Why does ordinary water have a triple point? Triple points and triple lines for pure substances and mixtures

My understanding is that pure substances can have triple points.  But mixtures of substances have triple lines, i.e. there isn't a single point but a one-dimensional path in $PT$ space where three phases can coexist.

This is understanding comes from the Gibbs phase rule that $F = C - P + 2$, where $F$ is the degrees of freedom, $C$ is the number of components, and $P$ is the number of phases.  At a triple point (or triple line) $P = 3$.   For a two-component system like a mixture of ethanol and water, $C = 2$.  If we apply the equation, then we find that in this case, $F = 1$, i.e. there is one degree of freedom.  If we pick $P$ then $T$ is defined for us by the constraints of any two-component, three-phase system.  That's a "triple line".  For a one-component system, the equation says that if $C = 1$ and $P = 3$, then $F = 0$.  A triple point.  We don't get to pick either $P$ or $T$.

Water isn't exactly a pure substance

But, ordinary water is not actually a pure substance.  It is a mixture of different isotopologues, e.g. $\ce{HDO}$ and $\ce{H2^{18}O}$, etc.  It is also a mixture of para- and ortho-isomers that differ only in the relative spin states of the two $\ce{H}$ atoms, as I learned from reading this very site the other day.

Does ordinary water have a triple point?

Pure $\ce{D2O}$ has a triple point distinct from that of $\ce{H2O}$.  For example, one web site says that $\ce{H2O}$ has a triple point of 612 Pa and 0.01 °C, while $\ce{D2O}$ reportedly has a triple point of 661 Pa and 3.82 °C.  But most (all?) water samples in the world are not 100% isotopically pure.  And very little work has been done separating para- and ortho spin isomers of water.

So given that water isn't a pure substance, how does it have a well-defined triple point?  "The" triple point of water was until recently fixed by the definition of SI units, so it seems that for most practical purposes the impurities in water don't matter much.  Why not?  What is the triple point of para-$\ce{H2O}$?  Of ortho-$\ce{H2O}$?  Of para-$\ce{H2^{18}O}$?  Etc. etc. etc.  Why doesn't the phase rule apply to these species?

Summarizing the relevant point (pun intended) of my answer to a related post: a mixture does not exhibit a "triple line" if its composition is constant. It is only a "triple line" insofar as you get to vary the proportion (mole fractions) of the components. Therefore different samples of water with different isotopic composition will differ in the value of the TP, but for a given sample the TP is just that, a single constant point (not a line).
This question however does point out a very important problem for the implementation of the standards adopted during establishment of an absolute temperature scale based on the triple point of water. A good discussion of this is provided in white papers (1,2) from Isotech (manufaturers of triple point measurement apparatuses):

There are 3 main sources of error in a water triple point cell[6,
7] (excluding errors associated with measurement):

The isotopic composition of the cell’s water may not be that of SMOW.

There may be air trapped inside the cell.

There will be impurities in the water.

Isotopic composition can be measured to great accuracy from a
small sample of the cell’s contents.

SMOW and V-SMOW are water samples of standard composition (this is amply discussed in Isotech white papers, referring also to the documentation behind the International Temperature Scale established in 1990 (ITS-90)):
\begin{array} {|r|r|}
\hline
\text{Component}  &  \text{Mole Fraction} \\
\hline
\ce{^1H} & 0.999 842 \\
\ce{^2H} (\text{deuterium}) & 0.000 158 \\
\ce{^3H} (\text{tritium}) & 0 \\
\ce{^{16}O}  & 0.997 640 \\
\ce{^{17}O} & 0.000 371 \\
\ce{^{18}O} & 0.001 989 \\
\hline
\end{array}
A second white paper provides guidelines on consideration of these effects (see the section "Isotopic Composition"):

We think of water as $\ce{H2O}$, more precisely we should write $\ce{^1H2^{16}O}$. This has been termed ‘light’ water and is unobtainable. If it were obtainable it’s triple point would be approximately +0.008°C. In practice what we drink from the tap, and what is in the ocean is a soup of $\ce{^1H}$, $\ce{^2H}$ with $\ce{^{16}O}$, $\ce{^{17}O}$ and $\ce{^{18}O}$. This means real water does not have a single triple point temperature. Its ‘combination triple point’ will vary as the isotopic composition varies. This  was  understood  more  than  50  years  ago  and  after  much  analysis  of  water  from  different  sources recent temperature scales state that  the water in a triple point cell shall have the isotopic composition  of ocean water.  Although not explicitly stated, there is a Standard Mean Ocean Water (SMOW) and it is assumed that the isotopic content of SMOW fulfils the ITS-90 requirements.

concluding:

Analysing the cell contents for isotopic composition is a useful and valid addition to the usual techniques of intercomparing cells and gives an accurate indication of the cells ideal triple point value.
[...]
Such cells, complete with isotopic analysis provide the closest association to the ITS-90 value of the water triple point and hence the KTTS Scale. It is now possible to produce cells, based on Stimpson’s original design which can be associated to V-SMOW within ±3mK and this is a smaller uncertainty than can be provided by traditional measurements.

References

The Establishment  of  ITS-90 Water Triple Point References  to  ±2μK,  and  the Assessment of 100 Water Triple Point Cells Made Between 2001 and 2006, John P. Tavener and Nick Davies, Isothermal Technology Ltd. (England).

The Water Triple Point - A Reference Cell Close to the ITS-90 Value. John P. Tavener & Anne Blundell. Northern Temperature Primary Laboratory, Pine Grove, Southport, Merseyside, PR9 9AG, England.

Supplementary Information for the International Temperature Scale of 1990, BIPM. ISBN 92-822-2111-3.


Does computational chemistry include neutrons in ground state calculations? I was under the impression that most particles in computational chemistry were treated as point charges. If this is true, wouldn't there be no need to include neutrons as a separate entity in ground state calculations? Or are particles not treated as point charges?

Depends on what you mean by 'computational chemistry'.
In the most common sense of the term, where the aim of a given calculation is determination of the electronic structure of a particular system, each entire nucleus is treated as a classical point charge/mass.  Thus, neutrons are not explicitly accounted for, aside from their effect on the various nuclear masses in, say, vibrational frequency analysis or in estimates of the kinetic isotope effect.
If you're referring to quantum models of the nucleus, though, then this statement of yours is incorrect:

... most particles in computational chemistry [are] treated as point charges.

In these nuclear computations, if the nucleons were being treated classically it would be correct to say that they are treated as point particles since, as you correctly note, neutrons carry no charge, and proper representation of both the neutrons and protons is necessary in order to obtain an accurate model.
However, as noted quite correctly by Mithoron in a comment, for quantum modeling of the nucleus, the protons and neutrons are not treated as particles but instead as contributors to the nuclear wavefunction.


Intuituve reason for the constant of proportionality 2/3 in pv=nRT=(2/3)U Why is there a $\frac23$ term?

I fully understand that it makes sense for pressure to be the volume weighted average kinetic energy. And the equation above shows that this is so (for an ideal gas). I also fully understand how to prove this with momentum and force. I'm wondering is there an intuitive reason why the for one cubic meter, the the pressure should be two thirds of the total internal kinetic energy, not some other value. Or is it just because the math says so?

The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), and the particles have equal probability of hitting a face in the x, y or z direction.  The pressure is exerted equally amongst the three orthogonal directions, hence a fact of 1/3.

Wikipdedia also shows the steps in deriving the kinetic theory.


Are buckminsterfullerene-corannulene complexes actually stabilised by π stacking? According to wikipedia and the references given therein, $\pi\cdots\pi$ stacking interactions are the result of interaction between the quadrupole moments of two aromatic rings, rationalising the stabilisation of perpendicular and offset-parallel association modes of the benzene dimer.

Conventional wisdom is that the supramolecular complexes of buckminsterfullerene and functionalised corannulenes (the so-called buckycatchers) are stabilised by $\pi\cdots\pi$ stacking. How can the quadrupole interaction explanation be reconciled with the pseudo-spherical symmetry of buckminsterfullerene? Surely the interaction must be entirely Van der Waals in nature?

In short, I think the answer is that although the overall symmetry of the buckminsterfullerene is almost spherical, the quadrupole–quadrupole interaction with other molecules with aromatic cycles are local (in addition to the ever-present van der Waals interactions).

I tried to think of a way to exemplify that, and drew the 2D picture below:

In this 2D model, the “double bonds” of the circular structure represent the individual aromatic rings of your buckminsterfullerene (which isn't superaromatic, i.e. electrons are not delocalized over its entire structure; think may be a key point), bearing each a local quadrupole moment (symbolized in the red/blue thingy). The other particle with a quadrupole is close to one of the ball's quadrupoles, and this local interaction dominates. In fact, other interactions maybe be stabilizing or not, depending on the orientation of the individual quadrupoles, but they are weaker that the local quadrupole–quadrupole interaction.


How can I have a concerted addition of bromines? I know how to add Bromines in the antiperiplanar position, but I have no idea how to get concerted addition:

Perhaps I am not seeing something obvious, but I would greatly appreciate any hint.

The proper term for concerted addition is syn addition.  I can't find any way of a direct way of producing a syn-addition product.  The only way I can figure you would have to do it is if you did the halogenation with water as the solvent you would get the halo-hydrin product with an anti-addition.  You could then follow this up with a bromination using $\ce{P,Br2}$ or $\ce{PBr3}$ which would invert the chirality and give you a syn-product.


Why is slush hydrogen at the triple point? Slush hydrogen is a mixture of liquid and solid hydrogen at the triple point considered as a possible vehicle fuel. What is the need of having it at the triple point? Couldn't any other set of thermodynamic conditions consistent with liquid–solid equilibrium work as well?

I'll have a stab at this one, despite the fact I've forgotten almost the totality of engineering thermodynamics, as I do recall seeing these papers before.

One of the references1 in the Wikipedia article describes the production of slush hydrogen through the auger technique, contrasting it with the freeze-thaw method. Specifically, the auger method can be performed above the triple point pressure of hydrogen, whereas the freeze-thaw method uses the latent heat of vaporisation to freeze some of the hydrogen. In this technique the vaporisation of some of the hydrogen (prompted by the application of a vacuum) at the triple point causes some of the liquid hydrogen to freeze, generating a crust of solid hydrogen which is then broken up to make slush.

As far as I can tell, the triple point conditions are only required to make the freeze-thaw process go, and are not required for storage.

The auger technique, also described in the report1, seems to work on the basis of shaving solid hydrogen which is refrigerated with liquid helium. Voth compares the technique to a conventional ice shaver (albeit much, much colder!). In this instance there's no requirement for triple point conditions, which is given as one of the advantages of the technique.

Finally, Haberbusch and McNelis2 give us this sad quote, in reference to a 1991 paper:

"According to Hans, projected annual demand for slush hydrogen in the year 2010 is 584 million pounds to support a fleet of 20 space-planes"

One day, maybe...

(1) Voth, R.O.; Producing Liquid-Solid Mixtures of Hydrogen in an Auger; National Bureau of Standards Department of Commerce; 1978, pp. 1-16

(2) Haberbusch, M.S. and McNelis, N.B.; Comparison of the Continuous Freeze Slush Hydrogen Production Technique to the Freeze-Thaw Technique; NASA Technical Memorandum 107324, 1996, p. 1


Thermodynamics, pressure volume work In thermodynamics work done by the system is negative but in case of expansion of gas it is positive because force and displacement are in same direction, why?

You are correct. The system does positive work. Textbooks introduce the negative to show that the system loses energy as it does expansion work. As the gas expands, it does positive work but it loses this energy in the process. Work, like heat, is basically a transfer of energy.


Is there any difference between de-Broglie wavelength and the wavelength of normal waves? Is de-Broglie wavelength different from wavelength of waves such as electromagnetic waves?

If yes, they differ in what respect?

And can we use the formula of wavelength of light ($c=v\lambda$) to calculate the de-Broglie wavelength?

Is de-Broglie wavelength different from wavelength of waves such as
  electromagnetic waves?

I would atually say no, there is no difference, since the wavelength is purely mathematical characteristic of a wave, the distance over which the wave's shape repeats. There is, however, an important difference between de-Broglie waves on the one side and usual physical waves (mechanical & electromagnetic) on the other:

The later consist of periodic oscillations of a physical quantity, such as, for instance, the pressure for sound waves or electrical and magnetic fields for electromagnetic waves.
The former consist of periodic oscillations of a mathematical quantity, the probability of finding a particle at some location.


Why does the 3rd electron shell start filling up with scandium? The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2?

What causes the 3rd shell to start filling up, rather than the fourth?

The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising:

Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$, the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$. Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.
In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine):

This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$, because the overall result is more advantageous for the atom as a whole.  But this is not something that can be predicted.  Why is it 2 electrons, rather than one or even none?  In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital.  In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$.
None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital.  To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.

So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms.


Etching a Screen Protector Film I have to etch a pattern on a plastic film similar to the ones used to protect phone screens. I want a very thin etch, one that is only visible on close inspection. How can I get that done?

Assuming by thin, you mean a narrow line (not depth). It all depends on how thin you want to go. For a fine width line, a relatively cheap $\ce{CO2}$ laser cutter will get you down to 100–200 µm, but depending on the plastic and how thick it is, the heat produced may warp the plastic near the cut. I think most phone screen protectors are PET film. I've personally made ~200 µm features with one, and if the plastic is thin enough, you can keep the cut width pretty narrow. Things like pulsed UV lasers will give you much finer lines on plastics without thermal effects (as they tend to ablate rather than burn plastics), but they're much more expensive. If you don't have access to either, many companies will contract for this kind of work, even for small jobs.

Another option for coarser cuts that don't have too complex of shapes is a vinyl cutter—basically a plotter with a cutting blade instead of a pen. These are quite cheap, but can't cut all shapes and resolution is on the order of a hundred µm or so, though they don't cause distortions along the edges in most materials.

If you're really serious about it, it's certainly possible to coat photoresist on most plastics and make a mask for reactive ion etching, which will get you down to 1 or 2 µm easily, but it sounds like overkill for you.


Nitrogen fertilisation expressed as nitrogen equivalent In a plant pathology paper I found levels of fertilisation applied in the experiment expressed as nitrogen equivalent.
Reporting the exact phrase:

  Nitrogen fertilisation with 8.6 g of nitrogen equivalent …

I found that equivalent weight for an element would be its atomic weight divided by its valency so, if I'm not wrong, equivalent weight for N would be ~4.67 g

What does this mean?
How much fertiliser was applied e.g. using ammonium sulfate?

This is referring only to the mass of elemental nitrogen available, rather than the total mass of fertilizer. In your case, ammonium sulfate is 21% nitrogen (see the table here for other fertilizers and here for more information on fertilizer composition), which gives ~40.95g ammonium sulfate.


What chemical test can differentiate secondary and tertiary alcohols? I have to perform a chemical test to determine whether the solution is a secondary or a tertiary alcohol. How can I do this? Please explain the procedure and the apparatus.

Or you can try the very simple Lucas' test using zinc chloride and concentrated hydrochloric acid:
$$\ce{ROH + HCl -> RCl + H2O}$$

It works best with tertiary alkohols, slow with secondary alcohols and not at all with primary alcohols.


What is the difference between a Chemical Formula and Formula Unit? In my textbook, a chemical formula is described as :

  "A formula that shows the kinds and numbers of atoms in the smallest representative unit of a substance"

While a formula unit is described as :

  The lowest whole number ratio of ions in an ionic compound.  Thus the formula unit for sodium chloride is NaCl.

Though the definitions seem to differ, the written formulas appear identical.  A chemical formula appears to be another way to represent a formula unit. For example, NaCl is both the chemical formula and formula unit for Sodium Chloride.
Is there ever a difference between the written terms?

Molecular compounds such as water exist as discrete particles, molecules. This is due to the forming of covalent bonds where each atom has a specific partner to which it is bonded. Each molecule of water contains one atom of oxygen and two atoms of hydrogen. So, H2O is its formula. This can be more specifically called a molecular formula.

In ionic compounds such as table salt, NaCl, the atoms (as ions) do not bond to specific neighbors. Surrounding each chloride ion in a salt crystal are six sodium ions. Likewise, six chloride ions surround each sodium ion. This attraction of oppositely charged ions extends throughout the entire crystal. There is no discrete bonding of a particular sodium ion to a specific chloride ion. So the formula for sodium chloride is expressed as the smallest whole number ratio between the Na and Cl ions which is 1:1 for sodium chloride. So, NaCl is also a formula but not a molecular formula.

The reason for the term "formula unit" is that it is useful when we talk about how much of one substance is required to combine with a particular amount of another substance. For example: to make the smallest amount of hydrogen carbonate we combine one molecule of water with one molecule of carbon dioxide.

But now suppose we want to combine silver nitrate with sodium chloride. Both of these are ionic compounds, they do not form molecules. In this case you would say that one formula unit (FU) of sodium chloride combines with one FU of silver nitrate.

The one-to-one combining in both these examples is simply coincidence. It may take 3 FUs of substance A to combine exactly with 2 FUs of substance B in a different example.


Why is the inductive effect for a protonated amino group greater than that of a nitro group? Why does $\ce{NH3+}$ have a higher $-\ce{I}$ effect ( inductive effect) than $\ce{NO2}$ even though oxygen is more electronegative? Does it have something to do with positive charge on $\ce{N}$?

The short answer is that $\ce{NH3+}$ is a cation, and any cation will always draw electrons towards it stronger than any neutral species.

The electronegativity of the species involved is only a secondary cause. However, one can safely say that $\ce{NH2}$, if nitrogen’s lone pair is (somehow) impeded from excersizing it’s $+\ce{M}$ effect, would have a lesser $-\ce{I}$ effect on the whatever than $\ce{NO2}$ would.


How can an ionic bond have partial covalent in character? I was going through the topic of 'ionic bond' and read this:

  No bond is 100% ionic in character. It has some percentage of covalent character.

I didn't understand how an ionic bond can be partially covalent in character. Please explain along with an example.

One way is Polarization. When a small cation approaches a large  anion it distorts the charge density towards itself thereby forming a type of bond which has directional characteristic i.e from ** (-) ----> (+) ** which is characteristic of a covalent bond.

This polarization has some rules given by **Fajan **

Smaller the cation better distortion/polarization.
More the amount of positive charge more is the polarization.
or more the charge : radius ratio of cation more is the polarization.

So in short every cation has the power of polarization and therefore no bond is 100% ionic.


Nomenclature of alkenoic acids - which group has priority: the alkene or the acid? Right, how do you systematically name them? Also, do you number them from the double bond end or the Carboxyl end?

Ie would it be something like.  2-methyl Hept-3-enoic acid?

Scuse the bad question and I made up the example, sorry! I'm really confused!

Thanks!!

The carboxylic acid moiety takes precedence over the alkene moiety, so the molecule has to be numbered starting from the carboxylic acid carbon. Hence, methacrylic acid, for example, has the systematic IUPAC name of 2-methylpropenoic acid.


Calculating molarity of a gas using ideal gas equation Is hydrogen gas at 1 atm pressure and 298 K temperature considered as a 1 molar solution? If yes then can I use ideal gas equation $$PV=nRT$$ to find molarity as $$\frac nV = \frac P{RT}?$$

Hydrogen at one atmosphere pressure and 298 K is certainly not a one molar solution. However, you can still calculate the concentration with the ideal gas. Note that this is a gaseous solution and since it is a pure gas, the solute is also the solvent.
$$
c(\ce{H2}) = \frac{n(\ce{H2})}{V(\ce{H2})}
           = \frac{p(\ce{H2})}{\mathcal{R}T}
           \approx 40.9~\mathrm{mol\,m^{-3}}
           \approx 0.04~\mathrm{mol\,L^{-1}}
$$


Why substitution and elimination reactions are favored respectively at lower and higher temperature? I read about the competition between nucleophilic substitution and nucleophilic elimination depending on temperature here. Though the webpage clearly says higher temperature favors elimination while lower temperature favors substitution, the explanation on the webpage is unclear.

Need a clear explanation regarding the phenomenon.

When the question is asking about favorability, it is really asking about the magnitude of the Gibbs free energy. The more negative this value, the more favorable a reaction is. Gibbs depends on enthalpy and a temperature scaled entropy: $$\Delta G=\Delta H-T\Delta S$$

When we think of nucleophilic elimination, we know that a hydrogen atom is removed by a nucleophilic group resulting in our final product. Since a hydrogen is removed in elimination, while left intact during substitution, there are more final species formed during an elimination reaction. According to general chemistry principles, this means that there was a larger increase in entropy compared to the substitution. As temperature increases this larger entropy becomes more noticeable as it is now scaled by a larger $T$.

Therefore at low temperature, the enthalpy term dominates Gibbs and the substitution reaction prevails. However as $T$ grows, so does the entropy term and at high $T$ this overrides enthalpy, making the elimination reaction more favorable.


Why does basicity go down along the group 15 hydrides? As we go down Group 15, the sizes of atoms increase and these are the central atoms in their corresponding hydrides. Shouldn't the large size of central atom (in case of $\ce{Bi}$) actually help the lone pair inside the $\ce{sp^3}$ hybrid orbital to easily be donated and thus increase the basic character down the group?

$\ce{N}$ is small and shouldn't it hold lone pair with much more strength than others, but $\ce{NH3}$ seems to be strong base while $\ce{BiH3}$ is the weakest one.

The very idea of $\ce{sp^3}$ hybridization you are referring to is pretty much not there, starting from $\ce{PH3}$. Look at its bond angles, they are almost exactly $90^\circ$. This means that phosphorus contributes to the $\ce{P-H}$ bonds almost exclusively with its $\ce p$ orbitals, and the lone pair just sits there hidden on $\ce s$ orbital, rather than stick out as $\ce{sp^3}$.

The same is even more true for the following hydrides.

Also, with nitrogen being the most electronegative of all, its partial negative charge in $\ce{NH3}$ attracts $\ce{H+}$, which is not the case for $\ce{PH3}$ and the rest.


Why can the lone pair not align with the phenyl moiety in 2,6-xylidine? I am told that for 2,6-xylidine (2,6-dimethylaniline), the amino group cannot line up in such a way that its p-orbital is parallel with respect to the p-orbitals of the carbons in the ring.

I've looked for papers and no one has actually proved to me why I should believe that two methyl groups are bulky enough to interfere with the amino group. I've come across various statements regarding this molecule, but no proof. For example, here's a Wiki article on steric hinderance and resonance.

  Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoic acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.

But there are no citations to back any of that up.

The closest I've found is:

  Steric effects of ortho methyl groups are base strengthening. This is not due to steric inhibition of resonance since the conformation remains planar in most derivatives. Two ortho methyl groups are necessary to distort the planarity; their steric effect is more than doubled compared with one methyl group.

http://onlinelibrary.wiley.com/doi/10.1002/poc.642/abstract

But note again – there isn't really any demonstration – just a statement.

Am I supposed to do some Euclidean geometry and use van der Waals radii and trigonometry or something?

There is no prove for this statement, as it is most likely incorrect for 2,6-xylidene. This is based on an electronic structure theory approach. However, there are non-negligible steric interactions, that affect the basicity as demonstrated in Ortho-effect in substituted aromatic acids and bases.

Whenever a distance between two atoms becomes (significantly) shorter than the sum of their respective van der Waals radii, this has to be considered. Because of the strain in the xylidene molecules, this is the case.

As a very rough estimate you can use the pen and pencil approach with tabulated values for van der Waals radii (german - much better overview) and covalent radii. Another approach is building the molecule with a molecular modeling set (Molekülbaukasten -  german wikipedia), rotate and wiggle bonds, to see if there are any close contacts.
There is also some free software that lets you build molecules and check them, they usually come equipped with a set of standard bond lengths and angles. It is always a good approach to visualise a molecule.
If you have access to quantum chemical tools you can perform calculations, but this is usually too much of an effort. For this educational purpose I ran some quick calculations, though.

The following table is composed from values from the german wikipedia pages (values in Å):
\begin{array}{rrr}\hline
             & \text{van der Waals} & \text{covalent}\\\hline
\ce{H}       & 1.10  & 0.32 \\
\ce{C}       & 1.70  & 0.77 \\
\ce{N}       & 1.55  & 0.71 \\\hline
\sum\ce{H-H} & 2.20  & 0.64 \\
\sum\ce{C-H} & 2.80  & 1.09 \\
\sum\ce{N-H} & 2.65  & 1.03 \\
\sum\ce{C-N} & 3.20  & 1.48 \\
\sum\ce{C-C} & 3.40  & 1.54 \\\hline
\end{array}

Now we can have a look at the optimised (DF-BP86/def2-TZVPP) geometry of the molecule. (Of course the pen and paper approach will result in a different geometry.)

Here you can see, that the $\ce{H\cdots{}H}$ distance is very close to the sum of the van der Waals radii. The $\ce{C\cdots{}H}$ distance with about $2.5~\mathrm{Å}$ and the $\ce{C\cdots{}N}$ distance with about $2.9~\mathrm{Å}$ are already shorter than the sum f their van der Waals radii. While these interactions are not strong, the will effect the behaviour of this molecule in solution (and gas phase). It is very important to understand, that these are only snapshot values. In reality the molecule is very flexible and bonds will rotate and vibrate. Steric interactions are basically electronic and dispersive interactions in disguise - they are most likely to influence the kinetic behaviour of a molecule.

The best evidence for this is the transition state of the rotation of the amine moiety. Here we can clearly see a bonding interaction between the nitrogen and the neighbouring hydrogen (see linked post), which is evidently influencing the basicity of this molecule.

This state is only about $7~\mathrm{kcal/mol}$ higher in energy, which means it is readily accessible at room temperature. This will mean that there are many different conformations present, there is no true one. This affects the reactivity of a molecule (in this case lowered basicity).

In conclusion, your gut feeling was correct. However, there are more effects at play and the whole picture is often not easy to see. It is usually a case by case decision/ interpretation how strong which effects affect reactions and molecules (properties and barriers). When substituents get larger the more dominant steric effects will obviously be.

I hope this answers your question sufficiently.


Reaction between sulfur and sodium hydroxide Sulfur reacts with sodium hydroxide in the following way:

$\ce{3S + 6NaOH->2Na_2S + Na_2SO_3 + 3H_2O}$

It is a disproportionation reaction: sulfur's oxidation changes from zero to -2 in sodium sulfide and from zero to +4 in sodium sulfite.

But how to remember the products of this reaction? By rote memorization?

The preparation of sodium sulfide is apparent. Is there a way to understand why sodium sulfite forms? We have the hydroxide ion, and it somehow reacts with sulfur to yield the $\ce{SO3^{2-}}$ ion - but how?

The reaction products depends on the conditions(temperature, pressure etc.) and stoichiometric ratio of reactants taken. Depending on that, you will get various products in each case if you slightly change the conditions or the ratio of sodium hydroxide to sulfur taken.

$$\ce{4S + 8NaOH ->[600 C] Na2SO4 + 3Na2S + 4H2O}$$

  Sulfur react with sodium hydroxide to produce sodium sulfate, sodium
  sulfide and water. This reaction takes place at a temperature of over
  600°C.(source)

$$\ce{4S + 6NaOH → Na2S2O3 + 2Na2S + 3H2O}$$

  Sulfur react with sodium hydroxide to produce sodium thiosulfate,
  sodium sulfide and water. Sodium hydroxide - concentrated solution.
  The reaction takes place in a boiling solution.(source)

Your reaction

When you compare the three equation, you can  see that though stoichiometric ratio of the reactants in the 1st and 3rd equation is same, the moles of sulfur and sodium hydroxide  is different and thus gives different products in each case.

Also, from reaction 2 we can say that sodium thiosulfate may have disproportionate into sodium sulfite and sulfur.

$$\ce{Na2S2O3 ->[220-300 C] Na2SO3 + S}$$

and thus one mole sulfur cancel from each side and gave reaction 3 (your reaction). This site  explains the disproportionation reaction.

  Chemical reactions of sulfur compounds have been studied in equimolar $\ce{NaOH-H2O}$ melt at 100°C by voltammetry and UV spectrophotometry. Disproportionation of sulfur is fast and quantitative according to:

  $$\ce{S8 + 6OH- -> 2S3^2- + S2O3^2- + 3H2O}$$

  On the contrary, addition of sulfur to sulfide is not quantitative and
  gives $\ce{S2^2-}$ besides prevailing sulfur
  disproportionation.$\ce{SO3^2-}$ reacts with $\ce{S2^2-}$ and
  $\ce{S3^2-}$ , while $\ce{S2O3^2-}$ decomposes $\ce{S^2−}$. A
  pseudoequilibrium is observed:

  $$\ce{S^2- + S2O3^2- <=> S2^2- + SO3^2-}$$

  with concentration quotient Formula Q = $\ce{10^{-0.8}}$.


Dependence of rates of neighbouring group participation on length of alkyl chain On the topic on neigbouring group participation, it is mentioned in Carey & Sundberg (2007)[1] that the effectiveness of the participation is dependent on on the ease with which the molecular geometry required for the participation can be attained. The example of the cyclization of ω-hydroxyalkyl halides was then given:

  The rate of cyclization of ω-hydroxyalkyl halides, for example, shows a strong dependence on the length of the chain separating the two groups.

To prove the point, they showed the data for the solvolysis rates for ω-chloro alcohols with various chain lengths. Clearly, when a 5-membered ring is form, the rate of reaction is fastest. This is reasoned with the fact that a 5-membered ring has relatively little ring strain and thus, the rate of forming it would be relatively higher. However, this analysis does not seem to be convincing. If this reasoning was indeed correct, then wouldn't the next member in the series, with the capability of forming the 6-membered ring, have a higher solvolysis rate?

\begin{array}{|c|c|c|c|} \hline
\text{Substrate} & k_\mathrm{rel}{}^\text{[2,3]} \\ \hline
\ce{Cl(CH2)2OH} & 2000 \\ \hline
\ce{Cl(CH2)3OH} & 1 \\ \hline
\ce{Cl(CH2)4OH} & 5700 \\ \hline
\ce{Cl(CH2)5OH} & 20 \\ \hline
\end{array}

In general, why doesn't the rate increase as the alkyl chain increases in length, based on the reasoning using ring strain in the intermediate? We see that 3-chloropropanol is also another outlier.

Additional information

It seems that the relative solvolysis rates of ω-methoxyalkyl p-bromobenzenesulfonates also follow the same trend, with the rate peaking with the formation of the 5-membered ring.

\begin{array}{|c|c|c|c|} \hline
\text{Substrate} & k_\mathrm{rel}{}^\text{[4]} \\ \hline
\ce{CH3(CH2)2OSO2Ar^*} & 1 \\ \hline
\ce{CH3O(CH2)2OSO2Ar} & 0.28 \\ \hline
\ce{CH3O(CH2)3OSO2Ar} & 0.63 \\ \hline
\ce{CH3O(CH2)4OSO2Ar} & 657 \\ \hline
\ce{CH3O(CH2)5OSO2Ar} & 123 \\ \hline
\ce{CH3O(CH2)6OSO2Ar} & 1.16 \\ \hline
\end{array}

*The first entry shows the rate of solvolysis without any neighbouring group participation from the methoxy group.

References

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.
Capon, B. Neighbouring group participation. Q. Rev., Chem. Soc. 1964, 18 (1), 45–111. DOI: 10.1039/QR9641800045.
Richardson, W. H.; Golino, C. M.; Wachs, R. H.; Yelvington, M. B. Neighboring oxide ion and fragmentation reactions of 1,3-chlorohydrins. J. Org. Chem. 1971, 36 (7), 943–948. DOI: 10.1021/jo00806a019.
Winstein, S.; Allred, E.; Heck, R.; Glick, R. Neighboring methoxyl participation in solvolytic nucleophilic substitution. Tetrahedron 1958, 3 (1), 1–13. DOI: 10.1016/S0040-4020(01)82605-3.

Thanks to orthocresol's suggestion to look at Clayden's (2nd ed. p. 805-807), I now have a satisfactory answer which I can now present.

We can explain these trends using the Gibbs free energy equation. Except that now, we are now interested in the quantities: Gibbs free energy of activation, enthalpy of activation and the entropy of activation, since we are analysing rate.

$$\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger $$

Explaining the decreasing of $\ce {\Delta H^\ddagger}$ as the ring size increase

$\Delta H^\ddagger$ is associated with the ring strain of the resultant ring formed. The larger the ring strain, the larger the value of the enthalpy of activation. Clayden, Greeves, & Warren (2012) explains$\ce{^1}$:

  Firstly, small rings form slowly because forming them introduces ring strain. This ring strain is there even at the transition state, raising its energy and slowing down the reaction. The activation energy for forming a three-membered ring is very high, due to strain, but decreases as the ring gets larger.

The data (from pg. 108 of ref. 2) presented here of the strain energy determined for cycloalkanes of varying sizes also supports this trend.

\begin{array}{|c|c|c|c|} \hline
\text{Ring size} & \text{Strain energy}^\text{2,*}/ \text{kJ/mol} \\ \hline
\ce{3} & 115 \\ \hline
\ce{4} & 110\\ \hline
\ce{5} & 26 \\ \hline
\ce{6} & 0 \\ \hline
\end{array}

*Reported to the nearest whole number.

Explaining the increasing of $\Delta S^\ddagger$ as the ring size increases

Think of it this way: a long chain has a lot of disorder, and to get its ends to meet up and react means it has to give up a lot of freedom$\ce {^1}$. However, for shorter chains, it is far easier for their ends to meet since these chains have fewer degrees of freedom and thus, the loss of entropy upon cyclisation is less negative.

Consolidating the trends

Clayden, Greeves, & Warren (2012) reasons that based on the above trends for $\Delta H^\ddagger$ and $\Delta S^\ddagger$, the activation energy for the formation of the 4-membered ring is the largest. This is because the formation of the 4-membered ring likely has similar $\Delta H^\ddagger$ to that for the 3-membered ring while its $\Delta S^\ddagger$ is relatively more negative.

This seems to agree with the solvolysis rates I have presented:

For the ω-chloro alcohols, we see that the relative solvolysis rate is lowest for $\ce {Cl(CH2)3OH}$. For the ω-methoxyalkyl p-bromobenzenesulfonates, the formation of the 4-membered ring also has the 2nd lowest rate ($\ce {k_{rel} = 0.68}$), with the rate for the 3-membered ring being the lowest ($\ce {k_{rel} = 0.28}$).

Why does the solvolysis rate not peak at the formation of the 6-membered ring and instead, peak for the formation of the 5-membered ring? This is probably a coincidence. There are 2 opposing trends here so it may be the case that the combined effects of the two trends give different results under different circumstances.

ron has provided a graphical representation of the data for strain energy and the entropy of activation here. We do see that in general, there is a peak in the rate of formation of the 5-membered ring. In other words, from the formation of 5-membered ring to formation of 6-membered ring, $\Delta S^\ddagger$ becomes significantly more negative.

References

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.
Anslyn, Eric V., and Dennis A. Dougherty. (2006). Modern Physical Organic Chemistry. Sausalito, CA: University Science.


Rate-determining step and steady state approximation failure 1) When can't the rate-determining step be applied, and why? How do you recognize these cases?

2) When can't the steady state approximation be applied, and why?

When can't the rate-determining step be applied, and why? How to recognize those mechanisms?

The rate determining step (often abbreviated as "rds") of a reaction is the slowest step in a reaction.  Every reaction has a "slowest step", therefore, every reaction has an rds.  The slowest step will be the step where we pass over a transition state with the highest energy, e.g. the highest energy point along the reaction coordinate. Look at the figure below, the reaction coordinate shown in the top left diagram

illustrates a reaction with a transition state, but no intermediate.  There is only one step in this reaction, passing over that transition state (marked by the double-headed arrow) is the rds for that reaction.  The drawing in middle and right side of the top row show a reaction with an intermediate, and consequently, two transition states.  In each case the double-headed arrow identifies the rds - the highest energy point along the reaction coordinate.  The 3 drawings in the bottom row illustrate the case of a reaction with 2 intermediates (and therefore 3 transition states) along the reaction coordinate - in each case the arrow identifies the transition state with the highest energy, the rds.

  When can't the steady state approximation be applied, and why?

Consider the following reaction
$$\ce{A <=>C[{k_1}]\   [B] <=>C[{k_2}]\ C}$$
$\ce{A}$ is the starting material, $\ce{[B]}$ is an intermediate and $\ce{C}$ represents the product.  Either the drawing in the middle or right in the top row represent the reaction coordinate for this reaction.  The steady state assumption applies to reactions that have one or more intermediates along the reaction coordinate.  The steady state assumption states that the rate of change of the concentration of the intermediate is approximately zero.  In other words, the intermediate is being created at about the same rate at which it is
$$\ce{\frac{d[B]}{dt} = 0}$$
being destroyed.  The steady state assumption is best applied to situations where the intermediate is present in low concentration (e.g. $\ce{k_2 >> k_1}$).  A general rule of thumb for the assumption to be valid is for
$$\ce{\frac{k_2}{k_1} > 10 }$$
Looking back at the figure, we now see that this condition is met in the top row, center drawing.  Here is a Wikipedia article on the steady state assumption, the section on "Validity" is nice and clear if you'd like to read further on the subject.


Role of Acetonitrile and Ammonium Acetate buffer in Leucomalachite Green stock solution? I am trying to repeat an experiment conducted by FDA on Malachite Green (MG).
According to this paper,
Prepare a series of LC calibrants by aliquoting into individual 15 mL graduated
glass or disposable polypropylene centrifuge tubes, 50 µL, 100 µL, and 200 µL
of the 0.1 µg/mL MG2 solution and 50 µL and 100 µL of the 1.0 µg/mL MG1 solution.
Dilute each to 5.0mL with (1:1 vol) acetonitrile/ammonium acetate buffer
(mobile phase A), vortex mix or stopper and invert to mix thoroughly.
Based on a 5 g sample weight, these solutions will generate standards with
concentrations of 1, 2, 4, 10 and 20 ng/g of MG.

If I am not wrong, Malachite Green used in this paper is from the salt, Malachite Green Oxalate.

Is Acetonitrile used here as a pH buffer?
What is the role of Ammonium acetate?

If I plan on using Raman Spectroscopy to detect MG in my samples, shouldn't I be using the pure forms of MG as my reference signal? If so, wouldn't the addition of these two chemicals affect my readings?
References
The pdf document originally linked in this post (see edits) is no longer available.
The most similar reference identified (available here via the Internet Wayback Machine) is:

Wendy C. Andersen, José E. Roybal, Sherri B. Turnipseed. Determination of Malachite Green and Leucomalachite Green in Salmon with In-Situ Oxidation and Liquid Chromatography with Visible Detection. Laboratory Information Bulletin (LIB) 4334: Malachite Green and Leucomalachite Green in Salmon. Volume 20, No. 11, November 2004.

Since they're using ACN/ammonium acetate to elute from solid phase extraction cartridges and that's what they're injecting into their LC analysis, they probably want to use the same mixture to avoid problems with unmatched matrices with their standards. The ammonium acetate is the pH buffer. Malachite green has different protonation states at different pH, so it's important to set the pH properly for elution from the SPE and LC analysis. MG is fairly soluble in water, but still has a lot of organic functionality, so adding an organic solvent would help elute from the SPE column and keep the analyte from being too strongly retained by the LC stationary phase.

As for raman analysis, water has rather weak raman scattering, so it's not usually a problem. Acetonitrile has one very strong peak at 2942 $\mathrm{cm^{-1}}$ and a few moderate peaks, so it might be an issue. Since you're not doing LC, you can likely prepare standards in buffer only. Unfortunately, ammonium acetate is also raman-active, but if you're not doing LC, you might be able to do without as well. Fortunately, one of the great powers of IR and raman spectroscopies is that there is often enough bandwidth to find one or two non-overlapping peaks in a complex mixture and use that for quantitation, so it may still work as written. They're both also very amenable to chemometric techniques like partial least squares when measuring complex samples.

The big problem is going to be sensitivity if you're interested in the concentrations they use in the paper.. Normal raman is not that sensitive and the smallest standard is only ~3 nM, so it's probably going to be hopeless. These guys do surface-enhaced resonance raman and are able to get to about 1 nM, and the raman setup's not actually that hard since MG has a huge absorbance at 620 nm, right where HeNe lasers emit. Unfortunately, MG doesn't fluoresce much in solution or it would be very easy to measure.


I have 100 mg of a proteinase K lyophilized powder and I need to make it to a working concentration of 25 mg/mL So I know that I need to add $4~\mathrm{mL}$ of solution to this powder, which will give me a concentration of the enzyme at $25\ \mathrm{mg/mL}$. What I do not understand is how to decide how much of each of these "ingredients" (below) to add. I have $\ce{TrisHCl}$ and $\ce{CaCl2}$ in granular form ($M = 157\ \mathrm{g/mol}$ and $111\ \mathrm{g/mol}$, respectively) and 100 % glycerol solution. Will I have to make a large batch of Tris and $\ce{CaCl2}$ just so I can actually weigh out the needed amount?

The specifications for the working solution are:

$10\ \mathrm{mM}\ \ce{Tris}$
$1\ \mathrm{mM}\ \ce{CaCl2}$
30 % glycerol
Final $\mathrm{pH} \approx 8$ (use $\ce{HCl}$ if necessary)
MilliQ $\ce{H2O}$ to volume

Simply adding 4 mL of solution to 100 mg of enzyme is not an accurate way of preparing this solution. I am going to describe what to do from an analytical chemist's point of view. If you don't need this level of precision, you could do this following the same steps without volumetric glassware.

The first thing you need to do is decide what volume of your 25 mg/mL solution you need. Ideally, you will have a volumetric flask corresponding to this volume. For now, let's go with the 4 mL that you mentioned.

Next you will need to make a stock solution of your buffer. This will include all of the components except for the enzyme:

Decide what volume of buffer to make. You will need to make at least enough for your working solution, and again you will want to use a volume for which you have a volumetric flask. Also, you want to be weighing out reasonable amounts of your reagents (because weighing out <10 mg of something can really suck). Let's assume you're making 1 L. You're reagents are cheap anyway and the larger scale will make everything easier.
Calculate how much $\ce{TrisHCl}$ and $\ce{CaCl2}$ you will need to obtain the desired concentrations in 1 L of solvent.
Weight out the $\ce{TrisHCl}$ and $\ce{CaCl2}$ and add it to your volumetric flask.
Add 300 mL glycerol to your flask using a graduated cylinder. 30% glycerol probably refers to v/v.
Fill nearly to volume with water. You need everything to be in solution so you can adjust the pH, but you also need to leave some room to work with to do the actual adjustment. I'd shoot for approximately 80% full.
Adjust the pH by adding concentrated $\ce{HCl}$ dropwise.
Fill to volume with water and thoroughly mix the solution.

Now that you have your buffer solution, you just need to weigh out the appropriate amount of your enzyme and dissolve it in your buffer. From the example above, you would weigh out 100 mg of the enzyme (don't just assume you have 100 mg, even if the bottle says so) and dissolve it in your buffer using a 4 mL vol. flask.*

*
If you don't have a 4 mL vol. flask and you actually need all 100 mg of your enzyme in this solution, you will need to mess around with dilutions. Start by by creating a 50 mg/mL (2 mL vol. flask) or 100 mg/mL (1 mL vol. flask) solution.


How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide? For the reaction
  $$ \ce{2N2O5(g) -> 4NO2 + O2(g)} $$
  the rate law is:
  $$ \frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k[\ce{N2O5}] $$
  At $\pu{300 K}$, the half-life is $\pu{2.50E4 s}$ and the activation energy is $\pu{103.3 kJ/mol}$.
  What is the rate constant at $\pu{350 K}$?

I know there is something fishy about the rate law, but I can't make sense of it.

\begin{align}
\frac{ln2}{k} &= \pu{2.50E4}\\
k &= \pu{2.773E-5}\\
\frac{\pu{2.773E-5}}{k_2} &= \frac{A\cdot\exp\left\{\frac{-103300}{8.314\times300}\right\} }{ A\cdot\exp\left\{\frac{-103300}{8.314\times350}\right\}}\\
\end{align}

Finding $k_2$ from this gives me a weird value: $k_2 = 0.0103$.
The answer for this question is $\pu{7.47E-8 s^-1}$.

Where have I gone wrong?

When using Arrhenius equation: you have to multiply the activation energy by $1000$, because it must be in $\pu{J}$ and not $\pu{kJ}$.
you must also divide by $RT$ and not by $T$ as you did.


Water in the ultrasonication bath Why something jelly forms in the water if it is kept in in a ultrasonication bath for long periods?
 When water is exposed to ultrasonication for say a month, some jelly substances form in the ultrasonicator bath. is this something related to some bioorganism growth?

To see these jelly flakes in an ultrasound bath is a normal thing -- referring to ``it happens regardless if the water was tap-water or deionized water, a just a matter of time''.  Perhaps this occurs faster in the former, than in the later, especially if the bath is used to clean glassware.  But it equally happens with glassware judged as clean (by eye) for example if the bath is deployed to initiate / support a reaction inside a reaction flask.  Ultrasound-assisted generation of Grignard reagents is one example.

The ultrasound bath works by sonication, and the cavitation simply detaches the oily/fatty/grease surface film covering quickly all what was touched once by naked hand (e.g. sweat), and breaks this into even smaller pieces.  Over time -- when the ultrasound bath is switched off --, this thermodynamically unstable emulsion decomposes, splitting into a water soluble phase, and a hydrophobic phase that then tends to agglomerate.  (I did not check the presence of organisms living in or on them since discharging the bath is a chore regularly done, too.)

Side note:  For me, these residues appear to be more a liquid, then a solid -- hence I used the term emulsion.  If they were a solid, the thermodynamically unstable mixture would be an suspension.


Amino acids: why is charge found by comparing pH and pKa and not pH to pKa + log [A-]/log[H+]? I am working on problems such as :

The charges on the side groups at each pH are found by the following rule:

at $\ce{pH < pK_a}$, $\ce{H+}$ on, protonated
at $\ce{pH > pK_a}$, $\ce{H+}$ off, deprotonated
at $\ce{pH = pK_a}$ , neutral

But why can we compare $\mathrm{p}H$ and $\mathrm{p}\ce{Ka}$ directly?
I just started my amino acids topic. In all my acid and base questions we always compared $\mathrm{p}H$ to $\mathrm{p}K_a$ using the Henderson Hasselbalch Equation:
$$\mathrm{p}H =\mathrm{p}K_a +\log \left( \frac{\ce[A^-]}{\ce{[HA]}} \right)$$
So why do we now compare $\mathrm{p}H$ to $\mathrm{p}K_a$  directly?

The properties you listed are correct, and all of them follow from the relationship described by the Henderson-Hasselbalch equation.
The equation shows a close relationship between $\mathrm{p}K_a$  and $\mathrm{p}H$. Note that $\ce{log ([A^{-}]/[HA])}$ equals zero when $\ce{[A^{-}]/[HA]}$ equals 1, and when $\ce{log ([A^{-}]/[HA])=0}$, $\mathrm{p}K_a=\mathrm{p}H$. This means that at the equivalence point, $\mathrm{p}H$ equals $\mathrm{p}K_a$.


Can sodium hypochlorite soak all the way through an unglazed ceramic cup and bleach blue paper wrapped around the outside of the cup? My question is regarding the ionic properties of sodium hypochlorite in house hold bleach. There is a debate among potters whether bleach will kill mold growing on a porous ceramic cup. It is my understanding that sodium hypochlorite cannot penetrate a porous surface of fired clay. Someone did an experiment where they put household bleach in an unglazed cup and allowed it to soak through. There was blue construction paper wrapped around the outside of the cup and the paper turned white after a while seemingly from bleaching. So my question is this: If the ionic properties of sodium hypochlorite will not allow it to penetrate through the cup to the outside, then how did the blue paper wrapped around the outside of the cup get bleached?

It is not the ionic properties that control the diffusion. Diffusion is a fundamental property of molecules dissolved in a liquid. A bleach solution stored in an unglazed container can potentially diffuse to the wall. The pore size of the material will mainly control the rate of movement of the fluid (viscosity, surface tension etc.) If the pore size and the connectivity channels are very small it will take ages to wet the wall under atmospheric pressure.


Acid catalyzed hydrolysis of ester During acid catalyzed hydrolysis of ester why does the carbonyl oxygen attack the $\ce{H^{+}}$ ? Why doesn't the alkoxy oxygen which is more electron rich and less electronegative not attack the $\ce{H^{+}}$ with its lone pair? Any ideas?

The difference between the carbonyl oxygen (red in colour) and alkoxy oxygen (blue in colour) is that the highly distorted pi-electron density (green shaded) towards to the carbonyl oxygen, due to more electro-negativity of oxygen that is the feature which is more on this oxygen than alkoxy oxygen . That feature provide the cavity for the H+ ion from acid, and the distortion of sigma- electron density at alkoxy oxygen is comparatively less than the distortion of  pi-electron density at carbonyl oxygen.


Solubility curve for sodium tetraborate in water What is the solubility curve for sodium tetraborate decahydrate ($\ce{Na2[B4O5(OH)4]·8H2O}$), otherwise known as borax, in water ($\ce{H2O}$)?

Additionally, what is the solubility of the aforementioned substance, borax, at $\ce{100°C}$?

The solubility curve of borax is taken from here.

We can infer that solubility of borax at 100°C is around 520g/L. It is observed that borax in water exhibits supersaturation . When the solution is at 100°C and then you drop at 40°C, then a clear solution is obtained when amount of borax is 60 grams per litre. More than that, there is a tendency of the solution to overflow. See ref.1 for the more detailed explanation of supersaturation and ref.2 for the dissociation reactions happening in water. First the tetraborate ion is formed and it is a weak base in solution and thus reacts with additional water to form boric acid, a weak acid and hydroxide ion.

$$\ce{[Na2B4O5(OH)4].8H2O <=> 2Na+ + B4O5(OH)4^{2-} + 8H2O}$$

$$\ce{B4O5(OH)4^{2-} + 5H2O <=> 4H3BO3 + 2H2O}$$

Reference

https://www.enotes.com/homework-help/how-much-borax-should-dissolve-hot-water-order-721345
https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://www.smc.edu/AcademicPrograms/PhysicalSciences/Documents/Chemistry_12_Experiments/Solubility%2520and%2520Borax.pdf&ved=2ahUKEwjr5b6W1tXdAhWRbisKHb1rBUAQFjAXegQIBBAB&usg=AOvVaw3c9IA0jqEROPS534lSedA8


What is the Major product in free radical halogenation of Alkanes. (Between 3°,2°,1°) In Free Radical Halogenation in presence of sunlight, a no. Of products are possible.

Due to higher stability of a 3°(Tertiary) radical, 3° products should be favored. But another factor that comes is that the formation of 3 ° radicals also leads to higher disproportion products. (Radical Disproportion- https://en.m.wikipedia.org/wiki/Radical_disproportionation)

Now, if i look at the intermediate, a higher no. Of Tertiary Radicals should form. But a lot of them will undergo radical disproportion as it is fastest in tertiary radical.

So is there any clear order of what product should be major or does it vary from case to case?

It depends on what halogen you use.

Chlorine has a reactivity ratio for $1^o$:$2^o$:$3^o$= 1 : 3.5 : 5.

To calculate the relative yield at that position, multiply the number of Hydrogens at that position to the reactivity at that position. This will give you a comparison of the yield at particular positions.

But in the case of bromine, the reactivity order is 1 : 250 : 6300. Therefore, without any further thinking, the bromine radical will always attack at the tertiary position if available.

In general, reactivity towards tertiary carbon increases down the group as in the case of halogens.


Calculating volumes of acid/base required to make buffer using pH and pKa How would we use our $\mathrm{p}K_{\mathrm{a}}$ of the acid component of the buffer to calculate $\large \frac{[\ce{A-}]}{[\ce{HA}]}$?

I know $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$$ rearranging gives me $$\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}$$
if for example the $\mathrm{pH}=4.5$, $\mathrm{p}K_{\mathrm{a}}=3.74$,

Is it correct to say
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{19}{250}$$
since
$$
\begin{align}
\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}F\\
\implies\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&=4.5- 3.74\\
\implies\frac{[\ce{A-}]}{[\ce{HA}]}&=\frac{\operatorname{e}^{0.76}}{10}
\end{align}
$$
if $\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{V_\text{b}}{V_\text{a}}= \dfrac{19}{250}$
how would I use this to find the volumes of acid and base needed to make my buffer?

First of all, note that:

$$\log_{10}\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)=4.5- 3.74$$
$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)\neq\dfrac{e^{0.76}}{10}$$

but

$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)= 10^{0.76}$$

This said, the volume will depend on the concentration of the acid and of the salt (if it's a solution) that you are provided with. This ratio will give you the relative concentrations required to get a buffer of the pH 4.5.


Graphitic acid-Its structure and properties? What exactly is graphitic acid? I know that it's obtained when you react graphite with concentrated $\ce{HNO3}$ and that its structural formula is $\ce{C11H4O5}$. But I can't find or deduce its structure. And also I am unable to find any of its properties, except that it's a yellow-green liquid insoluble in water.

Graphitic acid is a compound derived from graphite.

You may find it named "Graphite oxide" as well: as you see, it is obtained by reacting graphite with an oxidizing agent ($\ce{HNO3}$, for instance).

The structure resembles that of graphite, but it carries several oxygen-containing substituents, such as hydroxyl and carbonyl functions, thus having a less regular geometry, compared to that of graphite.

The exact features of the structure depend a lot on the method used for its preparation, and the layers may show some "defects", not described by this ideally planar model, but the key concept is this.

Here is a simple proposed structure for a single layer (source: Wikipedia)

Note that, just like graphite is made of several layers of graphene, graphite oxide is made of several layers that, upon reaction with a base, become graphene oxide.

Here you can find further information, and a nice image of a single graphene oxide film


Terminology: what's the difference between "monomer" and "protomer"? Using definitions from Wikipedia, a protomer is defined as:

the structural unit of an oligomeric protein. It is the smallest unit composed of at least two different protein chains that form a larger heterooligomer by association of two or more copies of this unit.

In terms of proteins, the term monomer may refer to the amino acid components. However, an oligomeric complex with one subunit can be called a monomer (according to the page on protein quarternary structure).
Are these two terms interchangeable?

Lacking an answer for 3 years, I'm going to attempt to answer this based on internet definitions and logic. Please, I welcome any insight from chemistry experts to improve this answer.

First of all, the Wikipedia link quoted actually gives two definitions for protomer. The one quoted is the "structural biology" definition rather than the "chemistry" definition. On the other hand, it seems as though monomer has one definition according to the Monomer Wikipedia article. However, as pointed out by the OP, another point of confusion arises in the definition of a monomer as an oligomer with just one subunit in protein quarternary structure. So we actually have two definitions of each term out there. Let's recap them and then compare.

Protomer (structural biology): In this case, the term protomer is reserved for the specific case of being part of a hetero-oligomer. The "hetero" part is important. The point is that the oligomer is made of a number of different macromolecules- glued together non-covalently. In this case, a protomer is defined as the smallest repeatable unit within this hetero oligomer, which may be the entire oligomer itself (in which case, confusingly, the oligomer is called a monomer). The term protomer in this case, as pointed out by Wikipedia, serves the sole purpose of disambiguation. When we say "blah-mer", as pointed out by Hanry, we are saying "this thing has blah parts". But what are "parts"? Are they quarks? Electrons? Protons? Atoms? Macro molecules? Or... repeatable sub-units. It seems as though the confusion was between the last two. You can break a hetero-oligomer into macromolecules and you'll get a bunch of different macromolecules. You might get 12 As, 12 Bs and 12Cs. So is this a 36-mer or is it a 12-mer? It depends on what you mean by "mer". If each "mer" is an ABC subunit, then it's a 12-mer. If each macromolecule is a subunit, then it's a 36-mer. The term protomer helps us here. The molecule can be viewed as a 12-protomer, meaning that it's 12 copies of the smallest repeatable subunit: ABC.
Protomer (chemistry): In this case, it looks like the term means something quite different. It looks like a protomer is defined as a specific type of tautomer, one in which the tautomerism arises as a result of variable proton position.
Monomer (standard): A monomer is a single molecule that forms part of a polymer. The polymer can either be a homo-polymer, consisting of repeated copies of that monomer, or it can be a copolymer (analogous to a hetero-oligomer), consisting of different monomers, according to Wikipedia's Polymerization article. The key here is that the monomer belongs to a polymer- meaning "many parts". Typically polymers are distinguished by the fact that they can theoretically continue forever in big long chains.
Monomer (oligomer special case): This is just an unfortunate use of terminology, but it kind of makes sense when you think about it. "Oligomer" means "Few parts". Typically the distinction between oligomers and polymers is that polymers can go on forever- they could keep growing through repeated polymerisation, but an oligomer will have a set number of parts, and then it's fully grown. Because oligomers can have a set number of parts, people decided to further sub-classify them by the exact number of parts. So a hexamer is an oligomer with six parts, for example. Following this naming convention, we have pentamer for 5 parts, tetramer for 4 parts... and all the way down to monomer for one part. Which just unfortunately intersects with the other definition of monomer, being a single part of a polymer chain.

So, now that it's been established that there are different definitions even of each term itself, let's try to compare them:

Protomer (structural biology) vs. monomer (standard): a protomer here is defined as a sub-unit of an oligomer, whereas a monomer is a building block of a polymer.
Protomer (structural biology) vs. monomer (oligomer special case): in this case, we need to think. A monomeric oligomer is one which has one "sub-unit". If by "sub-unit" what is means is protomer, then in this case the monomer, protomer and oligomer all coincide and are describing the same thing. If, though, "sub-unit" means the smallest sub-molecules that are non-covalently bonded together, then the oligomer is, trivially, a homogeneous oligomer and not a heterogeneous one: it consists of just one molecule so all the molecules are the same. And so, strictly speaking according to the above definition, the term protomer doesn't apply here.
Protomer (chemistry) vs. Monomer (standard): these two definitions are not mutually exclusive. By describing a compound as a protomer, in this way, we are just saying that there are different tautomeric configurations of this compound out there. On the other hand, a monomer is just something that can form a polymer chain. It's perfectly reasonable to assume that some tautomer could also be capable of undergoing polymerization, and therefore, would be a monomer. I don't know any examples but maybe someone reading can offer them.
Protomer (chemistry) vs. monomer (special oligomer): The monomer definition here seems to be a bit vague, but in the loosest sense, every single molecule is a monomer. But it seems as though what's meant is that a monomer, in the oligomeric sense, typically doesn't undergo polymerisation with other monomers to form a polymer - it just likes being alone. Still, even with that definition, it's likely that there are examples of things that are protomers, in the chemistry sense i.e. being tautomeric in terms of proton position, and that are also monomers i.e. oligomers that won't bond with each other any further.


Can inductive effect be the sole driving force for carbocation rearrangement? I've been taught that a carbocation mainly rearranges because of:

increasing degree (1 to 2, 1 to 3, or 2 to 3)
+M stabilization
ring expansion (in an exceptional case, ring contraction as well).

However, I've never been taught whether the following carbocation A would rearrange:

My professor would say that this carbocation would not rearrange, since at the alpha position we only have two hydrogen atoms, and either of the hydride shifts would still yield a two degree carbocation B:

However I believe that B is stabilized by the strong inductive effect of three extra methyl groups (at the alpha position) that were absent in A. I believe this is a strong enough driving force for rearrangement.

So, can we say A will rearrange into B? Is there experimental evidence to support the fact that A rearranges/does not rearrange into B? Finally, based on these, can inductive effect be the sole driving force for carbocation rearrangement?

The answer to this question is quite difficult to deduce logically. In the first place, the carbocation that is formed is secondary. Rearranging will only change the position of the carbocation, but it will be still secondary. But then there is also a stabilising +I effect.

Instead, if we take into account that that these carbocations are intermediates to form alkenes, a conclusion can be drawn. So, the best way is to consider the thermodynamic favour of the rearrangements.

Prior to rearrangement, the carbocation is secondary. If you consider that this carbocation is going to form an alkene, the major product from it (Saytzeff product) will be  4-methyl pent-2-ene, which is of the form $\ce{RCH=CHR}$. After rearrangement, this will be still a secondary carbocation, so there is a significant energy required for this hydride shift. This energy is not really compensated by the +I effect of the isopropyl group. But if we consider that this carbocation is an intermediate for the alkene formation, the major alkene (Saytzeff product) will be 2-methyl pent-2-ene, which is of the form $\ce{R2C=CHR}$, which is thermodynamically more favourable.

So it is seen that in the rearrangement, the energy of the intermediate is increased, but the overall energy of the product is decreased. So, I think that, at higher temperatures, this rearrangement can be possible, because the ultimate product (if we think this as an alkene) is thermodynamically favourable, while at normal temperature this rearrangement is energetically restricted a bit.


If an action increases the reaction rate in a particular equilibrium mixture, does it mean that both forward and backward rates are increased? Say for example we increase the temperature in a closed system containing gaseous reactants. The forward reaction is exothermic so the yield of the product will decrease as we increase the temperature. But increasing the temperature increases the reaction rate so does it mean that the forward and backward rates are both increased ?

The forward and reverse reaction rates both increase, but the reverse rate increases by more. By Le Chatelier's principle, we know that $K_{eq}$ for an exothermic reaction will decrease when the temperature is raised. We also know that $K_{\text{eq}}=\frac{k_{\text{forward}}}{k_{\text{reverse}}}$ and that $k=Ae^{\frac{-E_a}{RT}}$ (The  Arrhenius equation). Since this is an exothermic reaction, $E_a(\text{forward})<E_a(\text{reverse})$, which means any increase in $T$ will increase $k_{\text{forward}}$, but it will increase $k_{\text{reverse}}$ by more. This lead to a decrease in $K_{\text{eq}}$, consistent with what was predicted by Le Chatelier's principle.


Butane blowtorch condensing liquid on heating surface I have a small Butane blowtorch, and use it to heat an Aluminium plate. The plate is thick so takes 20 seconds or so to really start heating up.

As soon as I apply the heat, droplets of some liquid start "condensing" on the surface of the metal around the general area of the flame.

The liquid is colorless and odorless and is at room temperature, note this happens when the metal is still at room temperature as it hasn't heated up much yet.

What is this liquid exactly and why/how does it form? I suspect it could be a $C_{4}H_{10} + O_{2}$ reactant but not sure.  Thanks!

This is a guess - but I think it might be water formed as a byproduct of butane combustion precipitating on the plate:

$$
\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)}
$$

Aluminum has very high thermal conductivity, so it is possible that the plate is rapidly absorbing heat from the water vapor, causing it to condense. If this stops happening right around the point when the aluminum temperature reaches 100 C, that would be evidence to support this hypothesis.


How does hydrogen gas build up in hot water lines? Recently I was reading through the owner's manual of a washing machine; notably, the warnings in the front of the booklet. One warning stood out to me; it said that if the hot water in your house has not been run in a few weeks $\ce{H2}$ can build up in your lines. The warning of course stated that this can only occur under certain circumstances so I'm assuming it is quite rare.

Okay, so my question is, under what conditions does hydrogen gas build up in a hot water line?

I think that it is probably due to a reaction between an acid and a metal. These reactions can lead to the formation of hydrogen following this generic forumula:
$$\ce{M + 2H+ -> M^{2+} + H2}$$

For example iron could be a source of electrons. If you look at the corrosion of iron: $$\ce{Fe->Fe^{2+} + 2e^{-}}$$

At low pH (notice that there is a correlation between pH and temperature) hydrogen gas is formed:
$$\ce{2H^{+} +2e- -> 2H2}$$


What is the effect of oxygen on soap foam? I've been wondering the effect of $\ce{O2}$ on soap foam in a closed, test room. The room contains a candle and soap foam.

That is to say: if the room has lower and decreasing $\ce{O2}$ in its air, does the foam is effected (i.e., the soap produces lower foams)?

PS: I have very basic knowledge in Chemistry, and, excuse me if I didn't formalize the question in a good manner.

Hmmm. I don't know, but I'd say that if you significantly lowered the percentage of $\rm O_2$ in the air, the average molecular weight of the molecules in the bubble would drop (there'd be more low-molecular weight nitrogen and less higher-molecular weight oxygen), so the rate at which gas would diffuse out of the bubble would increase, and the foam would flatten faster.

Try blowing a foam with nitrogen, and compare it with the results of blowing a foam with air or oxygen---if I'm right, the foam blown with nitrogen will flatten a bit faster.

There'll be a bit more $\rm CO_2$ in the air from the candle, and that would diffuse more slowly out of the bubble (in fact, $\rm CO_2$ is used as a "blowing agent" to make foams, partially for that reason). But I'd think in this case that the small amount of $\rm CO_2$ produced by the candle would dissolve in the liquid, with gas dissolution rates enhanced by the lower surface tension and higher surface area of the foam.


Internal energy change during phase change Ice at $\mathrm{0\ ^\circ C}$ is converted to water at $\mathrm{0\ ^\circ C}$. If $\Delta H$ for the transition of ice to water is $\mathrm{1440~cal}$, calculate the change in internal energy.

Since internal energy $\Delta U=nC_V\Delta T$ and $\Delta T=0$, shouldn't $\Delta U=0$?

But if I use $
\Delta H=\Delta U +\Delta nRT$, I get $\Delta U=\Delta H\neq 0$.

You are considering the $\Delta H$ and the $\Delta U$ between the following two thermodynamic equilibrium states:

State 1:  1 mole of ice at 0 C and 1 atm.

State 2:  1 mole of liquid water at 0 C and 1 atm.

The relationship between $\Delta H$ and $\Delta U$ at constant pressure is:
$$\Delta H=\Delta U + p\Delta V$$where V is molar volume.  What is the molar volume of ice?  What is the molar volume of liquid water?  What is $\Delta V$?


What Type of Gas Would Vinegar and Epsom Salt Create Under Electrolysis? I'm restoring a motorcycle and have a small homemade DIY zinc plating operation which consists of vinegar, epsom salt, a piece of zinc and a power supply.

When I plate a steel piece such as a bolt or small hardware piece, the piece in the solution starts forming small bubbles and it appears to be "gassing off" something. I was wondering what type of gas this may be and if it's harmful to breathe.

You're most likely forming $\ce{H2}$ gas at the part.

If you're plating anywhere above about $2~\mathrm V$, that is probably sufficient overpotential to cause part of the passed current to go toward electrolysis of water.

There is negligible toxicity hazard to inhaling the small amounts of hydrogen that would be generated from an electrolytic cell on this scale.

There's a small ignition hazard if anything sparks near electrolyte of the plating cell as it's operating, but as long as you're not running it inside a tightly closed, small space you should have nothing to worry about with how small the cell is.


Examples of evolving networks in chemistry I'm a master student working on networks analysis in general. A network is something that has nodes and there are links between the nodes. Nodes and links could have attributes. An evolving network is one that changes overtime (new nodes and links are added..etc). An example of that is Facebook. Nodes are users and links represent the friendship relationship. Users have attributes (gender, age ..etc). A Facebook network as you know is an example of a social network.

The issue is that so many people studied traditional evolving networks like social networks, the web, or transportation networks. Currently I'm looking for novel examples of evolving networks to study them. So I thought there might be some examples in chemistry that could represent some kind of an evolving network.

So my question: Can you give me examples in chemistry for evolving networks?

Synthetic Pathways

... is the only example I can think of right now.

As far back as 1971¹ people have been positing the possibility of an AI to search for novel synthesis paths, but only a couple of years ago a decent one was developed² which could interpret literature and create novel sets of steps for particular reactions -- for example, given a particular reactant and product, produce a set of steps which would produce one from the other. The particular focus of the paper for the 2012 one was to produce single-pot (multi-step) procedures to replace awkward separated multi-step syntheses, but in principle the procedure is generalisable to any path.

Sridharan, N.S., An Application of Artificial Intelligence to Organic Chemical Synthesis, Doctoral Thesis, State University of New York at Stony Brook, New York, July 1971.
C M Gothard et al, Angew. Chem., Int. Ed., 2012, 51, 7922 (DOI: 10.1002/anie.201202155) but there's a good write-up by Philip Ball in Chemistry World, August 2012 issue.


Percent degree of dissociation of gas by pressure I stumbled across the following interesting problem:

Assume that you have started to live on a new planet where standard pressure condition is $\pu{2 bar}$, standard concentration is $\pu{1 M},$ and all types of gases behave as an ideal gas. On this planet, you are asked to determine equilibrium conditions for the reaction below:
$$\ce{XY4(g) <=> X(s) + 2 Y2(g)}$$
$K^\circ = \pu{1.1455E-9}$ (at $\pu{298 K}$).
Calculate the percent degree of dissociation for $\ce{XY4}$ at $\pu{298 K}$ where total pressure is $\pu{0.2 bar}.$

Since the equilibrium constant is very small, I approximated that the pressure of $\ce{Y2}$ at equilibrium is insignificant. Therefore I got:
$$p(\ce{Y2}) + p(\ce{XY4}) = \pu{0.2 bar}$$
$$p(\ce{XY4}) = \pu{0.2 bar}$$
I don't know how to proceed from here though and how to calculate the degree of dissociation.

If $\alpha$ is the degree of dissociation, at equilibrium $1-\alpha$ of XY is present and $\alpha$ of X and $2\alpha$ of Y making a total of $1+2\alpha$. The partial pressure of XY is $\displaystyle p_{XY}=\frac{1-\alpha}{1+2\alpha}\frac{P}{P^\mathrm{o}}$. The $P^\mathrm{o}$ is the standard pressure used to make the equilibrium constant dimensionless. In our world this is 1 bar, in your new world it is 2 bar, thus $P/P^\mathrm{o}=0.1$. If you calculate the partial pressures for the other species in terms of $\alpha$ and form the equilibrium constant then $\displaystyle K=\frac{4\alpha^3}{(1+2\alpha)^2(1-\alpha)}\left(\frac{P}{P^\mathrm{o}}\right)^2$.(please check).  As $K$ is small we will assume that $\alpha$ is also then $\displaystyle K=4\alpha^3\left(\frac{P}{P^\mathrm{o}}\right)^2$ which produces $\alpha \approx 3.10^{-3}$.


Masamune's rule of diastereoselectivity multiplicity explained In a reaction between a chiral substrate and chiral reagent, their asymmetric inductions could cooperate (matched pairs) or compete (mismatched pair). According to Masamune, in a matched pair the degree of asymmetric induction is approximately the product of the selectivities of the substrate and the reagent. In contrast, the asymmetric induction in a mismatched pair is equal to the quotient of the selectivities of the substrate and the reagent.

The following reaction perfectly proves Masamune's rule (syn enolate is produced during enolization):

Why is this rule valid and why is it so accurate? What's the mechanistic explanation behind the multiplicity trend?

First, we need to take a look at the possible transition states of the first reaction in your table using $\ce{9{-}BBN}$ as an enolising agent. There are two transition states shown in scheme 1: The upper, favoured and the lower, disfavoured one.

Scheme 1: Aldol reaction with $\ce{9{-}BBN}$ as an enolising agent. Both the favoured (above) and the disfavoured (below) Zimmermann-Traxler transition states are shown along with their respective products.

The difference in both transition states is in the steric clash of the α-carbon and the boron residue: If the aldehyde attacks from the si-side, the stereocentres on said α-carbon can be oriented favourably, namely with hydrogen pointing toward the most crowded area and the largest residue pointing toward the least crowded area. If the aldehyde attacks the enolate’s re-side, methyl must point towards the most crowded area to allow the largest residue to point away from crowded areas. A similar discussion can be made for the orientation of the aldehyde which could attack the enolate’s si-side with its si-side, resulting in an anti-aldol product.

It is important to note here that all four attacks are possible. However, they do not have the same probability of reacting. Based on the observed diastereoselectivity, one can say that $\mathrm{P}_R(F) \approx 92~\%$, while $\mathrm{P}_R(\bar F) \approx 8~\%$ ($R$ is the set reaction occurs and $F$ is the set favoured attack. I have subsumed the two anti-product giving attacks under $\bar F$, since they are nigh irrelevant).

Scheme 2: Aldol reaction with $\ce{(+){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s re-side. This is the same as the substrate’s inherent diastereoselectivity; a matched case.

A similar case can be made for $\ce{DIP}$-mediated aldol additions in the absence of substrate diastereospecificity ($\ce{Me} = \ce{H}$). Here, we need to construct a new set $A$ meaning auxiliary-favoured attack. This auxiliary-favoured attack is shown in scheme 2, although I used the matched case from your examples, i.e. substrate diastereoselectivity is present. The auxiliary-unfavoured attack (same caveat) is shown in scheme 3.

Scheme 3: Aldol reaction with $\ce{(-){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s si-side — but substrate diastereoselectivity prefers the si-side attack on the enolate (mismatched case).

Stochastically, we can consider both selectivities separately. Remember that all attack combinations are technically possible but some are just more likely to be productive. It is a good approximation to assume that $\mathrm{P}_R(A)$ and $\mathrm{P}_R(F)$ are independent of one another, which means that $\mathrm{P}_{R,A}(F) = \mathrm{P}_R(A) \times \mathrm{P}_R (F)$. Why? Well, the substrate’s diastereoselectivity is mainly based on the orientation of the residues on the α-carbon. This is pretty much independent of the orientation of $\ce{Ipc}$’s methyl groups, which cause the auxiliary’s enantioselectivity: in both methyl group orientations the same rotamer of the α-carbon has the same probability.

Having taken into account that both sets are independent, we see that we can multiply the probabilities of the different attacks to get the resulting attack’s probability. The substrate will favour (R,S) (counting from ketone to hydroxy group) in each case. To get the probability of (R,S), we multiply a large percentage (substrate selectivity) with a large percentage getting another large percentage (auxiliary selectivity). The probability for (S,R) will be a small percentage (substrate) multiplied by a small percentage (auxiliary) so even smaller. So favouring of the desired (matched) diastereomer is enhanced.

On the other hand, if the auxiliary favours (S,R), to get the probability of (R,S) we must multiply a large percantage with a small one giving a medium result. The same thing for the probability of (R,S). This means that both diastereomers have more equal chances of forming.

(A better discussion of this problem would involve relative reaction rates, as the article NotWoodward mentions does. I feel that the probability argument gets us the same conclusion without all the big kinetic calculations which I am not good at.)


How do you name a bicyclo compound with a fused benzene ring? What is the IUPAC rule for naming a bicyclo compound with a fused benzene ring? Are you required to state the number of double bonds by adding something like 2,4,6-ene to the name or do benzene rings have a specific name ?
For example:

What will this compound be named? Will it be bicyclo[4.3.0]non 1,2,4 triene or is there some specific rule for fused benzene rings, as I stated before?

Beside systematic nomenclature about organic structures, there are other systems in use.  One of them is the by Hantzsch and Widman which departs from a number of stem structures.  Some of these archetypical motifs are named systematically, others (still) retain their traditional name which does say little  to you about the structure (e.g., pyridine) until you learn them.  One (incomplete) compilation of these names, for example, is this one.
This being said, you either look up, or over time (i.e., training) recognize that the parent structure of your molecule is indene:

except that the double bond in the five-membered ring of the structure you showed is completely saturated.
You remember reactions on benzene, and recall about the importance to denote the sites in organic structures.  The approach to use numbers equally exists for larger compounds, too; a few positions are labeled.  While there are cases where you may shorten 1H-indene to indene, this no longer is the case to describe unambiguously the compound shown by you, which is 2,3-dihydro-1H-indene.


How to judge if a reaction is exothermic or endothermic Mix $\ce{HCl}$ and $\ce{NaOH}$ together (reactants are at the same temperature) in a calorimeter. The temperature increases, and the book says it's an exothermic reaction. I don't get it. The thermometer is in the solution which means it measures the temperature of the solution. It's the system. The temperature of the system increases, this means that heat flows in. How could this be an exothermic reaction?

This is one of the most commonly misunderstood principles in thermochemistry - if you can wrap your mind around it, you are most of the way there!

The good news is, there is an easy way to do that. In any physics problem (thermochemistry is really physics) you have to be very careful with how you define the system and the surroundings. This is important for two reasons.

It lets you write energy balance equations across the system boundaries (so we can use math to solve the problem)
By choosing the right system boundaries, we can make the math much, much easier.

Since thermochemistry at this level always involves the same types of problems, chemists have done the hard part for you and already selected the best system boundaries.

The system is whatever is in the reaction equation.

That's it. The entire rest of the universe is the "surroundings."

Looking back at your question:

  The thermometer is in the solution which means it measures the temp of the solution. It's the system. The temp of a system increases means heat flows in, how could this be an exothermic reaction?

If the reaction molecules are the only thing in the system, then that means the water, the calorimeter, the themometer, you, me, the room - everything else - counts as surroundings. Now, the water did heat up. We know this because the thermometer absorbed some heat and showed a temperature increase. Since they are all part of the surroundings, the heat must have come from the system. Since heat left the system, the sign of $q_{reaction}$ is negative, and the reaction is exothermic.

As a teacher, I have found that getting system boundaries and the signs right is one of the biggest problems students have. If you focus on getting that concept down early - and really understanding why they are that way, you are well on your way.


Why are antiaromatic compounds unstable? Antiaromaticity is so destabilizing that it can cause compounds such as cyclobutadiene to elongate or manipulate their orbitals so that the pi system is no longer aromatic. I understand that aromatic compounds are stabilized because they have a full HOMO and that antiaromatic compounds have a half filled HOMO, which according to this question is the reason that they are unstable. I don't agree with this explanation because in other compounds, half filled orbitals cause compounds to be stabilized. For example, triplet oxygen is significantly more common than singlet oxygen in nature. Also, some transition metals will promote electrons from their S orbitals to D orbitals so that they can have these shells half filled. I know that full orbitals are more stable than half filled orbitals, but are half filled orbitals destabilizing in alkenes while they are stabilizing in oxygen and metals? If not, why are antiaromatic compounds so unstable?

Antiaromatic compounds are not necessarily unstable - they are just less stable than a bunch of ethenes connected by sigma bonds. If the pi electrons in an antiaromatic compound were to delocalize across a large degenerate orbital, then the result would be a half filled shell. While a half filled shell is not the end of the world, it is a lot more stable to have a full shell of electrons. As a result, compounds such as cyclobutadiene undergo a Jahn Teller distortion (in this case, becoming a rectangle rather than a square) in order to form two separate (and non degenerate) orbitals. This creates a collection of separate, full, pi orbitals. The stability gained by having full rather than half full orbitals is more than enough to compensate for the energy required to break the symmetry in the first place. Cyclobutadiene is stabilized due to stretching not because it prevents antiaromaticity, but because it causes all the pi electrons to be part of complete shells. While it might seem weird that a physical distortion such as stretching can consume less energy than is released by going from a half to a full shell, let me remind you that this is not the case. Higher energy light (UV) is often required to stimulate electronic transitions while lower energy light (IR) can cause bond stretching in compounds.

I was able to figure this out thanks to the help of orthocresol.


Determination of metal state from protein X-ray structure Many proteins or enzymes contain metal ions within their active sites.
These metals are often directly involved in the reaction catalysed bu the enzyme. In certain cases the function is well known and understood, including the identity of the metal and the changes in oxidation state throughout the reaction.
I wonder how the metal and oxidation state is determined a priori when a new structure is obtained.  For examples if I solved a structure of a previously unknown enzyme by crystallography I may be able to determine that it contains Zinc (iron a better example), but knowing the density map tells me nothing about the oxidation state which may be important to function.
I think it must be possible to determine this, however I do not find much reliable information (inorganic complexes typically have other techniques in addition to X-ray). The situation is made harder as many protin structures in the pdb appear to be assigned without any real study.

X-ray diffraction experiments, regardless if about small molecules or proteins / RNA / DNA, are usually one element characterizing a sample.  On occasion, you may infer the oxidation state from the transition metals environment: how many closest neighbors, their distance, their number, their spatial arrangement.
You are free to complement findings by any other technique either using the same crystal sample as mounted on the diffractometer (e.g., UV-Vis), or with sample material from the same batch characterized at the same temperature as the crystal mounted on the diffractometer: spin characterization by EPR, Moessbauer, NMR (your are not constrained to CHN) and SQUID  to mention a selection. Because PubMed includes records about the structure characterization of large biomolecules, this database may serve as a source of inspiration how this was done, e.g., for heme


How can I know the colour change of this reaction? After you react rotundone ($\ce{C_{15}H_{22}O}$) with water it loses all of the double bonds (except the carboxylic bond, $\ce{C=O}$) and gets two of the functional group $\ce{-OH}$ (alcohol). Two compounds are formed, one (A) where both the alcohol groups are tertiary and another (B) where one is tertiary and the other is primary.

I'm asked to state and explain the colour change of both A and B when they react separately with acidified potassium dichromate(VI) solution.

Now I can see that compound A would have no colour change because the $\ce{-OH}$ groups are tertiary so the potassium dichromate won't react or oxidise them.
I can also see that there will be a colour change on compound B because one of the $\ce{-OH}$ groups is primary, so the potassium dichromate will oxidise and react with it.

Here's my problem:

I'm expected to know that when compound B reacts, it changes from orange to green. How would I know that rotundone is orange in the first place? And what makes it go green?
I know that I'm expected to use intuition as we definitely aren't required to know the colour of rotundone.

Potassium dichromate (VI) is the species that turns from orange and to green when reduced. Most of the organic substances I came across for testing purposes are colourless, so I would assume that compound B would be as well.

It is just like you are expected to know the colour change of potassium manganate (VII) from purple to colourless when it is reduced.

Those two oxidants were used frequently in my labs, and I'm pretty sure you'll be using them often as well.

I hope this answers your question! :)