text:
Why not an Avogadro's principle for solid matter?
Avogadro stated that for any ideal gas, 1 mole of particles will occupy a specific volume at a specific ratio of pressure and temperature. So why doesn't this apply to a liquid or solid?

I removed the "solid-state" tag. Tags are used for searching. If you tag a gas-law question as a solid-state one, it's misleading.

Isn't this question "why doesn't a gas-law apply in solid-state"?   That would seem to be a solid-state question...

Avogadro's law, which can be written as $V \propto n$, where $V$ is the volume of the gas and $n$ is the amount of substance of the gas (measured in moles), can be thought of as just another manifestation of the ideal gas law rewritten as follows,
$$
  V = (RT/p) n \, .
$$
Consequently, strictly speaking, Avogadro's law is applicable only for a hypothetical ideal gas, but since real gases under usual conditions (such as standard temperature and pressure) are not far away in their properties from an ideal gas, Avogadro's law usually works well enough for them also. Substances in condensed phase on the other hand are milage away from an ideal gas in their properties, and thus, Avogadro's law is inapplicable for such cases.
To clarify in which sense Avogadro's law is inapplicable for substances in condensed phases, note that for solids and liquids, as well as for gases, the volume is proportional to the amount of substance. But Avogadro's law states not just the proportionality. It is also insists that the coefficient of the proportionality is the same for each and every gas, and this is the thing which can not be generalised to substances in condensed phases. For real gases, the ideal gas description which depends on the following assumptions (too name just a few):

molecules are almost point-like (effectively zero-sized) particles;
molecules do not exhibit any intermolecular forcers;

usually works as a good approximation. But for liquids and solids such a description a priori is physically horrible: molecules here are relatively bigger and more tightly packed so that their size and intermolecular interaction between them couldn't be ignored anymore.

To be honest this question has made me wonder a bit. Let's take the case of water. If it doesn't follow Avogadro's law, i.e. the volume is not directly proportional to the amount, ie $V = kn$ for some $k = k(p,T)$, then since the molar mass of water is constant at 18 g/mol, that would mean that the density of water (an intensive property) is actually dependent on the amount of substance present?

@orthocresol, I think you misunderstood the statement that Avogadro's law does not work for liquids and solids. See, the Avogadro's law does not just require that the volume of each and every gas is proportional to the amount of substance, it requires that the coefficient of the proportionality is always the same. That last thing is what breaks for solids and liquids, not the proportionality itself, i.e. each and every solid or liquid has its own coefficient of the proportionality.

@orthocresol, but that is a good point, I'll try to incorporate this matter into my answer.

hmm, okay, in that case I definitely agree.

Would you consider a question about photons occupying a certain volume a related question?

gas molecules are small, and even largish gas molecules behave so much like small ones (example:  atoms of helium vs. carbon dioxide vapor) that Avogadro's models hold very well across all the spectrum of matter in gaseous form.  at a given temperature and pressure the individual molecules will have a certain statistically predictable average distance from one another.
in liquids, though the molecules are much closer together and the inter-molecular interactions--which vary widely from substance to substance--become the predominant phenomena of significance.  this is why one mole of water molecules occupies 18 cubic centimeters at STP, but one mole of ethanol occupies 58.4 cubic centimeters at STP, and one mole of mercury in liquid form occupies 14.8 cubic centimeters at STP.  One mole of the type of large, complex hydrocarbon molecules that make up the mixture we call "gasoline" or "petrol" may occupy around 162 cubic centimeters, if we use the specific example of 2, 3, 5 trimethylpentane.
in solids the molecules are closer yet, and we have the additional possibility of different allotropes or crystalline forms of a given solid substance which can exhibit very different physical characteristics, resulting in even further divergences from Avogadro's ideal gases, molecules of which maintain set average distances from one another at a given temperature and pressure regardless of their mass, charge, shape, etc., and which always collide with one another or with the walls of their container perfectly elastically.

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authors: نیما زکی زاده


text:
What is the difference between the KLMN and SPDF methods of finding electronic configuration?
I've always been puzzled by this because my teachers happen to use only the KLMN method, but what is the difference between the KLMN and SPDF methods of finding electronic configuration?

The KLMN(OP) method is based on electron shells, with the labels KLMN(OP) being derived from an experiment in which the spectroscopist wanted to leave room for lower energy transitions in case there were any.
K denotes the first shell (or energy level), L the second shell, M, the third shell, and so on. In other words, the KLMN(OP) notation only indicates the number of electrons an atom has with each principal quantum number ($n$).
The SPDF notation subdivides each shell into its subshells. For further information about the other quantum numbers, especially $l$, which defines the subshell, see the accepted answer to this question: What do the quantum numbers actually signify?

The K shell can hold two electrons: $n=1,\ l=0$

When $l=0$, we have an s subshell, which has one orbital $m_l=0$, with room for two electrons.

The L shell can hold 8 elections: $n=2,\ l=0,1$

When $l=1$, we have a p subshell, which has three orbitals $m_l=-1,0,+1$, with room for 6 electrons. The L shell also has an s subshell.

The M shell can hold 18 electrons $n=3,\ l=0,1,2$

When $l=2$, we have a d subshell, which has 5 orbitals $m_l=-2,-1,0,+1,+2$, with room for 10 electrons. The M shell also has s and p subshells.

The N shell can hold 32 electrons! $n=4,\ l=0,1,2,3$

When $l=3$, we have an f subshell, which has 7 orbitals $m_l=-3,-2,-1,0,+1,+2,+3$, with room for 14 electrons. The N shell also has s, p, and d subshells.
This table here summarizes the relationship between the two notations for all elements. Here is an example for scandium, $\ce{Sc}$:
Scandium has 21 electrons ($Z=21$). Its electron configuration in KLMN notation is
$$\begin{array}{cccc}
\mathrm{K} & \mathrm{L} & \mathrm{M} & \mathrm{N} \\
2 & 8 & 9 & 2
\end{array}$$
Its SPDF notation (based on the aufbau principle) is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$.

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text:
Generating mol files from a molecular structure image?
I have a question regarding mol files.For example I have this molecule-

The mol file for this is-
  Sample

 22 23  0  0  0  0  0  0  0  0999 V2000
   -2.5962    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -2.5962    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.7512    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    3.0047    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
   -4.6479    1.7042    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005    3.0047    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    4.5070    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958   -1.5023    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.6009   -2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.8967   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.8967    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.6009    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.1972   -2.2535    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0
    2.6009   -3.7559    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.8967    3.0047    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -5.1972    3.7559    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
   -3.1455    4.3052    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
    3.9014   -4.5070    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0  0  0  0
  2  3  1  0  0  0  0
  3  4  1  0  0  0  0
  4  5  1  0  0  0  0
  5  6  1  0  0  0  0
  1  6  1  0  0  0  0
  1 19  1  0  0  0  0
  5  8  2  0  0  0  0
  6  9  1  0  0  0  0
  3 10  2  0  0  0  0
  4 11  1  0  0  0  0
 11 12  1  0  0  0  0
 12 13  2  0  0  0  0
 13 14  1  0  0  0  0
 14 15  2  0  0  0  0
 15 16  1  0  0  0  0
 11 16  2  0  0  0  0
 14 17  1  0  0  0  0
 13 18  1  0  0  0  0
 19  7  1  0  0  0  0
 19 20  1  0  0  0  0
 19 21  1  0  0  0  0
 18 22  2  0  0  0  0
M  END

So,I have read that lines 5-26 are for atoms.first three columns represent x,y and z coordinates of atoms.So,my question is how are the coordinates of atoms decided?What is taken as the origin?Also,the atoms are large,so they may occupy more than one coordinate,so how do we decide what x,y,z value to take?Also,is H atoms not taken into account for creating mol files?

which program did you use to create the mol file?

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@PH13 I did not use any program.I got it from this site-
http://scitouch.net/imago/report/report.html#g2
It is the first image under image2structure(statistics)

because if you look closely, the protons of the CH3 and fluorine atoms of the CF3 group are missing

@PH13 fluorine atoms of CF3 are there but yeah no H are there.So,I thought may be in mol files we dont include protons

oh, didn't see them. they are wrong, though. it only produces coordinates in x-y-plane. origins may be the mass-center or just the center of coordinates.

Some programs for making 2D structures will store them in MOL files (with all atoms in a single plane), since the format stores all info needed to regenerate the (2D) structure. If you need 3D, you'll need something that can do basic molecular mechanics, at least geometry optimization. There are lots of options: I like Avogadro, which is free, but there are lots of other options (both free and commercial).

The MOL format specification doesn't specify an origin, so presumably any origin can be chosen, as long as the coordinate values are within the allowed range. (The format allows a range of roughly -10,000 to +100,000, but I don't know if all programs support that large a range.) For example, in one of the examples in the specification the atom closest to the origin is over 17.6 units away. Where the origin is actually set would depend on the software used to create the file.
The coordinates for each atom specify the nucleus of the atom. I think you have a misconception about atomic size. The size of an atom, and thus the space it takes up, can be defined in different ways. The representation of an atom as a sphere (with the nucleus at the center) is a convention, only approximating reality, and the size of the sphere is arbitrary (but usually based on measurable properties). The nucleus is the only point in the atom that has a well-defined location, and thus the only part that meaningful coordinates can be defined for.
In MOL files, the presence of hydrogen atoms can be implied from valence and number of other bonds. (They can also be specified explicitly if needed.) In the long set of zeros at the end of each atom line (just after the element), the fourth zero specifies to add however many single bonds to hydrogen atoms are needed for the atom's valence. The sixth zero indicates to use the element's normal valence. For example, the carbon of the $\mathrm{CH_3}$ group has only one explicit bond, so three bonds to hydrogen are implied by carbon's normal valence of 4. How implicit hydrogens are handled depends on the software (e.g., this molecule has four implicit hydrogens that, per the usual convention for skeletal formulas, are not shown in your image).
The specification for the MOL format is available from Accelrys. The this answer is based on the specification and my knowledge of chemistry. Also, it only describes the MOL V2000 format, which is what you sample file is (based on the header); there is a newer V3000 format.

Thanks a lot.I I understand that coordinates for atom specify nucleus of the atom.But how do I assign the coordinates if I have to create a mol file.I was trying to generate mol files from the jpeg images of structures.So,in that case what coordinates for the atom should I use?

If you're writing the file by hand, I'm not sure, but I suspect you could just use the pixel coordinates in the image. You'd just have to guess for fluorine and the CHO oxygen in this structure. Unless there's some specific reason not to (like writing an export plugin), you'd probably be better off using a molecule modeling (3D) or sketching (2D) program that can export MOL files.

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text:
Sodium by electrolysis through glass
In his 1926 thesis Robert Cady Burts describes the production of sodium by electrolysis of sodium nitrate through glass.
I can't find much else on this method. Are there any modern descriptions of the experiment? Videos maybe?

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See Shakashiri's chemical demonstrations. He carefully outlines the steps. As I recall it requires taking sodium nitrate into a molten state and applying a current. The elemental sodium is precipitated on the inside surface of an incandescent bulb.


text:
Stability of a carbocation, in part of camphor synthesis
Can someone explain to me why there is a rearrangement reaction of carbocation 1 to carbocation 2? To me the second carbocation appears to be less stable than the first because it is less substituted.

Thank you in advance

Can you elaborate on the context of this synthesis?  Is it a classical organic total synthesis, or enzymatic?  Also, the final product does not appear to be camphor, but rather the acetate ester of borneol, i.e. bornyl acetate.

Its a part of camphor synthesis, and its not enzymatic.

What are the reaction conditions for this step?

Just a guess but  I think, as you said, 2nd carbocation is less stable and thus it reacts easily. Steric hindrance also seems to be another factor because of which 2nd one reacts easily compared to the first one. There is an equilibrium between the two carbocations and the reaction is driven to the forward direction by the non reversible final step.

Could you add a source to this mechanism. It contains at least one obvious mistake. I.e. The two structures in the second row of the image are not equal. The numeration of the six-membered ring must be reversed and the charge is therefore on the left side. One could say the enantiomer is depicted.// On another note I highly doubt that there is a carbocation with an analogous geometry to 2. I also doubt a two step mechanism is the most economic. I'd rather guess it is concerted.

Yes of course its the solution of an exercize ! :)

Since mechanistic studies of the camphene/ bornyl acetate system are not available, I will focus on the parent compound norbonyl, and with it explaining the stability of the cation.
For quite some time there has been active debate over the structure of the 2-norbonyl cation. A very brief description of the two sides can be found in the Wikipedia article. A short history of the argument can be found at the homepage of the Royal Society of Chemistry (RSC): "The nonclassical cation: a classic case of conflict" by Mark Peplow.
In 2013 the case was finalized after I. Krossing's group was able to obtain a crystal structure of the compound. Another blog post at the RSC gives some more insight into the matter: "Crystal structure closes classic carbocation case" by Andy Extance. Finally we can speak of the 2-norbonyl cation as a true non-classical cation. The findings are also in good agreement with high-quality calculations. F. Scholz, D. Himmel, F. W. Heinemann, P. v. R. Schleyer, K. Meyer, and I. Krossing, Science, 2013, 341 (6141), 62-64.
This does in principle mean, that at very low temperatures, the 2-norbonyl cation is best described by the following resonance structures (courtesy of wikipedia):

At higher temperature the structure is quite disordered. In solution this means basically, that there is an ultra fast equilibrium (at the rate of molecular vibrations) between several structures. This would also explain the high reactivity of this compound.
Since camphene is a substituted norbornene, similar reactivity can be expected. The reaction is therefore most likely determined by the small steric effects. Since this reaction is highly selective, one can also safely assume, that the thermodynamic favoured product is formed. In this case it is the isobornyl acetate.

I have already mentioned it in the comments, but I am including it here nonetheless. The second row of your image is contains the wrong numeration and therefore depicts the mirror image. Correct is the following depiction:

Thank you very much, it make sence now, but its a bit more complex than I think. I though the final product was the cinetic and the thermodynamic favoured product. I will read the references you add. And can you tell me if you know a good software to do representation of organic molecules ?


text:
Does acetone eat through Thermoplastic polyurethane and Polycarbonate
I used nail polish remover on an expensive phone case to remove sticker residue (and it worked, no longer sticky). I didn't know it didn't mix well with plastic materials. Did the acetone damage it?

If there is no visual damage, e.g. cloudiness or crazing, the case should be OK. If there's some surface cloudiness, it might be possible to buff it out with mild abrasive, such as toothpaste or milk of magnesia.
In the future, try oil (olive, safflower, mineral etc., but not motor oil) to remove stickers. A bit slower, requiring subsequent cleanup, but safer.


text:
difference between solving and dissolving something
What is the difference between the terms 'solve' and 'dissolve'? Up until now I have used the two terms interchangeably, not having any sound scientific definitions for each. I have thought them both to mean to make a solid into a solution through the addition of a solvent (not sure if only for solids into solutions; is the mixing of two fluids also dissolving/solving because $\ce{CO2 (g)}$ can dissolve into water?).
I have not been able to find any scientific definitions on the internet that differentiate the two. it would also be helpful if you could clearly define any other terms related to the mixing of substances without a chemical reaction occurring (eg. immiscible, which I understand to be the state of two FLUIDS being unable to homogeneously mix)
Also, if both substances are liquids, which one is the solvent and which one the solute?

"Solve" is only used in "solving equations", it has an entirely different meaning from "dissolve". So searching for the difference between the two is kind of like searching for the difference between the Sun and 2-pentanone, you wouldn't get any answers.

but a solid can be said to have been solved by a solvent can it not?

No, you can't say that. (Well, you could, but nobody would understand you!)

fair enough then.

@ziggy I think you mean 'solvated' not 'solved'. Solvation is where solvent molecules coordinate to a solute.

@orthocresol, In older papers you will see "solved" used like ziggy is saying.  If you search on Google Scholar for solved in methanol, about 25% of the hits are for ziggy's usage.  The rest are mis-parsings of "dissolved", e.g. "dis- solved in methanol", etc.

If this blog were in German, Ziggy would be right.

According to the Merriam Webster dictionary, dissolve originates from the Middle English word solve, meaning to loosen or dissolve. There is no official modern English definition for solve other than 1) to find a solution, explanation, or answer or 2)  to pay (as a debt) in full.
The IUPAC Gold Book actually gives no definition for either solve or dissolve, though it defines solvation:

Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

When two miscible liquids are combined, they are said to form a solution.

Actually there is no difference between "To solve" and "To dissolve" in informal English language. But we can distinguish them in formal case based on scientific papers in chemistry. As "To solve" includes general types of finding a response; and "To dissolve" includes detailed meaning of solid solution in a ionic fluid like water.

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text:
Empirical and molecular formula
I was doing a homework assignment and I got the question wrong because I was supposed to multiply the empirical formula of $\ce{C6H10S2O}$ by 3 to get the molecular formula.
I put $\ce{(C6H10S2O3)3}$ and the teacher said it should be $\ce{C18H30S6O3}$.
My question is: isn't what I put the same thing?

$\ce{(C6H10S2O3)3}$ implies that there are $3 ~ \ce{C6H10S2O3}$ molecules bonded together instead of $1~ \ce{C18H30S6O3}$ molecule. Although they are the same thing, the correct notation is your teachers way because it shows the isometric formula for the molecule. Also if you were using that notation you should have put $\ce{(C6H10S2O)3}$ because $\ce{(C6H10S2O3)3}$ has 6 too many oxygens.

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text:
Why is condensation in dyed beverages colorless?
Does food coloring somehow evaporate and turn into nothing?   Or does the beverage itself being much more color intensive?
I had a bottle of Gatorade, colored Lemon-Lime (and flavored as such). I noticed that after leaving the bottle in the car during a hot day and returned to the car, that condensation had formed on the top parts of the bottle.
But the condensation is colorless; does this mean it is solely water that is evaporating from the drink itself as condensation occurs, leaving food coloring or the color of the beverage behind?

Yes. Food colorants generally do not evaporate; the condensate is mostly water.

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@Greg Curious, is there a reason liquid food coloring doesn't generally evaporate?

@thinlyveiledquestionmark The same reason the bottle doesn't generally evaporate: it has a much higher boiling point than water. It takes much more energy for molecules of the colouring compound to break free of the liquid than for molecules of water.

As can be seen in my answer, the food dye's melting point is considerably higher than that of water.

Given that the majority of beverages such as Gatorade is water, as stated in the comments, the condensate is predominantly water, as can be seen in the Patent Electrolyte blends providing reduced salty taste (which includes Gatorade), that water amounts to well over 99% by composition, of course this will vary. The dyes themselves don't make up much of the volume.
A common food dye used, according to the Indiana University online experiment webpage How Much Red Dye is in My Gatorade®? , is the red dye, also known as 'Allura Red'
its properties according to the Wikipedia page are stated as:

It usually comes as a sodium salt, but can also be used as both calcium and potassium salts. It is soluble in water; Its melting point is above 300 °C (572 °F).

It is likely that the majority of the food dye based salts would remain in solution.
Any salts (such as the example above), transported with the evaporating/condensing water would be too small of a concentration to cause any visible effects.
A similar question and answer is Does salt vaporise?

Yes, in a rough sense, components of the mixture evaporate separately (this is a simplification which ignores things like azeotropes) with their individual vapor pressures. This is why distillation works.
In your gatorade example, the condensate is indeed pure water. In fact, the same principle is used to purify water using something called a solar still. The remaining gatorade at the bottom will have a deeper color, and taste more strongly, in proportion to how much water evaporated - but you won't notice this without accurate assays because of how small the amount of evaporated water is (if you manage to evaporate say 90% of the water the remainder will be 10 times more concentrated, which you can notice).
If you want an experiment, try leaving a shot glass of gatorade in a well-ventilated area with low humidity. After days or weeks the level of fluid will be half of what it was or less. Now buy another fresh gatorade of the same kind and compare taste, color (put both in shot glasses) of both.

The experiment you suggest doesn't seem safe: I'd expect all kinds of bacterial and/or fungal contamination after leaving a sugary liquid out in the open for a week. A swan-neck flask would be a better bet but then we're into the whole drinking-from-laboratory-glassware thing.

@DavidRicherby There isn't a lot of bacteria or fungus that can just grow on sugar water, so I'd be surprised. Also, I doubt Gatorade is sterile, so chances are it's already contaminated.

I thought the answer was that condensation is on the outside. As such, it's the colorless water from the air, not the Gatorade, that lacks color.

wat :(  The condensation forms on the inside of a sealed Gatorade bottle in the hot sun.

This is condensation inside the bottle from being left in a hot place, not condensation on the outside of the bottle from the bottle being moved from a cold place (e.g., the fridge) to a warm, relatively humid place.

But still - Welcome to chemistry.SE!


text:
Is it possible to calculate energy of electron in any orbital and atom in Schrödinger wave model theory?
Is it possible to calculate the energy of electron in any orbital and atom in the Schrödinger wave model theory? If so, how? E.g. energy of the $3s^2$ electron of the $\ce{Na-}$ ion.

It is surely possible. Note though, that for many-electron systems orbital picture is an approximate description, but that is an approximation on which the whole general chemistry is based on, so that is fine from that perspective. The approximation is known as the Hartree-Fock (HF) model, and thus, to get the energies of electrons occupying some orbitals all you need to do is to solve the Schrödinger equation in this approximation.

As a quick exercise one can do HF calculation in, say, Gaussian, with the following input file:
#P HF/aug-cc-pVTZ Pop=Full

Na-

-1 1
 Na                 0.00000000    0.00000000    0.00000000

to get this picture of occupied orbitals:

The HOMO with the energy of $-0.01288 \, \mathrm{Hartree} = -0.3505 \, \mathrm{eV}$ is basically the $\mathrm{3s}$ orbital you're looking for (trust me). Note, however, that this energy is approximate since the basis set (aug-cc-pVTZ) is finite. We could do better than that, but that is a different story.

In response to permeakra (since he does not trust his colleagues) I quote an authoritative reference in the field which explicitly talks about individual electrons occupying individual spin-orbitals and orbital energies. As a reference I choose the infamous book entitled "Modern Quantum Chemistry" written by Attila Szabo and Neil S. Ostlund.
Quote #1 (Szabo & Ostlund, p. 50)

This Slater determinant has $N$ electrons occupying $N$ spin orbitals
  $(\chi_i, \chi_j, \dotsc, \chi_k)$ without specifying which electron
  is in which orbital.

Quote #2 (Szabo & Ostlund, p. 54)

$$   f(i) \chi(x_i) = \varepsilon \chi(x_i) \tag{2.52} $$ [...]
The
  solution of the Hartree-Fock eigenvalue problem (2.52) yields a set
  $\{\chi_k\}$ of orthonormal Hartree-Fock spin orbitals with orbital
  energies $\{\varepsilon_k\}$.

Below I also quote what Szabo & Ostlund have to say about Koopmans' theorem, because I already know that won't trust me.

Quote #3 (Szabo & Ostlund, p. 110)
The first theorem [Koopmans' theorem] constitutes an interpretation of
  the Hartree-Fock orbital energies as ionization potentials and
  electron affinities.

See, orbitals with their energies & occupancies do exist in the Hartree-Fock theory. It is the interpretation of orbital energies as ionization potentials which requires an additional theorem.

The approximation is known as the Hartree-Fock (HF) model  || HF model does NOT give energies of individual electrons, or even orbitals. HF wavefunction is antysimmetrical by electron position exchange, so there is no way to say if some particular electron occupies some orbital, only that 'orbital' is occupied or not. You are probably refering to Koopmans' theorem. It is not about energy of particular electron, but about energy required to build system with a 'hole' relatively to original one.

@permeakra, no, now I'm pretty sure, you're confused. Speaking of which particular electron occupies a particular orbital is meaningless, but speaking about the energy of an electron at particular orbital is perfectly sensible. Because an electron that occupies some orbital does have a well defined energy, no matter which particular electron it is.

@permeakra, the fact that electrons are indistinguishable does not forbid us to speak about individual electrons. For instance, in the HF approximation we talk about individual electrons occupying spin-orbitals, and that is perfectly legal (approximate, but legal). What can not be done is some sort of identification that a particular electron (say, twelfth one) occupies a particular orbital (say, $\mathrm{3s}$). But I surely can say that an electron occupies this orbital and that it will have a particular energy.

For instance, in the HF approximation we talk about individual electrons occupying spin-orbitals || Nope, HF approximation has nothing to say about individual electrons or orbital energy, it never appears in the formalism. There is so known 'Koopman's theorem',  which does not hold true and is about ionization potential, most likely it is the energy printed by provided input.  Again, it is not about individual electrons or even electron on the orbital. Read a good QC book.

@permeakra, which book is a good one? The one which you've read yourself to learn all this non-sense you're talking for already two days? No, thanks. I will better continue my road with reliable & authoritative like the one I quoted in the answer above.

Apparently you didn't read it, only quoted. A pity.

@permeakra, I got you. Only you read the right books. And only you understand them the right way. Everyone else is just quoting.

It would be great if you'd be a little bit nicer to each other. Every argument can be lead on a non-personal basis. Some readers might interpret this as a hostile argument.

No. In QM electrons are indistinguishable.
Still, it is possible to associate a somewhat characteristic value, using various spectroscopic methods, i.e virtually 'moving' some electron from one orbital to another (or removing it entirely) and calculating the corresponding energy. This energy, however, is unprecise, as other electrons 'feel' the move of the 'moved' one. Still, something is better than nothing, so people use what they can.

How does indistinguishability of electrons cause inability to calculate their energies?

@Wildcat since electrons are indistinguishable, one cannot attribute energy to some particular electron, since any exchange of two electrons gives raise to antisymmetric wavefuction with exactly same energy, that is, from observer's POV indistinguishable from original (and QM does not have internal variables). If you prefer more strict description, QM formalism gives only total energy of the system. It is possible, using some approximations, to give energy to some orbitals, but not individual electrons.

What you're saying sounds wrong for me. Look, even in the HF approximation, despite what you're saying, electrons are indistinguishable. That's why our trial HF wave function is antisymmetric product of spin-orbitals and not a simple product of them. Yes, I obviously couldn't say which particular electron occupies which orbital, but I know what is the energy of an electron is at each and every orbital.

Besides, systems of many non-interacting particles are considered in QM, and for them the many-particle wave function perfectly separates into the product of one-particle wave functions without any approximations. Yes, such systems are rather ideal, but QM in principle doesn't seem to mind that we speak about each and every particle as being in its own one-particle quantum state. There is nothing wrong with that. Of course, for interacting particles such description is approximate, but that's another story.

The "energy of an electron in an orbital" is equivalent to the energy of an orbital" so electron indistinguishability is irrelevant. Orbitals are distinguishable up to symmetry.

@Wildcat > What you're saying sounds wrong for me. || Well, you are incorrect. ||| but I know what is the energy of an electron is at each and every orbital  || Nope, you don't.  Neither HF, nor any other method provide a methodology to attribute any energy value to any electron or orbital. Koopman's theorem is about ionization energies (and does not hold true IRL), so it doesn't count, even if people may believe otherwise.

User permeakra's observation is right, but the inference is not. User Wildcat's answer is spot-on, but missed on one minor point, which I would explain at the end.
This is just a summary of Wildcat's explanation: The idea is that, we have a system comprising of indistinguishable fermions, which is why we need to use the antisymmetrized wavefunction. The LCAO-MO approximation results in a linear combination of AOs resulting in MOs, which can be allowed to occupy, given by electron densities. Variationally, the ground state would lead to the all the lowest levels being occupied, resulting in HF ground state wavefunction. The HOMO, which is $3s$ orbital, would be at a certain energy level which is definitely well defined for a given level of theory.
So, to answer your question, yes, you can calculate the energy of the $3s$ orbital for $\ce{Na^{-}}$. But, what is not possible is to calculate the energy of the $3s^{2}$ electron, because the electrons occupying the $3s$ (or any of the other) energy levels are indistinguishable. Another thing to note is that, when you talk of using a wavefunction to denote electrons, they are highly delocalized, and can occupy more than one energy level at once. Think about the case of Boron atom. Which orbital do you think the $2p$ electron occupies? The $2p_x$ or $2p_y$ or $2p_z$? They are all symmetric in nature, and hence you cannot point at one particular orbital. This turns into a multiconfigurational problem, which is beyond the scope of this discussion. Hence, the implication here is that, the electrons occupying these orbitals would be at specified energy levels, depending on which orbital they occupy.

No, I disagree. I do not miss anything. I know & perfectly understand what you're talking about in your answer and I have never said anything different. I agree that OP used a bit strange language ("the $\mathrm{3s^2}$ electron"), but that his problem and not mine.

So, I repeat my reasoning her point by point, and then you tell me what I'm missing. 1) We do have orbitals in the HF approximation. 2) HF orbitals are occupied by electrons. 3) I couldn't specify which particular electron occupies which orbital, but I could tell what is the energy of an electron on any orbital is. 4) This energy usually referred to as the orbital energy is also know as the one-electron energy, which clearly identifies its meaning. 5) Case closed.

Specifically, for the answer in question, what I say is that I could not tell which particular electron out of total 6 occupies $\mathrm{3s_{\beta}}$ spin-orbital, but I can tell what is the energy of this electron. And since chemistry is ultimately based on the RHF picture, this energy will be the same as the energy of $\mathrm{3s_{\alpha}}$ electron, thus, I simply say "the energy of $\mathrm{3s}$ electron": there are two of them but they have exact same energy in the restricted picture.

The meaning of the phrase "the energy of $\mathrm{3s_{\alpha}}$ electron" is perfectly well defined: it is the energy of an electron which occupies the $\mathrm{3s_{\alpha}}$ spin-orbital. One more time: I couldn't specify which electron it is, but whichever electrons occupies the $\mathrm{3s_{\alpha}}$ spin-orbital, I surely can tell its energy.

And another short remark since I notice you talked about "the lowest levels being occupied": conceptually there is no virtual orbitals in the HF approximation. Initially you have as many spin-orbitals and as many HF equations as many electrons there are in your system. In the RHF case where all spatial orbitals are doubly occupied you have as many spatial orbitals and as many RHF equations rewritten in terms of them as many pairs of electrons there are in your system.

Then in the Roothaan-Hall finite basis set treatment virtual (unoccupied) orbital arise once the number of basis functions is greater that the number of electron pairs. But! 1) These orbitals are just a byproduct of this particular way of solving RHF equations. 2) The quantities related to them do not have the same physical meaning as those attributed to the occupied orbitals. In particular, virtual orbital energies are not the energies an electron being placed here will have, they lack this physical meaning.

Your answer is absolutely fantastic, no denial, I just wanted to add a point which you missed, and hence I added an answer and not a comment. 2. The question was midway between a genchem question and a quantum chemistry question. It is our duty to clarify the distinction between electron and orbitals. 3. Note the stress on the 2 in the $3s^{2}$ orbital, posted by the OP. I wanted to make sure that the OP understands that except for basic VB theory expressions, $3s^{1}$ and $3s^{2}$ electrons are practically one and the same.


text:
Nomenclature of a poly-nitrile

For this compound, I think the ethyl group should be in the parent chain and the numbering should start from the right-most carbon. Hence, the compound should be named 2-methylhept-1-ene-1,3,4,5-tetracarbonitrile, instead of the name given above. Am I wrong somewhere?

Your approach is correct.
The given compound includes the characteristic groups $\ce{-CN}$, which are expressed as suffix ‘carbonitrile’. Since there is only one kind of suffix, the order of seniority of suffixes is not relevant in this case.
Functional suffixes are always attached to the name of the parent structure, which is modified by ‘ene’ or ‘yne’ endings if necessary.
When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. These criteria are applied, in the order listed, until a decision is reached.
For the given compound, there is a choice between two different principal chains:
    and
The first relevant criterion in this case provides that the principal chain has the maximum number of substituents corresponding to the principal characteristic group (suffix).

P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the principal chain of the given structure has all four substituents $\ce{-CN}$ corresponding to the suffix ‘carbonitrile’.

Furthermore, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound.
In particular, the order of seniority provides that low locants are given first to principal characteristic groups (here: carbonitrile) before low locants are given to multiple bonds (here: the double bond) and before low locants are given to detachable alphabetized prefixes (here: methyl).

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the resulting name is (1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile (not (6E)-6-methylhept-6-ene-3,4,5,7-tetracarbonitrile) since the locant set ‘1,3,4,5’ for the principal characteristic groups is lower than ‘3,4,5,7’.


text:
Is NaCl in water considered "solution" despite of NaCl's having some precipitation?
Is $\ce{NaCl}$ in water aqueous? For a certain amount of $\ce{NaCl}$, it can dissolve in water but if you add more $\ce{NaCl}$ it can't dissolve any more.
So, if in a certain reaction $\ce{NaCl}$ is formed with water:
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)}$$
My question is does $\ce{NaCl}$ in the previous reaction totally dissolve in the water formed? Or in other words is this equation correct?
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) + NaCl(s)}$$
Another question is why do we write (l) beside water rather than (aq) since it's part of the aqueous solution?

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here.

When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules  in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction.
Hence the first equation
$$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$$
will be the correct expression.

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here.


text:
What is the standard potential of carbon?
What's the value of standard potential of carbon?
The hydrogen and Carbon are in the electrochemical series despite being non metals hydrogen has zero potential difference what about carbon?

To look up the position of carbon in the electrochemical series, you would have to guess what species carbon is oxidised to. This is most likely carbon dioxide; you would then have to look up the value of $E^0$ for the $\ce{CO2}/\ce{C}$ couple in a textbook or some other data source. The standard reduction potential for the $\ce{CO2}/\ce{C}$ couple is given in Shriver & Atkins 5th ed as +0.207 V. Loosely speaking, this means that carbon is harder to oxidise than hydrogen, or in other words, carbon is below hydrogen in the electrochemical series (if you take Li to be at the top).
Just a note: The standard reduction potential for the $\ce{H+} / \ce{H2}$ half-cell is 0 V solely because we decided to define it as such.


text:
Stereochemistry of epoxide hydrolysis under acidic conditions
The epoxide on the top of the picture, in aqueous acid hydrolyzes to the compound below. What is the mechanism of this reaction? Standard epoxide opening in acidic conditions would happen at the more substituted end, therefore the secondary -OH group would end up below the plane of the paper, and the tertiary center would racemize to a significant extent. What is happening here?

The tertiary carbon of the epoxide is more substituted, so I would expect the C-O bond length be slightly longer after protonation and the antibonding orbital having a larger coefficient than the carbon in the other bond. Therefore I would expect an SN2 like attack at the tertiary carbon, giving the depicted product.

Isn't the quaternary C atom more substituted?

Standard epoxide opening in acidic conditions would happen at the more
  substituted end

This premise is false. Usually $\ce{S_{N}2}$ is faster than the cation-stabilized pathway (most textbooks don't even bother to explain the subtleties involved), yielding mainly the less substituted (less sterically hindered) product. It is not clear cut which of the two mechanisms will be predominant in an acid catalyzed epoxide cleavage given the reactants and reaction conditions.
See also Regioselectivity of acid-catalyzed ring-opening of epoxides (especially the second answer and then the excerpt from Clayden et al.).


text:
Does esterification require reactants to be present in the same plane?

I believe that the first step is the Cannizzaro reaction which will give (c) as a product.
But since we are also adding $\ce{H+}$ in the second step, shouldn't an esterification reaction take place? That would give (d) as a final product.
However the answer given is (c).
My only guess as to why esterification wouldn't take place is the fact that the phenyl groups are in perpendicular planes, therefore esterification is difficult?
Is it necessary to have all the reactants in a plane for esterification?

The nucleophile doesn't attack from the plane of the carbonyl group, it attacks below or above since this is where the $\pi^*$ orbital is (see http://www.chemtube3d.com/Nucleophilic%20substitution%20at%20the%20carbonyl%20group%20-%20Acid-catalysed%20ester%20hydrolysis%20and%20transesterification.html). Personally I don't see why esterification would not occur though.

structure d) lacks methylene groups so it can't be a product.

The two commenters are collectively correct. But to clarify and augment their answers I would say that structure (d) cannot be correct because it is missing 2 carbon atoms relative to (c), and if you tried to cyclize (c) by forming esters, the result would be highly strained. Try making a molecular model of it and see how it looks.
Below is the product you would get if (c) cyclized. What is the difference between this product and (d)?

Did not notice the missing methylene groups. So you say that even if the methylene groups were present, esterification wouldn't take place because of high ring strain?

That would be my back-of-the-envelope prediction. But I could be wrong. In any case it is not one of the choices presented in the question. Neither is the asymmetrical Canizzaro product.

I don't think the product will be overly strained, esp. when considering that the biphenyl parent is already rotated out of plane. I wouldn't be too sure about the transition states, though, and whether the intramolecular condensation would work well... anyhow, here's an image optimised with gfn2-xTB: https://i.stack.imgur.com/9uKLU.jpg


text:
Naming of substituted amines and amides
While naming substituted amines and amides, should we consider the "N" while deciding the order of substituents or simply place the substituents attached to nitrogen in the beginning?
E.g. Should the compound be named:

3-chloro-N-ethyl-4-methyl-N-propylhexanamide

or

N-chloro-N-propyl-3-chloro-4-methylhexanamide.

Those are two different compounds!

Substituents are listed alphabetically by the name of the substituent. The N in front of the substituent is a locant, and thus is not used when alphabetizing:

3-chloro-N-ethyl-4-methyl-N-propylhexanamide is correct because the substituents are in correct alphabetic order:

chloro, ethyl, methyl, propyl


text:
How does an hybridised orbital become unhybridised after bond cleavage?
In $\ce{CH4}$, $1 s$ and $3 p$ orbitals are hybridised into four $sp^3$ orbitals. But on homolytic bond cleavage an unpaired electron remains in an unhybridised $p$ orbital?
Similarly in heterolytic bond cleavage of $\ce{CH4}$, the methyl carbocation has an unhybridised empty $p$ orbital?

We have had this confusion many times before. See here for some detail. Hybridisation is simply a mathematical quirk to make valence bond theory work better. It doesn't exist in the real world. In methane, the atomic orbitals overlap to form a set of molecular orbitals. $sp^3$ orbitals do not come into the final bonding description of methane (at least in the MO theory version).


text:
How does SO2Cl2 react with alcohols?
Is the mechanism similar to the reaction of $\ce{SOCl2}$ with alcohols?
If not, then how is it different? We have been taught about the latter at school but my friend quizzed me about $\ce{SO2Cl2}$.
Please give the detailed reaction mechanism.

I was asking on high school level, we have been taught the reaction of Thionyl Chloride with alcohols to give a carbon with double-bonded =O and -Cl ( sorry I don't know the name for the functional group). How does SO2Cl2 react with alcohols, or other functional groups? @Mithoron

According to this article a mixture of products is formed, with alkylchlorsulfonates $\ce{Cl-SO2-O-R}$ being dominant products. These molecules are resistant to abstraction of sulfur trioxide, probably because trioxide is a very strong Lewis acid, unlike dioxide.

SO2Cl2 is source of Cl2 gas rather than Cl-. So it doesn't react the same with alcohols. It will form HCl when in contact with water - so a lot of its chemistry with organic molecules will originate from that process. Used with a radical initiator it will convert certain functional groups to alkyl chlorides.


text:
Why note the solvent level on chromatography paper when it is removed from the chamber?
Why is it important to mark solvent level on chromatography paper when you remove it from the chamber?

The distance traveled by a compound in thin-layer chromatography is a characteristic property of that compound, but it is also dependent on the size of the paper. Compounds will travel farther if the solvent travels farther on longer paper. The retention factor accounts for this by dividing by the distance traveled by the solvent.

This could be a good answer @BenNorris

Related: http://chemistry.stackexchange.com/questions/10054/how-do-i-find-rf-values-length-size-of-paper-in-chromatography/10055

Building on my comment, marking the distance that the solvent traveled allows us to calculate the retention factor $R_f$ (or apparently retardation factor, according to IUPAC meddlers who need to rename things with perfectly good names that everyone else uses). The retardation factor is the ratio of the distance traveled by the spot to the distance traveled by the solvent:
$$R_f=\dfrac{d_{spot}}{d_{solvent}}$$
If you are only performing one thin-layer chromatography experiment (TLC) ever, then you may be okay not marking the solvent distance. Since everything is on one piece of paper or one plate, you can compare readily:

From the above TLC picture, we can easily tell that unknowns A and B are probably the same compound as the standard, but unknown C is something else.
Now, what if we have two TLC plates/papers?

We can draw the same conclusions about the relationships between standard 1 and unknowns A, B, and C. Likewise, we can make some conclusions about the relationship between standard 2 and unknowns D, E, and F. However, we cannot easily determine if unknown F is the same compound as unknown A.
Determining the $R_F$ value lets us do that. Using the same solvent system, the distance traveled by the solvent and the distance traveled by a compound are both proportionally related to the length/height of the plate/paper. The solvent can travel twice as far on a paper that is twice as long. It turns out that the compound being analyzed also travels twice as far on a paper that is twice as long. The the ratio, $R_f$ is constant.
Let's drop in some numbers:

Now we can calculate the $R_f$ values for each spot:

After calculating the $R_f$ values, we can make some conclusions (assuming there are only two compounds in the entire experiment):

Unknowns A, B, D, and E are the same compound as Standard 1.
Unknowns C and F are the same compound as Standard 2.


text:
What factors contribute most to solubility of ionic substances in highly polar solvents, esp. ionic liquids and DES?
I am trying to determine a method to dissolve a mineral salt with a lattice energy of ≈18000 kJ/mol. I know that lattice energy is the most significant factor affecting solubility of an ionic substance in highly polar solvents, and my general understanding is that the more polar (or locally polar), the greater the ability of the solvent to dissolve materials with higher lattice energies, by contributing more hydrogen bonding and/or more efficient charge distribution.
Hence I have been testing solubility in an array of Deep Eutectic Solvents (DES) and doing research into solubility of this and similar compounds in other ionic liquids, and reached out to an expert in the field.
I have found that while these solvents are typically able to dissolve compounds with lattice energies unto around 12000 kJ/mol. in great quantity, solubility quickly drops off around 15000 kJ/mol. and the professor I contacted said that their research indicated that without other agents, this material would be insoluble in the ionic liquids they worked with because of it's high lattice energy.
My question then is what other factors would be most important in increasing the solubility of this compound in ionic liquids, anything from adding highly soluble cations, chelating agents, type of ionic liquid I might use, or any other other factors that play into solubility, and could be tuned to optimal conditions?

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text:
Dissolving ability of octane vs nonane on neoprene
I have been working with straight chain octane and nonane. The neoprene seals in some parts of the apparatus are allegedly far more vulnerable to the deleterious effects of octane than nonane - why? [I assume it's a solvent thing]

Which isomers of octane and nonane are you using? I'd expect the effect will depend on that.

No isomers, just the single chain

@DirkBruere That's still an isomer.

The difference between singular and plural - you said "isomers"

I can see someone else edited the question - could you please confirm you're asking about n-octane and n-nonane? Thanks.

Yes - nice straight carbons lined up in a row

Very interesting. Is there any documentation on this effect or have you done any tests yourself?

Using this guide: http://mykin.com/rubber-chemical-resistance-chart-4


text:
InChIKey generation from InChI
I am trying to find some good resources to understand generation of InChIKeys from the InChI. There seems to be no resource available online. I have gone through the Wikipedia article for InChI, but it hardly contains any information as to how the InChI keys are generated. Can anyone provide me with this information as to how we can generate InChIKeys from InChI. Also, if you happen to find any good resources for the same, please do share them with me.

InChI is the acronym for the IUPAC International Chemical Identifier. It is basically as system, to catalogue molecular information. Its prime advantage is, that it is just a text string. That basically means, that it can easily be understood by machines. The other advantage is, that it is also human readable. It consists of several layers: the molecular formula, connectivity of the heavy atoms, placement of the hydrogen atoms/ stereochemical information. This shall not be confused with the IChIKey, which is a string of 27 characters derived from the hash code of the InChI string.
A quite thorough presentation/ documentation of the software and its algorithms (including the generation of the InChI string and the InChIKey) is available as an open access paper (currently asap): Stephen R Heller, Alan McNaught, Igor Pletnev, Stephen Stein and Dmitrii Tchekhovskoi, Journal of Cheminformatics, 2015, 7:23. Since the abstract is not very comprehensive, I will quote a short passage from the introduction, which outlines the contents of the article:

We then describe the workflow of InChI generation (normalization, canonicalization, and serialization stages), as well as generation of the compact hashed code derived from InChI (InChIKey); the related algorithms and implementation details are briefly discussed.

In response to R.M.'s comment I am also including the introductory sentence of the linked paper, which describes what an IChIKey is:

InChIKey is a compact chemical identifier derived from InChI. The InChIKey is always only 27-characters long.

For a more technical description of the algorithms and the scope of the InChI string and the InChIKey, the reader is pointed towards the technical manual: Stein S, Heller S, Tchekhovskoi D, Pletnev I. IUPAC International Chemical Identifier (InChI) InChI version 1, software version 1.04 (2011) Technical Manual. The description of the InChIKey can be found in section V, pp 53-62:

The InChIKey is a character signature based on a hash code of the InChI string. A hash code is a fixed length condensed digital representation of a variable length character string. Providing a signature derived from an InChI string is helpful for search applications, including Web searching and chemical structure database indexing; also, it may serve as a checksum for verifying InChI, for example, after transmission over a network.

Please note that the InChIKey by itself does not contain enough information to restore the molecular structure. This is also mentioned in the open access paper:

A disadvantage of the InChIKey is that one loses the ability to algorithmically restore a structure from a textual label: InChIKey is a structure-based registry-lookup identifier [...].

But these are not the only sources. As an IUPAC member, most information relating to the developement and implementation can be accessed freely at iupac.org. Since itself is an non-profit organisation, it also has its own website, www.inchi-trust.org. This includes a very extensive technical FAQ (html and pdf) and since it is open source software also the source code.

The software is still actively developed by the chemical community and the trust itself. As of the current version (1.04, Sept. 2011) there are still several shortcomings, which have bean addressed in Structure that breaks InChI.

@Martin-マーチン I'm not sure I'd categorize 1.04 as actively developed, since there haven't been updates since 2011. I suspect there will be a InChI v2 at some point, but otherwise it's a great answer.

@Geoff Thanks. I agree, active development is something different. Take it as an indirect quote (and a little bit of a pun).

@Martin-マーチン Thanks a lot.I was actually looking for the actual algorithm which takes place internally in the softwares used for conversion.I got a good explanation in one of the links mentioned in the technical manual you referred.Thanks again.

This probably needs an update… we have inchi 1.06 from December 2020. So it'sn't really active development but not standstill either.

According to Wikipedia, the InChikey is generated by a hashing algorithm, like the SHA-2. The InChi key is designed for use in search, and thus is not necessarily designed to be comprehensible to the end user.
If you are looking for a way to generate InChi keys from InChis, there are a number of software packages that do so, including the open source program from IUPAC via the InChi Trust: http://www.inchi-trust.org/downloads/.
InChi is published under a more permissive variation of the GPL, so in addition to the standalone executable, you can download the code as well. Then, you can sift through the code for the hashing algorithm and figure out what it is doing. InChi even has a project on SourceForge. The provided documentation at these locations likely includes the workings of the algorithm, but remember that InChi key was designed to produce a string of characters from a structure that a computer can match; the end user is not necessarily supposed to interact with or parse the InChi key.
If you are looking for existing programs that can also produce InChi keys, IUPAC provides the following list:

http://cactus.nci.nih.gov/chemical/structure
http://www.chemspider.com/inchi.asmx
http://pubchem.ncbi.nlm.nih.gov/edit/
http://www.acdlabs.com/download/chemsk.html

The first three comfortably sit in your browser.

Lots of information in this link with references,
http://www.inchi-trust.org/technical-faq/

Since I recently looked into the matter myself, I know that your answer is the correct reference and in this case probably as valid as it can get. In general (for future references), however, it would be preferable to include the essential parts of the answer here, and provide the link for reference. This is probably why this was flagged as very low quality.

@Martin-マーチン IUPAC is the go to place for the OP's question.  Thanks for the link but the OP's question was quite broad.

I know, that is why I dismissed the flag and gave you and upvote, too.  I just wanted to inform you for further references.

@Martin-マーチン point taken, and thanks for taking the time and trouble to respond.

FWIW, link-only answers are not appropriate since the link might become invalid in the future. (Though unlikely in this case) The point is, we don't like someone to visit our Q/A and then be redirected to where they could've gone to with a Google search. It's really appreciable if you provide some of the relevant info in that FAQ here.


text:
Why are 'sec-' and 'tert-' prefixes printed in italics?
While reading about nomenclature of branched alkyl groups, I read that certain common names are also approved by IUPAC. Among those names, only few prefixes are italicised (namely sec- and tert-) and others like iso and neo are printed normally. These names are documented in the IUPAC blue book. What can be the reason for italicising sec- and tert- and leaving out the others?

No, no - I meant that sec and tert are used with "-" and that's why they are italicised.

From the ACD Labs page about IUPAC italicization,it is stated that

Italicizing mainly serves to mark letters which are not involved in the primary stage of alphabetical ordering.

According to the American Chemical Society Style Guide (p. 235), the use of italics for

positional, stereochemical, configurational, and descriptive
  structural prefixes when they appear with the chemical name or formula.

(There are several prefixes in the list).
These are separated from the chemical name with a hyphen and are not capitalised


text:
Why does hyperconjugation help for ring cleavage?

The question was to rationalize it the ring cleavage happens concerted or non-concerted.
The papers "The Mechanism  of  the Thermal Decomposition  of
1-Pyrazolines and Its Relationship  to
Cyclopropane Isomerizations" (Crawford 1965/1966; Original papers: here and here) have shown the loss of $\ce{N_2}$ is supposed to be concerted.
I think the C-C/C-D(H) bonds of the central position of the ring are involved in a hyperconjugative fashion, so its hybridization does matter.  The out-of-phase linear combination of the $\sigma^\star$ orbitals looks like a p-orbital, therefore I expect a through-space interaction.
I've drawn the the out-of-phase combination as following:

Now my problem is, that I don't see why the out-of-phase combination is involved in hyperconjugation, because I think the overlap is best in the educt, less good in the transition state and has no orbital overlap to the product because the out-of-phase $p$ orbital is kind of perpendicular to the $sp$ Walsh orbitals which lies in the C-C-C plane.
How can hyperconjugation help to understand the observed kinetic isotope effect results?
Summary:

How does the FMO picture (HOMO-LUMO) with all involved bonds for hyperconjugation looks like?
Can the ratio 66%:34% be explained?
What does reactant 3 make compared to 2?
How can the ratio 1.07/1.20 be explained?

Do you have any ideas?

Maybe these could help:Article1, Article2

That refers to this question: http://chemistry.stackexchange.com/questions/20405/inductive-effect-of-hydrogen-isotopes

A couple of questions. First, why do you think that hyperconjugation matters here in the first place? And secondly, what this 'trick' of making linear combination of MOs is all about? Honestly, this is the first time I encounter linear combination of molecular orbitals. I mean, is this even legal? :D

This is a bit late, but my instinct tells me you form the 1,3 diradical. The fusion to a cyclopropane seems pretty reasonable. The alkene is formed from a 1,2-hydrogen atom shift (I admit, this is a strange abstraction), but the formation of the alkene (and the fact that this is flash vacuum pyrolysis) make me think this is not completely unreasonable. Since the mechanism involves breaking the H/D bond, you would expect some selectivity 1:1 versus 2:1. The fact that there is essentially no primary KIE observed is consistent with diradical formation as the r.d.s.
The other reaction, I suspect, forms the cyclopentene with the deuterium substitution on at the 3- position. The secondary KIE would be consistent with diradical formation.


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Regioselectivity in Ziegler–Natta polymerisation of α-olefins
Why does the Ziegler–Natta polymerization display the regioselectivity shown in the image? Internal nucleophiles (coordinated to the metal) should attack the alkene on the less substituted end, like in the Heck reaction, whereas external nucleophiles should attack the more substituted end (like in the Wacker reaction). Why is this not the case for alkene polymerisation?

Could the regioselectivity be simply down to the sterics due to propene orientation towards the metal?

Good point. The palladium complex in Heck reaction shows different regioselectivety as it is planar, right?


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Periodic table groups - which grouping is "right"?
In searching online, I've noticed there are a lot of different ways to group the elements of the periodic table.
Take mercury in the two tables linked below, for example:

http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg
http://www.sciencegeek.net/tables/PToE_basic.pdf

On the first table, it's explicitly outside of the transition metals. On the second table, it's included in the "transition metals" group.
Mercury is an example of this inconsistent classification, but in general, why is there no canonical grouping for all elements, across the periodic table? I see some tables mentioning "metalloids," but others not doing so. Or, sometimes elements are classified in entirely different groups altogether (like zinc, cadmium and mercury in the above-linked tables). Or, the lanthanides and actinides being included sometimes in the transition metals group, and sometimes not.
So which classification is right? Or rather, which one is the "most right"? Really: which one should I focus my efforts on remembering?

There, removed it. As a reference for those who can reach it: http://sciencenotes.org/wp-content/uploads/2014/11/periodic-table-printable.png

I would choose the more recent one. Besides ... what's wrong with the one from Wikipedia?

or http://www.webelements.com/

Can you make your question more specific, in particular as to the elements you're interested in? 'Alkali metals', for example, are pretty unambiguous in my mind.  I agree that classification of elements like $\ce{Hg}$ is somewhat more challenging -- perhaps you should focus your question there, if that's what you're most interested in? Don't hesitate to post multiple questions, one for each grouping you want to learn more about.

I'm just looking to understand why all the periodic tables don't have the same groups. Yes, mercury is a good example, but in general why is there not a canonical grouping?

There is none because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.

You shouldn't focus on remembering anything that's in the periodic table. That's why we have the periodic table, so you don't have to remember.

There is no "most right classification scheme" for the elements of the periodic table, worth memorizing above all others.  This is because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.

As question is closed, I'll comment here: The specific reasons that Zn,Cd and Hg are usually not considered posttransition elements are: 1: they don't have partially filled p shells; 2: they're without doubt metals. On the other hand, they differ from other transition elements because they have full d shells. and therefore have different properties to typical transition elements. For example, Zn has only 1 common oxidation state (+2 corresponding to loss of its two s electrons) and generally forms colourless compounds (full d shell means there are no energy transitions in the visible reason)


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Trend in results for reproducibility of iron and phosphate
Periodically in our lab, we have a reproducibility test for the various samples that we analyze. One such sample is boiler water. Two of the parameters that we test are phosphate in boiler water and iron in boiler water. Random error aside (shaking of sample, measurement bias, etc.), why are my iron results usually lower than the original and my phosphate is usually higher? Is there any chemical basis for these trends?

Is your boiler water treated with phosphates?


text:
3-ethyl-2-methylpentane-2,3-diol with sulfuric acid
I was studying pinacol pinacolone reaction when I came across this. I believe the hydrogen cation will attack the hydroxyl group of carbon attached to two ethyl groups (since this carbocation is more stable due to the inductive effect), then methyl group will migrate and hence we get 3-ethyl-3-methylpentan-2-one, but the book says the answer is 2-ethyl-2-methylpentan-3-one. Is the answer in the book wrong or I am making a conceptual mistake?

The name ‘2-ethyl-2-methylpentan-3-one’ is not correct. The systematic name is 4,4-dimethylhexan-3-one.

With the correct starting compound, 3-ethyl-2-methylpentan-2,3-diol, predicting the product is tough! Here is the reaction scheme for this compound:

Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic.
Carbocations B and E are both tertiary carbocations that seem equally likely to form.
The rearrangement from B to C involves an ethyl migration, which seems less likely than the methyl migration involved in the rearrangement from E to F.

Compound F is 3-ethyl-3-methylpentan-2-one, which is what you chose as the major product. Compound C is 4,4-dimethylhexan-3-one, which is what the book says is the major product.
Without experimental evidence, I would have trouble choosing between C and F. If I had to choose, my first instinct would be F because it involved the methyl shift instead of the ethyl shift. Methyl shifts happen faster (methyl is smaller) than ethyl shifts.

My reasoning (and probably yours) is wrong, and here is why. We can only consider methyl > ethyl from the same carbocation. For example, in the following rearrangement:

In the rearrangment step at the end of the sequence, the methyl group is more likely to move than the ethyl group. We can make this comparison here because the pathway is identical up until the rearrangement.

In your reaction, the pathways toward the generation of the carbocations are not equivalent.

In other words, if carbocation B forms preferentially over carbocation E, then it might not matter that the methyl group migrates faster. We get more ethyl group migration because we have more of that carbocation formed. If carbocation B forms preferentially, one of my two assumptions about A, B, D, and E above must be incorrect. I will restate them below for convenience:

Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic.
Carbocations B and E are both tertiary carbocations that seem equally likely to form.

I'll leave it to you now. Which assumption is not correct? Why is that assumption incorrect? Why do cations A and B form preferentially over D and E?

I'll give you a hint: Have you considered steric factors?

Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.
A quick analysis of this reaction gives me the following:

3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation

The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.

Pinacol rearrangement from cation A

A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal.

Pinacol rearrangement from cation B

Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.
Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?

Which product is more likely the major product?

Rather than tell you, I will give you a hint:
One rearrangement produces a thermodynamically more stable cation than the other.

i made a mistake in the naming sorry.please see now

Thanks for checking. Rather than delete this answer, I am going to add a new one since the two are very different. This answer is still a good analysis for the wrong starting compounding.


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Why replace c with v in Broglie hypothesis
I am in XI and I am currently studying atomic orbitals, I stumbled up and am stuck at the following step of de Broglie's hypothesis.
I did a little search on-line and found almost same derivation everywhere, the only place I found a more complex and supposedly accurate formula was the Wikipedia itself, though it did not explain the derivation.
Few such sites:

http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength
http://entrancechemistry.blogspot.in/2012/09/de-broglie-equation.html

In the derivation of the formula $$\frac{1}{\lambda}=\frac{h}{m\color{green}c}=\frac{h}{m\color{red}v}$$
we replace $c$ with $v$. I want to ask why?
As of Wikipedia it gives a better formula with the Lorentz factor $\gamma$ :-
$$\frac{1}{\lambda}=\frac{h}{\color{red}\gamma m_oc}=\frac{h}{m_ov}  \sqrt{\color{red}1\color{red}-\frac{\color{red}v^\color{red}2}{\color{red}c^\color{red}2}}$$
Can anyone give the derivation or a hint or a logic that helps understand why the derivation from $E=mc^2$ and $E=h\nu$ gives the De Broglie equation.

http://physics.stackexchange.com/questions/101556/de-broglie-formula-inconsistency

http://physics.stackexchange.com/questions/1570/calculating-de-broglie-wavelength

C is for radioactive rays while v is for the matter waves..here c is for a particle travelling with light speed, i.e., radioactive rays, so we replace it with v velocity of any particle for its matter wave
$E=mc^2$ and $E=hv$.
Therefore,  $$mc^2=hv$$
Or $$mc^2=\frac{hc}{k}$$, where k=wavelength
$$mc=\frac{h}{k}$$
Or, $$k=\frac{h}{mc}$$
   Replacing c by v we get
$$k=\frac{h}{mv}$$
Note the difference between matter waves and electromagnetic waves. E.W. are associated with electrical and magnetic field while matter waves are not so E.w. travel at 3×10^2 m/s. While matter waves travel with different velocity. The equation of de broglie with c is valid, it can be treated for finding the wavelength of electromagnetic waves, or electrons. But the velocity of matter waves are quite smaller as compared to light speed. So he used a general v in place of c V can be equal to c when the particle is travelling at light speed

but c is the constant in the eqation $E=mc^2$ then how and why we replace it

We consider c as the velocity of the particle travelling in the wave, but a particle can travel at light speed only when it is electromagnetic wave. But for matter wave the velocity is variable. So just for those particles the velocity is v. We assume the velocity if particles in electromagnetic waves to be c so we write that equation

can we really replace a constant with a variable of course the speed of light is a property at which it goes and hence is confused with velocity. According to special relativity, c is the maximum speed at which all matter and information in the universe can travel. It is much like Planck'csconstant so I cannot and should not replace the Planck's constant with a variable of same units, or things can go serious wrong.

@Adityaultra c is not the maximum speed at which matter can travel. It would require infinite energy to do so. http://physics.stackexchange.com/questions/1557/accelerating-particles-to-speeds-infinitesimally-close-to-the-speed-of-light

de Broglie suggested that a moving body behaves in certain ways as though it has a wave nature. His conjecture of dual nature of matter was based on two points : $(1).$ dual nature of radiation, and $(2).$ nature loves symmetry.
de Broglie deduced the connection between particle and wave properties from the Einstein-Planck expression for the energy of the electromagnetic wave and the classical result for the momentum of such a wave. The two expressions are,
$E=hf$
and $P=\frac{E}{c}$
From these equations we get,
$\lambda=\frac{h}{P}$
de Broglie said that this equation is a completely general one that is applied to a material particle as well as photons. The waves associated with the moving particle are called matter waves or de Broglie waves.
de Broglie wavelength is given by,
$\lambda=\frac{h}{mv}$

That hypothesis worked & established as a theory later?

Yes, the concept of matter waves was used to justify quantization of angular momentum as proposed in Bohr's model. If we consider standing electron wave, then to maintain a standing wave over the circumference of a circular orbit, the wavelength must be an integral fraction of that circumference- 2πr=nλ=nh/mv ,which gives mvr=nh/2π


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Limits on contaminant size in filtration for extraction, and standard alternatives?
Following up on How to perform solvent extraction of a solid I realized that I haven't quantified the size of the contaminants.  I know they aren't soluble in acetone and the target is, but I don't know how small the contaminants are.
Once the target is dissolved presumably I must assume that suspended contaminants are as small as their molecular size.  In this case is a filter-based extraction still possible?  I don't see lab filters like extraction thimbles sold by filtration size.
In a case like this is the correct approach to run the solution through a centrifuge and then suck out the middle solution with a pipette?  (Or what is the lab best practice for mechanical extraction following spinning?)

There are two approaches regarding the size selectivity of solid to be filtered. Most common is some sort of "tortuous path", that is a fine weaving of some type of fiber having some average size cutoff regarding what gets filtered out and what passes through.  Secondly there are filters that are actually grids of a specific size of hole or square to restrict the solids from passing through. These look like a screen sieve through a microscope and I've seen them as small as 1 micron, though this is extremely small and usually used for sterilization.
In reality, you are more likely to have problems with your filter becoming clogged than with too much solids getting through. Your idea of centrifugation is good, but as a pre-filter step to help prevent clogging the filter.  In this case you could either decant or suction off the bulk of the liquid before the finer filter finish. The requirement for final filtration size is application dependent.


text:
Formation of magnesium bismuthide
Magnesium shows an oxidation state of $+2$ and bismuth shows an oxidation state of $+3$. Both do not show a negative oxidation state. Then how is magnesium bismuthide ($\ce{Mg_{3}Bi_{2}}$) formed?

Both do not show a negative oxidation state. || incorrect, bot are shown to have negative oxidation state in specific circumstances.

@permeakra while Mg3Bi2 almost certainly brings out an oxidatio state of -3 in Bi due to the great preference of Mg for +2, I am not convinced that there is a case where Mg truly forms a negative oxidation state. I would expect it to form simple alloys with the elements to the left and below it in the periodic table.

@steveverrill Oups! mistaken it for sodium. Sodium is shown to have -1 oxidation state in, for example, salts with potassium in crown esters. Mg, indeed, is unlikely to form negative oxidation state.... Though I wouldn't bet on it in long run.

I would not bet on magnesium most alkaline earth metals adopting a negative oxidation state. Negative oxidation state for an alkaline earth metal would require use of outer-shell p orbitals in significant bonding, and these orbitals seem too high in energy and diffuse for that to happen ... except maybe for Be. For now, no oxidation state is known below zero for any AEM.

Bismuth can have any oxidation state from +1 to +5 . As it is in the same group as nitrogen and phosphorous, its most commom oxidation states with strongly oxidizing substances such as halogens and oxygen are +3 and +5.
However with a strongly electopositive (reducing) element, bismuth can have an oxidation state of -3, just as nitrogen can. While it is true that the oxidizing strength reduces as we go down the group, the "softness" (tendency to of an element to adapt its oxidation state to the preferred oxidation states of other elements) goes up.
Without definite information I can't discount the possibility that $\ce{Mg3Bi2}$ is an alloy, but the fact that it has an even stoichiometric ratio of elements suggests that it is in fact a compound in which magnesium has an oxidation state of +2 and bismuth has an oxidation state of -3.

Mg3Bi2 is a semiconductor, thus dismissing alloy theory.

And, Springer Materials identifies Mg3Bi2 as having the La2O3 structure, thus only heteroatomic bonds.  A metallic alloy would have higher coordination number and some homoatomic bonds.


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Reactions involved in Titanium Anodisation
I am writing up a lab report investigating titanium anodisation and producing oxide layers. After some searching, all I can find is the reaction $\ce{Ti + O2 \to TiO2}$, which I am having no trouble understanding.
However, I seem to be confused/mislead by the simplicity of this. The whole unit is about electrochemistry and redox reactions but I cannot find any examples (or seem to produce any either - I have tried generalizing from the rusting of iron in water) of half reactions or anything similar.
I have come about as far as including the redox reactions for the hydrolysis of water and then jumping to the above reaction but am unsure where to go from here.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

I was able to locate what appears to be a useful summary of reactions and reduction potentials that are all titanium-specific here.
The reaction $\ce{ Ti + O2 -> TiO2}$ is illustrative of $\ce{Ti}$ being oxidized while oxygen is reduced. I agree with you that it is simplistic-looking and might not appear as traditional an example of redox chemistry as you require.
You might investigate the high-temperature reaction of titanium metal and water (steam), which produces $\ce {TiO2}$ and $\ce{H2}$ gas (titanium is oxidized, hydrogen is reduced), which might be a little clearer. This reaction (and several others) are given on the page I've linked above.


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Benzyne formation with Grignard reagents
Nucleophilic substitution through an elimination-addition mechanism (involving benzyne formation) occurs with a strong base. $\ce{NH2-}$ can show this reaction. Can Grignard reagents $\ce{RMgX}$, which form $\ce{R-}$ - an even stronger base - show this reaction?

Griniard reagent doesn't form R- and usually acts as nucleophile not base

Actually a Grignard reagent can act as a (protic) base.  Expose it to water and it's bye-bye Grignard reagent as the "R" part gets converted to a hydrocarbon.  Chemists handling Grignard reagents must exclude water and anything else with which the reagent can react as a base.

NAS reactions normally require strong electron withdrawing groups to facilitate the addition.  Normally these groups, nitro, carbonyl, cyanide, are preferentially good electrophiles for Grignard reagents (nucleophiles).
So no, organomagnesium reagents will not react via NAS.
You also mention the Benzyne mechanism (which is somewhat different).  In this case, where nucleophilic attack on other groups is not possible, I expect metal-halogen exchange due to aryl anions being slightly weaker bases than alkyl anions.  After said exchange benzyne formation is possible.


text:
Why can Grignard's Reagent be stored in ether?
Grignard's Reagent will generate a strong base and nucleophile, $\ce{R-}$. Can this not show nucleophilic substitution at the C atom attached to the O atom of the ether, causing $\ce{RO-}$ to depart as a leaving group? Since $\ce{RO-}$ is a better leaving group than $\ce{R-}$ this reaction should be feasible. Additonally, $\ce{R-}$ is a very good nucleophile.

The free nucleophile $\ce{R-}$ never really exists in a Grignard reagent. That's why we write it $\ce{RMgX}$. Ethers are used as a solvent as the oxygen coordinates to the $\ce{Mg}$, improving solubility and stability.
Regarding your mentioned substitution reaction, it probably would happen if a Grignard reagent was left long enough. There's a reason you make Grignard reagents immediately before use. They don't last if left on the shelf!
But Grignard reagents are usually reacted with carbonyl compounds. These are much more electrophilic than ethers, because the oxygen atom is pulling electrons away from one carbón atom instead of spreading its pull between 2, and because a double $\ce{C=O}$ bond is inherently weaker than two $\ce{C-O}$ bonds. Hence the desired reaction happens much faster than the undesired one.


text:
Feasibility of Nucleophilic Substitutions
Can $\ce{NH3}$ substitute an alcohol? The OH group will depart as a weaker base - some O is more electronegative - so will be a better leaving group than amine.
And what about using ammonium ions - such as in an ammonium salt? They will donate an $\ce{H+}$ to protonate the alcohol and make it a good leaving group, then show substitution followed by deprotonation to form an amine?
How is the the feasibility decided in case of nucleophilic substitution reactions? Is there any rule of thumb to know what happens, and what doesn't?

Short answer - No. See below for a workaround.

Consider your acids and bases:

Substituting an alcohol with ammonia directly

$$\ce{NH3 + CH3CH2OH <=> CH3CH2NH2+ + OH-}$$
If this were an acid-base reaction (and it is, at least in the Lewis sense), then which side of this reaction would be favored by equilibrium? Which side has the stronger acid-base pair? Hydroxide is a stronger base than ammonia, so this equilibrium would favor the reactants.

Using an ammonnium salt

The first step of this mechanism would be a proton transfer to form an activated leaving group and the $\ce{NH3}$ nucleophile.
$$\ce{NH4+ +CH3CH2OH <=> NH3 + CH2CH3OH2+}$$
Again, which side of this equilbrium is favored? Ammonia is definitely the stronger base compared with the alcohol. Estimates of the pKa of $\ce{ROH2+}$ are in the vicinity of $-2$, while the pKa of $\ce{NH4+}$ is $9.25$. The equilbrium constant of this reaction is on the order of $10^{-11}$.

Workaround

If you are needing to do the transformation
$$\ce{ROH -> RNH2}$$
there is a workaround in converting the alcohol into a molecule like the tosylate ester, which can then react with ammonia:
$$\ce{CH3CH2OH ->[\ce{TsCl,base}] CH3CH2OTs }$$
$$\ce{CH3CH2OTs ->[\ce{NH3}] CH3CH2NH2}$$


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Degree of association, van't Hoff factor and dissociation constant

$\ce{CH3COOH->CH3COO- + H+}$
  \begin{array}{c|c c c}
  \mathbf{Initial} &  \mathrm{1~mol} & \mathrm{0~mol} & \mathrm{0 ~mol} \\\hline
  \mathbf{Final} &   \mathrm{1-\alpha ~mol}&\mathrm{\alpha ~mol}&\mathrm{\alpha~ mol}
 \end{array}
  Where $\alpha$ is the degree of dissociation.

Therefore, $i = 1-\alpha+\alpha+\alpha =1+\alpha$, and $\alpha=(i-1)$.
This is what is written in my book. And I understand this.

$\ce{CH3COOH->CH3COO- + H+}$
  \begin{array}{c|c c c}
  \mathbf{Initial} &  \mathrm{c~ mol}&\mathrm{0~mol}&\mathrm{0~ mol}\\\hline
  \mathbf{Final} &   \mathrm{c(1-\alpha)~ mol}&\mathrm{c\alpha~ mol}&\mathrm{c\alpha~ mol}
\end{array}
  Where $\alpha$ is the degree of dissociation.

Therefore,
$$i = \frac{c\alpha+c\alpha+c(1-\alpha)}{c} =1+\alpha$$
 (same as in case 1), and $\alpha=(i-1)$.
$$K_a=\frac{c^2\alpha^2}{c(1-\alpha)} = \frac{c\alpha^2}{(1-\alpha)}$$
This is also written in my book, and I understand this, too.
But, my question is why can we not use the former method to calculate the dissociation constant, $K_a$, which will give me:
$$K_a =\frac{ \alpha^2}{(1-\alpha)}$$
as opposed to what I got in the second (and the correct) case, which was:
$$K_a = \frac{c\alpha^2}{(1-\alpha)}$$
I can see the difference of a $c$ in the two equations, but can someone explain why do we get the difference?

There is absolutely no difference in your two equations but you forgot the units which makes it seem like there is a difference.
Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$
since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?

 $$k_a=\frac{\alpha^2}{1-\alpha}~\ce{mol~L^{-1}}$$

How do I know that the c is the concentration?
Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$
and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.

The concentration of the solution can be expressed in many ways viz. molarity, molality, mole fraction etc. So, can c have any of those units (or no unit at all for mole fraction), or necessarily that of molarity? I ask this because many questions in my book have been solved using molality, and not molarity.

You can simply convert them. Molality $b_x=\frac{n_x}{m_L}$ (moles of x per kilogram solvent) and concentration $c_x=\frac{n_x}{V_L}$ (moles per liter) which gives you through $n=c_x~V_L=b_x~m_L$ how to convert from molality to molarity: $c_x=\frac{b_x~m_L}{V_L}$


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How does the stereochemistry of the enolate affect the stereochemistry of the aldol product?
Cis enolates usually produce syn aldols and trans enolates produce anti aldols. How do exemptions from this "rule" arise?


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How can rotamers of alpha-keto esters be named to distinguish between different geometries?
When talking about free rotation it is often helpful to determine about which geometric arrangement one is referring to. For example in the case of butane we talk about several eclipsed or staggered conformations when discussing the rotational barriers. This also often helps to explain the all trans (staggered) arrangement of fatty acids and surfactants.
In a wider sense even the pseudo rotation of cyclohexane has its own terminology, with boat, chair and twisted conformations.
I am interested in a way to name the two rotamers that can be thought of in α-keto esters (1, 2). My initial thought was calling these cis- and trans- configuration, in analogy of the partial double bond (blue, 3, 4), but for my taste this has a too rigid connotation. Has there been anything proposed in this area? Amides would also be comparable (5, 6).

With respect to the corresponding amides, my coauthors and I used the anti/syn designations in a paper, although we made sure to include a figure to make clear which conformations were anti and syn. http://pubs.acs.org/doi/abs/10.1021/jo025554u

@jerepierre Thanks for the practical example, nice little paper.

The IUPAC recommends s-cis and s-trans for the rotamers of conjugated dienes.

The nomenclature derives from having a "cis-like" or "trans-like" geometry about a sigma bond.
I feel like something similar should be appropriate, however that same IUPAC link recommends E/Z or sp/ap (syn-periplanar/anti-periplanar) for the N-alkyl amides (5,6).

Amide 5 would be Z (but ap w/r/t alkyl groups) and amide 6 would be E (and sp). E/Z is less ambiguous in my opinion.
Ketoester 1 is E, and ketoester 2 is Z. I suppose that ketoester 1 is also ap and ketoester 2 is also sp.

+1 for E/Z. All of the other variants really require defining "what's trans to what" or "what's anti-periplanar to what". It's a little weird to think of the "double bond character" to a dicarbonyl, but if we're really just thinking about the planar conformations, then E/Z are perfect designations.

Thank you very much, this is exactly what I was looking for and what a fast reply. I will use that and make sure to link it back once we publish something with it.