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Mechanism of synthesis of allene

I would like to know the mechanism of this reaction. In my first attempt to the question, I had replaced the two $\ce{Br}$s by $\ce{CH3}$s by $\ce{S_N2}$ of the organometallic compound but the answer given was the above allene. I think my reaction isn't possible because of too much hindrance. So if that's the case, then what is the mechanism of the actual reaction?

Lithium-halogen exchange

Yeah, it's definitely redox with rearrangement.

Just to be more explicit, replace one of the bromides with a lithium, i.e., a carbanion. Then, you should be able to draw a rearrangement that gives the allene.

@Waylander How to identify if lithium halogen exchange takes place in a particular reaction?

In this case it is because you have not added any carbons to the molecule. MeLi is not a great nucleophile for alkyl halide in general, principally because lithium-halogen exchange is an alternative pathway.

@Waylander Page 2  here shows carbene formation for this reaction, I am not able to understand what happens after that. Could you help me with that?

@abcd The simplest way to view the process is that one eq of MeLi reacts with the dibromide by Li-Hal exchange to give a carbanion at the carbon still bearing the second Br. This Br then leaves as Br- creating the carbene.

Note that the paper referenced states  " Tetra-substituted   allenes   can  not  be  prepared   by  the  reaction   of  tetraalkyldibromocyclopropane and   methyl   lithium. "

@Waylander Then, how are we proceeding with this reaction if tetra substituted allenes cannot be prepared by tetraalkyldibromocyclopropane + MeLi?

The OP asked about carbene formation. The carbene forms, but inserts in a C-H bond to form a bicyclobutane rather than collapse to the allene.

This is the Skattebøl rearrangement, the first step is of which is a lithium-halogen exchange to form a carbenoid. A mechanistic study of the rearrangement of monosubstituded lithium bromocyclopropylidenoids found that allene formation can occur either in a concerted (Scheme 2) or step-wise (Scheme 3) fashion depending on the ability of the substituent to stabilize the intermediate carbene$^{[1]}$.

The study found that electron donating groups (like the methyl groups in your example) lower the barrier to allene formation. For several EDG-substituted species, the free carbene structure 14 could not be found as an energy minimum and all attempts to locate these structures lead directly to allenes. This means the stepwise mechanism does not play a role in the Skattebøl rearrangement for EDG-sbustituded species$^{[1]}$.

$[1]$ Azizoglu, A.; Balci, M.; Mieusset, J.-L.; Brinker, U. H. The Journal of Organic Chemistry 2008, 73 (21), 8182–8188.


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Calculating excitation energies of a molecule
I am trying to calculate the excitation energies for the first three transitions of a molecule which is made up by a chain of k phenyl rings, where k shall be between 1 and 8. The angle between each rings shall be $\pm 30^\circ$. You can see a picture of the molecule with $k=8$ rings below.

I have calculated the first three transistion energies using Gaussian. You can see them in the table below.

Now I want to calculate them by hand using a very simplified model. Since there is a linear correlation of $E(\frac{1}{k})$ a good model would be the particle in a box. Therefore I tried to use such a model, but unfortunately the difference between my energies is way to high. I used $$E_{n+1}-E_n = (2n+1)\frac{h^2}{8m(kl)^2}$$ as formula where k shall be the number of phenyl rings and l ($\simeq 2.1\mathring{A}$) shall be the diameter of one ring.
You can see my results I obtained by using this formula in the table below.

How could I improve my calculations?

Do you mean the difference between levels of your particle in a box model? Or the difference between your particle in a box model and the Gaussian calculations? It would help to see a similar table of the particle in a box values. One thing I notice is your formula for the particle in a box energy difference seems wrong. I would think where you have $kl$ it should be $(kl)^2$.

I added the table with the results I obtained by hand as well as $(kl)^2$ which I just forgot. Thanks for the correction.

You seem to imply there is only one electron in your box.

The particle in a box is a very crude model. For example the choice for the length is somewhat arbitrary. Then there is no elecetron-electron interaction. you could try to approximate thjs by some additional energy offset, but how would one choose it? You could try to fit these parameters, but it will always be a very simple approximation to a complex system.

also the box would work best for a.completely conjugated pi system, Hückel could be more heloful here. but your 30 degree angle complicates things.

I have no experience in using Hückel method. How would one proceed with that?

wikipedia has actually a rather comprehensive article about it. You basically have to set up a matrix which says which carbon atoms are neighboring each other and diagonalize that. But this still won't consider the 30° angle, although you could try to put numbers $<1$ for these bonds in the HMO matrix.
It might be helpful to know why you need another method than DFT (I assume?). Because the excitation energies should converge with large $k$, which means you could just extrapolate.

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Why do we still need to know about the Rankine temperature scale?
I am learning process principles in chemical engineering and I was taught various temperature units like Fahrenheit, Celsius, Kelvin, and Rankine. I see the first three being used quite commonly, but I rarely see the fourth being used. Where is the Rankine scale used then?

I doubt even in the USA there are many who use Rankine. It's as useful as the Réaumur scale, in that some historical physics and chemistry research may have used them. BTW, C is 1.803e+12 furlongs/fortnight. See https://en.wikipedia.org/wiki/FFF_system

Speaking as a thermal engineer in the US, it is embarrassingly common here. I know of many of the largest and best known companies in the US do all their engineering in US customary units. Certainly not universal (I, for one, refuse to use it), but not at all uncommon.

The people I know in scientific fields in the USA generally use metric measurements, including Kelvin and Celsius for temperature. The engineers, on the other hand,... well...

It's there for completeness' sake and costs all of 10 seconds to learn.  Like learning Roman Numerals, it might come in handy some day.

@DrMoishePippik The trouble is that non-metric measures are common enough in the USA to cause confusion. NASA once crashed a satellite because the engineers accidentally mixed metric and imperial units.

@ericksonla Sometimes, I think that large US engineering companies use it just so they can blame their collaborators for the inevitable mistakes it causes! (I've been on the receiving end …)

@DrMoishePippik And then there's insanity like the Baumé scale(s) for liquid density.

I don't think this question should be closed as TOO broad. While there are multiple adequate answers, it seems they all follow the same thread of institutional inertia thus is not TOO broad.

Rankine is commonly used in the aerospace industry in the United States.
Rankine is to Fahrenheit what Kelvin is for Celsius. So when people in the United States were creating programs and using equations that needed an absolute temperature, they used Rankine before Celsius became dominate for scientific calculations. The reason people still sometimes use it in the aerospace industry is that there are a lot of programs that were developed using Rankine, so in order to be compatible with those old programs, it's often simpler to just use Rankine in the new programs too.
Thus, as an intern at NASA, I was writing simulations that used Rankine in 2009.

Actually, Thomas Jefferson favored the metric system, in 1875 ACE the USA accepted SI weights and measures as an alternative to the legacy British system, and by 1893, the old Imperial measures were defined in terms of the SI. It just seems there are reactionary industries which are a bit slow to adopt changes. See https://www.npr.org/sections/thetwo-way/2017/12/28/574044232/how-pirates-of-the-caribbean-hijacked-americas-metric-system

Rankine-in-space sighting: Does the NK-33 engine require subcooled kerosene so cold that it turns to wax?

Most of us in the world use the Celsius scale to measure temperature for day-to-day purposes. The Kelvin scale has been designed in such a way, it is not only an absolute temperature scale, but also 1°C change is equal to a 1K change. This makes conversion from Celsius to Kelvin pretty easy, involving just the addition or subtration of a certain constant (in this case, it turns out to be 273.16).
However, our friends who live in the United States prefer to use the Fahrenheit scale. The problem is, converting Fahrenheit to Kelvin is not as easy as it is for people using Celsius.
So what do they do? Here steps the Rankine scale. The Rankine scale is sort of like the Kelvin analog for Fahrenheit users. Rankine is an absolute temperature scale, and has the property of having a 1°R change equal to a 1°F change. This means Fahrenheit users who need to work with absolute temperatures will find it easier to use the Rankine scale instead of the Kelvin scale. Fahrenheit can be easily converted to Rankine by:
$$R = F + 459.67$$
This indeed is the case, as engineering systems using the Fahrenheit scale use the Rankine scale for absolute temperatures:
(source: Wikipedia)

The Rankine scale (/ˈræŋkɪn/) is an absolute scale of thermodynamic temperature named after the Glasgow University engineer and physicist William John Macquorn Rankine, who proposed it in 1859. (The Kelvin scale was first proposed in 1848.) It may be used in engineering systems where heat computations are done using degrees Fahrenheit.

"This means Fahrenheit users who need to work with absolute temperatures will prefer to use the Rankine scale instead of the Kelvin scale." In my experience, that's almost never true. At best, that's an opinion. I expect that the significant majority of people who have a need to use an absolute-zero referenced temperature scale, who also use Fahrenheit for day-to-day temperatures, will use Kelvin, not Rankine. That is, of course, just my impression of what happens. We'd need to survey usage in order to determine the relative popularity.

You answer the question "What is Rankine?", which is never asked. The question is, does some branch or area actually use it in practice for day-to-day work. This answer only assumes that's true. OP (and me!) is not so sure. Could you point to some examples of its usage?

@pipe I believe I did mention that Fahrenheit can be converted to Rankine very easily, and hence it is used by Fahrenheit users who need to deal with absolute temperatures.

Yes, it can be converted very easily, but how does that help if you still need Kelvins at the end because that's the SI unit? It sounds like you just assume that scientists use Rankine because of its simple relationship with Fahrenheit. If you could just show some example, or industry experience instead of just guessing, this answer would be good.

@pipe You can use the Rankine scale for thermodynamic calculations, you'd just need a different value for the universal gas constant. As far as me "guessing", please see my edit to the answer.

It says "may be used" - absolutely, it can be used like this, but is it used?

@pipe Well in that case, think about it this way. This scale has been so designed in a such a way as to be an absolute scale, and as well as to be interconvertible from Fahrenheit with just the addition/subtraction of a constant. Why? Why does this temperature scale have to be designed this way? For fun? Moreover, you may notice that ericksonla has mentioned that several top engineering companies in USA use the Rankine scale for thermodynamic calculations.

@pipe It certainly is used by major companies like Boeing, but I can't offer you any proof of that statement without showing you company confidential documents.

The difference between Kelvin and Celsius is 273.15, not 273.16. The triple point of water is at 273.16, however.

I am currently studying mechanical engineering in the US, and I have used Rankine.
It is used similarly to Kelvin. For example, in my thermodynamics class we used it to analyze various heat engines. Tables are available with properties of gases and steam using such units as BTU/R.
I can tell you that it is somewhat of a pain to use, because it is often coupled with units such as BTU or Horsepower, which don't line up with the rest of the customary system.

Mostly I use Rankine as a "way station" when converting between Fahrenheit and Kelvin, which often happens when working with manufacturing facilities in the USA.  Because the Rankine and Kelvin scales have no additive offset between them, the conversion route Fahrenheit <--> Rankine <--> Kelvin requires less mathematics than Fahrenheit <--> Celsius <--> Kelvin.

There are two kinds of situations where you want an absolute temperature scale

When you're working with cryogenic temperatures and don't want to look at negative numbers.  (Nice to have)
When you're doing multiplication or division with temperatures.  (Must have)

You probably won't run into Rankine much for #1, but if you're otherwise working in Fahrenheit and run into situation #2 it's a lot more convenient to convert to Rankine (simple addition) than to convert to Kelvin.  And Fahrenheit is still around, so it's good to know about Rankine.

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Why is the electronegativity of indium greater than that of thallium?
Why is the electronegativity of indium greater than that of thallium?
One possibility which I had assumed was the poor shielding effect by the d and f orbitals which leads to an increase in the effective nuclear charge and hence the anomaly in the 13th group. But with thallium having a f orbital (which indium doesn't) which has a much poorer shielding ability than 'd', why does indium have more electronegativity?

Going down any group of the Periodic Table the general trend is that electronegativity decreases. The relative electronegativities of In and Tl are simply a reflection of this fact. That's the end of the story; you don't need to bring in anything more complicated.
You are correct that the 4f electrons are poorly shielding, and thus the 6p elements (Tl, Pb, Bi, Po, At, Rn) would likely be more electronegative than otherwise expected. But there's no reason why this should make them so electronegative so as to reverse the general trend. None of these elements are more electronegative than their 5p counterparts.

Stuff like these are one of few things in Inorganic Chemistry that have an explanation in High School(unless you are willing to touch research material). Getting to the point this is a result of poor shielding by electrons in the f orbitals. Shielding effect/capacity decreases in the order s>p>d>f.After Indium the electrons are put in s,d,f orbitals and the latter two have little effect in shielding, also considering that the size difference between the two atoms is very small the outer electrons of Thallium are attracted more strongly by the nucleus than in Indium and hence the electronegativity is higher in the former.

Uh, didn't the question indicate that indium is more electronegative ?


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Is delocalisation of non-bonding electrons into the benzene ring possible in case of conjugate base of benzoic acid?
If we take a look at conjugate base of benzoic acid it is as given below:

Now my question is:
Is delocalisation of non bonding electrons(in the carboxylate part) into the benzene ring possible?
Critical Comment:I think that it won't be possible because we would end up having negative charge on both the oxygen which should make it less stable.
Reference supporting my claim:
Although the reason I gave may or may not be the actual,i don't know, but the reference also claims the same thing i.e delocalisation of non bonding electrons doesn't occur.
Link: https://www.asu.edu/courses/chm233/notes/acids/acids2017.html
scroll down and check this:
Factors controlling acidity and under that check 'substituent effects for some aromatic carboxylic acids'

https://chemistry.stackexchange.com/questions/19843/why-cant-the-conjugate-base-of-benzoic-acid-be-stabilized-by-conjugation-with-t

Well definitely not. The simplest answer would be that if you tried to do it, you would get a really absurd Resonance structure. What would actually occur is -
The non bonding electrons would delocalise only between the oxygen atoms and the benzene ring would have its own separate electron delocalisation(as in a normal benzene ring)which would have nothing to do with the former. This would be favored because-
1.The non bonding electron would be with only oxygen(which is more electronegative and hence gives better stabilizing effect to the resonance hybrid).

It gives rise to equivalent resonance structures which greatly enhance stability

Is any form of delocalisation possible where the benzene ring participates in resonance with carboxylate ion? If no such thing is possible then ortho effect would make no sense because if no resonance occurs in the first place between the ring and the carboxylate ion then we cant use the concept of steric inhibition of resonance

I knew this point would come up. Ortho effect is an experimental observation which we try to explain via theoretical Organic Chemistry. Hence there is more than 1 explanation to the problem, of which the conjugate base explanation is a bit weird(as said by my teacher and not justified). There's another one involving resonance in the acid itself which also involves SIR which seems more valid. Check it out if you want to. And my answer is pretty much valid according to the explanation I am referring to.

Isn't Steric inhibition of Resonance(SIR) effect similar to that of ortho effect.It's just that if SIR occurs in benzoic acid then it is Ortho effect right?...And moreover you provided no answer to my question about the existence of any other resonance structure involving benzene ring.It would be great if you explain the question in my previous comment as it would clear a lot of doubts for me.

This is wrong. Of course there is conjugation between the carboxyl group and the aromatic ring, they are after all in plane and belong to the same pi system. Just because you cannot draw a valid Lewis structure for this resonance contributor, it doesn't mean that the contribution does not exist.

@Martin-マーチン if there is conjugation what is it? Could you please provide the resonance structures

Yes I was wrong to not include the conjugated form as a contributing Lewis Structure because for the Reso Hybrid every possible structure has to be considered(irrespective of their feasibility).But with respect to the resonance hybrid its contribution would be very little and the hybrid would greatly resemble what I wrote in my main answer.To sum up what I believe is right according to my knowledge-The conjugated form would be a contributing Resonance structure albeit of limited contribution to the hybrid.Also ortho effect is also found in Aromatic amines(but there it reduces there basicity)

Your knowledge is wrong. Your answer is wrong. You interpret a Lewis structure as a literal resonance contributor, and you assume that just because you cannot fathom a Lewis structure for that kind of conjugation, the conjugation does not exist. Also it is impossible to judge from a Lewis structure alone whether it has little or large contribution to the bonding situation.

Well I guess I have to get back to studying again.Thanks for letting me know.


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Why do highly branched alkanes have higher octane numbers than their corresponding linear isomer?
The octane number of fuels are based on a set of primary reference fuels, which are mixtures of iso-octane and n-heptane. Now iso-octane being branched alkane has very less tendency to react and given ON=100 and n-heptane reacts easily and given ON=0. However n-octane that is the corresponding linear isomer of iso-octane has an ON<0. Now why does branching in a alkanes reduce its reactivity?

The reaction desired in an automobile piston is a rapid burning, not an explosion (very very rapid reaction). If the air + gasoline vapor burns much faster than the camshaft is turning, the explosion causes a ping, or high pressure rise in a smaller volume (because the piston has not moved far enough). This can damage the piston, or at least add an unusual stress.
Burning is a radical reaction and the radicals produced from linear alkanes are quite reactive (i.e., fast reactors). Branched alkanes might lose a hydrogen, even break apart, but the free electron is more stable, and the reaction rate is slower.
There are compounds with a higher octane number than iso-octane. E.g., toluene has an octane number ~120, and was highly sought after for airplane fuel during WW II. The benzylic radical which is the likely first product of a radical reaction is quite stable (but not inert).


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Is ortho-nitro benzoic acid more acidic than formic acid?
We know that the Electron Donating effect of the benzene ring in benzoic acid makes it less acidic than formic acid!
And we also now that whenever an ortho substituent is present in benzoic acid there would be ortho effect which makes benzoic acid more acidic.This is precisely because the resonance doesn't occur anymore because planarity of the molecule is lost. So,now since we don't have any electron donating effect due to resonance is ortho-nitro benzoic acid more acidic than formic acid?
Moreover can we now consider Electron withdrawing effects of benzene ring and the nitro substituent to support of claim?

o-NO2-benzoic aid pKa 2.19 1
Formic acid pKa 3.75 2
Nitro is a very powerful electron-withdrawing group and profoundly affects the properties of aromatics when it is present.
According to this paper 3 the ortho effect does also contribute to the acidity of o-nitro benzoic acid.

so,is o-nitro benzoic acid stronger acid than fromic acid only beacause of Electron withdrawing effect ?Is ortho efffect playing no role in increasing its acidic nature?...

@Banchin Answer edited to address this

Well actually all the nitrobenzoic acids are more acidic than formic acid // 3-nitrobenzoic_acid, pka = 3.46 // 4-nitrobenzoic_acid, pKa = 3.44

@MaxW Could you explain the reasons and is ortho effect one of them?

The pKa's speak for themselves. Compare similar acids. Are apples tastier than oranges?


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Why do electron donating groups decrease the acidity of aromatic carboxylic acids?
If we look at the resonance structures of benzoate anion, we would see that some amount of positive charge is present on the ring and this charge can get delocalised on the ring, like this:

Now, if we had an electron donating group (EDG) at the para position, then the EDG should stabilize the ring because it can donate pi electrons to the empty p orbital of carbocation (look at third resonance structure) and form a double bond, thus resulting in more resonance and more resonance implies more stability and this should imply that EDG increases acidity as conjugate base is getting more stabilized.
But, it is a well established fact that EDG decrease the acidity of the molecule (benzoic acid). For example: $\mathrm{p}K_\mathrm{a}$ (para methoxy benzoic acid) > $\mathrm{p}K_\mathrm{a}$ (benzoic acid)
Why am I facing such a contradiction?

Related: Why can't the conjugate base of benzoic acid be stabilized by conjugation with the aromatic ring?

You seem to pay too much importance to these individual structures. However, note that there are no individual resonance structure in existence, but only the resonance hybrid - a combination of all these resonance structures in proportion to their stability.
The final three structures that you've drawn have a really negligible contribution to the final resonance hybrid of the benzoate anion. Why, you ask? Because they exhibit charge separation: creation of an additional positive and a negative charge.
Hence, your conclusion is wrong, because it's based on the invalid premise that the third (and by extension, the fifth) resonance structures have a major role in the resonance hybrid. The resonance hybrid in fact has no appreciable positive charge on the ring carbons.

The correct way to explain this is that you should look at the resonance hybrid as a whole, which is negatively charged. Adding an electron donating group to the ring would increase the negative charge further, thereby destabilizing the anion.

If you look closely at your diagram, you will notice that once benzoic acid has released its proton, the conjugate base that is formed has 6 resonance structures out of which 4 have 2 more charges that offset each other. The overall charge of benzoate anion is -1 regardless.
The important part of this discussion revolves around the last 4 resonance structures. For the sake of the discussion, I will only refer to the 2 charges that offset each other. When a negative charge appears on the oxygen atom in the carboxyl group, a positive charge appears on the ring (the charge is spread across the nucleus).
At this point, we see that the conjugate base is quite stable because it has 6 resonance structures. The more resonance structures the conjugate base has, the stronger the acid is. This is how the acidity of benzoic acid is explained.
Now, if you placed a methoxy group in para position, it would doante electrons to the ring through a mesomeric effect. If that happens, the electron density around the nucleus increases, diminishing the positive charge. The methoxy group increases the electron density and so the positive charge on the nucleus is no longer “that positive” while the negative charge remains unaffected (“just as negative”). If the positive charge can’t properly balance the negative charge, the resonance structures involving charge separation are not advantageous to exist and so the strength of the acid decreases.
This happens only if the methoxy group is situated para relative to the carboxyl group. If it is situated meta relative to the carboxyl, something interesting happens.
We know that the methoxy (like the hydroxy group) exerts a -I inductive effect and a mesomeric (M) effect. That is why, if found para, it “donates” electron density through a +M effect (which manifests more than -I). When in meta, the +M effect is no longer a dominant effect and so, methoxy manifests -I effect mostly, withdrawing electron density from the nucleus. It is often said that hydroxy and methoxy groups are EDG, but only when it comes to electromeric effects. They also have -I inductive effects which makes them EWG.
Overall, the presence of a EDG in ortho or para destabilises the actual negative charge of the benzoate anion, thus making it less stable.

sorry, I can;t understand this part of your answer..."The methoxy group increases the electron density and so the positive charge on the nucleus is no ............ so the strength of the acid decreases".Before methoxy group was attached we had a ring with positive charge and once we attach methoxy group we have less positive charge on the ring.it should stabilize it right? the amount of negative charge,at the carboxyl(oxygen) remains same after attachment also .moreover what do you mean by positive charge not being able to balance the negative charge properly(line 14)?

@Gaurang Tandon, how do you know the last 4 structure have a insignificant contribution ? Though, your answer is beautifully couched.

@AndrewB The last three structures have an insignificant contribution because of charge separation. Charge separated resonance structures have a lower contribution to the resonance hybrid.

Benzoic acid ($\mathrm{p}K_\mathrm{a}=4.2$, PubChem) is actually more acidic than p-methoxybenzoic acid ($\mathrm{p}K_\mathrm{a}=4.47$, PubChem).
An electron donating group essentially destabilizes the conjugate base of the acid, because of which this phenomenon occurs. Your explanation speaks of the carbocation formed in some resonance structures which has a double negative charge and a single positive charge.
The electron donating group would increase the stability of that cation but there are two negative charges that aren't stabilised but destabilised instead.
Because of this, the overall stability of the conjugate base is lower than in case of no electron donating group.

I do agree that paramethoxy benzoic acid is less acidic than benzoic acid intuitively,but from my explanation(as mentioned in the question) I find some contradiction and my question asks you to explain it...


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Is steric inhibition of resonance, or of protonation, dominant in 0-substituted anilines?
There's a question about inhibition of resonance:
Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine?
If you look at Martin's answer he says that CH3 is not bulky enough to cause steric inhibition of resonance. What if I considered a group such as ethyl or propyl or even bulkier, then would we observe SIP (steric inhibition of protonation) as well as steric inhibition of resonance?
What would dominate?

You can make a follow-up question if it's significantly different.


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Is rearrangement possible in cyclic bromonium ion?

I wonder what the product of the above reaction is, that is, bromination of the given alkene in non polar solvent carbon tetrachloride (proceeds through cyclic bromonium ion intermediate).
Once the cyclic bromonium ion is formed, can the ring expand as shown or is the usual product formed? Five remembered rings happily expand to six membered ones if possible - so what's going on here?
It'd be great if you could quote sources with your answer - and help me understand whether or not species involving cyclic bromonium ions can rearrange in a reaction, and why?
Thanks!

I did a quick search, and it appears to be possible, at least for norbonyl type structures, but I haven't seen other examples yet. For example: https://pubs.acs.org/doi/abs/10.1021/jo401888f

This is apparently not a likely rearrangement, even in more strained circumstances. There are a number of reactions out there in the literature like the one below.

This result makes sense if we consider the relative stability of the bromonium ion vs. your proposed rearrangement. I've redrawn the mechanism to show all the formal charges. The bromonium should be much more stable (positive charge delocalized over two carbon atoms, all atoms have full octet) compared to a secondary carbocation.

Bromonium rearrangements are not entirely impossible though, check out this fun example (J. Org. Chem., 1998, 63, 2646 DOI: 10.1021/jo9722055)

As with all chemical reactions, the product(s) is/are determined by either kinetic or thermodynamic control.
Under thermodynamic control (where the transition-state activation energies are relatively low and all reactions are reversible) the products equilibrate to the lowest energy product. Under kinetic control (where the rate determining step is effectively irreversible), the difference in activation energies determines the product ratio. You might properly guess this is a kinetically-controlled reaction, since 1,2-dibromoalkanes do not spontaneously form bromonium ions and Br- in CCl4.
If you draw out the elementary steps of the reaction mechanism, following the rules for arrow-pushing (described, for example, here) you should be able to identify the step which determines the product:

The rearrangement product requires going through a non-stabilized secondary carbocation transition state (4) versus direct opening of the bromonium ion as in (1). For a deeper understanding, consider what changes to the structure or reaction conditions might favor the rearranged product.
Ref: Advanced Organic Chemistry: Part A: Structure and Mechanisms. Francis A. Carey, Richard J. Sundberg, pp. 362-371

What's your source, or is this just a guess?

First of all, the product you drew containing only one bromine atom is not correct since you added bromine to it, so it should have 2 bromine atoms ( there are some stages in this addition, but to skip to the most important ones, the bromine molecule suffers a heterolytic cleavage to form bromonium and bromide, the bromonium adds to the double bond according to Markovnikov’s rule. Now, the bromonium ion you drew where bromine is bonded to 2 carbon atoms is the result of the carbocation “attacking” one electron pair of bromine in an attempt to stabilise itself. Basically, once bromonium reacted with the double bond, you are left with a positive charge which will be offset by the bromide ion ). Under no circumstance can you have an organic compound with one bromine atom when you add bromine to an alkene. Probably you forgot to put a second atom of bromine there. Anyway, in every reaction where you encounter carbocations, there will be rearrangements occuring, but there are some exceptions such as this one. Normally, if you added hidrobromic acid, you would have rearrangements because there will not be any cyclic ions forming. In this case, like I told you, a carbocation will be formed when bromonium ion reacts with the double bond. That carbocation, however, will not rearrange itself because it has found another method to stabilise itself: it forms a bond with the bromine atom, generating a cyclic ion where the positive charge is more spread and thus more stable. And then the bromide ion attacks from the opposite side.

I haven't forgotten to put the second bromine atom, it's simply not there. If the reaction proceeds via ring expansion as proposed by me, there's no step in which the bromide ion would possibly attack and thus remains in solution.

@schrodinger_16, AndrewB is correct. There should be a carbocation remaining after the rearrangement, which could be attacked by bromide. That said, I think you have still asked an interesting question. I hope to have time to provide an answer.

If the compound with a cyclohexane ring has only one bromine atom, according to the rearrangement you proposed, there should be a positive charge residing on the tertiary carbon in the ring, so the compound is not electrically neutral ( if you said that the bromide stays in solution ).

Here is a reference.
http://research.cm.utexas.edu/nbauld/teach/alkenes3.html#bromination
Read the part “electrophilic addition of bromine and other halogens”. Technically, to refine my argument, the positive charge is mainly on the bromine.

Why don't you merge the answers? You can edit to do so.


text:
When does a pinacol undergo dehydration or rearrangement?

What happens when the above pinacol is treated with -
(a) conc. sulphuric acid
(b) dil. sulphuric acid
I'm confused between pinacol rearrangement and dehydration (E1 or E2?), as both possibilities exist. I think the concentration of the reagent might have a role to play, but I'm not sure. Please help.
P.S. It'd be great if you could quote data and sources with your answer.

Epoxide formation under dilute conditions?

I wonder if treatment with conc. sulphuric acid would form some octamethyl-1,4-dioxan (in analogy to glycol).

When you treat pinacol with sulfuric acid, an E1 elimination will take place ( having a carbocation as intermediate ) and it forms pinacolone ( 3,3-dimethyl-2-butanone ). I am not aware of how concentrated the acid should be. Do you require a photo depicting the mechanism ?

Is it possible for both the OH groups to eliminate?

No. If that happened, you would technically end up with 2 positive charges next to each other and that is not favourable. When a proton “attacks” one of the hydroxyl group, it leaves as water and a positive charge is left on the carbon and that rearranges. In all the books, this is the mechanism that appears.

That wouldn't happen, elimination could happen one after the other.

How could they happen one after the other when, after the first one, a ketone is formed ? How can a ketone eliminate water like that ?


text:
Rate of catalytic hydrogenation of alkenes
What does the rate of catalytic hydrogenation of alkenes depend upon? What'd be the increasing order of rate towards catalytic hydrogenation of the following alkenes, A (2-methylpropene), B (cis-but-2-ene), and C (trans-but-2-ene)?

I suppose this has something to do with sterics and orientation - as catalytic hydrogenation is a surface phenomenon (occurs on surface of metal catalyst with adsorbed hydrogen), but I'm not sure about the above order in particular (I don't think a clear comparison can be made without looking at data, as the alkenes are quite similar), and the factors on which rate of hydrogenation of hydrocarbons depends in general.

Related: Are geminal disubstituted alkenes more stable than their cis/trans isomers?

Stability (thermodynamic) does not dictate kinetics of a reaction.

Catalytic hydrogenation is more or less a surface phenomenon so sterics play the deciding role in the rate of hydrogenation.
From Wikipedia:

Here's a newman projection of the three Alkenes.

 A = cis-dimethylbutene, B = trans-dimethylbutene, C = 2-methylpropene
From here it is easy to see that the cis-dimethylbutene will have the least hindered approach towards the surface of the catalyst as both the methyl groups can be made to face away from the surface.
The rates of hydrogenation in the case of trans-dimethylbutene and 2-methylpropene will be decided by which one of them is more strained.

2-methylpropene is more strained than trans-dimethylbutene.
So, we can now establish the order.
$$\text{A > C > B}$$

References

Catalytic Hydrogenation:A Core Technology In Synthesis; Ryoji Noyori
Wikipedia article on Hydrogenation

This rationale doesn't seem entirely correct.. for proper syn delivery of both the hydrogens,one entire face of the alkene has to be easily adsorbed on the catalyst surface(like in B) .. even if it is easier for one side of the alkene to diffuse towards the catalyst surface(like in A)... The other side will have to come to the surface as well to receive the other H. So it seems that B will be more susceptible to complete hydrogenation than A or C

Yep,I was right. This answer is not completely correct. See this

The predominantly accepted Horiuti-Polanyi mechanism states that the rate depends on the three steps: (1) alkene adsorption on the surface (2) hydrogen migration to the β-carbon of the alkene with formation of a M σ-bond and (3) reductive elimination. There is a good discussion in J. Chem. Educ., 2013, 90 (5), pp 613–619.
You can infer the rate differences in reactivity (step 2 and 3) from first principles and experimental data on homogeneous systems. In hydrogenation experiments reported in Faraday Discuss. Chem. Soc. 1968, (46), p. 60 with homogeneous Wilkinson’s catalyst, the relative rates of hydrogenation normalized against 1-octene are 0.17 for trans-4-octene; 0.54 for cis-4-octene; and 0.69 for 2-methyl-1-butene. For heterogeneous hydrogenation, The transport phenomena associated with step 1 limit predictive ability for overall rates.


text:
Reaction of cyclopropane with hot potassium permanganate
Hot potassium permanganate is usually used as an oxidising agent, and cyclopropane being an unstable three membered ring undergoes reaction with several reagents such as bromine, $\ce{HBr}$, metal adsorbed with hydrogen, etc.
However, I don't know for a fact if cyclopropane undergoes oxidation with hot permanganate – and I couldn't really find much data online.

If at all cyclopropane doesn't react, what happens with its methyl derivative? Does it get oxidised to the corresponding acid?


text:
Is there a piece of software that can help me see the (hkl) planes in a crystal lattice?
I am doing research with calcium carbonate structures. I need a way in which I can familiarize myself with the lattice structure of calcium carbonate polymorphs, especially calcite.
I really struggle to see things in three dimensions. I borrowed an atomic model kit from my professor and that worked out OK, but I still wouldn't be able to tell which is the $(101)$ plane, for example.
I tried downloading software such as BallVIEW, but I found it quite confusing. I need something in which, I can load the structure, and then it can tell me where each $(hkl)$ plane lies. I don't know how to program so I would really appreciate it if anyone can direct me to something user-friendly.

Perhaps CrystalMaker.

Mercury (available on Windows, Linux, MacOS) has a free version and is arguably the best crystallographic viewer for beginners. It also allows to add as many $(hkl)$ planes as you want. I used the structure of calcite from COD (1010962.cif).
Open your CIF file; by default Mercury loads an asymmetric unit. Once the CIF is loaded, go to Calculate → Packing/Slicing... and tick Pack option. Alternatively, tweak the volume populated by the atoms if needed. (This also automatically populates the list of planes with the planes denoting BFDH crystal morphology.)

Navigate to Calculate → Planes... → New Plane... → hkl and add the planes by Miller indices:

and, if needed, BFDH planes one by one:

If you have CSD-Materials (paid option), then you'll be able to display BFDH theoretical crystal morphologies at once by going to CSD-Materials → Calculations → BFDH Morphology...; on export you'll get something like this:

Alternatively, there is KrystalShaper (Windows only), a free lightweight program made specifically for displaying crystal morphologies:

P. S. Thanks to Buttonwood for useful comments regarding graphical details and reminding about BFDH crystal morphology.

Sorry, incomplete. At least amend the display of the molecular structure and the calculated planes either by displaying the parallelepiped of the unit cell (Display -> Toolbars -> Display Options -> Show cell axes).  Alternatively, if the OP is interested to relate the orientation of the molecules to the habit of crystal, to compute e.g. the BFDH morphology of the crystal and to visualize this on top (e.g., https://bip.weizmann.ac.il/course/structbioinfo/databases/CCDC_Mercury/images/mercury.3.104.1.png) by the materials module of Mercury (enterprise version, typically with the csd data base).

@Buttonwood Ehhm, the unit cell is present and is packed (probably it's hard to see, I'll think about better representation, maybe grow the whole thing a bit more); the OP didn't ask anything about morphology, so I showed the basic principle of illustrating an arbitrary plane using Miller indices.

Hard to see, indeed, the three-colour red-green-blue of a, b, and c axis, but there is a white cage of an epiped.  A direct figure export (save as *.png) from the program about the second representation instead of a screen photo including the already presented GUI may be useful.  Perhaps a stick-only model instead of ball-and-stick is advantageous, too.  The morphology relation, introduced by "if" assumed the OP might become interested about this at a later point of research.

There appear to be a variety of programs these days. Most I haven't tried, so take with a grain of salt (the last program I used was for DOS and I bought it in the early 1990s. I guess I'm old now. Sigh).
VESTA is free for academic use.
CrystalMaker requires a paid license.
List of other programs can be found here.

Substitute "old" by "senior", please.  The senior Shelx program, for example, just got an functional update report (http://journals.iucr.org/a/issues/2019/01/00/ib5060/index.html), and among the authors is the initial program author, George Sheldrick (born 1942).


text:
What is the correct reaction of sulfur dioxide with chlorine?
I'm confused in reaction of $\ce{SO2}$ with $\ce{Cl2}$. Some source states the given reaction (1), while others state the reaction (2):
$$\begin{align}
\ce{SO2 + Cl2 &-> SOCl2 + Cl2O \tag{1}}\\
\ce{SO2 + Cl2 &-> SO2Cl2 \tag{2}}
\end{align}$$
While I read in NCERT Chemistry 12 part-II page-198 that chlorine oxidises $\ce{SO2}$ to $\ce{SO3}$. I wish to know the specific condition in which the particular reactions will occur.

http://www.sciencemadness.org/talk/viewthread.php?tid=23152

Which sources quote reaction 1 ?

Sulfuryl chloride ($\ce{SO2Cl2}$) is prepared by the reaction of $\ce{SO2}$ and $\ce{Cl2}$ over activated charcoal, details of the preparation are here 1.
Thionyl chloride may be prepared from $\ce{SO2}$ and $\ce{Cl2}$ but the reaction requires an equivalent of $\ce{SCl2}$ to be present.

Can you also add the equations of the two reactions?


text:
Thermal Conductivity of Calcium Carbonate?
Is Calcium Carbonate (Group 1 and 2) able to conduct heat more than other carbonate (Transition metal) allowing for them to be decomposed faster. The reason I am asking this question is because my experimental result from burning carbonates through a test tube and finding the exact time it starts to bubble (start of decomposition) shows that these metal carbonates decompose way faster than other metals. (Results are below)
(Substance Trial 1 (s),Trial 2 (s),Trial 3 (s),Trial 4 (s))
Zinc Carbonate, 30.70 s, 26.35 s, 25.24 s, 25.17 s
Nickel Carbonate, 30.65 s, 20.25 s, 20.27 s, 20.31 s
Lead Carbonate, 16.54 s, 17.66 s, 16.58 s, 16.73 s
Magnesium Carbonate, 10.06 s, 9.84 s, 10.19 s, 9.92 s
Calcium Carbonate, 7.66 s, 7.43 s, 7.57 s, 7.98 s

You have not measured thermal conductivity, you measured a rate of reaction under a rather poorly defined set of conditions.
If I was going to deal with this problem then I would first want to do some thermal gravimetric anaylsis on the carbonates to determine how hot I have to heat them to decompose them. I would also want to use an experiment similar to Lee's disk to get the thermal conductivtiy. Years ago I was shown this experiment being used on thin disks of uranium dioxide nuclear fuel.
https://www.researchgate.net/publication/234935832_Thermal_conductivity_of_uranium_dioxide_up_to_2900_K_from_simultaneous_measurement_of_the_heat_capacity_and_thermal_diffusivity
http://www.phox.at/upload/Laser%20Flash%20Method%20to%20Determine%20Thermal%20Conductivity.pdf
https://www.pragolab.cz/documents/Linseis_Laser_Flash_Thermal_Conductivity.pdf

Thank you for the response and I apologize for just replying to this. Sadly I do not have access to these apparatus as i am confined to using standard school laboratory apparatus . Can you point out exactly why these are poorly defined sets of conditions for me? I am still a high school student as such my knowledge is still very limited

Also my research/school project is trying to assess which of these carbonates decompose faster to show that they follow the trend of the reactivity series whereby less reactive metal decompose faster due to the weaker polarization, not really thermal conductivity. The reason why I am asking thermal conductivity is because I am confused as to why Group 1 and 2 metal decompose faster than the least reactive metals (Pb)


text:
Reactivity of carbonyl adjacent to cyclopropyl group

We know that cyclopropyl, because of $\ce{sp^{\sim 5}}$,  hybrid orbitals behaves much like a double bond. It even takes part in conjugation to stabilise positive charge on carbons adjacent to it. I would like to know if it conjugates with the carbonyl carbon too?
Are there cases of 1,4 addition and 1,2 addition with such compounds analogous to that observed in alpha-beta unsaturated carbonyl compounds?


text:
Where should we place alpha,beta unsaturated carbonyls in the reactivity order list?
In this question, reasons for this observed trend of reactivity of different types of carbonyl compounds are discussed.

I would like to know where we can place $\alpha,\beta$ unsaturated carbonyls in this list.
Initially I thought we can place them above acyl chlorides because now we have 2 electrophilic carbons but then I think the reactivity of both of the carbons should also be reduced because of conjugation so they should be placed somewhere in the middle, but where exactly?

What carbonyls? Ketones?

Both aldehydes and ketones @Mithoron

Well, they are already there, as ketones and aldehydes.

@Mithoron  Presence of adjacent unsaturation is going to affect reactivity, isn't it?

Everything does, but you seem to overestimate it.


text:
Manufacture of p-nitroacetophenone

Is it feasible to manufacture p-nitroacetophenone by reacting di(p-nitrophenyl)cuprate and acetyl chloride?

I think it should be possible but is there any chance that the NO2 being a highly deactivating group does not allow the substitution?

I'd rather worry about using acyl instead or alkyl.

@Mithoron Why ? My book says its not possible as one would have to make organocopper compound from onganolithium reagent here but the no2 group does not allow it. I could not understand it. Can you explain?

Well, that may be a problem, but in a different way then you mentioned in question.

@Mithoron But isn't it already an organocopper compound? I don't see why it should organolithium!

first https://chemistry.stackexchange.com/questions/32079/why-do-halogen-metal-exchanges-happen second metal exchange with Cu and finally reaction with AcCl but this doesn't make much sense in practice even if would work. One should just nitrate acetophenone, or maybe even buy nitrated.

Even if such a cuprate existed, the rxn. would give 50% yield at most of ketone along with nitrobenzene. Cuprates for manufacturing???

This is a crazy idea. Aromatic nitro groups are really not a good mix with organometallics. Acetyl chloride is a poor choice for the electrophile, acetaldehyde would give higher yields and the oxidation to acetophenone is easy.

If is it a bad idea to perform such a substitution, I think we should answer the question with reasons for being so, instead of closing it as "unlear".

In general, reactions involving organocuprate reagents ( Gilman reagents ) are used in nucleophilic aliphatic substitution ( example: chloromethane reacting with organocuprate containing methyl groups to obtain ethane ). In this case, you are dealing with a reaction that, if it worked, would have an addition-elimination mechanism. However, we can approach the synthesis in a different way. We can start from p-chloronitrobenzene. You know that this aromatic substrate is not electron rich ( because it has the nitro group which is extremely electron withdrawing ). In comparison to benzene, p-chloronotrobenzene is more “deactivated” when it comes to electrophilic aromatic substitution. At this point, I believe you know where this is going. Rather than making the aromatic ring a nucleophile like you proposed, we make use of its low reactivity in electrophilic substitution and its high reactivity in NUCLEOPHILIC AROMATIC SUBSTITUTION. Thus, we can use a nucleophile to displace the chlorine atom with another group ( a substitution which proceeds through a Meisenheimer ion ). What kind of nucleophile should we use to obtain an acetyl group in para position ? It would be idyllic if we could just substitute the chlorine atom with an acetyl anion, but that is impossible. We could use use instead reagents that “mask” this theoretical ion ( sinthon ). For more information regarding sinthons, read about retrosynthesis. So, in one method, we will react p-chloronitrobenzene with acetylide ion, yielding p-nitrophenylacetylene. Hydrating this alkyne in the presence of mercury ( II ) salt and sulfuric acid will result in the obtention of your desired compound. In another method, we take p-chlorobenzene and react it with the anion of nitroethane and then, using Nef reaction, we convert the nitro group to a ketone.
  In conclusion, your method is not really feasible due to the nature of organocuprate reagents. I have proposed alternative methods that I think they would work better ( note that they are lab methods ). It is recommended to find lab methods that do not imply oxidation reactions in order to ensure high yields. Let me know if you do not agree.

A lot of labs will not permit the use of mercury these days.

Well could you break your block of text into paragraphs? It is very hard to read.


text:
Ionproduct of Water
The ionproduct of pure water is well known to be
$$
K_{\rm W} = \left[ {\rm OH^-} \right] \left[ {\rm H_3O^+} \right] = 10^{-14}
$$
at some standard conditions (I think 25°C).
However having some acid HA in water as far as I'm aware this product is always the same i.e. $pK_{\rm a} + pK_{\rm b} = 14$.
How can this be derived?

I do not think this question is in any way answered there!
Equilibrium constants are defined for pure reactions, for instance
$$
{\rm H_2O + H_2O} \quad \substack{k_1 \ \longrightarrow \ \longleftarrow \ k_2} \quad {\rm OH^- + H_3O^+}
$$
is such a pure reaction and then $K_{\rm W} = \frac{k_1 \left[{\rm H_2O}\right]^2}{k_2}$. However adding additional reactions of the form
$$
{\rm HA + H_2O} \quad \substack{\longrightarrow \ \longleftarrow} \quad {\rm H_3O^+ + A^-}
$$
in principle changes the chemistry and thus also the equilibrium constant!

@Diger - You're sort of right but not in the way that your postulating. In reality it is the activity of H+ and OH- that are in equilibrium. In "dilute" solutions the activity is essentially the same as the concentration. However in more concentrated solutions the activity is less than the concentration.

@Diger - An "aqueous solution" should be about 1 kg per liter and have about 55 moles of water per liter for equilibriums like Ka and Ksp to work nicely.  (In other words, "dilute solutions.") As a poor example think of putting a drop of sulfuric acid in 100 ml water. That's fine. Now if you put a liter of sulfuric acid in 100 ml of water it isn't really an "aqueous solution" any more.

So in a general set of chemical reactions, does the equilibrium constant $K$ for one specific reaction change when I am adding further reactions to my system? In fact that is what I would expect.

For example consider the differential equation for $\ce{OH-}$ only of a dilute solution for the first reaction above in pure water
$$
\frac{{\rm d}}{{\rm d}t} \ce{[OH-]} = k_1 \ce{[H2O]^2} - k_2 \ce{[H3O+][OH-]}
$$
in equilibrium the derivative is zero and then you precisely get $K_{\rm W} = \frac{k_1 \left[{\rm H_2O}\right]^2}{k_2}$.

However if I now add the second equation to the system this equation changes to
$$
\frac{{\rm d}}{{\rm d}t} \left[ {\rm OH^-} \right] = k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm A^-} \right] - \left{ k_{\rm HA + OH^-} \left[ {\rm HA} \right] + k_{2} \left[ {\rm H_3O^+} \right] \right} \left[{\rm OH^-}\right] , .
$$
And then
$$
\ce{[OH-][H3O+]}=\frac{k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm A^-} \right] - k_{\rm HA + OH^-} \left[ {\rm HA} \right]\left[ {\rm OH^-} \right]}{k_2} ,.
$$

Of course in general/principle I would have to solve the entire system i.e. with the additional reactions
$$
\frac{{\rm d}}{{\rm d}t} \left[ {\rm H_3O^+} \right] = k_{1} \left[ {\rm H_2O} \right]^2 + k_{\rm HA + H_2O} \left[ {\rm H_2O} \right] \left[ {\rm HA} \right] - \left{ k_{\rm H_3O^+ + A^-} \left[ {\rm A^-} \right] + k_{2} \left[ {\rm OH^-} \right] \right} \left[{\rm H_3O^+}\right]
$$

and
$$
\frac{{\rm d}}{{\rm d}t} \left[ {\rm A^-} \right] = -\frac{{\rm d}}{{\rm d}t} \left[ {\rm HA} \right] = \left{ k_{\rm HA + H_2O} \left[ {\rm H_2O} \right] + k_{\rm HA + OH^-} \left[ {\rm OH^-} \right] \right} \left[ {\rm HA} \right] - \left{ k_{\rm A^- + H_2O} \left[ {\rm H_2O} \right] + k_{\rm H_3O^+ + A^-} \left[ {\rm H_3O^+} \right] \right} \left[{\rm A^-}\right]
$$
and then create the product $\ce{[OH-][H3O+]}$ with the solutions of steady state, but I do not expect this product to be precisely $K_{\rm W}$.

For some acid $\ce{HA}$, you have
$$\ce{HA + H2O <=> H3O+ + A-}$$
$$\ce{A- + H2O <=> OH- + HA}$$
We then define:
$$K_{\mathrm{a}} = \frac{\ce{[H3O+]\ce{[A-]}}}{\ce{[HA]}}$$
$$K_{\mathrm{b}} = \frac{\ce{[OH-]\ce{[HA]}}}{\ce{[A-]}}$$
Immediately, you see that $K_{\mathrm{a}}K_{\mathrm{b}}=K_{\mathrm{w}}$.
Now, if you take the negative log of both sides:
$$-\log\left(K_{\mathrm{a}}K_{\mathrm{b}}\right)=-\log K_{\mathrm{w}}$$
$$=\mathrm{p}K_{\mathrm{a}}+ \mathrm{p}K_{\mathrm{b}} = 14$$
The 14 comes from the value of $K_{\mathrm{w}}$.

I guess the problem I have with this is the way how to view it.
Say you start with very little $\ce{HA}$ so that $\ce{[A-]}$ is very small and the equilibrium hardly deviates from $\ce{[OH-]}=10^{-7}$ and $\ce{[H3O+]}=10^{-7}$. In this case you are right about the product being $K_{\rm W}$. Now you continuously increase the acid $\ce{HA}$ while both $\ce{[OH-]}$ and $\ce{[H3O+]}$ individually change their values. Is it because of $K_a K_b$ being a constant only dependent on temperature and also due to continuity, that this cannot change while the acid concentration is changed over magnitudes?

Another point worth mentioning which I'm not sure about is that: Aren't $K_a$ and $K_b$ by itsself so by each reaction above alone defined as I noted in the other comment? Do these values not changed upon mutual influence in the reaction?

Or put it another way: Does the value of $K$ not changed in equilibrium, regardless of how many reactions I might add to the chemistry which certainly will affect it.


text:
How to systematically find the reducible representation of a molecule? (eg: XeF4)
I am attempting to use the reduction formula to find the irreducible representation of $\ce{XeF4}$ to determine its IR stretching vibrations. I know the point group of $\ce{XeF4}$ is $D_{4h}$ and have the character table but I am having a hard time understanding how to get the characters for the reducible representation. I was taught to do it by considering moved/unmoved vectors emanating from the point of the molecule giving values of 1, 0 and -1 for each vector after the symmetry operations have transformed them.
Using this method I to find 2$E_u$ irreps, but I found in a paper (http://sces.phys.utk.edu/~moreo/mm08/penchoff.pdf) that $\ce{XeF4}$ also has other irreps including an Au irrep involved in IR activity. How do I find the irreps that I missed? I noticed the paper uses different reducible representations ($\Gamma_{xyz}$, $\Gamma_{unmoved}$ and $\Gamma_{Tot}$) and I don't understand why.

A way to do it is to put 3 axes ($x$, $y$ and $z$) on each atom (to represent the 3N degrees of freedom) and determine the characters for each axis.
After that, if you need the representations for the 3N-6 (3N-5 for a linear molecule) vibrational modes, you need to remove the representations for the 3 translational modes (what they call $\Gamma_{xyz}$), as well as the representations for the 3 (2 for a linear molecule) rotational modes. Those are usually stated in the character table for the space group, but you can also determine them by checking the characters for a vector.
Example with $\ce{H2O}$
Character table for point group $C_2v$
C2v | E  | C2 (z) | v(xz) | v(yz) |  lin  |    quad
-------------------------------------------------------
A1  | +1 |     +1 |    +1 |    +1 |   z   | x2, y2, z2
A2  | +1 |     +1 |    -1 |    -1 |   Rz  |     xy
B1  | +1 |     -1 |    +1 |    -1 | x, Ry |     xz
B2  | +1 |     -1 |    -1 |    +1 | y, Rx |     yz

Character table for all 3N = 9 axes
C2v | E  | C2 (z) | v(xz) | v(yz)
----------------------------------
Ox  | +1 |     -1 |    +1 |    -1
Oy  | +1 |     -1 |    -1 |    +1
Oz  | +1 |     +1 |    +1 |    +1
H1x | +1 |      0 |     0 |    -1
H1y | +1 |      0 |     0 |    +1
H1z | +1 |      0 |     0 |    +1
H2x | +1 |      0 |     0 |    -1
H2y | +1 |      0 |     0 |    +1
H2z | +1 |      0 |     0 |    +1
----------------------------------
Γtot| +9 |     -1 |    +1 |    +3

Reduction of $\Gamma_{tot}$ to irreducible representations
Using the reduction formula $n_i = \frac{1}{h}\sum_R \chi_{\Gamma_{tot}}(R) \chi_i(R)$ for each irreducible representation $i$ of the point group, we get :

A1 : $\frac{1}{4} \times (9 + (-1) + 1 + 3) = 3$
A2 : $\frac{1}{4} \times (9 + (-1) + (-1) + (-3)) = 1$
B1 : $\frac{1}{4} \times (9 + 1 + 1 + (-3)) = 2$
B2 : $\frac{1}{4} \times (9 + 1 + (-1) + 3) = 3$

Therefore, $\Gamma_{tot}$ = 3 A1 + A2 + 2 B1 + 3 B2.
We can check that there are the 3N = 9 representations that are associated with the 3N degrees of freedom of the molecule.
For degenerate representations ($\chi(E) > 1$), the representations correspond to $\chi(E)$ modes.
Isolation of the vibrational representation
Determine $\Gamma_{trans}$ and $\Gamma_{rot}$ (from the character table or by hand).
$\Gamma_{trans}$ corresponds to the linear terms in the table, $\Gamma_{rot}$ corresponds to the rotational linear terms in the table.
For $C_2v$ :

$\Gamma_{trans}$ = A1 + B1 + B2
$\Gamma_{rot}$ = A2 + B1 + B2

You can check that there are 3 modes for $\Gamma_{trans}$ and 3 (2 for a linear molecule) modes for $\Gamma_{rot}$
For degenerate representations, you will see things like $x + y$ in the character table. This means that this representation corresponds to both x and y and should be included only once
Finally, $\Gamma_{vib}$ = $\Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot}$ = 2 A1 + B2.
Activity of the vibrational modes in IR and Raman
IR activity is determined by the dipole moment and a mode will be active if the irreducible representation contains a linear part (x, y or z).
Raman activity is determined by the polarisability and a mode will be active if the irreducible representation contains a quadratic part ($x^2$, $xy$, ...).
For $C_2v$, all the representations correspond to active modes in Raman and all except A2 correspond to active modes in IR.
Therefore, the 3 vibrational modes of $\ce{H2O}$ (2 A1 and B2) are active in both Raman and IR.
Two modes have the same irreducible representation: A1. That does not mean that they have the same wavenumber. For $\ce{H2O}$, there are 3
vibrational modes with distinct wavenumbers.
For degenerate representations, there will be $\chi(E)$ modes with the same wavenumber.
In groups that have an inversion center (like $D_4h$), linear and quadratic terms are mutually exclusive. Therefore, an irreducible representation will always correspond to a mode active only in IR or Raman, not both.


text:
Is it possible for me to derive Avogadro's number?
To my understanding the mole is the unit used to translate between mass on the atomic level and mass in the macro level, defined as the number of atoms in $12$ grams of Carbon-$12$, which apparently turns out to be $\pu{6.0221409e23}$ atoms, and this was defined as Avogadro's Number.
How was this number derived? Is it something I can derive myself? If I had $12$ grams of Carbon-$12$, how would I be able to deduce that it had $\pu{6.0221409e23}$ atoms?

It would be hard, very hard. I'll point out that there is a gigantic different in stating  that $N_A \approx 6.022 \times 10^{23}$ and stating that $N_A = 6.0221409\times 10^{23}$. Determining a value to eight significant figures requires extremely good experimental technique.

@MaxW How did Avogadro do it in his time, to whatever degree of precision that he did?

This information is readily available on the web in sources such as Wikipedia. https://en.wikipedia.org/wiki/Avogadro_constant, "By dividing the charge on a mole of electrons by the charge on a single electron..." Look up Millikan oild rop experiment for charge of e-, Faraday and electrolysis for atomic mass/charge ratio etc.

@DrMoishePippik What is interesting to me is that apparently the oil drop experiment gave an answer that was technically wrong / inaccurate, but we still used it anyway to derive all these other numbers?

It's been refined over time. As @MaxW states, original values had a very large margin of error. It's like Galileo using a chandelier as a clock... instrumentation has improved since then.

But as those values get refined, is the same done for all other constants that depend on them, and are experiments re-interpreted using new values?

RE: But as those values get refined, is the same done for all other constants that depend on them, and are experiments re-interpreted using new values? // Now you are catching on. Recently there has been an effort to identify the "fundamental" values that are ones necessary to define all the others. So if I have a standard for length, a cube of that standard value has a defined volume rather than needing a separate volume standard too. So all the values have gone through a tortuous path to their current state.

A fascinatingly simple experiment based on spreading oil over a large water surface and assuming monolayer formation can actually be used to determine Avogadro's number. Not very precisely, of course...

The above info is obsolete now, as the latest SI definition is Avogadro number is defined as $\pu{6.02214076E23}$

@a-cyclohexane-molecule That link doesn't work, which is a pity as it the experiment is a school-level way to get a crude estimate of Avogadro's number.

The constant enters into the model for the MSD in Brownian motion, so if you're able to make those measurements, you could calculate an estimate.

To determine Avogadro's number you have to measure the same unit at the atomic and macroscopic scales.
This was first achieved by Millikan who measured the charge of an electron. The charge of one mole of electrons was already known and  is a Faraday. Dividing both, you get Avogadro's number.
Before that, Josef Loschmidt was the first one to calculate the number of particles in a cubic meter of gas, using the kinetic theory of gases.
This is, of course, also related to Avogadro's number.
Avogadro himself did not give any number. He just stated that equal volumes of different gases at the same pressure and temperature have equal amount of particles.
A more modern way of doing it is determining the density of an ultra pure element and then determine the number of atoms and their distances in a unit cell with X-ray diffraction.
So yes, you could, in principle, derive Avogadro's number with pure monocrystalline carbon 12. You will, however, need very good instrumentation.
As per the answer to this question on Chemistry SE it might become the other way round.
The Avogadro number will be absolute (defined, with 0 error) and then the kilogram will be redefined as a function of the number of atoms in a monocrystalline perfect $\ce{^{28}Si}$ sphere.

There is a school-level experiment that can estimate Avogadro's number
Avogadro's number is now defined to be a precise constant. But the question is whether simple experiments can be used to estimate it. The historic experiments designed to estimate it are not always useful as they usually involve techniques or equipment unavailable outside fully-equipped laboratories. Like the indirect method used by Millikan (measuring the charge on a single electron in the famous oil-drop experiment which, via known electrical measurements, can be converted into an estimate of the constant) or the estimates based on x-ray measurements of ultra-pure crystals which estimate density via measuring atomic distances in a known mass).
But a simple experiment can get a good approximate estimate given a small number of assumptions based on an 1890 idea from Lord Rayleigh. The experiment involves spreading a known quantity of a pure surfactant over a water surface and measuring the area it occupies. The assumption is that the surfactant occupies a compact monomolecular layer on the water surface (there are other ways to verify this). If you put in assumptions about the size of the molecule and the thickness of the layer and measurements of the mass of the surfactant that created the layer (easy to measure with some accuracy) then a decent estimate of avogadro's constant can be derived.
The process of doing the calculation is described in this Physics.SE answer (with the caution that assuming a surfactant occupies a cube is wrong).
In the 1980s this was a common school experiment. One description of that experiment is here.
So there are simple experiments that can be used to estimate the Avogadro constant. But more reliable estimates need a lot more equipment.


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What does “stoichiometric amount” mean?
Can anyone inform me what the term stoichiometric amount means? I know what stoichiometry means:

The relationship between the relative quantities of substance taking part in a reaction or forming a compound.

But what is “stoichiometric amount” supposed to be?

Stoichiometric just means element measure. With regard to a balanced chemical reaction equation, the reactants are in stoichiometric proportion if the reactants are in ‘just right’ proportion: no reactant is limiting or present in excess. In a sense, it is ‘least wasteful’. But, as one of the answers notes, real world situations may make it necessary to have non-stoichiometric conditions, e.g., synthesizing cisplatin from an expensive platinum reactant plus an inexpensive reactant. See wiki for cisplatin. Then the cheap reactant may be used in excess to avoid wasting the expensive reactant.

Sometimes, practical chemistry does not work like it's supposed to on paper. A perfectly balanced reaction equation may dictate that two reactants should be in a 1:1 stoichiometric ratio, but in practice, you get much better yields or reaction times if you use an excess of one.
When that happens, you are no longer using a stoichiometric amount, as specified by the balanced equation, but a super- or substoichiometric amount.
Welcome to chemistry in the real world.

IUPAC defines the term “stoichiometric amount” plainly yet rather vigorously on the basis of thermodynamic system [1] and employs it further in expression of chemical activity via chemical potential (emphasis mine):

2.1. The physico-chemical description of a component in a chemical system is based on two quantities:
(1) An extensive quantity: the amount of substance of the component added or removed in the process of formation of the system, positive when added, negative when removed.
This quantity is sometimes called the stoichiometric amount of substance of the component in the system, symbol $n.$
(2) An intensive quantity: the chemical potential of the component in the system, symbol $\mu.$
[…]
The component added to the system may dissociate or react with other components to form a series of derived components and only a fraction of the original component may actually exist in a free form in the system.
It is therefore essential to distinguish between the stoichiometric concentration and the substance concentration of the free form of the component in the system.
Sometimes stoichiometric quantities are indicated by subscript $(_\circ),$ e.g.
$n_{\circ,\ce{B}},$ $c_{\circ,\ce{B}}.$

The IUPAC Gold Book's definition of stoichiometry (even though there are several other definitions) also stresses out that stoichiometric relationship is merely a snapshot in time already for the simplest reaction such as $\ce{A -> B -> C}$.
Тhe majority of chemical reactions don't have constant relationship between the amounts of substances, and the stoichiometric equation is just a crude approximation of the real-life scenario.
Additionally, reactants not always combine in exact whole-number ratios.
It is therefore imperative not to (pre)define stoichiometric amounts based solely on a single stoichiometric equation.
There are several consequential features [2]:

It doesn't matter how such system reaches equilibrium or is “constructed” in general.
Stoichiometric amount depends on the choice of elementary components, and, since stoichiometric coefficient of an elementary reaction can be negative, the stoichiometric amount can also change not only its value, but also its sign, too, depending on which other non-derived components are present.

The author Ole Siggaard-Andersen, M.D. being a clinical chemist added a simple remark on his website:

Specifying the stoichiometric amount of the components in a system is a prescription for the system…

Note the terms “stoichiometric amount” and “equivalent amount” are often liberally used  interchangeably [3], and I suspect this is the meaning implied in your context (AoS is the amount of substance):

A stoichiometric amount of one reactant can mean the amount that will react completely with a given amount of another reactant, leaving no excess of either one.
In this usage, the amount may be a mass, a volume, or an AoS, and the adjective
stoichiometric refers not to stoichiometry in general but to stoichiometric equivalence in particular.
Thus, this sense is synonymous with equivalent amount.

I would recommend to read an entire Giunta's paper as during its proposition of a different meaning to the term suggesting it as a replacement for AoS, it also outlines pros and cons of several related terms, elaborates why “number of moles” is illiterate and does many other things right.
References

Siggaard-Andersen, O.; Durst, R. A.; Maas, A. H. J. Physicochemical Quantities and Units in Clinical Chemistry with Special Emphasis on Activities and Activity Coefficients (Recommendations 1983). Pure Appl. Chem. 1984, 56 (5), 567–594. DOI: 10.1351/pac198456050567. (Free Access)
Siggaard-Andersen, O. Stoichiometric Concentration and Chemical Potential. Scand. J. Clin. Lab 1977, 146, 7–13. DOI: 10.3109/00365517709098926.
Giunta, C. J. What’s in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount. J. Chem. Educ. 2016, 93 (4), 583–586. DOI: 10.1021/acs.jchemed.5b00690.

The amount of a substance produced or needed in the reaction of a specific amount of another substance with respect to a particular formula. It’s the amount your stoichiometric ratios tell you to use/expect.

Let’s take the simple example of the combustion of hydrogen $\ce{H2}$ in oxygen $\ce{O2}$. The equation of the reaction is:
$$\ce{2 H2  +  O2 -> 2 H2O}$$
It means that $2$ moles of $\ce{H2}$ $(2\cdot2\ \mathrm g$) react with $1$ mole ($32\ \mathrm g$) of $\ce{O2}$. These masses are in a ratio $4:32=1:8$. All mixtures containing $\ce{H2}$ and $\ce{O2}$ in a proportion $1:8$ are said to be “in stoichiometric amounts”.
If you use $0.7\ \mathrm g$ $\ce{H2}$ and $5.6\ \mathrm g$ $\ce{O2}$ to produce water, or if you have reacted $1.10\ \mathrm g$ $\ce{H2}$ and $8.80\ \mathrm g$ $\ce{O2}$, the components of the gas mixture are said to be “in stoichiometric amounts”.


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What is the product formed when 4-chlorobut-1-ene is treated with acetic acid?

My teacher did like this forming a really unstable cyclopropyl group but this is the correct answer instead. My question is why the product will be even formed?
In the mechanism that my teacher had drawn he formed a really unstable carbo anion there whereas he could have formed a primary carbanion and next may it have formed a four membered ring.
Why did not the neucleophile (acetic acid) did not approach to the carbon atom with which the chlorine is attached directly? Forming

I tried to justify the answer then it came to me that if the acetic acid does approach from right hand side may it be forming a stable carbocation by resonance

And this next form a very stable cyclo propyl carbocation. (Which is one of the most stable carbocations) So which explaination should I rely on?
Thank you.

See: resonance forms of cyclopropylmethyl carbocation

Acetic acid (not acetate anion) attacks an isolated double bond? Seems........unlikely. Anyone got reference for this reaction?

To get a nucelophile to attack a alkene double bond you need to bind the alkene to a transition metal such as mercury or palladium

OK, unless someone produces a reference showing it, I am of the opinion that this reaction does not occur.

So will it perform Sn1forming that stable cyclomethyl carbocation? Where as it should have easily performed Sn2


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Why does the compound formed after a coordinate bond have a net charge?
Why does the molecule have net positive charge?

Because the particles before the formation of the said bond had a net charge.

why not negative charge then?

Because they never had negative charge in the first place.

shouldn't positive charge of H get neutralised when it gets an electron? Also, what are the criteria for getting the total charge of the compound? Is it by adding formal charges? What is the formal charge of O after forming the coordinate bond?

It never gets an electron for free. Instead, it starts to share someone else's electrons.

Before you know a thing about formal charges, and even before you know a thing about atoms, you have to know that the total charge is always conserved. Now the total charge of $\ce{H2O}$ is 0, and the total charge of $\ce{H+}$ is +1 (we know that because it is written with that little $^+$). Just what their combined total charge might be?

+1, so every time a compound ionizes, is it because of coordinate bond?

No, this is not because of coordinate bond. The charge is conserved no matter what. Coordinate bonds are irrelevant here, as well as bonds of any other type.

@IvanNeretin I have feeling this is hopeless case here...

@IvanNeretin I finally understood it :)  Thanks for your patience

Consider the formation of the given hydronium ion: $\ce{H2O + H+ -> H3O+}$ (also more correctly stated as the auto-ionization of water: $\ce{2H2O -> H3O+ + OH-}$)
The positive charge you're seeing indeed belongs to the whole molecule, as any reaction conserves charge both on the left and right side of the arrow.
However, this is not true in general. If a neutral molecule itself forms from neutral atoms, which involves a coordinate bond, then the molecule won't be charged.
Although, do note that the distinction between normal covalent bond and coordinate covalent bonds is artificial.

The charges are produced only due to the nearness of the electrons with the nucleus of the respective atom.
Here the co-ordinate bonds produces charge because the electrons in this case is near the H atom which results in a partial positive charge on H that's why there is a charge overall in the compound.


text:
After boiling water in an electric kettle, should I empty it?
(I'm reposting my question from Physics SE where it got closed for being off-topic, but chat tells me that it might be on topic here.)
In the office I'm working in, there's an electric kettle for boiling tea water.  Every day, I boil water in it.
At one point, my boss asked me to empty the kettle after making the tea, that is, pour the unused water remaining in the kettle into the sink so that the kettle is as dry as I can easily get it.  His rationale was that this will stop the kettle from getting chalk residue from the hard water.
I obey my boss, both because he owns the office and so the kettle is his, and because he's an engineer.  But privately, I have doubts about this request.
I know from practice at home that it is very easy to clean chalk residue from an electric kettle: you just boil water with citric acid in it, then rinse the kettle properly, and this removes all the residue.  (Vinegar also works instead of citric acid, I prefer citric acid because it doesn't have the smell and is also cheap.)
On the other hand, I'm worried that if I don't leave any water in the kettle after boiling water for tea, the heating element may still be hot and its surface could become oxidized.  If that happens, that's much more diffictult to fix than the chalk.  Because of this, at my home, I always make sure that there is water in the kettle after boiling, including adding cold water from the tap if I used up all the hot water.   I have very hard water at my home, which causes a lot of difficulties cleaning the bathroom tiles.  (I do not know how hard the water is at the office.)
So what's the best practice at home?  Should I leave the kettle dry or wet after boiling water?

Chalk residue forms before and during boiling, rather than after it.

Rinse it out with cold water. Where I live, the issue isn't carbonates but silicates in the water - those don't clean up easily with mild acids, and HF isn't something to play with at home...

Simple matter of applying the Golden Rule - Him with the gold makes the rules.

Your boss is somewhat right. Probably not wrong enough to argue with him.
There are several types of scaling that can be caused by mineral content in a water supply. One of the most common is called "temporary hardness" or "carbonate hardness." At high or neutral pH, calcium and magnesium carbonates have quite low solubility in water: 139 ppm for magnesium carbonate, and a mere 6.7 ppm for calcium carbonate, both at 25°C. (NOTE: Until starting this reply, I hadn't realised how little agreement there is on the solubility of calcium carbonate. Every source that I checked agrees that it is quite low; but no two sources agreed to even one sig. fig.!)
However ordinary, clean rain water often has a pH as low as 5.6, through absorption of atmospheric carbon dioxide. (Actual acid rain is even lower!) When such water runs through deposits of these very common minerals, some of them dissolve through the reaction:
$\ce{CaCO_3 + H_2O +CO_2 <-> Ca^{2+} + 2HCO_3^{-} }$
and the similar one for magnesium. Calcium bicarbonate has an aqueous solubility of 166 g/L at 20°C -- some 13,000 times more soluble than the carbonate. Of course, the dissolution process from hard rock is slow, and most runoff will not actually be saturated.
The problem is that our reaction, above, is an equilibrium; anything which tends to drive it back to the left will cause precipitation of the insoluble carbonate. One of those things is just atmospheric outgassing; this is how stalactites and stalagmites form. But a much faster method is boiling the water. That decomposes the bicarbonate ion, and rapidly drives off the freed carbon dioxide gas. The resulting precipitate of calcium carbonate is called "temporary hardness" because, as you have observed, it is easy to remove with any mild acid. Some other forms of scaling are much less forgiving.
Now, if your water hardness problem is severe, a lot of carbonate will start to precipitate as soon as the kettle is heating up. This is why another commentator suggested that it is too late to do anything after the kettle has boiled: that the scale deposits have already grown. However, there are two complications. Firstly, the supersaturated calcium carbonate solution doesn't necessarily all precipitate at once; the scale crystals may continue to grow for some time. I can't say just how long, but a process whereby large, hard crystals tend to cannibalise small, soft ones can continue for weeks.
Secondly, you will not see hardness that bad from mnay common water sources. Well water, maybe, but most municipal supplies will already be softened to the greatest degree that it is economical. If the water hardness problem is mild, some of the calcium carbonate will precipitate immediately, but some (more than 6.7 ppm!) will stay dissolved in the very hot water, until the solubility falls with falling temperature and more of it precipitates.
Long story short, your boss's concerns are not unreasonable. On the other hand, as you note, temporary hardness is easily cleaned off, and only temporary hardness is related to boiling. On the third hand, it's his kettle and he's the boss, so do as you're told!


text:
1,1-Dibromo-2,2,3,3-tetramethylcyclopropane is treated with CH3Li

My approach:
The ring is extremely sterically hindered. And also due to ring strain ring gonna break .
Here methyl lithium ($\ce{CH3Li}$) acts as a base so it is going to take the most acidic hydrogen and then I drew like this but getting no way out there.

duplicate of https://chemistry.stackexchange.com/questions/99317/mechanism-of-synthesis-of-allene

According to this review paper 1 the answer is d - none of these
The reaction proceeds by initial lithium-halogen exchange followed by carbene formation. In some cases the allene (answer c) is formed but, as stated in the paper referenced, tetra-alkyl dibromocyclopropanes give bicyclobutanes by insertion of the carbene into one of the methyl C-H bonds.
This is further discussed in this posting Mechanism of synthesis of allene


text:
Why does a molecule with a Thiol group form Hydrogen bonds?
I'm doing the GRE Biology practice test, and question #1 asks:

The answer is E. I understand why II and IV are polar, but why is $\ce{R-S-H}$ polar? R is a side chain to the sulfur, sulfur's valence shell is full, and sulfur is highly electronegative. But without knowing what $\ce{R}$ is, how do we know the molecule's total polarity?

The question didn't ask about a molecule's "total polarity," it asked which would form Hydrogen bonds. The hydrogen on a thiol group  (R-S-H) can form hydrogen bonds with other groups like alcohols and amines. Thiols are sparingly water soluble so III is a bit strange but not impossible. Thiols only form weak hydrogen bonds between themselves.

@MaxW That answers it, thanks! If you'd like an accepted answer, please convert your comment to an answer and I'll accept. Either way, thank you.

@freeradical Could you please update the question title - in your own words - to reflect the actual question being asked?

@GaurangTandon Sorry, just saw this. I've updated the title, thanks.

Actually it's all four.  See here for I.  The article describes a hydrogen bonding interaction between methane and water.  It's weaker than the "usual" hydrogen bonds but still contributes to forming methane-water clathrates.
The larger lesson here is that hydrogen bonding is not just the electrostatic interaction often described in textbooks.  It has a molecular orbital component that works even with primary bonds of low polarity.

Fascinating, thanks! I'll report this to the GRE folks. I also recently read that the electronegativities of Hydrogen and Carbon are not equal (2.2 and 2.55 respectively), so it makes sense that even Methane is slightly polar and thus can make hydrogen bonds.


text:
Are secondary alcohols more easily oxidised as compared to phenol?

Which among the following compounds is oxidised most easily?

2-propanol ($\ce{CH3CHOHCH3}$)
phenol ($\ce{PhOH}$)
ethoxyethane ($\ce{Et-O-Et}$)
1-methyl cyclohexanol

The answer given in my workbook is option 1. It's obvious to me that options 3 and 4 can't be the right answer (ether, hindrance). But I am confused between options 1 and 2.
I have arguments that can support either of the 2 options. There's less hindrance on phenol so formation of an intermediate ester like chromate ester would be easier their as compared to secondary alcohol. However, phenol is losing aromaticity as well, therefore activation energy of oxidation should be higher as compared to secondary alcohol.
Since $E_\mathrm{a}$ matters more, we would say that secondary alcohol is oxidised more easily, is it?

Hindrance has little to do with this. Loss of aromaticity is a big deal, so the oxidation of phenol to quinone takes a high energy reagent (Ceric Ammonium Nitrate is a good one). Oxidation of a secondary alcohol by contrast is fairly easy.

Btw, what does "high energy reagent" mean?

Reagent in a high oxidation state e.g Cr(VI), Ce (IV), Mn (VII), hypervalent iodine salts; further reading here http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532008000800007

Also, choice #4 is a tertiary alcohol.  What would it be oxidized to?  Even if it were unhindered, oxidation is going to require breaking C-C bonds.

There's a good case to be made that #2 is the right answer.  If I leave isopropanol in one sealed container with O2 headspace at 80 C for a month, and phenol in another container under the same conditions, at the end of the month my guess is that the phenol would be substantially less pure than the isopropanol due to oxidative degradation.


text:
Rearrangement of Grignard reagent formed from exocycloproyl vinyl bromide

First, I prepared the grignard reagent then performed nucleophilic addition to the carbonyl carbon  and then did hydrolysis.
So basically in my product, $\ce{-Br}$ is replaced by $\ce{-CH2OH}$
But answer given is:

Why is my mechanism wrong?

Grignard reagents are capable of rearranging, especially in this situation where there is significant strain. The specific vinyl bromide you drew here has been studied: J. Org. Chem., 1974, 39 (10), pp 1411–1416. The table of data shown below is hard to read without including more of the text, but it was found that at 66 C in THF, after 5-10 minutes, 41% of the initial Grignard had undergone rearrangement, so this is a fairly likely process.

A potential mechanism for the rearrangement is:


text:
What hinders SN2 more: geminal alkyl group or alkyl group on the aromatic ring?

Which among these two (1-chloro-1-phenyl-ethane and 2-(chloromethyl)-1,3-dimethylbenzene) is more reactive towards SN2?

I strongly believe that it should be Y because of lesser hindrance on the electrophilic carbon.
However, the answer key of my exercise book states that X > Y. What's the reason behind that?

Which book is this? Could you please mention the source? Thanks!

@schrodinger_16 it's himanshu pandey's advanced problems in organic chemistry for JEE

The correct answer is X because the 2 methyl groups on the Y are bulky enough to cause steric hindrance. Surprisingly, from what I have seen, if you only had one methyl group on Y, the SN2 reaction would proceed.

Got a reference for that?

There's plenty of examples of SN2 on dimethyl substrates like Y. Rates are certainly going to be slower, but it's not too difficult of a substrate compared to others.

Needs a citation. In the reactive conformation, Cl–C–Nu should be parallel to the ring. In the absence of some data I'm skeptical as to how much steric hindrance those methyl groups actually produce. I would be happy to be convinced otherwise, of course.


text:
Difference in bonding between water and hydroxides
As we know, water is a covalently bonded molecule. However, the hydroxides - such as sodium hydroxide - are ionic compounds. Both of these are similar, although the difference between both is a sodium atom instead of a hydrogen atom.
Now, both hydrogen and sodium have one outer electron, which makes me come onto my question:
Could sodium share its outer electron with the OH group, like hydrogen does in water molecules, to form a normal covalent bond? Or, could hydrogen lose its outer electron to form a cation, and ionically bond to OH-, like what happens in sodium hydroxide?
Long story short: why is water covalent, whilst the hydroxides are ionic, if the difference is just another atom instead of hydrogen?

Ionic molecules is a contradiction in terms.

@IvanNeretin My apologies I have updated.

https://chemistry.stackexchange.com/questions/20511/why-isnt-water-an-ionic-compound?s=2|38.3041 https://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

Although I’m not to proficient in this chapter. I believe it’s because sodium and the other group 1 and group 2 elements in particular are electropositive in nature .it is due to this that they form ionic compounds .Hydrogen isn’t electropositive enough to form an ionic bond like you have stated .Just read up on Fajan’s rule and you’ll get the idea


text:
Stoicheometry questions: I've been having a bad time not understanding the questions and how the steps are to achieve the answer. please help

Cetyl palmitate (C32H64O2) is a waxy solid and is the main component of spermaceti, the once prized wax found in the skull of sperm wales. The heads of sperm whales can contain up to 1900 L of spermaceti. If the most famous sperm whale, Moby Dick, had a head containing 1514 L of spermaceti, consisting of 77.3% (by mass) cetyl palmitate, how many kL of liquid water would be produced from the combustion of the cetyl palmitate?  Assume there is excess oxygen and the reaction proceeds in a 68.2% yield. The density of spermaceti is 828.4 kg/m^3
Zinc metal can be dissolved in nitric acid (it actually reacts with the nitric acid to produce hydrogen gas and soluble zinc nitrate).  If a solution of nitric acid contains 2.6 mol nitric acid per liter of solution, how many mL of solution are needed to completely dissolve a piece of zinc metal that is 1.32 in wide, 4.66 in long and 0.436 mm thick? The density of zinc is 7.14 g/mL
When zinc metal is placed in an aqueous solution of copper(II) sulfate, the zinc is oxidized and goes into the solution as zinc ions.  At the same time, the copper is reduced to copper metal and deposits onto the zinc metal.  In effect, the zinc metal is slowly "dissolving" and being coated with copper metal at the same time.
A 1939.2 g strip of zinc metal is placed into a solution containing an excess of copper(II) sulfate and after some time has elapsed it is pulled out and its mass is determined to be 1927.3 g.  What mass of zinc metal was oxidized?

Homework! Please show your own attempt at a solution, and describe what difficulty you have. We're not a free homework service.

Hi Naomi, welcome to Chem.SE! We require you to show your efforts on this problem. What formulae/steps did you try? Where did you get stuck? Please add this to your question. Thanks!

@Gaurang Tandon I first have to balance the equations first. then convert my units. After that im not sure what the steps are called. The notes i have taken in my Chemistry class only help with non-lengthy questions, so the structure of these questions generally confuse me.

Also i came on this site for help. not for "free homework service" @Karl . that's why i asked for the steps so i could follow them on dozens of other assignments i was given.

It often helps to

ask yourself: What are they actually asking for?
reorder the information given in the question.

Let's try this for the first question:
Apparently, this is about cetyl palmitate ($\ce{C32H64O2}$). When the compound is burned in excess of oxygen, carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$) are produced. It's your task to calculate the amount of water generated upon burning a certain amount of cetyl palmitate.

Do you remember the concept of the unit mole? How many molecules of water can be formed from a molecule of cetyl palmitate that contains 64 hydrogen atoms? Knowing the number of molecules and the molecular mass of a compound gives the corresponding weight.

If the initial amount of cetyl palmitate was given in moles you'd be done. Obviously, that's too easy.
Instead, the Moby Dick example was given. This adds two other concepts:

density, which correlates mass and volume
percentages: if only x% of your starting is usable, what is the actual amount work with in further calculations?

Let this sink in for a while and then give it a try. You can do this by yourself!


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Is the kilogram still consistent with the old definition?
My old understanding was that a gram is the amount of mass in a cubic centimeter of water.
But then I recently learned that the second was redefined to be some 9 billion cycles of a Caesium atom. Then from this the definition of a meter changed from 1 ten millionth the distance between the equator and north pole, to $1/299 792 458$ the amount of distance that light travels in a second. From this I would have assumed that now a hundredth of that meter would redefine the gram, which would then be multiplied by $1000$ to get the kilogram. But I read that instead the kilogram is some reference mass kept in a safe somewhere.
Are these definitions nevertheless consistent with each other or are we just doing away with the linkage between the centimeter and the gram via the water cube?

Read about the history of the kilogram

You are really old school. Pt cylinders have been the reference standard since December 10, 1799.

The kilogram is likely to be redifined in the near future as a function of the number of atoms of a sphere of ultrapure $\ce{^{28}Si}$, as per the answer to this  Chemistry SE question.
That means that the standard will be the Avogadro number, and the mass will be a derived unit.
Then, any stored mass anywhere will just be an historical curiosity.
This does not mean that it has not been important.
We need exact measurements as Lavoisier taught to all future chemists.
At that time it was the best that was available.
Now we might go to a still better standard.

https://chemistry.stackexchange.com/questions/60767/effects-of-changing-avogadros-constant :)