text:
Is polylactic acid (PLA) soluble in ethylacetate?
I'm searching for polylactic acid (that used in 3D printing) solvent, that is less toxic. It's used to postprocessing to polish object.
Dichloroethane is toxic, acetone and dichloromethane less, but enough. Tetrahydrofurane and chloroform usage is limited in Russia.
As I know there is D- and L- polylactic acid, I don't know what type is it, but can test using acetone.
Can I use ethylacetate for this purpose? Or there is another solvents?
Sorry about my english
Useful links:
https://www.researchgate.net/post/What_is_the_best_solvent_for_polylactic_acid2/amp
https://www.researchgate.net/post/Solubility_of_PLA/amp
https://groups.google.com/forum/m/#!topic/tvreprapug/1wTmQZJMc9g

D- and L- polylactic acids come from both enantiomers of the monomeric lactic acids.
There will be no difference in solubility between both polymers.
Besides, lactic acid is normally produced by fermentation and the L- form is obtained, so the standard PLA used in 3D printing will be L- PLA.
I have tested the solubility of a piece of PLA and it is not soluble in ethylacetate in one hour.
It does even not seem to be attacked by it.

However, when you rub a 3D printed piece with a paper soaked in ethylacetate, some polishing is attained.

I make basically structural pieces, so I normally do not polish them and therefore I do not know if this is enough for your purpose.

@Mithoron Thanks for reducing the pictures. Can I do that after uploading them or must it be done externally?

https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know/3048#3048

also note it seems to help to have hot vapors instead of actually dunking it in the liquid. this method even allows for some softening of pla using acetone (something many 3d-printing folk claim to be impossible, just because it's way less impressive than ethyl acetate or what acetone does to ABS, but it might have some benefits if you want solvent welding without actually dissolving into the surface too much).

My experience with PLA and EA is the following:

Temperature matters a lot. At room temperature, almost nothing
happens. What I tend to do is put a print in a jar with some EA, and
hold it under running hot water while shaking it.
If this works at all also depends on lot on the 'PLA' used; and this
is true of printing materials and solvents generally; producers are
not in the habit of providing a complete list of ingredients, and
many filaments have additives that interfere with solvents. Ive tried
a bunch of brands and I had by far the best results with PLA/PHA from
colorfab.

But at the right temperature and with the right filament, EA works quite well as a smoother; typically done smoothing in about 20 seconds this way.
Also, dryness of your solvent may matter a lot. I used to be very disappointed by the action of acetone on ABS; untill I put a little CaCl in. I dont think EA naturally attracts water as much as acetone but in general the purity of your EA may matter; it might have things in it that compete with the dissolution of PLA, so try to keep those to a minimum.

How hot water you suggest? 60C (140F) degrees is maximum for PLA. After it shape degrades rapidly

Ive never measured the temperature of the EA; but vapor pressure goes up considerably so a substantial fraction of its BP. But PLA has low thermal conductivity and high latent heat; having it submerged around hot water tap temperatures for a few seconds isnt going to have any effect but at the surface, even if the internal temperature would go over 60C. But in the end there are so many variables that nothing beats experience I think; just experiment with different temperatures with some scrap pieces.

Note that the short working times that higher temperatures allow are infact helpful in keeping the same of your part intact. Ive also tried just waiting longer at lower temperatures; and what tends to happen then is that the EA penetrates deep into your part and makes it all rubbery (and delaminate), long before the surface actually smooths. So a short treatment under conditions that would in the long term completely destroy your part is actually pretty much a necessity.

authors: Евгений Новиков


text:
Why do halogens have odd numbers as oxidation number?
Halogens like $\ce{Cl}$ always exhibit $+7,+5,+3,+1,-1$ as their oxidation numbers. I found the following answer after checking several sources:

Halogen atoms have $7$ electrons in their valence shell, so $1$ of those electrons is spin unpaired, which is an unstable situation. If an even number oxidation occurs, said even number of electrons are taken away, leaving still an odd number of electrons with at least $1$ unpaired. Taking away an odd number of electrons leaves an even number of electrons with the opportunity of them to be all paired.

These oxidation numbers are observed in compounds like $\ce{HClO4}$ and $\ce{HClO3}$. We know that, before forming these compounds, the valence electrons of $\ce{Cl}$ are excited into the d orbitals. Thus now, they are not in ground state, but in the excited state instead.
The quoted explanation is giving the reason for my question on the basis of the ground state of halogens. (I especially consider $\ce{Cl}$). But we know that, before formation of compounds, the halogen atoms are in their excited state instead.
So is the quoted explanation correctly justifying why $\ce{Cl}$ exhibits odd oxidation states only?

Counterexample: https://en.wikipedia.org/wiki/Chlorine_dioxide
Further, d-orbitals need not be involved in even $\ce{HClO4}$, hypervalency is the more modern approach.

"We know that, before forming these compounds, the valence electrons of Cl are excited into the d orbitals." This statement is completely incorrect (like most of sentences in questions, which  start with "We know").

Because of the way we construct oxidation numbers.  Put an atom into a molecule with all electrons paired and all the bonds heteronuclear, and by assigning all the valence electrons to the more electronegative atom of each bond we render even number on each atom.  Combine that with an odd number required for a neutral atom and you count an odd difference, thus an odd oxidation number.
Note that I have assumed all electrons to be paired and all bonds as heteronuclear.  But, it is not always the case.  There could be homonuclear bonds as in the halogen element itself ($\ce{X_2}$), with an oxidation number = $0$.  Or there could be unpaired electrons on the halogen so we count an odd rather than even number of valence electrons per halogen atom.  $\ce{ClO}$ (chlorine oxidation number $+2$) and $\ce{ClO_2}$ ($+4$) are such examples.

Can you please give an example of what you wrote in the first para?

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sort: votes


text:
What will be the major product in the following reaction?

My approach-
Since a base is added the compound will release hydrogen ion.
My problem is from where will the ion be released?
Benzene ring containing bromine is more activated than the other benzene ring. So it should be difficult to extract hydrogen ion from it. But I am getting my answer right by doing so. Shouldn't hydrogen be removed from the other benzene ring?
Also how do we know how many hydrogen ions will release? Is it necessary that only one hydrogen ion is released?

The starting material will be heavily enolised. Consider the possible reaction pathways through the enolate, both through oxygen and carbon.

@Waylander  I can not understand which two pathways you mean. Please elaborate.

The Hydrogen will be removed from the benzylic carbon.And the produced negative charge will delocalise through the carbonyl carbon and there will be high conjugation between two benzene rings and oxygen atoms can get negative charge. The reaction then might proceed through benzyne mechanism and Oxygen can attack as a nucleophile to form a penta cyclic compound .

Re: "My problem is from where will the ion be released?" A proton will be released from the organic molecule. Since you're supposed to select the best
proton, the resulting carbanion should be the most stable. At which position do you get the most stable carbanion here?

Consider reaction pathways through the oxygen of the enolate formed by reaction of the starting material and base, remember that intramolecular reactions are favoured over intermolecular.

@GaurangTandon After enolizing the compound, all carbons seem almost similar to me. There is resonance in the whole compound.

@SoumikDas I think I understood you. But won't the benzyne mechanism happen first? Because a Hydrogen ion gets removed in the mechanism itself.

@VJ Have you been taught enolisation of carbonyl compounds?

@GaurangTandon Yes but not in detail. I read it from the net so I might be wrong somewhere.

@VJ Perhaps, that's why you're so confused on this otherwise decent question. Try solving a few basic cases on enolisation first. Then you'll immediately see the case here.

@GaurangTandon  Thank you. Will do that. I think I understood Soumik Das's explanation.

I don't think benzyne is involved here. Consider intramolecular SNAr by the enolate oxygen.

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text:
Study long term properties of two polymers
This is a material-science question, and I didn't know where else to put it.
I (and two others) are mechanical engineering students and are starting a bachelor's thesis which involves some material science. We are to study two polymers, PEEK along with an unknown elastomer.
We are doing the simulation using the software ANSYS, and there are different models of e.g. creep one can choose from, and they need some constants that describes the properties. We're about to do some tensile-tests to investigate the basic properties of these materials. However I believe we also need data about the long-term properties of these materials, which we'll have to get from previous studies. Can anyone point us in the right direction where we could find the relevant studies? We're not that familiar with searching for literature.
I'm about to dive down into textbooks about this, and if anyone has any advice that would be greatly appreciated!

A good start would be to know what the unknown elastomer is.


text:
How to characterise lipophilicity?
I was provided with a drug compound which is filling a hydrophobic pocket. The R group is changed and effects on the potency was measured; $\ce{R = -H, -CH3, -CH2F}$ were tested. As expected, the more lipophilic group $\ce{-CH2F}$ had the smallest dissociation constant. However, the question went on to ask me to suggest other groups that could be tested. Investigating more lipophilic compounds would be wise, to further prove that the pocket is indeed hydrophobic.
It was suggested $\ce{-OCH3}$ as one of the groups that could be substituted for R. I am confused, because wouldn't the oxygen be able to form hydrogen bonds, making it hydrophilic rather than hydrophobic?
My question is, how do we classify a compound to be lipophilic? i.e. How do we know if the molecules form hydrogen bonds or not?


text:
Meaning of electrochemical cell notation without a salt bridge
An electrochemical cell is shown here.
$$\ce{Ag(s)~|~AgCl(s)~|~KCl(1M)~|~Hg2Cl2(s)~|~Hg(l)~|~Pt}$$
I am used to seeing cell representations with a $||$ for a salt bridge between the two electrode solutions (generally with the anodic solution written first followed by the cathodic solution).
I am seeing the above type of representation for the first time. Can someone tell me what it means? (including which are the anodic and cathodic solutions). Also, how do we write the reactions at anode and cathode for the above cell?

This is an example of an electrolytic concentration cell.
Both the cathode and the anode have the same electrolyte present and both are in same solution. Thus, there is no need to separate the electrolyte in two different solution and connect by a salt bridge.
The anode and cathode both are metal-sparingly soluble salt-salt anion electrode. The $\ce {KCl}$ solution here is the common electrolyte.
The anode reaction is:$$\ce  {Ag(\text{metal})(s) + Cl^-(\text{salt anion})(aq) ->  AgCl(\text{sparingly soluble salt})(s) + e^-}$$
The cathode reaction is also similar: $$\ce {Hg2Cl2(s) + 2e^- -> 2\ce{Hg}(l) + 2Cl^-(aq)}$$
Observe that the $\ce {Cl^-} $ is present in both the reactions as a common electrolyte's ion.

is the order in which the species are written relevant in such a notation?

Yes, definitely to some extent, to represent which species is getting oxidized to which and which species is getting reduced to what.

Then how are cell reactions understood from a notation like this couple; $Cl^- | PbCl_2 | Pb $, or for cells like the Weston cell; $Hg | 3CdSO_4, 8H_2O(solid)|$ satd. solution of $CdSO_4|| Hg_2SO_4(s) | Hg$? These seem unrelated to the actual reactions I found on looking them up.

Just go through different types of electrodes used generally. There are many variations of them. For example the cell notation you have written above is an example of Metal-Metal Insoluble salt-Soluble anion type electrode.  Just look through what reactions happen in them actually. You'll understand.

Can we say $E_{cell}=E^°_{cell}$ for this example?


text:
Why is oxygen an oxidizing agent?
Why are diatomic oxygen molecules STILL reactive especially with metallic elements like sodium and copper even at room temperature?
You would think that since the two oxygen atoms already have the much needed 8 valence electrons when they bonded with each other that they wouldn't need to react with anything else but that doesn't seem to be the case.

Having a full octet for an atom is much like breathing for a man: important for sure, but it's not the only thing they need.

If you take at loo kat the MO diagram for Oxygen, you will see that oxygen is a di-radical as we call it. It has two unpaired electrons, while the oxide (so a single oxygen atom now) would have all orbitals (s and p) filled which is more stable.

I don't understand what you mean by "STILL reactive". "Still" suggests that you think the situation has been going on for too long ("I asked you to do X and you still haven't done it!") but that doesn't make sense here. What other scenario (even if hypothetical) are you comparing with?

I personally think the weird case is not that O2 is reactive, but that N2 isn't.

@Joshua it takes a lot to overcome a triple bond

@mbrig: And yet the triple bond e.g. in acetylene is quite reactive indeed, so that cannot be the full answer. Maybe that would make a decent question in itself. (Actually, it looks like we already have one.)

To estimate reactivity solely based on the the octet rule is a crude, (semi-)empirical approach in the first place. Atoms and molecules don't know anything about their number of electrons in shells, but the formation of molecules are based on the minimization of (free-)energy of the system.
The octet rule gives a hint, for what kind of systems the potential energy may be small, but does not say anything about the relative energies of two systems, both fulfilling the octet rule.
To "exactly" predict the reactivity, you would have to solve the Schrödinger equation for both states of your systems, find the eigenvalue of the energies, consider entropic contributions. Furthermore you would need to "sample" all possible pathways of your reaction, to check if the energy barriers are small enough to be overcome at a given temperature.
(To be really exact, you still would need to estimate entropic contributions)
Yet still a chemists' intuition often can predict reactivity very well. Concerning your specific problem, one possible qualitative explanation would be, that the octet rule is not fulfilled for copper, but only for oxygen, before the reaction happens. Taking a look on the structure of Copper(II) oxide on the other hand, shows that it fulfils the octet rule for both, copper and oxygen.(Ignoring d-electrons for copper)
Furthermore, the reaction enriches the electron density in the strongly electronegative oxygen, while reducing the electron density in the "electropositive" copper, which also stabalizes the structure of copper oxide.

I really can't agree with the explanation based on copper's octet being satisfied. A good model should have predictive power, and if the loss of the two 4s electrons is really thermodynamically favourable, then surely copper would react with - say - hydrogen to form $\ce{CuH2}$, or nitrogen to form $\ce{Cu3N2}$.

The octet rule has some predictive power. But basically it is as i said: u have to calculate the relative energies of concurent systems. The octet rule is widely applicable in organic chemistry, where u will (hardly) find themodynamic stable compounds, in which the octet rule is not satisfied. But on the other hand, not every compound, u can  write  down on paper satisfying the octet rule, will be easy to synthesize or stable in the presence of oxygen.

The octet-rule is a very inadequate rule for understanding overall reactivity.
Ultimately you need to understand two things to know whether reactivity is likely: the thermodynamics of any possible reaction; and the kinetics of any possible reaction mechanism for the reaction.
The formation of many oxides is thermodynamically favourable. That's why many compounds burn in an atmosphere containing oxygen. The formation of sodium oxide, carbon dioxide and many metal oxides is thermodynamically favourable. Carbon-containg things will burn in air, metals will often react (quickly, like sodium, or slowly, like iron). Some will burn spontaneously like caesium; others will take a little encouragement.
The second factor that matters is whether there is an easy way for the oxidation reaction to happen. Aluminium, for example, is very reactive but rapidly forms a protective layer of strong aluminium oxide (alumina) on the surface preventing further reaction; iron is relatively reactive but won't rust easily unless contaminants are present; gold won't react at all. Carbon needs to be heated to start the reaction but will burn by itself when it does start.
Where reactions have easy to access mechanisms and some thermodynamic advantage, oxygen is very reactive.
Another thing to remember about oxygen is that it has some reaction mechanisms that are less than obvious and easier than expected for a diatomic molecule. Unlike nitrogen, for example, where thermodynamically favourable reactions are inhibited by its strong triple bond (and filled octet) an oxygen molecule is actually a diradical: despite the apparent filled octets you would expect from simple electron counting, the molecule has two unpaired electrons (this requires a bit of molecular orbital theory to explain and is one of the observations that simple electron octet counting doesn't explain well). Radicals tend to be more reactive than paired electron orbitals (or filled octets).
In short, an octet-counting view is too simplistic to explain oxygen's reactivity and you need more sophisticated ways of looking at the electronic structure. And don't forget that you also need to know the thermodynamics and the kinetics of potential reaction mechanisms to get any useful predictions of whether reactivity is likely.

Oxygen (O2) generally exists as diradicals i.e. each oxygen bonded to each other through single bonds and the remaining two electrons remains on each oxygen atoms as radicals. So this structural feature makes oxygen act as a strong oxidizing agent. Even in our body oxygen coordinates through its radical and this structural arrangement makes oxygen can act as a strong oxidants in many places.

Saying that O2 is a diradical rationalises the fact that it is reactive, but that doesn't make the statement a true fact!!


text:
Does acrolein give Fehling's test?
Aromatic aldehydes do not give Fehling's test, which makes me believe that for the same reason acrolein should not. It has a double bond conjugated with the carbon-oxygen double bond as in benzaldehyde. Is my assumption correct? Or is extended conjugation a factor?

Glyoxal(CHO-CHO) and Glyoxalic acid(CHO-COOH) also don't give Fehling's test. Any reasons for that?

Does this answer your question? Why does benzaldehyde not respond to Fehling's test?

It has nothing to do with aromaticity. It is related to this answer.
Why does benzaldehyde not respond to Fehling's test?
Any aldehyde without an alpha hydrogen will not give Fehling's test.
Acrolein also doesn't have an alpha hydrogen, so it will not give Fehling's test.

Can you quote any book or paper that discusses this? I highly doubt that this is the correct reaction mechanism discussed in the answer that you linked. Also it does have an alpha hydrogen.

Please refer, Article by Ralph daniels (JCE, Vol.37, No.4, 1960 link: https://pubs.acs.org/doi/abs/10.1021/ed037p205). According to this paper, even acetaldehyde is not oxidised by Fehlings solution. They claim these are errors in the text books are carried over for many years.
In this journal, the authors observations are:
Category-1 :Positive fehling test: Glucose and other reducing sugars,glyoxal & pyruvic aldehyde (both of them does not have alpha H).
Category-II: Acetaldehyde, crotnaldehyde, cinnamaldehyde, these aldehydes show a colour change, but they have not observed the presence of Cu+1 in the reaction, (the aldehyde group is not oxidised,hence it is false positive test.)
Negative fehling test: Benzaldehyde other aromatic aldehydes, isobutraldehyde.

Do you have access to the free version of the article?

My best guess would be that the electron withdrawing groups attached to the aldehydes increase the reduction potential of the half reaction (make it harder to oxidize) since the electrons are more delocalized in the molecule. In acrolein, the inductive effect of the vinyl group isn't too big, plus there is an electron donating effect through resonance, so I'd expect it to be reactive to Fehling's reagent.

There is Resonance in Benzaldehyde as well. So shouldn't it also give Fehling's test by the same reasoning?

I'm thinking on how they react. Acrolein acts as an electrophile (Michael acceptor), whereas benzaldehyde acts as a nucleophile (EAS).

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text:
Understanding the basics of DFT
I have recently started doing DFT calculations as part of my PhD. The general approach to DFT in my department appears to be that it is regarded as a tool for doing calculations, which need not be understood in detail. While I understand this to some degree, I however still feel the need to at least understand the basic theoretical ideas behind a DFT calculation. I am therefore using this post to summarize a few of my questions about DFT starting from the basic theory.
As I understand it, the formalism of DFT rests upon the Hohenberg-Kohn theorems, stating that the ground state properties of a many-body system are unique functionals of the ground state density, which furthermore minimizes the total energy functional $E[n(\mathbf{r})]$ of the system:
$$
E[n(\mathbf{r})] = T[n(\mathbf{r})] + E_H[n(\mathbf{r})] + E_{ext}[n(\mathbf{r})] + E_{xc}[n(\mathbf{r})]
$$
Here $T[n(\mathbf{r})]$ is the kinetic energy functional, $E_H[n]$ the Hartree functional and $E_{ext}[n(\mathbf{r})]$ the energy due to external potentials while $E_{xc}[n]$ is the unknown exchange-correlation energy functional, which encompasses the exchange and correlation effects of the many-body problem and must be suitably approximated by e.g. the LDA.
In practice DFT calculations are based on the Kohn-Sham approach, which makes the ansatz that there is a non-interacting system, which reproduces the density of the many-body system one wishes to describe. Consequently, instead of solving a many-body Hamiltonian, one solves a set of single-particle equations, the Kohn-Sham equations:
$$
(\nabla^2 + V^{KS})\phi_i(\mathbf{r}) = \epsilon_{i}\phi_{i}(\mathbf{r})
$$
with the Kohn-Sham potential given by:
$$
V^{KS}(\mathbf{r}) = V_{ext}(\mathbf{r}) + V_{H}(\mathbf{r}) + \frac{\delta E_{xc}}{\delta n(\mathbf{r})}
$$
The density is then readily evaluated as a sum over the occupied orbitals:
$$
\rho(\mathbf{r}) = \sum_{occ}\lvert \phi_{i}(\mathbf{r}) \rvert^2
$$

My main question is essentially, if anyone can walk me through how the above theoretical ideas are used in a concrete DFT calculation. Say, e.g., I want to calculate the ground state energy of some molecule. How does a DFT calculation proceed using the equations outlined above?
My own (partial) answer: The first step is specifying the energy functional, which relies on choosing a suitable approximation for the exchange-correlation functional. My guess is then that the next step is to solve the Kohn-Sham single-particle equations. To do one needs the Kohn-Sham potential, which must be found from the functional derivative of the total energy functional. I am honestly not sure if this is how a DFT calculator proceeds, but it seems like the only way to get the Kohn-Sham potential. Having solved the KS system, one can then calculate the electron density, which should then equal the density of the interacting system. However, I speculate that the procedure may involve some sort of self-consistency such that one should in practice make an iterative procedure to find the correct interacting density.
Assuming everything is correct so far, how does one then get the ground state energy? So far we have only calculated the ground state density and the energies of the Kohn-Sham orbitals are not physical energies of the system.

Welcome to Chemistry! Take the tour to get familiar with this site. I'm having a hard time figuring out exactly what your question is. I'd re-format your last block to split it up since the question is buried in there. I think the answer is in your question: once you have self-consistently solved for $\rho(\mathbf{r})$ (which is the same thing as $n(\mathbf{r})$), plug back in to the energy functional, and you're done.

Thanks for the reply. I can understand it is a bit difficult to figure out what my question is. Essentially my main question is: How do you obtain the Kohn Sham potential to use in the Kohn-Sham equations? When I do a DFT calculation I specify the exchange correlation functional but not the KS potential.

Just from the equations themselves, you can see that $V_{xc}[n(\mathbf{r})] = \frac{\delta E_{xc}}{\delta n(\mathbf{r})}$: the potential is the functional derivative of the energy with respect to the density. You are correct that the potential is never specified, but in this case the potential is defined by the energy functional. In practice, the Kohn-Sham equations are identical to the Hartree-Fock equations, except the (exact) exchange contribution to the Fock operator is replaced with the Kohn-Sham matrix: https://chemistry.stackexchange.com/a/81832/194

Possible duplicate of How does one actually get the energy from the Kohn–Sham equations?


text:
How to determine if pressure surpasses capacity with the van der Waals equation of state?
I have to decide whether a $\pu{20 L}$ container with $\pu{0.8 kg}$ ethane can withstand the pressure at $\pu{373.15 K}$. The following information is given:
\begin{align}
  b &= \pu{6.448x10^-2 dm3/mol}\\
  a &= \pu{5.572 dm6 bar/mol2}\\
  p_\mathrm{c} &= \pu{49.64 bar}\\
  T_\mathrm{c} &= \pu{308 K}\\
\end{align}
The container can withstand any pressure up to $\pu{3000 kPa}$.
I am not sure how to solve this, so far I've tried inserting my values in the van der Waals equation of state, but I could not arrive at the right result with this approach. The correct answer should be $\pu{3.528E6 Pa}$ and I got $\pu{3.14E6 Pa}$.
I inserted the following but the result is wrong:
\begin{align}
  p &=\frac{\pu{26.60 mol} \times 8.3145 \times \pu{373.15 K}}
      {\pu{0.02 m3} - \pu{26.6 mol} \times 6.44 \times 10^{-8}}
     - (\pu{0.557 m6//mol2} ) \times \frac{\pu{26.6 mol2}}{(\pu{0.02 m3})^2}\\
    &= \pu{3.14E6 Pa}
\end{align}

I plugged the numbers and ended up within a rounding error from the correct answer. Not sure what else can be done.

did you type in the following: p=(26,60 mol8,3145373,15K)/(0,02m^3-26,6 mol6,44(〖10〗^(-8) m^3)/mol)-(0,557m^6)/(mol^2 )(〖26,6mol〗^2/(0,02〖m^3〗^2 ))=3,14〖10〗^6 Pa

Having edited the expression that caused all the problems I can now see that the units for $b$ were converted incorrectly:
$$6.448\times10^{-2}\frac{\text{dm}^3}{\text{mol}}\left(\frac{1\,\text{m}}{10\,\text{dm}}\right)^3=6.448\times10^{-5}\frac{\text{m}^3}{\text{mol}}$$
I got
$$v=\frac{\left(20\,\text{L}\right)\left(\frac{1\,\text{dm}^3}{1\,\text{L}}\right)}{\left(0.8\,\text{kg}\right)\left(\frac{1000\,\text{g}}{1\,\text{kg}}\right)\left(\frac{1\,\text{mol}}{30.070\,\text{g}}\right)}=0.75175\frac{\text{dm}^3}{\text{mol}}$$
$$P=\frac{\left(8.314\frac{\text{J}}{\text{K}\,\text{mol}}\right)\left(\frac{1\,\text{m}^3\,\text{Pa}}{1\,\text{J}}\right)\left(\frac{10\,\text{dm}}{1\,\text{m}}\right)^3\left(\frac{1\,\text{bar}}{10^5\,\text{Pa}}\right)\left(373.15\,\text{K}\right)}{0.75175\frac{\text{dm}^3}{\text{mol}}-6.448\times10^{-2}\frac{\text{dm}^3}{\text{mol}}}-\frac{5.572\frac{\text{dm}^6\text{bar}}{\text{mol}^2}}{\left(0.75175\frac{\text{dm}^3}{\text{mol}}\right)^2}=35.281\,\text{bar}$$
So convert units correctly and all should be well.


text:
Formaldehyde (37%) storage
Should Formaldehyde (37%) be stored at 5 degrees C or lower?
Will it damage the Formaldehyde?

https://chemistry.stackexchange.com/questions/75725/can-formaldehyde-be-stored-at-a-higher-temperature-than-room-temperature

In the cold some paraformaldehyde (formaldehyde polymer) or trioxymethane (a cyclic condensation product of three formaldehydes) may precipitate.
It will depend on how much methanol has been added to avoid the polymerization products.
Normally it will contain 10 to 15% methanol.
In reality, it is not necessary to store formaldehyde solution in the cold.

[deleted]...........

@Waylander   Right, we have it in a safety cabinet at room temperature.

The MSDS for 37% aq formaldehyde says this: "Store in a segregated and approved area. Keep container in a cool, well-ventilated area. Keep container tightly closed and sealed until ready for use. Avoid all possible sources of ignition (spark or flame)"
http://www.sciencelab.com/msds.php?msdsId=9924095
Combined with the answer below I would say not to store it cold.


text:
Bromination of ethyl substituted benzaldehydes
In a material that I use to study High School Organic Chemistry, the bromination of ethyl substituted benzaldehydes is given as follows:

Source: FIITJEE Grand Masters Package - Chemistry (JEE Preparatory Material)
$\ce{-CHO}$ is a meta directing group, and $\ce{-C2H5}$ is an ortho, para directing group. So isn't some of the products given wrong?
For example, let's take the bromination of ortho ethyl benzaldehyde. Shouldn't the 2nd product's bromine be in a way that is in position 4 wrt $\ce{-C2H5}$ and position 3 wrt $\ce{-CHO}$?
In the meta ethyl benzaldehyde case, why is the $\ce{-CHO}$ group not affecting the position of $\ce{Br}$? Even though it is meta directing, $\ce{Br}$ doesn't seem to substitute the $\ce{H}$ in the meta position of $\ce{-CHO}$
I searched the internet for more information regarding such reactions, and could not find any helpful results for this case.

Your first query is genuine and the 2nd product for the first case is wrong. It should be, as you noted as well, 5-bromo-2-ethyl-benzaldehyde.
Regarding your second query, you're wrong. Recall the general rule that:

when more than one activators are present on the same benzene ring, the substitution is governed by the best activator alone
when an activator and a deactivator is present on the same benzene ring, the substitution is governed by the activator alone
when more than one deactivators are present on the same benzene ring, the substitution is governed by the weakest deactivator alone

So, here $\ce{Br}$ will go to the o/p position of ethyl group.

Note that in the high school context, I've assumed we're only talking about the major product, because otherwise, all types of crazy products are generally formed, but in smaller ratios


text:
Why is it necessary to measure the NMR frequency relative to a standard such as TMS?
The horizontal axis of an NMR spectrum is "chemical shift". A standard environment such as tetramethylsilane is chosen as the zero of chemical shift. The frequency at which resonance occurs for other environments is then measured relative to the frequency at which resonance of the standard environment occurs e.g. a particular environment might resonate at a frequency 1ppm lower than standard environment.
Why can't we measure the frequencies at which resonance occurs absolutely, not relative to a standard environmnent? Then we would have a horizontal axis of frequency instead of chemical shift.

The use of an standard makes it possible to compare spectra. The frequency itself is dependent on the used magnetic field, while the the chemical shift is independent of the magnetic field and thus comparable using different spectrometers.


text:
What is the meaning of the term 'reserve acidity'?

In $\ce{CH3COOH/CH3COONa}$ buffer, the reserve acidity of the solution is due to the $\ce{CH3COO-}$ ions.

What is the meaning of the term of 'reserve acidity' in the above statement?

One would expect buffers to maintain constant pH. Thus, in that context, reserve acidity of CH3COO-  ions could mean that the CH3COO- ions help neutralize the base that has been added to the buffer solution. Though, I've been reading the mech here, and apparently my intuition doesn't seem to be correct. Will wait for someone to post a proper answer.

Reserve acidity, more commonly known as Buffer capacity, is the amount of acid/base (in gram equivalents) that must be added to a buffer solution to change it's pH by one unit.
In your question, the $\ce{CH3COO-}$ is able to react with any Hydrogen ions that are added to the solution, decreasing the $[\ce{CH3COO-}]$ while increasing the $[\ce{CH3COOH}]$. It takes care of any excess acidity, hence, it is called as the reserve acidity. Similarly, the $\ce{CH3COOH}$ should be called the reserve basicity, as it can reach with any base to prevent much pH change.
For a mathematical approach, we use the Henderson-Hasselbach equation for calculating the final pH.

Hi, thanks for your answer! You seem to suggest, by your statement - "Reserve acidity, more commonly known as Buffer capacity", that reserve acidity is the same as buffer capacity. I've been searching the literature but couldn't find any statement regarding this equivalence of buffer capacity and reserve acidity. Can you please cite one source that would affirm that statement of yours? Thanks!


text:
Why is Iron very rare in seawater compared to many other elements?
I don't know the associated technical words to research this question.
There was high presence of Iron in the oceans in anoxic precambrian times, and today copper is 5 times more abundant in the oceans than iron.
Is the solubility of $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ and $\ce{Cu^{2+}}$ very different?
Why is there five times more Copper in seawater than Iron? Surface minerals have 1000 times higher PPM values for the latter? It's precipitating?
Why is there 4000 times less iron in seawater than Strontium? Surface minerals have 100 times higher PPM values for the iron?
Here are some measurements stated previously:
https://web.stanford.edu/group/Urchin/mineral.html
https://en.wikipedia.org/wiki/Abundance_of_elements_in_Earth%27s_crust

There's  no minerals in seawater just dissolved salts. Why would you think density of metal has any importance here? It's not like colloidal iron  would be in seawater...

I do not see why to downvote. Why is not ok to ask why element abundance in seawater does not follow that in crust  mineral whatever? OP should also link to a source for the mineral data. Obviously the iron down the nucleus shouldn't not be counted as much as that on crust.

Why Mark Zuckerberg is about a 1000000 times richer than me, when I'm 30% heavier?

Come on OP id simply asking what makes the composition so different. It could have skipped any reference to weight as it was in the original answer.  The question is why the abundance of ions in the sea does not reflect the abundance of elements.  It would be why there is only one Zuckerberg on land and ten swimming in the oceans? To follow sarcastic comments

OP the various Kps of the possible combinations is one among the answers.  It is not a case that Sr is heart alkali and iron is not.  Nor hat the most abundant ions  in sea are sodium and chlorine.  If you take iron as example, it can be fixed as sulphide, etc. But as I said before you should compare abundances in seawater and a specific part of earth, probably the most outer part of the crust. The elemental composition of universe, stars, earth and earth portions is not the same.

I'll elaborate this answer as I learn.
It's an effect owed to the low solubility of iron in the modern oxic ocean and the tendency for iron to be ‘scavenged’ from the water column by sinking particles.
Brand, L. E., Sunda, W. G. & Guillard, R. R. L. Limitation of marine
phytoplankton reproductive rates by zinc, manganese and iron. Limnol.
Oceanogr. 28, 1182–1198 (1983).
Cooper, L. H. N. Iron in the sea and in marine plankton. Proc. R. Soc. Lond. B
118, 419–438 (1935).
Goldberg, E. D. Marine geochemistry. 1. Chemical scavengers of the sea.
J. Geol. 62, 249–265 (1954).


text:
What is the magnetic quantum number -1 vs +1?
In my book it states the magnetic quantum number “$m_\ell$" is the spatial orientation of the orbital with respect to a standard set of coordinates. I get what $n$ and $\ell$ is, but what does it mean visually or conceptually, if $m_\ell$ is $1$ vs $-1$? For instance if $n = 4$, $\ell = 1$ and $m_\ell = -1$ vs. $n = 4$, $\ell = 1$ and $m_\ell = 1$?

Related question - quoting: "Magnetic quantum number (ml) - indicates the orientation of a particular orbital in space. ... While each orbital is indeed a p-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space. " Does that answer your question?

Just like $n$ and $l$, the quantum number $m_l$ is an parameter of the wave function, which describes the orbital. The Hydrogen orbitals can be split into a radial part (parametrized by $n$ and $l$) and an angular part (parametrized by $l$ and $m_l$):
\begin{equation}
 \Psi(r,\theta,\phi)  = R_{nl}(r) Y_l^{m_l} (\theta,\phi) =  R_{nl}(r)  f_l^{m_l} (\theta) g_{m_l} (\phi)
\end{equation}
The angular part is given as
\begin{align}
 f_l^{m_l}(\theta) &= (-1)^{m_l} \sqrt{\frac{(2l+1)(l-{m_l})!}{4\pi (l+{m_l})!}} P_l^{m_l}(\cos \theta) \\
 g_{m_l}(\phi) &= e^{i {m_l} \phi}
\end{align}
with $P_l^{m_l}(\cos \theta)$ being the associated Legendre polynomials. As you can see for ${m_l}\neq0$ the angular part gets complex valued.
What does this mean conceptually?
For each possible value of $l$ there are $2l+1$ possible values of $m_l$. This is why $s$ orbitals always come alone, $p$ orbitals in groups of three, $d$ orbitals in groups of five and so on. For example we have $p_z, p_y, d_{z^2}, d_{xy}$.
They have different shapes (spatial orientations), but are degenerate in energy (in case of the hydrogen atom). Due to electron-electron interactions in many electron system, these energy levels can split.
The value of $m_l$ further specifies the orientation of the orbital angular momentum, while $l(l+1)$ is related to the magnitude of the orbital angular momentum.
The name magenetic quantum number refers to the splitting of the energy eigenvalues for different $m_l$ when an external magnetic field is present (Zeeman effect).
What does this mean visually?
If we plot for example for $l=1$ the angular part $Y_l^{m_l}$, we get $2l+1=3$ times the typical $p$ orbital shape, each one aligned along one cartesian coordinate ($x$,$y$ and $z$).
Note however, that we have a complex part $g_{m_l}(\phi)$ for $m_l\neq0$. Plotting complex valued 3-dimensional functions is not trivial. On Wolfram  I found a plot showing this thing.

Note that it looks like a donut, instead of the $p$ shape typically found in text books. To get there we can make a real superposition by employing  Euler's formula
\begin{align}
   \cos(\phi) &= \frac{1}{\sqrt{2}}(e^{i\phi} + e^{-i\phi}) \\
   \sin(\phi) &= \frac{1}{\sqrt{2}i}(e^{i\phi} - e^{-i\phi}) \\
\end{align}
So we combine two complex $Y_l^{m_l}$ with same $|m_l|$ to get two real valued new functions, one with $\cos(\phi)$ and one with $\sin(\phi)$. The real functions represent $p_x$ and $p_y$ orbitals, while the $p_z$ orbital is given by $Y_1^0$.

I would add in your conceptual section that they can split with a magnetic field  (the Zeeman effect) as well, thus explaining why it is called the magnetic quantum number.

I added 2 sentences about that, but it also covered in the answer by
Soumik Das.

The magnetic quantum number can be best explained if it's viewed through quantum mechanics.
The orbitals you have read are nothing but the solutions of Schrodinger's equation $\hat{H}\Psi =  E\Psi$. We can express the Laplacian $\nabla^2$ in spherical coordinates as such:
$$ -\frac{\hbar^2}{2m}\nabla^2=-\frac{-h^2}{2m}\frac{1}{r^2}\left\{\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin{\theta}}\left[\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)\right]+\frac{1}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}\right\}$$
or alternatively as:
$$ -\frac{\hbar^2}{2m}\nabla^2=-\frac{-h^2}{2m}\frac{1}{r^2}\left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)\right] +  {L^2}/{2mr^2}$$
where $L^2$ is the square of the angular momentum operator.
So, for solving Schrodinger's equation, one can independently solve the radial part and the angular part as the variables are independent. So, if you want to solve the angular part i.e $L^2 \Psi  = E_1 \Psi$. You will find the solution as spherical harmonics ($Y_l^m (\theta,\phi)$) and the eigen values related to the azimuthal quantum numbers as $l(l+1)\hbar^2$. The $\phi$ part depends on eigenvalues as $m_l\hbar$ where $m_l$ is your magnetic quantum number which can take value from $-l$ to $+l$.
This calculation also yields another special result. If the electron is introduced in an external magnetic field, the angular momentum vector can only make certain definite angles with the magnetic field which can be calculated as $$\cos\theta =  \frac{\langle \psi{\left|L_z\right|\psi\rangle}}{\left\langle\psi|L|\psi\right\rangle} = \frac{m_l}{\sqrt{l\left(l+1\right)}}$$
So  $l=1$ and $m=1$ means when applied to external magnetic field the angular momentum vector will orient at angle $\arccos(1/\sqrt2)$ or $45^\circ$. But $m=-1$ means that the angular momentum vector will orient with a magnetic field at angle $\arccos(-1/\sqrt2)$ or at $135^\circ$. Thus $m_l$ tells about the possible spatial orientation of the angular momentum vector in the presence of a magnetic field and that's why they are called magnetic quantum numbers.


text:
Polarity and acid strength in alcohols
I'm totally confused about the relationship between the polarity and acid strength of a given compoud.
Take the case of alcohols itself, it's said that primary alcohols have greater acid strength compared to secondary alcohols as the electron-releasing group in secondary alcohol decreases the polarity of $\ce{O-H}$ bond.
But why is this so?
I mean, less polarity will help in liberating $\ce{H+}$ ions more easily so acidity is meant to be increased.
That means secondary alcohols are meant to have more acid strength.
Some help to solve this dilemme is appreciated.

Why will less polarity help in liberating $\ce{H+}$?

Greater polarity means they are more tightly attracted, right?

I'm not sure why you even need to invoke "polarity" in this case. In fact, reduced polarity and high acidity seem to be separate cases arising from the root case of positive inductive effect of more carbon atoms towards the alcoholic group.

@Gaurang I didn't get how are both separate in this case aren't polarity and acidity linked if we are judging from the basis of inductive effects?

@AvatarShiny Yes, they are "linked" in the sense that they both arise from the same "cause". But, that doesn't mean the acidic strength is higher because of the reduced polarity. The relation is a correlation but not a causality. (video)

@Gaurang Then how would you explain the effect of inductive effects on acidity? (I used to justify it via polarity , seems like I have a conceptual doubt )

@AvatarShiny I have always justified that: as the degree of hydroxyl group increases from primary->secondary->tertiary, the inductive effect of the carbon atoms on the oxygen atom increases. This implies that when an anion would be formed on the oxygen atom (after the loss of $\ce{H+}$), the extra positive inductive effect would tend to destabilize the carbanion. Hence, the acidity is decreasing with increase in alcohol degree.

In analysing acid strength, bond polarity of the bond to the ionisable hydrogen atom is what should be considered, not the polarity of the molecule (i.e. the dipole moment). In the case of alcohols, the polarity of the $\ce {O-H}$ bond should be considered when assessing strength as an acid.
Why does the bond dipole matter? In the process of ionisation $\ce {ROH + H2O-> RO^{-} + H3O^{+}}$, the $\ce {O-H}$ bond is heterolytically cleaved. It is easy to observe that the greater the partial positive charge on the $\ce {H}$ atom, the more easy it is for it to be abstracted by the water molecule, due to stronger electrostatic attractions. Also, the more polar the bond, the lower the covalent character of the bond as the electron density shifts more towards the oxygen. Such a bond is also more easily cleaved. Recall that the covalent character of a covalent bond is highest when the atomic orbitals are of similar energies and that electron density is shared equally between the two atoms.
In the primary alcohol, there is one electron-donating alkyl group attached to the alpha carbon while in the secondary alcohol, there are two such groups attached to the alpha carbon. Due to the greater electron donation to the oxygen atom in the secondary alcohol, where $\ce {R = CHR'R''}$ , the oxygen atom actually withdraws less electron density from the $\ce {O-H}$ bond. We can think of it this way: The oxygen atom has a fixed demand for electron density. When this demand is met by an increased supply from the alkyl chain, the demand for electron density from the hydrogen atom would decrease. Thus, the $\ce {O-H}$ bond is less polar and less easily heterolytically cleaved.
Simarly, for the primary alcohol, where $\ce {R = CH2R'}$, there is also an increased supply of electron density from the alkyl chain R, as compared to $\ce {R = CH3}$. However, the increased supply from $\ce {R}$ is less so compared to that in the secondary alcohol. Thus, the primary alcohol would be a weaker acid than the methyl alcohol but a stronger acid than the secondary alcohol.
Conventionally, acid strength is assessed using conjugate base stability, rather than bond polarity. A brief overview of assessing conjugate base stability can be found here.

I'm so delighted to have a valid explanation. Can we apply the same reason in the case of acidity of phenol?

Water abstracts $H^+$ ion, so do water have a negative character? Forgive me, if this sounds absurd.

@SlayerDiAngelo This explanation is universal and works in every case. Of course, in different cases, there may be other factors which may have significant influences on the acidity as well. In the case of phenol, we would have to look at the delocalisation of the negative charge due to the pi system as well.

@SlayerDiAngelo The water molecule is able to abstract the hydrogen atom as it has a lone pair of electrons on the oxygen atom which can be donated to form a bond with the hydrogen atom. The abstraction of the hydrogen by water molecule is further enhanced by the electrostatic attraction between the partial charges on oxygen and hydrogen.


text:
What's the hybridization in diphosphorus molecule?
I read that diphosphorus has 2 phosphorus atoms joined by a triple bond (ie) it has one sigma bond and 2 pi bonds with 2 lone pairs ( 1 lone pair on each atom). I'm confused about the hybridization of the orbitals. Phosphorus in excited state has one 3s orbital, three 3p orbitals and one d orbital .
If sp2 hybridization takes place, it will form one sigma bond and what about the other 2 hybridized orbitals what would become of them? I'm confused on the lone pair part :/
If the above even took place, would the unhybridized p orbital form pi bond with the unhybridized p orbital of the other phosphorus. And the same would go for the 2 d orbitals. Am I correct on this??
note :https://socratic.org/questions/does-nitrogen-n2-and-oxygen-o2-goes-through-hybridization-how-can-i-know-when-hy
Here it says that "Diatomic molecules don't need hybridisation that is phosphorus would be using p pi p pi bonding which it could never do cuz of large diffuse orbitals !!

Note: diphosphorus is unstable. Natural phosphorus will instead exist in alltropes like white, red, or black phosophorus, which have entirely different structures.

Yes diphosphorus is unstable but I'm curious about it's hybridization . If the phosphorus undergoes sp hybridisation then 2 p orbitals and a d orbital will be left behind .

You can leave orbitals even $\ce{N2}$ has extra $p$ orbitals.

@AvatarShiny which orbitals ?

My bad we don't leave orbitals in N2 but you can leave put orbitals


text:
Does ethanol conduct electricity?
I am trying to find out about methylated spirts/denatured alcohol. As far as I can fine, methylated spirits is roughly 95% ethanol. Will it conduct electricity?
I have done some research, but I have not found much of anything. This reference discusses ethanol's pKa value (15.9, no source for value). I am aware that the reason why pure water conducts is because it dissociates into ions. Additionally, acid dissociation is written about on Wikipedia, but it always involves two reagents (something and water) - except for water self-ionization.
Water self-ionization is caused because of autoprotolysis. This led me to research autoprotolysis constants. Some specific research was conducted in this area by Faraji, Farajtabar and Gharib (Determination of Water-Ethanol Mixtures Autoprotolysis Constants and Solvent Effect). This roughly concurs with the pKa of 15.9 (Faraji et al. found a pKa of 16.62 with 90% ethanol and 10% water).
According to Wikipedia, the higher the acid dissociation constant the lower the chance that it will dissociate at any given pH. Since the pKa of pure ethanol would be higher, would it dissociate (I am unaware of what threshold is required). Even though I read the article, I was unsure about the autoprotolysis constant that it claimed to be investigating. It seems to me that the constant is just a pKa value, not anything more.
So, is it possible to determine whether high concentrations of ethanol conducts electricity or not? Will methylated spirits conduct electricity? In my research, I was not able to find anything like a conductivity or resistivity constant. Are these available for ethanol? What are they?

Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?

I hope that fulfils the requirements. If not, I suppose I could include more of the research.

https://chemistry.stackexchange.com/questions/31507/electrical-conductivity-of-organic-solvents

These individuals seem to say that ethanol does not dissociate, while Eric Griffin says that ethanol is an electrolyte. Curt F thinks that it will dissociate to a lesser extent than water, but it still may conduct a little. Basically, I have conflicting answers that aren't very clear. Can someone elucidate?

Avs Sharma seems to be saying that while ethanol may dissociate, it does not because it is highly unstable due to absence of resonance.


text:
Calculate the Magnitude of Nonadiabatic Coupling
I'm in the great position to be able to calculate nonadiabatic couplings with TDDFT.
The result of such a calculation is a displacement for every atom in x, y and z.
  Atom         X              Y              Z
---------------------------------------------------
   1      -0.000000      -0.000000      -0.160841
  ...         ...            ...            ...
   N       0.000000      -0.000000       0.009654
---------------------------------------------------

How do I calculate the magnitude of that coupling?
(It would be great, if you'd be as tutorial-like as possible.)


text:
Does titanium(II) oxide conduct electricity?
In my book, it is given that $\ce{TiO(s)}$ is an electrical conductor. But I'd think that since $\ce{TiO(s)}$ is ionic and in solid state, it shouldn't conduct electricity. So, why does it conduct electricity?

I'm not really convinced that $\ce{TiO_{$1\pm x$}}$ is best described as a mixture of Ti metal with oxygen.
The way I was taught it is that in $\ce{TiO}$ the oxidation state of $\ce{Ti}$ is $+2$; hence the $\ce{3d}$ band is partially filled and the electrons in that band can act as charge carriers. $\ce{TiO}$ has a notoriously wide range of stoichiometries, but the presence of extra or fewer vacancies doesn't affect the argument (as long as the average oxidation state of $\ce{Ti}$ is less than $+4$, the $\ce{3d}$ band will be partially filled).
For the $\ce{Ti 3d}$ orbitals to form a band, they need to have sufficiently good overlap; this is ensured by the fact that (1) $\ce{Ti}$ is early in the $\ce{3d}$ block, so $Z_\mathrm{eff}$ is low and the orbitals are relatively radially extended; (2) the ordered vacancies in $\ce{TiO}$ allow for a more compact structure and closer $\ce{Ti-Ti}$ contact. See e.g. Smart/Moore Solid State Chemistry, 4th ed., p 262.
I'm not a solid state chemist and I do not know if this is still a simplification, but I think it is an improvement over treating $\ce{TiO}$ as $\ce{Ti}$ metal with oxygen in it.

From Wikipedia,

Titanium(II) oxide ($\ce{TiO}$) is an inorganic chemical compound of titanium and oxygen. It can be prepared from titanium dioxide and titanium metal at 1500 °C. It is non-stoichiometric in a range $\ce{TiO_{0.7}}$ to $\ce{TiO_{1.3}}$ and this is caused by vacancies of either $\ce{Ti}$ or $\ce{O}$ in the defect rock salt structure. (citation given below)

Basically, for your purposes, this is basically titanium metal with oxygen dispersed throughout its crystal structure. Hence, the titanium metal constituent conducts electricity.
(Citation:
Holleman, Arnold Frederik; Wiberg, Egon (2001), Wiberg, Nils, ed., Inorganic Chemistry, translated by Eagleson, Mary; Brewer, William, San Diego/Berlin: Academic Press/De Gruyter, ISBN 0-12-352651-5)


text:
Difference between L&D and S&R in naming
What is the difference between L&D and S&R? Can we say S-alanine instead of L-alanine?

Read stereochemistry in any standard textbook of organic chemistry.

The main difference between L, D configuration and S, R configuration is that the first one is relative configuration while the second one is absolute configuration.
When you are distinguishing L-alanine from D-alanine, you only know that the $\ce{-NH2}$ group on the chiral carbon of alanine is on the left hand side, while in D it is on the right hand side relative in a Fischer projection. There you can't be specific about other functional groups (like $\ce{-COOH}$ or $\ce{-H}$) position.
But when you distinguish S form of Alanine from the R form, it is done by proper arrangement of other two functional groups and a hydrogen atom. If any one of those is switched to some other position, the absolute configuration may change from S to R or from R to S. So, the configuration is more rigid in case of R-S configuration.
As the L-D configuration doesn't specify other group's position, you can't always say S-alanine instead of L-Alanine unless you know total spatial arrangement of all functional groups of alanine.


text:
Rate of formation of hydrate in a carbonyl compound
When water in presence of any acid is added to a carbonyl compound, it leads to the formation of its hydrate. But how do we determine the rate of reaction? Is it done by checking the amount of partial positive charge on carbonyl compound  (by electron withdrawing group attached to carbonyl carbon)?

The formation of the hydrate of any carbonyl compound has the rate determining step as the nucleophilic addition of a water molecule to the electophilic carbon atom of the carbonyl group.
Hence, any electron withdrawing group, like fluoro or nitro, attached in the chain next to the carbonyl group will favor the product side of the equilibrium.
On the other hand, extra branching at the alpha position due to methyl/t-butyl groups will cause steric hindrance, while electron releasing groups will reduce the carbon's electrophilicity, both favoring the reactant side of the equilibrium instead.
It is important to note that while the hydration of alkenes goes to completion, the hydration of carbonyls does not. Hence, ignoring a few exceptions like formaldehyde, ninhydrin, chloral, etc., an equilibrium is setup during hydration of carbonyls. Therefore, it is better to talk about the position of equilibrium in this process rather than the rate of the reaction, as correctly pointed out by @Mithoron.


text:
Addition of acid to 3-methyl-2,3-dihydropyran

When a general acid, H-X reacts with the alkene in an addition reaction, which regiochemistry is expected?
In addition reactions X attaches to the most substituted carbon whilst H bonds to the carbon with fewest alkyl groups. In this case I would believe the third carbon to be the most substituted carbon due to it being bonded to three other carbons. And the second carbon to be where the hydrogen will bond. However, apparently it should be the other way around. And this would be the compound formed:

Why is the second carbon the most substituted in this case? Does it have to do with oxygen being more electronegative and thus it is seen as a higher prioritized substituent?

Apparently, you're using the Markovnikoff's rule here. Recall that it is only a basic rule intended for simple unsaturated hydrocarbons, and fails at more advanced examples involving heteroatoms or rings.
In this case, the first step is the addition of $\ce{H+}$ ion from the acid to the pi bond, which leads to the formation of a carbocation. The carbocation's positive charge is most stable next to the oxygen atom. Can you tell why?
In the second step, the $\ce{X-}$ ion will simply attach at the position where the positive charge was the most stable, leading to a product as you've shown in the second image.

Is the carbocation's positive charge more stable next to the oxygen because it has lone pairs which by resonance can donate a lone pair to form a new pi-bond (double bond) with X which is more stable than single bond?

@J.Se Yes, the donation of a single lone pair from oxygen completes the empty octet of the positively charged carbon atom, thereby stabilizing the carbocation. Correct answer!


text:
Why oxides with the element in higher oxidation state are more acidic
We see this trend of acidic-basic property of oxides in most cases:
$\ce{Mn2O7}$ is acidic, $\ce{MnO}$ is basic, the intermediate oxides are progressively less acidic and more basic, as the oxidation state of $\ce{Mn}$ decreases.
What is the reason of this general trend in almost all cases?
I have seen many explanations of this, most of them involving the proton releasing capacity of $\ce{E-O-H}$ bond, in the hydrolysed oxide. But most oxides of transition metals do not hydrolyze in water. So, I am seeking an alternative explanation without having to involve the hydrolysis of oxide. Also, most explanations on the internet lack any reference, so I prefer a reference.

https://chemistry.stackexchange.com/questions/37192/is-a-highly-oxidized-element-more-electropositive-than-a-lesser-oxidized-counter

Related: https://chemistry.stackexchange.com/questions/69282/manganates-ii

It does not appear to be always true.  According to Wikipedia, copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit lower than copper (II).  The higher oxidation state of copper qualifies as less acidic.

A simple answer for this would be to say that $\ce{Mn2O7}$ has +7 oxidation state. Thus it is electron deficient and behaves as a lewis acid.
$\ce{MnO}$ has +2 oxidation state and it is comparatively electron rich. Thus it behaves as a lewis base. So compounds in the middle are amphoteric.

In the bonds $M^{z+}-O-H$, where $M$ can be $Mn$ or $Cr$, the atom $M$ is positively charged. When this charge $z$ is large, $M$ repells the $H$ atom stronger than if z is small. So the molecule containing these bonds is a stronger acid. For example, $Z$ = $6$ in $H_2CrO_4$ and $Z = 7$ in $HMnO_4$. They both contain at least one $M-O-H$ bond. As $Z$ is high in these molecules, they are strong acids. Molecules containing $H, O$ and the same atoms $Mn$ and $Cr$ at a lower oxidation state exist. But they are not acidic in water. Examples : $Mn^{2+}$ makes $Mn(OH)_2$ which is insoluble in water and not acidic. $Cr^{3+}$ and $Cr^{2+}$ both produces $Cr(OH)_3$ and $Cr(OH)_2$ which are also insoluble and not acidic.

This is just wrong.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Reaction of electromagnetic radiations with the air pollutants
My friends, who are working on an IoT based technology project, explained their idea to me which is roughly like reducing air pollution by converting pollutants to oxygen ($\ce{O2}$) in any reaction, using waves and in particular microwaves. I have two questions in mind:

How do waves such as infrared, ultraviolet, microwave and others react with pollutants or in general with a gaseous mixture?
Which electromagnetic waves would be better in terms of factors such as time, temperature, rate of reaction and cost?

The pollutant mixture may contain all the pollutants described here, but the project is much more concerned with those that are more harmful than others to humans and life on earth.

I believe irradiation of $\ce{NO2}$ could convert it to $\ce{NO}$ and $\ce{O}$. Don't know how much that will help - it might make ozone.


text:
Can polypropylene be separated from polyethylene ?
What techniques are available for sorting on a large scale?

Practically only by mechanical means. Recycling numbers:
2 – HDPE (High-Density Polyethylene)
4 – LDPE (Low-Density Polyethylene)
5 – PP (Polypropylene)


text:
Hofmann rule and deviation from E2 mechanism
I've learnt that

The best reaction conditions for synthesis of alkene by dehydrohalogenation are those that promote an E2 mechanism.

(From "Organic Chemistry" by TW Graham Solomons)
But, in the same book it is given that when 2-bromo-2-methylbutane follows Hofmann elimination with tert-BuOK as follows:

Clearly, it isn't following E2 mechanism.
When given such reactions, should I go with E2 mechanism or Hofmann rule, and why?

The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.
Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.

Why not E1?
First let's consider the reaction as:
$$\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -> }$$
Clearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.
If you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.

Why not E2?
The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.

Why E1cb?
Now consider your above given case. Here, the base $\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.

So, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).

The anti-elimination is not controlled by the bulk of the base. What is controlled by the bulk of the base is the accessibility of the bases to either of the two methylene hydrogens or the six methyl hydrogens and consequently the distribution of the Hofmann and Saytzeff alkenes. From the data below you can see that an increase in the bulk of the base favors the Saytzeff alkene.
The selectivity values correct for the number of hydrogens available for each mode of elimination. For potassium ethoxide, assume that the ratio is 70/30 to make the math simple. The percent 70 is divided by 2 (70/2 = 35) and 30 is divided by 6 (30/6 =5). The ratio 35/5 simplifies to 7/1 selectivity. This means that the rate constant for Hofmann elimination is seven times greater than for Saytzeff elimination, both of which are irreversible. In the presence of potassium t-butoxide the rates of the two eliminations are about equal. The base derived from 3-ethylpentan-3-ol favors Saytzeff over Hofmann elimination by a factor of 3.

Hint: When the reactant is bulky , or it involves fluorine , then hoffmann rule is preffered over the normal saytzeff rule .

So, is there no relation between Hofmann rule and E2 mechanism?

@RinSam  This is another type of reaction also better known as $E_1cb$

Saytzeff  and Hofmann eliminations are competitive, irreversible E2 processes with your bromide and in general. The Saytzeff/Hofmann ratio of elimination products decreases with increasing bulk of the base. The potassium alkoxide of 3-ethyl-pentan-3-ol gives even more Hofmann product than potassium tertiary butoxide.

@user55119 It is understood that a bulky base triggers Hofmann elimination. But, if it was an E2 process, then won't it affect coplanarity?


text:
Can the change in the entropy of the surroundings always be obtained by dividing heat transferred by the temperature at which the transfer occurs?

Consider $\pu{1 mol}$ of an ideal monoatomic gas going through reversible isochoric heating from $\pu{100 K}$ to $\pu{1000 K}$.
Calculate $\Delta S_\pu{sys}, \Delta S_\pu{surr}.$

$$\Delta S_\pu{sys} = nC_v \int_{\pu{100 K}}^{\pu{1000 K}} \frac{dT}{T} = \frac{3}{2}R\ln10$$
Now, as I have
$q_\pu{sys} = 1350R$.
I can put $\displaystyle \Delta S_\pu{surr} = \frac{-q_\pu{sys}}{T_\pu{surr}}$ to get
$$\Delta S_\pu{surr} = -\frac{27}{20}R$$ but in a reversible process $\Delta S_\pu{total} = 0 $ but that gives me $$\Delta S_\pu{surr} = -\frac{3}{2}R\ln10$$
Where am I wrong?

Image source: Physical Chemistry, 10th ed. by Atkins and de Paula, p. 116.
Here, $\displaystyle q_\pu{sur} =$ heat supplied to surroundings.

The surroundings that are assumed in the question are not the same surroundings as in the image. The surroundings in the image are essentially at a constant temperature while it is assumed in the question that the surroundings are heating the gas reversibly. The gas is heated by some heat source which is in thermal equilibrium with the gas.
Let us assume that the surroundings which are not at a constant temperature are heating the gas. We assume that our universe consists of only gas plus the surroundings. The surroundings have some internal source of energy distributed uniformly so that no local hot spots occur. Now,
$$\Delta S_\pu{gas} = n C_V \int_{\pu{100 K}}^{\pu{1000 K}} \frac{\operatorname{d}T}{T} = \frac{3}{2} R\ln10 \tag1\label{(1)}$$
And for a reversible change, $\Delta S_\pu{total} = 0$ and thus $\Delta S_\pu{surr} = - \Delta S_\pu{gas}$.
We can also verify this in the following way:
$\pu{d}q_\pu{gas} = n C_{V} \cdot \operatorname{d}T$ and from the First law of thermodynamics, $$\pu{d}q_\pu{surr} = - n C_{V} \cdot \operatorname{d}T = - \pu{d}q_\pu{gas} \tag{2}\label{(2)}$$
Thus, $$\Delta S_\pu{surr} = \int^{\pu{1000 K}}_{\pu{100 K}} \frac{\pu{d} q_\pu{surr}}{T} = - n C_V \int_{\pu{100 K}}^{\pu{1000 K}} \frac{\operatorname{d}T}{T} = -  \frac{3}{2} R\ln10 \tag{3}\label{(3)}$$

I disagree with the explanation given in the text.  The entropy change of the surroundings is given by the equation presented in your title only for an ideal constant temperature reservoir.  Such an ideal reservoir is characterized by (a) an infinite capacity to absorb heat without its temperature changing and (b) infinite thermal conductivity, so that temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_{Res}$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible, and the entropy change of the system always satisfies $ΔS≥Q/T_{Res}$.
None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is thus always $-Q_{sys}/T_{Res}$, irrespective of whether the process is reversible. I should also mention that, although often not stated, the Clausius inequality is supposed to be expressed in terms of the temperature at the interface with the surroundings $T_{Res}$ where all the heat transfer is occurring.
Typically, all the surroundings reservoirs you will encounter in most thermodynamics texts will be ideal constant temperature reservoirs.
The description in your post is inappropriate because, if the reservoir has a constant volume (i.e., constant mass), its temperature can change between the beginning and end of the process.  Its entropy change will not be given then by the equation presented.  In addition, the description omits the requirement that the reservoir most have infinite thermal conductivity (so that the temperature in the reservoir is spatially uniform).

If I assume it to be constant temperature reservoir and use that equation then also there is a discrepancy in the two values of $\Delta S_{surr}$. The equation seems to hold only for irreversible processes.

Oh.  I forgot to mention the other key characteristic of an ideal reservoir:  the fluid is assumed to have infinite thermal conductivity, so that the temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs).  So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_R$.  In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible or not, and the entropy change of the system always satisfies $\Delta S\geq Q/T_{R}$.

None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is always $-Q_{sys}/T_R$, irrespective of whether the process is reversible or not.  I should also mention that, although often not stated, the Clausius inequality is always supposed to be expressed in terms of the temperature at the interface with the surroundings $T_R$ where all the heat transfer is occurring.

@ChesterMiller I think you should add those comments to the main answer. They're important.

@ApoorvPotnis OK.  I've done that.  I like the answer much more now.


text:
Can someone explain how do the mechanism for this reaction?

Do you protonate the oxygen first?

The dihydrofuran is an enol ether. You should try protonating it as if it were an enolate/enol to make a oxocarbenium ion first.

I think the mechanism will be something like this,
In first step the hydrogen attached to $C_2$ as the carbocation on $C_1$ is more stable.


text:
how to display diffusion to a young audience?
I need to display some science topic pertaining to the atomic bombs, or atomic history in some way to children of approximately middle school level. So I've decided that something interesting to cover would be gas diffusion (as used to isolate U-235 for the atomic bombs). I tried to see if there were any non-toxic, easy to obtain colored gases, but I couldn't find any. So the next thing I thought of is liquids.
Now I'm wondering if I can do something similar with liquids but I'm not sure how liquids would behave. Could I set up an effusion device, fill it with water and red food dye on one side, and water with blue food dye on the other and open a hole between the two and have them slowly diffuse? Will the diffusion of these two side be related to the molar mass or concentration of the dyes? Or would it follow some other law?

As you mention most colored gases (Cl2, Br2, NO2, I2) are toxic and being a gas difficult to handle. Therefore not suitable for such demonstration. Maybe, you could try two small colored smoke bombs in a transparent container although the colour is not really a gas (more like fine particulate) in this case.
I would use liquids with the most common diffusion demonstration being a drop of blue ink in a glass of water. However, we need something fancier than that so how about this: Prepare a solution of red dye in water and a solution of blue dye and sugar or salt in water as well. In a 120 mL volumetric cylinder (needs to be large and for people to be able to see) add the sugar or salt blue solution and then on top of that layer with the red water solution by adding it slowly and carefully. Due to the larger density of the blue solution the two solutions will not mix instantly. You should end up with a bottom blue solution and a top red solution. Diffusion will start where the two liquids come in contact giving you a purple band in the middle which will expand over time with the speed being depending on the concentration of the bottom solution in sugar or salt (you ll need to tune this according to your presentation). You could even show how diffusion becomes faster with temperature by heating with a hairdryer but the effect will be masked a bit by convection.
This is how we grow crystals of compounds in the lab as well: Dissolve your compound in a heavy solvent such as DCM and layer it with a light solvent the compound doesnt dissolve in such as hexane. As the hexane slowly diffuses in the DCM solution the solubility slowly drops resulting in slow formation of high quality crystals suitable for X-Ray diffraction studies.

Spray perfume in the air. Ask the people around you to raise their hands when they can smell it.
The people closest to you will raise their hands first. People farther away will raise then hands after that.
A perfect example of diffusion.


text:
What is the order of volatile nature of Group 15 hydrides?
Among NH₃,PH₃, AsH₃, SbH₃ the increasing order of volatility given in my textbook is as follows.

NH₃< SbH₃< AsH₃< PH₃

I know that among all these hydrides NH₃ alone is capable of forming Hydrogen bonds thus it is least volatile, but I’m not sure how to compare others.

Recall that volatility is inversely proportional to the boiling point. The boiling point itself is directly proportional to the intermolecular forces. In the above case, we have two major intermolecular forces: H-bonds and dipole-dipole attractions.
Now, it is factual info that ammonia's intermolecular forces surpass even stibine's intermolecular attractions, so its volatility is the least.
Thus, we are only left with phosphine, arsine, and stibine. Here, recall that intermolecular forces are van Der Waal's forces, and thus their magnitude is directly to their size. (read the "why" of it here). Hence, stibine is the second-least volatile (as it has maximum van der waal's force due to its bigger size), followed by asrine, and finally phosphine.


text:
How to determine crown ether cavity size?
I want to determine the cavity radius/ dia of crown ether. Which technique should I follow?

Knowing the crystal structure and van der Waals radii of the atoms forming the cavity it's a matter of finding distance between two points in 3D space. Many molecular editors and crystal structure software such as Avogadro or Mercury can do that via a couple of mouse clicks.


text:
How to decide what pathway a reaction will follow, elimination, addition to a carbonyl or nucleophilic substitution?
There is a substrate such as this,

and it is treated with alcoholic KOH.
I know that a hydrogen in $\alpha$-position to the carbonyl group will be abstracted, leaving a carbanion. My query lies in deciding what pathway will the substrate follow from hereon. I see three options: An E1cB elimination of the bromide, aldol condensation by nucleophilic addition to the carbonyl group and SN2 with the 1 degree bromide.
Which pathway will be favored and why?
I'd be glad if someone could specify the conditions for one pathway being predominantly favored over another (even if I haven't provided information regarding that, like temperature, solvent, etc) or some empirical data on one pathway being faster or slower. Any degree of help will be appreciated, thank you.
Edit: The answer in the book where this question is from says that nucleophilic addition should predominate (providing neither justification nor data), which contradicts my guess that elimination would dominate because of it being an internal mechanism.

Two of the possible pathway require the carbanion to interact with a second molecule, one does not. Which do you think will happen quicker?

I actually thought of that as well, but the textbook with this problem stated that addition would be faster so I am confused.
(Should I edit this into the original post?)

Yes, add that to the question

Can you tell the name of the book?

It's a problem book for preparation for IIT-JEE, Resonance Rank Booster.

Though a bit advanced, this book is very good and relevant. "Electron Flow in Organic Chemistry" by Paul H. Scudder

The nucleophilic substitution with another equivalent of the primary bromide will never be favourable as the nucleophilicity of the enolate generated is pretty low (the charge is delocalized!)
Even if E1cb was a possible route, you would end up with an extremely reactive vinyl aldehyde, which would likely participate in a Michael addition with an enolate in the reaction medium. So, isolating the product from this route would be very unlikely, even though I cannot back my claim up immediately.

If you ask me, both the Michael addition and the Aldol reaction should compete, given the nature of the base the problem has used. Had it been something like LDA (or any other directed Aldol conditions), the greater stability of the enolate would probably have precluded E1cb as a possible route.


text:
Expansion under isobaric adiabatic condition

A 3 mole sample of a triatomic gas at $\pu{300 K}$ is allowed to expand under isobaric adiabatic condition from $\pu{5L}$ to $\pu{40L}$. The value of change in enthalpy is:

$\pu{12.46 KJ}$
$\pu{-14.965 KJ}$
$\pu{-24.62 KJ}$
$\pu{-10.24 KJ}$

In case of adiabatic process, there is no transfer of heat so $Q=0$. Again for an isobaric process, pressure is constant that is $\Delta P=0$.
So, change in enthalpy:
$$\Delta H = \Delta U + \Delta(PV) = Q + W + P\Delta V + V\Delta P = W + P\Delta V = -P\Delta V + P\Delta V = 0$$
Thus, we see that change in enthalpy is zero. But the answer matches with none of the options. What is incorrect in my solution?

If, at time = 0, we suddenly drop the external force per unit area to $P_{\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:
$$nC_\mathrm v(T_2-T_1)=-P_{\text{ext}}(V_2-V_1)=-\frac{nRT_2}{V_2}(V_2-V_1)=-nRT_2\left[1-\frac{1}{8}\right]$$
Solving this equation for $T_2$ yields: $$T_2=\frac{T_1}{\left[1+\frac{7}{8}\frac{R}{C_\mathrm v}\right]}$$
But, for $C_v=3R$, this gives:$$T_2=0.774T_1=\pu{232 K}$$
So, $$\Delta H=3(4R)(232-300)=\pu{-6.78 kJ}$$
So the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.

I figured out the problem!
There is a misprint. The word isobaric should be deleted. Without it, the question makes complete sense. Here's how:
Since it is now a reversible process, apply $TV^{\gamma-1}=\text{constant}$, with $\gamma=1+\frac2f$ and $f=6$ to get $T_2=\pu{150K}$. Now, $\Delta H=n\cdot C_p\cdot\Delta T=3\cdot(1+\frac62)\cdot R\cdot(-150) \pu{J}=-\pu{14.965 KJ}$ as given a correct answer in your book.
Thank me later! ;)

Note: This answer has been completely rewritten. The previous version of this answer can be found in the revision history.

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


text:
Carbocation rearrangement involving three membered rings
Question:

Taking into account of various carbocations and, as well as the rules governing mechanisms of carbocation rearrangements, which reaction is most likely to occur during the given reaction?

My take on the question was this mechanism:

I'm torn between this mechanism and option (b). How do I judge between anchymeric assistance  and stabilization of the intermediate through 5 alpha hydrogens. How will the carbocation rearrange?
Source: MS Chouhan, Chapter: Alkenes, Q.17
This inadequate solution was given:

related https://chemistry.stackexchange.com/questions/32822/ring-expansion-in-cyclic-compounds

Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.
It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.
In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.
Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]

Disclaimer: This is a very simplistic view of the cyclopropyl methyl cation and its reaction given above with $\ce{Br-}$, and is completely wrong. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.

How will the carbocation rearrange?

The following carbocation:

will not rearrange. It is a cyclopropyl methyl cation. This is undoubtedly the best method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.
The ring will not expand to form cyclobutane. No alkyl shift occurs. The $\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).
PS: The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail here (remark by @ron: "The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane "banana bonds" and the adjacent cationic center").
These two links (1) and (2) are for further reading regarding exceptional stability of cyclopropyl methyl cation.


text:
Is it possible to determine the absolute value of entropy of a system?
I came across a question (in some assignments I had received) where it was asked:

Can the absolute value of entropy of a system can be measured?

and to my surprise, the answer said yes. Till now, I knew that entropy is a state function, just like enthalpy and internal energy, and that only change in entropy can be measured.
Am I missing something here?

Well, in practical I would say it is not possible. But there is a way to theoretically approximatly calculate an absolute value for entropy.
According to the third law of thermodynamics the entropy change can be calculated as
$$\mathrm d S = \frac{\delta q}{T}\, .$$
Per definition, a perfect crystal at $\pu{0 K}$ would have an entropy $S=0$.
So when you now heat that crystal and measure the change in entropy, since your initial $S$ was zero, the change is equal to the absolute value.
One problem is though, that the above equation only holds for constant temperatures. So the steps at which you measure the entropy changes have to happen at very (infinitesimal) small temperature changes.

"very (infinitesimal) small temperature changes." that means our crystal would always stay at almost $\pu{0K}$, won't it?

@GaurangTandon depends on how many measurements you would make :)

That's reasonable :P

Still, I wouldn't call it absolute by any means. It's just that it can't be reduced by cooling. If someone took into account freedom of movement of all electrons, I have a feeling that neutronium at o K would have much lower entropy then ordinary crystal.

Use the third law, 'the entropy of a perfect crystal is zero at the absolute zero of temperature' and then measure the heat capacity $C_P$ vs temperature  and use $\displaystyle S_T-S_0=\int_0^T \frac{C_P}{T}dT$. Many textbooks show plots of $C_P$ vs T. Additionally the entropy can be obtained ( for molecules in the vapour phase) by spectroscopic means for the internal energy and using the Sakur-Tetrode equation for the translational energy. Partition functions are calculated accurately from spectroscopic data for the vibrational and rotational energy. If $Z$ is the partition function $\displaystyle S=R\ln(Z)+RT\left(\frac{\partial \ln(Z)}{\partial T} \right)_V$ where the partition function is $Z=\displaystyle \sum_{levels}e^{-E_i/kBT}$ where $E_i$ is the energy of the $i^{th}$ electronic, vibrational or rotational energy level.


text:
Inductive effect comparison of benzylic and allyl group

I have to compare the acidity of the above two carboxylic acids. I wish to know which one has a stronger -I effect, the benzylic group or allylic group?

For future visitors, note that the difference in acidity of these two compounds is very small. The pKa of phenylacetic acid is 4.31, while that of 3-butenoic acid (page 3) is 4.34. One could thus argue that phenyl has a stronger -I effect than a simple alkene, which is expected as it has 6 sp2 carbons vs a single sp2 carbon in alkene.

Phenyl has a tendency to withdraw electrons.
see the link for the reasoning:
Inductive effect of phenyl ring
Also the allylic group would be donating via inductive effect, thus making the carboxylate ion unstable( after acid donates H+ ions)
But an sp2 carbon is more electronegative than an sp3 carbon, hence it will donate less.(more electronegative element tries to retain electrons)
So 1 has an electron withdrawing group,2 has an electron donating group which is not as good at donating as 3.This gives us the option  A

An important point to note here is that electron withdrawing groups increase H+ concentration. Think why?

Edit: Now that the Question has been edited, an edit for the solution is required.
Benzylic group has -I inductive effect as in the link. For the second compound in the question, with the allylic group, since less number of sp2 carbon are present(2 instead of 6). So we can infer that -I effect will be less. Infact it should show +I effect, especially since there is an sp2 Oxygen attached 2 carbons away, which will pull electrons as it is highly electronegative.


text:
Is there a way to determine if a compound is oxygen or water sensitive?
The synthesis of $\ce{CoH[P(OPh)3]4}$ requires the use of an inert atmosphere e.g. nitrogen as the compound is air-sensitive. Is there any way to tell whether it is the oxygen or water vapour in the air specifically? I thought it might be water sensitive seeing as metal hydrides are basic and may therefore react in an acid-base reaction:
$$\ce{CoH[P(OPh)3]4 + H2O ⟶ Co[OH][P(OPh)3]4 + H2}$$
Is rationalisation correct/is there more to it than this?

selectively expose it to pure oxygen or pure water and you will know whether only one of those is a problem.

I think that this will be water sensitive for the reasons you present but also oxygen sensitive for two reasons:

Due to the relatively low oxidation state of $\ce{Co(I)}$ in the compound which will react with $\ce{O2}$ towards formation of $\ce{Co(II)}$.
The $\ce{P^{III}(OPh)3}$ ligand is air sensitive as well and can oxidize to $\ce{OP^V(OPh)3}$ as well as reacting with water: https://www.alfa.com/en/catalog/A18662/.


text:
How to calculate the position of an element from its atomic number?
How to calculate the place of of a element if know their atomic number in a periodic table?

Try writing the electronic configuration of that element. It will give you some clues.

You will not fail following the numbers of periodicity from one electron in the outer electronic shell to the completely filled shells. Shells are filled

in the first shell with 2 electrons ($\ce{_2He}$),
in the second shell with 8 more electrons ($\ce{_10Ne}$)
in the third shell with again 8 more electrons ($\ce{_18Ar}$)
in the fourth shell with 18 more electrons ($\ce{_36Kr}$)
in the fifth shell with again 18 more electrons ($\ce{_54Xe}$)
in the sixth shell with 32 more electrons ($\ce{_86Rn}$)
in the seventh shell with - you guess - again 32 electrons.

The positions of elements were arranged empirical by their chemical properties by Mendeleev in 1865. Years later it became clear that atoms consist of electrons, arranged around the nucleus and the mentioned above electron configurations are inert or Noble gases.
Unfortunately physics claims shells to have $2n^2$ electrons (where $n$ is the number of the shell). Which brings you to the point to learn about Aufbau principle and Madelung rule.

Well, physics is perfectly correct, as long as you consider a shell to be represented by a single principal quantum number $n$. Nobody said that shells must correspond exactly to rows in the Periodic Table, which is what you seem to be talking about.


text:
Birch reduction of anisole
Hepworth, Waring and Waring (2002) mentions that:

In general, electron-withdrawing substituents make the arene more susceptible to reduction, while the opposite applies for electron-donating substituents.

They then go on to say:

Methoxybenzene (anisole) is reduced more rapidly than benzene under Birch conditions. This is an exception to the rule, the reasons for which are not clear.

Why is this the case? Does anyone know of any possible explanations?

Source: Wikipedia Commons
Also, I was thinking, since the methoxy group donates electron density mesomerically to the ortho and para positions, wouldn't it be unfavourable for the carbanion to form that the ortho positions? This seems to happen in the mechanism shown above.
Lastly, I understand that the dispute between whether it is "Mechanism O" or "Mechanism M" was settled a while ago but I've read that the evidence pointing to "Mechanism O" seems to be computational studies and experiments. However, there has yet to be any qualitative explanations, based on chemistry principles and concepts, as to why "Mechanism O" is the one which occurs to a greater extent. I was wondering if anyone knows of any such explanations?
Reference
Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

You have very good questions but in my opinion it would be better to split them up. Two of your three questions are research level (which may not actually have an answer!) and fully deserve to stand on their own.

@orthocresol I understand the first and third questions are somewhat research-level. I don't really expect complete answers for these questions but I do hope to get some insightful input from some users. As for my second question, I definitely do expect an answer as it is more of a beginner-learning-a-new-topic kind of question.

People can feel free to just answer one of the three questions when responding and I will accept the most comprehensive and insightful of them.

Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. Therefore, the rate-limiting step is the protonation of the radical anion 2 of anisole 1. Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating,3 favoring protonation rather than deuteration (isotope effect). The more basic anion, 4 or 7, which is less discriminating, will give rise to a higher D/H ratio. Employing deuterium NMR, the meta position in 8 was found to have 7-times the amount of deuterium than the ortho-position. Thus, the ortho-position (2a --> 3) is protonated first; mechanism O (ortho) predominates over mechanism M (meta). --continued--

Calculations showed that the methoxy ortho-substituted cyclopentadienyl radical is more stable than the meta- and para-isomers (red box). But your request was for a qualitative rationale for preferring radical 3 over 6. Radical 3 has an advantage over radical 6 because radical 3 may be written with the radical on the ipso-carbon, the same radical that would be expected to be generated by auto-oxidation of 5-methoxycyclohexa-1,3-diene. Cross-conjugated radical 6 does not have this option.
1) H. E. Zimmerman and P. A. Wang, J. Am. Chem. Soc., 1993, 115, 2205.; H. E. Zimmerman, Acc. Chem. Res., 2012, 45, 164.
2) V. D. Parker, M. Tilset and O. Hammerich, J. Am. Chem. Soc., 1987, 109, 7905.
3) Consider by analogy that chlorine atoms are less selective than bromine atoms in the radical halogenation of alkanes.

Thanks for your very comprehensive answer! It is truly very insightful and does respond directly to my queries. If I may clarify, the reason for why having the radical being able to be generated at the ipso carbon is favourable is that the methoxy group can better stabilise it through donation of its lone pair?

The radical is a resonance structure! A localized radical on the ipso carbon is just one of the canonical structures. I was trying to make an analogy to the ease with which diisopropyl ether or even diethyl ether forms hydroperoxides via a radical mechanism. Thank you for the complement.


text:
Why is the activation energy for 1,2-addition less than that for 1,4-addition to conjugated dienes?
1,2-addition proceeds faster than 1,4 addition making excess of 1,2-adduct at low temperatures. Also, since conversion of 1,2-adduct to 1,4-adduct is faster than the reverse, 1,4-adduct is in excess at high temperatures. Both facts account for the same thing, that activation energy for 1,2-addition less than that for 1,4-addition to conjugated dienes. But why?

I'm not sure if it's an outright duplicate, but I covered this topic in a previous question: Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control