text:
How to quantize the shape of a molecule?
I am looking for numbers which tell you something about the 3D shape of the molecule, and is also easily calculated.
One of these is the acentricity factor. I am aware of this one. My problem is you calculate acentricit, from thermochemical data, which are usually hard to find, and can contain a great degree of experimental errors.
A more suitable quantity would be constructed from the groups of the molecule. Is there such a thing? Or what are other quantifiers of molecular shape other than acentricit, factor?

This sounds like it could be a bit of an XY problem. What are you actually trying to do?

I am planning to relate some physical-chemical properties of large sets of molecules to their shapes. For example, trying to find a qualitative picture of what is the effect of the molecular shape on the critical temperature. I am also planning to characterize the efficiency of currently used descriptor calculation methods.

So you already know a bunch of descriptors, and we should figure out ones you may not know... Wouldn't it be helpful if you mention which ones you know? Also, it is nice to have an argument behind using a descriptor, not just some random parameter you tried and seemed to wok ona limited number of molecules.

There are lots of descriptors of molecular size and shape that come to mind:

Moments of inertia (e.g., how different are they)
Radii (e.g., enclosing the molecule in a best-fit box)
Surface area or polar surface area (there are existing group-based methods)
Volume (e.g., of the van der Waals surface)

Some people have also used 3D shape descriptors like "spectrophores".

Both the absolute magnitudes and relative values of the moments of inertia contain useful information, I think.  Absolute values give overall size, whereas the relative values (ratios close to $1$, or very different) give symmetry/asymmetry. The moments can't capture convexity or concavity, though.

I haven't seen any good measures of convexity or concavity, but those would likely be nice too.


text:
Synthesis of 3-Methylenecyclopentene from 1-methylCyclobutene
How do I make 3-Methylenecyclopentene

from 1-methylcyclobutene?

I thought to do ring opening reaction to add 1 carbon atoms but this won't add double bond in product. How to start please?

What sort of ring opening?

Here's a thought: 1) Simmon-Smith reaction, 2) bromine addition, 3) double dehydrobromination

I don't know  Simon-smith reaction. Can ring opening epoxide reaction work?

Here is my try.

Ozonolysis
Aldol condensation using a hindered base (like LDA)
Wittig reagent to convert $\ce{C=O}$ to $\ce{C=C}$

Step 1:

Step 2:
I used $\ce{LDA}$ which abstracts a proton from the least hindered carbon near to a carbonyl group. (So that the resulting carbanion is stable).
I am not drawing the enolate ion formation etc. You can find the reaction mechanism for aldol condensation here and the reaction using $\ce{LDA}$ here.

After protonation, you can heat the compound to remove $\ce{OH-}$ which is a simple dehydation. There are two possible alkenes. One of them (the one given below) is more stable because of conjugation.

Step 3:
We are almost done. To replace $\ce{C=O}$ with $\ce{C=C}$, use wittig reagent.

You'll need low temp for your intramolecualr aldol but otherwise, this is sold.

What are the chances of conjugate addition with a Wittig anion? Would Tebbe or Petasis be more likely to succeed? Well, if those things would be an issue, there's a ton of methylenation reactions.

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Why is tetramethylsilane (TMS) used as an internal standard in NMR spectroscopy?
I understand that TMS has a large proton number, giving a strong peak in the NMR spectrum. But apart from that is there any other reason why just TMS is used and not any other molecule?

In theory, you could use something like cyclohexane and get a similar result, but you have a much greater chance of overlap. Having a TMS singlet also allows you to shim your sample off the shape of the peak if you have no other strong, non-overlapping singlets. Most people just shim off the lock signal, but there are times that's not perfect.

TMS has 12 protons which are all equivalent and four carbons, which are also all equivalent. This means that it gives a single, strong signal in the spectrum, which turns out to be outside the range of most other signals, especially from organic compounds.
Although the chemical shift scales are still zeroed at the TMS peak, most spectra are now calibrated against the residual solvent peak. Typically, deuterated solvents such as $\ce{CDCl3}$ and $\ce{DMSO}$-$\ce{d6}$ are used and these contain a very small amount of undeuterated, or partially deuterated, solvent which produces a peak in hydrogen NMR. In carbon NMR the solvent peak is recognised by its splitting pattern, which is a triplet for $\ce{CDCl3}$ and a heptet for $\ce{DMSO}$-$\ce{d6}$ and the fact that it always comes at exactly the same place.

DMSO is a quintet in proton NMR, while chloroform is a singlet.  When you said heptet, I lost the carbon context, then I realized this comment should be added.

Other than for chloroform-d, it is not undeuterated solvent that is referenced, but rather the perdeuterated material. DMSO-d6 contains a small amount of DMSO-d5, which is why the 1H signal appears as it does. Undeuterated DMSO appears as a singlet. DMF is about the only common mulitply-deuterated solvent that usually contains completely undeuterated material

@long Yes you are right. I didn't think about that.

TMS is notoriously nontoxic too.

The newer IUPAC referencing standards actually do NOT reference the spectrum off the residual solvent peak.  Rather, all referencing is done relative to the 2H frequency of the solvent.  Modern instruments all use the deuterium signal from the solvent to lock the instrument (against magnetic field drift) and so this lock signal is now used for referencing all other chemical shifts (1H, 13C, 31P, 19F, etc.)  This is what "referencing by solvent" in modern NMR software is actually doing.

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was  used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one of the most common solvents used for running dilute solution samples.
TMS was proposed as a reference at the time becasue it offered the following advantages:

the chemical shifts are largely temperature and concentration independent
it offered a single phase method for referencing (That is, it was an internal standard. Most common methods at the time used an external reference, such as a coaxial insert; a method which is still acceptably used today)
The TMS peak lies outside the 'usual spectral region and is readily identifiable'
It is largely inert, and is not likely to react with most samples
It can be used for almost amy solvent (except H2O, D2O)

Today, TMS is still accepted as the single universal reference standard for NMR. For over a decade, the IUPAC recommendations have stated that, other than for 13C, all nuclei are referenced to the frequency of the 1H signal of 1% TMS solution in CDCl3. Yes, your 31P and 77Se peaks should actually be referenced to 1H of TMS. This unified reference scale is very easy to implement on modern instruments that incorporate a digital 2H lock (ask your friendly NMR custodian if they use it!). It is in fact arguably more accurate that referencing to internal solvent residues which are much more likely susceptible to concentration and temperature factors. Internal solvent residues are called secondary references, and are less accurate than the primary reference. However, they are accurate enough for the very vast majority of work done in the laboratory.
It has long been shown that the chemical shift of TMS is, in fact, both temperature and concentration dependent. But not much. There are actual corrections that should be applied to measured chemical shifts at extreme temperatures, but these are very small and rarely seen in the literature.
TMS is not water soluble, and in aqueous samples DSS is the recommended reference standard. It contains signals in the 1H spectrum at 3.1, 1.9 and 0.8ppm, all at a fraction of the intensity of the main reference signal, and these can interfere with analysis of other spectral regions. A deuterated version is widely used in the literature. An alternative, TSP, is sometimes used, although is more pH dependent. Other reference standards can also be used, however, and as long as the reference method and sample conditions are reported and verified, then all is good. For 1H NMR, compounds such as acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane are all examples of internal standards that are sometimes used, both for chemical shift referencing and concentration measurements, as they all largely satisfy that important list of reference characteristics above. They all have very simple 1H spectra, soluble in common deuterated solvents and fairly unreactive. Plenty of other compounds with multiple peaks can be used as well, such as toluene and dmf.
So, to specifically answer your question, TMS is the adopted primary standard for 1H chemical shift referencing for the reasons stated above. Any other material can be used as a secondary reference, either internal or external, provided its chemical shift can be accurately determined against the chemical shift for TMS. Results reported should include a clear description of how chemical shifts were determined. For example, 1H chemical shifts of all reported compounds were measured relative to the methyl peak of n-dodecane, determined as 0.88ppm. The disadvantages of using other materials is that it needs to be shown that the chemical shift of these signals is independent of sample conditions (concentration, temperature etc), and that they will often have peaks overlapping regions of interest.

Another reason that TMS was chosen as a good reference is that it's relatively volatile and, thus, can be easily removed from the sample later (back in the "old days" when sensitivity was very low, it was important to recover the material in the NMR tube for further analysis or further reactions).

Also, although TMS is technically the origin of the new IUPAC referencing standard, the new standard is designed to remove the need for putting TMS in your sample.  Every deuterated solvent that is in common use has had its deuterium chemical shift referenced against TMS.  Modern NMR software contains tables of these calibrated values, so that all NMR (1H, 13C, 31P, 195Pt, etc.) are now referenced directly off of the deuterium signal.  I haven't used Bruker software for a long time, but the Varian/Agilent software automatically references off the 2H signal and ignores the TMS peak.

@S.Burt - Both good comments. Yes, automatic referencing is alluded to in the answer above. And while, by definition, it is a secondary reference, it is incredibly easy to implement, especially when running obscure nuclei with rather hazardous reference materials, such as 199Hg or 205Tl. It does, however, require that the instrument is set up correctly.

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What is the chemistry behind heat produced when blood is bleached?
When I discard blood (20-40 mL) in a beaker containing bleach (400mL), I can feel that heat is produced in the beaker. What is the chemistry behind it? Will heat be produced due to Lysis of haem protein in RBCs? Or any other mechanism?

It is simply because of oxidation. Bleach is an oxidizing agent.

This reaction produces heat for the same reason as burning methane in oxygen. There's a lot of chemical energy stored in the reactants, and comparatively little in the products.
The hypochlorite is oxidising the organic content of the blood - not just breaking proteins but then breaking down the amino acids too, as well as anything else that happens to be in the sample.


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Why do some periodic tables have 18 groups while others have 8?
In school I have been taught about the periodic table with 8 groups, representing the number of electrons in the outer shell (with exceptions). But online almost all I can find is periodic tables with 18 groups. Why do both exist? (Is the one with 18 more modern?) What is the purpose of 18 groups?

The only official periodic table is one with 18 groups.

See this answer as well: https://chemistry.stackexchange.com/a/133169/72973

Unfortunately, life sometimes is more complicated than school :)
Using 8 groups works well to teach the basic principle, but higher shells have space for more electrons. (And lower shells for fewer - ever wondered why there are only two elements in the first period?)
Hence, you'll get 18 electrons in the next shell (the "transition metals"), then 32 electrons ("Lanthanides/Actinides") - the general rule is $2n^2$, where $n$ is the number of the period. This is unfortunately made even more complex, as the energies of the shells are not quite what one would expect and they are not filled "in order." One probably should stop speaking of "shells" at this point and move towards "orbitals."
Anyway, the community has pretty much settled on using 18 groups nowadays. The idea behind the periodic system is to find regularities (hence the name) between elements.
It turns out, that individual groups and trends within them are similar in the 8 "main" groups (as you have learned). In the transition metals, there are still a fair few regularities, that justify this arrangement - but they are already a lot more similar to each other than say alkali-metals and halides. The Lantanides/Actinides finally are so similar to one another, that they are "grouped" (although they in fact are part of a period) together - they in fact so similar, that they usually only occur in mixtures in nature, and separating them is quite hard.
Finally: no heavier atoms have been discovered so far that would have to go into an even bigger "shell"/orbital. Hence, one could justify up to 32 groups, but for practical reasons, 18 are conventionally used.

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text:
Le Chatelier's principle with pure solids and liquids
In the following reaction:
$$\ce{A(s) <=> B(g) + C(g)}$$
since adding a pure solid is not supposed to shift the equilibrium in any way, does that mean that I can essentially remove all of the solid except for 1 molecule, and the equilibrium will not shift at all since the concentration of the solid stays the same? Similarly, does this mean that I can essentially fill up the container with the solid, and no extra B and C will be produced?

Your first statement is right.
You just need a certain amount of solid or pure liquid for the reaction to occur, just like how putting in more solid in a saturated solution doesn't increase dissolution, while you need solid in there to call it an equilibrium between aqueous and solid phases.
However, the amount of solid or liquid does change the kinetics of the reaction: the rate at which it proceeds to equilibrium.
But if you put in some solid A, the Q is 0, so some B and C will form. It's the same reason why putting in a sugar cube in water increases aqueous sugar concentration: there is an equilibrium between solid and aqueous components, it's just that the constant means the maximum amount of aqueous concentration.


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Why does fluorine act as a leaving group in organophosphates?
C-F bond is the strongest single bond in organic chemistry with a bond energy of  $\ce{453 kJ\,mol^{-1}}$. And it is very difficult to break this bond and F does not act as a leaving group.
However, in organophosphates, F acts as a leaving group. For example, take an organophospate such as sarin(below). When this binds with acetylcholinesterase (enzyme), F is leaving as a fluoride ion. How can this be explained w.r.t. the P-F bond energy of $\ce{490 kJ\,mol^{-1}}$, which is even higher than that of C-F. Is this possible because of the catalytic activity of the enzyme?

According to David O'Hagana, Chem. Soc. Rev. 2008, *37 , 308-319, C-F bond is very strong due to the difference of electronegativities of C(2.55) and F(4.0), and the resultant ionic character of the bond. However, the electronegativity of P is 2.19 and according to this, P-F bond must be much difficult to be broken. But, this is not the reality. Sarin can be hydrolyzed (Andrew F.   Kingery  and  Herbert E.   Allen, Toxicological \& Environmental Chemistry 1995, 47, 155-184.) at pH 6 ($\ce{25\,^{\circ}C}$) with a half life of 312 hours in aqueous solution.

Well, did you at least check proper values? Bond energies vary... Not that this approach is terribly useful, as you can see.

So what approach should I use then? I tried looking for the actual bond energies, but couldn't find.

Enzymes can make miracles.

I am reluctant to post this as an answer because it is not really backed up by much research, but since nobody has said anything yet, here's my theory:

It is probably to do with the fact that the covalently bound sarin (which necessarily involves kicking out either $\ce{RO-}$ or $\ce{F-}$) mimics the transition state for acetylcholine hydrolysis too well, such that the expulsion of one of those two leaving groups is much more favoured than in aqueous solution.

Acetylcholinesterase (AChE) is very similar to a typical serine protease, and serine proteases are very well-studied. I am going to postulate that the mechanism of acetylcholine (ACh) hydrolysis in AChE is exactly the same as it is in a typical serine protease. So first, let's look at what the enzyme does normally. It's nothing special, just a nucleophilic acyl substitution:

(There's an additional glutamate residue, analogous to the aspartate in a serine protease, that fixes the conformation of His440, but it's not directly relevant to the mechanism. Also, there's an anionic site which binds to the positively charged quaternary nitrogen, but that's also irrelevant.)

The acyl-enzyme intermediate then goes on to be hydrolysed by water (not relevant here).
The reason why the enzyme can catalyse the reaction is because it stabilises the transition state for the ester hydrolysis, which means that the activation energy for hydrolysis is decreased and the rate is increased. (Strictly, it provides an alternative hydrolysis pathway in which the activation energy is lower - but the point is that since the catalysed hydrolysis is mechanistically similar to the uncatalysed hydrolysis, the enzyme must play a role in stabilising a tetrahedral intermediate or transition state.)
The transition state looks something like this (partial charges omitted):

Sarin is a potent AChE inhibitor because it covalently binds to the active serine residue as in the left-hand side of this diagram:

You can see that structurally, this is very similar to the transition state shown above. If the enzyme stabilises the TS above, this covalent binding of sarin to the enzyme must also be extremely stable. I do not pretend to know exactly how it is stabilised, but it must be related to the specific 3-D conformation of the enzyme. If you do not kick out the fluoride ion (right-hand side), then it does not quite resemble the transition state so well.
What this means is that there is a strong driving force for the fluoride ion to be expelled, in order for it to reach that stable, 4-coordinate phosphorus form.
That leaves one question that I feel is worth addressing, which is why fluoride is expelled instead of isopropoxide. Honestly, I do not know. As pointed out by jerepierre, fluoride is certainly a better leaving group than isopropoxide ($\mathrm{p}K_\mathrm{a}(\ce{HF}) = 3.17$, $\mathrm{p}K_\mathrm{a}(\ce{^{i}PrOH}) = 16.5$ - quite a large difference), and it might very well just be that. On the other hand, as with most biological systems, it is difficult to give a full explanation of why things happen the way they do, and applying (relatively) simple chemistry heuristics might not be the full picture. As to what other factors might influence this, I can only guess that is that there might be some hydrogen bonding arrangement that requires the specific size of the oxygen in the $\ce{-OR}$ group over the size of the fluorine (in the catalysis TS, there is an oxygen there as well). Essentially, I am saying that the LH diagram resembles the TS more than the RH diagram (don't quote me on it though...):

Note that, in case of tabun, the reaction also works with $\ce{CN-}$ instead of $\ce{F-}$.

"... why fluoride is expelled instead of isopropoxide." Fluoride is a better leaving group than an alkoxide.

@jerepierre Yes, I think it's not unlikely that isopropoxide picks up a proton from somewhere else when it's leaving though.

Wow It's so cool that this came up. I've been wondering the exact same thing for the last week! I asked a couple of my professors about it and this is the first time I've seen a reasonably satisfying answer.

I would point out (as did the author of the question) that sarin hydrolyzes in water in days to weeks without the need of an enzyme, and also that carboxylic acid fluorides (e.g. acetyl fluoride) also rapidly hydrolyze, despite the strength of the C-F bond. The ability of these acyl fluorides to hydrolyze is due to the relative ease of addition of water or hydroxide with formation of a tetrahedral intermediate (or pentacovalent intermediate in the case of phosphorus) - likely helped by the electron withdrawing characteristic of the fluorine atom - followed by expulsion of fluoride, which is a relatively stable anion and is a pretty good leaving group. Keep in mind that the term "bond strength" typically refers to the energy required to cleave a bond homolytically to form radical species. That is quite unfavorable in the case of P-F and C-F bonds.


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Isotopes and their percentage existence in atmosphere
When we talk about percentages of isotopes of an element in the atmosphere, is it percent by mass or volume or weight or something else?

Short answer: it's the percent by number of particles.
Longer answer: It's the proportion of particles of a particular mass for that element.  To chemists, a per particle basis is also commonly referred to as a per mol basis or a molar basis.
A way that it can be measured is by mass spectrometry - you pass the atoms through the apparatus which shows the mass to charge ratio. E.g. if you could pass $\ce{Cl}$ molecules, you'd see 2 peaks in the resulting graph - one is at 35 and the other at 37 marks which shows the mass/charge. First peak would be about 3 times larger (thus there were 3 times more particles). Thus chlorine would be about 75% isotope 35 and 25% isotope 37. So while this approach measures mass of each isotope, it also determines the relative molar amounts of each isotope.


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Reference request - Nomenclature of organophosphorus compounds?
Can someone please provide me a good reference for the nomenclature of organophosphorus compounds such as pesticides and chemical warfare agents? Something from the IUPAC would be great.
Thanks.

The usual substitutive nomenclature for organic compounds was extended to various other elements including phosphorus. The current rules for name construction of organic compounds containing phosphorus are included in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). These recommendations cover typical organophosphorus compounds that are used as pesticides and chemical warfare agents. However, they do not include the nomenclature for organometallic compounds, coordination compounds, polymers, and some natural products.


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Fraction of cube Volume empty
The unit cell in a body centered cubic lattice is given. Each sphere has a radius r and the cube has a side a. What fraction of the total cube volume is empty?

The fraction is known as the "packing fraction". https://en.wikipedia.org/wiki/Atomic_packing_factor

Assuming the $a$ is large enough to accommodate a BCC lattice with spheres of radius $r$, we may define
$V_s = \frac{4}{3}\pi r^3$ is the volume of the sphere
$V_c = a^3$ is the volume of the cube.
in a BCC lattice, there are 8 corner points, and one central. Because the lattice points on the corners are defined with respect to the center of each of the spheres, we know that each of the corner spheres will contribute $\frac{1}{8}$ of their total volume to the occupied volume of the unit cell. In addition, the sphere at the center will contribute its entire volume to the occupied volume. We may then write the total volume occupied in the unit cell as
$V_{occ} = 8\left(\frac{1}{8}\right)V_s + V_s =2V_s$
We know that the total volume of the unit cell is that of the cube, meaning that the unoccupied (empty) volume of the unit cell may be written
$V_{unocc} = V_c - V_{occ} = a^3 - \frac{8}{3}\pi r^3$
Wherefore we may write the empty fraction as
$\frac{V_{unocc}}{V_c} = \frac{a^3 - \frac{8}{3}\pi r^3}{a^3} = 1 - \frac{8\pi r^3}{3a^3}$


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Is the structure of a crystalized protein the native one?
When one finds the 3D structure of a protein by crystalizing it and then making a X-ray experiment, how does one know that the geometrical configuration of the crystal is the same (or even close) to the structure the protein adopt in aqueous solution?

Well, it doesn't need to be, be it's still something. NMR is getting structures in solutions.

@Mithoron And they're doing fancy stuff with proteins NMR, but I don't see how they would get a whole configuration. That would be amazing.

@DGS Significant percent of protein structures was obtained with NMR not X-ray. http://www.rcsb.org/pdb/static.do?p=general_information/about_pdb/nature_of_3d_structural_data.html

@Mithoron And to think I have trouble assinging something that weighs around 500 g/mol.

I'll use quotes from B. Rupp, Biomolecular Crystallography (p. 7-8) to answer.
Generally the structure is similar...

Comparison of many nuclear magnetic resonance (NMR) solution structure
  ensembles with crystallographic structure has shown that the core
  structure of protein molecules remains unchanged compared with the
  solution state during crystallization. In addition, enzymes packed in
  crystals even maintain biological activity.

with some parts not visible in the experiment...

The maintenance of the core structure and of enzymatic function shows
  that crystal structures are a very good approximation of the native
  protein solution structure. Nonetheless, highly flexible or mobile
  regions, frequently the amino- or carboxyl-termini of the protein chain or
  flexible loops connecting secondary structure elements, can be
  poorly defined or even absent in the electron density and thus can be
  modeled only with limited confidence.

but beware...

In certain situation flexible and dynamic regions of a protein
  molecule can be rigidly fixed in a specific conformation as a result
  of crystal packing interactions. In most cases this represents just a
  snapshot of one possible conformation out of many and it must be
  understood that such a specific conformation may not locally represent
  the protein structure in solution. A simple safeguard against
  misinterpretations -- which is usually assignment of certain
  biological relevance to regions where that is de facto not warranted --
  is to display all neighboring, symmetry-related molecules in the crystal
  structure and examine if any intermolecular interactions are
  present that are a result of crystal packing. Such packing induced
  artifacts can also hamper for example drug discovery by altering or
  blocking binding sites and thus preventing an otherwise active
  substance from binding.


text:
Why are hybrid orbitals more efficient at overlapping?
What is the reason for considering that hybrid orbitals provide more effective overlapping resulting in formation of stronger bonds?
Will this reasoning be correct that the hybrid orbitals have larger and smaller lobes than pure orbitals?

It is not so much about the size of the lobes but more about the direction. Hybrid orbitals usually have an orientation that allows them to overlap with the orbitals of their bonding partner much better, i.e. in a head-on manner.

Hybrid orbitals exist only in our head. It would be wrong to perceive an atom as a kind of amoeba which grows this or that kind of pseudopods (=orbitals) as it sees fit.


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Conversion of 4-nitrotoluene to 2-bromobenzoic acid

In the above conversion why doesn't bromine add to methyl group? Is the first reaction following free radical or polar reaction (Friedel-Crafts). If it's following free radical mechanism then Br should add to CH3, if it's F-C substitution then why there is no catalyst when NO2 is deactivating group?
What would be the change if I first do seconds step (i.e reduction to amine) and then bromination? Would it be wrong to do bromination at last?
Is the intermediate (2-bromo-4-nitrotoluene) the only major product of first step?

Too many questions in one! 1) You are right, ring substitution needs a Lewis acid catalyst. Without the catalyst nothing is going to happen. For free radical substitution of the methyl group you will need either UV light or a radical initiator. 2) and 3) You need to look up directing effects of substituents. 4) Yes.

At what temp or conditions was the bromine reacted in the first step? The methyl group may require a higher temp or specific conditions to be activated I'm only guessing tho , good luck.

@Swastik        okay no problem

I hope my answer will clarify your doubts @physics

Sorry but I would like to answer your questions from last to first,
(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place
(2) -COOH  is deactivating meta directing group (due to presence of  >C=O) so friedel-crafts bromination would take place at meta position.
(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2
(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced  one of the -H at ortho (to -CH3) of benzene.
(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)

preference of -CH3(activating) OVER -NO2(deactivating) will take place - I doubt this. $\ce{NO2}$ is a very strongly deactivating group whereas methyl is only weakly activating.

The directing effects of the methyl and nitro groups in p-nitrotoluene reinforce each other so that's not a problem in this specific case, but as bon said, the strongly deactivating nitro group is likely to exert a greater influence on the regioselectivity.

[for your reference}(http://www.utdallas.edu/~biewerm/17.pdf)  is the site which will answer your question.You can refer "Benzene and its derivative" chapter of any standard organic chemistry books like J.D. March or Paula Yukarinis or P.Bahadur  or O.P Tandonor Solomons(name of eminent chemistry book writer)   @ bon,orthocresol


text:
Benzenesulfonyl chloride dehydration of an amide to give nitrile
1) I am unable to understand how benzenesulfonyl chloride acts as a dehydrating agent in dehydration of amides? What is the driving force for losing chloride and taking the oxygen from the amide?
2) Why I can't use other dehydrating agents like $\ce{P4O10}$, $\ce{H2SO4}$, or $\ce{CaCl2}$?
3) How can I estimate what is suitable dehydrating agent for a particular reactant?
4) Is it necessary to use only those dehydrating agents that undergo chemical change? Or can I use dehydrating agents that adsorb water like silica gel?

Similar http://chemistry.stackexchange.com/questions/33742/what-makes-a-compound-a-dehydrating-agent

One key detail is that this reaction is performed in the presence of base. Typically pyridine is the solvent. Amides are nucleophilic at oxygen, explained by the resonance structure shown. Displacement of chloride by the amide oxygen gives the O-sulfonylated intermediate. After neutralization, the O-sulfonylimidate is set up for an E2 elimination across the C-N bond to give the nitrile.

In addition to sulfonyl chlorides, $\ce{P4O10}$ and $\ce{POCl3}$ are common reagents for this reaction. In general, the reagent should be electrophilic and be able to act as a leaving group.

The sulphur of the sulphonyl group is pretty electrophilic due to the two oxygens and one chlorine atom attached to it. So the nitrogen oxygen from the amide attacks on this sulphur atom, displacing the chlorine atom on it. The $\ce{Cl-}$ ion takes a proton from the amide with it and forms $\ce{HCl}$. The nitrogen then pushes its lone pair towards the former carbonyl carbon, causing the benzenesulphonyl group along with the former carbonyl carbon to take off, leaving a triple bond between the nitrogen and the carbon. The benzenesulphonate anion then picks up the proton left on the nitrogen, giving the products $\ce{R-C#N}$ and $\ce{Ph-SO3H}$.
Generally $\ce{P4O10}$ is a very strong dehydrating agent, so must be used where dehydration is difficult, or more dehydration than done by normal dehydrating agents is required. $\ce{H2SO4}$ is a good dehydrating agent, but probably isn't good enough for this particular reaction. It is difficult to predict which dehydrating agent must be used for any arbitrary reactant (atleast to my knowledge).
$\ce{CaCl2}$ or silica gel are physical dehydrating agents (more accurately drying agents) and cannot be used for chemical dehydration reaction like the above reaction.
Edit: My original mechanism was shown to be wrong by @jerepierre and @orthocresol, and hence i have corrected it above.

Do you have a reference for the proposed dehydration? I doubt that mech is correct.

I don't have a reference :/ But your mechanism does seem to make more sense...but I am not able to find concrete reference for the oxygen acting as a nucleophile, and not nitrogen. That is the part where I went off on a wrong track, otherwise your mechanism easily follows.

@jerepierre It seemed intuitive to me initially that the nitrogen would act as a nucleophile....that is the reason for me proposing the above mechanism.

The most nucleophilic part of an amide (aka the HOMO) is the carbonyl oxygen. The nitrogen lone pair is tied up by delocalisation into the C=O π*.

@orthocresol Okay. Thank you for correcting me! I have corrected the mechanism in my answer now!


text:
What is a hydrogen bond?
I was doing a little research on the Chemistry of water, and it said that water molecules have a hydrogen bond between them. What is it, and what does it do to water?

This answer should give you a nice short intro.

Firstly you should consider this simple model of atoms: There is a compact nucleus and smeared out electron clouds. In H this cloud is simply spherical. In many other atoms (e.g. O, C, N, S, etc.) the clouds are usually arranged like this (I did not find english site for some reason) (imagine tying 4 ballons together, forming a tetrahedron). This also explains the general geometry of water (and ice).
As you know in water the hydrogen is covalently bonded to oxygen. But due to the different properties of O and H the electrons of these bonds are not distributed symmetrically but much nearer the oxygen, leaving the hydrogen slightly positively charged. (Note that different elements cause different levels of asymmetry, a higher difference in electronegativity corresponds to a stronger asymmetry. This is why hydrocarbons ( C-H-bonds are rather apolar) are not very soluble in water.)
Now these positive H nuclei can bond to "full" electron clouds (e.g. the two of oxygen in water that are not bonded to hydrogen)
For something to be suluble in water you need to have many of those polar bonds and "free electron clouds", so e.g. sugars are easily soluble but e.g. alkanes are not.
These illustrations hopefully make now more sense. Sadly I could not find an english text about the tetrahedron model, which I consider didactically very useful.
Edit: Note that although hydrogen bonds are due to the electrostatic attraction, they are quite distinct from ionic bonds in the aspect that they have a definitive orientation, i.e. in ice the bonds still form tetrahedrons and not a cubic lattice or something like that.
I do not know a very good explanation but I think this is simply because the scale is much smaller; if the angle changes just a bit, the negatively charged part of the hydrogen atom interferes already.

Note that one of the defining features of a hydrogen bond is that it has a significant covalent component. It is not just an electrostatic interaction!

IIRC, the covalent component of a hydrogen bond makes up around 10% of the bond energy.


text:
Laws of electrolysis and quantisation of charge relation

The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by Faraday

I think about it a lot but could not find any relation, as the number of electrons involved is too large in electrolysis that the charge can be assumed to be continuous and discrete nature is negligible.
How did Faraday first suggested quantisation by this non microscopic process (electrolysis)?

Did you read https://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis? It's the number of  chemical valence electrons that is discrete, which is different for different metals. Some go into (or come out of) solution using one electron, others need two or three.

@CuriousOne      so discrete nature of valence electrons suggested quantisation of charge?

Yes. You are correct that one can't measure a single electron that way, but it's easy to distinguish 1 mole of electrons from 2 mole of electrons.


text:
What makes a species "fluorescence quencher"?
I just came across an article where they said that sodium fluoride is not quencher (in the context of fluorescence). Why is NaF not a quencher but NaBr is one?

quencher is any species that decreases the intensity of fluorescence when photochemically atoms collide

A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not behave the same way with every fluorophore. Some might be quenched by fluoride and not bromide. Some might be quenched by both. Some might be quenched by neither.
However, in the case of the specific fluorophore in the article, it is quenched by bromide and not be fluoride. The article likely provides an explanation or hypothesis about this behavior, or it cites something that does. Likely this article then is about a selective sensing application. Since bromide turns the sensor off (quenches fluorescence) and fluoride does not, the sensor is selective for bromide over fluoride.

Quenching by chemical reaction

Any chemical reaction will change the relative energies and energy spacing of the frontier molecular orbitals, which could lead to quenching. This chemical change could be something irreversible like oxidizing or reducing the fluorophore (probably not what is happening here). Other chemical reactions are possible, and it could be that bromide reacts and fluoride does not.
This change could also be a reversible one, like protonation/deprotonation, ligand binding, etc. If the fluorophore in your article is a transition metal complex, than it could be quenched by complexation with bromie anion (which displaces some ligand that has charge-transfer absorption/fluorescence with the metal). Fluoride might not do this.
This behavior could be more complex. Perhaps the fluorophore only fluorescences in the presence of some analyte. Maybe bromide reacts with the analyte (and fluoride does not) or bromide competes with the analyte but does not promote fluorescence.

Quenching by energy transfer

Fluorescence works as follows. The fluorophore absorbs a photon and is excited from its ground state to its excited state. When the excited state relaxes back to the ground state it emits another photon.
Fluorescence can be quenched if the excited state can be intercepted. An example is as follows. You have a system with your fluorophore and another molecule. Your fluorophore has a smaller HOMO-LUMO gap than this other molecule, so the fluorophore selectively absorbs photons of a certain frequency range and is excited. However, if the other molecule has a lower energy excited state than your fluorophore, then energy transfer can occur between them if they are close. Now your fluorophore is back in its ground state and the other molecule is in its excited state. If this molecule does not fluoresce, then fluorescence is quenched. This is probably not happening in your example. Note this phenomenon can be given various names like "resonance energy transfer" or "Forster energy transfer."
This type of quenching can get more complex as well. The energy acceptor could go and do chemistry or it could be so eager to be reduced that it not only steals the energy from the fluorophore, but it grabs an electron as well. This is not likely what is going on with bromide and fluoride.
The other type of behavior that can go on here is analyte-induced aggregation. Aggregation of the fluorophore causes the electronic properties to smear out over the aggregate, which likely means no more fluorescence (and perhaps no more absorbance) at the frequencies you care about. This phenomena could be happening in your example. Bromide could induce aggregation while fluoride does not.

Fluorescence (or phosphorescence) quenching is the process by which an excited state looses its energy to another species during a collision (in solution or vapour phase) and it always occurs in competition with fluorescence (or phosphorescence) and non-radiative transitions of internal conversion and intersystem crossing. Thus quenching always results in reduced fluorescence yield and lifetime.
Let S$_1$ be the excited singlet S$_0$ the ground state, T the triplet  and Q the quencher. The mechanisms are (a) heavy atom quenching (b) energy transfer (c) electron transfer and (d) chemical reaction.
In (a) The heavy atom, bromide ions, iodide ions, krypton, xenon etc. causes increased non-radiative transitions so  S$_1$ crosses to T with a greater rate constant and so the fluorescence yield falls as does the fluorescence lifetime.The quenching is caused by increased spin-orbit coupling from the quencher. The quencher does not need to remove any energy from the fluorophore since this is left as vibrational and rotational energy in the triplet, which then transfers it to solvent. The triplet can then suffer quenching from another quencher molecules to form S$0$ the ground state. This also reduces any triplet lifetime and phosphorescence yield.
The extent of spin-orbit coupling is proportional to the quenchers mass so is far greater for heavy species hence the name. This explains the effect of Br vs F in your question. Paramagnetic species such as O$_2$ also quench by causing extra spin-orbit coupling but O$_2$ can also quench by energy transfer as it has two low lying excited states.
(b) Energy transfer. This has two forms (i) Dexter transfer and (ii) Forster or dipole dipole transfer. Dexter is generally short range but Forster can be long range transfer in which S$_1$ can be a few nanometers away from Q for quenching to occur. The condition for transfer is one of energy conservation, the emission spectrum of the donor (S$_1$) must overlap in energy with the absorption of the acceptor. In dipole-dipole transfer the acceptor is electronically excited. This is usually written as D$^*$ + A -> D + A$^*$
(c) In electron transfer the donor and acceptor have to be stable in their ground states. Either the donor or acceptor has to be electronically excited and thus its redox potential is changed to be favourable ($\Delta G$ negative) and reaction occurs. D$^*$ + A -> D$^+$ + A$^-$ or A$^*$ + D -> A$^-$ + D$^+$. After this the D$^+$ and A$^-$ can recombine to D and A in their ground state. The rate constant of energy depends exponentially on their separation and only occurs at close to contact.
(d) In a reaction (other than electron transfer) the excited state is deactivated as bonds are broken or made. The subsequent mechanism depends on the exact chemical nature of the excited molecule and quencher.


text:
How to calculate pH of a solution when H3PO4, NaH2PO4, Na2HPO4 and Na3PO4 are mixed together in certain amounts to form a solution?
How to calculate pH of a solution when $\ce{H3PO4}$, $\ce{NaH2PO4}$, $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ are mixed together in certain amounts to form a solution?
I know how to calculate pH when they are present individually but not when all of them together. I searched the net also but couldn't find any such example.

Let's start from a simpler one: do you know what to do with a solution of $\ce{CH3COONa;/;CH3COOH}$?

Yes its a buffer.Henderson equation. @IvanNeretin

So you may do the mixture of (say) $\ce{Na2HPO4/NaH2PO4}$ in the same manner, I suppose?

But there are 4 mixtures isn't it ?? And I will get a different pH for each combination of two mixtures.I'm confused @IvanNeretin

Wait a sec. You can deal with any neighboring two of them, right?

Yes I can.Then ? @IvanNeretin

Usually doing calculations of this kind is not hard.  Roughly, you start from some idea of where the pH is going to end up.  For example, if you are only adding these phosphates, count the number of phosphate ions and the number of sodium ions.  Divide the number of sodium ions by the number of phosphate ions.  You (hopefully) get a number $x$ between $0$ and $3$, say $x=2.5$.  Almost all phosphates will then be either $\ce{NaH2PO4}$ or $\ce{Na2HPO4}$.  Pretend that these are the only ions and treat the problem like an ordinary two component buffer.  Then you are done.  This is an approximation but it will work rather well, as long as the $\mathrm{p}K_\mathrm{a}$ of the various acids are different enough.  If this is not good enough, you just treat this as a bunch of acids, with a bunch of $\mathrm{p}K_\mathrm{a}$ and use the acid-base equilibrium equation over and over, that is
$$\frac{[\ce{H}][\ce{B}_i]}{[\ce{A}_i]} = \mathrm{p}K_\mathrm{a}$$
where $[\ce{A}_i]$ is the concentration of an acid "$i$" and $[\ce{B}_i]$ that of the conjugate base.  Given the pH, this gives you the ratio of the concentration of each base to each acid.
But, really, for phosphate buffers, you are very well off if you just use one acid base pair.  For acids like terephthalic acid or isophthalic acid, which have two $\mathrm{p}K_\mathrm{a}$ values that are rather close to each other, that is they differ by less than $1.5$ or so, finding and solving such polynomial equations makes possible sense.  Otherwise, other approximations we routinely make are more important.
For an explicit example, suppose that we consider a compound $\ce{BH2}$ that can give up two $\ce{H+}$ ions, making both $\ce{BH-}$ and $\ce{B^2-}$.  Suppose that we have a total molar concentration of $\ce{B}$,
$$B_t=[\ce{BH2}]+[\ce{BH-}]+[\ce{B^2-}],$$
and that we have also added strong bases or strong acids to change the $[\ce{H+}],$ so that the total possible molar concentration of $[\ce{H+}]$ is
$$H_t=[\ce{H^+}]+[\ce{BH^-}]+2[\ce{BH_2}].$$
We then also have the relationships
\begin{align}
[\ce{H+}][\ce{BH-}] &= K_{\mathrm{a},\ce{BH_2}}[\ce{BH_2}]\\
[\ce{H+}][\ce{B^2-}] &= K_{\mathrm{a},\ce{BH-}}[\ce{BH-}]\\
\end{align}
If we now call $H=[\ce{H+}]$ and $B=[\ce{B^2-}]$ we find the equations
\begin{align}
H_t &= H + B (K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH2}} H))\tag{1},\\
B_t &= B (1 + K_{\mathrm{a},\ce{BH^-}} H (1 + K_{\mathrm{a},\ce{BH2}} H))\tag{2}.\\
\end{align}
We need to solve these equations for $B$ and $H$. Equation $(2)$ is easy to solve for $B$ and we can then substitute into equation $(1)$ to find
$$H_t = H + B_t
  \frac{K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH_2}}H)}
  {(1 + K_{\mathrm{a},\ce{BH-}} H(1 + K_{\mathrm{a},\ce{BH_2}}H))}\tag{3}.$$
We need to solve this formula for $H$, and (of course) $\ce{pH}=-\log_{10}(H)$.  Some algebra from this point gets us to a cubic equation for $H$.  Of course this has three possible solutions.  But, only one is sensible, provided we insist on "physical sense", e.g. all the equilibrium constants and concentrations are positive real numbers.  Somehow, this must be easy to see from the rules about roots of polynomials.  I am not currently willing to do that proof.
Physically, however, the total amount of hydrogen bound to the base will increase with increasing (positive) $H$.  So the right hand side of $(3)$ increases with increasing $H$ - from $H$ for $H=0$ to $2B_t + H$ as $H\to\infty$.  As the right hand side of $(3)$ increases with increasing $H$ there is exactly one positive solution.  However, this right hand side is also not a very "nice" function - the rational fraction is approximately constant except when we are near a "buffer" condition, where it changes quickly.  So, probably an effective way to solve the equation would be to use a computer and "regula falsi" or something like that.  A little thought will allow sensible guesses for any set of $H_t$ and $B_t$.  Then regula falsi will converge quickly.
Or, from a good enough initial guess even Newton-Ralphson.
Or, possibly it would be better in many cases to figure out which acid/base pair is closest to being in the "buffer" region.  And then, suppose that you have $H$ in that buffer region.  Solve for $H$ with only that acid-base pair.  Then, figure out what you did wrong - there can be $\ce{H+}$'s that are not bound to the base in the buffer pair you have chosen or $\ce{H+}$'s that are bound to stronger acids.  Assume that $H$ calculated from a single buffer pair is correct.  Calculate how many $\ce{H+}$'s are unexpectedly bound or unbound from your assumed buffer pairs.  Use this to calculate a new $H_t$.  If $H_t$ has not changed much, which will be the case unless there is another acid-base pair with very nearly the same $\mathrm{p}K_\mathrm{a}$ - differing I suppose by less than an integer from that of the pair you are considering, then use this value of $H_t$ to solve again for $H$, and recurse.
For the stated problem, all the $\mathrm{p}K_\mathrm{a}$ are very different so the $K_\mathrm{a}$ values differ by orders of magnitude.  And, really, all you need do to is figure out which buffer region you are near.  In principle, you can then find tiny errors using the recursion.  But, this is silly in most circumstances.  This will get a clear, continuous curve which always has the correct slope.  But, it will vastly overstate what you know about the problem.  The changes will be smaller than (I expect) lots of stuff: deviations from ideal solutions, errors in $K_\mathrm{a}$ values, etc..
).

I can provide an example which is useful:

$1,2,3$ and $4$ moles are taken of $\ce{H3PO4}$, $\ce{NaH2PO4}$, $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ respectively with $K_{a.1}=10^{-3},K_{a.2}=10^{-7},K_{a.3}=10^{-13}$.
Find the resulting $pH$ when these are mixed together to form an aqueous solution.

First write the $3$ ionic equations that are relevant here for the given $K_a$ values and then pair them up in such a way that you get rid of $\ce{H^+}$ terms on both sides and the $K$ value is of a very high magnitude such that the reaction shift almost completely in the forward direction.
For eg. in the given question, we can pair up the ionic equation for $K_{a.1}$ and reversed equation for $K_{a.3}$ such that the net $K=10^{10}$ and then we put in the value of the moles of each substance:
$\ce{H3PO4 + PO4^{3-}<=>H2PO4^{-} + HPO4^{2-}}$
 1        4            2          3
 _        3            3          4

Similarly we can pair $K_{a.2}$ and reversed $K_{a.3}$ ionic equations, and are finally left with $10$ mole of $\ce{HPO4^{2-}}$ for which pH can be evaluated with the given data as $10$.


text:
How to label atoms or groups "backwards" in ChemDraw
I'm trying to draw L-tyrosine with the hydroxyl part of the phenol as HO instead of OH as it appears in the ChemDraw image. Everytime I type in HO, it automatically corrects to OH. I have the settings in ACS Document 1996 style. Is there anyway to make the format render my "backwards" type?

I have version 14.0.0.117 and it works as expected: "HO" instead of "OH". It should automatically rotate the labels.

I have 15.0.0.106 and I don't think it worked when I had version 14 either.

The problem with the question is, that it becomes a software issue, that might have a very specific solution. Support might be the best option there. However, as it is working with my version and you have a newer one, I suppose it should work with yours, too. You can find me in [chat] to talk about it. Have you ticked "File > Preferences > Building/Display > Automatic Atom Label Alignment"? Have you tried entering "OH" instead?

Okay, so apparently typing it in as "OH" works.

If you type "HO". Click on it. Go to Text>flush right and it will also fix it.

ChemDraw is a piece of software that attempts to be as intelligent as possible. In many cases, that is truly helpful, in others it can be annoying.
When labelling an atom in text mode, ChemDraw will automagically assume a label from left to right (i.e. in standard North American/European writing direction) and interpret the label as such. If it finds that the atom has bonds to the right but none to the left, as tyrosine’s hydroxy group has in your example, it will group the label into ‘sensible’ fragments[1] and reverse them. This makes sense if you think of rotating a molecule: If you had drawn tyrosine rotated by $180^\circ$, you would have wanted to add OH on the rightmost atom, not HO. If you then manually rotate the entire molecule within ChemDraw, you will notice the OH flipping into an HO as soon as it graphically makes sense to do so.
Note that if you just hover over the atom in lasso mode or bond adding mode and then hit o (lowercase!) it will automatically add OH (or HO, depending on the orientation) at that position. Hitting o again will remove the additional hydrogen (this holds true for all chemical elements that are attached to single letters regardless whether the atom is attached to a lowercase letter (i.e. o) or an uppercase one (B for boron)).

[1]: Consider OTBS: ChemDraw automatically recognises TBS as an abbreviation for the $\ce{TBS}$ group so when reversing, it becomes TBSO. However ChemDraw does not standardly recognise the triethylsilyl group $\ce{TES}$, so when reversing OTES it arrives at SETO. The preferred way would be to add an abbreviation to ChemDraw.

Note that you can view the known nicknames using File>List Nicknames. You can define your own nicknames using Structure>Define Nickname. Also note that ChemDraw breaks atom labels (not nicknames) into tokens that consists of an uppercase letter followed by any numbers or lowercase letters. Therefore, $\ce{-OH}$ is flipped to $\ce{HO-{}}$, whereas $\ce{-Br}$ is not flipped to $\ce{rB-{}}$.

@ortho How write you are in assuming a righting error there … ;)


text:
How can eosin be acidic and negative at the same time?
The broadest acid-base definition is done by Lewis and he states that a lewis acid is:
"a compound or ionic species which can accept an electron pair from a donor compound."
And all the sources say that eosin is an acid and a negatively charged molecule.
I think eosin cannot accept a pair of electrons if it's already negatively charged, thus, it should be considered as a base.
A clarification on this subject would be perfect.

There are many examples of negatively charged acids. Look at polyprotic acids. As examples with biological relevance, $\ce{H3PO4}$ is an acid, but $\ce{H2PO4^{-}}$ and $\ce{HPO4^{2-}}$ can also act as acids. Similarly, $\ce{H2CO3}$ is an acid, but $\ce{HCO3^{-}}$ can also act as an acid.

There is nothing unusual in being an acid and a base at the same time. Many organic molecules (for example, pretty much any peptide) are like that.

@NicolauSakerNeto That is true, but what proton do you see on that molecule that is acidic when the molecule is negatively charged? These things look more like Lewis acids to me (Michael acceptors, maybe, to be specific).

@SendersReagent do you mean the molecular structures of eosin? (at https://en.m.wikipedia.org/wiki/Eosin)

There are a lot of hydrogen atoms on aromatic rings. Can't they be ionized?

@m1clee Indeed I do. Aromatic protons are usually very acidic. I don't really see any in there that I would describe as acidic.

Aren't "Aromatic protons" the hydrogen atoms on ch=ch-ch=ch-ch=ch  benzene ring for instance? So there are aromatic protons? And you say that they are very acidic.  So I can't see any problem here.

By the way I am not a chemist and right now I am just trying to understand your question. So if you think I don't have some basic  knowledge based on my last two replies, I'd be very happy if you could adress them.

Geeez, there's only an bad wording on Wikipedia - the picture shows anions of the acid as they are important - it's an acid so it is present in dissociated form in diluted solutions.

And no those anions aren't acidic in water solution. I guess @Senders wanted to say "aren't usually". Also edited this article you're referring to. Check it as it is now.

@Mithoron If that's true biologists misuse the term "acid" all the  time. For instance they call dna, "nucleic acid" whereas it is always in an aquaus solution with tons of negative po3(-2) in it which are conjugate bases. So it must be nucleic base, if we use the terms in a correct, unconventional way.

@Mithoron Aren't usually* correct.

DNA is nucleic acid and caring only about it's conjugate base doesn't change that. Still it's usually more or less neutralized with various counterions, so one could say it's in form of a salt or tight ionic pair, like bonded eosin.


text:
Which species gives electron in the following reaction?
In a given reaction like this:
$$\ce{FeO4^2-  + 3e^-  + 8H+ <=> Fe^3+ + 4H2O}$$
From where does the electron come  in the acidic medium, as there is no other species to get oxidised in the medium? Also if this were in basic medium then why there is no oxidation of any other species in the medium?

This is not a reaction, but a half-reaction. The electron comes from the other half-reaction.

Expanding on Ivan's comment, what you have here is a reduction half-equation (electrons are being accepted). As the name suggests, this only shows half of the actual reaction. You also need to have the oxidation half-equation (electrons being donated) and add them together to get the entire equation for the reaction. When you add these two half-equations together, the electrons should cancel out.

What you have here is a reduction half-equation as the $\ce{Fe(IV)}$ is accepting electrons to get reduced to $\ce{Fe(III)}$. As the name suggests, this only shows half of the actual reaction as if something is getting reduced, something else must be getting oxidised. So what you need is the oxidation half-equation which shows the species that is getting oxidised and donating electrons. When you add these two half equations the electrons will cancel each other out and you will get your equation for the reaction.
For example, one possible oxidation half-equation is the following:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
Now if you times the co-efficients of the reduction half-equation by 2 and times the co-efficients of the oxidation half-equation by 3 and then add them together you will get the following equation:
$$\ce{2FeO4^{2-} 3Mg + 16H+ -> 2Fe^{3+} 3Mg^{2+} + 8H2O}$$
As you can see all the electrons balance out and that is your complete equation for the reaction.
Sorry, I don't think I really understand your second question about basic medium. Could you please expand on it to clarify what you mean.


text:
Understanding effect of ph on solubility of salt
In the following picture why Ksp is not simply S2- ? Why is f included with one species only? Please explain the last part.

What does f represent and why it is used in molar solubility?

I am also very confused on what is happening here. I have no idea what f is suppose to be and I don;t know how they got that expression for Ksp. I will will write up an answer on how I would do this which is hopefully much clearer

$f$ is defined in the question as the fraction of $[\ce{X}]{tot}$ that is ionized.  It's just a shorthand abbreviation for $\frac{[\ce{X-}]}{[\ce{X-}] + [\ce{HX}]}$.  By charge conservation, the concentration of metal $\ce{M+}$ is equal to $f S$, so $K{sp}$ can be written as $[S] * f[S] $.

Let assume we have the following equilibria:
$$\ce{MX \rightleftharpoons M+ + X-}$$
$$\ce{HX \rightleftharpoons H+ + X-}$$
The expression for their respective equilibrium constants are the following:
$$\mathrm{K_{sp} = [M^+][X-]}$$
$$\mathrm{K_a = \frac{[H^+][X^-]}{[HX]}}$$
We also know the following (let S be solubility):
$$\mathrm{S = [M^+] = [X^-]_{tot}}$$
Now we just need one more equation before we can start solving for S which is a mass balance. Since all the $\mathrm{X^-}$ that is produced either reacts with water to form $\mathrm{HX}$ or remains as $\mathrm{X^-}$, we get the following equation :
$$\mathrm{[X^-]_{tot} = [X^-] + [HX]}$$
Now lets put $\mathrm{[X^-]}$ and $\mathrm{[HX]}$ in terms of $\mathrm{[M^+]}$:
$$\mathrm{[X^-] = \frac{K_{sp}}{[M^+]}}$$
$$\mathrm{[HX] = \frac{[H^+][X^-]}{K_a} = \frac{K_{sp}[H^+]}{K_a[M+]}}$$
Now let plugs those values into our mass balance and replace $\mathrm{[X^-]_{tot}}$ and $\mathrm{[M^+]}$ with S:
$$\mathrm{[X^-]_{tot} = \frac{K_{sp}}{[M^+]} + \frac{K_{sp}[H^+]}{K_a[M+]}}$$
$$\mathrm{S = \frac{K_{sp}}{S} + \frac{K_{sp}[H^+]}{K_a\times S}}$$
$$\mathrm{S^2 = K_{sp} + \frac{K_{sp}[H^+]}{K_a}}$$
$$\mathrm{S = \sqrt{K_{sp} + \frac{K_{sp}[H^+]}{K_a}}}$$
This gives you the exact same expression as the one on the sheet but hopefully this is much clearer for you.

Could you please explain why X- tot=X- + HX ?

@JM97 That is because the total X- that is produced either reacts with water to form HX or it just remains as X-. Therefore X-tot = X- + HX


text:
Determination of structure by 13C and 1H NMR spectroscopy

I'm trying to interpret a $\ce{^{13}C}$-NMR and $\ce{^1H}$-NMR spectrum. So far, I've decided that $\ce{CH3-C}$, $\ce{CH2-CH3}$ and $\ce{CH-CH}$ exist in the structure. I'm not sure but I guess there should be different seven carbons too.
I've come up with the structure below but I feel like there is not enough information to be sure. What should I need to consider to go further at this point?

Hi and welcome  to Chemistry SE. The structure you are proposing is not correct. You should be able to work back through your proposed structure and see evidence for each part of the molecule. For instance, your molecule has 4 non-equivalent methyl groups in it, but the 1H spectrum only shows 2 distinct methyl groups.

As a clue to get you thinking in the right direction (we don't want to just give you the answer here) count up the total integral of signals in your 1H spectrum - I count 10. Your molecular formula indicates there are 20. This would suggest that each 1H group is duplicated in the molecule.

This is a tricky little problem, given there is an element of symmetry to consider. As mentioned in the comments, once you come up with a proposed structure, and your one is not a bad first attempt, you should go through and test the spectrum against what you would predict from your structure. Unfortunately, in your suggestion, there are 4 non-equivalent $\ce{-CH3}$ environments, and the spectrum only shows two, so we can eliminate that structure as a possibility. But don't be disheartened — that is the process of structure elucidation! Propose a stucture, test it, refine or eliminate it. And then start again.
Looking at the $\ce{^1H}$ spectrum, we can see:

$\ce{-CH3}$
$\ce{-CH2}$
$\ce{-CH3}$
$\ce{-CH}$
$\ce{-CH}$

We can construct the $\ce{-CH2CH3}$ spin system based on coupling, and the chemical shifts of the two downfield protons indicate a double bond (and, using some old tricks of the trade, looks to be cis — the splitting here is much less than about $14~\mathrm{Hz}$, which would be the expected splitting between the outside lines of the methyl triplet). The $\ce{CH3}$ at about $3.3$ is in the general ball park for a methyl ether, I'd say.
That gives us 10 protons, but the molecular formula says $\ce{H20}$ … so we must have symmetry in there somewhere which gives us:

2 $\ce{-CH2CH3}$
2 $\ce{-CH3}$ (probably $\ce{-OCH3}$)
2 $\ce{-CH=CH-{}}$

So, looking at the carbon spectrum:
I count 7 carbon peaks.
From the proton information, we have 5 $\ce{C}$ environments, each appearing twice, which gives a total of 10 carbons. The other 2 carbon environments occur once only in the molecule.
From DEPT experiments (which nobody in the real world should ever run when they can instead run a phase-edited HSQC instead) we can see we have

$\ce{CH3}$ at $9.9$ and $51.4$
$\ce{CH2}$ at $38.0$
$\ce{CH}$ $132.8$ and $133.2$
$\ce{C}$ at $47$ and $110$.

We know (or should know) what the non-quaternary carbon cantres all are. We worked that out from the $\ce{^1H}$ spectrum. So let's concentrate on the quaternaries. The peak at $110$ is interesting, and is typical of a carbon centre attached to 2 oxygens. Many sugars have carbons that come around this area. The other quaternary at about $47$ is typical of an aliphatic quaternary; not overly exciting.
So somehow we need to construct a molecule that has 2 non-equivalent quaternary centres that have a plane of symmetry through them, with the $\ce{-CH2CH3, -OCH3}$ and $\ce{-CH=CH-{}}$ attached. Each of these three spin systems must be isolated, otherwise we would see coupling to the vinylic protons.
As this is clearly a homework question, I'll leave it to you to piece together, and ask further questions if you need, but my answer gives the exact same predicted spectra in ChemDraw as your spectra.

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


text:
DFT Code for Atoms : Sources
Some time ago I implemented the restricted and unrestricted Hartree-Fock methods. I had a lot of fun and I learned a lot. It goes without saying that Szabo and Ostlund's book "Modern Quantum Chemistry" has been fundamental to achieve a working code.
I would like to repeat the experience implementing a DFT code. I would like to implement it for atoms with all-electron non-relativistic calculations (scalar relativistic calculation will maybe be an interesting extension as well as the generation of norm-conserving pseudopotentials). Unfortunately I did't came across a "DFT bible" such as Szabo and Ostlund's book.
Does anybody knows good references (at the same level of Szabo and Ostlund's for in-depth discussion of the practical implementation) for the practical implementation of DFT for atoms (radial SE, logarithmic grids, different functionals, ...)?
Note that I am not searching a general DFT introduction or an explanation of real PW+NCPP implementations.

This question is similar to this one but I don't think it is a duplicate. In the previous question I was asking about a book explaining the implementation details of DFT for periodic systems. DFT applied to periodic systems usually involves a PW basis set as well as pseudopotentials and FFTs. This corresponds to a simplified version of the PWSCF code of QuantumESPRESSO.
In this question I am asking for the references that will help me to write of an atomic code, i.e. no basis set expansion and no pseudopotentials (all-electron calculation, logarithmic grid, ...). This corresponds more and less to the atomic code of QuantumESPRESSO which is usually used to generate pseudopotentials.

you might find a few ideas here: http://www.etsf.eu/system/files/users/SottileF/file_101.pdf

Surprisingly, you may look into quantum espresso code again. It contains atomic code, used for pseudopotential generation, though I don't remember the exact location in the source tree.

@permeakra The source code is in espresso-5.x.x/atomic/scr ! ; ) I know everything has already been implemented, but ATOMIC contains around 100 files (relativistic calculations, PP generation, ...)... I would like to start to write a minimal code in order to understand the underlying theory. ATOMIC is already too complex for my needs. In addition the risk is that I will start to "copy" QE routines; I would like to write them from scratch based only on the theory of the references.

Maybe my answer there would have been more suitable here, but anyway, I think it answers partly both topics.

Marx, D.; Hutter, J. Ab initio molecular dynamics: basic theory and advanced methods; Cambridge University Press: Cambridge, 2012. This has some info on implementation of algorithm in CPMD code including flow chart of algorithm.
Thijssen, J. Computational Physics, 2nd Ed.; Cambridge University Press: Cambridge, UK ; New York, 2007. Chapter 6 comes with a short computer program downloadable from companion site written by author of the book.

Maybe these two help a bit.


text:
Spontaneous and Favoured Chemical Reactions
I have some problems in understanding what is the difference between a reaction which is spontaneous and a reaction which is termodynamically favoured. I'm looking at the equation of Gibbs Free Energy:
$$\Delta G = \Delta G^0 + RT ln \prod a_i^{\nu_i}$$
I know that a reaction is spontaneous if $\Delta G <0$ and a reaction is Termodynamically favoured at a given T, by convenction, if $\Delta G^0<0$.
So the condition of $\Delta G^0<0$ is a necessary, but not sufficient condition to be spontaneous?
Thanks a lot!!

You are missing a ln before the product.

Your equation shows in what direction an equillibrium  reaction will move to achieve td. equillibrium. A "spontaneous" reaction must of course be thermodynamically possible, but it also requires a viable kinetic pathway, which cannot be predicted from such simple thermodynamics.

Thanks for Your reply!  Therefore a reaction which is thermodynamically unfavoured in the considered direction will have a small amount of products with respect to the reactants but it will occur anyway. The condition of unfavoured reaction can be overcome with kinetics considerations (i.e. catalyzers)?


text:
Rate of Evaporation of solution
Why does rate of evaporation of solution decreases on adding a non volatile component?
Also what happens if we add a volatile solute?

The rate of evaporation decreases because the activity of the solvent in the liquid phase decreases, while the activity in the vapor phase (or solid, for that matter) stays the same. This is also the reason for the increase in boiling point and the decrease in melting point of the solvent.
If we add a volatile solute, it will evaporate too, thus changing the solvent activity in the vapor phase as well. The resulting behavior can't be characterized universally; you'll have to consult the binary phase diagram of boiling for the particular system, and these can be pretty diverse.


text:
Number of stereoisomers of 1-(2-methylcyclohexyl)ethan-1-ol

How many stereoisomers are possible for the compound in the picture? I tried to draw a rough picture and indicated the chiral carbons. I'm getting 8 stereoisomers are possible but my book gives the answer as 2. Can someone explain where I'm going wrong?

I'm wondering this as well - the only other interpretation I can see is if you had to specify how many types of stereoisomers there are. That would give 2, because you'd have diastereomers and enantiomers but not meso compounds.

Can't there be geometrical isomerism ?

Geometrical Isomerism doesn't apply in this case since there are no double bonds.

But one could be above the plane and one below...there need not be a double bond

True, but that would not change the name of the compound regardless of whether the particular group in question takes an axial or Equatorial position.

The comments on the question are incorrect [edit: the offending comments seem to be gone now], as is the answer given in the book. In general, answers printed at the back of a book should never be taken as the absolute truth. There are typos in all books. Even well-known and well-written textbooks have many errata (usually corrected in subsequent printings).
As you said, there are three chiral centres and therefore 8 stereoisomers.
There are no other elements of symmetry present in this molecule which render any of those 8 stereoisomers equivalent.


text:
knowing that a strong base is a strong electrolyte, when we add a weak acid in water, will its conjugate strong base be a strong electrolyte too?
I am not sure if the questions makes sense. What i am trying to say is that, if weak acids form strong conjugate bases, why aren't they strong electrolytes?  Thank you in advance.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Note that your question should ideally be posted in the body of your post. You can [edit] it in.

You know that basicity/acidity and conductivity are unrelated topics?

A strong acid is a strong electrolyte and a weak acid is a weak electrolyte no?

Not necessarily. Sulphuric acid is a very strong acid, but $96~%\ \ce{H2SO4}$ is a lousy electrolyte.

that explains a lot. i mixed up some concepts thank you for your help.


text:
Possible Products in the Addition To a Double Bond in a Ring

I don't understand why both these products should be different. I know hydrogenation is syn addition and both the hydrogens add from the same side, but why should the plane in which both the hydrogen atoms are present even matter?
Also, bond rotation is possible around sigma bond which makes both these products identical.
Please explain why the two products are different and where am I wrong with this logic?

You are right, these two are identical.

Those two are identical but the rotation of that sigma bond is not possible.

@orthocresol what is your logic of calling them identical?

Just flip the one on the left upside-down, it is the same

@orthocresol is it okay to flip molecules like that?

Of course, if you really don't like flipping the molecule you can always flip yourself and look at it from the bottom up instead of looking at it top down. Why don't you try using a molecular modelling set and make the two molecules. That will also show you why that bond can't be rotated.

That pretty much answers my question. Thank you.

The two molecules are not different, they are identical. You can prove this in a number of ways:

Build both with a molecular model kit and see if you can orient them in the same way (spoiler: you can).
Use your imagination and 3D-vision to rotate one of the two structures until it looks like the other one.
Look for decisive symmetry elements. In this case, there is a plane of symmetry traversing the molecule through the $\ce{C-Br}$ bond. Therefore, the molecule cannot be chiral. Therefore, two versions of it that differ only in stereochemical orientation must be identical.

I can only speculate why your source drew it that way. One possibility would be to stress that the hydrogenation is not only syn-selective but can also happen from both sides (even though it gives the same product here). That doesn’t change that they’re identical, though.

why should the plane in which both the hydrogen atoms are present even
  matter?

In this case it doesn't.  The author of the example is laying the groundwork for the future.
The area above and below the double bond is often referred to as a "face". In your molecule, the two faces are not equivalent. If you build two models of your compound and in both color the top of the double bond red and the bottom blue, you'll find there is no way to superimpose one model onto the other where the red face is superimposed on the blue face.  These two faces are therefore not equivalent, they are said to be enantiotopic (mirror images) in this molecule.
In this case, addition of (an achiral reagent such as) hydrogen to the two faces produces the same molecule.  In more complex molecules, or when using chiral reagents this will not be the case.
Consider the case where there is a methyl group attached to the methylene group adjacent to the $\ce{C-Br}$ carbon in your molecule.  Now the faces are different (diastereotopic) and addition of hydrogen to the top and bottom faces will produce two different products (methyl and bromine on the same side or opposite sides of the molecule).
The concept of faces is important in chemistry.  Always consider if the faces are equivalent, enantiotopic or diastereotopic. Depending on what you find, the products from attack at the two faces may be the same, enantiomers or diastereomers.  The concept of faces is introduced in your problem.


text:
What are the implications of having different conformations?
I have seen videos explaining what conformations are. I am curious about whether different conformations can give rise to slightly different properties. Particularly of interest is whether you can modify the smell of something by just rotating some bonds.

They are rotating all the time so you could say that compounds are described by their conformational ensembles. You could bind a molecule to manipulate conformations, but then you have complex of molecules not a single one.

That's an interesting question. To start off, what Mith said in the comments is absolutely right. Let's say that two conformations of a molecule have different smells (which is in fact not an absurd proposition by any means). The problem is that, because there is a very low barrier to the interconversion of the conformations, the individual conformations do not really hang around long enough to be individually detected; you can only really detect a weighted average of the conformations.
To give you an example, you can devise di-substituted cyclohexanes that have two predominant chair conformations. These can be investigated by proton NMR spectroscopy, since axial-axial coupling constants are larger than equatorial-axial or equatorial-equatorial couplings. From the integrals of the relevant peaks you can determine the relative proportions of both conformations.

However, if you somehow managed to synthesise a sample of the pure diequatorial conformer (the one in blue) and put it into a NMR tube, some of it would have converted into the diaxial conformer (the one in red) before you could even run your NMR. In fact, there's probably no way you can synthesise a pure conformer to begin with.
Likewise, if you replace the NMR machine with your nose, there's no way you can open a bottle of a perfume X, with conformation X1, and smell something, then open a bottle of the same perfume X with conformation X2 and smell another thing. They would both smell the same because conformers aren't isolable. (Well, not unless you are working in an argon matrix at temperatures near absolute zero or something like that!)

Mith then goes on to say that you can bind to a molecule to manipulate its conformation. That's the more interesting part, in my opinion at least! If you think about it, it's going to be very hard for us to devise some kind of surface that particularly favours one conformation over another. You'd need a very complex binding partner, with a very specific 3D shape, that "fits" with one conformation but not another conformation.
Does that ring a bell? It might, because nature has made those binding partners for us already. That previous paragraph is a perfect description of an enzyme: a protein that catalyses a biological reaction. Lysozyme is one of the most famous enzymes; it was the first enzyme whose structure was successfully solved by X-ray diffraction. It is also a great example of an enzyme that works by manipulating conformations.
Lysozyme binds to a series of carbohydrates in the peptidoglycan cell walls of some bacteria. Together with water, it hydrolyses the glycosidic bonds (a fancy name for an acetal) in these carbohydrates:

(image from Wikipedia)
I'm not going to go into the full mechanism of this, but the striking thing is that upon binding to lysozyme, the "D ring" (which is the NAM residue on the left-hand side of the diagram above) is distorted from its usual chair conformation into a half-chair conformation. This is because it "fits" better into the exact 3D conformation of the enzyme (not shown in the diagrams):

(image modified from Wikipedia)
This makes the ring more unstable and prone to hydrolysis (strictly speaking, more prone to nucleophilic attack by the aspartate-52 residue). Without this manipulation of the conformation, it's likely that the hydrolysis cannot proceed.
So, there you have it! One way in which the conformation of a molecule, upon binding to another molecule, affects its properties.

I really like when my ideas are so nicely used :)


text:
Estimating stability and mean life of anomalous tautomers
A pair of bases in the DNA, say, A-T, have a tautomerized form A*-T* (resulting from switching the sides of both protons along the hydrogen bridges).
I have studied how, by means of DFT computations, one can compute the energy difference between the normal and the anomalous forms. But, another point of interest is the stability of the tautomer.
I have read several times that those anomalous tautomers "decay" to the normal A-T form very quickly (much quicker than the time it lasts a round of DNA replication), which is interesting because it implies that tautomerization is not likely to be a source of spontaneous mutations.
What I would like to learn is how can one compute this expected life-time of the anomalous tautomer
I have an idea (which I don't know if it could be rigth or if maybe is completely worthless), which involves making a Born-Oppenheimer separation for the two protons (''after'' the DFT computation of the electronic ground state) and then study the non-adiabatic transitions between the different vibronic states of the couple of protons. I am happy to provide details, if someone whises so.
But perhaps "my method" is too messy, or naive (or both) and / or there are other standar approaches to compute approximately the expected mean life of anomalous tautomers such as the mentioned A*-T*.
I would be very grateful if someone could apport some insigth and/or references into how are these expected life-times calculated typically in computational chemistry!

I am by no means  an expert, but molecular dynamics simulations are sometimes used to study lifetimes. A lot of force fields exist, both for unreactive and reactive systems. Seeing as how tautomerism involves breaking and reforming of bonds, perhaps the ReaxFF could be fitted, validated, and then used?

find TS for tautomerizarion -> compute rate constant using TS theory -> compute half-life for tautomer

I had not heard of the ReaxFF, thanks for the note. @Jan Jense: could you elaborate further your answer, if you have time to spare?

Have a look at http://molecularmodelingbasics.blogspot.dk/2009/08/get-half-life-from-transition-state-to.html

Thank you very much, Jan Jensen. So, a first guess would be to compute the expected half-life from a reaction rate constant $k$ given by $k=\frac{\text{k}\text{B}T}{h}\exp\left( -\frac{\Delta F^{\ddagger}}{\text{k}\text{B}T} \right),$ where $\Delta F^{\ddagger}$ is the free-energy difference between the tauotomer and the transition state... even accepting this, a doubt remains: how do I "choose" a transition state? There a two coordinate reactions (there are two protons in hydrogen bonds) and thus tons of possible paths to the tautomer!

The protons might tunnel, so kinetics vs. barrier height might have a different relationship.

TLDR: With a quantum/semi-classical transition state theory.
I order to determine the lifetime of the tautomer, you need to know the rate constant for the decay from the tautomer to the conventional dimer.  This is a kinetics question rather than a thermodynamic question so you won't be able to answer this by comparing single point DFT energies.  The rate will also vary with temperature.
A complicating factor is that the nuclei moving are protons in hydrogen bonds.  Hydrogen bonds are frequently double well potentials with two minima, one near each end of the bond - in this case the tautomer and conventional dimer.  In these case the physical distance between the minima is small and the proton may tunnel through the potential barrier between the two minima - hence nuclear quantum effects are likely to be important, at least at low temperatures.
Methods that can calculate nuclear quantum effects can be very computational expensive and may require you to work in a reduced dimensional represention - only considering the two protons quantum mechanically and potentially simplifying the surrounding structure.
Two potential methods that could calculate a rate taking account of nuclear quantum effects are  semiclassical transtition state theory, or ring-polymer molecular dynamics which has been applied within pockets of larger systems such as proteins.

For more information about his class of reactions - often referred to as proton-coupled electron transfer - see this review.


text:
References for Transition State Theory and Quantum Transtion State Theory
Having asked recently this, I started to look for good references and notes on Transition State Theory and, also, modern versions of it including things like tunneling (which isn't taken into account in Eyring's equation).
Unfortunately, I haven't been able to find any satisfactory reference. I have downloaded a bunch of lecture notes and gained conceptual understanding of TST, but nowhere do I find a explanation with all the mathematical details.
What could you recommend for a thorough and detailed exposition of the TST?
Also, I would like to find stuff about how the theory has been improved over the years (the theory is mainly classical). I guess there are more accurate quantum mechanical versions of the theory.
Thanks a lot.

I am unaware, that there has been any development for an improvement of TST over the original publication by Eyring. The many flaws of his equation are being ignored to a certain extent and it still works. A short summary of the critical points can be found at Wikipedia or in this question. I think starting with the original publication is your only chance to understand it from a rigorous point of view.

@Martin-マーチン If my memory serves me correctly,  Eyring refered to the coordinate inaccurately  as a fourth translational freedom degree. I am unsure if would be simple to catch his idea without a very solid background. Glasstone's "A Treatise on Physical Chemistry" maybe is more friendly as an starting point.

There are many sources that you can check out, in particular the reviews of Truhlar:

Current Status of Transition-State Theory by Truhlar et al.

This article was used in my undergraduate course of chemical kinetics, so I think that it could be a good source. Also we used:

Variational transition-state theory by Truhlar and Garrett

Maybe they are a bit outdated, but they are good as start points. The abstract of this newer article make it seems to be useful for you

Transition state theory: Variational formulation, dynamical corrections, and error estimates by Vanden-Eijnden and Tal

To deeply understand the theory statistical physics is a must. A good source (that also deals with TST) is the Ralph H. Fowler's book.

Thank you! That last reference is exactly the type of thing I was looking for.  It has the details and it clearly shows, for instance, how the ergodic hypothesis is crucial for TST, which is never clear in the chemistry literature, as far as I've seen. Also Atkin's "Physical chemistry" seems to be great for all of this stuff.

Your welcome. You should search for articles on Google Scholar, Scopus or similar sites.  ;-)


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Difference between Excess property and property changes of mixing
I have some confusion in my mind about the difference between an excess property and a property change of mixing. I know that the Excess property is defined as the difference between the actual value and those of an ideal mixture:
$$\ M^E(T,P,x_i)=M^{actual}-M^{id}(T,P,x_i)$$
On the other hand the property changes of a mixture is:
$$\ M^{mix}=M{actual}-\sum n_i*M_i$$
So the latter seems to me that the whole change of a mixing process is with respect to the pure, ideal gas, species.
Excess and mixing are the same for enthalpy and volume but are different for entropy and Gibbs free energy.
Now come the problem: property changes of mixing and excess properties are related each other by the residuals properties (deviations from ideal gas)?
Otherwise I do not understand why they are different.
Thanks a lot

The difference is subtle, however they are different as you suspect.
You can look at the property of mixing as the change in property resulting from mixing the pure components (with everything at the same temperature and pressure) together to form the mixture.
The excess property is deviation between the actual property and the property in an ideal mixture at the same temperature and pressure.
They are the same for volume and enthalpy because the volume of mixing and the enthalpy of mixing in an ideal solution are both 0. This is not the case for G or S.
In other words the difference arises because $M^{id}=\Sigma x_i M_i$ is not true for all properties. It is only true for enthalpy and volume. So the second term in your expression for the excess property above, cannot in general be written as $M^{id}=\Sigma x_i M_i$ (if it could, the excess property and mixture property would be the same). It should instead be $M^{id}=\Sigma x_i M_i + M_{mix}^{id}$.
The extra term for mixing the components would be be given by $$
\frac{G^{id}_{mix}}{RT} = \Sigma x_iln(x_i)
$$
and
$$
\frac{S^{id}_{mix}}{R} = -\Sigma x_iln(x_i)
$$
for Gibbs free energy and entropy, respectively.


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Chemical processes and chemical reactions difference
Whats the difference between chemical processes and chemical reactions?
Ive heard that the chemical industries only uses chemical processes to produce new substances, but isnt that also what happens when a chemical reaction occurs? I couldn't find a lot on this topic online except for the wikipedia aricles that only made me more confused.

According to IUPAC:

Chemical reaction
A process that results in the interconversion of chemical species.
Chemical reactions may be elementary reactions or stepwise reactions
(It should be noted that this definition includes experimentally
observable interconversions of conformers.) Detectable chemical
reactions normally involve sets of molecular entities as indicated by
this definition, but it is often conceptually convenient to use the
term also for changes involving single molecular entities (i.e.
'microscopic chemical events').

A chemical reaction is a kind of process in which interconversion of chemical species takes place. Because of this, we can say that a chemical reaction is a chemical process.
Chemical process can be a broad term, which describe any process that is commonly studied by chemists because they are closely related to chemical reactions. It is somewhat diffuse, as boundaries of the field of chemistry.
Industries use a variety of procedures to obtain substances. Chemical processes/reactions take place. If your interest is realising what is done in industries, do not get stuck in these subtleties. Most of the times they are not intentionally distinguished, and their usage is customary in the field.

While a reaction is a kind of process in the broader sense of that word, in the chemical industry the term process usually refers to an engineered system for producing a chemical product, including one or more reactions, the equipment, the feedstocks, the separation methods, etc. An example of how a process is not the same as a reaction is that a given substance may be produced from the same precursors using either a batch process or a continuous flow process.

In other words, the chemical industry uses the term "process" primarily to refer an engineering process.