How can titanium burn in nitrogen? I was going through the properties of titanium when a certain thing caught my eye: It was the reaction of burning of titanium in nitrogen. I was astonished to read it as I knew that neither is nitrogen a supporter of combustion nor does it burn itself. But then how can titanium burn in nitrogen? Moreover it is one of the few elements to do so.

It was written that titanium burns in pure nitrogen.

There are two questions one has to answer when asking ‘will $\ce{X}$ react with $\ce{Y}$?’

Thermodynamics

This basically asks ‘will the products of a hypothetical reaction be more stable than the reactants?’, i.e. whether the entire process will release energy. A quick web search tells me that $\Delta_\mathrm{f} H^0 (\ce{TiN}) = - 337.65~\mathrm{\frac{kJ}{mol}}$ and therefore the reaction is thermodynamically possible.
Kinetics

This is asking ‘can I overcome the activation energy required for said hypothetical reaction to occur?’ And this is where it gets complicated.

The process of titanium burning can be split into the following sub-equations:

$$\ce{2 Ti + N2 -> 2 TiN} \\
~\\
\ce{Ti(s) -> Ti(g)}\\
\ce{Ti(g) -> Ti^3+ (g) + 3 e-}\\
\ce{N2(g) -> 2 N(g)}\\
\ce{N(g) + 3 e- -> N^3- (g)}\\
\ce{Ti^3+ (g) + N^3- (g) -> TiN(g)}\\
\ce{TiN(g) -> TiN(s)}$$

Broken down in this way, we should immediately realise:

The first step is generally okay-ish but may require some activation. (We don’t need to make the entire material gaseous in one step; an atom a time is enough.)
The second step requires energy (no surprises there).
The third step is extremely hard to do, since the $\ce{N#N}$ is so strong.
The fourth step also requires energy since nitrogen has an endothermic electron affinity.
The fifth and sixth steps are the ones where huge amounts of energy are liberated; especially the sixth one where lattice enthalpy is liberated.

Now if we consider a block of titanium, as DavePhD mentioned, the the first step is already rather hard to do. The block’s crystal lattice is rather stable as is so we cannot easily extract atoms from there. However, the larger the surface to volume ratio gets the easier this can happen, especially since we have multiple ‘attacking’ points at the same time. Thus, solid metal blocks are usually hard to burn, while metal powder often burns just by putting it in pure oxygen. (Iron wool doesn’t even need to be a powder, the surface to volume ratio is small enough already to oxidise in pure oxygen at room temperature.)

As Jon Custer mentioned in a comment and as I mentioned earlier, the cleavage of dinitrogen molecules is also extremely difficult. It will take a lot of activation energy to do so. This means, that a lot of energy must be gained in the formation of the ion lattice so that at least the dinitrogen cleavage can be overcome and combustion can continue.

Thus, the kinetics part of the question can be answered with: ‘it can, but.’

The question remains how to supply activation energy. Since we are most likely dealing with powder, it could be too slow to ignite titanium in air and then immediately immerse in nitrogen. Since nitrogen cleaving is so hard, many other ignition ideas that rely on, say, a match’s flame won’t work (the match is too weak). Maybe the activation energy is ‘low’ enough that heating to a few hundred degres will suffice. Unfortunately, I do not know the experimental details.


Hooke's Law for absorption spectra Does anyone happen to know who was the first to use Hooke's law to model vibrational absorption peak frequencies/wavenumbers?

I know Hooke's law is first outlined here:

R. Hooke, Potentia restitutiva, or spring, in Lectiones cutlerianae, or a Collection of lectures: physical, mechanical, geographical, and astronomical, made before the Royal Society on several occasions ... to which are added divers miscellaneous discourses, ed. by J. Martyn (John Martyn, London, 1679), p. 4

But I'd like to know about whose idea it was to use it for spectra.

Okay, I think I found an answer. It's a set of four papers which were published consecutively in Physical Review 36 (1930). Interestingly, each was submitted on June 23rd, 1930, which makes me wonder about the back story.

The first comes from General Motors Corporation Research Laboratories in Detroit (Kettering, Shutts, and Andrews). The other three papers are out of Johns Hopkins (Donald Andrews, from Chemistry, and Yates, from Mathematics). I suppose Andrews could have graduated from Hopkins and found that GMCRL was working on the same thing; or perhaps he had a dual appointment.

As a side note, there appears to be a patent suggesting that L. W. Shutts invented shock absorbers, so it makes immediate sense that he would propose the spring law as an explanation.

So Donald Andrews (or at least Johns Hopkins) is the obvious link between GMCRL and Robert C. Yates. Did Andrews know Yates? Was someone about to be scooped? The answer to that is probably beyond the scope of this question.


Ground state oxidation potential and Ionization potential from density functional theory (DFT) Can someone please tell me how to calculate the ground state oxidation potential (GSOP)?

I understand that according to Koopmans' theorem, the negative of HOMO can be considered as the GSOP. Is there any other way to calculate GSOP?

Also, are GSOP and ionization potential same?

Short answer - compute the relative energies of the neutral and cation species

Typically, when people discuss the "oxidation potential," they're considering electrochemistry, so there's solvent around. Keep this in mind if you're trying to compare a calculation done in gas phase - the best comparison would be with an appropriate solvent calculation, although calibrating between a calculation and an experimental electrochemical potential requires work.

On the other hand, ionization potential is typically used for a gas phase measurement - for example, with XPS or UPS spectroscopy, where an electron is removed from a species.

Finally, there's a difference between the vertical ionization potential when an electron is ripped out of a species, but the cation hasn't had time to relax (change geometry) and the adiabatic ionization potential, which is the difference of energy between the neutral and the relaxed cation.

That's a long way of saying that if you want to compute an accurate ionization potential, you need to:

1) Optimize the geometry of the neutral species using an accurate method.

2) Take the optimized geometry, and run a geometry optimization with one fewer electron. (You will get both the vertical ionization energy as the first SCF energy, and the adiabatic ionization energy of the optimized cation.)

Subtract the total energies and convert to your unit of choice.


Overpotential for photocatalytic water-splitting with TiO2 What is the meaning of overpotential in the context of photocatalytic water-splitting? That is light drives charge separation inside TiO2 and the generated electrons and holes react with water to produce hydrogen and oxygen.

I have seen many papers talk about the high overpotential associated with the oxygen evolution half of the reaction. I understand overpotentials as the potential you apply to (an electrochemical) cell in order to drive the reaction without energy barriers. However, in a photocatalytic system, there are no bias voltages. So, what is the physical significance of overpotential here?

Thanks!

Photon absorption by titania results in an excitation of an electron from the valence band to the conduction band.  The difference in energy of an electron in the valence band and the conduction band is the band gap, which for titanium dioxide is about 3.3 eV.

What matters for electrochemistry, though, is the absolute voltage of (i) the excited electron, and (ii) the absolute voltage of the "hole" left in the valence band, not the just the difference between them (gap).  That is far more difficult.

The effective electrochemical voltages of the excited electron (or hole) can differ from what the desired electrochemical reaction is.  So there could still be an overpotential.  For titania, one paper I found estimates of the absolute energies of the valence band "hole" and the conduction band electron of -7.8 eV and -4.5 eV, respectively (at pH 0).  The thermodynamic potential for water oxidation is -5.8 eV, 2 eV higher than the hole energy.  That energy is lost if the hole is used to power water oxidation.  You could thus view excited titania having an overpotential of 2 eV for the water oxidation reaction.  In contrast, the theoretical potential for proton reduction is -4.5 eV, close to the conduction band electron energy.

Thus the energetics of bulk titania provide lots of overpotential for the water oxidation reaction but the tradeoff is that a high band gap means only high-energy photons can power the electrochemistry.  Lowering the band gap would make more and longer wavelengths of radiation available to power water photolysis, but the band gap can only be lowered in certain ways.  For titania, it would not be very desirable to lower the energy of the conduction band, because that would mean excited electrons would not be able to reduce protons to hydrogen .  Thus, the only way to lower the band gap is by raising the energy of the valence band, which would both lower the overpotential and and make more types of radiation accessible for powering water photolysis.


Why is the name 'propylene oxide' preferred for epoxypropane? This is the molecule I know as epoxypropane:

In my neck of the woods (example), this molecule is used as one of the standard examples of small chiral molecules. The name 'epoxypropane' makes a heck of a lot of sense to me $-$ an epoxy bridge  spanning a single bond of what would otherwise be propane.

It seems, however, that most everyone thinks that a better name for this is propylene oxide, from Wikipedia upwards. Now, I'm an atomic physicist, so bear with me:

I can see how the name 'propylene oxide' fits that structure. You've taken propylene and you've oxidized it.
However, I cannot see why the name 'propylene oxide' doesn't also apply to other ways to oxidize propylene, say, sticking that oxygen over at one end or in the middle,

i.e. what Wikipedia calls propionaldehyde (also propanal) and acetone, or making that bridge over the (1,3) combination in what Wikipedia calls oxetane,

none of which is chiral.

Now, I imagine that IUPAC has all sorts of rules to suss these things out, so:

why is 'propylene oxide' the preferred name for epoxypropane? and
why is epoxypropane the only isomer encompassed by the designation 'propylene oxide'?

why is 'propylene oxide' the preferred name for epoxypropane?

Propylene oxide is not the "preferred" IUPAC name (PIN): Wikipedia lists the PIN as 2-methyloxirane, which is in line with the 2013 IUPAC recommendations.

  why is epoxypropane the only isomer encompassed by the designation 'propylene oxide'?

It's more of a tradition than anything else. Note how the two ring carbons in propylene oxide are the same as the two doubly bonded carbons in propylene: this is not a coincidence. A very common way of synthesising an epoxide is to oxidise the corresponding alkene (you can think of it as simply adding an oxygen atom across the double bond), and so "alkene oxide" has become a rather conventional way of referring to an epoxide (see also https://en.wikipedia.org/wiki/Styrene_oxide for another example).

So, the meaning of "propylene oxide" is not really "an oxidation product of propylene". It should pretty much be read as "the epoxide derived from propylene", or "the epoxide formed when you add an oxygen atom across the double bond of propylene".

In the strict sense of oxidation/reduction being addition/loss of oxygen, oxetane is an oxidised form of propylene, but there is no simple chemical oxidant that will perform this transformation. So, calling oxetane "propylene oxide" would never have - and has never - caught on.


Iodine clock reaction I have been investigating the iodine-clock reaction for my chemistry project. I have done a great deal of research on the titration of iodate ions, iodide ions and sodium thiosulfate using HCl. But, I still don't get the use of this reaction. Where can this type of reaction be used in our lives.

I read articles that said iodine clock reaction can be used to determine the concentrations or pH. But where can it be used? What are the real-life applications of the iodine clock reaction?

This iodine clock reaction project can apply to the real world in different ways, but I think the biggest way is concentration. After doing some research, I found out that it can teach you about concentration, because it shows you that once you add any bigger amount to a substance, it changes reaction time, or in a real world case, how a substance feels, acts, looks, and possibly how it smells.

I know this is really late and you probably don't need this and you are done your project, but I'm just a bored grade 7 doing the same science project.


What would follow in the series sigma, pi and delta bonds? I realise, that this question is a stretch, but I was wondering, how would a bonding orbital be called if it was formed from two $f_{x(x^2−3y^2)}$ or $f_{y(3x^2−y^2)}$ orbitals. Have there been any suggestions on this, was it anywhere proposed or discussed? I am not arguing about the existence of such a thing, but as a thought experiment, how would it be called? Phi, $\varphi$?

I am also throwing in a reference request here, since I am interested in any literature related to the topic.

Some background:

A $\sigma$ orbital is an orbital, that has no nodal plane between the bonding partners. It is $C_\infty$ symmetric with respect to the bonding axis (in first approximation - of course $d_{xy}$ orbitals can align in a $\sigma$ fashion, too, but for the sake of the argument, let's not consider this.) The point group of this orbital can further be assumed as $D_\mathrm{\infty h}$. It shares therefore most features of an $s$ orbital, hence $s$igma, $\sigma$.
The most prominent example for this is probably the dihydrogen, $\ce{H2}$, molecule.
A $\pi$ orbital has one nodal plane and the bonding axis of the involved atoms is part of this plane. It is therefore asymmetric, $C_1$, with respect to this plane. However, it is $C_\mathrm{s}$ symmetric with respect to the plane that is perpendicular, also sharing the bonding axis. The orbital has further the point group $C_\mathrm{2v}$. It has therefore most features of a $p$ orbital, hence $p$i, $\pi$.
The most prominent examples for $\pi$ bonds are probably ethylene, $\ce{C2H4}$, and acetylene, $\ce{C2H2}$.
A $\delta$ orbital has two nodal planes between the bonding partners. These are perpendicular to each other. The bond orbital is asymmetric with respect to both of them. As a result, it is $C_\mathrm{s}$ symmetric with respect to the bisecting mirror planes. The point group of this orbital is $D_\mathrm{2h}$. Therefore it has shares features with a $d$ orbital, hence $d$elta, $\delta$.
One example is the $\ce{[Re2Cl8]^{2-}}$ ion, see Wikipedia's quadruple bond.
I imagine two $f$ orbitals aligning with the bonding axis being simultaneously a $C_3$ symmetry axis, giving it an overall $D_\mathrm{3h}$ point group. Since there is no greek letter starting with an $f$, I think the logical thing would be to choose $\phi$ as it sounds a like. (I cannot be the only one thinking that.)

The following graphic should somewhat demonstrate what I mean:

I have to admit, that I have not done much research about this. It was more like an impulsive question for me to ask.

tl;dr The next in the series is called φ bond. There is even a tiny Wikipedia article about it.

Nicolau pointed me to the Wikipedia article, that had at the time a tiny section about the φ symmetry of the bond. Ben also kindly agreed with my naming proposition. I'd like to back up just a little bit an quote one sentence of this article:

  The type of interaction between atomic orbitals can be further categorized by the molecular-orbital symmetry labels σ (sigma), π (pi), δ (delta), φ (phi), γ (gamma) etc. paralleling the symmetry of the atomic orbitals s, p, d, f and g. The number of nodal planes containing the internuclear axis between the atoms concerned is zero for σ MOs, one for π, two for δ, etc.

Basically this was what I was already thinking, but since the section is quite tiny overlooked it accordingly. But since this is a reference request, I'll go a little deeper.

The concept of the bond is actually quite old and I was able to track it down to at least 1984, where Bruce E. Bursten and Geoffrey A. Ozin "demonstrate[d] for the first time the existance of φ bonds between metal atoms."[1] A little further back, 1978, N. Rösch and A. Streitwieser already performed SCF-Xα calculations on thorocene and uranocene, where they came across f orbital contributions in bonding. Yet, they did not coin the phrase φ bond.[2] Following that, there have been more theoretical studies on the diuranium molecule, with increasing level of theory.[3] One of the more recent studies is by Laura Gagliardi and Björn O. Roos, which is available free of charge (see below).[4] Xin Wu and Xin Lu also studied the phenomenon in endohedral matallofullerenes, i.e. diuranium in a Buckminster fullerene.[5]

It is quite safe to assume, that there are more studies about this research topic or in a more general aspect the research on molecules with high bond orders. For this purpose I will include a couple of good reads.[6]

References and further reading

Bruce E. Bursten, Geoffrey A. Ozin, Inorg. Chem., 1984, 23 (18), 2910–2911.
(a) Rösch, N., Streitwieser, A., Jr., J. Organomet. Chem., 1978, 115, 195-200. (b) Rösch, N., Streitwieser, A., Jr., J. Am. Chem. Soc. 1983, 105, 7237-40.
Melanie Pepper, Bruce E. Bursten, J. Am. Chem. Soc., 1990, 112 (21),  7803–7804.
Laura Gagliardi, Björn O. Roos, Nature, 2005, 433, 848-851. Available free at Archive ouverte UNIGE.
Xin Wu, Xin Lu, J. Am. Chem. Soc., 2007, 129 (7), 2171–2177.
Without order: (a) Gernot Frenking, Nature, 2007, 446, 276-277.
(b) Gernot Frenking, Science, 2005, 310, 796-797.
(c) Gabriel Merino, Kelling J. Donald, Jason S. D'Acchioli, Roald Hoffmann, J. Am. Chem. Soc., 2007, 129 (49), 15295–15302.
(d) Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, Inorg. Chem., 2013, 52 (6), 2838–2843.
(e) Jia Zhou, Jason L. Sonnenberg, H. Bernhard Schlegel, Inorg. Chem., 2010, 49 (14), 6545–6551.
(f) Awal Noor, Rhett Kempe, Inorg. Cimica Acta, 2015, 424, 75-82.
(g) Frank R. Wagner, Awal Noor, Rhett Kempe, Nature Chemistry, 2009, 1, 529-536.
(h) Björn O. Roos, Antonio C. Borin, Laura Gagliardi, Angew. Chem. Int. Ed., 2007, 46 (9). 1469-1472.


Emission spectra (Be and Mg) The line says

Alkaline metals and their salts impart characteristic colour to an oxidising flame. This is because the heat from the flame excites the outermost orbital electron to higher energy level. When excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum.
Beryllium and magnesium do not impart colour to flame because electrons in beryllium and magnesium metal are too strongly bound to get excited by flame.

What if reason that Be and Mg do not impart colour to flame is not that the electrons in Be and Mg are actually bound too strongly to be excited by flame but that the radiation emitted does not fall in visible spectrum?

You are right and whoever wrote this quote "Alkaline metals and their salts impart characteristic colour to an oxidising flame. This is because the heat from the flame excites the outermost orbital electron to higher energy level. When excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum. Beryllium and magnesium do not impart colour to flame because electrons in beryllium and magnesium metal are too strongly bound to get excited by flame." is mostly wrong about Mg and Be.

The first process in flame spectroscopy is to decompose the compounds in the flame into atoms. /This requires a lot of energy. In ordinary lab flames, like that of the Bunsen burner (methane+air) the temperature is so low that alkaline earth (Ca, Ba, Sr) emit light because of their compounds in the flame not by their atoms. Recall magnesium and beryllium forms refractory oxides, so the correct explanation is that the flame temperature of ordinary oxidizing Bunsen burner is so low that it is not able to break magnesium compounds and excite beryllium in the flame.

Now if you use a proper flame, say that of acetylene-air for Mg or better acetylene-nitrous oxide flame for Be, you should see emission mostly in the UV but also very low intensity visible lines. The visible lines could be detected by a sensitive detector rather than the eyes. Nitrous oxide flame is highly colored itself (sunset pink color) and one cannot see visible emission easily without using a monochromator.


IUPAC name, phthalic anhydride Could someone help me please (preferably talk me through not just provide the answer) with the IUPAC name of phthalic anhydride?  I'm sorry for not posting attempts but I'm afraid I have absolutely no idea how to begin here...

Benzofuran is named by the position of the oxygen, hence 2-Benzofuran.  (below is 1-benzofuran)

The =O s are then added on the 1 and 3 carbons, hence, 1-3-dione

Compare, cyclopentan-1,3-dione

The end result is

2-benzofuran-1,3-dione


Magnetic stirrer with hot plate I did not know there are products such as a magnetic stirrer with a hot plate, enabling one to apply heat and stirring to a solution with relative ease. Did I understand correctly, that device needs a magnetic bar to be added to the solution the device to work? Does that pose any limits when it can be used (if the magnet would react with something..) How much do these devices increase productivity?

Most of the things to be told about magnetic stirring are already in Rons comment. A rotating magnetic field is induced by a rotating permanent magnet inside the hot plate magnetic stirrer. This field drives the magnet inside your reaction flask. To ensure chemical inertness in most situations the magnet is coated with PTFE (Teflon) or protected by a thin layer of glass. Magnetic stirring can replace traditional mechanical stirring is most situations and has many benefits over it: it does not require an additional inlet to the reaction flask, which is very handy for reaction under reduced pressure and is also more easily adjustable to different scales due to the possibility to replace the magnet in the reaction flask. Two exceptions are 1) very large/very viscous reactions or reactions in which excessive amounts of solid are present 2) reactions that involve ferromagnetic compounds that interfere with the rotational movement of the magnet inside the flask (as in Urushibara nickle catalyst preparation).

Vogel has a section on reaction heating and stirring in which this and more is discussed extensively. See pages 71-80.


How to correct the pH and composition of bottled water? For some time I've been buying a brand 'A' of (non-sparkling) bottled water.
It wasn't easy to find one that I liked, because I make tea with it, and most other brands I tried made murky, smelly, dark tea, with a disgusting film on top, whereas this one was perfect.
Unfortunately, I just found that the supermarket doesn't stock brand 'A' anymore (I know, it's not a big problem, by any standards; just discussing a bit of science, for fun).
So I picked the closest one I could find (brand 'B'), naively thinking that the main 'factor' to look at was pH.
Brand 'A' had a pH pf 6.7; brand 'B' was 6.9. I thought: close enough.
Actually: wrong.
I made tea with brand 'B' water, and while it wasn't as horrible as with higher-pH ones I had tried before, it was nowhere as good, it made that slimy film, and smelled a bit off.
So I looked at the composition of the two brands side by side, converting the concentration from mg/L to mmol/L:
$$
\begin{array}{lcc}
\hline
\text{Component} & C_\mathrm{A}/\pu{mmol L^-1} & C_\mathrm{B}/\pu{mmol L^-1} \\
\hline
\ce{HCO3-} & 0.38 & 4.72 \\
\ce{SiO2}  & 0.32 & 0.00 \\
\ce{Na+}   & 0.13 & 0.18 \\
\ce{K+}    & 0.02 & 0.16 \\
\ce{Mg^2+} & 0.04 & 1.60 \\
\ce{Ca^2+} & 0.10 & 1.07 \\
\ce{F-}    & 0.01 & 0.00 \\
\ce{Cl-}   & 0.02 & 0.06 \\
\ce{NO3-}  & 0.03 & 0.00 \\
\hline
\end{array}
$$
Clearly there is much more 'stuff' in water 'B' than in 'A', and in fact the residue is 289 mg/L in the former, 43.6 mg/L in the latter.
One thing that caught my eye was the charge balance, which seems a bit off for 'B' (+0.91 mmol/L), whereas it's a bit closer to 0 for 'A' (-0.02 mmol/L).
But OK, my main question at this point is: can I add something to water 'B' so that it 'behaves' a bit more like 'A', for the use that I mentioned?
My hypotheses so far are that:

the pH should be lowered
$\ce{Ca^2+}$ and $\ce{Mg^2+}$ are responsible for the formation of that 'film', and I might prevent that by adding something that forms soluble coordination compounds with these ions (and more stable than the ones they form with whatever it is in tea that reacts with them)

Obviously I am not going to use stuff that is not edible, like EDTA; and for the pH correction, I'll have to rely on calculations, not going to stick a pH-meter into it.
So my best bet at the moment is: citric acid.
Though I am not sure how I should calculate the correct quantity to add to the 5 L bottle I buy (surely I can't aim to convert most $\ce{HCO3-}$ to $\ce{H2CO3}$, or the final pH would have to be too low).
What do you think?
Any ideas/comments/suggestions?

Long time ago, I supervised an undergraduate on a summer tea project (for fun). In South Asia there is a natural strawberry colored tea made from slightly matured green tea. You boil the leaves in water with a small amount of sodium bicarbonate. Ice cool it, and boil again and aerate it with a ladle. The tea becomes reddish on aeration and after adding milk, it becomes pinkish. So I can recall your experiences because I tried this recipe in another country. No matter what I did, the right color would not develop. It remained murky brown. It turned out that the chlorinated water was ruining the natural colors, it was particularly destroying the reddish pigment. Passing the tap through a carbon filter+ion-exchanger did not work as calcium also seemed important for the right color.
Finally a mineral water bottle worked and the undergrad saved a project for his poster presentation.
Anyway, it seems your tap water is of bad quality or it is hard. This tea "scum" are Ca/Mg salts. You guessed it right. My suggestion is do not add any acid such as citric acid. Tea is not meant to be made in the presence of citrate salts especially if you add milk to it. I do not know what the taste be like. Most likely horrible!
I do not know how easy it is find potable distilled water in stores. Ask the manufacturer if it is fit for human consumption. Distilled water is not ultrapure water as it may have organic impurities or volatile impurities. It is only free from water soluble salts. All these tea scum problems should vanish.


Is it possible to calculated the concentration of hydroxide ions from the known concentration of hydronium ions? I came across the following question in Chapter 15 (Acids and Bases) of Holt High School Chemistry:

  Neutralizing $\pu{5.00 L}$ of an acid rain sample required $\pu{11.3 mL}$ of $\pu{0.0102 M}\ \ce{KOH}$. Calculate the hydronium ion $[\ce{H3O+}]$ concentration in the rain sample.

This is a reasonably straightforward question requiring substitution into the equation
$$c_1V_1 = c_2V_2,$$
which lets us derive the concentration of $\ce{H3O+}$.

I initially mis-read the question (which does not mention $\ce{H3O+}$) to read it as wanting the concentration of $\ce{OH-}$. However, had I not mis-read the question, would it be a valid question?

For example, can we (after getting the concentration of $\ce{H3O+}$) via the equation above, use the equation with the ionization constant of water
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]$$
to also get the $\ce{OH-}$ concentration?

Yes, that works. In fact, some textbooks and teachers ask this type of question (asking for the lesser ion of the autodeprotonation of water) to test which students actually read the question and which gloss over details. I fell into the trap of calculating the wrong one and stopping too early, too.

So if your question had asked for the hydroxide concentration, you would indeed calculate the hydronium concentration first (because that is actually accessible) and then invert and multiply $10^{-14}$ to get the desired one as you suggest.


Shelf-life of solutions If I prepare the silver nitrate solution or any other solution, how can I know the shelf life of this solution? I know that it depends on the stability of the substance, but maybe there is some document which give information and is interesning how realy it was found.

You won't find a general formula for the shelf life of a solution.  A silver nitrate solution will have a very short shelf life if it is kept in the sunlight, whereas potassium nitrate could last for years.  The shelf life is also going to be very strongly dependent on the quality of the container.  Think about what happens to the aging of a wine bottle if the cork does not provide a sufficient seal .  Interactions between the solution and the container have an impact on the solution shelf life as well.  To stick with the wine theme, corked wine is due to contamination of the wine with 2,4,6-trichloroanisole which is generated by natural fungi in the cork that have been exposed to chlorine.

So it is important to understand the chemistry of the solution with which you are working in order to determine its shelf life.


Practical differences between storing 2-electron integrals and calculating them as needed? Some texts that I read suggest calculating all of the 2-electron integrals at once (taking into account permutational symmetry) and some suggest calculating them on demand when building the Fock matrix.

Which of these is the most computationally efficient, and when should one pick a certain approach over another?

It depends on system size. There is a crossover point where it becomes faster to calculate the integrals as needed because as a system becomes larger and larger, the number of disk operations required to store the integrals becomes very large which greatly slows down the calculation. This is because disk operations are relatively slow compared to memory manipulation.

Thus, most software packages have an option to calculate the integrals directly and not store them. In Molpro I believe this command is gdirect.

To be more specific, there are two aspects of the storage which are slow when there are a large number of integrals. One is that disk operations themselves are slower than just doing the calculation which is only memory manipulation. The other is that the integrals are stored in some fancy way so as to minimize the amount of time it will take to find the specific integral you are looking for, but when there are a very large number of integrals, some non-negligible amount of time will go into finding the value of the integral. When these two components (plus maybe others I am missing) are larger than the time required to just do the integral calculation, then it is faster to compute the integrals directly.

I'll let someone else be more quantitative with the timing of disk operations and the lookup tables as these are more of computer science questions. But quite interesting ones.

For a useful reference on how many integrals we are talking about storing on disk, see this question.

Also as a practical guide, in my experience this crossover point was around 14 hevay atoms with a triple-zeta basis set, but this means almost nothing because this will be heavily dependent on the computing architecture you are using. I would expect the crossover point would be much higher for most supercomputing systems as I did not have much disk to work with comparatively.


Do entropy changes have an equivalent of bond enthalpy? Is it possible to calculate entropy changes in a way similar to enthalpies from bond enthalpies? For example, to work out the enthalpy change of a reaction you can get the sum of all bonds in the products and minus the sum of all bonds in the reactants to get a reasonable approximation of the enthalpy change of reaction. Does such a thing exist for entropy?

Thanks!

Please note that you are calculating the $\Delta U$ with your bond breaking/forming method. If there is low volume work done, then you can approximate:
$$
\Delta U \approx \Delta H
$$

For the entropy: Yes and no. I have no idea for liquids but qualitatively I would guess that it is not possible to have a quick and dirty method as you wish for this phase. And even for gases and solids it is quite a long work.

For gases you can calculate the entropy for every component itself and then add/substract.

Translational (Sackur-Tetrode):
$$
S_\text{trans}= R \ln \left( (2\pi m k_B T)^\frac{3}{2} \frac{V_m}{N_lh^3} \exp(5/2) \right)
$$

Rotational:
$$
S_\text{rot}=R\left(\ln\left(\frac T\Theta_\text{rot} \right) +1 \right)
$$

Vibrational:
$$
S_\text{vib}=R \left(\frac{\Theta_\text{vib}}{T} \frac{1}{\exp\left( \frac{\Theta_\text{vib}}{T}\right)-1} - \ln \left(1-\exp\left( \frac{-\Theta_\text{vib}}{T}\right) \right)\right)
$$

For solids you could calculate the heat capacity $C_p(T)$ with the Debye modell and then integrate from 0K.

I am really sorry, but I do not think, that there are easier methods.


How can nitrogen achieve +5 oxidation state? Nitrogen has no available d-orbitals. Yet, it forms compounds where its oxidation state is +5, like in nitric acid. How?

The concept of oxidation states is just that: A theoretical concept. You assign electronegativity values based on one of the numerous scales available, split the bonds heterolytically according to those numbers, and then count electrons.

As such, it is very easy to make a cation like @permeakra stated in the comments that satisfies the octet rule and still gets to state +5 (the molecular cation in question is $\ce{NF4+}$).

There are no $d$ orbitals necessary (which are also theoretical, but slightly closer to reality) to explain how a +5 state is possible.


Which branches of chemistry and in what order to learn (self study)? Hi I have decided to give chemistry another shot (it didn't take so well way back in high school). My main interest is for reading up on brain chemistry and pharmacology.

I currently have a textbook on inorganic chemistry [Introduction to Organic Chemistry 3rd edition: William Brown.Thomas Poon] that I picked up and am working my way through - well I have made it through chapter 1 at least [Covalent Bonding and Shapes of Molecules]. And am also making use of The UC Davis (Organic) ChemWiki.

So after finishing the chapter today I decided to take a break and play around with an experiment I sore on youtube to make copper nanoparticles using copper sulphate and ascorbic acid. In the video he was pretty lenient with the measurements [1g sulphate, 5g ascorbic, 200ml water] and I wanted to scale the reaction up a bit so I had a go at trying to figure what the reaction was so I could get the true ratio... Obviously that duck wouldn't fly (I am only just starting on chapter 2 after all) so I just went ahead and multiplied his recipe by a factor of 5.

So where I was going with that last paragraph is that even after I finish reading the organic chem textbook I'm not seeing how that's going to help me with something like $CuSO_4.5H_2O$ since i'ts got nothing to do with carbon. Should I put the organic book to the side and start of with something like the (Inorganic) ChemWiki? Is it necessary to go through both of them or would it be enough to learn one subject and fill the gaps as there encountered [this exercise is a personal mission - I don't envision any possible future where I become a full fledged working chemist]? Better yet are there any broader texts that teaches a unified chemistry that doesn't make artificial distinctions? Or should I go ahead and invent my own syllabus?

Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.

I agree with LDC3 that inorganic chemistry should definitely be your starting point; it builds foundations that apply to all of chemistry, which is why they teach it first; organic chemistry is just a branch, albeit a very large one, of chemistry that can also be very interesting once you know the fundamentals. Unfortunately using conventional textbooks etc, you will most likely be constricted to this order of learning, because of the level of these courses in the education system, where inorganic chemistry is very traditionally taught first, with organic chemistry only afterwards.

  Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.

I don't know if there are any shortcuts that you are seeking, Wikipedia can be pretty advanced because it is (usually) written by experts who know exactly what they're talking about; and especially on those high-level concepts, there are a lot of "things" (for lack of a better word) that you have to already know to understand fully what the big picture is. Trying to pick and choose would be maybe even more time-consuming than learning things in a progressive order.

Even though this sounds pessimistic, learning chemistry can be rewarding if you work at it, not to mention you have this site and tons of other online resources to help you out along the way.

-Best of luck should you choose to continue, and I hope you do!


Which electrons will be taken when oxidations take place and how does the resulting orbital diagram look like? In detail, what I really mean is which electron in which orbital is being taken when a oxidation happen? Let me give a example: chromate ion(II)

Two electrons are taken but from which orbital:

For transition metals (specially for those on the second and third rows), like chromium, or iron, which electrons are taken away in oxidation is not straightforwardly deduced. The ones that will be lost are those which are higher in energy and lead to a more stable situation.

For iron, which is $\mathrm{[Ar]3d^6 4s^2}$, the higher energy electrons are $\mathrm{4s^2}$, and those that will be removed on oxidation, leading to a $\mathrm{d^6}$ configuration for iron (II).


Order of Acidic strength: Phosphorus oxoacids I was asked to compare the acidic strength of the following phosphorus oxoacids,

I thought the order (of strength) would be $\ce{H3PO4 > H3PO3 > H3PO2}$, because they have three, two and one ionizable $\ce{O-H}$ bonds (that furnish protons / $\ce{H^+}$ ions).

However, looking up the $\pu{pK_a}$ values on the respective Wikipedia pages of each of those acids, suggests the exact opposite order of acidic strength:
$\ce{H3PO2 > H3PO3 > H3PO4}$. Apparently the acid with one $\ce{O-H}$ bond is a far stronger than the acid with three $\ce{O-H}$ bonds!

This is horribly counter-intuitive :-(

So, why is the correct order of acidic strength  $\ce{H3PO2 > H3PO3 > H3PO4}$, and not the other way around as I believed? What was wrong with my analysisi ( acidity in oxoacids $\propto$ number of $\ce{O-H}$ bonds)?

Acidic strength of the acids is determined on the relative tendency to donate one proton.

The source where you have seen then $\mathrm{p}K_\mathrm{a}$ value has considered the tendency of first deprotonation of the acids ($\ce{H3PO2}$ deprotonates once, $\ce{H3PO3}$ can deprotonate twice and $\ce{H3PO4}$ can deprotonate thrice).

The double bonded oxygen has negative inductive effect (-I). And this electron withdrawing effect is experienced by one $\ce{H}$ of the one $\ce{-OH}$ group in $\ce{H3PO2}$. Hence it has greatest acidic strength.

Similarly, the -I effect of $\ce{O}$ is experienced by two $\ce{H}$ of the two $\ce{-OH}$ groups in $\ce{H3PO3}$, hence electron withdrawing effect is reduced because it is getting distributed between two groups. Hence, it has lesser acidic strength. Similarly, in $\ce{H3PO4}$, it is distributed between 3 groups and so the acidic strength is least.

Thus the acidic strength, is:
$$\ce{H3PO2 > H3PO3 > H3PO4}$$
(I hope it's clear.)


Conversion of barbituric acid to veronal This is just out of curiosity. I have no intention or authority to do conversions.

I am studying chemistry of tranquilizers, in that an important class is of barbiturates, which are also hypnotics. The structure of Barbituric acid and veronal, are similar, except veronal has two ethyl groups instead of hydrogen at lone $sp^3$ hybridised carbon. How can I convert Barbituric acid to veronal?

I thought that making carbocation or carbanion will help, but I don't think carbon will get negative or positive charge until nitrogen has hydrogen. Probably we'll have to protect nitrogens, am I thinking right?

This paper from Tetrahedron Letters here details synthetic procedures using carbonyl compounds in solution with barbituric acid and Pt or Pd hydrogenation catalysyts to give reductive di-C alklyation. This includes examples of N-unsubstituted barbituric acids R1 = R2 = H. Using barbituric acid plus acetaldehyde should give you veronal in a single step.


Are there examples of crystals with random but static mixtures of two unit cell configurations? The image in this question turned out to probably be wrong, but it still makes me wonder. Are there examples of crystals where the atoms within a unit cell can have two (or more) different configurations or orientations who's energies are so close that a sample crystal at room temperature could have a random but static mixture of the two?

With sufficiently large organic molecules (e.g. proteins) this could be true and possibly even common, but I'd like to know about a simple example with a small number of atoms in the unit cell. It might be produced by rapid cooling or crystallization from solution in unusual conditions, and it might be detectable (at least in some cases) by a slight broadening in high resolution X-ray diffraction peaks. (sort of a solid-state version of Raleigh scattering perhaps?)

Possibly retracted soon, because I'm not aware how to include a picture in a comment to clarity your question.  Do you mean instances like the following, where sites seemingly are occupied by two atoms simultanously -- as in analogues of Prussian blue -- just because the model deposit in the .cif is the average of the unit cells' content?

(Chen et al. in Angewandte Chemie,  129, June 2017, 9151-9156, doi 10.1002/ange.201702955)


Why is the recipe of Coca Cola still a secret? Why is the recipe for Coca-Cola still a secret?
I think that given the current state of technology, it should be proficient enough to find any of the secret ingredients in Coca Cola.

Any thoughts?
Can we make a 100% identical clone of Coca-Cola next week?

There are different angles this question can be answered:
Chemical point of view:
A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution...  Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components.
That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor:
Business point of view:
The recipe of Coca Cola itself worth nothing.
Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.
The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow DIFFERENT than the other ones!


Low viscosity, low surface tension liquid for volume measurement General Case
This question concerns the general problem of accurate volume measurement of a given object. Among other techniques, that of liquid displacement is sometimes used for this purpose.

In case of a solid material, the latter may be immersed in liquid, and its volume calculated by measuring the amount of fluid displaced in a graduated cylinder.

In case of a hollow object, the latter may be filled in with a liquid which amount is accurately determined, either while filling in, or by emptying the object in a graduated container a posteriori (or - maybe more accurately - by weighing).

In both cases, the choice of the liquid is of crucial importance for the accuracy of the measurement.
Specific Issue
I have a CO2 gas sensor with a complex hollow embodiment, also containing electronics, and since I wish to perform response time measurements, I would like to precisely know its inner volume. I came up with the idea of filling it with a liquid, as stated above. I do not have and cannot have accurate drawings of the inner sensor.
Questions
1. Is there other methods to perform such a volume measurement which may be preferable?
Indeed, I came up with this idea because I find acceptable to sacrifice a sensor to perform such volume measurements. Alternatively, I also thought of placing the sensor in a closed, accurately known volume, along with some amount of chemical reagents that would produce a given amount of CO2. Then, by dilution and measurement of the CO2 percentage inside the enclosure, I may know the volume occupied by the sensor. However, I believe that there are far too many sources of errors in such procedure for it to yield accurate results. I may be wrong, though, and other variations of the latter idea may work well...
2. Which liquid may be the best for such measurement?
This is the main issue here. I would tend to say that such a liquid should have both a low surface tension and a low viscosity. Additionally, low volatility would be a plus, since it would prevent the amount of poured liquid to decrease with time. I find methanol or ethanol to be good canditates despite their high-volatility. I cannot use acetone because of the risk of plastic dissolution and the resulting leakage of the sensor's embodiment. Would you know of a more suitable liquid, possibly readily available in the lab?
3. Can surfactant be added to ethanol or methanol? If so, do they further decrease surface tension? Which are they?
A quick search in the literature revealed that methanol or ethanol are themselves considered as water surfactant. Their mixture with other surfactants is often analysed in emulsions of oil, or other chemicals, but not alone.
Related topics:

Non-wetting fluids for precise volume measurements
Looking for a non-volatile liquid less viscous than water

Edit:

As pointed out by Poutnik, low volatility and low viscosity are mutually incompatible with each other. Indeed, low viscosity tends to be positively correlated with high volatility, both of them being the consequence of low intermolecular forces. More resources on the topic are available here (saved version). The requirement on low volatility was not mandatory though, as it was just to ensure that not too much of the liquid evaporates during the filling of the sensor...

Air would be a convenient fluid. Fill your sensor with air at a known (atmospheric) pressure - well, it is probably filled with air already.
Then connect it to a vacuum chamber tube of known volume with a pressure sensor, i.e., a tube with two valves and a pressure sensor. Attach a vacuum pump to one valve and evacuate the tube; attach the $CO_2$ sensor to the other valve and then open it and measure the new pressure in the closed system. All the air will have been supplied by the sensor volume, and the new system volume will be the sensor volume plus the (known) tube volume. Doing it this way is likely to be less messy and more accurate than using liquids, and it won't sacrifice a sensor - and you can check every sensor!
A problem with liquids in a blind experiment (you cannot see the internals) is that there could be blind alleys that would be hard to detect and hard to fill with liquid (or which would not empty easily).


The heat content of the products is more than that of the reactant in an ............. reaction The heat content of the products is more than that of the reactant in an ............. reaction.

In this question the answer given is "exothermic"; yet I thought the answer would  be endothermic as the products have more heat content and energy cannot be created nor destroyed. Therefore, from the reactants' side, more energy should have been put in for the heat content of the products to be greater.

Also mathematically endothermic make sense as: $$\Delta_\mathrm{r}H = \sum \Delta_\mathrm{f}H_\mathrm{P}-\sum \Delta_\mathrm{f}H_\mathrm{R}$$

And as: $\sum \Delta_\mathrm{f}H_\mathrm{P} \gt \sum \Delta_\mathrm{f}H_\mathrm{R}$
,$\Delta_\mathrm{r}H \gt 0$

Could someone please explain where am I wrong and why the answer is 'exothermic'?

Don't blame yourself.  This is a badly-worded question. Products and reactants don't have "heat content", they have thermal energy. I.e.,  heat is not a property of substances—they don't 'contain heat.'  Instead, heat is a property of processes; specifically, it is the flow of thermal energy across a boundary. So, for instance, in an exothermic reaction in a system with diathermal walls (walls that allow free flow of heat across them), thermal energy flows out of the system, and thus the heat (q) for that reaction is negative.

Having said that, it seems what they mean by "heat content" is the thermal energy of the system after a reaction if the system is adiabatic (no heat flow in or out, i.e., perfectly insulating walls).  Thus, if you have an exothermic reaction in an insulated container, the products you get will be hotter than the reactants you started with.

A much better way of asking the question would be:  Suppose a reaction takes place at constant pressure in an adiabatic (insulated) container, and the temperature increases.  Is the reaction exothermic or endothermic?  Answer: exothermic.


Nucleophilic nature of ions Consider the following species:

  $$
\underset{(\text{A})}{\ce{\overset{⊖}{O}H}}
\quad
\underset{(\text{B})}{\ce{CH3-\overset{⊖}{O}}}
\quad
\underset{(\text{C})}{\ce{\overset{⊖}{C}H3}}
\quad
\underset{(\text{D})}{\ce{\overset{⊖}{N}H2}}
\quad
$$

  Arrange these species in their decreasing order of nucleophilicity.

  (a) C > D > A > B
  (b) B > A > C > D
  (c) A > B > C > D
  (d) C > A > B > D

My teacher taught in polar protic solvent nucleophilicity is inversely proportional to basic character. I assumed solvent to be $\ce{H2O}$ as solvent was not mentioned in the question.
As acidity order is

$$\ce{H2O} > \ce{CH3OH} > \ce{NH3} > \ce{CH4},$$

so basicity order should be

$$\ce{CH3-} > \ce{NH2-} > \ce{CH3O-} > \ce{OH-},$$

therefore using above proportionality I arrived at the nucleophilicity order

$$\ce{OH-} > \ce{CH3O-} > \ce{NH2-} > \ce{CH3-},$$

but you can see the option is not even given. Why is it so?

What you probably missed was that the order you mention is for the SAME GROUP elements. For example, down the group basicity of halide ions is-

F->Cl->Br->I-

But the the nucleophilicity order in polar protic solvents is-

I->Br->Cl->F-

This reverse in order is due to higher charge density of smaller halide ions and thus better interaction with polar protic solvents.

But along a period the charge density is similar and thus the interaction between solvent and nucleophile is almost similar. Thus, along a period the order follows basicity order.


What does an equal sign mean in chemical equations? I have been doing a bit of literature review, and have come across a few equations which use an equal sign (=) instead of an arrow, for example:

$$\ce{Fe^2+ + 1/2O2 = 2Fe^3+ + H2O}$$

What does the equal sign mean? Is the same as a single arrow $\ce{->}$ or the double arrow $\ce{<=>}$?

The IUPAC Green Book (Quantities,
Units and Symbols in Physical Chemistry, Third Edition, 2007) reads:

  2.10.1 Other symbols and conventions in chemistry
  [...]
  (iv) Equations for chemical reactions

  (a) On a microscopic level the reaction equation represents an
  elementary reaction [...] A single arrow is used to connect reactants
  and products in an elementary reaction. An equal sign is used for the
  "net" reaction, the result of a set of elementary reactions. [...]

  $\ce{H + Br2 -> HBr + Br} \quad$ one elementary step in HBr formation
  $\ce{H2 + Br2 = 2 HBr} \quad$ the sum of several such elementary steps

  (b) On a macroscopic level, different
  symbols are used connecting the reactants and products in the reaction
  equation, with the following meanings:

  $\ce{H2 + Br2 = 2 HBr}\quad$ stoichiometric equation
  $\ce{H2 + Br2 -> 2 HBr}\quad$ net forward reaction
  $\ce{H2 + Br2 <--> 2 HBr}\quad$ reaction, both directions
  $\ce{H2 + Br2 <=> 2 HBr}\quad$ equilibrium

  The two-sided arrow $\ce{<->}$ should not be used
  for reactions to avoid confusion with resonance structures [...]

  2.12.1 Other symbols, terms, and conventions used in chemical kinetics
  [...]
  (ii) Composite mechanisms
  A reaction that involves more than one elementary reaction is said to occur by a composite mechanism. The terms complex mechanism, indirect mechanism, and stepwise mechanism are also commonly used. Special types of mechanisms include chain-reaction mechanisms, catalytic reaction mechanisms, etc.

    Examples
    A simple mechanism is composed of forward and reverse reactions
    $\quad\quad\quad\quad\quad\quad\ce{A -> B + C}$
    $\quad\quad\quad\quad\ce{B + C -> A}$
    It is in this particular case conventional to write these in one line
    $\quad\quad\quad\quad\quad\quad\ce{A <--> B + C}$

  However, it is useful in kinetics to distinguish this from a net reaction, which is written either with two one-sided arrows or an "equal" sign
  $\quad\quad\ce{A <=> B + C}$
  $\quad\quad\ce{A =  B + C}$

  When one combines a composite mechanism to obtain a net reaction, one should not use the simple arrow in the resulting equation.
  $\quad$Example
  $$\begin{align}
\ce{A &\to B + C}\quad\text{ unimolecular elementary reaction} \\
\ce{B + C &\to D +E}\quad\text{ bimolecular elementary reaction} \\
\hline
\ce{A &= D + E}\quad\text{net reaction (no elementary reaction,no molecularity)}
\end{align}$$


What gives brown butter its color? Brown butter is butter that has been heated until the water content has boiled off and the remaining liquid has turned brown. It is often called for in cooking and baking, for example in recipes for financiers.

Discussions of the process typically explain the brown color as browning of the milk solids in the butter due to the Maillard reaction. However, even when these milk solids are removed, as is done in most recipes by filtering or careful pouring, the brown color remains.

I believe I understand how the flavor components of the brown butter survive filtering - the volatile products of the Maillard reactions remain dissolved in the liquid fat - but is there a similar explanation for the brown color? Or is it down to the coarseness of the filters allowing small suspended particles through with the liquid?

Any details on the particular compounds responsible for the color, or the chemical processes involved in creating them, would be much appreciated.

The Maillard reaction is also called the non-enzymatic browning reaction. The reaction occurs when there are reducing sugars and amino acids present. The reaction is controlled by solvent choice, time and temperature considerations.
When browning butter, certainly, the Maillard reaction is occurring. There may also be other reactions happening such as Strecker degradation and caramelization.
As noted, browned butter is made by essentially heating butter until the water boils off at 100C and then the temperature ramps upward. So, what is causing the browning once the browned particulates are strained out? Some are chemicals that are formed from the reaction.
The color comes about due to production of melanoidins. As the Maillard reaction produces a range of chemical compounds, some of the components may have a lot of color on their own, such as furfural and its derivatives.
Professor Chi Tang Ho at Rutgers University has done extension work on Maillard reactions in model systems. Heath & Reineccius in their book Flavor Chemistry and Technology write about the Maillard reaction in more general terms that includes more on color production as part of the process.


IUPAC systematic name of Carminic Acid Every IUPAC Systematic name I have come across, I have understood up until now. This is Carminic Acid ($\ce{C22H20O13}$):

(source: chemspider.com)
This is the IUPAC Systematic Name for Carminic Acid:
(1S)-1,5-Anhydro-1-(7-carboxy-1,3,4,6-tetrahydroxy-8-methyl-9,10-dioxo-9,10-dihydro-2-anthracenyl)-D-glucitol
I have 3 main questions:

I do not understand the numbering system inside the brackets. The numbering system implies that the numbers assigned to the carbon atoms is the following:

I would have thought the numbers assigned to the carbon atoms would be the following:

Why is the numbering system the way it is? (Why does the numbering system not go around the molecule, in a normal cycle, counting off each carbon?, Why is the number 10 carbon opposite the number 9 carbon?, Why are 4 carbons [circled in the picture below] left out in the numbering system?)

I also do not understand why Carbons 9 and 10 in the anthracene group are referred to as '9,10-dioxo-9,10-dihydro'. Surely, they only need to be referred to as '9,10-dioxo'? And what is a '-hydro' group?

Doesn't '1,5-Anhydro...-D-Glucitol' imply the following: Hexan-2,3,4,6-tetrahydroxy-1-yl, a chain, and NOT a ring structure of: Oxan-3,4,5-trihydroxy-6-hydroxymethyl-2-yl, considering D-Glucitol (shown below). Also, as well as '1,5-Anhydro...-D-Glucitol' not accounting for the ring structure of that section of the molecule, it also doesn't account for the Oxygen.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name (PIN) for the parent ring component consisting of three benzene rings fused in a linear manner is the retained name anthracene.

Subsection P-25.1.1, Table 2.7, Entry (11) shows that anthracene has a special numbering:

According to Subsection P-14.4 (a), this fixed numbering must be used in PINs and in general nomenclature.

The prefixes ‘hydro’ and ‘dehydro’ are used to indicate addition and subtraction, respectively, of hydrogen atoms (see Subsection P-31.2.1).  Hence, the PIN for the ring component of the structure given in the question is 9,10-dihydroanthracene:

The corresponding monovalent substituent groups derived from such ring systems by the removal of one hydrogen atom are named by replacing the final ‘e’ of the name of the hydrocarbon by ‘yl’ (see Section P-29). Thus, the name for the substituent group derived from 9,10-dihydroanthracene is (9,10-dihydroanthracen-2-yl):

(Note that locants are placed immediately before that part of the name to which they relate (see Subsection P-14.3.2); i.e. ‘…anthracen-2-yl’. Thus, the name ‘…-2-anthracenyl’ given in the question is not correct.)

The substitutive prefixes for the remaining groups are
‘methyl’ $(\ce{-CH3})$,
‘hydroxy’ $(\ce{-OH})$,
‘oxo’ $(\ce{=O})$, and
‘carboxy’ $(\ce{-COOH})$.

The name for the complete substituent group is (7-carboxy-1,3,4,6-tetrahydroxy-8-methyl-9,10-dioxo-9,10-dihydroanthracen-2-yl):

The bases for carbohydrate names are the structures of the parent monosaccharides in their acyclic form (see Subsection P-102.2).
Alditols are named by changing the ending ‘ose’ in the name of the corresponding aldose into ‘itol’ (see Subsection P-102.5.6.5):

The prefix ‘anhydro’ expresses the corresponding intramolecular ether, which is formally obtained by elimination of water from two hydroxy groups of a single molecule of a monosaccharide or monosaccharide derivative (hence the structure is commonly called an intramolecular anhydride). The prefix ‘anhydro’ is preceded by a pair of locants identifying the two hydroxy groups (see Subsection P-102.5.6.7.1).


Energy levels in non-hydrogen-like atoms The energy $E_n$ for a hydrogen like atom is given as

$$E_n = -hcR_\ce{H}\frac{Z^2}{n^2}$$

However, aside from on wikipedia where there is

$$E_n = -hcR\frac{Z_\text{eff}^2}{n^2},$$

I can't find anywhere that relates to the energy of an atom that has more than one electron. Is there an equation that gives the specific (or approximate) energy or is there an 'easy' way to calcuate a value for $Z_\text{eff}$?

Consider an atom to consist of an electron orbiting a positvely charged ion core (with one or more electrons associated to it). If you excite the electron to larger and larger values of the principal quantum number $n$ the classical electron orbit becomes larger and larger as well (in fact it scales as $n^2$). At sufficiently large distance, the postively charged ion core behaves as a point charge (like a single proton) and at long range you basically have the hydrogen problem with a different mass (assuming perfect shielding of the core electrons). In order to find the overall solution to this problem, one has to connect the long-range and short range behavior of the system. One way of doing this, is is by considering the system as a collision problem at negative energy (that is, a bound system). The effect of the nonhydrogenic ion core then results in a scattering phase shift between an incommming and outgoing electron. It can be shown that the solutions to this problem have a very similar appearance as the Bohr formula, namely:

$$
E_{n,\ell}=E_\text{IE}-\frac{hcR_A}{(n-\delta_\ell)^2}
$$
where $R_A=R_\infty(1-m_e/m_A)$ is the mass-corrected Rydberg constant, $E_\text{IE}$ is the ionization energy and $\delta_\ell$ is the so-called quantum defect for electrons with orbital angular momentum $\ell$. The quantum defect is related to the phase shift induced by the core and is different for $s, p, d, f$ etc. electrons. In principle, the quantum defect is a function of the binding energy of the electron to the ionic core, but in many cases this dependence may be neglected (in particular for large $n$).

What does this mean in practice? Let us for example look at the $n$s levels of the sodium atom. We find that $R_\text{Na}=109734.697205$ cm$^{-1}$ and $E_\text{IP}(\text{Na})=41449.451$ cm$^{-1}$. The second column in the table below gives the energy (in cm$^{-1}$ from the NIST website) and the third column gives the binding energy of the electron (also in cm$^{-1}$). The last column gives the so-called effective quantum number, that is $n^*=n-\delta_\ell$, with $\ell=0$ for $s$ orbitals

 3s         0.000   -41449.451  1.63
 4s     25739.999   -15709.452  2.64
 5s     33200.673    -8248.778  3.65
 6s     36372.618    -5076.833  4.65
 7s     38012.042    -3437.409  5.65
 8s     38968.510    -2480.941  6.65
 9s     39574.850    -1874.601  7.65
 10s    39983.270    -1466.181  8.65
 11s    40271.396    -1178.055  9.65

As you can see the quantum numbers take the form of the hydrogen solution if we start counting the principal quantum number form the 3$s$ levels and take $\delta_0\approx -0.65$.

To calculate the quantum defect is difficult as it requires ab initio calculations and then converting the energy levels to effective quantum numbers from which the defects can be extracted. However, experimentaly these numbers are very convenient, and one may use this quantum-defect theory for instance to determine the ionization energy of atoms and even simple molecules very accurately by extrapolating Rydberg series.


Partial pressure and chemical equilibirum Suppose I am given following reaction in a vessel:
  $$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$
  $P_\ce{NH3} = P_\ce{H2S} = \pu{1 atm}$ at equilibrium; if volume is doubled at this instant, then what would be $P_\ce{NH3}$ at new equilibrium?

I know that as soon as volume is doubled, pressure would drop to half at this instant, and $K_\mathrm{c}$ will have the same value at new equilibrium, but how to actually use this to calculate the partial pressure of ammonia at new equilibrium?

No, at new equilibrium $K_c$ would actually is the same value. This constant stands for reaction between these two compounds no matter the pressure, it is only affected by temperature.

As ammonium hydrosulfide is a solid, the reaction is only relevant at the surface of the crystals, and the "concentration" of ammonium hydrosulfide at the relevant areas will remain constant. Therefore same partial pressure of both ammonia and hydrogen sulfide is needed to maintain the equilibrium. Hence, at equilibrium, the partial pressure of both ammonia and hydrogen sulfide should be the same as the initial partial pressure.


How can I be safe and protect myself using trimethyl tin chloride? I will be using trimethyl tin chloride (1M in THF) in a few days for the first time. Looking at the MSDS, it looks like it is very toxic and has noxious vapors. Are there any more standard precautions for working with it other than the usual?

E.g.,

Check for proper airflow in the hood
Wear lab coat, safety glasses, and double gloves

In addition to the safety precautions mentioned in the question, the following can increase the safety for the experimentator:

Keep the fume hood as closed as possible with the head outside and the arms inside. Usually this keeps fumes from entering the body, since the fume hood airflow goes the other direction.
Work quickly. While this not only minimizes side reactions of the reactant with the air and other things, it also ensures less exposure time, minimizing vaporization.
As Aesin mentioned in the comment, if you have concerns about the infrastructure, you could always perform the experiment in a glovebox.

In the end, you have to feel confident that the safety measures are effective. If you don't, you will have fear of the reaction, and inevitably be less concentrated and more prone to errors. In combination with "work quickly" this may well become a disaster.

Maintain a healthy portion of respect, keep quenching agents close by and execute the reagent transfer with a steady hand.


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

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

My computation goes like this:

$$V_2=\frac{T_2V_1}{T_1}=\frac{(\pu{261 K})(\pu{400 mL})}{\pu{305 K}}=\pu{342.295082 mL}$$

As per the given, the least number of significant figures is one (from $\pu{400 mL}$). How do I approximate $\pu{342.295082 mL}$ to one significant figure?

I didn't know how to create a search query about this so I went here. Thanks.

Assuming the precision of the given volume, $V_1 = \pu{400 mL}$, is only to 1 significant figure (i.e. only the hundreds place digit of "4" is precise), then your final answer would be rounded to $V_2 = \pu{300 mL}$.*

If we were to say that $V_1$ had 3 significant figures, then the final answer would round to the ones place, resulting in $V_2 = \pu{342 mL}$. However, the value would technically need to be written as "$\pu{400\!. mL}$," including the decimal place to denote the precision going out to the ones place, in order to imply 3 significant figures instead of 1.

In general, zeros (to the left of the decimal place) not surrounded by non-zero values are not denoted as "significant." For further rules and guidelines for identifying what numbers are significant, how many significant figures there are, and how addition/subtraction or multiplication/division impact the final answer's significant figures, check out this web page from Northern Kentucky University (recommended search terms for future information on this subject would be "understanding significant figures" or "significant figures rules").

*If difficulty was also found in the process of where to begin rounding the answer from, the key is finding the "rounding digit," which, if you have only 1 significant figure to work with, will be the furthest numerical value to the left. See this page for further tips.


What is the correct number of stereo isomers in case of 3,6-dimethylpiperazine-2,5-dione? What is the correct number of stereo isomers in case of the following compound, i.e. 3,6-dimethylpiperazine-2,5-dione?

The answer key gives three, but I think there should be four.
I recognised the two planes of symmetry, one through the methyl groups and the other through the keto groups.

Three is the correct answer.

As Ivan pointed out, there are no planes of symmetry, for example:

Ivan is also correct - the "RS" is the same as the "SR". You can prove this to yourself most easily using molecular models, but here is how to see it on the page. You just need to rotate the molecule by 180° in the plane of the page:

So, despite having chiral centres, this compound is achiral because it is superposable with its mirror image. "But wait, I was told that a molecule needs to have an internal plane of symmetry to be a meso compound" I hear you say. Unfortunately, this is an incomplete truth that many sources repeat. To be a meso compound, it can alternatively have a centre of symmetry, also known as an inversion centre. This is simply one point in space through which any atom is reflected through - ie. each atom is diametrically opposite an equivalent atom at an equal distance from this point.

The other two stereoisomers are the (R,R) and the (S,S). These are non-superposable mirror images, and hence enantiomers. Note these isomers have no plane of symmetry or centre of symmetry.

Once again, this is easier to see with physical models.


Mechanism of rearrangement reaction found in synthesis of retinol Today I was solving a problem that was about the synthesis of retinol. This is a step I didn't understand:

In the solution, it's written that the mechanism goes through this intermediate:

I have not seen anything like this yet.

When trying to propose a mechanism, especially one where the bond changes are not obvious, I like to start by numbering the atoms of the product and trying to map them to the reactants. Here, I've taken the numbering given by ChemDraw for the product and translated it to the reactant.

Carbons 1-3 are pretty logical. In the product, carbon 4 is alkyl with 2 hydrogens and between to other carbons. In the starting material, the carbon connected to C3 is a carbonyl with zero hydrogens. It seems unlikely to me that this would be C4.

So I look for another, more logical mapping. For example, carbons 7 and 9 are methyl groups in the product, and there are two methyl groups in the starting material, which I label as 7 and 9. In both structures, they are bridged by a single carbon, which I call 6.

In the product C5, is part of an alkene and so is the carbon connected to C6 in the reactant, so that seems logical. And the next carbon in the starting material has two hydrogens, so that seems a better mapping as C4. The last carbon of the reactant must be lost as carbon dioxide, so I've labeled it C10.

With a hypothetical atom mapping, now I can identify bonds broken and formed. C10 and its adjacent oxygens are being lost as carbon dioxide, which makes it easy to identify the bonds broken as C3-C10 and C6-O, with a shift of the pi-bond from C4-C5 to C5-C6.

Without being aware of [3,3]-sigmatropic rearrangements, or the Claisen rearrangement specifically, you might be stuck. The Claisen rearrangement is a cyclic movement of electrons converting an allyl vinyl ether into a gamma,delta-unsaturated carbonyl compound. This is driven by the formation of a carbonyl at the expense of an alkene.

The given starting material isn't quite an allyl vinyl ether, but as a 1,3-dicarbonyl, it is equilibrium with one by tautomerization. From the tautomer shown, the Claisen rearrangement is possible, which forms the C3-C4 bond we noted earlier and shifts the pi bond into the desired position. Also note that the resulting intermediate is the one shown in the solution manual. Decarboxylation through a similar cyclic movement of electrons gives the tautomer of the product along with carbon dioxide.


Why is degree of dissociation used to estimate strength of an acid? Problem

Which of the following will give strength of acid?
A) pH
B) Normality
C) Degree of dissociation
D) Dissociation constant

Answer

 C, D

Question
If pH is not considered as a metric to assess the strength of an acid because it depends on dilution (even strong acids can have high pH), then why is degree of dissociation a reliable metric for the same? Doesn't degree to dissociation depend on dilution as well?

I would answer only the constant of dissociation $K_a$, because it is a thermodynamical constant and you can calculate it from potentiometric measurements without dealing with activities and so on. If you have the $\Delta G$ of reaction, which is related with the difference of electric potential of the cell, you can calculate the constant as $\Delta G = -RT \ln K$. The pH is not an indicator of the acid strenght, but only its consequence. If you have a very diluite solution of sulfuric acid, its pH will be near to 7, but it does not mean that sulfuric acid is weak. Also the degree of dissociation can give you incorrect information if you are dealing with very diluite or very concentrate solutions.


Can I equalise osmotic pressure using two different solutes? I understand that there is a pressure gradient between an aqueous solution, e.g. salty water, and, say, pure water due to osmosis. My question is, is it possible to match or almost match, the osmotic pressure between two solutions of different solutes, but using a solution with a different solute. For example, would it be possible to make the osmotic pressure gradient between salty water and some other solution zero, assuming both are at the same temperature and pressure?

I am not a chemist, so purely qualitative answers are helpful.

Yes, I think so. Based on our comments above, I believe you want to prepare two solutions that are isotonic with each other (have the same concentrations of different solutes) and allow them to communicate across a membrane that is selectively-permeable to water only.

Tonicity is described here as:

  Tonicity is a measure of the effective osmotic pressure gradient (as defined by the water potential of the two solutions) of two solutions separated by a semipermeable membrane. In other words, tonicity is the relative concentration of solutions that determine the direction and extent of diffusion. It is commonly used when describing the response of cells immersed in an external solution. Unlike osmotic pressure, tonicity is influenced only by solutes that cannot cross the membrane, as only these exert an effective osmotic pressure. Solutes able to freely cross the membrane do not affect tonicity because they will always be in equal concentrations on both sides of the membrane.

I found this nice explanation with illustrations and one example where it shows that you could prepare a 10 mM solution of albumin and a 10 mM solution of glycine, separated by a water-permeable membrane, and no net flow of water occurs in either direction.


Double Deionized Water and Superheating I was told that dd-water will superheat when heated up in a microwave because of a lack of ions.

1) Doesn't water auto-ionize? I asked this; the answer was it auto-ionizes "too slowly" in dd-water to prevent superheating.

2) This begs the question: what are the kinetics of the auto-ionization of water? What's the rate?

3) This also begs the question - what is superheating? What does ions have anything to do with superheating?

This has a lot to do with nucleation.  Just like crystallization and initial rain drop formation, for boiling to occur an initial site within the solution for a bubble of vapor to form on is required. So if you have very clean water that has been deaerated in a highly polished vessel there is an increased chance for superheating to occur due to the reduction in potential nucleation sites.  Tap water can superheat too, it is just less likely to do so because it typically contains more impurities that can serve as nucleation sites. This why boiling chips are commonly used in lab distillations.  They provide a site for vapor nucleation and usually prevent superheating of the material to be distilled.  Introducing other sites for nucleation, such as dissolved bubbles, solid contaminants, rough spots on the wall of the container all serve to lessen the probability of superheating.  I'm not sure what ions would have to do with the superheating process other than they are often representative of microscopic solid impurities that can provide that surface for the initial nucleation event.


Can amide bonds fragment in ESI-MS? Electrospray Ionization Mass Spectrometry(ESI-MS) is a technique in mass spectrometry to get an idea about molecular ion peak. In the ESI-MS spectra of an amide compound, I observed that the molecular ion peak is much less abundant as compared to two other peaks, which correspond to the carbonyl and amine fragments between which the coupling reaction was carried out. But, TLC confirmed that there was no amine or acid remaining in the mixture after the reaction(EDC coupling).Hence my question is, although amide is a very stable bond, can it fragment in ESI-MS?

Yes, amide bonds can fragment during electrospray ionization.

Intentional fragmentation in a collision cell

Amide bonds are intentionally fragmented in mass spectrometers all the time, during proteomics analysis.  In that case, peptides are ionized, particular ions are isolated (by the quadrupole of the mass spectrometer), and then the intact peptide ions are fragmented intentionally in a collision cell inside the instrument.  After collision with high-energy gas molecules, peptides fragment in mostly predictable ways giving rise to y, b, a, and z ions.

"In-source" fragmentation

The key step of the fragmentation process is collision with high-energy gas molecules.  That can happen accidentally during electrospray ionization in the ion source.  That's because the ion source, being at atmospheric pressure, has many many more gas molecules around than a collision cell, which might be at 10$^{-5}$ torr or lower in pressure.  So after your ions of your peptidic molecule form in the ion source, some of them might bump into these atmospheric gas molecules and be fragmented in the ion source, before they even make it into the mass spectrometer.

Controlling in-source fragmentation

The propensity of this "accidental" fragmentation to happen is somewhat tunable by ion source parameters -- nebulizer and capillary voltages, gas flows and temperatures, etc.  The details depend entirely on your exact model of MS.  If the carboxyl and amine fragments are both more intense than the "molecular ion", it could be a sign that the ion source is tuned too aggressively -- trying lowering a voltage or two and see if you boost the molecular ion signal.  (Alternately, for some applications, in-source fragmentation could be desired.)


Is there a difference between oxidation state and valency? I just looked up peroxide on Google, and I saw that the oxidation state of oxygen in $\ce{R-O-O-R}$ is $–1$, and not $–2$ (like $\ce{H2O}$), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong?
I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each oxygen forms $2$ bonds so, if I remove the other atoms, won't the imaginary charge left still be $-2$? Then how, does $-1$ come into play?

I think it will help you understand if you know the historical context. The concept of valency existed well before chemical bond, electrons, protons, etc. were known. It is a primitive concept from late 1860s. It referred to the capacity of an element to bind with hydrogen or any other element.
For example, if we say carbon has a valency of four, all it means that it can combine with 4 hydrogen atoms. If we have say oxygen has a valency of 2, it means it will combine with two hydrogen, the famous H$_2$O! Does this make it clear? One would determine valency by pure elemental analysis. If I have FeCl$_3$, iron has a valency (=combining power) of 3, because it can combine with three chlorine atoms. I am avoiding the word ion, because ions did not exist then! The valency of oxygen in peroxide is still 2 because each oxygen atom is combining with two atoms via single bonds. In this case, the other atom is another oxygen itself.
Now read the Wikipedia article Valence. Note that I did not assign any sign to valency
Oxidation number or state is a newer notion which came about when electrons were very well recognized to play a role in bonding and the atomic structure become relatively well understood. I quote it's definition from the OED

a number representing the degree of oxidation of an element in a
particular ion, molecule, or substance, being a notional electric
charge equal to the number of electrons lost (or gained, if the number
is negative) by an atom of the element in forming the species or
substance, all bonding being assumed to be completely ionic

Oxidation state/ number has a sign associated with it. Sometimes, valency and oxidation number can match in terms of their absolute values.
NH$_3$
Valency of nitrogen= 3
Oxidation state of nitrogen = -3


Basic chemistry of the Shinoda test for flavonoids? In this test, four pieces of magnesium fillings (ribbon) are added to the ethanolic extract followed by a few drops of concentrated hydrochloric acid. A reddish colour indicates the presence of flavonoid.

What is the basic chemistry of the test?  What are the chemical reactions taking place in the test that lead to the formation of red precipitate, and also, why have we used magnesium and concentrated HCl in the test?

I found one source indicating that the reaction involves the conversion of the flavonoid to the corresponding anthocyanidin. The structure of that resulting anthocyanidin contains an extended system of conjugated $\pi$-electrons, which constitute a chromophore. As it happens, the electrons of molecules containing conjugated $\pi$ systems have energy level transitions corresponding to absorbance in the visible (and also ultraviolet) region of the electromagnetic spectrum. This accounts for their perceived color.

As for the precise mechanism involved, I was not able to find any sources whatsoever. It's obvious that the flavonoid is reduced, with the magnesium metal being the reducing agent. It could be the process occurs by single electron transfer (SET) with radical and anionic intermediates, or via a cyclic transition state and carbenoid intermediates with coordination of the magnesium, or something else entirely. In any case, the purpose of the acid is likely to activate the carbonyl of the flavonoid to reduction by protonating it, and providing a source of $\ce{H+}$ ions for the subsequent formation of water.

(I find it unlikely that a definitive mechanism for this reduction has been determined, if only because the mechanisms of many similar reductions involving metals and mineral acids widely used in synthetic organic chemistry have yet to be entirely resolved.)


Maximum Local Value for the Electron Density So the electron density is the function $\rho(\vec x)$ that associates a value of electrons per cubic angstrom, to each point $\vec x$ in the 3D space.
This information tells us how likely it is to find electrons at specific regions of space within a system. Integrating $\rho$ over a particular volume $\mathcal{V}$ gives us the number of electrons likely to be found, at any given moment, inside $\mathcal{V}$, denoted $n[\mathcal{V}]$.
$$
\tag{2}
\label{eq:rho-integrate-volume}
n[\mathcal{V}] = \int_\mathcal{V} \rho(\vec x) \mathrm{d}\vec x
$$
Integrate over the whole space $\Omega$ where the system is and you get $N$, the total number of electrons in the system.
$$
\tag{3}
\label{eq:rho-integrate-omega}
N = n[\Omega] = \int_\Omega \rho(\vec x) \mathrm{d}\vec x
$$
Electrons repel one another and are attracted to the atomic nuclei. I believe that particularly high values of $n[\mathcal V]$ are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more.
So my question is Is there a maximum value for $n[\mathcal{V}]$ above which the mutual repulsion of electrons would mean the density was unreal?
I understand that such a value would vary depending on promixity to the atomic nuclei (which would shield electrons from their own mutual interactions) and the size of the integrated volume $\mathcal{V}$. What I mean is if there is a known relationship between these factors that would give us the maximum amount of electrons allowed inside a particular region of space.

Single electron systems

I believe that particularly high values of n[V] are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more.

The main constraint on high electron density is not the repulsion between electrons, but the Heisenberg uncertainty principle. Even a single electron has a limited average electron density, given its average momentum (see e.g. here). For single electron systems (hydrogen-like ions), you get the highest electron density by increasing the nuclear charge. According to the virial theorem, as the potential energy of the electron increases (the magnitude, so tighter binding), the kinetic energy also increases. This will go along with a higher momentum, allowing the electron to occupy a smaller volume (i.e. have a higher density).
Multi-electron systems
For multiple electrons, theory gets complicated. X-ray diffraction gives experimental data related to electron density, but you need extremely high resolution (and extremely low static and dynamic disorder) to be able to map the electron density near the nucleus. In this study on LiF and NaF, the electron density at the lithium nucleus is estimated as 50 electrons per cubic angstrom using maximum entropy methods (panel b).

Fluorine, with a higher atomic number, will have higher electron density at the nucleus, but the density is cut off in the figure so you can't see the maximum value.

I understand that such a value would vary depending on promixity to the atomic nuclei (which would shield electrons from their own mutual interactions)

The innermost electrons would not be shielded, so the maximum electron density for a uranium nucleus with one electron would not be much different from one with 92 electrons. In fact, the binding energy of the inner electrons was used to figure out atomic numbers using Moseley's law.

Is there a maximum value for n[V] above which the mutual repulsion of electrons would mean the density was unreal?

Electron repulsion is not the main constraint on high electron density. To get high electron density, you need a high nuclear charge. At some point, relativistic effects become important, and for super-heavy nuclei, electrons get captured.
There was a comment from Martin about the use of the word "unreal". I agree with Martin that if you observe a certain high electron density, that is as real as it gets. You might run into a calculation claiming a ridiculously high electron density, and then might conclude that results are not plausible and warrant further research.
So where can I find high electron density?
That would be near the nucleus of the heaviest elements as long as they are stable at least for a while. Using the arguments made above using the uncertainty principle, the binding energy of inner electrons, and the associated momentum of the electron (or looking at the Bohr model), the radius $a_0$ should be proportional to the reciprocal of the atomic number, meaning the electron density increases with the cube of the atomic number.


Why does carbocation stability increase in the order 1°, 2°, 3°? I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms. 1° carbocations are unstable to the point of not having been observed in solution, ever. 2° are more stable, and 3° carbocations are the most stable.

I know that higher stability carbocations do also require less activation energy in their formation. Why is that? Why exactly are they more stable? And is this another way of saying they are also less reactive? Does this in turn mean they require a better nucleophile compared to less stable carbocations?

A small side question. How should I think about resonance structures? You don't view it as changing between the forms, spending more time being the more stable form, but rather as a seamless blend of the forms, with more characteristics of the more stable form, right?

One word: hyperconjugation! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.


Reduction of carboxylic acids by LiAlH4 I've read that when a carboxylic acid reacts with $\ce{LiAlH4}$ the corresponding alcohol is formed:

But when I try to think of the mechanism, I get stuck here:

$\ce{LiAlH4}$ produces $\ce{H-}$. Since $\ce{H-}$ is a strong base it should immediately abstract a proton from the carboxylic acid to give the corresponding carboxylate ion (just like in the reaction of carboxylic acids with Grignard reagents), instead of undergoing nucleophilic addition to give the alcohol.

How do I resolve this?

The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:

Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:

I found the pictures here.


IUPAC name of the tert-pentyl group What is the IUPAC name of tert-pentyl? I know that isopentyl is 2-methylpentane and neohexyl would be 2,2-methylhexane, but what about tert?

The retained names “tert-pentyl”, “isopentyl”, and “neopentyl” were still included in the 1993 IUPAC recommendations.
However, according to Subsection P-57.1.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these retained names are no longer recommended. The preferred prefix names are as follows:

“tert-pentyl”: 2-methylbutan-2-yl
“isopentyl”: 3-methylbutyl
“neopentyl”: 2,2-dimethylpropyl

P-57.1.4 Retained prefixes no longer recommended
The retained names phenethyl (2-phenylethyl) for $\ce{C6H5-CH2-CH2-{}}$; benzhydryl (diphenylmethyl), for $\ce{(C6H5)2CH-{}}$; isobutyl (2-methylbutyl) for $\ce{(CH3)2CH-CH2-{}}$; sec-butyl (butan-2-yl, 1-methylpropyl) for $\ce{CH3-CH2-CH(CH3)-{}}$; isopentyl (3-methylbutyl) for $\ce{(CH3)2CH-CH2-CH2-{}}$; tert-pentyl (2-methylbutan-2-yl, 1,1-dimethylpropyl) for $\ce{CH3-CH2-C(CH3)2-{}}$; and neopentyl (2,2-dimethylpropyl) for $\ce{(CH3)3C-CH2-{}}$; are no longer recommended; the first name in parentheses is the preferred prefix name.


Numbering in a derivative of azulene I was wondering if anyone can tell me how to number this molecule:

This molecule is derived from azulene by addition of two sulfurs and two oxygens at positions 2, 6, 1, and 5 respectively. I would like to learn the proper procedure of numbering. I tried googling around and reading papers but could not find the correct terminology for this process.

The rules for numbering of bicyclic compounds such as azulene are given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

First, you need to generate a "preferred orientation" of the molecule:

  P-25.3.2.3.2 Polycyclic fused ring systems are oriented in accordance with the following criteria considered in order until a decision is reached:
  (a) maximum number of rings in a horizontal row;

Because of this, the two rings need to be horizontally aligned with each other in the preferred orientations. Note that this rule alone does not distinguish between the two orientations on the left.

Then, the numbering proceeds as follows:

  P-25.3.3.1.1 The numbering of peripheral items in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter 'a', 'b', 'c', 'd', etc.

This leads to two possible numbering systems for the two orientations above. Which one is preferred is given by a further rule

  P-25.3.3.1.2 If alternative numberings for a ring system remain [...] the following criteria are applied in order until a decision is reached: [...]
  (c) low locants are assigned to fusion carbon atoms;

The ketones in your molecule have greater seniority over the thiols (P-41), and hence the ketones are expressed as the suffix (leading to a name ending in "dione") instead of the thiols (a name ending in "dithiol"). Note that if the thiols are deprotonated this seniority ranking would be reversed, as anions have greater seniority over ketones (hence you would have a name ending in "bis(thiolate)").

The quinone derived from azulene is named as azulene-x,y-dione, where x and y are the locants of the two carbonyl carbons (P-64.2.2.2.3). The asymmetry in the quinone leads to ambiguity in the numbering, which is resolved by the next rule:

  P-14.4 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);

Finally, the –SH groups are inserted as prefixes into the name, as they can no longer be suffixes. The appropriate prefix for –SH is 'sulfanyl'. The multiplier used is "bis" and not "di" to avoid ambiguity:

  P-16.3.6 The prefixes 'bis', 'tris', 'tetrakis', etc. are also used to avoid ambiguity:
  (a) before a mononuclear subset of a polynuclear acyclic structure;
  Examples: bis(sulfanyl) (preferred prefix; defines two –SH groups, see P-63.1.5; whereas disulfanyl defines the –SSH group; see P-63.3.1)

As you can see, the manual generation of such names is in no way trivial. Most people use software to automatically generate names, which is generally sufficiently clear for the chemical community to understand them. However, it should be mentioned that these automatically generated names are not always the preferred IUPAC names: for example, ChemDraw 16 gives "2,6-dimercaptoazulene-1,5-dione" as the name, even though the prefix "mercapto" is no longer recommended (P-63.1.5).


Can plastic be used as fuel for vehicles? I don't have much of a background in chemistry, but I know that plastics are largely made of carbon, as is oil. Is there any way to "covert", i.e. a plastic water bottle, into a substance that can be used as fuel for ? Thanks.

There are many ways to recycle different types of plastic, and conversion to liquid fuels suitable for internal combustion engines is currently done on an industrial scale. The key to converting the carbon contained in long-chain polymers that make up the plastics is called pyrolysis, which is essentially a way to thermally decompose these products into shorter chain hydrocarbons suitable for use as liquid (or even gaseous or solid) fuel.  According to this Wikipedia article:

Anhydrous pyrolysis can also be used to produce liquid fuel similar to diesel from plastic waste, with a higher cetane value and lower sulfur content than traditional diesel.

An example of a current commercial waste-plastic-to-fuel operation is the Canadian corportion Plastic2Oil, Inc., who claims to "convert waste plastic to ultra clean oil" in their 250,000 gallon fuel production and blending facility.
The following is an excerpt from the description of the patent-pending process used by Plastic2Oil:

The processor accepts unwashed, unsorted waste plastics, composites and commingled materials. Although many sources of feedstock are available, we are focusing initially on post-commercial and industrial sources, since these are readily available in large supply, and present a cost-effective solution for companies. We currently have several companies providing plastic feedstock.
The hopper is loaded with up to 4,000 lbs. of plastic feedstock per machine per hour using a forklift. The plastic is loaded into the pre-melt reactor by a continuous conveyor between the hopper and the reactor. The plastic is then heated using the off-gases produced by the process. After the plastic has been liquified in the pre-melt reactor it passes through a solids-liquids separator before going into the main reactor.
In the main reactor, the liquified plastic hydrocarbons are cracked into various shorter hydrocarbon chains and exit in a gaseous state. Plastic2Oil's proprietary catalyst and unique process engineering enables us to capture an average of 86% of the hydrocarbon content of plastic. The Petcoke residue produced at this stage (approx. 2-4%) remains in the processor chamber and is automatically removed while the processor is in operation.
From the main reactor, the gases that are fuel oil and diesel are condensed and separated, then proceed into temporary fuel tanks. All of the gaseous "light fractions" or off-gases (approx. 10-12% of process output), such as methane, ethane, butane and propane, exit the system and are used to fuel the furnaces.

Note: I have no affiliation whatsoever with Plastic2Oil or their industry. I chose their company as an example largely at random, or more accurately, due to their very usable and easily found website.