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Does quantum mechanics play a role in protein folding?
Protein folding takes a very long time (relatively speaking) when thinking of quantum mechanical effect. However, for the initial micro-steps of folding, when an atom, or a configuration of atoms, can go (freely speaking) right or left (that is, to other possible configurations), in processes that take less than a picosecond, isn't QM relevant? that is, the atom has a probability to go to either places, in a QM sense, that it goes to both places, and due to interaction with the world, it's wave-function collapses or localizes and that is how the folding progresses?

related http://chemistry.stackexchange.com/questions/53703/how-does-entanglement-explain-myoglobins-preference-for-o%e2%82%82

There's several aspects to your question.
Does quantum mechanics play a role in protein folding? Yes. The origin of the van der Waals interaction is ultimately a quantum mechanical one. (At least the induced dipole portions: how the electrons move with respect to each other and an external electrical field is driven by quantum mechanics.) Also, while part of hydrogen bonding can be described by classical dipole-dipole interactions, there's a non-negligible portion which is due to non-classical electron interaction. (How much this last bit contributes, if at all, is a bit controversial, though.)
But that's all electrons. The other part of your question is asking implicitly about the quantum behavior of nuclei.
In molecular modeling we often use a viewpoint termed "the Born-Oppenheimer approximation". That is, while we allow the electrons to delocalize and exchange and be their quantum mechanical self, we treat the nuclei pretty much like classical particles. Using this approximation, we can get pretty decent correspondence with experimental results.
So the best we can tell, the quantum nature of atoms as a whole (that is, nuclei as opposed to electrons) isn't needed for things like protein folding. The nuclei are probably not substantially delocalized, they're probably not interfering with one another, and they're probably not quantum tunneling.
Probably. There's some evidence that quantum tunneling of nuclei in the process of enzyme catalyzed reactions. This is mainly hydrogens, though. In most of these experiments the quantum tunneling drops off dramatically when you go to deuterium or tritium nuclei. I'm unaware of any system where carbon/oxygen/nitrogen tunneling is seriously proposed - they're just too heavy.
So there may be some contribution to the quantum nature of nuclei to protein folding, but all indications at this point indicate that it would be minor. The quantum properties of electrons, though, are definitely important.

Here is a lecture containing examples of reactions where heavier atoms like nitrogen and carbon tunnel and the author stresses on the fact that not only the individual atoms that tunnels but the system containing these atoms is tunneling as well.
https://www.princeton.edu/chemistry/macmillan/group-meetings/DEC_tunneling.pdf

@M.ghorab Can you send another link? This is not working. I'm interested in reading

Protein folding takes a very long time (relatively speaking) when
  thinking of quantum mechanical effect.

First note that, in principle, for the time being, quantum mechanics is considered to be universally valid at all size and time scales. So, in general, it does not really matter how big a system is or how long a process takes: every system is a quantum system and every process is a quantum process.
Yes, in some circumstances we use classical theories instead for the description of physical systems and processes. But we do so only for the reasons that in some cases quantum theories provide unnecessarily complicated descriptions of such systems and processes, while simpler classical theories lead to the very same predictions (taking the desired accuracy into account), and thus, can serve as good approximations. For example, as the number of particles in a system grows, the smaller "quantum effects" become and the better classical mechanics describes such a system. In the infinite number of particles limit quantum mechanics simply reduces to classical.
Note, however, that it is quite difficult to say in general when you can safely switch from quantum description to classical one1, so that in practice we often do the switch not when we can, but when we are forced to, i.e. the classical modeling is usually used when the quantum one is simply impossible due to enormous computational cost and not necessarily when the former is certainly a good approximation to the later. This is indeed the case of any simulations at atomic level that are performed classically, including protein folding simulations: currently we have computers at which at best we can do only semi-classical/semi-quantum treatment (AIMD) and of only very small proteins. This is just a technical limitation.
To summarize, there is no distinct boundary between classical and quantum worlds simply because classical world is not real, it is just a mental construction that serves occasionally as a good approximation to quantum reality. The world we live in is a quantum world.

that is, the atom has a probability to go to either places, in a QM
  sense, that it goes to both places, and due to interaction with the
  world, it's wave-function collapses or localizes and that is how the
  folding progresses?

This is an interpretational question that, strictly speaking, lies outside of the domain of science. Protein folding is described by the laws of quantum mechanics, the way you interpret these laws is irrelevant.

1) At best you can have some fuzzy criteria such as, for instance, that quantum theories are unavoidable when action is of the order of the Planck constant.

This answer, however nice, could use a little of more direct approach ;)

I was curious to see what kind of research had been done on this topic, and didn't expect to find much, but I was quite wrong. So, I'll summarize some articles that I found and give the references at the bottom.
Why would we ever consider quantum mechanics in protein folding?
First, an introduction to why someone would even try to come up with a quantum mechanical model for protein folding. Well, protein folding can be thought of as a chemical reaction, and should thus be described by the various laws we have for rates of chemical reactions. Specifically, the Arrhenius rate equation relates temperature to the rate of a reaction. It is often seen written as follows, $$k(T)=Ae^{\frac{-E_a}{RT}}$$ where $k(T)$ is the rate constant, $A$ is the pre-exponential factor (not a very descriptive name), $E_a$ is the activation energy, $R$ is the gas constant and $T$ is the temperature. This equation is where we get the general idea that as temperature gets larger, the rate constant will also get larger.
Another key is that this equation predicts that if we were to plot $ln(k)$ vs. $\frac1T$, we would get a straight line. This is indeed true for most all chemical reactions. These so-called Arrhenius plots have been made for many proteins, and they are very unusual. Here are a whole lot of these plots which I found in an article on non-Arrhenius kinetics of folding and unfolding.

The details of these plots are beyond my understanding of the experiment, but you can read more in my first citation. One detail which should be pointed out is that they are using some type of absorbance parameter to make the experiment possible but this absorbance is directly related to the rate of at that temperature, so this still demonstrates non-Arrhenius kinetics. The point is, something weird is going on here (rate decreasing with temperature non-linearly)! What makes it even weirder is that something else which defies chemical intuition happens: the unfolding process is not simply the reverse of the folding process. In fact, the unfolding process generally produces a fairly linear Arrhenius plot, but the folding process does not.
Finally, the last reason one might expect quantum mechanics to be important in a process like this is the very fact that it ever happens! It's been pointed out (I can't find where I read this, but it was in a journal article) that if a protein found the minimum energy configuration by sampling all possible states, it would never finish folding! Even if every sample was less than a femtosecond, proteins of any reasonable length (say 100 amino acid) could not fold nearly as quickly as they do because the number of possible combinations is astronomical. Thus, this cannot just be random motion like how we normally think of these types of things.
How do we explain this?
Needless to say, lots of people have tried to explain this phenomenon with varying levels of success. Most of the early interpretations ascribed it some kind of solvent effect which has to do with the non-linear change in hydrophobicity (I hate that word) of portions of the protein with temperature. Alas, the physicists caught wind of this, and they were not convinced. Somewhere in China, a physicist said to himself, this sounds like quantum mechanics to me! Why not? Super weird things happening? Probably quantum mechanics.
So, he came up with a quantum mechanical model for protein folding. How is this even possible? Well, I'll explain the various assumptions he made (as far as I understand them), and why this could even work.
First, he makes a model which can only describe two-state proteins. These are proteins which have two minima on the potential energy surface where the protein is almost always found, namely, folded or unfolded (see ref. 2).
Second, he knows from other literature that the most time consuming portion of protein folding is any kind of torsional motion. Thus, this is the portion of the motion which he treats explicitly.
Third, because one must have a potential energy function to describe these torsions, he assumes that they behave like harmonic oscillators.
Finally, because the exact form of the partition function for a harmonic oscillator is known, he is able to put this quantum mechanical model in terms of various thermodynamic quantities such as Gibbs and entropy. This is key because the model must be compared to previous experiments.
Feel free to look at the math in references 3 and 4. It's not impossible to follow, but isn't easy either...
Now, I will show two comparisons of their quantum mechanical theory with experiments performed in the literature (this is from ref. 4). The first figure shows Arrhenius plots for protein folding (notice their unusual shape), while the second shows Arrhenius plots for unfolding (mostly linear). It is very hard for me to understand this difference between folding and unfolding, but their model predicts it, so there is precisely no reason to reject the quantum mechanical aspect of this. Of course, one can never prove this sort of thing (nothing is proven in science), but they've basically shown that any truly predictive theory of protein-folding must account for quantum mechanical effects somehow.
Figure 1 (protein folding):

Figure 2 (protein unfolding):

Conclusions:
Well, despite what we might have expected, when one devises a quite simple model using quantum mechanics (simple compared to what it could've been), one gets very good comparison between theory and experiment. Interpret that however you please, but there's no denying that it's true. Expect for people to make models which are mor robust, or ae capable of describing multi-state proteins. Should be interesting to watch this whole thing... unfold.
References:

Matagne, A., Jamin, M., Chung, E. W., Robinson, C. V., Radford, S. E., & Dobson, C. M. (2000). Journal of molecular biology, 297(1), 193-210.
Zwanzig, R. (1997). Proceedings of the National Academy of Sciences, 94(1), 148-150.
Luo, L. (2014). Quantum theory on protein folding. Science China Physics, Mechanics and Astronomy, 57(3), 458-468.
Lv, J., & Luo, L. (2014). Science China Life Sciences, 57(12), 1197-1212.
Onuchic, J. N., Luthey-Schulten, Z., & Wolynes, P. G. (1997). Annual review of physical chemistry, 48(1), 545-600.

I don't see how the papers you reference and then show pictures tell us anything about how quantum mechanics influences protein folding. First folding is not ' a chemical reaction' in the normal sense of the word and there seems no reason why the Arrhenius equation would describe such a complex process. It is observed in numerous simple reactions A+B->C etc. where one barrier is crossed from reactants to products. But proteins are much more complex and many barriers are involved between the many unfolded states and folded protein. Arrhenius does not work, it should not, its not applicable.

The comment you make about proteins searching configuration space and never finding the folded state as this takes too long is called the Leventhal paradox. It means that, since proteins can fold spontaneously after being unfolded, folding must be a co-operative process.

@porphyrin They tell you how quantum mechanics could potentially be involved because someone made a purely quantum mechanical model for protein folding (solid lines) and compared with actual experimental rate data and found quite impressive agreement. And clearly the Arrhenius equation does sometimes apply because many of those unfolding plots are nearly linear. I believe that it works because all you're comparing is the initial and final state, so they're being intentionally naive about what happens in the middle, so one should still expect Arrhenius behavior.

And also protein folding as far as considering it a chemical reaction is nothing but very complicated isomerization. And isomerization is definitely a chemical reaction.

Hi jheidel, I looked at your reference 1, it describes non-arrhenius behaviour of the Lysozyme. The fits to the data come not from any quantum calculation but initially from normal (phenomenological) chemical kinetics equations (their eqn. 1) which leads to their eqn. 7. They use the Arrhenius/Eyring equation on the way to this result. They fit the data by varying several parameters unit a fit is found, not from any  QM calculation. The Arrhenius equation is not based on quantum mechanics per se but is based on the Boltzmann equation and classical thermodynamics. It was developed before QM.

I'm quite aware of what they did in the first reference. The fourth and fifth references are the one where an actual quantum mechanical model was developed which is what the last two figures show. The first figure is there only as an example of non-Arrhenius kinetics in protein folding. That's why after those figures I point out that something unusual is happening and then go on to explain that some people have tried to give meaning to non-Arrhenius kinetics through a quantum mechanical treatment of torsional transitions in protein folding (references 4 and 5).

Yes it is true that protein folding takes a long time compared to bond vibrations (few femtosecond) or rotations of a amino acid (picoseconds to nanoseconds) and for small proteins can take of the order of a few tens of nanoseconds. Typically proteins with many amino acid residues, say 100, this time can reach to microseconds and seconds and in fact cannot be characterised by a single lifetime but a range of these. As to quantum mechanics, I would reiterate other answers in that as far as we know quantum mechanics describes the world we observe, but it does not mean that every thing has to be analysed using QM in order to make sense of it. Having said that, in proteins the place where classical behaviour breaks down is always to be found in tunnelling, which the H atom in hydrogen bonds does. So in the sense I think you mean, hydrogen bonding is the place where it is most apparent.  However, experiments suggest that in folding, van-der waals interaction in the normal usage of polarisable electrons, as well as pi-pi interactions play an important part in forming intermediate (temporary, short lived) structures that allow further folding to take place by reducing the configuration space available. They also play a vital part in the final structure.  The pi-pi interaction is quantum in nature and can, for example, be between pi bond electrons in the peptide C=O bond and those in aromatic residues such as tryptophan. We should not also forget that the chemical bond is inherently quantum in nature; bonds would not exist in a classical world.  If you don't restrict yourself to protein folding but to proteins in general then in photosynthesis there is overtly quantum behaviour both in energy transfer and in electron transfer.


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Determine volume of methane used to heat x water to x degrees?
I have 126.39 L of water. What volume of methane gas would I need to combust to heat it (at room temp, 21 degrees Celsius) to 38 degrees Celsius? How much $\ce{CO2}$ gas would be produced?
I calculated the molar enthalpy of combustion of methane and got −890.27 kJ/mol, and I know that factors into my calculation. I'm just confused as to what I actually have to do to determine exactly how much methane I need to heat the water.

A great place to start would be the heat capacity $C$ of water. It is defined to be the "amount" of heat change divided by the resulting temperature change. In symbols,
$$C=\frac{q}{\Delta T}.$$
Heat capacity is an extensive property; the size of $C$ depends on how much matter there is. We take this into account by dividing the equation by mass $m$.
$$C_s=\frac{q}{m\Delta T}$$
$C_s$ is called specific heat (capacity). Taking $\rho_{\ce{(H2O)}}\approx1$,
$$C_s \approx \frac{q}{V\Delta T}.$$
Can you take it from here? (Be careful with units.)

Note: It is probably assumed that

the combustion of $\ce{CH4}$ is complete, yielding only $\ce{CO2}$ and $\ce{H2O}$;
density of water does not change over the temperature range;
heat capacity is constant as $\Delta T$ is rather small;
vaporisation is neglible;
no heat lost to surroundings;
decreasing solubility of $\ce{CO2}$ does not contribute to the overall amount of gaseous carbon dioxide (joking...)

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Is (high) pressure required to form (quartz) crystals?
I've been reading a lot of sources about the hydrothermal synthesis of quartz using autoclaves. I want to try doing this myself, but before I can I need to know the minimum pressure required.
If you need a refresher, hydrothermal synthesis uses steam as a solvent, and creates a supersaturated solution that places the quartz molecules onto a seed to grow a pure crystal.
I've gotten many different numbers in terms of how much pressure [from the steam] is required for the process, ranging from 7500 psi to 30000 psi. I want to work with as low of a pressure as possible.
So then my question is: is high pressure necessary for quartz crystal formation? My understanding is that the pressure in the case of hydrothermal synthesis is created from having large amounts of steam in order for more material to be dissolved in said steam, so the crystal will form faster. So is this the case, in which having a lower pressure will work, or is pressure required for the crystal to form properly? As in, does the pressure actually affect the crystal structure in some significant way?

What would happen if you just melted SiO2 sand and cooled it down very very slowly. It would allow crystallization and all quartz is is SiO2So I don’t see any reason it wouldn’t work. Also it shouldn’t take too long because it’s just cooling into a shape but exposed sides would crystallize.

You're offering an alternative way of making quartz crystals. Whether or not it's correct, it doesn't seem to answer the question and it would best serve as a comment.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

Not sure you get quartz.  My understanding of silica (assuming the 4-coordinate-silicon, 2-coordinate-oxygen varieties) is cristobalite and then tridymite are the first stable firms upon cooling the liquid, and only after that quartz.  What if you form one of the first two phases and then things freeze up, so to speak, kinetically?

Pressure does not affect the crystal structure (unless it gets insanely great, so that stishovite formation kicks in; that's not going to happen in an autoclave, though). Pressure, however, does affect the temperature (indirectly), and hence solubility, and hence the rate of crystal formation. There is no lower threshold, that's why you keep getting wildly different estimates for the "required" pressure. It is just that the higher you get, the faster your crystals grow. They would grow at normal atmospheric pressure as well, just very slow (so that only your great-grandchildren may possibly be able to see them).
I suggest that you aim at something less extreme. Calcite, maybe?

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NaI in acetone v/s acetone and water
When NaI is in acetone and we have R-X, then I replaces X by Finklestein mechanism (Sn2 => Inversion). But when we have acetone and water, why do we get a racemic mixture?

NaX (X = Cl, Br) is insoluble in acetone, but is soluble in water. So, if you have water/acetone system, the solubility of NaX is increased comparing to acetone. Solvation effect doesn't have a big role on iodide ions and does not greatly impact it's nucleophilicity. So, the mechanism is still going to follow SN2 path. So, after one halogen is displaced by I-, you have introduced a better leaving group in the substrate. Then, another iodide is going to attack the newly formed R-I, so the result is: 2 x inversion = configuration retention. As no NaX precipitate forms, it is better for I- to attack RI than RX. It turns out that after equilibrium is reached, you end up with racemic mixture, or something close to racemic mixture. Note that free iodide ions keep attacking R-I and that this is a dynamic process which constantly repeats in the solution.

Good catch, now I know +1.

Wikipedia says sodium iodide is soluble in acetone. Data, please.

The Wilipedia data indicate that $\ce{NaI}$ has between one-third and one-fourth as soluble in acetone as in water.  Another factor could be dissociation of the salt (thus availability of the iodide ion for multiple exchange reactions), which is less in most nonaqueous solvents than in water.


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Does ∆Gº exist?
To the best of my understanding, the quantity of Gibbs free energy of an entity depends on its concentration in the system. Equilibrium is reached when ∆G = 0, or when the Gibbs free energies of all the entities are equal to each other.
If Gibbs free energy is dependent on concentration, then at standard conditions, the concentrations of all entities is 1M—this is fixed by convention. How then can you have a ∆Gº value? All the concentrations are fixed and there is no change in concentration nor energy. Is ∆Gº then just a rudimentary concept for those who do not know calculus? Should it be better written as dGº?
In a similar vein, though slightly unrelated, I would also want to know if enthalpy is related to concentration, similar to how Gibbs free energy is. Why or why not? If enthalpy is also concentration dependent, then does ∆Hº exist?

Yes, $\Delta G^0$ is defined for standard conditions, including 1M concentrations of everything. It is very real and has a lot of practical consequences. Whether or not the system is stable, is irrelevant for the purposes of this definition.

@IvanNeretin Could you explain why ∆Gº is defined when there seems to be no change in energy of the system at standard conditions?

OK, I got it. Say, we have a reaction $\ce{A + B->C + D}$. Now, $\Delta G^0$ of this reaction is the energy of the change (real or imaginary, this is not important) from one system ($\ce{A + B}$ in standard state) to another system ($\ce{C + D}$ in their standard state).

It exists in the sense that it is a well defined concept, but it does not exist as something that is directly physically observed.
The IUPAC Gold Book defines standard reaction quantities as follows:

Infinitesimal changes in thermodynamic functions with extent of reaction divided by the infinitesimal increase in the extent when all the reactants and products are in their standard states. For the quantity X they should be denoted by $Δ_r X °$, but usually only Δ X ° is used. For specific types of reactions the subscript r is replaced by: f for formation, c for combustion, a for atomization and superscript ‡ for activation.

So though delta is capitalized, it still represents an infinitesimal concept.
Also, the requirement that all the substances are in their standard states requires that they not be mixed together, and for gases and solutes requires fictional ideal states, so it is not directly observed.
Particularly, standard state is defined as:

State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p°, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p°. For a solute in solution it is the (hypothetical) state of solute at the standard molality m°, standard pressure p° or standard concentration c° and exhibiting infinitely dilute solution behaviour. For a pure substance the concept of standard state applies to the substance in a well defined state of aggregation at a well defined but arbitrarily chosen standard pressure.

$\mathrm G$ itself is not a physical observable, either.

@Brian I agree, but I'm trying to say that beyond that, to the extent that you can observe an equilibrium constant and corresponding delta G, it isn't going to be the standard equilibrium constant and standard delta G, because that involves the concept of the each substance being in a pure state and/or a fictitious state.

Oh, I'm not disagreeing with your answer at all! Just noting that nothing about any $\mathrm G$ or $\Delta\mathrm G$ is experimentally measurable, standard state or no.

You don't measure $\Delta G^\circ$ directly.  You measure concentrations, which gives you the equilibrium constant, which gives you $\Delta G^\circ$.
You can measure the enthalpy change by, for example, measuring the heat released or absorbed during the reaction.  The heat exchanged will depend on the concentrations used but when the heat is converted to $\Delta H$ (using the heat capacity) it is typically converted to units of energy per mole, which effectively is the 1 M reference in this particular example, making it the standard enthalpy change ($\Delta H^\circ$).
If there is a volume change during the reaction the heat exchanged will also depend on the pressure and if the pressure wasn't 1 bar then the $\Delta H$ value needs to be corrected before it can be called ($\Delta H^\circ$). One usually assumes that the volume change is negligible if neither product nor reactants are gasses.

I think I understood what you were asking, and the people here just explained to you what $\Delta G$ is.
$\Delta G^\circ$ is defined as the $\Delta G$ in the natural conditions, at $298\ \mathrm K$ and with every concentration at $1\ \mathrm{mol/l}$. But think of any reaction. Are they at equilibrium when the concentration of reactants is equal to that of products? Probably not. For example, a strong acid dissolving in water has a very high equilibrium constant, meaning that it reaches equilibrium when there is very little reactant and a lot of product. Another example is a very insoluble salt dissolving in water. The equilibrium is reached when there is mostly reactant, but a tiny tiny fraction of it will dissolve. In both cases, when all the concentrations are $1\ \mathrm{mol/l}$, the system will be trying like crazy to turn the reactants into products, or vice-versa. Since:
$$\Delta G=\Delta G^\circ+RT\ln Q$$
in equilibrium:
$$0=\Delta G^\circ+RT\ln K$$
$$\Delta G^\circ=-RT\ln K$$
If $K$ (the equilibrium constant) is high (the case of the strong acid), $\Delta G^\circ$ will be very negative; meaning that when all concentrations are equal, there is a very strong tendency to form products.
If $K$ is small (much smaller than 1, the case of the insoluble salt), $\Delta G^\circ$ will be very positive, meaning that when all concentrations are equal, there is a strong tendency to form reactants back.

Note that the definition of standard state makes no reference to fixed temperature. It is possible to have various standard states of a substance as the reference temperature varies. However, the most favoured value for the reference temperature is indeed 298.15 K.


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Is the plane close packed?
Is  the (200) plane of fcc crystal structure(the plane parallel to the face of unit cell and passing through the body centre of it) a close packed plane?
A close packed palne is a plane where the neighbouring atoms touch each other.
Eg: (100) of simple cubic
(111)of fcc
But in thiscase when i draw the plane, i got four half circles on the middle edges of the square, touching the atoms on each side. But, they donot touch the atom infront.
So, i have a doubt whether it is close packed?
Please explain.

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Close packing does not mean that the atoms 'touch'. It is just the highest areal density plane in a given crystal structure. And, no, (200) is not it for fcc.


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How phenyl shift is possible in Benzil-benzilic rearrangement?
We know $\ce{C-C}$ bond is much stronger than $\ce{C-OH}$ bond (because similar sized atoms have maximum overlap) as in case of phenyl-carbonyl carbon bond. Also due to resonance there is double bond character in phenyl-carbon linkage. Then how does that bond break to give phenyl shift?

The rearrangement is driven by the (effectively irreversible) deprotonation of the carboxylic acid, not by bond strengths

I have a feeling you didn't read https://en.wikipedia.org/wiki/Benzilic_acid_rearrangement which describes mechanism well.

I have a feeling you didn't read my question well as I asked why phenyl shifts instead of oxygen linkages as oxygen have lone pairs to form three member cyclic transition state easily than phenyl whose bond strength is much high even to form that trasition ring easily than -OH  @ Mithoron

A hydroxide anion attacks one of the ketone groups in 1 in a nucleophilic addition to form the alkoxide 2. The next step requires a bond rotation to conformer 3 which places the migrating group R in position for attack on the second carbonyl group. In a concerted step, the migrating R group attacks the α-carbonyl group forming another alkoxide with concomitant formation of a keto-group at the other carbon. This migration step is rate-determining. (1/2)

This sequence resembles a nucleophilic acyl substitution. Calculations show that when R is methyl the charge build-up on this group in the transition state can be as high as 0.22, and that the methyl group is positioned between the central carbon–carbon bond. (2/2)

https://upload.wikimedia.org/wikipedia/commons/thumb/f/f6/Benzilic_Acid_Rearrangement_Mechanism.png/750px-Benzilic_Acid_Rearrangement_Mechanism.png - the above description is with reference to the above image and the description itself is Wikipedia sourced.

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Ring compounds of non-carbon atoms
Are there compounds that are rings of non-carbon atoms, say a ring of six oxygen atoms, or 5-6 nitrogen atoms with attached hydrogens?
Or are these too unstable to exist for long (if at all), like long oxygen chains ($\ce{HO_{n}H}$)?

This one’s pretty close ;)

Are you looking for a single element around the ring? Otherwise borazine would count: https://en.wikipedia.org/wiki/Borazine

Sulphur forms many cyclic allotropes, and phosphorous has some as well.

There are many common rings with more than one element in them and no carbons. Single element rings seem harder.

Good answers so far. Anyone know anything about rings of Oxygen? It wouldn't be aromatic since there wouldn't be any bonds left over. Right?

You can have oxygen atoms in aromatic rings (furan, pyrilium ion), but making such a ring out of oxygen alone would force it to have too much concentrated positive charge.

Non-carbon cyclic structures can also be found in 3D; boranes yield several examples of cages, as well as some few-atom metal clusters.

see this:-http://chemistry.stackexchange.com/questions/42612/stability-of-isomers-of-n4o

It seems there is theoretically an O4 square ring https://en.wikipedia.org/wiki/Tetraoxygen, but it doesn't actually exist. But there is an O8 square block if you squeeze hard enough: https://en.wikipedia.org/wiki/Solid_oxygen#Red_oxygen

Pentazole exists - an $\ce{N5H}$ ring. It is stabilised by aromaticity, with 6 $\pi$-electrons in a cyclic, planar system.

Sulphur forms many allotropes which are rings but these do not contain hydrogen. $\ce{S8}$ and $\ce{S7}$ are the most common.
Phosphorus also forms cyclic allotropes such as white phosphorous, $\ce{P4}$.

DavePhD also mentioned in his answer that a silicon analogue of benzene has been synthesised recently but it is not $\ce{Si6H6}$ because it has other groups attached to some of the silicon atoms.

Pentazole does not quite exist. I mean, well, it kinda does, but that's not quite the same kind of "exists" as benzene or anything...

I doubt that counting lone pairs in pentazole is good idea here. Also many substituted ones are more stable.

Not 10 $\pi$ electrons.  Only six, isoelectronic with pyrrhole.

@OscarLanzi Oh yes sorry my mistake.

You can also have $S_8$: https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/Molecular_Sulfur_%28S8%29_V.1.svg/2000px-Molecular_Sulfur_%28S8%29_V.1.svg.png

See Silicon goes aromatic :

Chemists in the UK have constructed a structural analogue of benzene made from silicon atoms. The molecule is not flat like benzene, but it reveals a new type of aromatic stabilisation.

See Why isn't the Silicon analogue of Benzene flat? for more details.

Do not forget the trihydrogen cation (https://en.wikipedia.org/wiki/Trihydrogen_cation, https://en.wikipedia.org/wiki/Triatomic_hydrogen).  This has been found in the interstellar medium and in hydrogen-rich planetary atmospheres, and is believed to be responsible for the formation of early-generation stars.

The trihydrogen cation is an interesting case. Though you can draw a triangular ring between the nuclei, as far as I understand the highest electron density is actually at the centre of the resulting triangle, so you could think of it more like three atoms joined by a Y-shaped bond than a ring.


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How much lactose is there in milk (mechanism)?
Iodometric titration is sometimes used to find out how much lactose there is in milk. The overall oxidation reaction is written as follows.
$$\ce{C12H22O11 + I2 + 3NaOH ->[\ce{CuSO4}]C12H21O12Na + 2NaI + H2O}$$
How does this reaction proceed? What is the role of $\ce{CuSO4}$?
The solution is then acidified with $\ce{HCl}$. Excess $\ce{I2}$ is determined via
$$\ce{2Na2S2O3 + I2 ->2NaI + Na2S4O6}.$$

What I think might be going on (with references)
Using basics of inorganic chemistry, iodine reacts with cold sodium hydroxide to give sodium hypoiodate.
$$\ce{I2 + 2NaOH <=>NaIO + NaI + H2O}$$
$\ce{NaIO}$ is the main oxidiser of lactose.$^{[1]\ [a]}$
$$\ce{C12H22O11 +NaIO + NaOH ->C12H21O12Na + NaI + H2O }$$
I have no clue as to what the complete mechanism might be here.
Alternatively, $\ce{Cu^{2}^+}$can be reduced. The following picture$^{[2]}$ shows the general transitions for (semi)linear lactose.

Sadly, neither reference mentions the other option, nor are there arrow-pushing mechanisms which is what I am after.
Sodium iodate is prone to decomposition even under room temperature, but the reaction is vigorous from $80 \ ^\circ \text{C}\ ^{[3]}$:
$$\ce{3NaIO->NaIO3 +2NaI}.$$
This is partly why the acid is added. To get an accurate reading of "unreacted" iodine we must somehow re-release $\ce{I2}$ from the ions $\ce{IO^-}$ and $\ce{IO_3^-}$.$^{[1]}$
$$\ce{NaIO + NaI + 2HCl -> I2 + 2NaCl + H2O\\ NaIO3 + 5NaI + 6HCl -> 3I2 + 6NaCl + 3H2O}$$
Thiosulphate is a widespread reagent that is subject to inaccuracies$^{[4]\ [5]}$ due to
$$\ce{Na2S2O3 + 4I2 + 10NaOH -> 2Na2SO4 + 8NaI + 5H2O}.$$
This reaction is less favoured in acidic conditions.$^{[1]}$

Clarification
I am asking for

arrow-pushing mechanisms that show the role of $\ce{NaIO}$ and/or $\ce{CuSO4}$ with cyclic lactose (if possible).

$^{[1]}$ A Study of the Method for Titrating Aldose Sugars with Standard Iodine and Alkali.  G. M. Kline, S. F. Acree. Bureau of Standards Journal of Research. Link [Comment: the study actually deals with aldose sugars. The equation for cyclic lactose might thus be a stretch.]
$^{[a]}$ Agu-Tõnis Talvik. Organic chemistry. (1996). page 490 (as far as I know, not available in English)
$^{[2]}$ Dairy Chemistry and Biochemistry. Second Edition. P. F. Fox, T. Uniacke-Lowe, P. L. H. McSweeney, J. A. O'Malhony. page 62. Link
$^{[3]}$ Иодат натрия. Russian Wikipedia. Link
$^{[4]}$ (G. Topf). Zeit. anal. Chem., 26, p. 137; 1887
$^{[5]}$ (Bougault). Compt. rend., 164, p.; 949; 1917

This also got me wandering... What is the go-to software for chemists who wish to draw reaction mechanisms on the computer?

@LinearChristmas: wondering*... :(

ChemDraw - if you can afford it or get it free from your institution.

@bon OR ChemDoodle which is very good and not too expensive.

There are a number of free software tools around for Linux. Didn’t find one I would gladly recommend yet, though. They all have their shortcomings somewhere …

Using the basics of organic chemistry, solutions of iodine and sodium hydroxide retain enough free $\ce{I2}$ to do other things. I would disocunt $\ce{NaIO}$ as a significant component of this process.
What if instead, we had exactly what your picture shows (except the last few steps are prevented - we'll see how in a few lines). You are worried abut the cyclic form, but the cyclic form and the acyclic form are in equilibrium in aqeous solution, especially if acid or base is present. Note that in the reactions equations below, I am using the notation $\ce{R}$ to refer to parts of the molecule that are not involved.

Isomerization to the ene-diol

$$\ce{R-CH(OH)-CHO <=>[\ce{OH-}] R-C(OH)=CHOH}$$

Oxidation by copper(II)

$$\ce{R-C(OH)=CHOH + 2Cu^2+ +3OH- -> R-CH(OH)-CO2^- + 2Cu+ + 2H2O}$$
Now, if that copper(I) cations stumbles into a hydroxide anion, then it's all over since $\ce{Cu2O}$ is not soluble. However, it's iodine to the rescue! Iodine oxidizes $\ce{Cu+}$ back to $\ce{Cu^2+}$.
$$\ce{2Cu+ + I2 -> 2Cu^2+ + 2I-}$$
These last two equations add to give the one you quote (only I left out the sodium spectator cations and simplified the carbohydrate).

Good answer! I do agree that there will be free iodine; and it can be iodine itself that reacts. However, so far I have only been able to find references for $IO^-$. Since posting, I found even another source that supports $IO^-$ over $I_2$ (edited post). Also, the question asked for the mechanism. So is this answer to say that the mechanism is not known, or not useful in understanding the process? (I am interested in the $Cu^{2+}$/$IO^-$ intermediary steps.) Definite +1 and thanks for pointing out the iodine oxidation of $Cu^+$ which I had completely overlooked.

If the terminal oxidant is $\ce{IO-}$, then its reaction with  $\ce{Cu+}$ is $$\ce{2Cu+ + IO- + H2O -> 2Cu^2+ + I- + 2OH-}$$ However, the mechanism of a redox reaction is a series of one-electron transfers.

Could $\ce{Cu(OH)2}$ also participate in the process instead of or in parallel to $\ce{Cu^2^+}$? This is also what a few sources suggest.


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Why are beta-ketoacids better at decarboxylation than gamma or delta?
My book claims that beta-ketoacids and beta-dicarboxylic acids are better at decarboxylating than their gamma and delta counterparts. Why is that?
I would assume that since gamma and delta keto-acids make a larger ring intermediate (less ring strain) and thus the energy barrier for reaching that intermediate would be lower...

It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that participates in the bond-breaking event).


text:
d-orbital splittings in WS2 monolayer
The formerly degenerate $d$ orbitals of the tungsten atoms in the $\ce{WS2}$ monolayer are split into three groups: (1) $\mathrm{d}_{z^2}$, (2) $\mathrm{d}_{x^2-y^2}, \mathrm{d}_{xy}$ and (3) $\mathrm{d}_{xz}, \mathrm{d}_{yz}$ in the order of increasing energy with the large band gap between the first and the second groups.

Now the valence band maximum consists dominantly of $\mathrm{d}_{x^2-y^2}, \mathrm{d}_{xy}$ orbitals, whereas the conduction band minimum is composed mainly of $\mathrm{d}_{z^2}$ metal orbital.

So, the order of $\mathrm{d}$ orbitals in the first two splitting groups now is reversed, since the valence band is lower in energy than the conduction band. Where is the correspondence between these two facts?

I think this link might help: https://www.researchgate.net/figure/282068894_fig5_Figure-2The-electronic-band-structures-of-monolayers-1L-and-doublelayers-2L-of-MoS2

One vague explanation I came up with is the following.
A prismatic arrangement of atoms (first figure) does not capture fully all symmetries of a hexagonal arrangement in a crystal lattice (which is the case of the second figure). However, the most overlap (in terms of the number of symmetries) between the two occurs at the $\Gamma$ point, since it is the point of the highest symmetry. Thus, the first figure resembles mostly the orbitals at the $\Gamma$ point in the first Brillouin zone of the crystal lattice.
Please, if I am thinking in a right direction, refine my answer to make it more strict and rigid.


text:
Can electrostatic interaction affect nucleophilic attack by thiolate anion
Suppose a protein side chain has a $\ce{-SH}$ group which is partially deprotonated to thiolate anion $\ce{-S -}$. This thiolate can perform a nucleophilic attack on a certain substrate.
Will the propensity of nucleophilic attack be reduced if the thiolate anion is in electrostatic interaction with a $\ce{NH3+}$ group of a basic amino acid? Or is the likelihood of nucleophilic attack independent of electrostatic effects?
P.S.- The $\ce{-S-}$ will probably be in eletrostatic interaction with H+ in the solution also. Maybe electrostatic interaction with $\ce{NH3+}$ will be more stable?
Core question: If a negative nucleophile is in electrostatic interaction, will it affect its nucleophilicity?

It would since the S atom is now negatively charged. However, S is a very poor hydrogen-bond acceptor as evidenced by its weak basicity (thiols are more acidic than alcohols). So this electrostatic effect will be greatly diminished if you compare with more basic nucleophiles like F- and O-. This is because sulfur larger and will be more polorizable making the localization of the negative charge more disperse around the atom and much weaker to form ion-ion interactions. It will still remain as a relatively powerful nucleophile, so you can expect side reactions.


text:
If I mix 10 mg of salt in 1 liter of water... Will I drink exactly 0.01mg of salt per ml?
If I dissolve 10 mg of salt in 1 liter of water, will every milliliter of that water contain exactly 0.01 mg of salt? In other words, is a solution uniform in composition, or will there be areas within the solution where the salt has a higher concentration than other areas?

Erm … yeah? So … what is the question?

Not exactly because the salt as a volume

@Jan The question is : are every 1ml of water containing exactly 0.01mg of salt ?

@Clippy it depends if your salt is or not soluble in water, what do you know about it. ? It depends on the temperature too. And Your salt has at least a volume. So if 10mg of your salt is equivalent to 1mL the concentration will not be 10mg/L ..

Clippy, see Benzene's edit -- if it doesn't match what you're actually trying to ask, please roll it back.

If I mix 10 mg of salt in 1 liter of water

If mixing means dissolving the salt in water...

Will I drink exactly 0.01mg of salt per ml?

Yes. Not exactly $0.010000~\mathrm{mg}$, but $0.01~\mathrm{mg}$

So if I drink 100 ml I will also drink 1mg of salt ?

Yes.
Note that the answer assumes that the water used did not contain the same salt before adding the 10 milligrams.

There are a couple of factors at work here. Assuming that the addition of the salt does not change the total volume of the solution and that it dissolves completely, you are correct. The definition of a solution is a "homogeneous mixture composed of two or more substances." Since a solution is homogeneous, this means that the salt is evenly distributed within the water, and thus you will consume exactly .01 mg of salt per mL, implying in turn that you will also drink 1 mg of salt per 100 mL. If, however, the volume of your solution changes by adding the salt, you will consume slightly less than your calculated ratio of 1 mg per 100 mL, as there is more volume over which the salt is uniformly distributed. In addition, if your salt solution becomes saturated, leaving some undissolved salt at the bottom, your concentration will once again become lower than 1 mg per 100 mL. Exactly how much the concentration would change depends on the amount that the salt changes the volume as well as the amount of salt that fails to dissolve.


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Why is there no Activity Series for general non-metals?
In school, for determining whether single displacement reactions occur, we use an activity series. There is one for metals, and one for halogens. Why is there none for non-metals in general? Are halogens the only ones that will participate in single displacement reactions?

I think, periodic trends is the key word. For the halogens, they are homologue elements from a single group. Therefore, they display very similar overall reactivity and typically either fluorine or iodine is best at something, either because a large size and low electronegativity is good for a process or because of the opposite.
I don’t know which metals are included in your series. But I think that if they are from the first two groups and maybe including aluminium, then the reactivities are still similiar enough to compare along the lines of general trends.
However, once you start going to different non-metals — or worse, to metalliods — things become a lot less predictable. You can no longer draw a line down (or up) a group because you have different groups involved. And you can not draw a line along a period because there will be multiple periods. Quick and dirty methods are bound to fail soon.

Long story short, metals are relatively uniform and nonmetals are diverse.
Indeed, metals have much in common. Many would react with acids, and nearly all would react with salts of some other metals, so the position of a metal in the activity series tells us a good deal about its possible reactions.
With nonmetals, it is not that simple. True, you can order them by electronegativity, and this is quite useful, but not that useful as it was with metals. While the displacement reactions are not limited to halogens, they do not constitute a majority of all reactions. For example, I expect this to happen: $\ce{K2S + Br2 -> 2KBr + S}$, but this tells us nothing about the oxygen-containing compounds of sulfur. Will bromine replace sulfur in $\ce{K2SO4}$? It won't. How do you know that from electronegativity alone? You don't.
So it goes.


text:
Can equilibrium be reached in a non-reversible chemical reaction?
Can chemical equilibrium only be achieved in a reversible process?
I have only seen equilibrium in processes that are reversible so I was wondering if it were possible in reactions that are not reversible.

In any chemical reaction there is a generally a potential energy barrier between reactants and products. How easily this is crossed from reactants to products or vice versa depends primarily on the temperature. This is because to react, molecules have to gain energy that is more that the average to be able to cross the barrier. (This is called the activation energy). The energy is accumulated randomly from solvent in solution and by collisions in gas phase, which is why some chemical reactions can be slow. Barriers are often high compared to thermal energy so it can take ages to get enough energy by chance to cross a barrier. The chance of having a certain energy E is determined by the Boltzmann distribution $exp(-E/k_bT)$ where $k_B$ is the Boltzmann constant, T temperature. As Boltzmann distribution also depends on temperature reactions are generally observed to be faster at higher temperatures.
If a reaction is called 'reversible' then one expects to see reactants and products in the mixture at the same time at amounts depending on the equilibrium constant. The product is reverting to reactant and reactant to product, assuming that all product is not continuously being removed, say by precipitation or otherwise.  In an 'irreversible' reaction the barrier from products to reactants is so high that in normal experience enough of the product never returns to reactant to be measurable. The effect can be very dramatic (orders of magnitude) with a relatively small change in activation energy because of the exponential dependance of the Boltzmann distribution.  All this means that there is no fundamental difference in 'reversible' and 'irreversible' reactions, just a matter of time-scale. However, it is obviously practically useful to distinguish these cases and so we can look at equilibrium constants to guide us. Note (a) That thermodynamics does not inform us about time-scales but only the relative free energy of reactants and products. (b) The words ' reversible' when used as 'reversible process' has a special meaning in thermodynamics and needs to be used with care. (c) I have excluded the special case of 'barrierless' reactions.

In terms of a dynamic equilibrium where the rate of formation of products and reactants are the same, no, because the reaction will proceed to products completely and $K \gt\gt 1$. It is highly unlikely that the reaction will go back to reactants.
Irreversible reactions in chemistry are considered to achieve a static equilibrium because the reaction stops after a certain point where reactants are converted to products and the ratio of products:reactants stops changing.
A perfect example of this is the conversion of graphite to diamond.
For further info see this.

What about static equilibrium? Isn't an irreversible reaction going to end upon a static equilibrium, whereas a reversible reaction lands upon a dynamic equilibrium?

Interesting, I have never heard or considered that. I would say yes, it would happen upon a static equilibrium where there are no forces to act on the PES to go either direction from R -> P or P -> R. The chemical system is at rest. Thanks, @timaeus222!

You are aware that use of the terminology reversible/irreversible in connection with processes is fundamentally different from how we use this terminology with respect to chemical reactions, correct?

@ChesterMiller I will defer to you. I am not completely aware of the correct usage of the terminology behind reversible/irreversible processes in areas outside of chemical systems. If you can help improve my explanation, that would be great for the question and myself as general knowledge. I am not good at physics...

@ChesterMiller Are you referring to how a typical irreversible chemical reaction is technically a very, very skewed dynamic equilibrium and generally difficult to actually reverse? Chemwiki does note that the graphite example is just an example. My understanding is that not all irreversible reactions end upon a static equilibrium (that was my bad, but some do).

We refer to an irreversible chemical reaction as one that is capable of very nearly going to completion, irrespective of whether it is allowed to occur spontaneously, or whether it is controlled to proceed very slowly (by using semipermeable membranes or other devices) so that it is always very close to equilibrium.  A reversible process is one that is always controlled to progress very close to equilibrium, and ideally represents a continuous sequence of thermodynamic equilibrium states.

You should add some > something like K >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>1 :P


text:
Alkalide compounds
I have read much more about metallic hydrides but I am totally confused about "inverse alkali hydrides" or "hydrogen alkalides" while reading refer to this. What are these compounds and how do they form? Is it possible for two elements to form two distinct binary compounds with one containing ions reversely charged with respect to other? I have also heard about "calcium natride" ($\ce{CaNa2}$) and similar compounds. Can someone guide me about them?

Here is a related answer from another question.

Reading modern research papers is probably a great way of learning chemistry, but it is also a great way of confusing oneself quite profoundly, since they typically do not bother to tell you what is common and what is exotic.
Indeed, two elements are pretty unlikely to form two distinct binary compounds with reversed charges. But that's not the case here anyway. The thing you are talking about is not a binary compound, it contains not just the two elements, but also a lot of other stuff, and it is not surprising at all that we can do pretty much whatever we want with these two elements. A simpler example can be thought of. Say, in $\ce{COCl2}$ the oxidation state of $\ce{C}$ is positive and that of $\ce{Cl}$ is negative; in $\ce{(CH3)4N+ClO4-}$ it is the other way around. Finding a pair of true ionic compounds related in the same way presents some challenge, since most elements do not readily form ions of both signs, but it can be done too, as you see.
Look at the picture from that paper:

See, it is pretty much like all other alkalides: $\ce{H+}$ is hidden inside the cryptand so that it wouldn't grab that extra electron from $\ce{Na-}$ (which it otherwise would).
In the same manner, one can get natrides of other metals, and probably calcium as well. It is not like you simply mix $\ce{Ca}$ and $\ce{Na}$ (that would produce just an alloy, or maybe nothing at all).
Then there is an ionic salt $\ce{CsAu}$ (without any cryptands or anything), but that's another story.

Oh no quite not confused about about aurides. I am getting your point @Ivan but if there are cryptands all the time then why they are called inverse sodium hydride??

Because that's a catchy name that sounds good.

Got it. I was reading about these compounds with your answer keeping in mind and now its quite clear.


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Can someone explain the unintuitive ordering in the metallic activity series?
I would understand it if the metallic activity series Li, Cs, Rb, ... Pt, Au corresponded to electronegativity or ionization energy. However, the way it is written, lithium will displace caesium, while caesium will displace rubidium (it goes from the top of family 1 to the bottom and then up again).
This order seems so random. Is there any order to it or is it just determined experimentally through trial and error? In addition, wouldn't different valences on the metals affect their placement in the series?

There is an order but not all metals obey that order. We need to experimentally verify the theoretical order. Of course,  the electronegativity and ionization energy are closely related to the order. The higher the tendency of the metal to get oxidised, the more reactive it will be and hence a more active metal. The ability to get easily oxidised depends on the factors you mentioned.

The metals in the activity series are ordered by their oxidation potentials; there is nothing random about it. The displacement reactions, though, are but one of the applications of these potentials, and not the most important one at that. Moreover, with active metals it does not make much sense anyway. The oxidation potentials are defined for water solutions; their significance ultimately stems from the fact that we live in a world dominated by water. Well, you can't react lithium or rubidium with anything in water, because they would rather react with water itself. And if you try a reaction in other media than water, the oxidation potentials do not apply and the outcome can't be predicted that easily. For example, magnesium would reduce sodium from its oxide, despite being nominally "less active".
Indeed, the very concept that some things are "more active" than others is just an approximation; admittedly, a pretty good one when we talk about metals, but still an approximation.


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Is there a such thing as a gas that stay a gas?
We know if you heat a liquid up it becomes a gas. But I need to know if there is a such thing as a element that is already a gas And will stay that way. If there is a such thing how could one keep it that way? Is there a certain temperature it must stay at what?
If there is no such gas, is there something that comes close to it?

Look around you. Oxygen and nitrogen are such elements: they are already gases and will stay that way, unless you cool them down to extremely low temperatures.

The closest you can get to is probably helium, which does not have a solid phase at ambient pressure

A gas cannot be condensed to liquid no matter how high a pressure is applied if the temperature of the gas is higher than the critical temperature of the gas.


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Are all concerted electrophilic addition reactions of alkenes syn?
The Organic Chemistry book I am using (8th edition Carey, et al.) doesn't say it straight out, but I am wondering if it's a universal rule that all concerted electrophilic additions of alkenes are syn.
In every concerted example I have seen so far in the textbook, the bonds that are being made wind up on the same side, but I don't know if it's a rule. For example: epoxidation, transition metal catalyzed hydrogenation, and hydroboration are all concerted and all syn.

If they are concerted, there is a very, very large probability of them being syn-specific. For explantion, let’s look at the π orbitals of ethene (scheme 1). (Remember that double bonds are planar and have two distinct sides when viewed across an entire molecule.)

Scheme 1: π and π* orbitals of ethene, viewed from the side.
Reactants have one of two possibilities, typically. They are either electrophilic and will thus attack the highest occupied molecular orbital (HOMO). Or they are nucleophilic and will attack the lowest unoccupied molecular orbital (LUMO). It just so happens that the π bond is the weakest bond in a typical alkene and thus the bonding molecular orbital is typically the HOMO and the antibonding one is the LUMO.
To attack this orbital, you somehow need to interact with both lobes. If you do so, you will either attack syn-stereospecificly, i.e. arrive from one side due to simplicity. Or you are a very large attacking molecule and are able to grab around the double bond, interacting with one atom on one side and with the other one on the other side.
Most reactants are not that large. Typically, there will be a single attacking atom (e.g. halogens, meta-chloroperbenzoic acid), two atoms bonded together directly (e.g. borane) or two atoms that are still in very close proximity (e.g. $\ce{OsO4}$ — two bonds — or a diene for a Diels-Alder reaction — three bonds). None of these even have the chance to reach across, especially if you consider that the largest of these, the diene, is in fact a large π system and inherently restricted to being planar (within very small possible deviations).
Of course, that does not make it impossible. If you use a large enough π system in a Diels-Alder like reaction, that may be able to wind itself downwards like a spiral staircase. I haven’t seen that type of reaction yet, though.


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Chemical Formula for Barium Chloride
Barium Chloride is represented as $\ce{BaCl2}$.
Since chlorine is a diatomic molecule, It should be denoted as $\ce{Cl2}$.
Formulating, we get
\begin{array}{|c:cc|}\hline
\small \rm Element & \ce{Ba} & \ce{Cl2}\\
\small \rm Valency & 2       & 1       \\\hline
\end{array}
Crisscrossing the valencies of Barium and Chlorine we get $\ce{Ba(Cl2)2}$ — as opposed to the accepted formula of $\ce{BaCl2}$. How is it so?

In barium chloride, the chlorine is not in the form of a diatomic molecule. Think about how ionic compounds are formed.

@TIPS How are they formed in this case?

Valency is a property of an element, not a molecule.

@IvanNeretin How do you obtain the formula for Barium Chloride then?

Much like you did, except don't mention $\ce{Cl2}$ at all. There is just Cl, its valency is 1, and then there is Ba with valency 2, so...

The compound barium chloride is not the same thing as barium and chlorine mixed together.
When they react, a barium atom will give up two electrons to form a action, and a chlorine molecule will pick up two electrons to form a pair of chloride ions: $$\ce{Ba -> Ba^2+ +2e^-}$$
$$\ce{Cl2 +2e^- -> 2Cl^-}$$ When you have both of those things at once, the electrons are "consumed" as fast as they are "produced", so they don't appear at all in the result: $$\ce{Ba +Cl2->Ba^2+ +2Cl^-}$$ which forms an ionic lattice when solid. Since this lattice has overall neutral charge, its ionic charges must balance with integer coefficients.
Funnily enough, these coefficients are $1$ and $2$ for $2+$ and $1-$ respectively, so these are applied to the ions which carry those charges. Thus: $$\ce{Ba1Cl2}$$ or more simply and directly: $$\ce{BaCl2}$$

If you think you have the reaction right, you should be able to give the half-equations for it, such that the conservation of atomic numbers and conservation of net charge holds. Your proposed formula would require something to break at least one, probably both of those.

When chlorine is in its free state it is diatomic. But when it reacts with barium it is not in the form of $\ce{Cl2}$. It will be in its ionic state which is $\ce{Cl-}$. The same goes for Barium. Barium is mono-atomic and its ionic state is $\ce{Ba^2+}$. Barium gives one electron to a chlorine atom and another electron to another chlorine atom, as valency of chlorine is 1, so it is $\ce{BaCl2}$.


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Is sodium borohydride really a better base than an alcohol?
I have learnt that sodium borohydride can reduce aldehydes, ketones and acyl chlorides into alcohols. However, it fails to react with esters or acid anhydrides.
On the other hand, alcohols — supposedly weaker nucleophiles — can successfully react with both acyl chlorides and acid anhydrides to give esters. Moreover, my book (Solomons and Fryhle) mentions that acidic conditions are not required for the mentioned reactions.
Why are alcohols able to do what sodium borohydride cannot, even in the absence of acidic conditions?
I have come up with a possible explanation, but I am not sure if it is correct: When an alcohol reacts with an acid anhydrides to produce an ester, a carboxylic acid is formed as a side product. Maybe this provides the required acidic conditions for further reaction?
In general, can a reaction be catalyzed by the formation of a side product initially not added in the reaction mixture?

Under appropriate conditions sodium borohydride can also reduce esters to alcohols.

There are three parts to this answer:

Sodium borohydride is more basic than alcohols
Sodium borohydride reacts with anhydrides under some conditions
Reactions can be autocatalytic

Sodium borohydride is more basic than alcohols

Sodium borohydride will react with water and alcohols to form small amounts of hydrogen gas. If you doubt this, add some to water or ethanol or try to find a video of this phenomenon.
$$\ce{NaBH4 + ROH -> NaOR + BH3 + H2 ^}$$
If sodium borohydride can deprotonate an alcohol, it is more basic than an alcohol.

Sodium borohydride reacts with anhyrides under some conditions

Here is an exmaple In this article in the Journal of the American Chemical Society, sodium borohydride in THF is used to reduce cyclic anhydrides to lactones (esters), not your expected product! However, this reaction might not be general to all anhydrides. Introductory textbooks tend to try to make things simpler to help you learn the material. Think about how maddening it would be to try to learn all of the conditions in which sodium borohydride reacts with anhydrides and all of the ones that do not work.

Reactions can be autocatalytic

If a reaction produces a species as a product that can catalyze the reaction, the rate may increase over time instead of decrease. These reactions are autocatalytic.
Consider the reaction of an alcohol with an anhydride.
$$\ce{ROH + (R'CO)2O -> ROCO2R' + R'CO2H}$$
As you mention, acid is not necessary. However, the substitution reactions of carboxylic acid derivatives are all accelerated by acid.
Rate law without acid
$$\mathrm{rate}=k[\ce{ROH}][\ce{(R'CO)2O}]$$
Rate law with acid
$$\mathrm{rate}=k[\ce{ROH}][\ce{(R'CO)2O}][\ce{HA}]$$
As the reaction proceeds, a carboxylic acid is produced. While it is not a particularly strong acid, the reaction mixture does become more acidic over time, and the reaction rate increases as a result. This is one reason why alcohols (seemingly poor nucleophiles) react with anhydrides on their own.

By the way, are hydrides considered good nucleophiles or not?

The hydride anion $\ce{H-}$ is not a good nucleophile. The hydride equivalent of borohydride is a different matter.

General considerations
It’s not the basicity that is the important factor here, it is the nucleophilicity. The hydride ion, whether present as tetrahydridoborate or as a lone hydride is a very lousy nucleophile while an alcohol is an okay-ish nucleophile. The general mechanism of the addition is given in scheme 1.

Scheme 1: General mechanism of a nucleophilic attack on a carbonyl.
Now in the case of a hydride — a very weak nucleophile — this reaction is generally slow. It is faster, the more reactive — electrophilic — the carbonyl species is. If $\ce{X}$ is a carbon or hydrogen atom, then all we have is a localised double bond without any significant reduction in reactivity.[1] Reactions with those should thus be fast. If $\ce{X}$ is an element that has a lone pair, such as oxygen or chlorine, then there is a notable mesomeric stabilisation as shown in scheme 2.

Scheme 2: Mesomeric stabilisation by lone pairs next to a $\ce{C=O}$ double bond.
This stabilisation is not strong if a chlorine atom is $\ce{X}$ since chlorine is much larger and its lone pairs are more disperse. The stabilisation is stronger in the case of acid anhydrides; however, since a single oxygen has to stabilise two carbonyls it is still not too strong. In esters, carboxylic acids (and their anions) and most so in amides this stabilisation is very strong ad nucleophilic attacks difficult. This explains the general electrophilicity trend:
$$\text{aldehydes} > \text{ketones} > \text{acid chlorides} > \text {acid anhydrides} > \text{esters} > \text{amides}$$
Now let’s look at the reactions in detail. Throughout this post I will assume neutral conditions only; no acid or base catalysis.
The hydride attack
As I already noted, hydride is a bad nucleophile. It makes sense that it can attack the first members of the series — aldehydes, ketones and acid chlorides — but fails to do so for more stabilised members. For completeness, the mechanism of a hydride attack onto an acid chloride is given in scheme 3.

Scheme 3: Mechanism of a hydride attack onto an acid chloride.
We see another key point here: Once the carbonyl has been attacked, the oxide anion has a tendency to ‘rebound’ a lone pair and displace one of the bonded atoms. Generally, the displaced anion must be reasonably favourable for the rebound to occur. Chloride ions are good leaving groups and thus a displacement of chloride is likely to happen. Hydride ions are absolutely absymal leaving groups, so the hydride attack is not reversible.
After the inital attack and displacement of the chloride, we are left with an aldehyde which of course will react quicker and lead to full reduction to the alcohol.
Acid anhydrides are not attacked this way because the initial step does hot happen. If you accidentally got a hydride to attack, the $\ce{^{-}OOC-R}$ leaving group is great and thus a reaction would occur.
The alcohol attack
Alcohol hydroxy groups are still not the best nucleophile but they are better nucleophiles than hydride is. Thus, they even manage to attack acid anhydrides and — to a limited extent — esters and acids. The general mechanism of attack is shown in scheme 4.

Scheme 4: Reaction mechanism of an alcohol attacking an acid chloride including a potential second attack.
The key difference here is that the alcohol is a neutral nucleophile, not an anion like hydride is. Thus, when the rebound occurs, it is susceptible to redisplacement. This is shown in the final equilibrium of the mechanism above. This is also the reason why — under neutral conditions! — an alcohol may attack an aldehyde or a ketone but the attack is not productive. The alcohol will simply be rebounded back out since the hydride is a terrible leaving group.
There is a productive reaction mechanism for alcohols with aldehydes or ketones — the formation of acetals and ketals — however, that generally requires acid catalysis.
Finally, there is also the possibility of base catalysis when alcohols react. It would deprotonate them to give the alcoholate which is a much better nucleophile and even able to attack esters. However, base catalysis is outside of the scope of this question.

Why is hydride a bad nucleophile? I thought it was one of the best..

@Kalyan Recheck your sources ;)

It has high charge density, and least steric hindrance possible. Doesn't that mean it is a good nucleophile ?

@Kalyan Its charge density is not that high. And it is really not that good of a nucleophile. If it were, it would easily displace halides. However, SciFinder is full of examples of hydrides that do not displace the alkyl bromide. It gets even worse if we talk about tetrahydridoborate, which is technically a 2+2 cycloaddition mechanism.

Hydride is an excellent nucleophile, dear. It so happens, that it rarely occur in solutions, though. However, hydride solutions stabilized by TMEDA are strong reducing agents, capable of reducing carboxilic acid esters http://pubs.acs.org/doi/abs/10.1021/jo00228a027

@permeakra Unfortunately, the database is still missing many hydrides including just about everything alkaline metal hydride and outside of DMSO or water as a solvent. Note that reducing agent $\ne$ nucleophile.

Hydride anion is a bad nucleophile-- true.  Sodium borohydride is not the same nucleophilie as hydride anion, and is not nearly as bad.

@Kaylan regarding the hydride anion:  One proton, two electrons.  It is 'big and fuzzy'-- roughly the size of a bromide.  The charge is not very concentrated.

@Lighthart Hydride anion is a good nucleophile - it has small size, high charge density and hydrogen has low electronegativity. But it very rarely occurs in synthesis - it is unstable in protic solvents, adds to many polar solvents and hydrides do not dissociate in nonpolar solvents, and, frankly speaking borohydride is much easier to work with. Still, if hydride is forced into nonpolar liquid it becomes strong enough nucleophile, comparable to alkyllithium.  I gave the link. Borohydride, on the other hand, is a relatively weak nucleophile.

@permeakra:  Hydride does not have a small size.  I do not have full access to the ACS journal to vette your paper, but TMEDA changes things considerably.

In acidic conditions sodium borohydride is neutralized.  Some of the reactions indicated are acidic environments.
In general, sodium borohydride cannot attack/reduce esters in neutral conditions.  Neither do alcohols.  However, in acid, esters become much more nucleophilic such that alcohols can react/transesterify.  Sodium borohydride does not react because it gets consumed by the acidic reaction conditions.

Please view the edited question.


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Can the regions of 1s and 2s subshells overlap?

I was studying atomic orbitals and always had this question lingering.
As we can see in the image, the 1s and 2s subshells both are known to closely surround nucleus. The overlapping peaks in the graph (of probability vs radius) lead me to believe that the part region of 2s before the node should overlap with the region under 1s.
And by 1s and 2s I mean their contours ( regions which contain 90% probability of electron occurring).
Am I committed a blunder here or do subshells overlap?
Additionally, does the same logic apply when comparing the spatial distribution of say 2p and 2s aswell? Do they have overlapping regions of high electron probability density as well?

Subshells are not a real physical objects in the sense that we can't measure them, can't designate them. There are just a mathematical abstraction. So, there are meaningless in terms of physics to ask about their overlap. It can lead you to misunderstanding.
Let's look at the region where 1s-electron can be found with 90% probability. There are non-zero probability of 2s-electron being found there. Even if you'd consider the region where 1s-electron can be found with 10% probability, there are still non-zero probability of 2s-electron being found there. So, at first you should understand that 2s (2p, 3s, 3p and so on) electron have non-zero probability to be found anywhere except where its probability density function equals to zero.
Now let's say we have performed infinite number of measurements of 1s and 2s electrons coordinates. Let's discard 10% outliers in both cases. Now we have clusters which contain 90% probability of electron being found there. Will these regions overlap? Maybe. The answer is one can calculate how much outliers one should "remove" to obtain "non-overlapping regions" (I can't calculate this, unfortunately). For 1s and 2s it may be 95%-regions, or may be 89%-regions that not overlapped. And theoretically, one can perform such calculation for any pair of orbitals he wants.

There is nothing physically special about 90%. And if we don't think of 90%, then of course all orbitals of all atoms overlap, since each of them is infinite and spreads over all space. True, most of them overlap in the regions of extremely low probability, like $10^{-100}$, so for all practical purposes we might just as well consider they don't. But orbitals of one atom (pretty much all of them) would overlap in the areas of significant probability.
One might wonder why wouldn't they interact the way the orbitals of different atoms would. Well, there is a reason to that. Being eigenfunctions of the same operator, they are orthogonal,  that is, the overlap integral equals zero if we consider the signs of $\psi$.

I'm not sure that thinking about sub-shells is very useful at all. They are not distinct entities but just part of the shape of the wave-function which in turn depends on the quantum numbers. You can see from your graph and figures (which are different ways of displaying the same thing) that there seems to be a chance that electrons in 1s, 2s & 3s occupy the same region of space. However, the overlap integral between total wavefunctions (radial and angular parts) from the same atom e.g. $\int\psi_{1s}\psi_{2s}d\tau =0$ shows that they are orthogonal. If n, l, m are the quantum numbers for one orbital and n', l', m' for the other orthogonality is found from the product $\delta_{nn'}\delta_{ll'}\delta_{mm'}$ where $\delta_{ab} $ is the delta function, = 1 if $a=b$ else zero. (If orbitals are on different atoms, the wavefunctions now have to be represented with respect to a common set of coordinates, i.e. the equations for orbitals, say, on atoms A and B, must use the same zero (origin) for radial distance. In doing this the wavefunctions are no longer orthogonal so chemical bonds can be formed.)


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Is there a way to faciliate remembering which are alpha/beta hemiacetals of sugars?
I understood the concept of hemiacetals quite well, but have considerable trouble remembering which are called $\alpha$ or $\beta$ respectively.
Is there some concept/mnemonic/common reaction/something else to faciliate remembering this convention?

The short answer is no, not really. Or, more precisely, not that I know of. Probably because $\alpha$ and $\beta$ depends on whether the sugar is a D or an L. I simply learned that for D, $\alpha$ means down. Then everything else logically follows.

When you draw the Haworth projection, the alpha anomer is written with the -OH group pointing downwards (see the image), when you write the symbol alpha, you start by writing it downwards. For the beta anomer the -OH group is written upwards and when you write the symbol beta, you start writing it upwards. This mnemonic applies only to D-monosaccharides. For L-monosaccharides the opposite applies.
Actually, a better definition is that the alpha isomer has the -OH group pointing in the direction opposite to the -CH2OH group (on carbon 5), the opposite applies for the beta isomer, therefore for L-monosaccharides the -CH2OH group should point downwards and the alpha isomer will have the -OH group pointing upwards.

I think, this is only true for pyranoses.

CC what I just posted under Ringo’s post.

I think this one is easier to remember, but make sure to read ringos post too.

When I was studying for my organic chemistry class last semester, I ran across an ACS published paper that provides several useful mnemonics for interconverting Fischer and Haworth projections of monosaccharides. I believe number four is what you are looking for specifically:

Start the ring with the anomeric carbon on the right, and continue to draw the remaining atoms in the ring in a clockwise direction, ending with the ring oxygen at the top of the drawing.

Hydroxyl groups on the right (Dextro-) are drawn Down; those on the left (Levo-) Levitate up.

The last -$\ce{CH2OH}$ group is drawn beLow for an L-sugar; above for a D-sugar.

The hydroxyl group attached to the anomeric carbon and the last -$\ce{CH2OH}$ group are drown Both on the same side of the ring for the $\beta$-anomer; for the $\alpha$-anomer each is Alone.

Same comment I gave Marko. I think, this only applies to pyranoses.

@Jan the title of the article says monosaccharides. I don't see why it would only apply to hexoses.

Ah, there’s the problem. It seems like it’s only strictly true for sugars that have a $\ce{CH2OH}$ as final, extracyclic group. It also doesn’t work for galactofuranose.


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Is acetone saturated or not?
This was a question in a chemistry exam:

How many moles of hydrogen is required to convert 1 mole of this compound into a saturated compound?
     $\ce{CH3COCH3}$ (acetone)

Some teachers said that it's saturated so the answer is zero.
Others said that one mole is required to break the bond between $\ce{C}$ and $\ce{O}$ and convert it into an alcohol.
So which answer is correct and please mention a reference if available.

I do not know any official definition of the term. If you define a saturated compound as having only single bonds (whatever that means) there would not be any saturated fatty acids, what contradicts the normal use of the word. So I guess the answer to your question depends on the definition that was given in your class.

Willfully "defining" your own chemical nomenclature does not help. The question in this exam is badly posed, and your teachers are afraid to admit that.

Related: https://chemistry.stackexchange.com/questions/180681/is-acetophenone-a-saturated-compound/180754#180754

Typcially, the terms saturated and unsaturated only apply to the carbon chain in itself. A saturated compound would be one whose systematic name ends in -ane while unsaturated ones would end in -ene or -yne — grossly simplifying. Going by this rather simplistically strict definition, at first sight acetone (propanone) is saturated. But.
You may have heard of the keto-enol tautomery. That is, a ketone is always in equilibrium with its corresponding enol form, for acetone that would be propen-2-ol — an unsaturated compound. It is shown in the scheme below. The first indication of the distinction being difficult.

Furthermore, acetone can be reduced by molecular hydrogen to isopropanol. That means that given the correct conditions, it can basically react much like an unsaturated compound would. It also reacts with elemental bromine similarly to how an unsaturated hydrocarbon would, although only the first mechanistic step is identical; only one carbon is brominated in the product.
There is also the measure of double bond equivalents, that is often used in the process of structure elucitaion. Taking the molecular formula of a compound (which can be deduced well from high-resolution mass spectrometry) and performing a simple calculation allows one to arrive at a number of double bond equivalents for a compound. The formula is:
$$\tag{1} \text{DBE} = \frac{2 \ce{C} - \ce{H} + \ce{N} + 1}{2}$$
Where $\ce{C}$ is the number of carbon atoms (including other tetravalent atoms such as silicon), $\ce{H}$ is the number of hydrogen atoms (including other monovalent atoms such as chlorine) and $\ce{N}$ is the number of nitrogen atoms (including other trivalent atoms such as phosphorus).
However, having 1 double bond equivalent or more does not automatically mean that a molecule is unsaturated, no matter by which definition. Take cyclohexane, $\ce{C6H12}$: It has one double bond equivalent which is caused by the fact that it is cyclic. It will not react with hydrogen or with bromine and no sane chemist would call it unsaturated. So this measure is not of much help.

Concluding, I can only say that both sides of the discussion are wrong. It is wrong that acetone will not react with hydrogen to give isopropanol. It is equally wrong that that makes acetone an unsaturated compound. The concept of unsaturation simply does not extend well to $\ce{C=O}$ double bonds.
The exam question should not be graded.

It seems pretty clear to me that acetone is unsaturated. The question for me is whether isopropanol is unsaturated.

I like the example of cyclohexane. Think about the reverse process. Starting from hexane, how do you make cyclohexane? By removing H2, i.e. oxidation. So by that definition, cyclohexane is unsaturated. This is reflected in the DBE calculation, often called 'degrees of unsaturation'.

I definitely agree that it's a poorly worded exam question and shouldn't be graded.

@jerepierre  You're free to invent your own chemistry, but the established textbook OC nomenclature thinks otherwise.

The degrees of unsaturation for a hydrocarbon of formula $\mathrm{C_xH_y}$ is equal to $\dfrac{(2x+2-y)}{2}$. The addition of an oxygen atom does not change the number of degrees of unsaturation, so acetone has 1 degree of unsaturation. This is confirmed by looking at acetone's structure, which contains a double bond between the carbon and the oxygen. Reacting acetone with $\ce{H2}$ will result in 2-propanol according to the reduction reaction found here, and thus, 1 mole of hydrogen should be required to convert acetone into a saturated compound.
While acetone is not a saturated compound according to the aforementioned criteria found from this source, the degree of unsaturation comes from the carbon-oxygen bond, meaning that within some contexts, it is considered a saturated compound. According to the wikipedia article on saturated vs. unsaturated compounds, unsaturation only counts when it is the result of double or triple carbon-carbon bonds. All of the carbon-carbon bonds on the compound in acetone are single, implying that the compound is saturated according to these criteria.
Either way, this is a poorly written question, and either answer should be considered acceptable.

Acetone is not a hydrocarbon. Saying that it is "unsaturated" is pointless, you are, as are the OPs teachers, misapplying the term.

I think it depends on the context. Check out this link. Given the way the problem is worded, the original question seems to be asking about how many moles of H2 will a mole of acetone react with. Regardless, the original question is ill defined, so either answer should be right.

The only correct answer to a stupid question is explaining why it is stupid.

Also, Acetone can tautomerize to make an unsaturated compound.

That is ridiculous.  The enole of acteone  is EXtremely unstable.

But it will react with the H2. Even though it is a minor contributor, it is a source of unsaturation.

Have you got a source for such predication? And you are overstretching the term "unsaturated". It is an on/off statement, there are no vague shades of unsaturation.

I am just saying that a solution of acetone contains some unsaturation on a carbon-carbon bond. A large number of carbonyl reactions occur due to the enolate tautomer. One example can be found here

Let us continue this discussion in chat.


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Why are osmoles not considered SI units?
The Wikipedia article for osmoles is pretty clear on what they are and how they are used. Right off the bat, though, it states that the osmole is not an SI unit, and indeed the usual lists of derived SI units don't mention it. This seems weird to me - it looks like a pretty reasonable thing to be talking about and it looks reasonably uniquely defined for each solute to me.
Is there some fundamental reason why this unit did not make the cut? Is it down to historical accident? Or did the unit stop being useful by the time the SI was designed? Or is it simply not that well defined, changing with respect to exactly what situation one is talking about?

I think the reason is ‘what the heck would I use it for?’

The unit osmole (symbol: $\mathrm{osmol}$) is a coherent derived unit with a special name and symbol like the similar unit equivalent (symbol: $\mathrm{eq}$). Both units have not been accepted for use with the International System of Units (SI).
The main reason why these units have not been adopted is probably that such use is neither necessary nor convenient. All concerned quantities have the same dimension as the amount of substance $n$ $(\dim n = \mathsf{N})$. Therefore, all values can be expressed in terms of the base unit mole (symbol: $\mathrm{mol}$).
Note that, according to the definition of the mole,

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

Thus, the amount of substance can always be expressed in terms of the unit mole, regardless of the considered elementary entities. For example,

amount of mercury atoms: $n(\ce{Hg})=1\ \mathrm{mol}$
amount of hydrogen molecules: $n(\ce{H2})=1\ \mathrm{mol}$
amount of chloride ions: $n(\ce{Cl^-})=1\ \mathrm{mol}$
amount of electrons: $n(\ce{e-})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/2H2SO4}$ corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction: $n(\ce{1/2H2SO4})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/5MnO4-}$ corresponding to the transfer of an electron in a redox reaction: $n(\ce{1/5MnO4-})=1\ \mathrm{mol}$
amount of solute that contributes to the osmotic pressure of a solution: $n_\text{solute}=1\ \mathrm{mol}$

Nevertheless, certain coherent derived units with special names and symbols have been adopted for use with the SI. The main reason is convenience. The special names and symbols are simply a compact form for the expression of combinations of base units that are used frequently, for example “one kilogram metre squared per second squared” may be written “one joule” $(1\ \mathrm{kg\ m^2\ s^{-2}}=1\ \mathrm J)$. However, his reason does not apply to the units osmole and equivalent, since “osmole” and “equivalent” are not shorter than “mole” and $1\ \mathrm{osmol}$ and $1\ \mathrm{eq}$ are not more compact than $1\ \mathrm{mol}$.
In many cases, coherent derived units with special names and symbols may also serve to remind the reader of the quantity involved. For example, a value expressed in the unit newton may help to remind the reader that the considered quantity is a force. However, the unit symbol should not be used to provide specific information about the quantity, and should never be the sole source of information on the quantity. Therefore, even if the special name “osmole” or “equivalent” is used, the special nature of the considered quantity would still have to be explained in the text.
There are, however, a few important exceptions. For the quantity plane angle, the unit one is given the special name “radian” (symbol: rad), and for the quantity solid angle, the unit one is given the special name “steradian” (symbol sr). For the quantity frequency, the unit reciprocal second is given the special name “hertz” (symbol: Hz), and for the quantity activity, the unit reciprocal second is given the special name “becquerel” (symbol: Bq) in order to emphasize the different nature of the quantities. For the quantity absorbed dose $D$, the unit joule per kilogram is given the special name “gray” (symbol: Gy), and for the quantity equivalent dose $H$, the unit joule per kilogram is given the special name “sievert” (symbol: Sv) to avoid the grave risks of confusion between these quantities (e.g. in therapeutic work). Apparently, the concept of “osmotic concentration” (formerly known as “osmolarity”) is simply not important enough to justify a similar exception for the unit osmole.

eq or meq are frequently used in soil chem to be clear that a quantity is charge specific rather than an ion or undissociated electrolyte count. Mainly because the work is with complex solutions with a mix of ion charge levels, several weak or poorly soluble electrolytes, and interest in the ion exchange capacity of a solid matrix with surface charge.

eg. A soil test indicates a cation exchange capacity of 9 milliequivalents per 100grams and currently contains 6meq of various(non-proton) cations leaving room to add 3 meq of some fertilizer. (ie 1.5 mmol Mg or 3mmol Nh4, or n mmoles of some more complex mixture)  So I would say eq is much more compact in this case.

Maybe because "osmole" has the same measurement units as "mole" and the latter is the SI unit with that measurement unit.  For example, one mole of sodium chloride gives two osmoles in water solution; the conversion between them is
$2$ osmol/mol
showing the two quantities have the same measurement unit.
The mole was included into the SI system in 1971 (https://en.wikipedia.org/wiki/Mole_(unit)).

Osmol would be specifically referencing the total particle count as dissolved. Both to avoid confusion with moles of undissolved salt molecules and because weak electrolytes do not fully dissociate creating an osmol count that is not a fixed ratio to the moles of dry electrolyte.


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Sodium permanganate from manganese dioxide and sodium hypochlorite
This page states that sodium permanganate can be formed by the reaction of manganese dioxide with sodium hypochlorite and sodium hydroxide:
$$\ce{2 MnO2 + 3 NaClO + 2 NaOH -> 2 NaMnO4 + 3 NaCl + H2O}$$
I assume sodium carbonate ($\ce{Na2CO3}$) will also function in this reaction as a substitute for $\ce{NaOH}$.
Does this reaction take place at standard conditions, or are increases in temperature/pressure etc. required? Approximately how long will it take (at STP)? Also, will sodium carbonate be sufficient, or is sodium hydroxide required?

I wouldn’t be too sure about the carbonate working. It may well be too weak a base …

Different question: I am slightly confused about the [tag:home-experiment] tag. Are you planning to do this at home? If so, adding a note to the question would be great.

@Jan I added the home-experiment tag because this is a small experiment which can easily happen at home (bleach, batteries, baking soda).

I performed this reaction, and the answer is yes, concentrated sodium carbonate can be used in place of sodium hydroxide, though the latter works better. The reaction goes under normal pressure, but it requires heating. The rate of the reaction also depends on the quality of manganese dioxide - freshly prepared works best. You will notice the formation of permanganate as soon as you boil the reaction mixture, but complete conversion of dioxide may take several minutes to hours. If you want a complete conversion, better use the sodium hydroxide.


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Why might PM2.5 concentration increase or stay the same when SO2 is reduced?
The question is in the title. I've made some research, but in every paper I found the correlation between $\ce{SO2}$ and $\mathrm{PM_{2.5}}$ is described as positive. Thus if $\ce{SO2}$ is reduced, $\mathrm{PM_{2.5}}$ should also be reduced.

Your research is correct: increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$ concentration.  Gaseous $\ce{SO2}$ is photochemically oxidized to $\ce{SO3}$ which becomes sulfuric acid on contact with water vapor. This is the process for producing the atmosphere's finest aerosols, as they grow from aggregates of a few molecules.
Your final sentence is also correct: decreasing $\ce{SO2}$ concentration decreases $\mathrm{PM_{2.5}}$ concentration. The logic here is simply the reverse of that for the first statement (that increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$).
Although the overwhelming source of atmospheric aerosols is salt particles from dried ocean spray, and there are other sources like dust storms, these produce much larger particles than those produced by the oxidation of $\ce{SO2}$. So, in a pristine ocean environment far from natural or anthropogenic $\ce{SO2}$ sources for example, you may have more larger aerosols than would be found nearer to $\ce{SO2}$ sources, so that the total particulate matter would be greater in such a region of low $\ce{SO2}$. But, the $\mathrm{PM_{2.5}}$ portion would be very low such that $\mathrm{PM_{2.5}}$ would still be decreased in this region of low $\ce{SO2}$ emission.
For the reasons stated above, the answer to your title question is that there is no significant process by which $\mathrm{PM_{2.5}}$ would increase or stay the same when $\ce{SO2}$ concentrations are reduced.


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Why are the densities of europium and ytterbium anomalously low?
Why do europium and ytterbium have lower densities than expected in comparison to other lanthanides? I know it has something to do with the fact that they have half-full and full 4f subshells in the +2 oxidation state, but how exactly does it relate to density? And how does it affect the melting point?

Why do you think that properties should vary monotonically? Given the wide variety of room temperature crystal structures across the lanthanides, one really should not be all that surprised that things change.

Mostly because these two tend to be 2+, while the rest is 3+.

Expanding upon @Ivan's comment: Eu and Yb can access the +2 oxidation state instead of +3, due to the +2 ions having a relatively stable half-filled or filled f subshell.
That includes the embedded ions in the structure of the metal. Eu and Yb have $\ce{M^2+}$ instead of $\ce{M^3+}$ ions in the metal. The extra electron and reduced effective nuclear charge on the outer subshells make those $\ce{M^2+}$ ions larger, therefore these metals have lower densities than the surrounding lanthanides. The lower ion charge also means fewer electrons binding the ions together, making Eu and Yb easier to melt.
The difference in ion charge can also show up in other ways. Eu and Yb, for instance, resemble heavy alkaline earth metals more than other lanthanides by dissolving in liquid ammonia.  Europium is also known to appear in some rocks alongside alkaline earth metals rather than alongside other lanthanides (see Wikipedia):

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths. This ion-exchange process is the basis of the "negative europium anomaly", the low europium content in many lanthanide minerals such as monazite, relative to the chondritic abundance.

"Eu and Yb do favour the +2 oxidation state instead of +3" is a bit strong I think, in most situations the 2+ ions are very good reducing agents, Yb2+ in particular. But yes, at least part of the difference is due to these two lanthanides having a chemically accessible 2+ state. The interesting case is Sm, which is the next most stable 2+ Lanthanide, but doesn't show the above anomalies.

As you likely know, the structure of metals is a lattice of cations surrounded by freely moving electrons. The structure of most lanthanides consists of +3 anions, but the electronic structures of europium ($\mathrm{(4f)^7 (6s)^2}$) and ytterbium ($\mathrm{(4f)^{14} (6s)^2}$) mean that the +2 state is more stable. The lower charge difference gives weaker metallic bonding and therefore lower density.