text:
Utility of proton vs carbon NMR
I've just learnt about both types of NMR, and I haven't ever used NMR in real life, so anything I say may be wrong.
The crux of this question is that, in my own opinion, proton NMR gives you more information than carbon NMR, which I'd imagine to be much more helpful when analysing compounds. Because of this, can anyone tell me whether carbon NMR is ever used instead of proton NMR (which seems to be more 'powerful')? If it is, when?

Good question! The first spectrum an organic chemist will take after isolating the product of a reaction is the 1H NMR. It is quick, routine and informative.  But a synthesized natural product is better .compared with the natural product by the decoupled 13C NMR as well as 1H NMR. The 13C NMR is all singlets and and a field of chemical shifts that is ~20 times that of a 1H NMR spectrum. Chemists are fortunate to have many different tools to identify structures.

TL;DR Yes, sometimes. However, in general 13C data is much less useful than 1H (and 2D) data and the truth is we can often do without it.

I will limit my answer to the context of organic chemistry. Firstly, I directly answer your question: is 13C NMR ever used instead of 1H NMR? The answer is, yes.
However, I can only think of one case where one would do so. This is when there is severe peak overlap in the 1H NMR, such that one cannot reliably use it to ascertain whether one has (for example) a mixture of compound, or just a single compound. In such a case the 13C NMR (which has much more dispersed peaks and hence nearly no overlap) is more informative than the 1H NMR. This is generally not very common.

I want to go beyond a direct answer, though, and talk about the usefulness of 13C data in general. You are correct about the 1H NMR being more helpful than the 13C. This is generally true, and the main reasons for this are twofold:

13C NMR simply does not give as much information as 1H NMR. The only useful thing that one can obtain from the 13C NMR is the list of chemical shifts – but in turn, these are virtually impossible to interpret for all but the simplest molecules. On the other hand, 1H NMR gives you information about proton environments (shifts), number of protons (integrals), and limited information about molecular connectivity (couplings).
As porphyrin wrote, 13C NMR takes significantly longer than 1H NMR, because (a) the natural abundance of 13C is much lower (b) the gyromagnetic ratio of 13C is smaller.

As such, out of all the NMR data that organic chemists collect, the 13C NMR is by far the least useful and most time-consuming spectrum. On its own, its only real use is to provide a "fingerprint" of the molecule, and perhaps some information about functional groups present. So, in most cases, the 13C NMR is indeed less helpful than the 1H NMR.

The question then becomes, "why do people still take 13C NMRs?" Before going on, though, I should make it clear that people do not take 13C NMRs preferentially over the 1H, but rather in conjunction with it.
It primarily comes back to the fingerprint idea – when everybody tabulates 13C shifts of molecules they make, this gives chemist A the information they need to figure out whether they were successful in making what chemist B previously made. You might argue that the 1H is good enough as a fingerprint, but this is not always true, and anyway the consensus is that two is better than one: all major journals require submission of 13C NMR data for novel synthesised compounds.
The only other real case where the absolute 13C shifts are useful are when one needs to resolve very close peaks in 2D spectra (e.g. HSQC, HMBC), or in compounds with very few protons (thus rendering H–H or H–C correlations in 2D spectra less useful). Even then, it is not true to say that the 13C data is preferred over the 1H data.
All in all, your question is perfectly justified. In most cases, the 13C NMR is less useful as the 1H NMR, and only in some limited cases is it as useful as the 1H NMR.
In fact, there is a pretty recent perspective article by Liermann and Schlörer[1] arguing that the 1D 13C NMR should not be recorded, except for in the special cases described in the previous paragraph. The authors also make the point that the 2D spectra serve as much more useful fingerprints and are also much better at validating a proposed structure for a molecule. Even though in crowded spectra the 13C fingerprint may be more useful than the 1H fingerprint, it's still less useful than the 2D data, where peaks are dispersed in two dimensions.

Reference

Liermann, J. C.; Schlörer, N. E. Data handling in NMR facilities and assignment of NMR spectra in synthetic chemistry labs: Why electronic structure validation should become part of the routine. Magn. Reson. Chem. 2017, 1–7. DOI: 10.1002/mrc.4675.

Many thanks for the detailed answer. I asked the question simply because I have to analyse both proton and carbon spectra in exams, and I always end up coming to the conclusion that proton NMR is much 'easier' to analyse due to the information proton spectra give you compared to carbon spectra.

In principle the method is the same for both nuclei but there are practical differences. Proton NMR is approx 4 times more sensitive due to the difference in magnetogyric ratios but more importantly 99.98% of $^1$H are present in a molecule but only 1.1% of $^{13}$C. So unless you synthesise with $^{13}$C, (v expensive etc.) proton nmr is most commonly used.  $^{13}$C can then be used as well if proton NMR is not definitive. Note also that NMR does not generally give you a structure (bond lengths, angles) just the topology, i.e. this atom next to that group etc.

It is possible to get bond angles from proton NMR, if you consider the torsion angle between H-C-C-H then it is possible to use the Karplus equation, it is also possible to use NOESY and other NOE experiments if you deoxygenate the sample before doing the experiment.

I agree, that is why I used the words 'not generally',  some angles using empirical equations and NOESY etc, but its not like doing x-ray diffraction.

Practical aspects are very important when deciding which experiments to run, and Proton 1Ds are by far the simplest and fastest experiment you can run for common organic molecules. 13C experiments take much more measurement time, that alone is a severe drawback. There might be exceptions, but in almost every case a Proton 1D is the first experiment you measure, and the only question is which experiment you run in addition to that one.
There are cases where the 1H experiments are not all that useful, for example if you molecule has very few Protons. You'd probably still run that experiment, just to make sure that there isn't other stuff in your sample that has more Protons.
But one important aspect is that those experiments give you different kinds of information. And then there are many more advanced NMR experiments that give you much, much more than those simple 1D spectra. What you measure depends on how difficult your problem is, and how sure you need to be that you're right. You can get a lot of information from a 1H 1D spectrum, but a lot of it is somewhat indirect. With 2D experiments like COSY, TOCSY, HSQC or HMBC you can directly measure the connectivity between nuclei in your molecule. Each experiment gives you different information, no single experiments gives you all the information about a molecule.

http://www.chem.ucla.edu/~harding/notes/notes_14C_cnmr.pdf
Take a look at this, for example.
When an organic chemist decide to make an NMR analysis, it is mostly to characterize a new synthesized molecule. And (s)/he will do both H1 NMR, C13 NMR, IR, MS and try to understand by looking all of them.
But if you still want to compare or only select one of them, H1 NMR will be much faster and in many cases gives many information, yet for instance when there is no hydrogen bonded to an atom, for especially bigger molecules, not just for the questions of organic chemistry or NMR courses, you will mostly not be able to see the entire molecule by merely H1 NMR. Well, again noone will guarantee that C13 NMR will make you completely solve the structure.
Funny it is, that even there are professors who say that looking solely on MS, if you know enough you will get the entire molecule in the end. In theory you can even say this to IR spectrum since neighboring atoms really have some effect on place and strength of absorption. However, these are some marginal thoughts, consumes very high amount of time if you try to apply and much prone to err compared to having as many analysis data as possible.

1H is probably the most common one, it is fast and it is quite intense for small quantities. You can use it as a fingerprint to compare results or you can determine at least some information from it. Hence it is preferred. But whenever we make a new compound or whenever we are unsure we also make a 13C. The problem is that you require more substance or a higher concentration or you won't be able to see much later. In 13C you cannot 'hide' your carbons as easily as in 1H. And it is sometimes useful because you can see if something is quarternary like in a substituted benzene ring. I think it is less used because it takes more time to measure and is often done during the night so the NMR will be available for others during the day. But that is just an operational factor so I wouldn't count that here. In general, although many groups do not do it when they publish their results, it is of interest to always mention the retention factor for the column chromatography or the boiling point under vacuum, the 1H NMR, the 13C NMR, the IR and the mass spectrum. 13C is also less used in practical courses at universities so students often believe it would be a problem. Same for 2D NMR. The option is always available and shouldn't be a problem if you ask the operator. But this common myth that it is something you wouldn't do because of costs or difficulty is just not true. If you cannot determine your compound you can always ask for the more advanced option. So in the end it depends, yeah 1H is standart because in most cases you know what you want to do but 13C is quite common as well.


text:
Reversibility of the binding of carbon monoxide to haemoglobin
I recently learnt the basic mechanism of haemoglobin: oxygen forms coordinate bonds with the $\ce{Fe^2+}$ ion in haem groups, and this reaction is reversible to allow oxygen to be released. I also learnt that carbon monoxide does the same thing as oxygen, however, this reaction is said to be irreversible due to larger bond strengths etc. meaning that it can't be released.
What isn't clear to me is what happens to carbon monoxide if it does bind to a haem group. If the reaction is irreversible, then how is carbon monoxide removed from the body if you were to breathe it in?

That's just a matter of what you'd call irreversible in this case it's just a lot harder to unbind then O2

The reaction is not irreversible, but hemoglobin adheres to $\ce{CO}$ about 200 times more tightly than it does to $\ce{O2}$. Thus, part of the treatment for $\ce{CO}$ poisoning is to expose the patient to oxygen, which will slowly displace $\ce{CO}$ from carboxyhemoglobin.
For more on the chemistry involved, see this qestion.

Thanks for clearing that up. My exam specification says that the haem-carbon monoxide bond is irreversible, which I thought was a bit strange.

You might politely point out to the examiner the references citing reversibility, e.g. Wikipedia, and that people do recover from CO poisoning more rapidly than it would take for RBC's to be replaced (their lifetime is on the order of three months), depending on how open the examiner is to criticism.

It's not a single examiner, it's a full exam board. For me to want to mention this to them, I'd have to ask my teacher to send some form of notice to them. That of course would take time etc. haha

Sorry about that. As a teacher, I occasionally noticed mistakes in a text and would point them out to students, but an exam with an error is downright, frustratingly, unfair.

Maybe it was intentional. For the A Levels we only need to know the concept of coordination chemistry, so maybe haemoglobin is just an example for us to know; they might've said it's irreversible and left it at that. Just like we need to know that cisplatin is a chemotherapy drug but we don't need to know its full mechanism of action.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Glowing stone - Scarab carving
I have an intriquing question and, after searching whole Google, found almost no information. I simply don't know where else to look for answer and so I am asking you.
When I was a child, my parents travelled with me to Egypt and there, they bought me a little scarab beetle carved of some stone, about 5 cm long, 2 cm width and height. The stone is white-ish and translucent and glows green in dark (for some time). I only remember that it might (just a speculation) have been in Luxor in Alabaster workshop. But, according to online sources, Alabaster is not supposed to glow in dark, so what is it then?
Can you please tell me?
Oh yea, right. Here are some images. The fluorescence is really weak and doesn't last longer than 5-10min so I am sorry for a bad photo, just couldn't make it better.

Daylight 01

Daylight 02

Daylight with flash

Daylight against sun 01

Daylight against sun 02

Fluorescence
Well, it doesn't look like Fluorite to me(according to Google images) but then again, I'm no mineralogist either so, maybe it's some form of it... what do you think?

Your description matches fluorite, you can see many pictures in Internet, and compare with your stone. To determine what it is precisely, you have to post a picture. EDIT. After the pictures are posted, I take back my suggestion: the key difference is that fluorite is not transparent, at least those pieces that I possess, and those which I can see on Internet. But in the dark, it looks exactly of the right color. There are very many fluorescent stones, look for example, here: http://www.galleries.com/minerals/property/fluotabl.htm

Earth Science SE will take rock identifications questions if properly presented.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What really are the units of enthalpy change?
A reasonable definition of standard enthalpy change of formation would be:
'The enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions'
Enthalpy change itself is understood as 'the exchange of energy between a chemical reaction and its surroundings at constant pressure.'
I am a little confused on what the units of enthalpy change should be. For example, for the enthalpy change of formation of water, this is shown: ΔHfo = -285.8 kJ mol-1
Now, if enthalpy change already contains 'one mole of a compound' in its definition, do we really need to include the per mole (mol-1) part in the enthalpy change? If enthalpy change is the change in just energy, how are the units kJ mol-1 and not just kJ ? I am sorry if I am overthinking here.

You grab a shovelful of your compound, it reacts and gives you some kJ, you measure those and write the number down for later use. Next time you have a whole dump truck of the same compound. Will the number of kJ be the same?

I agree with your logic, but enthalpy change of formation itself describes me taking only a mole in my shovelful of compound (as far as I know).


text:
Term symbols: counting electrons. How?
I am having some difficulty in writing the correct term symbols for diatomic molecules in their ground states.
For instance, the $\ce{S2}$ molecule has a MO configuration in its valence shell $\ce{3s\sigma^2_u 3s\sigma^2\mbox{*}_g 3p_z\pi^2_u 3p_y\pi^2_u 3p_x\pi^2_u 3p_y\pi^1\mbox{*}_g 3p_x\pi^1\mbox{*}_g}$
Now, for its term symbol I would say (erroneously) that I have two unpaired pi electrons, hence $\ce{\Lambda}=2$, so I would obtain a $\ce{\Delta}$ symbol.
As for the spin multiplicity, it would be 3. And the spin-orbit coupling term would be therefore S+2=3. Concerning simmetry, I would say that it's g, since the last occupied shell is g (still, I don't understand why Ar2 should be g if the last shell is u. Maybe because having wto electrons in a u shell I should consider u*u=g? If that is the case, it's correct to say that every molecule with an even number of electrons in the last energy level, and hence every homonuclear diatomic, is g?)
Actually, it looks like the correct term symbol for the $\ce{S2}$ molecule is $\ce{^3\Sigma _g}$, but I don't understand why. Can you help me please?

Because $\mathrm{g \otimes g = u \otimes u = g}$, all closed-shell configurations have a g term symbol (assuming, of course, that the molecule itself is centrosymmetric)

Thank you, a new trick in my toolbox! But, if I may ask, which is the correct mental process to derive the "g" term for an open-shell molecule like S2?

Also, I corrected the question: I forgot the "homonuclear" word here: "If that is the case, it's correct to say that every molecule with an even number of electrons in the last orbital, and hence every homonuclear diatomic, is g?".
But maybe "energy level" instead of orbital fits better my question: I was still referring to S2

Finding the parity of the term is nothing more than multiplying the parity label of each orbital together, so all you need to know is $\mathrm{g \otimes g = u \otimes u = g}$. That wasn't a trick or a shortcut, it was precisely the way to derive it. And yes, all ground state term symbols of homonuclear diatomics are g.

Have a look also at this answer, https://chemistry.stackexchange.com/questions/37038/how-to-find-the-number-of-atomic-microstates-for-a-given-electronic-configuratio/54333#54333

This example is probably described quite thoroughly on the internet for the case of $\ce{O2}$ (which is isoelectronic), so it may be worth searching a bit for alternate explanations.
You cannot have a $^3\Delta$ term symbol, because of the Pauli exclusion principle. The π* orbitals come as a set of two: essentially, one of those two is associated with +1 angular momentum, and the other is associated with −1 angular momentum.
Here, to be specific, "angular momentum" refers to the projection of the angular momentum onto the internuclear axis. The "angular momentum" quantum number of each MO is denoted $\lambda$, so in other words, the π* orbitals come as a set of $\lambda = +1$ and $\lambda = -1$.
The quantum number that measures the "angular momentum" of the term is denoted $\Lambda$, and because $\Lambda$ is a projection quantum number, the value of $\Lambda$ is simply the sum of the individual $\lambda$ values. The Greek letter that represents it in the term symbol depends on the absolute value of $\Lambda$: if $\Lambda = 0$ then it is a $\Sigma$ term, if $|\Lambda| = 1$ then it is a $\Pi$ term, and if $|\Lambda| = 2$ it is a $\Delta$ term.
Now, in the ground state configuration of $\ce{S2}$, we can ignore all the orbitals below the π* orbitals because all those $\lambda$'s sum to zero. The only thing of interest is the π* electrons.
If you want to have a triplet state, i.e. the electrons have the same spin, then you cannot put them into the same π* orbital: one will have to go into the $\lambda = +1$ orbital, and the other will have to go into the $\lambda = -1$ orbital. Hence you get $\Lambda = 0$, which leads to a $^3\Sigma$ term symbol, not a $^3\Delta$.
Finding the parity of the term symbol is nothing more than multiplying all the individual parities of the electrons (i.e. for doubly occupied $\mathrm{g}$ MOs you need to multiply by $\mathrm{g}$ twice). This is useful, because $\mathrm{g \otimes g = u \otimes u = g}$, which allows you to immediately discard all doubly occupied MOs. You only need to look at the two π* electrons, which are both $\mathrm{u}$, leading to an overall $\mathrm{g}$ parity and a $^3\Sigma_\mathrm{g}$ term symbol.
For $\Sigma$ term symbols there is an additional detail in that you need to specify the symmetry with respect to reflection in a plane containing the molecule. In this case it turns out that the molecular wavefunction is antisymmetric with respect to this reflection, so the proper term symbol is $^3\Sigma_\mathrm{g}^-$.


text:
Approximated formula of acid dissociation constant
$K_\mathrm{a}$ can be expressed as $K_{\mathrm{a}} = \frac{[\ce{H^{+}}] [\ce{A^-}]} {[\ce{HA}]}$, however, it can be approximated that $K_{\mathrm{a}} = \frac{[\ce{H^{+}}]^{2}} {[\ce{HA}]}$.
I usually use the approximated equation when I'm dealing with questions that don't provide [$\ce{A-}$]. What I don't understand is why/how can we approximate it to this?
To approximate to $\ce{[H+]^2}$, then $\ce{[H+]}$ and $\ce{[A-]}$ must be pretty much equal, and so this must assume that the acid completely dissociates. However, in practice, do acids dissociate to the point that $\ce{[H+]}$ and $\ce{[A-]}$ are equal? If not, then how can we approximate to $\ce{[H+]}^2$, and are there any mathematics that demonstrate this approximation?

As an approximation, it's valid when 1) the amount of $\ce{H3O+}$ from the dissociation of water is negligible compared to that from the dissociation of the acid, and 2) dissociation of the acid is almost complete ($\alpha \approx 1$). So in general it works for relatively concentrated ($1 < \mathrm{p}C < 5$) solutions of moderately weak ($3 < \mathrm{p}K_\mathrm{a} < 8$) acids.

The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.
We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.
We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.
Here is some mathematics that you requested:

$K_{\mathrm{a}}$ can be calculated using equilibrium concentrations:
  $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$

Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)

Here is an extension to fully show how the two equate. Taken from the same textbook:

$$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$
  Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$


text:
Calculating the pH of a solution of sodium carbonate
I've seen similar questions on this site, but I'm having trouble as my book said something else. For a $\pu{0.09989 M}$ solution of $\ce{Na2CO3}$, my book said:
$$\ce{Na2CO3 + 2H2O -> H2CO3 + 2Na+ + 2OH-}$$
So , $\mathrm{pH} = 14 + \log(2 \times 0.09989) = 13.3006$ .
The problem is that I haven't been given a value of $K_\mathrm b$ as I saw in other questions on the site. We usually don't use any of them if they're not supplied in the question. What should I do?

You should apply the correct formula  of hydrolysis of salts( here the salt is of strong base and weak acid) which is, $$\ce{pH = 7 + \frac{pK_a(HCO_3^-)}{2}  + \frac{log (c)}{2}}$$ You should look  for the correctness,  justify by your own which should be correct. You should be confident enough to say the thing wrong if it seems wrong by all possible logics even though it is written in book. By this formula you will find that the answer is close to $11.66$.

If you are given, say $K_b$, then you need to know that $K_b$ and $K_a$ are related as $K_w=K_aK_b$ .

https://chemistry.stackexchange.com/questions/99012/how-to-calculate-ph-of-the-na2co3-solution-given-ambiguous-ka-values/99814?sfb=2#99814


text:
How to tell whether lone pairs in heterocycles contribute to aromaticity?
How can I decide whether lone pair electrons take part in the continuous overlap of electron or not?
For example, the lone pair in pyridine is not a part of the conjugated system, whereas the lone pair in pyrrole is.
I had read that this was due to the lone pair in pyridine being localised in an sp2 orbital, while the lone pair in pyrrole is delocalised.
How do we identify this?

What would it look like if the pyridine N would put its pair to the p-orbital and the remaining electron to the sp2 that points outward? A wannabe-aromatic ring with 7 electrons and a lone unpaired electron on sp2.

Related: Why is only one lone pair in imidazole delocalised?; Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?

For the most common situations use this rule: a conjugated ring atom needs either a ligand or a localized electron pair to replace a missing ligand.  Any remaining electron pairs are then part of the conjugated pi system.
Thus pyridine needs a localized lone pair on nitrogen to replace a missing ligand, and that leaves only three bonding pairs for the pi system.  Pyrrole has a ligand on nitrogen so the lone pair is conjugated into the pi system along with two carbon-carbon bonds.  Furan has two lone pairs on oxygen but a missing ligand, therefore oxygen contributes one pair to the pi system.

Okay so whenever the contribution of lone pair may result in aromaticity, it will always participate in the conjugated system ? Also thanks!

The ability to form an aromatic ring is favorable, but in most cases the sigma orbital requirements come first.  Check that, then count what's left.

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


text:
Reaction of dihalides with excess of NaNH2
Gem-dihalides react with $\ce{NaNH2}$ to give terminal alkynes. Why is a terminal alkyne prefered?
For example,

I had thought that 2-pentyne would be preferred over 1-pentyne as it would be more stable due to hyperconjugation, but it says the final product is terminal alkyne.
Why?

It's all about how the reaction is run. 2-Pentyne can form under milder conditions. More vigorous conditions induce the "zipper reaction" producing the anion of 1-pentyne. 1-Alkyne (pKa ~ 25; NH3, pKa ~ 35.

this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.
Trusted source: Jerry March Organic Chemistry 6th edition.

When the base is NaNH2 1-alkynes predominate (where possible), because this base is strong enough to form the salt of the alkyne, shifting any equilibrium between 1- and 2-alkynes. When the base is $\ce{OH-}$ or $\ce{OR-}$, the equilibrium tends to be shifted to the internal alkyne, which is thermodynamically more stable.

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


text:
How come when it rusts iron can undergo oxidation without the prescence of a another metal to undergo reduction?
It seems to me rusting is an electrochemical process where oxygen, which has particularly high electronegativity, fills the role of cathode or is that an incorrect way to put it?

The main problem you're having is that electrochemical processes don't have to have a cathode/anode.

Rusting is a redox reaction.
$\ce{4 Fe + 3 O2 -> 2 Fe2O3}$
Redox reactions mean one chemical is being reduced and one is being oxidized. Reduction = electron gain, oxidation = electron loss. The reason it is called oxidation is because oxygen is so good at reducing other chemicals (it wants electrons). How do you tell what gets electrons in a compound? Oxidation numbers.
Oxidation numbers signify how many electrons each atom has. -n means it has n more than neutral state, +n means it has n less than neutral state. In molecules which have only one element, oxidation state is 0. So Fe/O2 both have oxidation state 0. There are some rules on oxidation states in compounds which are complicated, basically usually group 1 is +1, group 2 is +2, O is -2, and group 17 is -1. In this case each O in the end is -2, and Fe is +3. So each Fe lost 3 electrons.

Some rules: https://socratic.org/questions/how-do-you-calculate-the-oxidation-number-of-an-element-in-a-compound


text:
obabel doesn't split molecules
I have a .xyz input file
8
Energy =
C     0.0354972    0.7111560   -1.7500000
C    -0.0354972   -0.7111560   -1.7500000
C     1.1823040   -1.4277968   -1.7500000
C     2.3594716   -0.7301614   -1.7500000
C     0.0354972    0.7111560    1.7500000
C    -0.0354972   -0.7111560    1.7500000
C     1.1823040   -1.4277968    1.7500000
C     2.3594716   -0.7301614    1.7500000

(this is shortened, problem stays the same) which I would like to split with obabel. However, with
obabel coord.xyz -o split.xyz --seperate -m

I only get one outputfile split1.xyz containing the whole supermolecule. This is also the case if I enlargen the z-distance between the two molecules.
Does anybody either:

Know why this doesn't work
Has a different program which I can use

I don't have the time to manually split 500 molecule files.

Not knowing which version you are running, it is difficult to actually help you.
I am using Open Babel 2.4.1, and had absolutely no problem separating the two ethylene fragments from the following file (super.xyz):
12
symmetry c1
C       -0.498388809     -1.662363855     -2.140955132
C       -0.498388809     -1.662363855     -0.810797132
H       -0.498388809     -0.739022855     -2.713905132
H       -0.498388809     -2.585704855     -2.713905132
H       -0.498388809     -0.739022855     -0.237847132
H       -0.498388809     -2.585704855     -0.237847132
C        1.730729269     -0.175065928     -2.629761741
C        1.730729269     -0.175065928     -1.299603741
H        1.730729269      0.748275072     -3.202711741
H        1.730729269     -1.098406928     -3.202711741
H        1.730729269      0.748275072     -0.726653741
H        1.730729269     -1.098406928     -0.726653741

Using the command
$ obabel super.xyz -Ofrag.xyz --separate -m

results in frag1.xyz and frag2.xyz containing one ethylene each.

I'm using 2.3.2. Your input file with your command works. This helps me debug the problem at least, thank you.

@Fl.pf. your input worked for me, too (after fixing the atom count). Maybe it's the lower case o which should indicate the file type, not the output location.

yeah that, and the fact that I'm an idiot (it's --separate, not --seperate). Thanks for being my rubber ducky though :D


text:
What's the relation between molarity and molality?
I am trying to derive the relation between the molarity and molality of a solution

Mole concept $\displaystyle n = \frac{M'}{M''}$

Here, ($n$) stands for number of moles
$\ce{(M^{'})}$ stands for mass of substance in gram
$\ce{(M^{''})}$ stands for molecular/atomic mass of substance

Let $V_{\text{sol}}= $ volume of solution
Molarity (M) $= n/V_{\text{sol}}$

The above equation of molarity ($M$) shows the concentration of solution of substance, i.e. ($M$) is the ratio of number of moles ($n$) with respect to it's volume ($V_{\text{sol}}$)

let $W$ = mass of solvent in kilogram
molality $(m) = n/W$

The above equation of molality ($m$) shows the concentration of solution of substance
i.e. ($m$) is the ratio of number of moles ($n$) with respect to mass of it's solvent ($W$)

What is the relation between molarity ($M$) and molality ($m$)?

Molarity ($M$) = $n$/$V_{\text{sol}}$
Molality ($m$) = $n$/$W$

I'm confused on how to relate the $W$ with $V_{\text{sol}}$.

$W$ and $V_{\text{sol}}$ can be related via the density of solution. We have, mass of solution = mass of solute + mass of solvent.
Hence,
$$ M_{\text{sol}} = W + M’ $$
Let us assume the density of the solution to be $\rho$. Since, density is equal to mass over volume, we have,
 $$ \rho=  \frac{M_{\text{sol}}}{V_{\text{sol}}} = \frac{W + M’}{V_{\text{sol}}} $$
Thus, you have relation between $W$ and $V_{\text{sol}}$.


text:
Is there an equivalent unit cell that can describe a bcc unit cell where all points are the same element?
Similar to how an NaCl-structure can be described as just a primitive cubic structure if all the lattice points were the same element or ion. I'm asking if there any such alternative for the CsCl-structure.

(comment made into answer)

You certainly can construct what is called a primitive cell containing one atom for a bcc lattice.  There could be more than one way, but here is my method.
Start with the usual cubic representation.  Pick any face and connect the midpoints of its edges to make a smaller square.  Do the same with the opposite face.  This gives a pair of squares which are then the bases of an elongated square prism, whose lateral edges connect the two opposing faces you "contracted".  This cell, which contains just the central atom of the original bcc structure, is a primitive cell.  In this cell the lateral edges, which we did not "contract", are about 1.41 (actually, the square root of 2) times as long as the basal edges.
How does this replicate to make the lattice? If you just scoot the primitive cell over to "catch" a neighboring atom on either side of the lateral faces, you find that you must move the adjacent cell "up" or "down", in the direction perpendicular to the bases, to get the second atom centered properly.  The "up"/"down" move is half the height of the prism.  But once you see this, you can repeat this process and get all the atoms involved in the entire cubic representation.
Last question:  I said the primitive cell is an elongated square prism.  It no longer has cubic symmetry.  Where did the cubic symmetry go?  The answer is you can't make a primitive cell in the bcc lattice with the full cubic symmetry.  Instead, note that when you modified the cube you had the choice of three pairs of opposite faces to "contract".  Thus there is really a combination of three primitive cells with different orientations.  The cubic symmetry emerges from the combination rather than any one primitive cell.

Yes there is. In your particular case, it is also the primitive cubic lattice.
A hint is in the name - BCC = Body Centered Cubic Any arrangement that is BCC is also primitive cubic, for the isolated constituent building blocks.

I guess I just don't see that though, where can I see the primitive cubic lattice in a BCC system?

I can't see a primitive cubic lattice either.  I end up with a tetragonal one instead.

@OscarLanzi I have perhaps misunderstood the question. If you take the lattice of element A in an AB bcc cell, the elements A and elements B form primitive cubic lattices with each other. Not elongated prisms and not tetragonal - however if the question supposes you replace A with B I might understand your comments. (?)


text:
White compound, insoluble in water, turns pink and then white again on adding water

Which white, ionic compound(s) that is insoluble in water turns pink when you add only a little of water but white again when you add more water?

I know that the cation is one of the following:
$$\ce{Mg^2+, Ca^2+, Sr^2+, Ba^2+, Fe^3+, Ni^2+, Co^2+, Zn^2+, Al^3+, or NH4^+}$$
I was physically examining such a compound and determined it does not change the color of litmus paper (red or blue). I thought that it could be zinc carbonate because it also precipitates hydroxide (when then compound is added to dilute nitric acid).
EDIT: I found out that the compound is lithium carbonate, and it just took a long time to dissolve. How would lithium carbonate change color like this?

Reversible color change reminds me immediately complexation, and in this case maybe a complex only stable when there is very less amount of water around. However, if it is really insoluble as stated I can't know.
And the question is a riddle like one, do you have any effort on solving this question, if so edit and share it after the question, please.

I landed here because i am looking for some ingredient of ash (Wood fire, possibly some metal) that turns rose/pink if a fresh twig smoulders on top (my guess was that this is due to water vapor), but stays ash-colored in the rain. So i hope somebody will answer this.

I found out it was lithium carbonate. Why did this happen?

@Bob Where'd you get the answer from? The sources may help us.

I was told. The compound was known by someone else.

https://chemistry.stackexchange.com/questions/96539/white-compound-insoluble-in-water-turns-pink-and-then-white-again-on-adding-wa?rq=1


text:
Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?
Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?
I thought that since it is a tertiary halide it will proceed via SN1 reaction. So it must be faster in water as it is protic solvent.
But the answer given is Acetone. Why?

It  is simple this is the famous finkelstein reaction.

As you can see in the image NaCl is formed which is insoluble in acetone hence it precipitates and the reaction cannot go reverse.

But if we take water the reaction will shift in backward direction as NaCl is soluble in water.

 That is the reason we take acetone.

plus 2-Cl-2-Me-pentane is not soluble in water so that hinders the reaction further to the point where I doubt it would proceed at all


text:
Why does iron lose exactly 3 electrons in oxidation?
I'm not a chemist, but I have this question. Why does iron (Fe atom) lose three electrons when exposed to oxygen for the first time? Why doesn't it lose less than or more than three electrons?

The electronic configuration of iron is $\ce{[Ar] 3d^6 4s^2 }$.
It can lose two electrons to form the ferrous $(\ce{Fe^2+})$ ion with E.C. $\ce{[Ar] 3d^6}$. This is one stable ion of iron. The second (and more stable) ferric $\ce{(Fe^3+)}$ ion is formed by losing another electron to form $\ce{[Ar] 3d^5}$. This has half-filled d subshell, and hence imparts extra stability.
Though your question is not clear. Iron can exist perfectly in both the oxidation states. Both the oxides $\ce{FeO}$ and $\ce{Fe2O3}$ are well known, although - as expected - the first one is rarer.

Thank you. But why would it become more stable if it loses three electrons instead of two ?

@hellothere Half filled subshells provide more stability to an atom/ion as compared to non-half-filled subshells.


text:
'Alcohol Free' toiletries
Sometimes I see 'alcohol free' written in big letters on toiletry bottles, such as on deodorants and shaving foams.
You would think that the toiletries bearing these words contain 100% no alcohol, but looking at the ingredients suggests otherwise - I still see alcohols in these products.
So does 'alcohol free' only refer to one type of alcohol not being present in the toiletry, such as a type of alcohol which causes irritation etc?

Might answer the question: https://www.fda.gov/Cosmetics/Labeling/Claims/ucm2005201.htm

Consumer product labeling is not chemistry. Don’t get me started on ‘organic’...

@Tyberius If you turn that into an answer, you'll get my upvote :3

@JonCuster Yeah, chemists and non-chemists perceive 'organic' products slightly differently...

Summarizing from the FDA website (some of this info could vary by country, but I think the general ideas should hold anywhere) commercial use of the term "alcohol" doesn't coincide with its use in chemistry. Cosmetic products use "alcohol" to mean ethanol, so you may find other alcohol compounds in a product that is ostensibly "$100\%$ alcohol free".


text:
What forms the emulsion in chocolate?
Some days ago, before buying a chocolate I read the ingredients and saw that it contains a natural emulsifier. A Google search didn't reveal anything about of what the emulsion formed is, do you have any idea?
The general composition of chocolate:

The main components in milk chocolate is cocoa butter, sugar, and milk
  powder.

Then, a concrete example:

Hershey’s chocolate is made of sugar, chocolate, cocoa butter, cocoa
  processed with alkali, milk fat, lactose (milk), soy lecithin, PGPR
  (emulsifier), vanillin, artificial flavor, and milk.

This is taken from The Chemistry of Chocolate on ChemistryIsLife.com
I continue wondering what the emulsion is formed by.

Liquid chocolate is emulsion, afair lecithin emulsified.

@Mithoron yes, I'm asking why chocolate is an emulsion...

Only molten. And what else could be pretty homogeneous liquid containing fat and water?

There isn't enough information in the question to provide a definitive answer. Do you have an ingredients list?

Chocolate consists of tiny clumps of cocoa solids, sugar, and milk solids dispersed in cocoa butter.  Because the solids aren't soluble in the butter, they tend to clump up, which reduces the quality of the texture of the chocolate, it feels less "smooth."  When you add an emulsifier, the hydrophilic end tends to stick to the relatively more polar surface of the solids (the proteins in the cocoa solids, for example, or the carbohydrates) and then the hydrophobic end interacts favorably with the cocoa butter.  The coating of emulsifier tends to help the clumps stay small and well dispersed in the fat.  That makes it easier and faster to process the chocolate, and it also lets the chocolate last longer before the fat and chocolate solids start to phase separate on a macroscopic scale.


text:
origin of use of differential equations for modeling chemical reactions?
what are some of the original examples of uses of differential equations for modeling and analyzing chemical reactions, particularly those relevant to biochemistry, involving proteins and enzymes? Michaelis and Menten's work in 1910s is one example.
What are the earlier examples or antecedents to this work? (references to good historical accounts of this modeling endeavor would also be helpful).

This might be better suited for History of Science/Mathematics

The inversion of sucrose by strong acids by Wilhelmy in 1850 is considered to be the first  kinetic study . Daniel Bernoulli studied the spread if disease (reported in 1766) using differential equations effectively the same as the rate equations we use today.


text:
What is the difference, if there is, between this two distillation methods?
Reading about water vapor distillation, I found two different approaches:

Direct
Indirect

The direct approach has an arrangement like this:

Here the chamber contains the water and the plant. It is cooled and as the water vapor flows away it carries the low volatile compound vapor of the plant/vegetable. Both condense and the oils can be obtained.
The plant matter is in the same pot as the water.
The indirect approach is setup like this:

Here the water vapor is generated in the first container (the one furthest to the left) and when flowing through the second flask, it carries the oil vapor. I should remark: the plant matter is in a different pot than the water.
Question
Do you see any fundamental differences in these processes or something related?
It might have to do with the aggregation state of the material we need to extract, but I'm not sure.

Larger diagrams where we can actually read the text and/or clear labels would be really helpful here...

The images are fine, but I have no idea what's going on because I don't know what the substances are.

The upper picture is very small and grisly. Upon a second look, I get it, and the problem is that extracted stuff drops back in the water vessel below.

Of course not. The idea with the second setup is that you don't have your vegetables or whatever in the high-powered steam can, which walls might be considerably hotter than 100°C. Steam distillation is used to distill non-volatile, high-boiling (>100°C) compounds without heating them above 100°C. Picture 1 looks cute but is a technical nonsense imho. More AD1500 alchemy-style.

Partial pressure of what? I said the wall of the steam can can be hotter than 100°C. Unless you run it at low power, in which case you will never finish distilling your product, which has a low vapour pressure at 100°C. You need to go through a lot of water for a successful steam distillation, unless your compound actually boils below 100°C. In which case I wouldn't usually do steam distillation.

There is no fundamental difference between the two methods.
In both cases steam is formed and it is passed through the plant material.
The intention is to extract the non volatile water insoluble essential oils from the plant.
The method is called steam distillation.
On the other hand, normally it is not done in this way.
Since generally only the essential oil is of interest, the integrity of the plant does not matter, it is placed in water and the whole is heated and the water is distilled.
The students do this in the lab with cinnamon and orange peels, for instance.
It is much easier because you only have to heat one vessel.
In these two step setups, you must isolate the part containing the plant very well so that the steam does not condense, once the whole system is hot enough.
Preferably, you should also heat the part containing the plant, especially in your second example.

Thanks! Possibly the direct method is what you described, I choose that image because I suppose them almost the same...

Well, not exactly. In your first image the plant is above the water. What we normally do is to put it inside the water.

But we can call it direct method

Nice, thanks. I have tried the second option today on a lab, and it works quite good. But it is really nonsense if can be done as you suggest, I suppose...; The only advantage is that, in my opinion, it is visually nicer...

It complicates the setup but it might have sense if you need the plant for a second different extraction, maybe with solvent. When you boil the plant you get a mass that can not be handled very well. If you need to extract it after the distillation and then filter it, it might be very complicated. Then it is possibly better to use a two stage steam distillation to maintain the plant as whole as possible.

I like how people here think they know who downvoted them. :-/ And I still say method one and even more so the all in one pot variant are inefficent both speed-wise and in being dangerous to your precious product, which comes in contact with the hot (>100°C) wall of the vessel. It's a lousy technique, which might work for some separation problems, but is also slow. The one advantage is that it doesn't need the separate steam can. Poor people's steam distillation, who become even poorer when the energy bill comes in later.


text:
How to apply Henry's law down the sea?
Background: Henry's law states that:

At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

And mathematically:
$$S=k\times p_{\text{gas}}$$
where $S$ is solubility, $k$ is Henry's constant and $p$ is the partial pressure of the gas.
Question: I want to calculate the oxygen solubility down the sea. Is partial pressure the atmospheric one? If it is so, we have same solubility on all sea deepness, which is contradictory. What am I missing?

Edit
I know this law can be thought as depending on this equilibrium:
$$\ce{A {(g)} <=> A {(s)}}$$

What does it even mean to calculate or measure solubility of something down the sea?
Why, there is a precise and well-defined meaning, and it is hinged on equilibrium between the gas and the solution, which supposedly should be established at the said conditions. Our imaginary gas bubble, like everything else, must withstand the immense pressure of all those layers of water up above.
So no, the partial pressure is not just the atmospheric one.
Deep abyss of the seas is undersaturated in all gases, including oxygen, and that by a wide margin.

Do you mean deep on the sea there is no equilibrium with atmospheric gases? (I don't know why) so this law is just valid for a small surface layer of the sea?

To be in equilibrium, two things have to be in touch to begin with. The gas of the atmosphere is not in touch with the deep waters, hence no luck.

I don't think so, we can suppose that there is a dynamic equilibrium between the oxygen down the water and on the top...; Suppose the oil water system, there you have no touch between liquid and water vapor, and a equilibrium, isn't it?

OK, well, skip the "in touch" part. (Indeed, I have mentioned the oil/water system in a closed vessel on an earlier occasion.) The counterparts in equilibrium have to be in the same conditions, which they are not, thanks to the gravity potential. In a laboratory setup yon can neglect it; in 100m down the sea you can't.

Just in case, do you know any expression to estimate solubility in the example I ask? Taking int account gravity maybe

Taking into account gravity is the same as taking into account pressure, just like I said.

Yes but there are no gases there, that's what stuck me...

Oh, Im sorry, what I wanted to ask is if there is a way to estimate oxygen concentration on deep sea, but i'll ask on a diff question


text:
Gas concentrations along the gradient from sea level to deep-sea
Studying scuba diving-related phenomenas I thought about how to estimate gases concentration down the sea.
Of course, solubility of gases at, for example, $5$ atm of pressure, can be estimated with Henry's law, i.e
$$S=k\,P_g$$
But this is not the case down the sea, cause there is no gas in equilibrium with sea-water. There should be a way to relate solubility on the deep with the surface pressure, but can't find how.
So, is there any way to estimate gases concentration deep into the sea?

Gas concentrations down the sea can't be estimated; instead, they must be measured. Equilibrium is of no help here, just like I said.

Yes, and your opinion is valuable @IvanNeretin, but I don't see there is a reason for that. There is no eq but maybe it can be estimated.

It can't. It is below the equilibrium pressure (and well below it, in fact), but that's about it. You can't tell any more than that from the first principles. Can there be a sea of water not containing any gas at all? Why, sure. Can there be a sea nearly saturated with oxygen? Why, sure. Where is the real, actual sea? Someplace between.

@IvanNeretin There should be something like an exponential decrease...(I'm just joking). If there isn't any valuable answer or opinion I'll remove it, but let me wait.

Says who? Why an exponential decrease? In fact, there actually is a decrease, and probably it can be well approximated by a downward exponent, but the reason for this (if any) lies far away from chemistry.


text:
Measure of partial pressure on blood
Here is the issue of blood gases partial pressure.
On medical reports is common to see partial pressure of gas on blood, usually oxygen, but there is no gas in equilibrium with blood, so it is confusing. What they mean with partial pressure of a gas on blood is, according to this answer:

the partial pressure which would prevail in a gas phase in equilibrium
with the liquid [blood] at the same temperature.

I suppose they use a partially filled vessel with blood and a manometer or something like that, but I would like to know the way they measure it.
Question
How do they measure $p(\mathrm{O}_2)$?

In principle you could measure it by getting a sample of blood and putting it in contact with gaseous O2 at various pressures.   Whatever external pressure of O2 results in the amount of O2 dissolved in the blood neither increasing nor decreasing would equal p(O2) in the blood.
As for how it's actually measured, I believe the usual answer these days is spectroscopically.  That is, you shine light through a sample of blood and look for the absorption at certain frequencies which are typical of hemoglobin with O2 bound to it.  The amount of absorption is proportional to the concentration of oxygenated hemoglobin, and your software has some internal chart that converts that to partial pressure.
As for why it's measured that way, in part it's probably just tradition, and medicine changes very slowly and conservatively, for obvious reasons (you're experimenting with health and life).  In part, though, it may be that it directly relates to something important: the partial pressure of O2 in the atmosphere, which is about 160 mmHg.  That's the concentration of O2 in the lungs.  What fraction of that reaches the blood at various points in the circulation is a natural indicator of how well oxygen is being taken up, transported, and delivered.  That is, you can directly relate this particular measure of O2 concentration to the form of concentration usually used to describe a mixture of gases, in this case the air you breathe, which is the origin and destination of the blood gases.

I believe you and nice reasoning, but any resources about what you say on second paragraph?

BTW because of a higher concentratiom of water vapor on lungs oxygen partial pressure is 150 approx...

For the casual setting, like the ER, I believe you use one of those pulse oximeters.  Wikipedia has an article on those.  I looked around for what's done for serious blood gas measurement, and this article: https://www.ncbi.nlm.nih.gov/books/NBK371/ suggests it's actually done with electrochemistry, by reducing the dissolved O2 and measuring the current required.

I'm not sure about your idea of measuring O2 spectroscopically as it does not have an ir spectrum (no dipole) . Oxygen uptake is measured (in your finger) by using  visible (red)  light and the absorption change O2 binding makes to the electronic absorption spectrum of haemoglobin vs that without O2.

Oh excellent point, I made an edit to make that clear, thanks.


text:
Intermolecular interaction between aniline and dichloromethane
Short background
As it is known there are interactions between:

Charges
Charge and dipole
Hydrogen bonding
van der Waals (VDW) forces

From stronger (1) to weaker (4).
VDW forces are divided into dipole-dipole, dipole-induced dipole, induced-induced dipole; also from stronger to weaker.
The problem
Reading about solutions I found interesting the intermolecular forces involved on aniline and dichloromethane mixing. Below both structrures are shown:
Aniline
 Dichloromethane
Question
What is the most important intermolecular force between those molecules and why? It might be interesting to see an AIM analysis of aniline, but I have no idea how to do that. Also, I'm not sure about what experiment could give valuable information on this case.
My attempt
Those previous interactions aren't mutually exclusive. $\ce{-NH2}$ might be a negative density region of the molecule (inductive attraction), but it is also donor by resonance, so the ring has partially a negative charge density. In consecuence, it is not easy to see the dipole. Dichloromethane, on the contrary, is an easy-to-see dipole.
It might result that London dispersion is the most relevant interaction here, but I can't very well justify it.

According to this NIST publication (caution: link leads directly to a biggish PDF file) the dipole moment in the gas phase of dichloromethane and aniline are quite similar (1.6 and 1.5 D respectively).  These are respectable dipole moments, and neither molecule is very big, so offhand I would say an argument can be made that the dipole-dipole interaction is most important, particularly because it's orienting.
Note by the way that rating dispersion forces as the weakest should be done cautiously. It works for small molecules, but since dispersion forces scale with the number of electrons in a molecule and no other interaction does, once you get above a fairly modest size of molecule dispersion forces (along with excluded volume forces) dominate.  That's why eicosane is a solid at room temperature while water is a liquid.

Very interesting, do you find it easy to see the dipole direction on aniline? I'm in trouble with that...

Nope.  That's why I looked it up.  For what it's worth, I could rationalize it by arguing that the entire C6-N system has some pi delocalization going on because it lowers the energy of the pi electrons, to be able to spread out over all 7 atoms.  But that would necessarily mean the N has a partial + and the ring a partial - because the lp is not always on the N, roughly speaking.  Hence, a dipole.  But that's all ex post facto hand waving.  One would have to do a quantum chemistry calculation to know whether it checks out.

From a VB point of view it would suggest a contribution from the resonance structure with NH2+ on one end (with a double bond to C1) and a carbanion at the para position of the ring.  My o-chem is rusty, but isn't that res str the reason NH2 is para (as well as ortho) directing for electrophilic substitution?

Cool, thanks :) What is VB? Yes you are right with substtutions, that could be seen in some calculations also, I suppose. So the dipole is possibly positive on N and negative on para... interesting.

Sorry, VB = "valence bond theory, " I mean if you are trying to explain the electronic structure by some combination of Lewis structures.


text:
When to use iso and sec while naming compounds in alkyl halides?
I am having confusion on when to use iso- or sec- while naming the alkyl halides?

Preferably, you don’t use “iso” and “sec” at all when naming alkyl halides for two reasons:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefixes “isobutyl” and “sec-butyl” etc. are no longer recommended (“isopropyl” is still retained, but only for general nomenclature).
Preferred IUPAC names of alkyl halides are formed using substitutive nomenclature, i.e. using the prefixes “bromo”, “chloro”, “fluoro”, and “iodo” (e.g. iodomethane instead of methyl iodide).

If you cannot avoid using functional class nomenclature with traditional prefixes,

the “sec”-compounds are the simplest isomers that contain a secondary functional group, i.e. the halogen is bonded to the second carbon of the chain, e.g.
“sec-butyl chloride” is 2-chlorobutane,
“sec-pentyl chloride” is 2-chloropentane, etc.;
the “iso”-compounds are the simplest isomers that are branched at the end of the chain, i.e. all carbons except one form a continuous chain, e.g.
“isobutyl chloride” is 1-chloro-2-methylpropane,
“isopentyl chloride” is 1-chloro-3-methylbutane, etc.

http://www.chem.ucla.edu/~harding/IGOC/C/common_name.html According to this link,sec shoulld only be used for 4 member chains. Is this always the case?

iso- is used when the all carbons form a straight chain except one methyl at "2nd Carbon", like 2-methyl-R (where R refers to straight chain alkane)

And, sec- is used when any functional group is linked to a secondary carbon.

Then why is the name for iso-octane 2,2,4 trimethyl pentane rather than 2 methylheptane?

look up isoctane it will show up as  2,2,4 trimethylheptane... is that incorrect?

I am sorry, actually, 2-methyl-heptane should be the isooctane. But, since 2,2,4-trimethylpentane is the most important isomer (by application) of octane. Therefore, traditionally it is known as isooctane. Note: All other iso- compounds follows the general rule of 2-methyl-R. After all, exception is a common term in Chemistry ;)


text:
Does activation energy depend on steric factors?
I thought that activation energy depends on steric factors.
By Arrhenius equation it must depend on steric factors since $A$ depends on steric factors. $$k=A\cdot\mathrm e^{-\frac{E_\mathrm a}{RT}}$$ (for bimolecular reaction)
But my book says it doesn’t.
Does it depend on steric factors or not and why?

The Arrhenius model of a rate rate constant is very basic so has to be modified in many cases. One of these is when steric factors (generally undefined exactly) are used as an empirical  multiplier to A and used as a rationale to fit experimental data.


text:
What is the effect of oxygen in welding?
Welders use a shielding gas to avoid oxygen interact with the electrode. I supposed is might be because of ferric oxide formation as the electrode is steel-made. But fusion temperature of ferric oxide is quite the same than iron so I'm not sure what is the reason for the shielding gas (usually argon). I have gone through some works but found nothing about it.
Any idea what is the effect of oxygen? Is there any reaction involved which plays a role against a good welding or just physical issues?

The notion is to have iron in the weld seam not iron oxide which is very much weaker mechanically.

Consider - welders add oxygen to turn a torch into a cutting tool.

According to https://www.bernardwelds.com/mig-welding-shielding-gas-basics-p152080 the purpose of shield gas in steel welding is:

...to prevent exposure of the molten weld pool to oxygen, nitrogen and hydrogen contained in the air atmosphere. The reaction of these elements with the weld pool can create a variety of problems, including porosity (holes within the weld bead) and excessive spatter.

There is also a picture of a weld that either didn't use a shield gas or didn't have enough. It's pretty obvious that such a weld isn't going to be very strong.

Edit: From this Wikipedia article explaining the purpose of inert gas in welding:

Their purpose is to protect the weld area from oxygen, and water vapour.

And in what appears to be a bit of a contradiction:

Oxygen is used in small amounts as an addition to other gases; typically as 2–5% addition to argon. It enhances arc stability and reduces the surface tension of the molten metal, increasing wetting of the solid metal. It is used for spray transfer welding of mild carbon steels, low alloy and stainless steels... Oxygen causes oxidation of the weld, so it is not suitable for welding aluminium, magnesium, copper, and some exotic metals.

When you consider that atmospheric oxygen is almost 21%, the two to five percent used in inert gas welding seems pretty small. However, though I can find no definitive reference, I'd have to say that the prevention of oxides of the metal(s) being joined is a consideration only when extremely reactive metals are being welded. Instead, it seem that the prevention of gaseous materials (probably mostly water vapor) within the weld seem to be of the most importance. Also, considering the temperatures involved in welding, the creation of oxides of nitrogen from atmospheric nitrogen and oxygen would have to be a concern.

As someone who has run out of gas while welding, I can confirm... it doesn't work for beans.


text:
Why is sulphuric acid added in the ripper titration?
Ripper titration can be used to determine free $\ce{SO2}$ in wine by titration with iodine.
The method requires sulphuric acid to be added to wine before titration. Why is this the case?

There are 3 forms:

sulfur dioxide (molecular) $$\ce{SO2}$$  ,
Bisulphite $$\ce{HSO3-}$$
Sulphite $$\ce{SO3^2-}$$
(Additionally, bisulphite can become oxidized with water to form sulphuric acid)

These species exist in equilibrium, and shifts in pH with shift the concentration of each species.
The addition of sulphuric acid lowers pH and favors the molecular form. The molecular form can react with iodine and water. When there is no longer any of the molecular form to react with the iodine, the end point its reached and the color changes.
$$\ce{SO2 + I2 + 2H2O → H2SO4 + 2HI}$$


text:
What is the more stable conjugate base of cyclohex-2-en-1-one?
We have to identify the structure of its more stable conjugate base obtained by proton transfer of cyclohex-2-en-1-one.

In the following question, there are two possibilities:
We can either remove a hydrogen nearest to carbonyl group or remove a hydrogen next to carbon carbon double bond.
I think the later should be correct as it involves higher conjugation, but the answer is the opposite, they have removed hydrogen which is nearest to carbonyl group. Why is this position preferred?

The level of conjugation is the same, so inductive effects are most important here.

Under kinetic conditions, deprotonation occurs at C6 to give an enolate that is not conjugated with the double bond. Under equilibration conditions, deprotonation occurs at C4 to give an extended, vinylogous enolate. The more stable conjugate base is the one formed under equilibrating conditions (leveling effect).

You may want to look at: https://chemistry.stackexchange.com/questions/43007/what-is-the-most-acidic-proton-of-2-methylcyclopent-2-en-1-one?rq=1


text:
why do the complex [Cu(NH3)5NO2]SO4 show ionisation isomerism?
In ionisation isomerism, secondary valencies can be replaced by primary valencies on the condition that net charge going outside the coordination sphere must be equal to net charge coming inside it . Here NO2 ligand has charge - 1 on it . So how can it be replaced by SO4 which has -2 charge? Thanks!

You have two complexes.

$\ce{[Cu(NH3)5(NO2)]SO4^{-1}}$
$\ce{[Cu(NH3)5(SO4)]NO2^{-1}}$

Both $\ce{NO2^-}$ and $\ce{SO4^{2-}}$ are necessary for the overall complex to have a -1 charge overall. It just depends on which is attached directly to the copper ion.


text:
In Bohr's model what is angular momentum in the lowest orbital?
Angular momentum is given by: $n\frac{h}{2\pi}$ where $n=1,2,3...$
And angular momentum is also given by: $\sqrt{l(l+1)}\frac{h}{2\pi}$
By the first equation angular momentum should be $\frac{h}{2\pi}$ but by the second it should be $0$.
And if you say that in first equation $n$ is also $0$  then first radius of hydrogen should be $0\pu{A^\circ}$ instead of $0.53\pu{A^\circ}$.

You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular momentum is thus given by
$$
l_n=m_e|\vec{r}_n\times\vec{v}_n|,
$$
where $m_e$ the electron mass. Bohr postulated that the angular momentum must be quantized (this is the semi-classical part) resulting in a set of orbitals.
Though Bohr's model works remarkably well it is not perfect because electrons are not really particles in the classical sense, but also behave like waves. A better treatment (but not the best) of the hydrogen atom is therefore to use quantum mechanics. In quantum mechanics the electron is described by a wavefunction that is spread out over all space and has a certain shape. For $s$ orbitals, such as the lowest orbital in the hydrogen atom, the wavefunction is spherical symmetric and has no angular momentum. The second expression you mention is the quantum mechanical operator to obtain the angular momentum.
To conclude, you cannot compare the angular momentum of the Bohr atom to the Schrodinger atom because they are based on different assumptions. Note further that for principal quantum number $n$, $\ell=0, 1, \ldots, n-1$.


text:
Why are 4s electrons removed before 3d electrons?
According to Aufbau's Principle energy of 3d orbital is greater than 4s orbital, so which filling 4s if filled beforehand 3d. Whereas while removing electrons 4s electrons are removed first. Why?
I somewhere read that Aufbau's Principle is true only for ground state. When both 3d and 4s orbital get populated due to some interaction 3d shell is raised a to higher energy level than 4s. It it the case or it has some other explanation.

This might be interesting: https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html (see also "Reference" Section of the page).

That's what I already read, lol


text:
How does the number of electrons increase in beta decay?
So, I am having trouble understanding beta decay. Particularly the number of electrons.
For example; Carbon-14 decaying into Nitrogen-14.
Carbon-14 has 12 neutrons, 6 protons and 6 electrons.
When it decays 1 neutron is converted into: 1 proton, 1 electron and 1 anti-neutrino.
So after the neutron breaks down, and before emission, there should now be:
11 neutrons, 7 protons, 7 electrons and 1 anti-neutrino.
But now 1 electron is emitted from the atom (as a beta ray), alongside the anti-neutrino. Which means the numbers should now be:
11 neutrons, 7 protons and...6 electrons.
Since 1 electron is created in the process and 1 is emitted out of the atom, then surely the number of electrons shouldn't change from the original 6 of carbon-14.
But Nitrogen-14 has 7 electrons?? Can someone explain?
Thanks.

The world is full of electrons. Go steal one from something else, like any other nitrogen anion.

Surely you must have heard of such thing as ions. What are they? Guess what? To put it bluntly, they are atoms with a wrong number of electrons.

Beta decay refers to nuclei. It does not take into account shell electrons.

The two commenters have pretty much answered your question: after beta emission, which is a sufficiently energetic process to tear apart a molecule, the resulting fragments/ions would likely undergo a series of additional reactions with any nearby molecules to finally produce stable species. The emitted electron also would be captured by nearby molecules, leading to a cascade of reactions, also eventually producing stable species. In any case you do indeed end up with a nitrogen atom and an extra electron.
Example: if decay occurred in a bottle of liquid methane, you would get ammonium cation (isoelectronic with methane) and an extra electron loosely attached to methane - a methane radical anion. These could recombine to give neutral ammonia and a hydrogen atom, and then further reactions are possible.
I would also point out that carbon 14 has 6 protons and 8 (not 12) neutrons, but in any case your question remains valid.


text:
How is laughing gas produced in the Nef reaction?
I can't think of a mechanism for this reaction to give laughing gas.
Attempt:
I used the base to take out the acidic hydrogen at alpha position with respect to nitro group, but that makes the intermediate formed even more unstable. What should be the proper mechanism?

https://www.organic-chemistry.org/namedreactions/nef-reaction.shtm https://en.wikipedia.org/wiki/Nitroxyl

The Nef reaction does not produce nitrous oxide directly. It produces nitroxyl as a reactive intermediate, which rapidly decomposes via a bimolecular reaction to nitrous oxide and water$^{[1]}$. You are correct that the base removes the alpha hydrogen; the nitro group is more than capable of accepting this negative charge. The crux of the Nef reaction is this:

The mechanism for the formation of nitrous oxide is as follows:

$[1]$ Wikipedia: Nitroxyl, Reactions


text:
Dependency of rate of diffusion on mean free path and velocity of molecules?
What is the dependency of rate of diffusion ( Graham's law) on mean free path, molecular size , average velocity, collision frequency and pressure? I have searched everywhere on internet but just found that rate of diffusion is inversely proportional to square root of density or molecular mass.

A proper textbook on physical chemistry should explain this in detail. You might have better luck searching in one than on the internet

My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.
Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.
Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)
Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.
All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$
The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.
By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$


text:
Why can't NH5 form?
In ammonia a nitrogen atom forms 3 covalent bonds with hydrogen atoms and has one lone pair which can be used in a dative covalent bond with another hydrogen atom to form an ammonium ion. Instead of using both electrons for one dative covalent bond why can't they both individually be used for 2 separate covalent bonds forming NH5?

This violates the octet rule.

As the person whose answer supposedly answers this question, I really don't see how this is a duplicate, the obvious difference being that nitrogen is a Period 2 element whereas sulfur is Period 3. Furthermore, $\ce{NF5}$ is just as hypothetical as $\ce{NH5}$, so the argument I used doesn't apply here. So I have reopened this. Please think and look at the questions carefully before closing.

@orthocresol I would say that, besides the comments about size effects for period 2 elements, the currently accepted answer here doesn't differ much from the reasoning given in your prior answer. I definitely see now that it is not a duplicate, but I think at the very least it covers some of the groundwork for this question.

I have pursued such questions regarding the hypervalency of particular molecules before and thus, I would like to provide some insights to the matter.
A common superficial answer that one can provide to answer the question is that there are ten electrons around the central nitrogen atom in $\ce {NH5}$ and thus, it cannot exists based on the octet rule. And we do know that the octet rule must be strictly abided by for the period 2 elements.
However, let us look at the deeper reasons why period 2 elements behave in this way by looking at the characteristics of hypervalent molecules.
One very crucial reason why particular hypervalent molecules and molecular ions, such as $\ce {PCl5}$ and $\ce {SO4^{2-}}$, can exist is due to the presence of electronegative ligands around the central atom which can draw electron density away, reducing the inter-electronic repulsion in the molecule and hence, increasing the stability of the molecule significantly. Due to the significant ionic character resulting from forming multiple polar covalent bonds, there is actually significant ionic character within the molecule. Computational studies have shown that some of these central atoms are, in fact, bearing significant charge, with some even having a greater charge than +1. Although you may expect that central atoms in hypervalent molecules have more than 8 electrons, the reality is that they likely have less! This was shown in a study conducted by Noury, Silvi and Gillespie.[1] This goes to show how electron-withdrawing are the electronegative ligands.
Another important factor to consider is sterics. Simply put, we cannot put multiple large atoms around a small atom. Atoms of the elements of period 2 may not have reached the "critical size" for it to accommodate more atoms around it while period 3 elements seem to be able to accommodate more atoms much more easily. This is the likely reason why the molecule $\ce{NF5}$ does not exist. While the electronegative fluorine ligands may help to reduce electron density accumulating in nitrogen, the space constraints simply are too limiting. You can read more about the molecule and its associated instability at its Wikipedia page. The references provided on the page are also very insightful.
I hope it is now clear why $\ce {NH5}$ cannot exists based on considerations of the two factors above.
References

Noury, S.; Silvi, B.; Gillespie, R. J. Chemical Bonding in Hypervalent Molecules:  Is the Octet Rule Relevant?. Inorg. Chem. 2002, 41 (8), 2164–2172. DOI: 10.1021/ic011003v.

Nice. Regarding the positive charge on the central atom, it makes perfect sense too that the Period 2 elements - which are more electronegative - don't quite tolerate it as well.

Just a hint: the term hypervalent is misleading (one could say deprecated) and such molecules should better be referred to as hypercoordinated instead.


text:
Are crown ethers "recycled", typically?
I know that crown ethers have the interesting property of strongly complexing alkali metals, which allows them to be used in organic solvents (eg. as "dissolving-activity enhancing agents" for potassium permanganate or potassium tert-butylate).
Having never used such molecules myself, I was wondering: is there a procedure employed in order to remove the complexed metal, hence allowing a recycling of that ether?
I am sorry if the question might be trivial, but I couldn't find an aswer online.

In my experience it is rare to try to recover chemicals used in the lab. I've only known it to be done with precious metals, silver gold and platinum.

You can always recover the lighter ones by (vacuum) destillation. Most chemistry labs would just scale down the reaction until recycling in no longer worthwhile. But, if you need liters of the stuff, one would sure think about recycling. Mind, worthwile would mean that over a year, it would save thousands of Euros or dollars. Because you already cost ten times as much. ;-)

Given that crown ethers are very expensive (edit: apparently not quite so), I imagine if they were economically recyclable the procedure would be widely divulged, but it doesn't seem to be the case. That said, the Wikipedia article for 18-crown-6 says that it can be dried using NaK, which destroys any water and forms the electride [K(18-crown-6)]Na. This effectively "spends" the crown ether, so it makes no sense to use this process if there weren't a way to recover the metal-free crown...

Crown ethers are typically not recycled because they remain in the organic phase where normally also all the byproducts are (so the purification is very complicated) and they are still complexing the alkali metal after the reaction.
You want to recycle them to again complex some alkali metal, but then you have first to break down the previous complex, which is not straightforward.
To recycle crown ethers you must use a different approach.
The trick is to immobilize them on some solid support.
Typically, you have some crown ether functionalized monomer and polymerize it, or you have some reactive polymer and introduce the crown ether at the end.
A review can be found here.
These solid phase crown ethers are used in the same way as the better known ion exchange resins.
The problem is that you will have to synthesize them, because they are not readily available for purchase.


text:
How do we know that we have isolated a substance?
In The Historical Background of Chemistry by Henry M. Leicester page 131, the author says:

Carl Wilhelm Scheele (1742-1786) [...] isolated many organic compounds, including hydrogen cyanide, lactic, citric, and malic acids, and glycerol.

But how did Scheele know that those substances were isolated?
According to me, we can’t be sure of that without doing some very high-tech analyses.

Who says he was sure?

There is another consideration. What "pure" meant in 1780 versus what "pure" means today. Even today we can't isolate absolutely pure compounds.

@Mithoron leaving the philosophical debate on certainty aside, one has to be sure enough to claim to have a substance isolated. the answer is in @MaxW‘s comment and the answer: the chemical rigor was much less and there was still sufficient evidence.

While modern tools make this easier, there are plenty of physical characteristics that help identify that a substance is pure
Modern tools like NMR, X-ray structure determination and others make identifying pure substances and their structures relatively easy, there are plenty of other characteristics than can support the identification of a pure substance.
Pure compounds tend to have sharply defined boiling points or melting points, for example. So if you can generate crystals, that is a good start. Crystallisation from solution, for example, tends to yield relatively pure compounds and this can be checked by measuring the melting point. Repeated fractional crystallisation can confirm this if the melting point doesn't change. For liquids you can distill them and observe the boiling point. Fractional distillation can separate mixtures and you can check that the results boil at a single temperature. Gases also have physical properties that can be checked after fractional distillation or other separation techniques.
Here is an example where attention to detail yielded the discovery of a new element. Rayleigh and Ramsay were experimenting with the separation of gases and noticed that nitrogen isolated from air (by removing all the other known gases) seemed to have different properties to nitrogen formed from chemical reactions (it was about 0.5% denser, for example). By careful fractional distillation, Ramsay was able to isolate a much denser gas from the nitrogen which we now call Argon. This was verified to be an element by observing its discharge spectrum which had new lines in it not present for other elements. So careful separation techniques and carful observation of physical properties could yield a new substance with different properties to nitrogen.
Before we had modern instrumentation this careful application of known chemical and physical techniques was the basis for the development of the laws of chemistry and the identification of new pure substances. It takes more time, but it works.

Another example for a discovery by attention to detail: Rainer Ludwig Claisen discovered his well known rearrangement when he tried to find the melting point of his newly synthesized crystals of allyl 1-naphthyl ether (1-allyloxynaphthalene).


text:
Ring Expansion And Final Structure
[If you don't like to read the question just look at the 3 pictures; the first one has the carbocation which has to be stabilised by ring expansions. The other 2 are my predicted structure so please confirm which one is correct or if both are wrong please provide the correct one]
Please help me in expanding the rings in the following structure :

I tried a lot expanding the the rings twice to get this :

But the intemediate in this structure is unstable as it has a bridge carbocation.
Instead we can have a intermediate with sigma-pi resonace stablising the carbocation by expanding the ring once as follow  :

I can't wrap my head around which one is correct?

There are mainly two factors governing this phenomenon of ring expansion and rearrangement of carbocation, mainly Stability factor and Thermodynamic factor. When you have a carbocation which can rearrange within itself, it will definitely prefer a rearranged structure which has the highest stability. Not only that, the carbocation also needs some energy for the rearrangements to occur. If it needs multiple steps of rearrangements to attain a structure which has decent stability, but the loss in the potential energy can't compensate the requirement of energy for rearrangement, it is not also possible for the molecule to undergo that many number of steps thermodynamically.

Based on these two factors, we can at least logically conclude which will be the final rearranged product. In both of your predicted compounds, first there will be a rearrangement by Hydride shift. Now it is highly difficult for a molecule to undergo two rearrangements at once unless it gets  extreme high stability by the second case. In this case, it does by forming a cyclopropyl methyl cation which is unusually highly stable due to non-classical resonance and thus ring-expansion occurs in this case. So, by these two steps you have got your first possible carbocation.

Now , if we look for the third possible rearrangement of ring-expansion, here energy requirement is extremely high as it becomes more and more difficult for a molecule to undergo rearrangement as its number increases as a single reaction intermediates. Also, the newly formed carbocation is tertiary which is obviously less stable than cyclopropyl methyl carbocation. Thus, the last ring-expansion is energetically less favourable.

So, I think the correct one should be the image you have drawn at the end in your question, as further more rearrangements on that system is both thermodynamically and stability wise less feasible.

But carbocation stability isn't the only factor consider - relief of the torsional and bond angle strain may provide the additional driving force for the final ring expansion step?

But that factor isn't strong enough to compensate for the lose of stability of cyclopropyl methyl cation , which is much more dominating.

I'm not sure why you think the final compound is so unstable. It's still tertiary, although the adjacent substituents are maybe marginally less electron-rich. It absolutely is not a bridgehead carbocation, since the compound is a fused rather than bridged bicyclic.


text:
Deperoxidation of ethers via alumina: what happens?
I was reading about the procedures for the removal of peroxides from ethereal solvents, and a procedure which seems to have survived to our days is filtration through a column of "activated alumina" (which, if my understanding is correct, is alumina in its "gamma" form).
What I was wondering is: what happens to alumina? My guess is that it could form a sort of stable complex with (hydro?)peroxides, just like urea or carbonates do.
Yet, I couldn't find an explanation.
If a complex is formed, how stable is it? For instance, if the same alumina employed was dried, either via evaporation at room temperature, or at high temperature in an effort to obtain, again, its gamma form, would it decompose violently? Or maybe would it release oxygen, or gaseous peroxides?
And, if alumina has to be handled with care after use, is there a way to dispose it safely, or (better) to "reduce" it, maybe making it usable again for the same process?


text:
How do significant figures work on this computation?
This isn't a homework question although it may seem like one. We've never really discussed significant figures and it was just introduced to us when we were learning how to solve word problems regarding gas laws.

Find $V_2$ when $V_1=\pu{400 mL}$ and $T_1=\pu{32^\circ C}$ and $T_2=\pu{-12^\circ C}$.

My computation goes like this:
$$V_2=\frac{T_2V_1}{T_1}=\frac{(\pu{261 K})(\pu{400 mL})}{\pu{305 K}}=\pu{342.295082 mL}$$
As per the given, the least number of significant figures is one (from $\pu{400 mL}$). How do I approximate $\pu{342.295082 mL}$ to one significant figure?
I didn't know how to create a search query about this so I went here. Thanks.

400 is three significant figures. 0 is as good a figure as any other.

342 the number with least figures determines result figures in this operation. Usually 400 should be written with a bar above the zeros. If we took 400 with just one figure, the result would be 300.

@IvanNeretin No it isn't. The zeros might be significant. And to solve this question (i.e. report a sensible answer), you'll have to decide for one or the other.

Technically $400$ has 1 significant figure. $400.$ has 3. Though for this problem I think assuming 3 significant figures in $\pu{400mL}$ is reasonable

Assuming the precision of the given volume, $V_1 = \pu{400 mL}$, is only to 1 significant figure (i.e. only the hundreds place digit of "4" is precise), then your final answer would be rounded to $V_2 = \pu{300 mL}$.*
If we were to say that $V_1$ had 3 significant figures, then the final answer would round to the ones place, resulting in $V_2 = \pu{342 mL}$. However, the value would technically need to be written as "$\pu{400\!. mL}$," including the decimal place to denote the precision going out to the ones place, in order to imply 3 significant figures instead of 1.
In general, zeros (to the left of the decimal place) not surrounded by non-zero values are not denoted as "significant." For further rules and guidelines for identifying what numbers are significant, how many significant figures there are, and how addition/subtraction or multiplication/division impact the final answer's significant figures, check out this web page from Northern Kentucky University (recommended search terms for future information on this subject would be "understanding significant figures" or "significant figures rules").
*If difficulty was also found in the process of where to begin rounding the answer from, the key is finding the "rounding digit," which, if you have only 1 significant figure to work with, will be the furthest numerical value to the left. See this page for further tips.


text:
How to separate two liquids with both having the same properties?
I was curious if it was possible to separate such liquids e.g. two vegetable oils that happened to have different ingredients, so it is a different liquid, but the same density etc.

Is this one mixture your inquiring about, that has two different vegetable oils (such as soybean and grapeseed) and different solid products mixed all together? If so, then the answer is yes, the oils can be separated regardless of the solid ingredients in them. However, if both oils are actually the same, i.e. both being grapeseed, then there is no way to "separate" them as they are not different liquids, no matter what solids (or other items) are mixed into them. The oils themselves are then equivalent (density, flash points, smoke points, cloud points, etc...).

Strictly having the same properties no, at least not as I know. But they are many great technics for separating mixtures of liquid having very similar properties, i.e., isomers like ortho, meta and para-xylene. But having very different properties is not always a guarantee of separation, if you have an azeotrope, for example, you can't separate the components.

If they have the same properties, they are the same for all practical purposes. Then why bother separating?

@Len_sprague Sorry for being a little vague. They are different oils just as soybean and grapeseed. How could they be separated then?

If they have the same physical and chemical properties then you cannot separate then. If their boiling points are so close they will not distill apart, if they have the same solubilities in all solvents then extraction is not going to work. If they co-elute on different chromatography media then you're pretty stuck with the mixture.

Thank you for your additional information, @Peter. Karl's answer therefore is comprehensive to your question. Hope it helps you!

You could theoretically separate two fatty acids with the same set of physical properties by chemically reacting one to change its physical properties and then reversing the chemical process, for example a polyunsaturated fatty acid and a saturated fatty acid could have the same melting and boiling point, density, solubility etc, but the polyunsaturated fatty acid would be easily oxidized, and the oxidized product would then have different properties and could be removed from the saturated fatty acid.

Also, I believe this to be true.  When you mix two liquids with very similar mepting points, the melting point of the mixture is supressed, so for example, mixing two liquids with MP of 290 might yield a mixed melting point of 280.  The solid mixture resulting from the freezing process has a greater entropy than the two substances would be if frozen separately.  As a result, the mixed solid will TEND to be more stable if like particles are next to like particles.  If the mixture were melted and frozen repeatedly, there would be a progressive trend toward the substances being analyzed.

@Karl where did I say such a thing?

@Karl then we can remove our comments?

Typical vegetable oils are not just one chemical compound, but a mixture (after removing everything that is not) of glyceride triesters. The chance that there is some overlap between your two oils is big, so the answer to your immediate questions likely is no, completely impossible.
Even if there is no overlap in the chemical composition, you would probably still have to separate the oil mixture into all its components, and then know how to mix them together again to get the original two oils.
The latter is possible, of course, with a reasonable analytical effort, the former is close to impossible to perform on a macroscopic sample. A GC(-MS) can easily separate (and identify) the compounds, but that's a few billion molecules or so. Doing liquid column chromatography would be very tricky, with any reasonable apparative effort would give you a few milligrams of each compound, and perhaps there are still a few you can't separate.
Distillation is impossible (boiling points are high and too similar), fractionation by crystallisation would only work well for some of the triglycerides, etc. All in all it's a bad mess. Don't mix two oils and try to separate them afterwards. ;-)

I agree with major point which is that such oils are a mixture. So separating all the individual parts and then trying to put the oil back together is a futile effort.

By saying that you have different materials ("different ingredients"),
an analytical technique which shows a difference could be scaled up to effect a separation. Density was rejected as not giving a difference; chromatography was not recommended (but might work).
The problem with two (different but very similar) vegetable oils is that they are not simply two different materials, but rather two different mixtures of very many of the same or similar compounds. If you could separate the compounds in the mixture (e.g., by chromatography), then you would have to figure out how to apportion the separated individual compounds to form mixtures like the original oils.
In the 1970s, Canadian farmers grew strains of rapeseed with significantly lower amounts of (toxic) erucic acid through crossbreeding. The new plant was called canola, suggesting “Canadian oil, low (erucic) acid.” Canola oil and olive oil are vastly different in source and manufacture, although their composition is similar (from Wikipedia: Saturated fatty acids: 7.4%/13.8%; Oleic acid: 61.8%/71.3%; Linoleic acid: 18.6%/9.8%; smoke point: 238 C/193 C.).
Now if you mixed canola oil and olive oil, you could determine the proportion of each by analysis of all the constituents, but as far as separating the individual compounds and reconstituting the original olive and canola oils… Nah, I don’t think so.
On the other hand, two practically identical materials might be separable if they were only single entities. For example, H2O and D2O are regularly separated even though they are very similar.