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Rank the following radicals in order of decreasing stability
Question

Rank the following radicals in order of decreasing stability

Aromaticity makes a cyclic compund more stable. Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?). So, I thought maybe 3 is most stable followed by 1, then followed by 2 and 4(I dont know which is more stable out of 2 and 4). But, that's not the case. 3 is most stable, but the next stable one is 2 and not 1. I don't get how.

Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?

Yes you're wrong over here, for applying the Huckel's rule the compound must be a closed circuit of delocalized electrons. In both the cases the circuits are not closed

If you're not aware of Conjugation then give that a read .
Now
The most stable compound is 3 since the extent of conjugation is the greatest i.e 2 pi bonds.
3 is more stable than 4. Even though 4 is aromatic it is an aryl free radical the p orbital containing the radical is out of the plane and can't be stabilized through resonance/Hyperconjugation.
In fact I would say that option 4 is the least stable compound .
The second most stable compound is is 2 since it is stabilized via resonance which is more effective than Hyperconjugation; 1 is stabilized by Hyperconjugation.
3 is more stable than 2 since the extent of conjugation is greater.
The order of stability will be
$$\mathrm{3 > 2 > 1 > 4}$$

Sorry, but I didn't get what closed circuit means. Would you please explain? I would be really grateful!

I meant that the loop should be closed see this PDF for more details

So they are open. How? I mean, what is exactly being closed. I am really sorry for bothering, but since I am still at basics, I am confused at petty things as well.

For the flow of delocalized $\pi$ electrons to take place a complete loop must be the compounds 2 and 4 do not have complete loops you'll notice that the carbont atoms are sp3 hybridized, they are all $\sigma$ bonded . No $\pi$ orbitals are present for the flow of electrons

Oh, now I get it. I knew about that fact, but just didn't get what loop you meant. Thanks!

@DivyankaS.Chaudhari It may be otherwise stated as "complete conjugation throughout the ring"


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Directing nature of the -SOR substituent in electrophilic aromatic substitution
This is (alkanesulfinyl)benzene:

I didn't draw it, but we know that the sulphur atom also has a lone pair. So, the sulphur atom can increase electron density in the ring by donating a lone pair. However, oxygen can also withdraw electron density from the ring by pulling back an electron pair. They both are contradictory, so which effect dominates? What is the final directing effect of the alkanesulfinyl group?
That said, I am not entirely sure if what I said above is 100% correct. The reason being that since the sulphur atom is sp3 hybridized, it is tetrahedral in shape, and not planar. So, either the sulphur's lone pair or the oxygen's pi bond will be out of the plane (but not both at once), and won't take part in any delocalisation. But we can't theoretically predict which one it would be, can we?
If there could be experimental evidence to support the claim whether it is meta-directing or o/p-directing, that would be even more beneficial. I have checked both Wikipedia and masterorganicchemistry, and neither even mentions this functional group, let alone its directional nature.

https://chemistry.stackexchange.com/questions/88111/is-methyl-phenyl-sulfoxide-optically-active confirms it's tetrahedral

Situation may be similar as in https://chemistry.stackexchange.com/questions/8509/electrophilic-aromatic-substitution-are-protonated-oh-groups-ortho-para-or-me or in protonated amines

I think it will be o-p directing. I don't have any theoretical reasoning for it but just a comparison with Ph-NO. In this molecule too there is a lone pair on N and a pi bond. As a fact I know that NO is o-p directing. So, I'm assuming that the compound you gave might also be o-p directing.

@user26732 Thanks for dropping by! Though I think the distinction here between Ph-NO and Ph-SOR is that - correct me if I'm wrong though - the former is linear but the latter is tetrahedral.

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Chemistry of Benedict's Reagent
I want to understand what is the exact mechanism by which "enediols"  reduce cupric to cuprous of  Benedict's reagent.

The principle of Benedict's test is that when reducing sugars are heated in the presence of an alkali they are converted to powerful reducing species known as enediols. Enediols reduce the cupric compounds (Cu2+) present in the Benedict's reagent to cuprous compounds (Cu+) which are precipitated as insoluble red copper(I) oxide(Cu2O).
  [source]

Any reference would be helpful too.

https://chemistry.stackexchange.com/questions/46577/why-is-vitamin-c-called-an-antioxidant answers it

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Neutral anions in salts and strong acids are similar - is it a coincidence?
Recently I was studying salts and my professor provided a list of anions which are considered neutral in salts (that is, they do not change the $\mathrm{pH}$ of an aqueous solution when added as a salt): $\ce{Cl-}$, $\ce{Br-}$, $\ce{I-}$, $\ce{NO_3^-}$, $\ce{ClO_4^-}$.
Now I'm looking at this and I realized the list seems to be almost identical to a list of strong acids: $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HNO_3}$, $\ce{HClO_4}$, $\ce{H_2SO_4}$.
Is this a coincidence? If it's not, what's the reason, and why is $\ce{SO_4^2-}$ not on the list of neutral ions?

The conjugate acid of a weak base is a strong acid, and the anions you have mentioned are weak bases.

Though couched in different terms, the first part this question is essentially the same as this other one:
Does a strong acid's conjugate base act as Brønsted-Lowry base?
No, this is not a coincidence.  The conjugate base of a strong acid will always be a weak base.  The 'neutral salt anions' of your professor's nomenclature are just weak inorganic Brönsted bases.

The reason why $\ce{SO4^2-}$ doesn't appear in this list of 'neutral salt anions' is because it actually can noticeably affect the $\mathbf{pH}$ of a solution when added as a salt. $\ce{H2SO4}$ is often treated as though it's a diprotic strong acid, but in reality only the first deprotonation is strongly acidic.  This can be seen by examining its acid dissociation constants; from Wikipedia:

$\mathrm pK_{\mathrm a1} = -3$
$\mathrm pK_{\mathrm a2} = 1.99$

This $\mathrm pK_{\mathrm a2}$ is about the same as the first $\mathrm pK_{\mathrm a1}$ of $\ce{H3PO4}$ $($$2.148$$)$, and is actually somewhat less acidic than nitroacetic acid $($$1.68$$)$.
So, in some cases sulfate may act more or less as a 'neutral salt anion', when the $\mathrm{pH}$ is sufficiently high that it will all stay doubly deprotonated, but in others it can definitely influence the solution $\mathrm{pH}$. I've actually used sulfate/bisulfate before as an acidic buffer for various projects at work—works great!

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Can a molecule be achiral whilst it's resonance form chiral?
I'm trying to work out whether methylphenylsulfoxide is a chiral molecule.

It is my understanding that the sulfur uses sp2 hybridised orbitals. Why is this? The central sulfur atom is bound to 3 atoms and then has a lone pair, so surely it has to use sp3 hybridised orbitals?
Or does it use sp2 hybrid orbitals so the empty p orbital allows for extended delocalisation of the benzene ring through the sulfur and onto the oxygen?
Apparently there are resonance forms of this molecule which are chiral.

This molecule looks odd to me, would it make more sense to have electrons resonate out of the benzene ring?

Unfortunately the sloppy depiction of sulfoxides will cause such misconceptions. Your analysis is quite good:

The central sulfur atom is bound to 3 atoms and then has a lone pair, so surely it has to use sp3 hybridised orbitals?

The only thing one should keep in mind is that the model of hybridisation starts to break down for these compounds (third period and below).
In this case it also is not strictly necessary to employ hybridisation, but as a first (very rough) approximation it can give you an educated guess.
In any case, the sulfur is approximately tetragonal coordinated.
Both depictions are indeed missing the lone pair at sulfur.
As such the first one would has 10 electrons around sulfur, which used to be regarded as correct, but is now a deprecated viewpoint. Sulfur would need to use d orbitals, which has been proved wrong.
The charge separated structure is superior to the above, because it does not suggest the use of d-orbitals.
The issue whether or not these compounds are chiral has been answered by Jan (Can heteroatoms with lone pairs be chiral centres?):

The sulfoxide case. Sulfoxides $\ce{R1−SO−R2}$ have a free electron pair and a trigonal pyramidal configuration on the sulphur atom. The lone pair is in an orbital with a high s-contribution meaning that the nitrogen inversion highlighted above is rather complicated. Thus, sulfoxides can be assumed to be chiral on sulphur (although in practice they are often generated racemicly).

You have asked more general though:

Can a molecule be achiral whilst it's resonance form chiral?

It is very important to realise that resonance only affects the electronic structure. Therefore in all resonance contributors the position of the nuclei is the same. Since chirality is determined by the spatial arrangement of the atoms, the question itself is void. There cannot be a set of resonance contributors which has chiral and achiral structures. Please also see What is resonance, and are resonance structures real?

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Difference between 'Preferred IUPAC Name' and 'Systematic IUPAC Name'
When looking at different chemicals, I come across the 'preferred IUPAC name' and the 'systematic IUPAC name'. I've noticed that when I name chemicals, I name them by their 'systematic IUPAC name', as I've been taught to name them using this system.
When is the 'preferred IUPAC name' used? I sometimes come across it when looking at aldehydes, ketones and carboxylic acids. However, the fact that it's 'preferred' makes me wonder whether chemists use this method more often than the other system.

The problem with systematic IUPAC names is that many compounds can have more than one systematic IUPAC name. A compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC. Thus, asking for the systematic IUPAC name does not make sense for such compounds.
Therefore, current IUPAC recommendations (currently only for organic chemistry) include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including also the many retained names.
Whether the preferred IUPAC name is also actually preferred by chemists cannot be generalized and depends strongly on the context.


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Equilibrium change in sealed container
Question:

Decreasing the product's concentration seemed to follow the logic of Le Chatelier's principle, so a decrease in volume will force the reaction to the left to increase the pressure (increasing the number of molecules). However, decreasing nitrogen gas concentration also has the same effect. I chose volume compression over the other because I assumed. I am probably stretching what it means by "sealed"  though. Apparently I got the question incorrect.

Don't hold me on this, but I'd say that $N_{2}$ was removed from the system. Removing reactant will shift the equilibrium to the left to oppose the change, so more $NH_3$ would have to decompose into $N_{2}$ in order to do this. However, I'm intrigued by the volume increasing. If we're abiding by the ideal gas laws, pressure is inversely proportional to volume, so increasing the volume decreases the pressure. There are less gaseous moles of product, so decreasing the pressure would also shift the equilibrium to the left.

By the way, the 'sealed container' is used to provide a closed system, so that an equilibrium can be achieved.

The given graph indicates that $\ce{[NH3]}$ remains constant till the 15 minute mark which implies that the equilibrium was established till then. Now, after the 15 minute mark, $\ce{[NH3]}$ decreases and reaches a constant value which means, the equilibrium was disturbed and re-established.
$$\ce{N2 + 3H2 <=> 2NH3 + heat}$$
Since we observe a decrease in $\ce{[NH3]}$, the reaction must have shifted in the backward direction according to the given equilibrium.
Out of the five options, only match for this is (C) removing some amount of $\ce{N2}$. Let us analyse the other four options,
(A) Decrease in temperature: Note that the above reaction is exothermic and hence according to Le Chatlier’s principle, the reaction shifts in the forward direction for a decrease in temperature.
(B) Decrease in volume: Since we are dealing with a gaseous(ideal) equilibrium, a decrease in volume implies an increase in pressure and hence, according to Le Chatlier’s principle, the reaction shifts in the direction where there is less no. of moles ie., forward direction here.
(D) Removal of $\ce{NH3}$: As Gaurang pointed out in the comments, removal of $\ce{NH3}$ will cause a sharp fall along a vertical line, but the graph indicates a gradual fall in $\ce{[NH3]}$. Hence this is also not valid.
(E) Increase in volume: Assuming a sudden increase in volume, say n times, $\ce{[NH3]}$ should also fall by n-times. Clearly, the graph doesn’t match with our condition. Hence this is also incorrect.
So (C) must be the correct answer, as removing some $\ce{N2}$, shifts the reaction in the backward direction, thereby decreasing $\ce{NH3}$ gradually till equilibrium is re-attained.

Let us continue this discussion in chat.


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N,N-Dimethyl-2-(indol-3-yl)glyoxylamide from indole
I would like to know how to obtain the following molecule, whose technical name should be something like N,N-dimethylindol-3-ylglyoxylamide (let's call it DMIG for short)

One of the constraints being that I'd like to limit myself to compounds that are relatively easy/cheap (and, of course, legal) to find.

I figured out that indole (A) may be a good starting point. Its very reactive spot on position 3 would make an electrophilic aromatic substitution likely to take place there.
Then, oxalyl dichloride (B) and dimethylamine (C) would be good candidates to construct the "dimethylglyoxylamide" leg.
And... my technical knowledge stops right there. Would it be possible to obtain DMIG from compounds A, B and C by "simply" putting them in solution in an adequate way?
As a layman in experimental chemistry, it seems to me that the most reactive spots in the system are the position 3 of the indole, the chlorine endings of the oxalyl dichloride and the N-H bond in the dimethylamine.
If this premise is correct (which may not be the case), then I have the intuition that not much would be required to lead to the formation of DMIG.
What process would you suggest? What other substances may be necessary? Are there other paths or more practical starting points for this synthesis?
Thanks in advance

Add an ethereal solution of indole (1 equiv) slowly to ethereal oxalyl chloride (1.1 equiv). Substitution occurs at the 3-position of indole. The acyl chloride is yellow and crystalline. Dissolve acyl chloride in a solvent and react with the amine. Welcome to SE!

I have improved confusing name "N,N-dimethylindol-3-ylglyoxylamide" to N,N-dimethyl-2-(indol-3-yl)glyoxylamide (in the title), correct preferred IUPAC name would be 2-(1H-indol-3-yl)-N,N-dimethyl-2-oxoacetamide. Also note that your formulæ have "typo", miss one degree of unsaturation in indole.


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Hyperconjugation in ethane conformers
I refer you to this set of presentation slides. On slide 8, the author explains the preference for the staggered conformation of ethane by saying that there is a favourable hyperconjugative interaction between $\sigma^{}_{\ce{C-H}}$ of the $\ce{C-H}$ bond in one methyl group and $\sigma^{*}_{\ce{C-H}}$ of another $\ce{C-H}$ bond in the other methyl group.
However, I am puzzled by the diagram. Since both $\ce{C-H}$ bonds are similar in terms of bonding, their bonding and antibonding $\sigma$ molecular orbitals should have the same energies. Instead, the diagram shows that one of the $\sigma_{\ce{C-H}}$ bonds is lower than the other in energy. How can this be right? I have shown a screenshot of the diagram below.

Also, how can a $\sigma_{\ce{C-H}}$ donate electron density to a $\sigma^*_{\ce{C-H}}$ effectively? After all, the energy difference between them is rather significant since the orbital interaction between the atomic orbitals of $\ce{C}$ and $\ce{H}$ is large.

I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $\sigma_{\ce{C-H}}$ MOs and a $\sigma_{\ce{C-C}}$ MO, lower in energy.
My best guess would be the position of the $\sigma_{\ce{C-H}}$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $\sigma_{\ce{C-H}}$ of one $\ce{C-H}$ bond and the $\sigma^{*}_{\ce{C-H}}$ of $\ce{C-H}$ bond anti to it - you don't really need to be worried about the $\sigma_{\ce{C-H}}$ of that anti bond because it cannot overlap with the other $\sigma^{}_{\ce{C-H}}$.
That brings us to the second part of your query. A $\sigma_{\ce{C-H}}$ can donate electron density to an empty $\sigma^{*}_{\ce{C-H}}$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference.

You are right. But if the energy difference is large, interaction between the orbitals of differing energies would be very small. In this case, this should be the case.

You're talking about energies of the order of ~1 kcal/mol. You think that's large as compared to the bond enthalpies of the various C-C or C-H bonds?

I agree that I don't know what the third MO in the diagram is meant to be; the cited Nature paper doesn't include it either and it could just have been something that was elaborated on during the presentation itself. [I didn't check the Angew paper.] I also agree that the hyperconjugative stabilisation is tiny compared to some other similar things we see (e.g. anomeric effect, (Z)- vs (E)-ester conformation preference). I don't agree, though, that the σ(CH) cannot overlap with another σ(CH). The overlap seems to me to be very poor, but perfectly symmetry allowed.

One should be very careful assigning energies to hybrid orbitals, as these are not the eigenfunctions of the Hamiltonian. (Hybrid orbitals do at the very least break wave function symmetry.) As such the image is just a very tempting illustration of that concept.

Just to build on the very good answer from Sagnik, the purpose of the σC-H bond on the right side of the diagram is to show that the C-H bond is destabilized by electron density in its antibonding orbital. But the energy of the overall system is reduced.
The red orbital structure isn't in the Angew. paper either.


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Comparing acidic strength in oxan-2-one and oxan-4-one
I need to compare the acidic strength in these molecules:

My Approach:
In compound I: Removing a proton would lead to conjugation between -ve charge, pi bond and lone pair of O. Moreover the inductive effect due to 'O' will stabilize the charge.
In compound II: Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation, but relatively less than one. Moreover 'O' is far away, so inductive effect does not play a role here.
From this I concluded that compound 'I' must be more acidic.
But the answer says that the second compound is more acidic.
Where am I going wrong?

Btw, In the second compound O is as far away as in first compound.

This has a simpler answer than you expect. First of all, your reasoning is wrong because inductive effect is much less dominant than conjugation of electron pairs. So, if the latter is unequal in both cases, we should first prioritize it over inductive effects.

Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation

It seems you already know that the beta hydrogen of carbonyls is most acidic. Both molecules have carbonyl groups, and both have beta acidic hydrogens. So, you're on the right track!
Whichever molecule is more acidic, it will have a more stable conjugate base.
Note that in the second anion, the excess negative charge is completely delocalised over to the carbonyl group. Compare that with the first anion, where both the excess negative charge and the oxygen's lone pair are delocalising with the same carbonyl group. Thus, the excess negative charge is unable to completely delocalise with the carbonyl group.
Can you now identify why the first molecule is less acidic?


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Calculating Kp in two different ways

The degree of dissociation of $\ce{NH3}$ at $\pu{1 atm}$ is 20 % as follows:
     $$\ce{2NH3 <=> N2 + 3H2} $$

I follow two ways but end up with two different answers.
Here, I assume that initial amount of reactant mole is 2 and that of product is 0. Then I take the amount of ammonia in equilibrium to be $2-2\alpha$, nitrogen to be $\alpha$ and hydrogen to be $3\alpha$. Then, I find $K_\mathrm{p}$ (where $\alpha=0.2$).
Here, as dissociation is 20 %, I assume that amount of ammonia in equilibrium will be 0.8 and that of nitrogen and hydrogen will be 0.2 and 0.6 respectively. Then also, I find $K_\mathrm{p}$, but it differs from the initial one.
Where am I mistaken?

First solution is perfect.
The issue is in the second solution.
If we start with 1 mole of $\ce{NH3}$ then with 20% dissociation we are left with 0.8 moles as 0.2 moles react these 0.2 moles give o.1 mole of $\ce{N2}$ and 0.3 moles of $\ce{H2}$
Hence $$K_\mathrm p = \frac{[\ce{H2}]^3[\ce{N2}]}{[\ce{NH3}]^2} = \frac{[\ce{0.3}]^3[\ce{0.1}]}{[\ce{0.8}]^2}$$

That means for $2 mol$ , I should say $1.6 mol$ $NH_{3}$ remains in equilibrium ...?

@NehalSamee yes that is perfectly correct

...But, if we take that total mole at equilibrium be 100 , then equilibrium contains 80 mol $NH_{3}$ , 15 mol $H_2$ and 5 mol $N_2$ ...Then , the calculation doesn't match ...

that is not the definition of degree of dissociation degree of dissociation is % of reactant that undergoes reaction @NehalSamee

But I'm not finding any way of mistake in my argument nor in yours ...@KarmanyaGB

https://en.wikipedia.org/wiki/Dissociation_(chemistry)#Dissociation_degree read this try to apply here. @NehalSamee

@NehalSamee Karmanya's answer is right. If you lose 0.2 moles of the reactant, you get only 0.1 moles of dinitrogen.

@GaurangTandon ... OK ...Then , if I say that 20% of product remains in equilibrium , then will my argument be applicable ?

What say @KarmanyaGB .

There are two products. 20% of which one?

@NehalSamee lets start from scratch. what according to you is degree of dissociation?

@GaurangTandon ... I said , like if only 80% reactant remains at equilibrium ...

Then , mole fraction is $0.8$

@NehalSamee correct if 80% remains how much reacted?

The question does not state that 2 mol reactant is given ...

take it as X. solution is still same

Then , amount of $N_2$ is $\frac{1}{4}×20% $ ...


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London Dispersion Forces and Dipole-Dipole forces together?
How do London Dispersion Forces and Dipole-Dipole forces exist at the same time between polar molecules?

Let's take a polar, covalently bonded molecule, like $\ce{HCl}$.
In this molecule, chlorine is more electronegative than hydrogen, so chlorine attracts the bonding electrons more than hydrogen in the covalent bond. This means that chlorine will always be slightly more negatively charged than hydrogen. This gives way to permanent dipole-dipole interactions between other molecules, since slightly negative chlorine atoms will attract slightly positive hydrogen atoms in other molecules.
However, the electrons are constantly moving around both atoms in the molecule. This means that at any point in time, there may be an imbalance of electrons on one side of a molecule, meaning that side of the molecule will be slightly negatively charged, whilst the other side will be slightly positively charged. This gives way to a temporary dipole that's induced between molecules, and it works in the same way as the permanent dipole-dipole interactions, where the slightly negative side of a molecule will attract the slightly positive side of another molecule. Since this dipole is temporary, the electrons may redistribute themselves so that there's a new temporary dipole, or no temporary dipole at all.


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How can we obtain very pure hydrogen?
Very pure hydrogen (99.9) can be made by which of the following process ?
(A)Electrolysis of water
(B)Reaction of salt like hydrides with water.
As per my knowledge the answer should be (B) but I am a bit confused and not exactly sure about the answer.
Any explanation for the correct answer would be appreciated

Hi there!  welcome to Chem.SE! We require you to show your efforts on this problem. Share your thoughts on why it should be (B). Please add this to your question. Thanks!

@MollyCooL I read the answer in the book once.However I am not aware of the reason behind it.That’s why I asked this question.Can you please tell me the reason for -1 vote.Where did I go wrong?

It must’ve been because your question didn’t fit into the rules. The one who downvoted should have left a comment. But as stated above it has been put on hold clearly stating that reason.

It really depends on how you implement it.
(A):
If you use deionised water (There are also different levels of purity),add sulphuric acid to it(to make it electrically conducting) and use appropriate electrodes you will get very pure hydrogen. You just have to isolate the two electrodes, otherwise you will get a mixture of hydrogen and oxygen, which is definitively not pure and also extremely dangerous.
Just take a look at the Hofmann voltameter on how to do the spacial isolation.
(B):
The reaction of hydrides with water generates also pure hydrogen, but you still have to use deionised water and (sometimes pure) hydrides (they need not to be pure if the impurity doesn't liberate a gas in any way in any reaction).
You should take into account that hydrides are very reactive and thus dangerous.
I would prefer (A) because deionised water is cheaper than hydrides ( pure or not) and this way is safer.
The purity of the two means is (if correctly implemented) very high, there is no major difference in purity.
So which way gives the most pure hydrogen depends on some parameters you didn't specify in your question.

What about water vapour which may be formed due to the heat of reaction?

Good point! I think it can be prevented, by only adding small amounts of hydride at once or cooling the whole thing.

You could add a cold trap or several cold traps. If you used liquid helium or liquid hydrogen to cool it with several cold traps in a series you would freeze or condense every other gas first before the hydrogen liquifies as well. Perhaps, as hydrogen adsorbes to some metals you could filter everything else of by temporarily binding it to some metal sponge and the releasing it again.

@EashaanGodbole It would be present in both cases and the gas should be dried before use. IRL pure hydrogen is usually produced with electrolysis because it is cheaper (and dirty hydrogen is produced via methane partial oxidation)


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Is there any obvious relation between sodium bicarbonate and ammonia; (how) can sodium bicarbonate be involved in producing ammonia?
Using sodium bicarbonate as a deodorant sometimes can produce a strong ammonia smell. Is there some obvious chemical explanation for this?

Ammonia is present in variable amounts in perspiration, especially after exercise, but since the pH of perspiration is usually acidic (4.5 - 7), the amount of free NH3 is small. Adding NaHCO3 can raise the pH to 8.5 and  free up NH3 from its ammonium (NH4+) ion.


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What is the difference between dissolution and absorption?
Let us first look at the definitions from Gold Book:
Absorption:

The process of one material (absorbate) being retained by another (absorbent); this may be the physical solution of a gas, liquid, or solid in a liquid, attachment of molecules of a gas, vapour, liquid, or dissolved substance to a solid surface by physical forces, etc

Dissolution:

The mixing of two phases with the formation of one new homogeneous phase (i.e. the solution).

I've re-read them again and again, but have been unable to come up with the terminology to exactly differentiate between the two. In fact, I've reached some interesting counter-examples.
Here's one: we always says "sugar dissolves in water" and never "sugar absorbs in water" (Ngram). But, going by the first definition which reads: "physical solution of a gas, liquid, or solid in a liquid", can't we say that the physical solution of solid sugar in liquid water is an example of absorption as well? The water is retaining sugar particles in its bulk after all, so why couldn't we call water an an absorbent and sugar an adsorbate?
Another one: we say that "cotton absorbs water" but never "cotton dissolves water" (Ngram). But, going by the second definition: "mixing of two phases with the formation of one new homogeneous phase (i.e. the solution)" can't we say that the solid cotton mixed with liquid water to form a new homogenous phase? After all, the wet cotton needs to be dried to separate the two components, so they clearly are homogenous.
Interestingly enough, the Gold Book definitions seem to have been created to deliberately introduce confusion. Notice that the definition for "absorption" mentions both the words "solution" and "dissolution". So, why did the Gold Book need to refer to dissolution to explain absorption?
Dissolution and absorption definitely are different from each other, otherwise some explicit relation should've been mentioned in the Gold Book. For example, the definition for "chirality centre" reads "A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element" clearly highlighting their relation (the former is a superset). Such isn't a case in the definitions for dissolution and absorption. So, they are different.
So, to summarize, my questions are:

What exactly is the difference between absorption and dissolution? And, as a direct consequence, given a physical process: how would I identify if it is a dissolution or absorption?
In the two examples (sugar and cotton) I highlighted, where is my reasoning wrong in applying "dissolve" in place of "absorb" and vice-versa?
Is there any subset/superset relation between absorption and dissolution? If so, how's that apparent from the Gold Book's definition?
Why did the Gold Book need to refer to dissolution to explain absorption? Couldn't it have explained them both in separate terms without creating confusion?

"Absorption" is for chemical engineers.

A bathing sponge absorbs water, but we cannot say it dissolves water, nor water dissolves the sponge. OTOH, liquid absorbs water vapor or gases, but it also dissolves them. I would say dissolution is a very specific kind of absorption.

Casual use of "absorption" is different from the normal, technical, use of the term
The key difference between "absorption" and "solution" is confused in your examples because you use the casual use of the term not the technical one. Even so, the two processes are very distinct.
Solutions are homogeneous. But when things are absorbed, either in casual use or technical use, there are distinct phases. Absorption is usually a technical term for a surface phenomenon where, for example, gas molecules stick to the surface of a catalyst. This is a widely studied process as it is a vital part of many major industrial processes. Water absorbed onto cotton wool is a more casual use of the term absorption but the result is still not a homogeneous mixture. Liquid water sticks to the surface of the fine fibres of cotton wool by surface tension. There are still two phases: liquid water and solid fibre.
Only in the most careless and casual use should you ever describe the process of dissolving sugar in water (or sugar taking up water) as one "absorbing" the other.


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Force Constants for Angles From Gaussian To AMBER: which value should I put in the matrix equation?
I am trying to make a new force field in AMBER for a novel compound including boron (which does not exist in GAFF). For this, I calculated freq and I found bond force constants but I couldn't find force constants for angles.
For example I tested $\ce{H2O}$: I calculated $\ce{H2O}$ frequencies in Gaussian and I got force constants from the log file. In the log file, the $\ce{O-H}$ stretching force constant was $\pu{9.1465 mDyne/\require{mediawiki-texvc}\AA{}}$ and I converted this to approximately $\pu{1315.65 kcal mol^{-1} \AA{}^{-2}}$. In AMBER, $v = 0.5k(l-l_{o})$, so I multiplied this by 0.5 and I found
ow-hw 657.825

which is very close to the GAFF2 forcefield:
ow-hw  612.98

which can be found on line 87. However, I couldn't find a way to calculate force constants from Gaussian for angles. In GAFF2 for $\ce{H2O}$, the values are
hw-ow-hw   43.276     104.520       AMBER              1    TIP3P_water

I searched and I found a lot of answer like this, but in this answer, for the Seminario approach I do not know what value to put in the matrix equation. Could you provide an example on a Gaussian log file?

Hi all, I don't think this should count as homework. Not only because it's unlikely to be asked as homework, but because the translation of quantum chemical calculations to simple force fields is highly desirable.

I really don't know exact answer but I have an idea for you it may help: You can use some "force field developer toolkits". One of them is: VFFDT. It parametrizes (for amber forcefield *.frmcd files) bond and angle force constant from gaussian out/log file.

Could you edit your answer to include a link?

https://pubs.acs.org/doi/abs/10.1021/acs.jcim.5b00687

https://www.dropbox.com/sh/hlefz8xy8o23q44/AADZGTBN-aJ07flZ-kM0s4U6a?dl=0

@pentavalentcarbon baris's answer is true. YOu can download VFFDT from his link.

First of All thanks for your answer Erdem Caner. After your answer I found VFFDT download link and I created my *.frcmod file with this program.
Here is the VFFDT link: https://www.dropbox.com/sh/hlefz8xy8o23q44/AADZGTBN-aJ07flZ-kM0s4U6a?dl=0
After download the program be careful your OS language must be English. Then run program according to manual which is included in the link I provided. Then you can create your forcefield very easily.
For dihedral parameters use gaussian scan and amber manual7 appendix C (http://rmni.iqfr.csic.es/HTML-manuals/amber7.pdf appendix C page 287). Then combine all parameters in a *.frcmod file and then you can run MD simulation for your new molecule.

How did you get the dihedral parameters because the program couldn't build dihedral parameters


text:
Does isopropyl chloride undergo SN1 or SN2 with sodium acetate?
What's the reaction of isopropyl chloride with sodium acetate? (The solvent is acetic acid)
Is it SN1 or SN2?
My thoughts:

It's not that difficult to approach isopropyl chloride's carbon, for there isn't much steric crowding. It's just a 2 degree alkyl halide with two methyl groups. This favours SN2.
The carbocation formed by the 2 degree alkyl halide, here in this case, is stabilised by the +I effect of two methyl groups and also six hyperconjugative structures. This should favour carbocation formation, hence SN1.
Acetic acid is a polar protic solvent, and can solvate sodium acetate.
Acetate ion isn't a great nucleophile (or is it?). I'm not sure, here. I know that acetic acid is a weak acid, so acetate ion is a strong base but I believe its nucleophilicity is suppressed by solvation.
Chlorine isn't an amazing leaving group.

So, does SN1 take place or SN2? Could someone also give some experimental data to support their reasoning?

Here's something I found, it says that both SN1 and SN2 can occur. Now which one occurs here, is probably decided by the nucleophile? I'm not sure. Anyway, thought I'd put this table up here for reference:

Source:
Organic Chemistry - Clayden - Oxford University Press - 1st edition - Page 424
Edit: Someone in the comments section said that you can't predict which mechanism is major, and have to look for experimental facts. So, for the sake of the completeness, where are the facts? Could someone please post the relevant details as an answer?

A lot of related questions have already been asked, though none is a duplicate, am putting them here for the purpose of quick linking: https://chemistry.stackexchange.com/q/90534/5026  https://chemistry.stackexchange.com/q/40555/5026 https://chemistry.stackexchange.com/q/68814/5026 https://chemistry.stackexchange.com/q/88579/5026 https://chemistry.stackexchange.com/q/90013/5026 https://chemistry.stackexchange.com/q/87735/5026 have a look at them if you wish to!

Thanks, I've already had a look at them though - none have answered my question.

Might help https://chemistry.stackexchange.com/a/83120/51182

Actually in these type of reactions, where the reactant can both possibly undergo $\mathrm{S_N1} $ and $\mathrm{S_N2} $, the reactions actually proceeds through a mixture of both type of reaction mechanism. In these type of reactions, rate is related as $$\text{Rate} = k_1 [\ce{R}] + k_2 [\ce{R}][\ce{X-}]$$ So, there is also the stereochemically inverted product and also racemised product.
Fraction of inverted product is $k_2/(k_1+k_2)$ and the fraction of racemised product is $k_1/(k_1 +k_2)$. But the actual values of $k_1$ and $k_2$ is experimentally determined and probably very difficult to determine by logical approach only.

Which mechanism is major?

I think you can't determine it logically, you have to look for experimental facts.

@schrodinger_16 "Which mechanism is major?" This question is ill-defined. Which mechanism dominates is highly dependent on reaction conditions, including concentration, solvent, and temperature.

In this case the transition state of the reaction is much polar than ground state if the reaction follows the SN1 Pathway. So in presence of polar medium like acetic acid the transition state will be more stabilised than ground state and hence activation energy of the reaction will decrease. And the rate of the reaction will be faster. But if the molecule follows SN2 pathway the transition state of the reaction will be less polar than ground state. As a result in presence of polar medium like acetic acid the ground state will be more stabilised than transition state.As a result the rate of reaction  will decrease. So by analysing all aspects we can predict in acetic acid medium isopropyl chloride will wish to follow up SN1 reaction readily than SN2.


text:
What is spatially and chemically heterogeneous materials?
Follow this paper,

Structural analysis of humins formed in the Brønsted acid catalyzed dehydration of fructose
They explained "humins are spatially and chemically heterogeneous materials."
I found "spatial" means "about space" but I think this meaning doesn't correlate with this context"
In summary,
What is spatially and chemically heterogeneous materials? and How did they know humins are spatially and chemically heterogeneous materials?

Spatially heterogeneous means that the composition varies as a function of position within the material (varies "in space").
Chemically heterogeneous means that the chemical composition just varies in general. In this case, I assume they desire to emphasize that the composition varies among species of widely differing molecular weights—from large polymers down to small molecules.
So, it appears to me that "spatially and chemically heterogeneous" was used to emphasize that the chemical composition varies significantly, in particular as compared to prior assumptions; and that this compositional variability itself varies as a function of position within a given volume of the material.

In general, "spatially and chemically heterogeneous" would be rather a redundant term: a material can't really be spatially heterogeneous without also being chemically heterogeneous: if a substance is chemically pure, there is no possibility of any spatial distribution of compositional variations. So, I think the particular history of this material, which apparently had been assumed to consist only of polymeric material, is what led the authors to use the phrase.


text:
Baeyer-Villiger Oxidation
Is there any logical reasoning behind the migratory aptitude for different groups in Baeyer-Villiger Oxidation?
Like in pinacol-pinacolone rearrangement, hydride ion is a better migrating group than an aryl which is better than an alkyl group. But why isn't this kind of order in Baeyer-Villiger Oxidation?


text:
Why does radical chlorination and bromination of propane occur at different positions?
When propane reacts with chlorine, we will mostly (theoretically) get an isomer with a chlorine on the end. ($\ce{CH_3-CH_2-CH_2Cl}$). However, when propane reacts with bromine, we will mostly get an isomer with bromine attached to the second carbon ($\ce{CH_3-CHBr-CH_3}$). Could anyone please explain why this happens?

https://chemistry.stackexchange.com/questions/64255/what-determines-the-high-degree-of-selectivity-in-free-radical-halogenations https://chemistry.stackexchange.com/questions/26949/what-is-the-major-product-on-chlorination-of-2-methylbutane

So it is explained with Hammond's postulate?

Hmm. Could you please expain to me how it is possible to explain this fact by using Hammond's postulate?

I guess what we have to do is to explain why bromine is more selective than chlorine.

Almost duplicate: https://chemistry.stackexchange.com/questions/33199/predicting-the-products-of-the-radical-chlorination-of-propane

The Hammond Postulate (HP): Exothermic reactions have early transition states (TS). Endothermic reactions have late TS's. [The HP may easily be demonstrated with a retractable, metal tape measure. Open several feet of the tape with the left hand holding the case and the other hand at the end of the tape. Bend the tape upward so that it looks like a TS. Raise your left hand above the right. This is an exothermic reaction. Note how the high point (TS) moves to the left. Make an endothermic reaction. Watch the TS move to the right.]
To compare the radical chlorination and bromination of propane, the following chart is instructive. Only the first propagation steps need to be examined because this is where the 1o vs. 2o choice is made. Chlorination (A & B) is exothermic (earlier TS) for both C-H bonds with the secondary more favorable by virtue of the weaker 2o C-H bond. Bromination (C & D) is endothermic (later TS) and more reversible than chlorination. The difference in the heat of reaction in both chlorination and bromination is 2 kcal/mol with the secondary radical formed in favor of the primary radical. But this equal difference does not explain the difference in the greater selectivity in the bromination vs. chlorination. The difference in activation energy (Ea) in the TS between secondary C-H bond cleavage vs. primary is greater in the bromination than in chlorination. [ADDENDUM: The ΔΔH‡ (ΔEa) for chlorination is 1 kcal/mol; for bromination, 3 kcal/mol. Therefore, bromination is more selective.][2] [continued]

The data for methane in the chart below is instructive. Fluorination has a very low Ea and it is more exothermic than chlorination. A fluorine atom is virtually indiscriminate in choosing among primary, secondary and tertiary C-H bonds. The less reactive chlorine atom has a higher Ea and shows greater discrimination. While these two halogenations are readily conducted at ambient temperature, bromination is conducted at elevated temperature to obtain useful rates of reaction. With a high Ea selectivity is the greatest with bromination.
To put a folksy spin on the discussion, a bromine atom has more "time" to make choices as to what C-H bonds it is capable of breaking while a fluorine atom is virtually a bull in a china shop.

[2]: Streitwieser and Heathcock, Introduction to Organic Chemistry, 4th ed., pg. 114.


text:
Is the shorthand electron configuration of Br- just [Kr]?
Is the shorthand electron configuration of $\ce{Br-}$ just $\ce{[Kr]}$?
Also what about $\ce{Ca^2+}$—is it just $\ce{[Ar]}$? And, is $\ce{Na+}$ just $\ce{[Ne]}$?

I meant is Na+ just [Ne]

Welcome to Chem.SE! I edited the post to reflect your change; you could have edited it yourself, though -- see the 'edit' link in gray at the bottom-left of your post.  Please take a look at the [tour] and [help] for more information about our site and community.

You could render bromide ion that way.  But the outermost electrons in the "krypton core" are potentially reactive, valence electrons; therefore $\text{[Kr]}$ is not an accurate description of how the electrons really act.  Only the argon core acts like a noble gas, so you are better off rendering the bromide ion as $\text{[Ar]3d}^{10}\text{4s}^2\text{4p}^6$.
Cations stripped to the noble gas core, however, really are stripped to unreactive electrons, so calcium ion can accurately be rendered as $\text{[Ar]}$.


text:
Pourbaix diagram and the order of the species

Figure 1. Pourbaix diagram of iron (Source)
The line between $\ce{Fe}$ and $\ce{Fe^2+}$ is there because the potential $\ce{Fe/Fe^2+}$ is $\pu{-0.44 V}$. The same applies for $\ce{Fe^2+/Fe^3+}$ with $\pu{0.77 V}$. But $\ce{Fe/Fe^3+}$ is $\pu{-0.03 V}$, but it's not on the graph as a line.
How do I know which species have a line, and which don't? And if I'm drawing a graph, how do I know that I also shouldn't include this line?

As a hint, it's easier to understand a Pourbaix diagram (or any other phase diagram) not as a set of lines where transformations take place, but as a set of areas where one species is thermodynamically the most stable and therefore preponderant - when two neighbouring areas meet, their boundary happens to lie at the line that describes their equilibrium. However, while you can write standard reduction potentials for any reaction that you can imagine, that doesn't imply that the species involved are preponderant at both sides of the line that describes said equilibrium.

In other words, while at $\mathrm{pH=0}$, the $\ce{Fe^{3+}/Fe^{0}}$ standard potential  of -0.03V means that below it $\ce{Fe^{0}}$ is more stable than $\ce{Fe^{3+}}$, and the opposite is true above that standard potential, $\ce{Fe^{2+}}$ is actually more stable than both between -0.44V and +0.77V - which is what the area labelled $\ce{Fe^{2+}}$ represents in the diagram.

Pourbaix Diagram In Environmental Chemistry
Pourbaix diagrams are very useful to identify the nature chemical species in different pH environment.
The figure contains additional information on the stability of water. The lines at 1.23 V and 0.00 V represent the redox chemistry of water. At 1.23 V water is being oxidised to oxygen. At 0.00 V the water is reduced to hydrogen. As labelled in Fig.1, there are 3 regions where behaviour of water is represented. The middle region is where the water in "stable state".
Note: These voltages are obtained at 298 K. In order to derive these lines, you can use Nernst Equation (where gradient is: -RT/nF, as the both reactions involve 2 electrons the gradient is the same in both cases.)

Figure 1: Stability of water.
Now moving on to iron. Overlapping water behaviour and iron as in Fig.2. (The dashed lines - redox chemistry of water). In the case of water, knowing the pH given you can predict the behaviour of iron. For the sake of understanding how to utilise the Pourbaix diagrams.
Assuming you are analysing alkaline lake (high pH: 10-12) and assume there is not much oxygen content (to utilise also the y-axis, low oxygen level - E decreases). In Fig.2, the region of your interest is pH 10-12 and since low oxygen, and region around 0.00 V. The Pourbaix indidicates that at these conditions you can have some amounts of Fe(OH)2, Fe(OH)3.

Figure 2: Analysis of Iron species in 1 M of water (298 K).
To see different behaviour of other species:
http://www.crct.polymtl.ca/ephweb.php

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text:
How to write a reaction that you have never seen?
I have been studying chemistry for a while but I feel a total noob about it, so please be clear when answering.
I was studying and I encountered this exercise:

Write the reaction between copper and concentrated nitric acid.

Now, I know what copper is and also what nitric acid is, but how do I figure out that the reaction gives $\ce{Cu(NO3)2 + NO + H2O}$ ?

You could only partially figure  it out. You need to know when nitric acid is reduced to NO, or NO2, or even N2O.

related https://chemistry.stackexchange.com/questions/59059/how-to-write-reaction-equations-for-the-reaction-of-calcium-and-dilute-nitric-ac https://chemistry.stackexchange.com/questions/27357/why-does-nitric-acid-release-hydrogen-only-when-reacting-with-mg-and-mn

even more https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri?s=1|47.0736

Possible duplicate of How can I predict if a reaction will occur between any two (or more) substances?

There is no general method for predicting the products of a chemical reaction, given only the starting materials and no other information.
You will learn certain trends, such as, for example, the increasing reactivity down the Group 1 metals with water, or dilute acids, but this often isn't enough to generalise to more complex situations.
You'll also learn about relative reactivity, that allows you to predict that, for example, although (because zinc is more reactive than copper) zinc reacts with copper sulphate to form zinc sulphate and copper, copper does not react with zinc sulphate in the same way.
Often, even after the products of a reaction are known, it can still take a good deal of research to establish exactly how it took place - how the starting materials interacted to form intermediates, for example, and how those intermediates interacted to form the products.  This is known as the reaction mechanism, and although many reaction mechanisms are known, plenty are not.
You should continue to study, and learn as much as you can about the principles of chemical reactivity, and you'll find that in more and more cases, there are patterns that let you predict what will happen in similar cases.  But there are often exceptions, that are only explainable once you have an in-depth understanding.
There is often no alternative to actually doing the reaction, and seeing if what is predicted actually happens.  This is why chemists do research!

I see, then I should not blame myself for not being able to solve this exercises since a lot of experience (that is what I am lacking of) is required. Thank you for your answer!


text:
How to balance this manganese peroxide reaction with hydrogen peroxide?
I have the following equation, already written, with a few coefficients to write.
$\ce{2MnO4^- + a H2O2 + b H+ -> 2Mn^2+ +c O2 + dH2O}$
I have put a = 1, b = 6, c = 3 and d = 4.
The number of atoms and the charges appear balanced to me. It is wrong, so where am I mistaking?
EDIT:I have just noticed that the exercise says "under acidic conditions": may it help?

If they were expecting a unique set of coefficients, there is some bad news.  See the answer(s).

@OscarLanzi Yes, they expected just a set of coefficients; that's what I can guess from the answers who only gave me a=5, b=6, c=5 and d=8.

The solution is not unique.
$\ce{2 MnO_4^- + 1 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 3 O_2 + 4 H_2O}$
And
$\ce{2 MnO_4^- + 3 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 4 O_2 + 6 H_2O}$
And
$\ce{2 MnO_4^- + 5 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 5 O_2 + 8 H_2O}$
You really have two reactions:
$\ce{2 MnO4- + 6 H^+ -> 2 Mn^2+ +} \frac 52 \ce{O2 + 3 H2O}$
And
$\ce{H2O2 ->} \frac 12 \ce{O2 + H2O}$
The coefficients you get by adding these together depend on how much permanganate and hydrogen peroxide are assumed to react.


text:
DFT calculation of solids with different periods
I would like to simulate a layer of graphene on a solid substrate, say Cu or quartz. I'm interested in their binding energy and other physical or chemical properties. However, their periods do not match each other, so if I want to use some standard DFT package like VASP or Quantum Expresso, I cannot use a single period for the whole supercell. I had some experience with molecules absorbed on some specific surface, which is an easier case because the period of the supercell is just that of the substrate.
Does anyone have experience with this situation? Thanks.

I have found a useful reference here
Basically the idea is to avoid the mismatch by best fitting their primitive cell unit vectors.

Another useful reference may be found here. The authors model MoS2 monolayers on a range of periodic support materials and therefore have to address this issue as well.


text:
The reactions of grignard reagent
Question

Find y in this question :

The $\ce{ -CH2-Cl}$ below according to me will not take part the reaction (or please correct me) thus my answer is y=2 but actually it's y=3

This is surely from an older version of MS Chauhan. I remember this being an error in the book which was corrected in the recent edition.

The third Grignard reagent molecule could extract a proton from the methylene group which is in an electron withdrawing environment.  This type of enhanced acididity on carbon is referred to by the term "carbon acid" .
While this reference deals with carbonyl groups, conjugation with a phenyl ring can do the same thing to the methylene group.
We do not see any change of the methylene group in the final product because it's just deprotonated, and the acid workup brings the proton back.  We would, however, see the third Grignard reagent being protonated to its parent hydrocarbon, in this case methane which bubbles off.

What about the third equivalent of MeMgBr coupling with the benzyl chloride to produce an ethyl group. Alternatively, ethoxide may cause SN2 displacement of chloride or, less likely, the intermediate magnesium alkoxide of the product may displace chloride and effect polymerization.

Possible, but apparently not considered by the text.  Grignard reagents are known to preferably act as Bronsted-Lowry bases instead of nucleophiles at the drop of a hat, and conjugation of the methylene group with an aromatic ring drops the hat.  The resulting deprotonation would kill the coupling.  Here It is noted that benzyl chloride +Mg gives 98% Grignard reagent and only 2% coupling.

If allyl MgX is prepared under standard conditions, the yield is modest owing to coupling. Must be run dilute.


text:
Why is benzenediazonium fluoroborate stable at room temperature and insoluble in water?

I think that as it is diazosalt it shouldn't be stable at room temperature. But it is stable. How is that possible?
And it is ionic hence insoluble in water.
Why?


text:
Why does the SnAr reaction proceed by the Meisenheimer complex?
I am studying SNAr mechanism from this webpage. I am confused as to why this convoluted mechanism occurs. It seems as though we simply create an anion when the hydroxide attacks in order to quench it directly after by pushing out the chloride. Why does the hydroxide not push out the chloride in a typical substitution reaction? Also, what evidence do we have for the Meisenheimer complex actually forming?

Basic level explanation:
SN2 occurs only at sp3 carbons, where you can get backside attack.
Intermediate:
Addition to the double bond generates a highly stabilized anion. This is a lower energy transition state than the concerted SN2.
Advanced:
The reaction occurs at the site of maximum frontier molecular orbital overlap. The HOMO (highest occupied molecular orbital, the hydroxide nucleophile) adds to the LUMO (lowest unoccupied molecular orbital,  π* of the arene). To do SN2, one would need to interact with σ*, which is not the LUMO, and has an unreachable trajectory (points towards the middle of the ring).

Well, you do not typically find substitution reactions occurring at olefins. Think about it.
You would come up with a sort of situation where the 'leaving group' would be a lone pair of electrons (C-C $\pi$ electrons). This has been done with organometallic reagents like lithium organocuprates, involving oxidative addition and reductive elimination sequences. But that is typically not the nucleophilic substitution that organic chemists like to think about.
So, in this case, 'pushing out' the ipso chloride in an SN2 fashion is not an option. In fact, the ipso position will often have a fluoride to stabilize the Meisenheimer complex.
And because this is an equilibrium process, you often find that the reaction is being driven forward by some means, because a lot of the intermediates are pushed back to the starting material. In this particular case, I think deprotonation of the final product (p-nitrophenol) would be a suitable way.
As for the evidence of existence of the Meisenheimer complex, I'll point you to the first compound/intermediate here. The steric factors of the ethoxy groups prevent the collapse of the adduct.

Oh I see. So is it because the Sn2 transition state is sterically hindered by the benzene ring that it cannot proceed?

No it's got nothing to do with sterics. As I said, normal Sn2 on an olefin is simply not done. It's orbitally unfavorable (no $\sigma^*$ and $\sigma$ overlap between the electrophile and the incoming nucleophile). As for the isolated intermediate, the ring does not even remain planar due to the interactions between the o-nitro and ethoxy groups.

I've just been reading an article in the September 2018 issue of Chemistry World "Textbook aromatic substitution mechanism overthrown". Apparently there's a recent paper (I haven't read it yet, but I intend to) claiming that in most cases the reaction is concerted, without formation of a discrete Meisenheimer intermediate. (Ref E E Kwan et al, NAt. Chem. 2018, 10, 917 (DOI: 10.1038/s41557-018-0079-7)


text:
What is "time-dependent evolution of the complex noncovalent interaction networks" and why this is important?
From Maximilian Scheurer, Peter Rodenkirch, Marc Siggel, Rafael C. Bernardi, Klaus Schulten, Emad Tajkhorshid and Till Rudack Biophysical Journal, Volume 114, Issue 3, 577-583:

One of the most relevant molecular properties that can be investigated
  by all-atom MD simulations is the time-dependent evolution of the complex noncovalent interaction networks governing
  such fundamental aspects as molecular recognition, binding strength, and mechanical and structural stability.

What are noncovalent interaction networks? And why their time evolution is important in structural biology. I tried also to use the authors' library pycontact, but it's not clear to me how to use it.

@Mithoron, sigh fancy name for nothing new. I don't get what is so special about this approach. Applying a threshold to a pidst 2D matrix, ins't it trivial thing to do?

@Mithoron why the downvotes? Is the question not legitimate?

@ToddMinehardt, as opposed to grilled cheese an "interaction" based on contact map is a bit arbitrary selection. For example 2 amino acids are 0.2nm apart do they interact what if they are 0.21nm what if the threshold is 0.3nm and they are 0.30002nm apart have they stopped to contact?

This is not homework; please don't flag it as such.


text:
What are the consequences of having an element with an atomic number above 137?
In other words, is an element such as Feynmanium the last chemical element that can physically exist on a planet such as Earth?

137 is in no way special.

I think he is referring to a predicted limit for Bohr atoms. When you look at the shell model of atoms 137 is the limit, because 138 is supposed to have electrons that are moving faster than the speed of light. I think when you add modern quantum mechanics you can show that the number should be higher but I can't find that reference anymore.

Almost duplicate: https://chemistry.stackexchange.com/questions/8073/the-last-elements-atomic-number

The critical value "137" occurs in calculations based upon relativistic theories that treat the nucleus as a point, such and Dirac theory or Sommerfeld theory.
However, when the nucleus is correctly treated as having a non-zero size, a significantly higher value is calculated.
The most recent calculations I see are in New method for solving the $ \bf Z > 137$ problem and determining hydrogen-like energy levels (2014) Phys.-Usp. 57, 189.  (alternative link to pre-print):

Following the development of the Dirac theory in 1928, Dirac [reference 2], Darwin [reference 3] and Gordon [reference 4] obtained expression (1) as a result of exact solution of the Dirac equation in the Coulomb field of a point charge (-Ze). ...  Formula (1) [results in a complex number] if Z [is greater than 137]...the complexity of energy levels in (1) is often called the “Z>137 catastrophe”.
In 1945, Pomeranchuk and Smorodinsky [6] considered
an atomic system with [the nucleus being a sphere of radius $0.8 \times 10^{-12}$cm and having uniform charge within the nucleus and found 175 to be the critical value].

The article goes on to explain that since the 1945 calculation, various calculations, with various nuclear charge distribution models, have calculated critical values in the 170-175 range.  The article's own calculations find 178.

I believe you are overthinking it, and that by a wide margin. The "critical" value 137 occurs in calculations based upon non-relativistic (and not even completely quantum) Bohr theory, when you equate the electron velocity and the speed of light. Refuting it does not require anything nearly as sophisticated. Its reappearance on a different level of theory is hardly relevant, though of course not accidental.

@IvanNeretin I think I'm biased because of a previous question that asked whether the 137 still occurs in relativistic theories.  https://chemistry.stackexchange.com/questions/41645/relativistic-effects-in-element-137-feinmanium-and-above

@IvanNeretin https://chemistry.stackexchange.com/questions/41645/relativistic-effects-in-element-137-feinmanium-and-above?noredirect=1&lq=1#comment67038_41645


text:
In what approximations, if any, can we assume a constant temperature in a Bomb calorimeter?
In most, if not all, general chemistry books, you will find that in from constant volume calorimeters, $\Delta H = \Delta U + RT \Delta n_\mathrm{g}$, which is, of course derived from $H = U + PV$.
My question is, given that the measured quantity is the change of temperature of the calorimeter, how can we assume that $T$ is constant to get to $\Delta H = \Delta U + RT\Delta n_\mathrm{g}$?
Edit:
As said in the comments, I found this link where it says that this approximation is valid for a small $\Delta T$ value. But if we have the values, why not just use them to get a more accurate number? Furthermore, for example, let's assume an experiment we measure a $\pu{10 kJ mol-1}$ heat from a $T_\mathrm{i} = \pu{298 K}$ to $T_\mathrm{f} = \pu{303 K}$, and a $n_\mathrm{i} = 0$, $n_\mathrm{f} = 2$. Then the approximation gives:
$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$
$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(2)(\pu{303 K}) - (0)(\pu 298 K)\right] = \pu{-4.96 kJ mol-1}$$
And the error is, indeed, small ($\pu{80 J}$) (side question, isn't $\Delta n$ supposed to be in moles? my dimensional analysis doesn't match if it does). However, if the reaction was from $n_\mathrm{i} = 2$ to $n_\mathrm{f} = 4$, and we keep the other values, we get:
$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$
$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(4)(\pu{303 K}) - (2)(\pu 298 K)\right] = \pu{-4.88 kJ mol-1}$$
And the error is, yes, still small but it doubled (of course). So I'm guessing the $\Delta T \approx 0$ to be valid only for small values of $\Delta T$, but also dependent on the actual values for $n_\mathrm{i}$ and $n_\mathrm{f}$. Is there a rule I'm missing?

On the contrary, I have a book that writes $\Delta U$ is the enthalpy change at constant volume, thus indirectly skipping the very problem posed in your question. But it does seem right given that $\Delta U=nC_\mathrm{v}\Delta T$. I am not sure though.

How can $\Delta U$ be equal to $\Delta H$ at constant volume, if there is necessarily a $V\Delta P$ term in the definition of $\Delta H$? I agree $\Delta U$ is the heat at constant volume, but the definition of enthalpy is at constant pressure.

Indeed, I do not know :( I am a sad victim of the frequent misuse of the terms "heat" and "ethalpy" in my textbook.

Here says that as long that $\Delta T$ is small, this is a decent approximation. I believe that, but why not just use the $\Delta T$ value if it's known?

But then what's the right one? And goes back to, if we have the data, why don't we calculate the more precise one?

PS: "(side question, isn't Δn supposed to be in moles? my dimensional analysis doesn't match if it does)" That $\pu{-10 kJmol^−1}$ has to actually be $\pu{−10 kJ}$; internal energy has the SI unit Joule (related

did you ever get an answer to this question?

You have more or less answered your own question. Enthalpy is a function of state, so you can write an expression of the form $\Delta {H} = {H_f}-{H_i}$, and we know that this depends only on the initial and final states, and not the path taken between them.
You also have the relation, $H = U + PV$, using this and the previous equation, and the fact that $U$ is also a state function (so we can write $\Delta {U} = {U_f}-{U_i}$)
$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i \tag{1}$
One comment, in the link that you posted equation (6) namely,
$\Delta {H} = \Delta {U} + \Delta(PV) = \Delta {U} + P\Delta V + V\Delta P $ seems absurd, for finite, non-infinitesimal differences.
For infinitesimal changes, i.e if we write the differential
$\mathrm{d}H = \mathrm{d}U + \mathrm{d}(PV) = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P $ this holds true; (chain rule from mathematics)
Anyway, for solids in liquids, the $PV$ term can be neglected on account of their small molar volume. I guess this would be one approximation/assumption you are interested in.
Second, assumption/approximation would be that a gas behaves ideally, and thus follows the equation of state: $PV = nRT$
Plugging it into (1);
$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i = \Delta {U} + R(n_fT_f-n_iT_i) $
This expression is true for finite differences. Now, the finally approximation is indeed $T_f \approx T_i \equiv T$, i.e $\delta T$ is "sufficiently" small, then indeed we can write
$\Delta {H} = \Delta {U}+ RT\Delta n $
Note however, that we don't necessarily require $\Delta T \to 0$ in a strict, mathematical sense. What can be considered "sufficiently" small depends on your system, your problem, and the kind of solution you need.
What I mean to say is in some cases/applications $-4.96\ \mathrm{kJ}$ is just as good as $-4.88\ \mathrm{kJ}$
There are many other simplifying assumptions in calorimetry when it is taught such as heat-capacities are independent of temperature for example; this only, really true over narrow ranges of temperature and in general, heat capacity is in fact a function of temperature.

With the second edit, you're basically wavering towards answering your own question i.e. the approximation is wrong and may not always yield the right result. $\Delta T$ is not always small, and when the above formula if used for large $\Delta T$, the answer may end up being way off.

"Why don't we calculate the more precise one?"

It doesn't matter. You would also have encountered a similar situation in ionic equilibrium calculations (the most popular neglection is taking $K_\mathrm{a}=C\alpha^2$ instead of the actual $K_\mathrm{a}=\frac{C\alpha^2}{1−\alpha}$) Ultimately, it boils down to how many decimal places you want in your calculations. In competitive exams, time is more important than accuracy, so a few digits off doesn't matter for a reasonable estimate. That is probably the reason why the books use the neglection in this formula


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Why do diamond and graphite have different structures, when they are both composed of carbon?
Both diamond and graphite are covalent networks and are both made entirely from carbon, but why does diamond have a three dimensional network of strong covalent bonds which makes it hard, whereas graphite has flat layers of carbon atoms which makes it a weak object and breakable.
This seems strange to me, because they both are made of the same atoms and both are covalent networks.

Why do USA and Cuba have different governments, when they are both composed of people?

Diamond has $sp^3$ hybridisation where as graphite has $sp^2$ hybridisation.

@IvanNeretin people are not identical, but carbon atoms are identical, that's why te OP is confused. The answer is because: 1) they were formed in different conditions, 2) the convertion from one to another is very slow..

@santimirandarp OK, forget people, look at two very different buildings built from similar bricks. Identical parts do not imply an identical whole, not even tentatively; that's my point.

Many elements exhibit more than one stable crystal phase as a function of temperature. For carbon, the diamond phase is never stable at normal pressure, regardless of the temperature. Mn exhibits 4 different crystal structures before melting. Pu has 6. Many elements have stable high pressure phases. Nothing particularly special about carbon.

Why are you asking this question? When a simple Google search would give you like a thousand answers have you researched this at all? Or maybe you know the answer but your just looking for points. This is far too basic to be on the Chemistry Stack Exchange, surely...

@santimirandarp : OP doesn't specify isotopically pure carbon...

@IvanNeretin : Perhaps LEGO blocks would be an ideal analogy.

Why do humans and ice cream have different structures, when they're both composed of mostly hydrogen, oxygen, carbon and nitrogen? (Hint: It's not because of the small amounts of other elements mixed in. Actually, the elemental compositions of humans and ice cream should be pretty similar in most respects.)

Chemical structures are a tradeoff of several factors, including the conditions on how they were formed.
The stability of any given chemical structure depends on the ease with which any specific reaction can turn it into something else. Both graphite and diamond are very stable structures which basically means they are hard to easily convert into something else. Another way to think about this is that to convert either into a different allotrope (different versions of the same element) of carbon requires a lot of bonds to be broken and rearranged and this requires a big input of energy. That's why both are stable.
But why is there more than one stable allotrope of carbon? Note that there are more allotropes than just graphite and diamond: buckminsterfullere is also a stable allotrope of carbon with a completely different structure that involves molecules of 60 carbons not a covalent network solid. The basic answer is that there are a lot of ways to arrange the atoms to give a stable structure (partially because there are a lot of options for stable C-C bonds given the chemistry of carbon). This is related to the same reason that organic chemistry is a big subject: C-C bonds are very "stable" and can be organised to give a huge variety of structures.
Not every arrangement of C-C bonds is equally stable: many will fall apart into other structures because there are mechanisms that make the rearrangement easy. But there are no easy mechanisms for rearranging the bonds in diamond or graphite into something more stable. Buckminsterfullerene is only formed from a plasma of carbon atoms when it is allowed to cool: you can't just make it by simple routes from other allotropes of carbon.
Both graphite and diamond have very similar energies of formation at normal conditions, though graphite is marginally more stable (the particular arrangement of bonds has a slightly lower energy than the arrangement in diamond under normal conditions). But this is not true at all conditions. Under extreme pressures, diamond is more stable. And this is how diamond is formed: it crystallised from solution in hot melted volcanic rocks deep in the earth containing lots of carbon and under high pressure. In these conditions, diamond is the more stable way to arrange carbon atoms and the conditions are such that the less stable form will covert to the more stable form because there is enough energy around to break the C-C bonds until the most stable structure is created.
Normally (by which I mean at atmospheric pressure) carbon rich compounds can be pyrolised to give pure carbon in the form of graphite. Here the heat drives off the non-carbon atoms and the carbon atoms rearrange to the form stable at lower pressures.
Once formed neither major allotrope is easy to convert to the other without extreme conditions. Which is why you can wear diamond jewellery without constant anxiety.
In short the stable allotropes of carbon are a function of the way they were made. And there are a variety of ways to join carbon atoms up to give different structures. Once formed they don't easily interconvert. Hence we find a variety of allotropes depending on the variety of conditions that led to their formation.
NB carbon isn't the only element with multiple allotropes. Tin, for example, changes crystal structure and properties depending on the temperature (and, unlike carbon, this interconversion is fast enough to have been a problem in the era when tin cans were actually made of tin and stored in very cold conditions).

Good explanation, but the tin cans part is misleading.  Tin cans are actually steel cans with tin plating.  The tin plating still may crumble if cooled, but the effect is to render the steel more prone to corrosion over time, not a catastrophic failure.

Nice answer..Also sulphur has many allotropes

@OscarLanzi That's what some cans are made of today (some are aluminium). But the original cans were made of tin.

Another relevant issue: "stable" is in most cases relative to time scale and precise definition of the system in question. A diamond sitting in an idealized heat bath at room temperature and under 1 atm of inert atmosphere will, according to thermodynamics, spontaneously convert to graphite eventually, because graphite has slightly less free energy. But the time scale is so staggeringly massive that no one cares.

There are various stable structures for pure carbon.  As the 1st answerer said, we now know that Buckyballs and Carbon Nanotubes are other pure carbon structures.  Carbon nanotubes come in both single-walled and multi-walled varieties and can be both conductors and semiconductors [https://www.sigmaaldrich.com/technical-documents/articles/materials-science/single-double-multi-walled-carbon-nanotubes.html].
So, graphite forms sheets of graphene, 1 atomic layer thick, with the carbon atoms in a flat hexagonal lattice. There are also interactions between the graphene layers due to stacking energies. Hexagonal lattices were by far the most common form of atomic packing seen in early images from the STM - the scanning tunneling microscope.
And diamond has a tetrahedral covalent packing, which is reasonable for an atom that typically forms 4 covalent bonds and is the essential element for all of organic chemistry.
This piece in Scientific American has a nice description of the structural and other differences between graphite and carbon
https://www.scientificamerican.com/article/how-can-graphite-and-diam/


text:
Truncated CI Wavefunction
In a calculation of for example a Li-Atom in a DZ basis we have 4 MO.
For the alpha string (containing two electrons), that means we have for a FCI calculation the determinants:
|12>, |13>, |14>, |23>, |24>, |34> (electrons in MO 1 and 2, 1 and 3 and so on).
If I wanted to do a truncated CI calculation (CIS for example) only singly excited wavefunctions would contribute.
My (maybe extremely dumb) question now is: Are the determinants |23> and |24> singly excited or doubly excited? Is it the excitation
$$a^\dagger_3a_1|12> = |32> = -|23>$$
 or is it
$$a^\dagger_2a_1 a^\dagger_3a_2|12> = |23>$$
 Or is it both? And the coupling (1,-1) so to say determines whether this in used as a single or doubly excited determinant?
EDIT: From the "Molecular Electronic Structure Theory" book by Helgaker, Jorgensen and Olsen, from the chapter 11.8.5 (Direct CI algorithms for RAS Calculations) I read "In CISD theory, there are three such classes (string classes): class 0, with no electrons in virtual space; class 1, with one electron in virtual space; and class 2, with two electrons in virtual space".
This for me reads as if |23> would count as a class 1 string and not a class 2.

Both. If you think about it from the context of CC theory, the first determinant will come from $\hat{T}{1}$, and the second will come from $\hat{T}{2}$ and $\hat{T}_{1}^{2}$. The prefactors handle overcounting.

As for the book statement, without any more context, I would view this as a partitioning that avoids the overcounting by simplifying the number of strings before generating the determinants.

Also, to be more exact, I know $\hat{T}{2}$ is not the same as $\hat{T}{1}^{2}$, but I don't think it's important here.

@pentavalentcarbon yes you're right, my error was in thinking it's both |23>, but one is |32> and the other |23>. Writing the question actually made myself realise my error 5 minutes later :D

But isn't that a separate issue? I have to admit that because I don't do CI/CC I can't immediately see what you're talking about. I also don't see there being a problem with with question, it's a good one.

I also don't know about the coupling, maybe if I tried to work it out...

@pentavalentcarbon Well one problem would be that it kind of answers itself.

In configuration interaction (and in coupled cluster) we typically think of excited determinant being caused by single particle excitations of the form $\hat{t}^a_i = \hat{a}^\dagger_a\hat{a}_i$ where $i$ labels a virtual orbital and $a$ labels an occupied orbital, causing an excitation of the Hartree Fock reference determinant $\Psi_0$.
The resulting CI expansion would go:
$$
\Phi_\text{CI} = \Psi_0 + \sum_{i\in occ\\ a \in vir} \hat{t}^a_i \Psi_0 + \sum_{i,j\in occ\\ a,b \in vir} \hat{t}^{ab}_{ij} \Psi_0 + \sum_{i,j,k\in occ\\ a,b,c \in vir} \hat{t}^{abc}_{ijk} \Psi_0 + \: ...
$$
Hence, in the calculation on Li mentioned above, we would not consider the excitation $\hat{t}^2_1 = \hat{a}^\dagger_2\hat{a}_1$ in the first place as it would annihilate the Hartree Fock determinant $|12>$ containing a particle in orbital 2.
By this logic the determinant $|23>$ is a singly excited determinant caused by $\hat{t}^3_1= \hat{a}^\dagger_3\hat{a}_1$, or class 1 as described by Helgaker.
Note:
As determinants $|23>$ and $|32>$ only differ by a multiplicative constant (-1) only one needs to be considered and the resulting coefficient/cluster amplitude will take care of the relative sign.


text:
Shift in the position of equilibrium by a catalyst to favor the forward reaction
We were looking at the factors that affect dynamic equilibrium today. The teacher explained how catalysts only affect the rate at which dynamic equilibrium is achieved, and not its position.
My question is: will a catalyst specific to the forward reaction shift the position of equilibrium to favour the forward reaction? Or will it, still, only affect the rate?

Wikipedia says

Catalyzed reactions have a lower activation energy (rate-limiting free energy of activation) than the corresponding uncatalyzed reaction, resulting in a higher reaction rate at the same temperature and for the same reactant concentrations.

This means that a catalyst will reduce activation energy as seen here

We clearly see that the activation energy of both forward and backward reaction equally. Hence by Arrhenius equation both the rate constants change by same factor thereby there is no change in equilibrium constant.


text:
Assignment of the 13C NMR of 2-heptanone
The 13C NMR of 2-heptanone is given below:

I understand the assignments of the carbons labelled 7, 6, 4 and 1.
My question is: Why is the shift of environment 5 higher than that of 3, despite being further away from the electron withdrawing carbonyl group?

For open chain alkanes, chemical shifts in 13C are dependent on substituents at the α, β and γ positions. α substituent effects are largely correlated to electronegativity. β substituents cause a shift of similar magnitude, but are not related to electronegativity. In fact a number of studies have shown an inverse relationship to electronegativity, but this is not consistent. The origins of the β effect are unclear. The downfield shift is larger for primary/secondary carbons than for tertiary/quaternary (hence the larger downfield shift for C-5 which has 2 β carbons). The γ effect is small and has a dependence on sterics of the bonding pathway.
The chemical shifts for straight chain alkanes can be calculated quite accurately using additive tables (such as in Pretsch, or used by Chemdraw etc). As a β substituent, an alkane -CH2- has a shift increment of +9.4, versus -C(O)- having an increment of +0.5. Hence your difference of about 8ppm for carbons 3 and 5.
From Chemdraw
Node     Shift    Base + Inc.   Comment (ppm rel. to TMS)
CH2     22.9         -2.3      aliphatic
                     18.2      2 alpha -C
                      0.5      1 beta -C(=O)-C
                      9.4      1 beta -C
                     -2.5      1 gamma -C
                     -0.4      general corrections
CH2     31.3         -2.3      aliphatic
                     18.2      2 alpha -C
                     18.8      2 beta -C
                     -2.7      1 gamma -C(=O)-C
                     -0.7      general corrections


text:
Nucleophilic addition of grignard reagent

Here why isn't the phenyl group attacking the carbonyl carbon? Isn't the first step in addition of grignard reagent the attack by the carbanion on the carbonyl carbon ?

Hmm, it should... what's the source of this picture?

@Mithoron It's from  A Guidebook To Mechanism In Organic Chemistry by Peter Sykes page 201.

The attacking of the Grignard reagent in these kinds of $\alpha$-$\beta$ unsaturated carbonyl compounds depends on two factors mainly. $1$) Hardness of the acid centre $2$) Steric repulsion between the surrounding groups and incoming nucleophile.When the attacking nucleophiles are Grignard reagents,as per your equation, they are generally hard bases.According to Pearson's HSAB Theory, Hard bases prefer to bind to the Hard acid centre and soft bases prefer to bind to the soft acid centre.  In your compound i.e in $\alpha$-$\beta$ unsaturated carbonyl compounds, the carbonyl carbon centre is the Harder acid centre, whereas the $\ce{C=C}$ double bond centre is the softer acid centre.So, Grignard reagents try to first bind with the carbonyl carbons unless there is no other destabilising factor.The second governing factor is the steric crowding, which can sometimes make the attacking nucleophile attach to the least sterically crowded region even though it is violating HSAB theory. So, the addition compound actually depends on these two competititive factors.  Now, in your given compound the attacking nucleophile is $\ce{PhMgBr}$.It is undoubtedly Hard Base. But if it were to bind to the carbonyl carbon i.e. $\ce{C=O}$ double bond, there will be huge repulsion between the lone pair of oxygen and $\pi$-electron cloud of Phenyl ring which actually makes the previously stated second factor to dominate. So, due to this huge steric repulsion, The attacking nucleophile has to attack the less sterically crowded(Note: The repulsion is also there at the double bonded carbon centre , but repulsion between methyl groups and phenyl rings are only due to inefficient spatial orientation which is definitely smaller than inter electronic strong columbic repulsion forces in close vicinity in the case of $\ce{C=O}$) softer basic $\ce{C=C}$ centre which infact gives the product.


text:
Stereoselectivity of Roush crotylboration
One can use a tartrate derived boronate to enact enantioselective crotyl transfer. Eg in scheme below:

I do not understand how the stereoselectivity arises. We have a chiral boronate, true, and we have facial selectivity on addition to boron centre. But I don't understand why one face is selected over the other - With one ester group on the ring, I could see that would sterically block one face of the boron centre, but with 2 ester groups it seems that each side of the boron is equally blocked.

This figure from this reference (p. 170). might help you visualize things.
As you can see, the stereochemistry is given by electronic repulsion between the carbonyl group and the incoming aldehyde.


text:
Predicting Ka/Kb values using values for solubility in water
Given that $K_\mathrm{a}$ values are determined by the position of equilibrium for a species such as $\ce{HCl}$ where $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$, I was wondering whether it would be possible to work out $K_\mathrm{a}$ values using 'Solubility in Water' values.
For example: $\ce{HCl}$ has a solubility of $\pu{720g/L}$ in water at 20 degrees Celsius. Is this possible or is there no direct correlation between solubility in water and $K_\mathrm{a}$?

I'm no expert, but I found this link when googling your question.

This question doesn't ask us to do OP's homework for them,  so please don't VTC it even if you don't like it.

There's no correlation.

Your example, $\ce {HCl}$, is considered to be a strong acid. Thus, it it dissociate $100\%$ in water. Thus, the molarity of the given solution (if it exists):
$$\frac{\pu{720 g}}{\pu{1.0 L}}\times\frac{\pu{1.0 mol}}{\pu{36.5 g}} = \pu{19.7 mol L^-1}$$
That means, since $\ce {HCl (aq) -> H+(aq) + Cl-(aq)}$, $[\ce {H+}]$ of the solution is $19.7$, and $[\ce {Cl-}]=19.7$ as well so you can find $\mathrm{pH}$ of the solution. But, since no $\ce {HCl}$ remains in the solution, $K_\mathrm{a}$ has no meaning.

Hey nice answer! But, isn't $K_\mathrm{a}$ defined for $\ce{HCl}$. I mean, it's of the order $10^6$ (very high), but it is defined nonetheless, isn't it? So, why do you say that "$K_\mathrm{a}$ has no meaning"?

By definition, $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$ for  $\ce {HCl}$ as you asked. But there is no $\ce {[HCl]}$ in solution because all dissociate in water. Thus, $K_\mathrm{a}$ for $\ce {HCl}$ has no meaning even though $\ce{[H^{+}]}$ has a finite value.

A simple Google search for "pKa HCl" will bring up the standard results. So, I am still not sure why you feel it has "no meaning"?

A simple Google search gives you the approximate "pKa HCl" value, which is given as $-7$, meaning $K_a=~10^7$.  That would give you the $\ce {HCl}$ concentration of said solution the value of $3.88\times10^{-5}$. I think I made my point clear now.  So long!


text:
Is the index 1 in the substituent name "1-methylethyl" a redundant index?
I want to identify the correct IUPAC name for this compound:

I would say this is 2-methyl-5-(methylethyl)-5-(1-methylpropyl)-nonane. However, the solution manual says that it is 2-methyl-5-(1-methylethyl)-5-(1-methylpropyl)-nonane.
I don't understand why the first "1-" index is necessary. It doesn't add any information, as "(2-methylethyl)" would actually be just a propyl.
Am I missing something?

Awesome question. Possible reference: http://www.acdlabs.com/iupac/nomenclature/, iupac blue book online (although they don't have the 2013 version).

The suggested name “2-methyl-5-(1-methylethyl)-5-(1-methylpropyl)-nonane” seems to be based on obsolete IUPAC recommendations from 1979. The preferred name according to current (2013) IUPAC nomenclature is 5-(butan-2-yl)-2-methyl-5-(propan-2-yl)nonane and many people would probably still call it 5-(sec-butyl)-5-isopropyl-2-methylnonane.

Related: IUPAC rules on omission of locants in carboxylic acid

I'd not subtract a point for your "mistake," because there is no point to prove you are wrong. However, suppose the group is different than methyl. Then the position matter at that point. Say its bromo-group. So, 1-bromoethyl- and 2-bromoethyl- are two different things. That's why that rule is for. Can't give any favors just for methyl! :-)


text:
Thermodynamics energy balance question help

I made the following attempt to solve this question.However I don't know how to find final pressure.I don't know how to find final temperature for both sides.Can you give me a hint?

I don't understand what does "water quality" means.

It is liquid vapor mixture .Vapor ratio is %15 and liquid ratio is% 85.@Hexacoordinate-C

The total volume is constant.  The air compression takes place adiabatically and reversibly, so you know the relationship between its pressure and its volume.
You are told the initial volume of the water container, the mass fractions of liquid water and water vapor, and the temperature of the saturated mixture.  From this, you can determine the mass of the water in the container.  What value do you get?  This doesn't change.
You are told that the final state of the water is 100% saturated vapor and 0 % liquid water.  So, from the steam tables, for any given pressure, you can determine the specific volume.  And, since you know the mass of water, you can determine the actual volume.  If you guessed the final pressure correctly, then the pressures of the water and air on the two sides of the piston must match and the sum of their volumes must add up to the original total volume.  This is how you determine the final pressure and temperature of the water vapor.
No work is done on the combined system, so, knowing the final states of both the water and air, you can calculate the change in internal energy for each of them.  The sum is then equal to the amount of heat added.


text:
What is the function of nitrobenzene as a solvent in Friedel Crafts alkylation reaction?
I would like to know what is the function of nitrobenzene as a solvent in Friedel Crafts alkylation reaction.
Answer to this question suggests that Friedel Crafts reaction is not possible with nitrobenzene and I understand that it's because of the big loss of electron density of the benzene ring due to the highly deactivating nature of $\ce{-NO2}$

The aluminium halide complexes with nitro compounds (e.g., nitrobenzene) are known to display catalytic properties in some organic reactions, including that in Friedel craft alkylations. Those complexes are believed to be more soluble in the solvent (nitrobenzene).
Reference: Russian Journal of Coordination Chemistry 2001, 27(7), 469–475.

Although I agree solubility of the $\ce{AlCl3}$ is good in nitrobenzene (they sell solutions of this), perhaps due to the formation of these complexes, I wouldn't go as far as saying that the complexes are catalytic. The referenced paper cites two references to support this claim, none of which actually say this (one doesn't even use nitrobenzene). If you have another references that shows this catalytic activity, by all means, would be very interesting to read!

@ralk912: I'm sorry for the late reply but like to verify my comment. The Russian group is talking about complexes of  aluminium halide with nitrocompounds as a whole. They claim N...O...Al complexation. I didn't go through whole paper so I wasn't sure about how specific is this claimed catalytic activity. Yet, I'm thinking to do little literature survey and put them here if I found solid evidence. Thanks for your interest.

Without any more information, it's just that, a solvent. Since it is a poor substrate for Friedel Crafts alkylations, it can be considered inert in these conditions. The choice of nitrobenzene would probably be to carry out the reaction at a relatively high temperature, since it boils at 210.9 °C.


text:
Is the new equilibrium total pressure of this equilibrium reaction always more than its initial total pressure?
In the case of the reaction $$\ce{2NO2 <=> N2O4}$$ at equilibrium and considering Le Chatelier's principle, if we add more $\ce{NO2}$, the pressure of $\ce{NO2}$ will increase acutely and then begin to decrease until the system reaches its new equilibrium. Why is the new equilibrium have a higher pressure of $\ce{NO2}$ than the original equilibrium? How can we tell this just by looking at it? Will it always be larger?

The higher pressure of $\ce{NO2}$ even after reaching equilibrium after adding some amount of $\ce{NO2}$ can be shown mathematically.
Suppose initially before adding $\ce{NO2}$,the pressure of $\ce{NO2}$ was $p^0_{\ce{NO2}}$, and that of $\ce{N2O4}$ is $p^0_{\ce{N2O4}}$.So, the equilibrium constant is $$K_\mathrm{p} = \frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}$$ Now assume that the addition of $\ce{NO2}$ just increases the pressure instantaneously to $p_x$ but immediately after that due to dissociation to achieve equilibrium, the pressure decreases by $\ce{p_y}$ atm.
$$\ce{2NO2 <=> N2O4}$$
$$p^0_{\ce{NO2}} + p_x - p_y \ce{<=>} p^0_{\ce{N2O4}} + \frac{p_y}{2}$$
So, the increase of pressure $\ce{N2O4}$ will be $p_y/2$ atm. So, now a new equilibrium is achieved where the equilibrium constant $K_\mathrm{p}$ will be still the same assuming that during addition of the gases, temperature is not changed( or the change is very much negligible in real cases). Thus, equating $K_\mathrm{p}$ on both the sides we have,
$$\frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}= \frac{p^0_{\ce{N2O4}} + \frac{p_y}{2}}{(p^0_{\ce{\ce{NO2}}} + p_x - p_y)^2}$$
Taking the squares on numerator and denominator and taking square root we will find:
$$1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}}= \sqrt{1+\frac{p_y/2}{p^0_{\ce{N2O4}}}} > 1$$
So, we see that $1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}} > 1$. Hence it is proved that $\ce{p_x - p_y > 0}$. So, the net pressure of $\ce{NO2}$ after achieving equilibrium is $p^0_\ce{NO2} + p_x - p_y > p^0_{\ce{NO2}}$. So, its pressure has increased than the previous conditions even after reaching equilibrium.

The total pressure of the system is, according to Dalton's law, $p_\text{T} = p_\ce{NO2} + p_\ce{N2O4}$. From here, $p_\ce{NO2} = p_\text{T} -
 p_\ce{N2O4}$.The equilibrium expression for this system can be written as:
$$K_\text{p} = \frac{p_\ce{N2O4}}{p_\ce{NO2}^2} = \frac{p_\ce{N2O4}}{(p_\text{T} -
 p_\ce{N2O4})^2}$$
So we get:
$$p_\ce{T} = \sqrt{\frac{p_\ce{N2O4}}{K_\text{p}}} + p_\ce{N2O4}$$
Assuming the process is isothermic ($K_\text{p}$ is constant) and isochoric, then introducing moles of a gas that participates in this reaction will increase the pressure of the system, hence $p_T$ will increase. Since $p_T$ increases, the previous equation shows that $p_\ce{N2O4}$ must increase. Since $K_\text{p}$ is a constant, the ratio $\displaystyle \frac{p_\ce{N2O4}}{p_\ce{NO2}^2}$ is constant. Hence, if $p_\ce{N2O4}$ increases, so does $p_\ce{NO2}$.

Sweet! I am just linking my attempt as well, so the ease of getting tangled in the maths (like I got :P) is clear, and as well the brevity of this marvelous solution is more visible!

Yeah I started there, but once I got to the double quadratic with four unknowns, it became apparent it was very likely to lead nowhere. PS. There's a mistake on your quadratic formula, I think

I solved it using WolframAlpha, where's the mistake? :/

Oh well, then it probably it's more efficient, I was just trying to solve by hand. I trust WA :)

Thank you! This makes a lot of sense! I did not realize that pressure and concentration themselves did not affect K, so K is constant in an isothermic process and therefore the pressure of NO2 must be higher than its original value.


text:
Inorganic Chemistry: the order of MO diagram
I understand that to create a molecular orbital diagram, the following rules are supposed to be met:

number of A.Os = number of M.Os
Symmetry of the binding orbitals
Similarity in energy

They will eventually create bonding, anti-bonding, and maybe non-bonding orbitals. However, I am having trouble stacking them up in a correct order. For instance, if you refer to the attached diagram below, I don't know how it first starts with 4a1' and goes up to 5a1'. The only clear ones for me are the non-bonding orbitals, which have the same energy as the original group orbitals.

Could you please help me understand this? The coefficients that come in front of letters are also confusing me as well...
Thank you so much!

The numbers that come in front of each molecular orbital irrep are simply referring to the energy order in which the MO's are arranged, with the maximum number (for example, $5a_1'$ for $BF_3$) corresponding to the highest energy MO that has an $a_1'$ symmetry.
For understanding the molecular orbital energy order, you need to explicitly consider the combination of the fluorine ligand group orbitals and the boron AO of the same symmetry. If you do that you'll find $4a_1'$ has perfect bonding interactions between all the ligand p-orbitals involved in $\sigma$-bonding and the boron s orbital (both transform as $A_1'$ in $D_{3h}$). The identity of the p-orbitals would depend on the co-ordinate axis you choose to define as the molecular axis. $3e'$ would arise because of $\sigma$ interactions between the same ligand p-orbitals, this time slightly differently oriented and the boron p-orbitals. The $a_2''$ MO is pi-bonding in nature, so on and so forth.
But without further ado, I'll point you to this excerpt from Purcell and Kotz's Inorganic Chemistry, which is self contained and provides a detailed pictorial description of the various ligand group orbitals very nicely.
Best of luck!

Thank you so much. You confirmed some ideas I had in mind. Have a wonderful evening.

You're welcome! And have a nice day. (It's 10 AM here, in India :P)