurTipot, especially to avoid the sharp transition above $[\ce{Na+}] > 2$.  And, for future reference, learn how to use mhchem in the body of a question / answer / comment on ChemSE (not in the title), e.g. here.

The exact answer, including the exact equation and titration curve, are already here: https://chemistry.stackexchange.com/a/43422/79678. I just found them. Also, the equation at the link is the same, except for trivial notation differences, as the one I have used for many years and use in an Excel spreadsheet here: https://chemistry.stackexchange.com/a/136203/79678. See the second spreadsheet.

So the answer I linked is one you already linked in your question! Argh! So I will do penance by telling you how to deal with that equation. You solve the equation just like in my spreadsheet (linked above). You can generate the titration curve two different ways. Easy way: pick a series of pH values in your desired range. Evaluate the equation for volume of added NaOH for each pH used. Then simply plot pH versus base volume added. This is the standard trick. Hard way: do algebra on the equation to yield a 5th order polynomial in hydrogen ion concentration. Solve that, using a root finder.

The hard way is how I make titration curves, since I use software with a nice root finder. Of course, you can use the equation with the solver in the Excel spreadsheet, but this is tedious because you would compute the pH for each base volume.

Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$.
Calculate fractions of respective phosphate forms as the function of $\ce{[H+]}$ and acidity constants.
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^3$$
$$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = [\ce{H+}] \cdot K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$a_4 =  K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot K_\mathrm{a3} = a_3 \cdot \frac {K_\mathrm{a3}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 + a_4$$
.. and then reuse it and its additive terms(*):
$$[\ce{H3A}]=c_0 \cdot \frac {a_1}{CD}$$
$$[\ce{H2A-}]=c_0 \cdot \frac {a_2}{CD}$$
$$[\ce{HA^2-}]=c_0 \cdot \frac {a_3}{CD}$$
$$[\ce{A^3-}]=c_0 \cdot \frac {a_4}{CD}$$
$c_0$ is the total phosphate concentration. From the fractions, you can get $\ce{[Na+]}$:
$$\ce{[Na+]} = \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]}$$

There is one important thing to consider: Titration dilutes the acid.
If  there is the volume $V_0$ of $\ce{H3PO4}$ with molar concentration $c_\mathrm{0,init}$, and if we spend the volume $V_1$ of $\ce{NaOH}$ solution of concentration $c_1$, then:
$$[\ce{Na+}]=c_1 \cdot \frac {V_1}{V_0 + V_1 }$$
$$c_0=c_\mathrm{0,init} \cdot \frac {V_0}{V_0 + V_1}$$
So perhaps the best approach can be:

There are given concentrations of respective solutions
For given volumes of respective solutions calculate $c_0$ and $\ce{[Na+]}$
For given $c_0$ and $\ce{[Na+]}$, calculate $\ce{[H+]}$ by finding the solution of the equation below ( $f([\ce{H+}]) =  0$ ) by numerical methods, like Regular falsi, or Newton one.

$$ \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]} - \ce{[Na+]} = 0$$
For $y_1=f(x_1)$, $y_2=f(x_2)$, $y_1 \cdot y_2 \lt 0$, there is the Regula falsi iteration:
$$x_3 =  \frac{x_2 \cdot y_1 - x_1 \cdot y_2  }{  y_1 - y_2}$$

(*) Note that the formulas are quite common, I have just rearrange them. They should be able to be found on internet. I have once derived them myself, but not this time, because I remember them.

Excelent! Thanks :) I'll post the code in another answer and accept yours.

Hi again! I've been through this again and thought I'd try getting an analytical solution too. Do you think this equation is right?  Thanks!

The R code equivalent of Poutnik's answer, for refrence:
# Sumamente facil

Ka.1 <- 7.1 * 10^-3
Ka.2 <- 6.3 * 10^-8
Ka.3 <- 4.5 * 10^-13
Kw <- 10^-14

P_ca <- 1

pH.seq <- seq(from=0,to=14,length.out = 1000)

a1 <- function(H) H^3
a2 <- function(H) H^2 * Ka.1
a3 <- function(H) H * Ka.1 * Ka.2
a4 <- function(H) Ka.1 * Ka.2 * Ka.3
cd <- function(H) a1(H) + a2(H) + a3(H) + a4(H)

H3A <- function(H) P_ca * a1(H) / cd(H)
H2A <- function(H) P_ca * a2(H) / cd(H)
HA <- function(H) P_ca * a3(H) / cd(H)
A <- function(H) P_ca * a4(H) / cd(H)

OH <- function(H) Kw/H

Na <- function(H) H2A(H) + 2*HA(H) + 3*A(H) + OH(H) - H

Na.seq <- Na(H = 10^-pH.seq)

plot(Na.seq, pH.seq)

Also as suggested by Poutnik, a volume correction method was implemented here (it's not Newton nor regula falsi, I got lazy).
Plots
Get your plots here!

https://github.com/naikymen/titration#3-experimental-data

Note that [H+] giving [Na+]=0 means [H+] of H3PO4 solution at c0. Negative [Na+] have no chemical meaning, it is just mathematical extrapolation. There will be need of iteration as c0=f([Na+]). You dilute the solution by titration.

See the A update, addressing dilution.

R code syntax note: They could not use boring  =, that had to use <- . I would hate to type it 1000 times.

Sorry about the arrows "<-" indeed assignment is not the same as equality :P (Alt + - is handy). I wrote an approximation method for the correction, calculating [Na] as before, but using it as an estimate for updating solution volume, then repeat until convergence. Seems to not-crash, but I'm not sure it converges where it's supposed to.

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Entropy and Gibbs Free Energy of Chelate Complexes
Of [Cd(NH2CH3)4]2+ and [Cd(en)2]2+ the stability of the latter is higher than the former attributing to chelation in the en complex .This is evident from the fact that the Gibbs Free Energy change for the latter is -60.7kJ/mol-1 whereas for the former it is -37.2 kJ/mol-1.
But the T*deltaS do not all provide any correlation between the stability and the entropy values as the values are -20.1kJ/mol-1 and 4.2kJ/mol-1 respectively . If the second complex has a positive entropy implying lesser stability and more randomness why do we agree on to the Gibbs value that confers stability to the second complex?
Does positive entropy changes not imply greater randomness and hence lesser stability?

Review proper units and values for entropy, unless you mean the $T \Delta S$ term of the $\Delta G$, which is not entropy.

Do not think of entropy as randomness in the way you suggest,  instead think about it as the number of configurations and energy levels that can be accessed by a molecule.

@Poutnik Yes,I corrected it. Now how should I proceed?

@porphyrin Had my interpretation of entropy been correct I woudn't have posted this question. Could you provide me link to text material that could be helpful?

have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entrop.html#e3 for a start. A good phys. chem. textbook or a textbook on statistical mechanics should explain the statistical basis for entropy in a lot more detail as well as the thermodynamic one.

@porphyrin Thanks! I will go through it.


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What is the most appropriate (and easy-to-use) all-atoms force field for liquid electrolytes?
I need to simulate a salt diluted in a liquid electrolyte.
My experience is with LAMMPS, but so far I only had coarse grained toy models, and setting up the force field was straightforward.
Now I have a real molecule (>20 atoms), which I optimized quantum mechanically with MP2 and I have tried to run various tools to automatically get the parameters: VMD/topotool, ATB+moltemplate. Nothing really works and I need to tweak too many parameters by hand.
So I am wondering whether or not to use an all-atoms design MD software such as Amber or Gromacs. Which one would be the best candidate for my study case considering the physical system? which one is more easy to use when it comes to infer the force field from QM results?
Thank you
Marco

A good question for matter modeling SE, probably better suited there than here.


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Is the case of the linear chain of atoms one-dimensional?
I am currently studying Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Mashanovitch. In chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS, the authors say the following:

On the other hand, in a covalently bonded solid like the semiconductor materials we use to make diode lasers, the uppermost energy levels of individual constituent atoms each broaden into bands of levels as the bonds are formed to make the solid. This phenomenon is illustrated in Fig. 1.4. The reason for the splitting can be realized most easily by first considering a single covalent bond. When two atoms are in close proximity, the outer valence electron of one atom can arrange itself into a low-energy bonding (symmetric) charge distribution concentrated between the two nuclei, or into a high-energy antibonding (antisymmetric) distribution devoid of charge between the two nuclei. In other words, the isolated energy level of the electron is now split into two levels due to the two ways the electron can arrange itself around the two atoms.$^1$ In a covalent bond, the electrons of the two atoms both occupy the lower energy bonding level (provided they have opposite spin), whereas the higher energy antibonding level remains empty.
$^1$The energy level splitting is often incorrectly attributed to the Pauli exclusion principle, which forbids electrons from occupying the same energy state (and thus forces the split, as the argument goes). In actuality, the splitting is a fundamental phenomenon associated with solutions to the wave equation involving two coupled systems and applies equally to probability, electromagnetic, or any other kind of waves. It has nothing to do with the Pauli exclusion principle.

If another atom is brought in line with the first two, a new charge distribution becomes possible that is neither completely bonding nor antibonding. Hence, a third energy level is formed between the two extremes. When $N$ atoms are covalently bonded into a linear chain, $N$ energy levels distributed between the lowest-energy bonding state and the highest-energy antibonding state appear, forming a band of energies. In our linear chain of atoms, spin degeneracy allows all $N$ electrons to fall into the lower half of the energy band, leaving the upper half of the band empty. However in a three-dimensional crystal, the number of energy levels is more generally equated with the number of unit cells, not the number of atoms. In typical semiconductor crystals, there are two atoms per primitive unit cell. Thus, the first atom fills the lower half of the energy band (as with the linear chain), whereas the second atom fills the upper half, such that the energy band is entirely full.

My question relates to the following section:

In our linear chain of atoms, spin degeneracy allows all $N$ electrons to fall into the lower half of the energy band, leaving the upper half of the band empty. However in a three-dimensional crystal, the number of energy levels is more generally equated with the number of unit cells, not the number of atoms.

So this explanation seems to be distinguishing between how the energy bands get filled in the case of a linear chain of atoms vs how the energy bands get filled in the case of a crystal. I understand that the case of the crystal is three-dimensional, but is the case of the linear chain of atoms one-dimensional? I'm wondering what the correct/accurate way to conceptualize this is.

I don't believe the author is trying to compare 1D and 3D rigorously. A perfectly linear chain of covalently bonded atoms doesn't exist in isolation past 3-4 atoms, to my knowledge. (longer polyynes and cumulenes have very slight bond angle deviations and are unstable).
So the 1D description is merely didactic. It's using a hypothetical linear chain of atoms to keep the explanation simple, so you can focus on picturing how the band gap arises.
In reality, semiconductor lattices consist of repeating unit cells that form a band gap.
It seems to me the passage's intent is to communicate that the real-world case of being 3D would still yield a band gap analogously to their hypothetical example. The approach in the text might be summarized as:

Here's a hypothetical example of linear atoms to make it easier to conceptualize how the band gap arises from orbital splitting.

However in reality, these aren't actually linear atomic chains in semiconductors but 3D chains of unit cells.

Nevertheless, 3D unit cells still lead to an analogous splitting effect, so you can safely extend your conception from the hypothetical example to the real world cases.


text:
Can % location of electron on metal or ligand be negative (EPR)?
Given an EPR spectrum of [Co(3-methoxysalen)(py)] :

and equations:
A(x) = A(iso) + B
A(z) = A(iso) - 2B
%Co in SOMO = B/B0
where A(x) and A(z) are hyperfine coupling constants measured from the spectrum and B0 = 7.6 mT, we have to calculate %Co. This is straightforward, but returns a negative percentage answer (-43%) - does this mean that there is no %Co in SOMO, or the value should be taken as an absolute and SOMO is in 43% based on the cobalt?
Thank you in advance for help.


text:
cis- and trans-isomers in equilibrium

Rotation of groups about double bonds are usually restricted due to large torsional strain and it is due to this reason that cis- and trans-isomers have their independent existence. However, certain structural modifications in the compound makes the rotation of groups about double bond possible and hence, for such compound, cis- and trans-isomer exist together in equilibrium.
Identify which of the following compounds has such structural feature and its cis and trans isomers exist together in equilibrium:
(A) $\ce{CH3-CH=CH-CHO}$
(B) $\ce{CH3-CH=CH-CH3}$
(C) $\ce{cyclohexyl-CH=CH-cyclohexyl}$
(D) $\ce{CH3-CH=NH}$

Could someone explain the answer step-by-step and the theory behind it?
I think that rotation of groups about a double bond can be possible due to conjugation of double bonds, so the answer should be (A). But I am afraid I haven't completely understood the logic behind it.

Have you considered keto-enol tautomers?

I would suggest D as the H atom can easily tunnel from one side on the N atom to the other. Tunnelling cannot happen in any of the other molecules as the groups are too massive.

See also https://chemistry.stackexchange.com/questions/60800/how-can-an-%CE%B1-%CE%B2-unsaturated-ketone-tautomerise

Option (D) is the answer. This may be predicted by redistribution of hybrid orbital theory or simply "Bent's Rule".
Clearly, during movement of the lone pair from one sp2 hybrid orbital to the other(to interconnect between the cis and trans isomers), there has to be an intermidiate transition state, during flipping when the lone pair resides in an unhybridized p-orbital. However the activation energy to achieve this intermidiate transition state is low for option (D).
Between N and H (as in the question), the electronegativity of N is higher than H. This pulls the shared electron pair in the N-H bond towards the N atom. In order to stabilize the high electron density, there is a redistribution of hybridization observed. The s character of the sp2 hybrid orbital of N (forming bond with the H ) increases ( eg from sp2 to sp1.8). This causes a corresponding decrease of s character of the other sp2 hybrid orbitals of N, such as the one containing the lone pair. The decrease in s character is associated with a a corresponding increase in p character of the lone pair containing orbital( eg from sp2 to sp2.3). Since it is easier to achieve transition from sp2.3 to unhybridized p orbital (having 100% p character), in the transition state it is easier to mediate between geometric isomer, possibly even existing in equilibrium.
For further reading see this answer:
Why are oxime geometrical isomers stable?


text:
Why is oxalic acid more acidic than benzoic acid?
I came across this fact while reading my book and was confused. Looking at the structures, I had the intuition that benzoic acid would be more acidic as the conjugate base formed will resonate all over the benzene ring.
Why does this happen?

Compare the induction effect of phenyl and carboxyl.

"the conjugate base will resonate all over the benzene ring"- Its not necessarily a good thing. Charge instead of resonating completely with a more electronegative oxygen is also being shared with carbon which doesn't handle it that well making it less stable.


text:
Confusion regarding dehydration of 1° alcohols
My lecturer told me that 2° and 3° alcohols prefer to dehydrate through E-1 mechanism. On the other hand, 1° alcohols tend to prefer the E-2 mechanism, which does not involve the formation of a carbocation intermediate, and thus rues out scope for rearrangement.
This is backed by online sources:

Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). Source

Another site tells me:

You will find two versions of the mechanism for the dehydration of primary alcohols on the web and in various textbooks. One of these is exactly the same as the mechanism for the reaction involving propan-2-ol and other secondary or tertiary alcohols (known technically as an E1 mechanism), but the other is different (known as an E2 mechanism).The more reliable sources give the E2 mechanism for the dehydration of primary alcohols including ethanol. I am going to treat this as the "correct" version, and have added a note as to why I think is better following the mechanism. Sourced from chemguide.

The note goes on to read:

[...]you would get a primary carbocation formed, CH3CH2+, but this is much less stable than a secondary or tertiary carbocation. That would lead to a very high activation energy for the reaction.
The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy.

Hence my first question:

What should be the preferred mechanism?

But wait a minute. That wasn't my original question.
My original question is derived from this other post: Benzene from cyclopropane-1,2,3-triyltrimethanol in acidic medium. The comments seem to suggest that this reaction is not known, while another writes that "I strongly doubt it is real".
The top voted answer suggests the formation of a carbocation intermediate leading to ring expansion. And here's where my questions coalesce. The above mentioned compound consists of three $1° OH^-$ groups. This seems to suggest that an E-2 elimination reaction should occur, which rules out the possibility of rearrangement, or for that matter, ring expansion.
TL, DR;

What should be the preferred mechanism for 1°, 2° and 3° alcohols?
If the mechanism for 1° is E-2, why should carbocation induced ring expansion occur in the dehydration of cyclopropane-1,2,3-triyltrimethanol?

I couldn't find a suitable image for the aforementioned compound; if someone does, be so kind as to edit it in.

That's a stabilized primary cation. It's debatable if it's even a primary cation at all, since the system would be homo-allylic.

@Zhe so is the answer mentioned in the other post is correct? From what I could gather, there was quite a bit of discrepancy.

The question is kind of moot because it's a hypothetical reaction. But as proposed, it would not violate the rule.


text:
Reactivity of group 13 elements with nitrogen
Do all group 13 elements react with nitrogen to form nitrides of form $\ce{MN}?$
J.D. Lee's Inorganic chemistry book mentions the following:

Aluminium burns in dinitrogen at high temperatures forming $\ce{AlN}.$ The other elements do not react.

However, a reaction of boron with nitrogen is mentioned elsewhere:
$$\ce{2B + N2 -> 2BN}$$
I'm assuming these boron and aluminium do react. Do gallium, indium and thallium react with nitrogen as well?
Please post your answer mentioning a reliable reference or source.

Boron nitrides are a real thing. Gallium and indium are as well (and wide band gap semiconductors). not sure if they are focusing on the 'react' part. Not sure about thallium...

Yeah I've seen gallium and indium nitrides been referred to at some places too, but my question is whether they can be formed by direct reaction with dinitrogen.

Well, it turns out N2 is really hard to break up for a reaction. For example, you can't react iron with N2 - you need to use ammonia (or a plasma source to get atomic N). You will note that air is 80% N2, and many metals are perfectly happy with N2 (not so much O2). In general, both gallium and indium are perfectly happy sitting around in air.

@JonCuster https://www.sciencedirect.com/science/article/abs/pii/S0254058419300914

Cool. I would not consider reactive sputtering to be reacting with N2 - the plasma provides atomic nitrogen.

All form nitrides of the formula $\ce{\overset{+3}{M} \overset{-3}{N}}$

All group 13 elements (can technically) form nitrides from the direct combination of the elements (elemental reaction).

Boron: See @Oscar's answer. Also, see references(1,2)
Aluminum: There is a paper(3) which discussed about combustion of aluminum in a high-temperature and high pressure (up to 300 MPa) nitrogen atmosphere to form aluminum nitride.
Gallium: Gallium nitride is formed by the direct combination of elements at 1200 °C.
Indium: Direct combination of indium and nitrogen cannot be done due to factors like dissociation of $\ce{InN}$ at around 550 °C and high-equilibrium partial pressure of nitrogen over $\ce{InN}$ films. So, a technique called ion-implantation is used to obtain indium nitride nanoclusters over silica matrix. There is also a certain low-temp reaction (250°C) of indium iodide and nitrogen to obtain indium nitride.
Thallium: Thallium nitride has been synthesized in a film form by DC reactive sputtering of metal thallium in nitrogen.

References

https://www.osti.gov/servlets/purl/591090
Direct elementary reactions of boron and nitrogen at high pressures and temperatures
C. S. Yoo, J. Akella, H. Cynn, and M. Nicol, Phys. Rev. B 56, 140 (1997), DOI: https://doi.org/10.1103/PhysRevB.56.140
Loryan, V.É., Borovinskaya, I.P. Aluminum Combustion in Nitrogen. Combustion, Explosion, and Shock Waves 39, 525–533 (2003). https://doi.org/10.1023/A:1026105600954
Nitrogen Compounds of Gallium. III, W. C. Johnson, J. B. Parson, and M. C. Crew
The Journal of Physical Chemistry 1932 36 (10), 2651-2654
DOI: 10.1021/j150340a015
P. Santhana Raman, K.G.M. Nair, Jay Ghatak, Umananda M. Bhatta, P.V. Satyam, S. Kalavathi, B.K. Panigrahi & V. Ravichandran (2013) Formation of embedded indium nitride and indium oxide nanoclusters in silica samples sequentially implanted with indium and nitrogen ions, Journal of Experimental Nanoscience, 8:7-8, 957-964, DOI: 10.1080/17458080.2011.630034
Indium nitride from indium iodide at low temperatures: synthesis and their optical properties by Changzheng Wu et.al., New J. Chem., 2005,29, 1610-1615, https://doi.org/10.1039/B510426B
Synthesis and properties of thallium nitride films, L.R.Shaginyan, Materials Chemistry and Physics, Volume 227, 2019 , Pages 157-162, https://doi.org/10.1016/j.matchemphys.2019.02.005

Good summary here +1.  Given the relatively low stability of Tl(III) and nitrogen being only a modest reducing agent, the formation of thallium nitride as a 1:1 compound is fairly interesting.

In the case of boron, a reaction with nitrogen does occur under some eye-popping conditions:

Combustion of boron powder in nitrogen plasma at 5500 °C yields ultrafine boron nitride used for lubricants and toners.[1]

The characteristic molecular thermal energy $kT$ at this temperature is roughly half an electron volt, which is enough to at least partially break a significant fraction of the nitrogen-nitrogen bonds.  The boron is also vaporized at this temperature, and both of these factors facilitate the reaction.  Vaporizing the boron also favors forming the fine particles which are desired for the lubricant and toner applications noted above.
Most boron nitride, not requiring the ultrafine particles, is synthesized by metathesis reactions under more typical industrial conditions, which are also given in the WP article noted above.  These commonly involve boron oxide, which both becomes fluid at much lower temperatures than the element and can act as an acid towards basic nitrogen sources such as ammonia.
Cited Reference
1.
Paine, Robert T.; Narula, Chaitanya K. (1990). "Synthetic Routes to Boron Nitride". Chemical Reviews. 90: 73–91.


text:
Fischer's projection, enantiomers and diastereomers
Given Fischer's projections of a few stereoisomers of some compound, how can we find out if they are enantiomers, diastereomers, etc.? From what I understand, we can rotate the projections to have the same groups on the top and bottom and then compare (non-superimposable mirror images --> enantiomers, non-mirror image --> diastereomers) and we can convert to eclipsed sawhorse projections and vice-versa, could anyone explain further?
EDIT: I have a doubt in the following MCQ question from VMC's JEE archive -->
Q. Which of the following forms a pair of enantiomers?

(A) I and II
(B) III and IV
(C) I and IV
(D) I and III
The answer is given as (D), that is, I and III. I also got this as an answer, but I also got III and IV, which is (B) as enantiomers, as I and IV were identical in my solution

Assign Cahn-Ingold-Prelog designations to each stereocenter and compare the related centers.

You don't need the eclipsed sawhorse. As for the rest, you explained it yourself pretty well; what's to explain further?

@IvanNeretin I just edited the question...my doubt is actually in an MCQ question, which I have now included

On my end, the question shows no signs of having been edited.

Maybe its some connectivity issues or something on my side, please check again in some time maybe


text:
Why would substitution of H with D alter chemical shift?
The $\ce{^{119}Sn}$ chemical shift of the dimeric tin hydride shown below (R=terphenyl) changes on addition of $\ce{D2}$ from a 1:2:1 triplet at $\pu{657.9 ppm}$ (due to coupling between $\ce{Sn}$ and bridging $\ce{H}$) to a doublet at $\pu{650.6 ppm}$. I assume that's because of reversible dissociation of one of the bridging $\ce{H}$, which allows formation of a mixed species where both $\ce{H}$ and $\ce{D}$ are present (the doublet being due to coupling with the remaining $\ce{H}$). However, I'm not quite certain why that alters the $\ce{^{119}Sn}$ chemical shift (I know the $\ce{Sn-D}$ bond will be shorter/stronger due to the lower ZPE). I would be interested in any thoughts/comments.

You can probably find some relevant literature if you search for "nmr isotope shift". Not sure what level of theory you're looking for: some of the older papers can get really theoretical!

OP's main question: Why would substitution of $\ce{H}$ with $\ce{D}$ alter chemical shift?

OP has correctly indicated that the $\ce{Sn-D}$ bond will be shorter/stronger than $\ce{Sn-H}$ due to the lower ZPE, which may cause the chemical shift difference. That is the major reason for isotope effect in chemical kinetics mainly due to bond strength (shorter the bond distance, stronger the bond). However, here, how change of isotope effected the chemical shift is caused by increasing the electron density of the adjacent atom ($\ce{Sn}$). Since $\ce{Sn-D}$ distance is shorter than $\ce{Sn-H}$, the sharing electron pair in $\ce{Sn-D}$ bond is closer to $\ce{Sn}$ tan that in $\ce{Sn-H}$ bond. Thus, it is safe to say that $\ce{Sn}$ of $\ce{Sn-D}$ is shielded more than that of $\ce{Sn-H}$. Therefore, we can expect up-field shift (red-shift) in relevant $\ce{Sn}$ chemical shift. That's what you have experienced $(\pu{650.6 ppm} - \pu{657.9 ppm} = -\pu{7.3 ppm})$ when you have changed the $\ce{H}$ with $\ce{D}$ of the dimer. This phenomena is common among deuterated organic compounds (Ref.1):
$$
\begin{array}{llrrr}
\hline
\text{Hydrated cpd} & \text{deuterated cpd}  & \delta \ \left(\ce{^{13}C-H}\right)/\pu{ppm} &  \delta \ \left(\ce{^{13}C-D}\right)/\pu{ppm} & \Delta \delta/\pu{ppm}\\
\hline
\ce{CHCl3} & \ce{CDCl3} & 77.36  & 77.16 \pm 0.06 & -0.20  \\
\ce{CH3C#N} & \ce{CD3C#N} & 1.79  & 1.32 \pm 0.02 & -0.47  \\
\ce{(CH3)2C=O} & \ce{(CD3)2C=O} & 30.60 & 29.84 \pm 0.01 & -0.76 \\
\ce{C6H6} & \ce{C6D6} & 128.62 & 128.06 \pm 0.02 & -0.56  \\
\ce{CH3-OH} & \ce{CD3-OH} & 49.86 & 49.00 \pm 0.01 & -0.86 \\
\hline
\end{array}
$$
These chemical shifts are also depend on the other factors such as temperature and solvent. For comparison purposes, all chemical shifts of hydrated compounds have been recorded in their deuterated versions (e.g., $\ce{CHCl3}$ in $\ce{CDCl3}$) at a constant temperature (Ref.1).
Note: I also realized that $^1\!J_{\ce{Sn-D}}$ constant must be zero or negligible according to your observation. I tried to find the value of $^1\!J_{\ce{Sn-D}}$, but failed. Note that in deuterated organic solvents, $^1\!J_{\ce{C-D}}$ varies between $\pu{20 Hz}$ and $\pu{30 Hz}$ (Ref.1).
References:

Hugo E. Gottlieb, Vadim Kotlyar, Abraham Nudelman, "NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities," J. Org. Chem. 1997, 62(21), 7512-7515 (DOI: https://doi.org/10.1021/jo971176v).


text:
Does the SDF format have any benefit over SMILES?
The eMolecules page for example offers a complete dataset containing ~26 million structures. They're providing it either as a 300 MB SMILES archive or 6 GB SDF archive (1.5 GB and 50 GB uncompressed respectively).
Does the SDF file contain any information about the chemical structures, that the SMILES format does not contain?
First record in SDF:

  Mrv1722001051816042D

 37 40  0  0  0  0            999 V2000
    4.3268    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.6124    0.3220    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.8979    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.1834    0.3220    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.4690    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.7545    0.3220    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0400    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.6744    0.3220    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    4.3268    1.5595    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
    5.0413    0.3220    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
    5.7558    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.4703    0.3220    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    7.1847    0.7345    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.7558    1.5595    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.4703    1.9720    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.4703    2.7970    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    7.1847    3.2095    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    7.8992    2.7970    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.6137    3.2095    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.7852    4.0165    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.6057    4.1027    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.9412    3.3491    0.0000 C   0  0  2  0  0  0  0  0  0  0  0  0
   10.7482    3.1775    0.0000 C   0  0  1  0  0  0  0  0  0  0  0  0
   11.3002    3.7906    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   12.1072    3.6191    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   12.6592    4.2322    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   13.4662    4.0607    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   14.0183    4.6738    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   13.7212    3.2760    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   11.0031    2.3929    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.3281    2.7970    0.0000 C   0  0  2  0  0  0  0  0  0  0  0  0
   10.0818    2.4615    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.3281    1.9720    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.6137    1.5595    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    7.8992    1.9720    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    7.1847    1.5595    0.0000 C   0  0  1  0  0  0  0  0  0  0  0  0
    7.8992    1.1470    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0  0  0  0
  1  9  2  0  0  0  0
  1 10  1  0  0  0  0
  2  3  1  0  0  0  0
  3  4  1  0  0  0  0
  4  5  1  0  0  0  0
  5  6  1  0  0  0  0
  6  7  1  0  0  0  0
  7  8  1  0  0  0  0
 10 11  1  0  0  0  0
 11 12  1  0  0  0  0
 11 14  1  0  0  0  0
 12 13  1  0  0  0  0
 13 36  1  0  0  0  0
 14 15  1  0  0  0  0
 15 16  2  0  0  0  0
 15 36  1  0  0  0  0
 16 17  1  0  0  0  0
 17 18  1  0  0  0  0
 18 19  1  0  0  0  0
 18 35  1  0  0  0  0
 19 20  1  0  0  0  0
 19 31  1  0  0  0  0
 20 21  1  0  0  0  0
 22 21  1  0  0  0  0
 22 23  1  0  0  0  0
 22 31  1  6  0  0  0
 23 24  1  0  0  0  0
 23 30  1  1  0  0  0
 24 25  1  0  0  0  0
 25 26  1  0  0  0  0
 26 27  1  0  0  0  0
 27 28  1  0  0  0  0
 27 29  1  0  0  0  0
 31 32  1  6  0  0  0
 31 33  1  0  0  0  0
 33 34  1  0  0  0  0
 34 35  1  0  0  0  0
 35 36  1  0  0  0  0
 36 37  1  6  0  0  0
M  END
>  <EMOL_VERSION_ID>
9986

>  <EMOL_PARENT_ID>
9986

>  <EMOL_LINK>
https://www.emolecules.com/search/#?query=9986&querytype=emoleculesid

Letting MarvinJs draw it:

The same record in SMILES:
CCCCCCCC(=O)OC1CC[C@]2(C)C3CC[C@]4(C)[C@H](CCC4C3CC=C2C1)[C@H](C)CCCC(C)C 9986 9986

Letting MarvinJs draw it

Obviously it is mirrored. Since I am no chemist, would it make any difference for you? Should I expect more discrepancies like that, even more complicated ones?
The question stated differently: Does it give a scientist searching for a catalogue of structures a real benefit when using the vastly more explicit SDF format? From a programmer's perspective, carrying around gigabytes is far more difficult than megabytes, so the benefit should be reasonable.

Is is worse than just mirrored. Looks like the stereocenters are simply assigned some random configuration.

SMILES ≅ SDF − (atomic coordinates).

The SDF allows storage of additional data/metadata - supplier SDFs often contain things like CAS, catalogue info, names, salt form info etc. If you JUST need a structure there is no benefit

@Marc Out of curiosity, what text editor did you use to open that 70.7 GB version.sdf? I only managed to do it with EmEditor, but it's a proprietary tool. Sublime took about 15 min and eventually silently crushed. VSCode, Notepad++, emacs and vim refused to open the file.

Unzip the downloading file on the fly using the Linux shell: curl https://downloads.emolecules.com/free/2021-03-01/version.sdf.gz | zcat > emolecules.sdf.txt and interrupt it after a few seconds by ctrl+c gives you "enough from the beginning, but not all of it to crash your editor"

On Windows you could check out https://github.com/RolandPheasant/TailBlazer - it's intended for log files, but I assume SDF would work pretty well, too.

@MarcWittke Thank you for the info. TailBlazer doesn't display the file for me. I found that glogg loads the file for a couple of minutes, but then allows for snappy navigation. And MarvinView opens the file instantly and lists all the structures.

@andselisk, if you need a read-only editor - less will open it instantly (f to move forward, b to move backward, g - jump to the beginning, G - jump to the end).

@StanislavBashkyrtsev Thank you very much, less works like a charm. File opening, navigation and search are instant.

In my experience, the most relevant type of information that a (classic) SMILES does not encode, whereas a SDF does, is about stereochemistry.
Most commonly, the explicit stereocenters in a molecular representation with no further information can be interpreted as 'absolute', i.e. fixed in the configuration formally encoded by the literal SMILES, or as belonging to an 'and enantiomer' group, i.e. all the stereocenters in the molecule must be interpreted as the represented ones + their mirror image.
This distinction is made in an SDF by a specific 'flag'.
See 'chiral flag' in this Wikipedia page:
https://en.wikipedia.org/wiki/Chemical_table_file#Counts_line_block_specification
Going even further, one can decide to have within the same molecule separate groups of 'absolute', 'and' and even 'or' stereocenters, all of which can be encoded in special versions of SDF's.
See 'enhanced stereochemical representation' in this document, for instance:
http://help.accelrysonline.com/insight/2016/content/pdf_files/bioviachemicalrepresentation.pdf
A classic SMILES can't tell you any of this, so when you only have a SMILES string and no additional information, you may 'assume' that it represents the depicted species and nothing else; but I can assure you that doing that can be the source of endless trouble, especially when the representation is used to buy real chemical samples from vendors, or even just for statistical modelling.
Imagine for instance a case where the substance that the originator of the SMILES 'means' to represent is in fact an 'and' mixture (thus of at least 2 enantiomers). If you had an SDF, you would know that for sure, because the 'chiral flag' would be 0. But with only a SMILES you don't know.
Say you have biological activity data associated to this SMILES. The activity might have been caused by the depicted species, or by its mirror image.
But if you do not consider that, you will associate one specific 'absolute' species to activity, maybe the wrong one, just because you don't know any better.
PS
Unfortunately I must disagree with some of the comments above.
SMILES can perfectly encode salts and solvates. Fragments in a SMILES are separated by '.'.
As for the inversion of the stereocenters when converting a SMILES text to representation, well that for sure should never happen, so if it does, there is something wrong in the software. There are very clear and unambiguous rules to decide how to convert '@' and '@@' to the correct configuration - otherwise it would be pretty useless, to be honest.

+1 Agreed that SDF can provide enhanced stereochemical information not in a SMILES. I'm not sure that particular example does .. I'm pretty sure it's auto-generated from the SMILES by eMolecules.

@GeoffHutchison : you're probably right about this particular catalogue. I was just answering the OP's question, which was more general: "Does it gives a scientist searching for a catalogue of structures a real benefit when using the vastly more explicit SDF format?". In fact, in the example he showed, the molecule has multiple explicit stereocenters, and the chiral flag is 0, so the face-value interpretation a cheminformatician would give of the SDF is: racemic mixture; the SMILES instead would probably be assumed to be absolute. So depends really how much one cares about this aspect.

In my work as a cheminformatician, I did not want to make assumptions, so I asked one by one the chemical vendors who provided SDF's to us how I should interpret their chiral flag (intentionally set to 0 or 1, or not?), and all the one who provided SMILES to us if we should assume their molecules were the literal SMILES ones or mixtures. Some of them knew what I was talking about and answered; many others frankly were quite unprepared and cooked up some unsatisfactory and unconvincing story. But when your research can be severely harmed by such assumptions, you let someone else make them :)

These two formats have different purposes. SMILES encodes only covalent connectivity while SDF also (in principle) specifies the conformation of the molecule (by including atomic coordinates).
SMILES doesn't attempt to represent the three-dimensional structure of the molecule. For instance, the alkane tail of your molecule is quite flexible and is likely to occupy many different rotamers at standard temperature. SMILES does not specify a rotamer. On the other hand, SDF includes the three-dimensional coordinates of each atom. While you can generate atomic positions from a SMILES representation, that would require energy minimization using some model of bonded and nonbonded forces or a quantum chemistry calculation. SDF explicitly specifies the conformation, which could come from an empirical force field, quantum chemistry, or experiment, like NMR or x-ray crystallography. Or the atomic coordinates could just be generated automatically from a two-dimensional diagram (as in your example) or just be complete garbage.
So you should note that the atomic coordinates in the file you include all have z components of zero. This is not at all a realistic conformation for your molecule. At minimum the cyclohexane rings would be puckered and there is no way for most of the hydrogen atoms to be in the same plane as everything else. Just because an SDF file includes atomic coordinates, doesn't mean these coordinates are any good.

Re «z components of zero», this is like running a conversion from smiles to .sdf with OpenBabel but omitting its --gen3d toggle.  In addition, often energetic minimizations of conformers are about isolated, individual molecules.  These geometries need not be the ones about molecules in the solid (or even crystalline) state.  Here, the intermolecular interactions may render the question for «what is the most favourable geometry?» with multiple shallow minima more difficult, despite public blind tests.

@Buttonwood I totally agree. Thanks for the complementary information!

tldr; SMILES is often used as a compact format for chemical database searching
The SMILES representation is a compact string format indicating the atom-bond graph including stereochemistry.
In some cases the SDF may include additional information not possible to store in a SMILES. It will almost always include 2D or 3D coordinates for layout, so presentation is helpful.
Because the SDF allows arbitrary key/value property storage, there have been several extensions. Most notably there are efforts to support "zero-order bonds" for metal coordination, e.g.:
J. Chem. Inf. Model. 2011, 51, 12, 3149–3157
In this particular case, there doesn't seem to be a lot of chemical information in the SDF beyond what's in the SMILES. (I'd guess that the SDF are auto-generated from the SMILES.)
One other caveat. The typical "v2" SDF indicated above is limited to 999 atoms and 999 bonds. Many programs support the v3000 SDF with larger molecules, but YMMV.
In principal, there's no limit to the length of a SMILES, although I suspect many programs would choke on something really big (e.g., >32k atoms).


text:
Solubility of metal Hydroxides in excess Sodium Hydroxide (excess NaOH)
I recently came across a question which asked me to find out which metal hydroxides are soluble in excess $\ce{NaOH}$ solution.
Number of hydroxides which are soluble in excess of $\ce{NaOH}$ among the following given hydroxide?
$\ce{Al(OH)3, Zn(OH)2, Cr(OH)3, Pb(OH)2, Mn(OH)2, Fe(OH)3, Ni(OH)2, Sn(OH)2.}$
Is there any way to determine rationally which of these hydroxides will be soluble or should I rely on remembering them?

The only hydroxides which are insoluble in an excès of $\ce{NaOH}$ are $\ce{Mn(OH)2, Fe(OH)3, Ni(OH)2}$. The other hydroxydes are soluble in a concentrated solution of $\ce{NaOH}$. I don't know whether there is a way of predicting this result, apart from experiment.

At least $\ce{Al(OH)3}$ is soluble. Look here for explanation.

There are three kind of oxides/hydroxides: basic (e.g., $\ce{NaOH}$, metal hydroxides), acidic (e.g., $\ce{CO2}$, non-metal oxides), and amphoteric (e.g., $\ce{Al(OH)3}$, Group 13 and 14 hydroxides). Amphoteric Hydroxides act either as Bronsted-Lowry bases (accepting protons) or as Lewis acids (accepting an electron pair), depending on reaction conditions. For example, $\ce{Al^3+}$ in $\ce{Al(OH)3}$ is a Lewis acid, which can accept an electron pair from either Lewis or Bronsted-Lowry base:
$$\ce{Al(OH)3 (aq) + OH- (aq) -> Al(OH)4- (aq)}$$
The resulting neutral compound, $\ce{Na[Al(OH)4]}$ is much more soluble in water (it is a sodium salt) than Aluminum Hydroxide. Besides $\ce{Al(OH)3, Pb(OH)2}$ and $\ce{Sn(OH)2}$ are also dissolved in excess $\ce{NaOH}$ (Group 14 hydroxides):
$$\ce{Pb(OH)2 (aq) + 2OH- (aq) -> Pb(OH)4^2- (aq)}$$
$$\ce{Sn(OH)2 (aq) + 2OH- (aq) -> Sn(OH)4^2- (aq)}$$
$\ce{Zn^2+}$ in $\ce{Zn(OH)2}$ is also a strong Lewis acid, thus, $\ce{Zn^2+}$ can accept an electron pair from either Lewis or Bronsted-Lowry base (Wikipedia):
$$\ce{Zn(OH)2 (aq) + 2OH- (aq) -> Zn(OH)4^2- (aq)}$$
It is well known that $\ce{Cr(OH)3}$ is also an amphoteric hydroxide (Wikipedia and Ref.1 & 2), which dissolve in excess $\ce{NaOH}$:
$$\ce{Cr(OH)3 (aq) + OH- (aq) -> CrO2- (aq) + 2 H2O}$$
Ref.1 also reports that following reaction with even water is evident:
$$\ce{Cr(OH)3 (s) + H2O <=> Cr(OH)4- (aq) + 2 H+ (aq)}$$
These facts leaves only $\ce{Fe(OH)3, Ni(OH)2},$ and $\ce{Mn(OH)2}$ are the hydroxides, which are not soluble in excess of $\ce{NaOH}$.
References:

Dhanpat Rai, Bruce M. Sass, Dean A. Moore, "Chromium(III) hydrolysis constants and solubility of chromium(III) hydroxide," Inorg. Chem. 1987, 26(3), 345–349 (https://doi.org/10.1021/ic00250a002).
Richard A. Reinhardt, “The interaction of chromium(III) ion with hydroxide ion. An experiment for the undergraduate inorganic laboratory,” J. Chem. Educ. 1966, 43(7), 382-383 (https://doi.org/10.1021/ed043p382).


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Understanding Joule-Thomson expansion apparatus
I am reading the Joule-Thomson experiment and although I can follow the mathematics I can't understand what is going on. Here is the apparatus

taken from Atkins. So initial the gas is at $P_i$ and because of the opening and the difference in pressure it will eventually reach the right compartment with $P_f$. What I can't understand is what the role of porous plug and how this setup differs from an adiabatic expansion. I mean couldn't we let the the gas expand from $P_i$ to $P_f$ and then record the difference in temperature? Also what is the role of porous plug in this setup if the system doesn't exchange heat and the work is done on the environment in right and left compartment respectively.

The purpose of the porous plug is to cause a viscous frictional pressure drop, resulting from the gas flow through the tiny interstitial pores of the porous medium.  This results in a dissipation of mechanical energy (pressure and kinetic energy) to internal energy, which translates into a "viscous heating" to essentially offset the adiabatic expansion cooling of the gas.  The net result is no change in temperature for an ideal gas, and small change in temperature for a real gas.


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Comparing the zero point vibrational energies of DCl and HCl

(a) Compare the zero point energy (i.e. $E_0)$ in joules for $\ce{H^{35}Cl}$ and $\ce{D^{35}Cl}$ (assuming both are anharmonic oscillators).
(b) Explain the effect this will this have on their chemistry (you may wish to use a diagram to help here).

I have calculated a value for the vibrational frequency and have a value of the anharmonicity constant for $\ce{H^{35}Cl}$ (but not the anharmonicity constant for $\ce{D^{35}Cl}).$ I am unsure on how to proceed with these as all I know is an equation linking vibrational energy to the vibrational frequency and anharmonicity constant. I would be so grateful for any insight into this.

Hint: look at the values of H$_2$ and D$_2$ in the X state and see if you can find the mass scaling of $\omega_e$ and $\omega_e x_e$.

Marina, I have edited the answer a little bit, the previous answer was slightly wrong btw.

I don't know which equations you are using, but this is the one I was taught:
$$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$
Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity constant:
$$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$
where $D_\mathrm{e}$ is the depth of the lowest point of the potential well (not the depth of the $v = 0$ vibrational level!).
The zero-point energy is the energy when $v=0$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $(D_\mathrm{e})$ of the $\ce{H-Cl}$ and $\ce{D-Cl}$ bonds are same.
The only difference between $\ce{H-Cl}$ and $\ce{D-Cl}$ would be in the value of the vibrational wavenumber $(\tilde{\nu}),$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.

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


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Why is the energy levels in hydrogen independent of the quantum numbers l and ml?
I've been looking for an answer for this for quite a bit but I'm still quite confused. On the independence from ml, I think that it originates from the symmetry of the molecule. However, the independence from l, is what confuses me. Others have mentioned that it is due to the fact that the potential energy is purely Coulombic and related to the 1/r aspect. However, my textbook shows that the effective potential energy (related to the radial wavefunction) actually depends on two things: the Coulomb potential energy and the centrifugal force from angular momentum. The latter has the l quantum number, so when l = 0, the effective potential energy is purely coulombic. However, when l is not 0, the effective potential isn't purely Coulombic, which contradicts my earlier statement. So I'm a bit confused on whether the potential energy is purely coulombic/not, or if we are referring to different potential energies?

Your thinking is correct. As it happens there exists a cunning ploy to confuse us all :) In a Coulomb potential there is an accidental degeneracy which means that the energy does not depend on $l$, The degeneracy wrt the magnetic quantum number $m_l$ depends on symmetry, i.e. an atoms properties are independent of orientation in space ( in the absence of external fields)


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Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid despite its intramolecular hydrogen bonding?
I was reading about $\mathrm{p}K_\mathrm{a}$ values of 2-fluorobenzoic acid and 4-fluorobenzoic acid and found that the latter had a higher value which was surprising to me. I think that intramolecular hydrogen bonding should dominate all the other effects in 2-fluorobenzoic acid as mentioned in the following figure and thus it should have a higher value:

What is wrong with my logic?

See ortho effect in benzoic acid.

@LightYagami I don't think ortho effect has a role to play here. Fluorine has too small an A Value of 0.15 compared to the groups which prominently show ortho effect, and the crystal structure here shows the cooh group almost in the plane of the ring.

@Rishi Also note that 2-Fluorophenol is more acidic than phenol even after having intramolecular H-bonding because of the fact that F has a strong -I effect. So I think even here the "distance-dependent" inductive effect has taken over the H-bonding.

@LightYagami ortho effect won't come into play here because of the small steric size of fluorine.

You seem to be asking the question "Why does this intramolecular hydrogen bonding not cause a considerable effect to the $\mathrm pK_\mathrm a$ values of these compounds?"

TL;DR That is because this type of bonding is only present in a specific conformer only and this type of conformer of the compound 2-fluorobenzoic acid exists in low population levels, 2%$^{[1]}$ to not affect the ${\mathrm pK_\mathrm a}$ of the compound considerably.

Regarding the conformers of the 2-FluoroBenzoic Acid

A throughout conformational search on the B3LYP/6-311++G(d,p) potential energy surfaces of the
studied molecules (the mono- and di- ortho fluoro- or/and chloro- substituted benzoic acids, plus the
parent benzoic acid, for completeness) was undertaken. The identified conformers are presented$^{[2]}$

Here both the cis conformers are planar and low energy while only one trans conformer is planar while both of them are relatively high energy of which Trans II is of relatively lower energy.

The two cis conformers have rather similar energies, with cis-II (where
the fluoro-substituent stays the same side of the molecule as the acid oxygen atom and a C-H . . . O=
stabilizing interaction exists), being the most stable form. The cis-I conformer has a relative energy of
2.82 kJ·mol−1
. In this conformer, the C-H . . . O= interaction present in cis-II is replaced by a weaker
stabilizing C-H . . . OH interaction. The C-H . . . O= interaction is more efficient in stabilizing cis-II than
the C-H . . . OH interaction in stabilizing cis-I due to the more favorable localization of the interacting
carbonyl oxygen lone electron pair, which stays in the molecular plane, whereas in cis-I both lone
electron pairs of the acid oxygen atom are out of the molecular plane. On the other hand, the interactions
between the fluoro-substituent and the oxygen atoms, which are of repulsive nature, also favor a lower
energy for cis-II compared to cis-I. This can be rationalized in similar terms as for the interactions
involving the ortho hydrogen atom discussed above. In cis-I, the F . . . O= repulsion is stronger than the
F
. . . OH repulsion in cis-II because the interacting lone electron pair of the carbonyl oxygen atom is in
the same plane as the fluorine atom in the former conformer, while the lone electron pairs of the acid
oxygen atom in cis-II are both out of the plane in relation to the fluoro-substituent.$^{[2]}$

It is important to note that though the population level of Trans II are low, it is still a stable compound and nevertheless has been detected.$^{[3]}$

So.. Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid?

I could not find any research for this but I could make an educated guess that the acidity of 2-fluorobenzoic acid comes from the similar stabilization as that in salicylic acid.$^{[4]}$ in the conformer cis II as shown below:-

References:

Kuş, N.; Fausto, R. Effects of the matrix and intramolecular interactions on the stability of the higher-energy conformers of 2-fluorobenzoic acid. J. Chem. Phys. 2017, 146 (12), 124305. [DOI: 10.1063/1.4978794]
(https://doi.org/10.1063/1.4978794).
Ogruc Ildiz, G.; Fausto, R. Structural Aspects of the Ortho Chloro- and Fluoro- Substituted Benzoic Acids: Implications on Chemical Properties. Molecules 2020, 25 (21), 4908. DOI: 10.3390/molecules25214908.
Daly, A. M.; Carey, S. J.; Pejlovas, A. M.; Li, K.; Kang, L.; Kukolich, S. G. Gas phase measurements of mono-fluoro-benzoic acids and the dimer of 3-fluoro-benzoic acid. J. Chem. Phys. 2015, 142 (14), 144303. DOI: 10.1063/1.4917031.
Dunn, G. E.; Penner, T. L. Effect of intramolecular hydrogen bonding on the relative acidities of substituted salicylic acids in benzene solution. Can. J. Chem. 1967, 45 (14), 1699–1706. DOI: 10.1139/v67-274.


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Green deposition on the anode after electrolysis of sodium bicarbonate solution in tap water
I use $\pu{12 V}$ battery with copper wires and tap water with $\ce{NaHCO3}$ dissolved in it. I connected the battery and everything went as expected but after 5 minutes bubbling stopped. I took out the copper electrodes and saw green deposits on the anode that were blocking the current.
I think green stuff was malachite formed by the reaction
$$\ce{2 Cu + O2 + H2O + CO2 -> Cu2(OH)2CO3}$$
because there is oxygen in the anode water in solution copper in the wire but how did the $\ce{CO2}$ get there? I know bicarbonates hydrolyze to give carbonic acid which gives $\ce{CO2}$ but doesn't that only happen at the surface? Any ideas what that green thing might be?

Tap water leaves room for imagination due to possible side reactions occurring with the dissolved salts and even organics. Ideally you want to use distilled water for any electrochemical experiment.

If sodium bicarbonate is dissolved in water, it produces $\ce{HCO3-}$ ions which are in equilibrium with $\ce{CO2}$ according to the following equilibrium $$\ce{2HCO3- <=> CO3^{2-} + CO2 + H2O}$$ This is the origin of the $\ce{CO2}$ engaged in your equation.

So are there $\ce{CO2}$ molecules everywhere in the solution?

Yes. CO2 molecules are everywhere in a $\ce{NaHCO3}$ solution.

https://chemistry.stackexchange.com/questions/49578/copper-chloride-color-in-electrolysis-of-salty-water https://chemistry.stackexchange.com/questions/138612/electrolysis-with-baking-soda-solution-a-penny-as-the-cathode-and-graphite-for


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Ordered and random energy
One of the books refer to potential energy being ordered whereas heat energy being random. Could someone explain what is ordered and random energy?

A falling object moves in one direction, and only one : downwards. A pendulum may move in two directions : North-South and/or East-West or a combination of both. It has two degrees of freedom. Its movement is more random than a falling object. If now the pendulum hangs at a rubber band, it gets a third degree of freedom : up-downwards. Its movement is still more random. A hot body is made of particules that can move or vibrate like my elastic pendulum. It has three degrees of freedom. Its movement is more random.

Ordered and random energy is merely an abuse of language.

Imagine a group of children at a summer camp trying to pick up a heavy wooden log/trunk a) orderly by command b) randomly. The present energy is the same, but in the latter case of less use or even useless. That is like the case of the thermal energy, than can be directly use as the energy source only by balancing temperature with a colder object. For other cases, you need other primary energy source to pump this energy in thermal pumps.

Such a textbook should not be used which propagates wrong ideas.

Could you provide passages (quotes) from the textbook that explicitly refer to "ordered" and "random" energy?

Consider the ability to predict something in a detailed manner as synonymous with "ordered", and the inability synonymous with "random". For instance, a perfect crystalline solid is more ordered than a gas because given the position of one particle in a solid and a few lattice constants we can derive to a good extent the positions of all particles in the solid. In a gas, knowing the position of one particle constrains the maximum distance of all others (given the temperature and pressure) and with some additional parameters provides perhaps a distribution function for the particles, but we can say little more about where the other particles actually are (we might say something detailed however about where they might be).
Consider an analogy useful to get physical intuition into why we might really refer to a system as "ordered" or "random". Consider a system of many balls colliding elastically. Up to some number of balls we can describe the motions of the balls with a high level of detail and maybe even predict future trajectories. We would consider this a relatively orderly process. But if the balls become too many, then we give up and instead try to describe the statistical properties of the collection of balls, and say the balls move randomly.
Usually one speaks of a system being ordered or a configuration being random, but not about energy itself. However, heat can be understood as energy being transferred through the random motion of particles, whereas work involves coherent motion of the same particles (in a gas some of the molecules happen to move coherently, that is, in what looks like an orderly parallel displacement. These are responsible for work). In that sense one might describe heat as involving a transfer of "random energy" and work as being a transfer of "ordered energy".
So what decides whether you call something random versus ordered (or heat or work) is then how much information you have to describe the system. The idea is that describing the detailed motion (positions and momenta) of a very large numbers of things, on the order of $\pu{6E23}$  particles (moles of stuff), and on top of that as a function of time, is difficult (actually it's impossible). So instead of even trying we speak of "heat" as the energy that is exchanged by all those particles when they collide with each other in ways we can't measure or predict in detail. We say that their motion is random, which is shorthand for saying we can't describe their motions except perhaps in a statistical sense, using measures such as entropy. Work (and change in potential energy) on the other hand is something we can measure in detail, because we use things such as volumes and distances of large objects to describe it. For instance when a gas expands we can measure how much the volume changed against an externally applied pressure (force per unit area). So we call this an orderly process, even though the molecules in the gas that do the work are moving about in individually undescribable ways.
Another example is the solar system versus a galaxy. The solar system does not appear random because we can describe the motions of the sun, planets and other large objects in great detail, and sometimes predict their motion far into the future. OTOH when we look at a galaxy the large number of objects make a detailed description more challenging, and it can be useful to use statistical measures to describe the collection of objects and just say that the distribution of stars looks pretty random. See for instance Ref 1 (the following from the introduction):

Except for interacting galaxies, it is reasonable to suppose  that  elliptical  galaxies  are,  at  least,  objects  in  thermodynamic quasi-equilibrium (the best evidence being the regularity in   the   luminosity   distribution).   Hence   we   will   proceed   by computing the entropy of a self-gravitating system using classical thermodynamics. Let us assume that the stars in a galaxy behave like   an   isolated,   self-gravitating   gas   of   equal-mass   particles. Consequently, we start by taking the differential of the entropy: ...

Reference

G. B. Lima Neto D. Gerbal and I. Marquez. The specific entropy of elliptical galaxies: an explanation for profile-shapedistance indicators?Mon. Not. R. Astron. Soc.309, 481±495 (1999)


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Why is the molozonide adduct thermally unstable?
In the mechanism of alkene-ozonolysis, the intermediate molozonide adduct, which is a cyclical but symmetric molecule (shouldn't symmetry about the C-C single bond bring stability?), decomposes thermally into a ketone and a peroxy-zwitter ion via cleavage of the O−O and C−C bonds.
However, it seems unlikely, and is against the symmetry argument. Is there some better explanation? Or we don't know the correct mechanism? Please add some references.
Knowing that there are two symmetrically placed O−O bond in the adduct, the cleavage of the only one bond is weird.

The O-O single bond is weak (142 KJ/mol). See this answer https://chemistry.stackexchange.com/questions/100110/bond-dissociation-energy-order-for-c%E2%88%92c-n%E2%88%92n-o%E2%88%92o-and-f%E2%88%92f-single-bonds

Why should symmetry be stable? The "symmetry argument" doesn't exist. Symmetry does not contribute anything to the energy.

@orthocresol, I disagree. The two bonds are equivalent . So, the breaking of either bond is equally probable. Why the hell only one of them cleaves.

Look into the vibrational modes and look for any that might suggest an assymmetric cleavage path.

You don't need to use that kind of language here; and what exactly is wrong with one out of two bonds being cleaved? Water has two O–H bonds which are equivalent. According to your logic, when coming into contact with a base, water shouldn't just lose one proton to become $\ce{OH-}$, it should lose two protons at once to become $\ce{O^2-}$. That's obviously not true. So you have to reevaluate your idea that equivalent bonds must either all remain intact or all be broken at the same time.

That said, bond breaking is quite a different thing from stability. You seem to also claim that symmetry should confer on a molecule some kind of stability, such that a symmetric molecule is always more stable than its asymmetric isomer (for example). However, this is also easily falsifiable: nitrous oxide, for example, does not have oxygen as the central atom. The stability of a molecule comes from various thermodynamic considerations such as bond strengths and so on. There is no stabilisation term to be added simply based on whether a molecule happens to reflects onto itself.

Are you meaning the mechanism of alkene-ozonolysis (not mechanism of alcohol-ozonolysis)?

@Mathew Mahindaratne , yes . It's alkene (not alcohol) - my mistake.

@AweKumarJha The two bonds might be equivalent. But that doesn't mean both will break. One will break first, randomly. As soon as it does, the bonds are no longer equivalent and the argument can tell you nothing about what happens next.

Why is the molozonide adduct thermally unstable?

The molozonide adduct (1,2,3-trioxalane; the initial adduct by 1,3-dipolar cycloaddition of alkene and ozone) is thermally unstable because it contains two weak $\ce{O-O}$ bonds (Ref.1). Mechanistic studies reveal that the ozonide (1,2,4-trioxalane) is formed by a sequence of three pericyclic steps involving 1,3-dipolar cycloaddition, retro-1,3-dipolar cycloaddition, and again 1,3-dipolar cycloaddition as shown in following scheme:

Also, I'd like to point out that the initial molozonide adduct is not always symmetric. It is symmetric only if the alkene is symmetric. And, the rearrangement of 1,2,3-trioxalane to 1,2,4-trioxalane is not about the stability of $\ce{C-C}$ in 1,2,3-trioxalane as OP argued. It's about thermal stability of consecutive $\ce{O-O}$ bonds in 1,2,3-trioxalane (recall that one $\ce{O-O}$ bond in peroxy compounds is not thermally stable, e.g., $\ce{HO-OH}$).

Knowing that there are two symmetrically placed $\ce{O-O}$ bond in the adduct, the cleavage of the only one bond is weird.

The accepted mechanism (Criegee Mechanism) says one of either $\ce{O-O}$ bond in the initial adduct can be broken to give a corresponding carbonyl oxide (a dipole) and carbonyl compound (dipolarophile), which undergo second 1,3-dipolar cycloaddition to give stable ozonide (see above scheme). None of the mechanism says one $\ce{O-O}$ bond cleaves selectively. it could be either one:

However, the second 1,3-dipolar cycloaddition of either set of carbonyl oxide (a dipole) and carbonyl compound (dipolarophile) gives identical ozonide. This second dipolsar cycloaddition $([_\pi2_\mathrm{s} + _\pi\!4_\mathrm{s}])$ has also been confirmed by $\ce{^{17}O}$-NMR studies (Ref.2):

The mechanism of ozonolysis was revisited with the use of $\ce{^{17}O}$-NMR spectroscopy. In a crossover experiment with $\ce{^{17}O}$‐labelled benzaldehyde and the ozonides of styrene and ethylidenecyclohexane it was shown that only the ether bridge of the secondary ozonides is carrying the $\ce{^{17}O}$ label. This is contrary to results reported earlier and confirms the Criegee mechanism.

References:

Dipak K. Mandal, “Chapter 4: Cycloadditions 1: Perturbation Theory of Reactivity, Regioselectivity and Periselectivity,” In Pericyclic Chemistry: Orbital Mechanisms and Stereochemistry; First Edition, Elsevier Inc.: Amsterdam, Netherlands, 2018, pp. 107-190 (ISBN: 978-0-12-814958-4).
Christian Geletneky, Stefan Berger, “The Mechanism of Ozonolysis Revisited by $\ce{^{17}O}$-NMR Spectroscopy,” Eur. J. Chem. 1998, (8), 1625–1627 (DOI: https://doi.org/10.1002/(SICI)1099-0690(199808)1998:8<1625::AID-EJOC1625>3.0.CO;2-L)).

That's a nasty DOI...


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What is inside nanobubbles?
A recent article (ACS Omega 2021, 6, 8021−8027) confirms the stability of nanobubbles with theoretical principles. Nanobubbles, having a size in the range of 50-500 nm, are reported to be metastable, i.e., stable in various liquids for times ranging up to weeks. Is the inside of the bubble filled with a gaseous form of the liquid surrounding the bubble, or with some other gas? Are there repeatable observations of nanobubbles? Is the theoretical derivation even correct?

Well, the normal big bubbles can't be born big, can they?

@Ivan: my ignorance about nanobubbles concerns the distinction between ordinary bubbles, whose behavior is reasonably well understood, and this new (?) class of bubbles, which is claimed to be so different. This resembles "nascent" hydrogen, etc., which isn't really thought of anymore. Maybe there is some connection. But I'm wondering about freely suspended bubbles.

Is it true that the reference you give does not mention what the bubble are made of, or supposed to be made of? Obviously there should not be vacuum, so either liquid vapors, or air /gases bubbled inside (unless one think of a robust cage extra resistant to inwards collapse, but how could they form at first?)

@Alchemista: one other reference suggested that nanobubbles have a negative charge that stabilizes a skin structure against intrusion and expands it to be less dense inside. It is not clear to me whether all nanobubbles are assumed to be in water; sometimes other liquids are mentioned, but whether as the body or the contents of a bubble is not defined. I wonder if there is any similarity to defects or vacancies in solid metals. Water does have an extended hydrogen-bonded structure.

Inside nanobubbles would probably be "a high-pressure gas". Whether it be an inert gas or vaporized surrounding liquid depends on the temperature, pressure, components, and many other experimental variables.
Regarding theory on nanobubbles, and starting from a classical, macro-scale standpoint, the Laplace pressure describes the pressure inside a bubble due to surface tension on the curved vapor-liquid surface. The pressure will be necessarily higher inside the bubble in order that it doesn't collapse. The back-of-the-envelope calculation on that Wikipedia page indicates around 10 bar for a 300 nm bubble in water.
For smaller nanobubbles, things get more interesting. The surface tension is not constant with curvature and at the extreme curvatures necessary for nanobubbles this will change things. A first-order approximation to surface tension corrections is the Tolman length. Also, at the higher range of pressures involved gases would not be ideal, so some equation of state is likely necessary to describe the interior phase. As things get even smaller, then you get into the details where the continuum approximations like surface tension break down entirely.
Song et al. comes to my mind of a paper where they look at nanobubbles in an actual experiment. I'm sure there are many other experiments in this area.

This seems like conventional thinking. Micro bubbles, ~10 microns, are observable in carbonated beverages. That's as small as they get; surfactants don't make them smaller. Song's bubbles are pinned to a surface - I'm more interested in free-standing bubbles that don't even float in a liquid. The Tolman length is new to me, but seems to say a chain can have different tensions along its length - interesting theory, but hard to use here. High pressure vapor or other gas should dissolve in the liquid; nanobubbles are claimed to be stable for weeks. What structural elements could exist?

@JamesGaidis then it is some kind of cage.... But still the question remains about what the interior is... It is a nice point to think of. It made me think of fullerenes as nanobubbles.... They certainly are, though covalent. I know that my last sentences aren't related to the bubbles at discussion, but the point of view / thinking are.


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Possibility of removing two water molecules from glycerol
Not a chemist, but from a chemical formula perspective, how hard would it be to create acetone by acidifying glycerol?
$$\ce{C3H8O3 + 2 H+ <=> C3H6O + 2 H2O}$$
Has anyone ever heard of this being attempted? If it isn't possible, why?

You do know that charge is conserved? Dehydration of glycerol yields acrolein.

Patent for forming acrolein by dehydration of glycerol  https://patents.google.com/patent/US8378136B2/en

Your equation is not valid, because it is not balanced. There are 2 positive charges on the left hand side, and none on the right hand side.

The chemical formula doesn't tell you much that is useful for judging a reaction's feasibility. You need to look at the structure and plausible reaction mechanisms. In this case, a reaction giving acetone is not remotely likely.

Thank you. So to balance the equation neutral H2 would need to be substituted for the 2H+. And you point out also that the dehydration of glycerol results in acrolein or propenal in the above patent. A secondary reaction would then be needed to add H2 and produce propanal. How unlikely is the conservation of the C2 oxygen if you remove 2 out of the three from glycerol? I understand that it couldn't exist without H2 addition if you remove two H2O.


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Mechanism for condensation reactions of amino acids
It is well known that peptides are synthesized by a so called "condensation" reaction between an amine and a carboxylic acid group to form the final amide moiety (because it releases water). In its general form, it resembles the picture below.

However, what is the actual chemical mechanism behind this reaction? My thoughts are that the amine can deprotonate the carboxylic acid and form a carboxylate - which is a very bad electrophile - so how would it form the amide from nucleophilic attack? I am aware that in the body, it is possible that this reaction is mediated by biochemical catalysts but I am still not sure of the overall mechanism. If this reaction does indeed involve biochemical catalysts, is it possible to complete the synthesis without them?

Getting a carboxylic acid to react directly with an amine is very difficult, that's why the body uses a huge ribosome to facilitate it. In a lab these couplings are also very complex, everything is protected apart from your target carb. acid and amine, which are then joined with coupling agents like DCC or HATU. Either way you need to activate your carboxylic acid for anything to happen.

Here is a proposed mechanism in a 2005 paper. One thing to point out is that this is a proposed mechanism but it is all I could find:

link to pdf: https://www.pnas.org/content/pnas/102/35/12395.full.pdf
Typically, these reactions involve the nitrogen attacking the partially positive carbonyl carbon as a form of nucleophilic substitution. However, OH is a bad leaving group so often, reagents such as DCC can be used to make the OH into a good leaving group before reacting. This is why it is easier to make nylon using a diacid chloride instead of a diacid.

In addition to this being a proposed mechanism, this reaction is slighly different from the one asked in the question, in that the leaving group is -O-H rather than -O-tRNA (correspondingly, the by-product is H2O instead of HO-tRNA). Nevertheless, I agree with your answer - the reaction goes by a mechanism where the nitrogen attacks the partially positive carbonyl carbon.

@Quantum I believe the reason why I chose this example/proposed mechanism was that the t-RNA acts like DCC. Usually, I believe, you don't see carboxylic acid reacting with an amine spontaneously since OH is a bad leaving group. You often would use some sort of reagent to make OH a good leaving group before proceeding in the reaction.


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Examples to show how useful or not useful of dimerization, trimerization, tetramerization and their likes in synthesizing alkenes and other compounds?
this answer shows that dimerization is a synthetic method
On wikipedia, 1-octene is showed to made from numerous dimerizations or a tetramerization.
When I read that piperazine  $\ce{C4H10N2}$ is made with ammoniation of chloroethanes or alkanolamines, I wonder if few repeating units synthesis is a general trend for making polyenes and heterocyclic saturated compounds. If dreaming is allowed, could dimerizing ethylamine $\ce{C2H7N}$ with eliminating hydrogen or aziridine $\ce{C2H5N}$ make piperazine?
For the title, I am hesitant to use telomerization or oligomerization because I am unsure about them in the contexts of this question.

It is not so simple. You may of course add or subtract atoms as you want. But this approach is not sufficient. It is much more important to take care of the way the atoms are present in space and react. This is not as easy as counting  atoms.

You can't judge whether such reactions will happen by looking at the composition of the potential monomers. You need to have realistic mechanism to make the oligomerization happen.


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Why does contact glue get stickier (to itself) as it dries?
I'm thinking of that type of glue that you apply to both sides, let dry and then press together. Like this. It seems counter intuitive.
Other glues behave in the opposite manner, and get less sticky as they dry. So does duct tape. So does jam. And actually everything else I can think of. The container recommends letting the two parts dry to touch dry. It grabs really well even if you leave it quite a bit longer.
So what goes on inside contact glue?

What I can say is to take drying and gluing (curing) distinct. First the solvent - that is in the formulation because we must be able to apply the glue - must be let evaporate. Then polymerization and crystallisation take place. The example of jam or glue as in post-it is different as for there is not a real chemical bonding within the glue. Adhesion relies on weak interaction only.

@Alchimista It's my sloppy language, sorry.  I do understand the difference between curing and drying. Cement and Gorilla glue both cure rather than dry, yet neither becomes stickier as they do so...

Still, it seems the point. When the solvent is there, no bonding and crystallisation take place. Some glues works on weak intermolecular force and cannot glue once dry. They need solvent to arrange between them and the parts to be connected. Other glues work on polymerization and the reaction can be effective from the first seconds but also take up to ours. One type does glue because of drying, the second one needs to dry to start curing.


text:
What is the product of reaction between 2‐cyclobutyl‐2‐isopropyl‐1,3‐dioxane and hydrogen with Raney nickel?
I have 2‐cyclobutyl‐2‐isopropyl‐1,3‐dioxane (alcohol protection) and I have to react it with $\ce{H2}$ and Raney nickel, but I don't know what the product would be.

I have seen a similar reaction that uses sulfur atoms instead of the oxygens and at the end the entire protection is removed and leaves two $\ce{H}$ atoms in the end. I tried doing that, but my professor told me that I got it wrong. I'm thinking that there wouldn't be a reaction.

I don't think there will be any reaction

Raney nickel desulfurization applies to sulfur not oxygen deoxygenation. It is possible your professor is expecting reduction of the cyclobutane ring.

@user55119 that would have been my gut shot too


text:
How to find normal boiling temperature if vapor pressure is known in any temperature?
I can calculate the vapor pressure of the Acetone at any given temperature using the Antoine equation. Is it possible to find the normal boiling temperature of the Acetone?
(It has to be done by just using the vapor pressure at various temperatures without using any constant specific to Acetone)

The boiling temperature is simply the temperature at which the vapor pressure equals the ambient pressure.  That's because the vapor pressure equaling the ambient pressure is what allows the liquid to boil.


text:
How to find the normal boiling temperature if vapor pressure at any temperature is the only thing that can be usable?
Here is my program for calculating the vapor pressure of the Organic Chemicals using Antoine's Equation:
def Psat_Antoine(temp,A,B,C):
    return numpy.exp(A-B/(temp+C))

A_Actn, B_Actn, C_Actn = 14.31451662590449, 2756.0411263712826, 228.03915724014055
A_Mthnl, B_Mthnl, C_Mthnl = 16.572099163421438, 3634.3383035408656, 239.33433646852495

temp = input("Give temperature in C:")
vapor_pressure_actn = Psat_Antoine(temp, A_Actn, B_Actn, C_Actn)
vapor_pressure_mthnl = Psat_Antoine(temp, A_Mthnl, B_Mthnl, C_Mthnl)

I want to calculate the normal boiling temperature of the Acetone, Methanol, or any kind of chemical that I know A, B, C constants. But there is a challenge, I can't use if, while or for loop. I have to use some kind of equation that calculates the normal boiling temperature of Acetone, Methanol etc. And the Psat_Antoine() function is the only thing that I can use to figure out the normal boiling temperature. Is it possible to write such program?
Thanks for the help.

Not quite sure what you're asking. The Antoine eqn. is $T=\frac{B}{A-\log_{10}p}-C$. So you just substitute in the values for A, B, and C, and the std. pressure in whatever units correspond to your constants (e.g., 760 mm Hg), and solve for T. This can be easily done in any computer algebra system, e.g., Mathematica, Maple, or Matlab, without needing if, while or for. If there's an additional requirement (which you didn't specify) that you must use Python, that's a programming question rather than a chemistry question, and thus should instead be posted on an appropriate programming site.

@theorist Why bother with computer algebra systems? The equation can easily be reversed by hand. Moreover, you already did that.

@IvanNeretin  The OP's question isn't clealry written, so when he wrote "the Psat_Antoine() function is the only thing that I can use", I took that to mean "the Antoine equation is the only thing I can use."  So I provided it in terms of T, at which point it becomes trivially easy to create code that spits out bp given A, B, and C for various solvents without using if, and, or for using a CAS.   Yes, it can be done by hand, but the OP didn't want to know how to do this by hand, he wanted to know how to do this with code.  And the easiest way to do this with code is with a CAS....

...it's the best tool for the job.  Now if what the OP instead meant is that he had to start with the Antoine eqn in terms of p, and use code to reverse it, my answer is still the same: use a CAS, since that's still clearly the best tool for the job, because it has built-in functions that do just that (reversing the eqn adds only one more line of code in a CAS).  OTOH, if the OP's requirement is that it specifically has to be done in Python, then he should have said that explicitly.  But his title doesn't say Python, it say he needs to do it using p and the constants.


text:
Does the entropy of a chemical system increase in an endothermic reaction?
Since entropy increases when temperature increases, the entropy of surroundings increases during an exothermic reaction. When I apply the same logic, it seems like the entropy of a chemical system increases in an endothermic reaction. Is this true?

Not necessarily.
Exo- and endothermic refer to the direction in which heat is exchanged between system and surroundings, not to changes in the entropy of the system, or of the temperature of either system or surroundings. Your logic though is partly correct: exothermic means the entropy of the surroundings increases, as it would do with an increase in T - the catch is that we usually assume that the surroundings has constant T, but that's because we say it has infinite heat capacity.
It is a recurrent subject of confusion to conflate changes in the entropy of the surroundings, of the system, and of the sum of both. Always keep the second law in mind. What determines whether a reaction will be spontaneous is the sum of the changes in entropy of the surroundings and the system.
When a reaction occurs reversibly, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}=0$.
When spontaneous, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}>0$.
When a reaction is exothermic,  $\Delta S_{\mathrm{surr}} >0$.
When a reaction is endothermic,  $\Delta S_{\mathrm{surr}} <0$.
Nothing about the thermicity (whether the reaction is endo or exothermic) tells you what happens to $\Delta S_{\mathrm{sys}}$, unless you already know whether the reaction is spontaneous or not (in which case the sum of the entropy changes is positive).

All spontaneous chemical reactions are described by the following formula:
$$\Delta_\mathrm{r}G < 0$$
If the reaction is endothermic, it means that
$$\Delta_\mathrm{r}H > 0$$
As a consequence, $\pu{T\Delta_\mathrm{r}S}$ is the sum of two positive terms:
$$T\Delta_\mathrm{r}S = - \Delta_\mathrm{r}G + \Delta_\mathrm{r}H > 0$$
and as the temperature is always positive, it means that for all spontaneous endothermic reactions, the entropy of the system increases:
$$\Delta_\mathrm{r}S > 0$$

See Which symbols are written in roman (upright) font and which are italicized?

One way to carry out a reversible ideal gas process where you start with stoichiometric quantities of pure reactants at T and P and end up with corresponding stoichiometric quantities of pure products at T and P, would be to use a three step sequence.  The middle step would involve using an van't Hoff equilibrium box in which a reaction mixture is already at chemical equilibrium within the box at T and P, and very slowly injecting stoichiometric quantities of pure reactants at the same temperature and partial pressures as in the reaction mixture while, at the same time, very slowly removing corresponding stoichiometric quantities of pure products at the same temperature and partial pressures as in the reaction mixture.  In this open system step, the heat you add is equal to the enthalpy change of the reaction, and the entropy change is equal to the enthalpy change divided by T:  $$\Delta S_{middle\ step}=\frac{\Delta H}{T}$$
The first step of the overall process is very different from this.  In the first step, you take each reactant individually and expand it isothermally at T from its initial pressure to its partial pressure in the equilibrium box.  This step is not at constant pressure P, and the enthalpy change of the gas is not equal to the heat added.  Since it is at constant temperature, the enthalpy of the gas is constant, but its entropy increases (because of the pressure change).  So in this step of the process, the enthalpy changes of the gases are zero, but not their entropies increase.
Similarly, for the final third step of the process, the pure product  gases are compressed isothermally from their partial pressures in the equilibrium mixture to the total pressure P.  Here again, the enthalpy change is zero, but, in this case, the entropy changes of the gases are negative.
Overall, the entropy of the gases changes in all three steps of the process, but the enthalpy of the gases only changes in the middle step.  So, overall, the entropy change for the entire process is not equal to the enthalpy change divided by the temperature.  We say that the overall process, the pressure is constant (i.e., the same at the end as in the beginning) even though in steps 1 and 3, the pressure does change.