Software/server for calculation of protonation states of small molecules depending on pH For docking calculations at different values of pH, I am working with flavonoids and would like to know about any software/server that can prepare ligands depending on the pH. Up to now I read about Maestro (Schrodinger) but it is too expensive for us. I wonder if you could suggest other alternatives.

This is a challenging problem and there are many programs that do this.

One review was published a few years ago: "Comparison of Nine Programs Predicting pKa Values of Pharmaceutical Substances", J. Chem. Inf. Model., 2009, 49 (12), pp 2801–2812

  Knowledge of the possible ionization states of a pharmaceutical substance, embodied in the pKa values [...] is vital [...] We compare nine commercially available or free programs for predicting ionization constants. [...] We compared their performances by applying them to 197 pharmaceutical substances with 261 carefully determined and highly reliable experimental pKa values from a literature source. The programs ADME Boxes 4.9, ACD/pKa DB 12.0, and SPARC 4.2 ranked as the top three with mean absolute deviations of 0.389, 0.478, and 0.651 and $r^2$ values of 0.944, 0.908, and 0.894, respectively. ACD/pKa DB 12.0 predicted all sites, whereas ADME Boxes 4.9 and SPARC 4.2 failed to predict 5 and 18 sites, respectively.

Surprisingly, the quantum program did poorly.

There's even a full book on the subject: Computational Approaches for the Prediction of pKa Values by George Shields and Paul Seybold. It covers both quantum and QSAR approaches.

Some of the best seem to be:

ACD/pKa
ADMET Predictor
Propka Also available as open source code: https://github.com/jensengroup/propka-3.1
Marvin / ChemAxon

But there are dozens out there. I'd check the review articles and try a few.


Is the SN1 reaction faster with an axial or equatorial leaving group? Why does compound 1 undergo the $\mathrm{S_N1}$ reaction faster than 2 even though both proceed via the same carbocation intermediate?

In compund 2, both substituents can be placed in equatorial positions, whereas in 1 the $\ce{Cl}$ group is forced into an axial position since the bulky t-butyl group has to be placed equatorial. This makes compound 1 more unstable (higher energy), leading to a faster rate of formation of the carbocation.


Does the order of protonation/deprotonation in the formation of hemiacetals matter? In the mechanism depicted above, deprotonation occurs first and then protonation.
Can this order be changed? If not, why?

The first (productive) step of this mechanism in acidic medium MUST be protonation of the carbonyl.  Alkoxides and hemiacetal-alkoxides have no appreciable concentration in acidic solutions, and the carbonyl is not a strong enough electrophile to be attacked by alcohol without protonation.


Is there a piece of software that can help me see the (hkl) planes in a crystal lattice? I am doing research with calcium carbonate structures. I need a way in which I can familiarize myself with the lattice structure of calcium carbonate polymorphs, especially calcite.

I really struggle to see things in three dimensions. I borrowed an atomic model kit from my professor and that worked out OK, but I still wouldn't be able to tell which is the $(101)$ plane, for example.

I tried downloading software such as BallVIEW, but I found it quite confusing. I need something in which, I can load the structure, and then it can tell me where each $(hkl)$ plane lies. I don't know how to program so I would really appreciate it if anyone can direct me to something user-friendly.

Mercury (available on Windows, Linux, MacOS) has a free version and is arguably the best crystallographic viewer for beginners. It also allows to add as many $(hkl)$ planes as you want. I used the structure of calcite from COD (1010962.cif).

Open your CIF file; by default Mercury loads an asymmetric unit. Once the CIF is loaded, go to Calculate → Packing/Slicing... and tick Pack option. Alternatively, tweak the volume populated by the atoms if needed. (This also automatically populates the list of planes with the planes denoting BFDH crystal morphology.)

Navigate to Calculate → Planes... → New Plane... → hkl and add the planes by Miller indices:

and, if needed, BFDH planes one by one:

If you have CSD-Materials (paid option), then you'll be able to display BFDH theoretical crystal morphologies at once by going to CSD-Materials → Calculations → BFDH Morphology...; on export you'll get something like this:

Alternatively, there is KrystalShaper (Windows only), a free lightweight program made specifically for displaying crystal morphologies:

P. S. Thanks to Buttonwood for useful comments regarding graphical details and reminding about BFDH crystal morphology.


How do you melt metals with super high melting points? At the Renaissance fair a few years back I was watching a smith forge metal into shapes. During this time a very odd question came to me. I was wondering what the furnace was made of. My logic stated that whatever the furnace was made of must have a higher melting point than the materials he was melting. This quickly turned into an elemental arms race resulting in an odd question of how do we melt stuff like refractory metals (more specifically the one with the highest melting point) so we can melt other things inside of it.

Now I know that (for some odd reason I don't understand) rapid cooling can manipulate the strength of an item. Is there a similar property to manipulate the melting point?

Note: My current best guess (like can be done to make weapons harder) is that we take two elements, melt them, and the resulting compound has a higher melting point.

Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon (3550 °C) among the elements. This is why tungsten is used in rocket nozzles and reactor linings.  There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum carbide at 3800 °C.

Carbon cannot be used to hold molten tungsten because they will react to form tungsten carbide.  Sometimes ladles and crucibles used to prepare or transport high melting point materials like tungsten are lined with the various higher melting ceramics or alloys.  More typically tungsten and other refractory materials are fabricated in a non-molten state.  A process known as powder metallurgy is used.  This process uses 4 basic steps:

powder manufacture - a variety of techniques are available to generate small particles of the material being worked
powder blending - routine procedures are used to blend the constituent particles into a uniform mixture
compacting - the blended powder is placed in a mold and subjected to high pressure
sintering - the compacted material is subjected to high temperature and some level of bonding occurs between particles.


What is the liquid which has the most hydrogen atoms per volume? What is the liquid which has the most hydrogen atoms per volume at STP ?
Is there anything better than water?
This answer does not answer this question because none of the compounds listed in it are liquids at STP, which contain more hydrogen per volume than water.

Hydrazine has more hydrogen atoms per mL than water
Some simple calculations give the moles of hydrogen per mL in some possible alternative liquids. Taking into account density and molecular mass we get the following results:
Water 0.056 mol/mL 2 Hydrogens -> 0.11 mol H/mL
Hydrazine 0.0313 mol/mL 4 hydrogens -> 0.125 mol H/mL
symmetric dimethylhydrazine 0.0138 mol/mL 8 hydrogens -> 0.111 mol H/mL
methyl hydrazine 0.0191 mol/mL 6 hydrogens -> 0.115 mol H/mL
So, while some other liquid at STP hydrazine derivatives beat water, it looks like hydrazine has notably more hydrogens per unit volume. This is probably the record.


Condition for formation of azeotropes Consider a binary azeotropic solution of two volatile components $A, B$. I wanted to try and find out the required concentrations of $A, B$ in the solution in terms of their vapour pressures. I tried using Raoult's Law to do this, but the I got an unexpected answer:

Let the vapour pressure of $A, B$ at the current temperature be $P_A, P_B$, respectively. Let their concentrations be $\chi_A, \chi_B$ and their partial vapour pressures over the solution be $p_A, p_B$.

Then, by Raoult's Law,

$$p_A = P_A\chi_A$$

$$p_B = P_B\chi_B$$

By Dalton's Law of partial pressures, the mole fractions of $A, B$ in the vapor phase are given by $\frac{p_A}{p_A + p_B}$ and $\frac{p_B}{p_A + p_B}$ respectively. Since the solution is azeotropic, these two are equal to their mole fractions in solution.

That is:

$$\frac{P_A\chi_A}{P_A\chi_A + P_B\chi_B} = \chi_A \implies \frac{P_A}{P_B}(1-\chi_A) = \chi_B\tag1$$

Similarly,

$$\frac{P_B}{P_A}(1-\chi_B) = \chi_A \tag2$$

Substituting $(1)$ in $(2)$, yields

$$ \frac{P_B}{P_A}\left(1-\frac{P_A}{P_B}(1-\chi_A)\right) = \chi_A$$

$$ = \frac{P_B}{P_A}-1+\chi_A = \chi_A$$

$$\implies P_B = P_A$$

But there are numerous azeotropes where this condition is not satisfied. How come?

It is not that surprising that you got this answer: the existence of azeotropes is due to non-ideal behaviour of mixtures (see for instance the Wikipedia page and the associated diagram). It is not surprising that when you impose an ideal behaviour of the solvent that follows Raoults law, you get some unexpected results. Basically, what your calculation leads to it that for ideal solvent mixture, the only situation in which the gas phase molar ratio can be equal to the solution molar ratio is when both solvent have equal vapour pressures.


How are atoms encapsulated in buckminsterfullerene molecules? According to nature.com, buckminsterfullerene molecules (bucky balls) can be formed by vaporizing graphite by saturating it with laser irradiation.

I haven't entirely understood why the carbon forms into a regular truncated icosahedron, but my real question is, how do scientists go about doping these "bucky balls" with molecules or atoms?

There’s a great summary to your question in the introduction of this article:

  Levitt, Malcolm H. "Spectroscopy of light-molecule endofullerenes."
  Philosophical Transactions of the Royal Society of London A:
  Mathematical, Physical and Engineering Sciences 371.1998 (2013):
  20120429.

It’s well worth a read, but I’ll summarize the basic ideas for you since I think it’s behind a paywall.

There are several techniques to get atoms and molecules in to fullerenes (or buckyballs). Note that the nomenclature for endofullerenes is A@C$_n$, where A is the atom or molecule trapped inside the fullerene and $n$ is the number of carbons making up the cage.

Early techniques involve vaporizing carbon and a metal or noble gas, and hoping that a) a fullerene would form, and b) a metal or noble gas would get trapped inside. For this reason, early endofullerenes were metallo-endofullerenes, which were fairly easy to create in low yield by laser vaporizing a metal oxide and graphite composite. This is how La@C$_{82}$ was first synthesized and reported. Fullerenes themselves are often made by a high temperature vaporization of carbon (arc discharge). If you do this in the presence of a metal, you can get metallo-endofullerenes. Some noble gases can be trapped inside a fullerene this way. They’ve actually found He@C$_{60}$ trapped in ancient rocks, which are hypothesized to have been formed before the formation of the Solar System in a similar process. These extreme techniques are limited to metals and noble gases.

Another way to get noble atoms into fullerenes involves heating fullerenes and nobles gases at high pressure. It’s thought that temporary openings in the fullerenes arise, and this allows the noble gases to insert and become trapped. You are limited to noble gases here. Yet another way is to use ion beams to “blast” atoms into the fullerenes. You can get P@C$_{60}$ and N@C$_{60}$ this way.

The above techniques give you low yield, and you are limited to metals and noble gases.

The latest techniques are often called “molecular surgery” and involve chemically creating a hole, or orifice, in a fullerene. The orifice may be enlarged by chemical means, depending on the size of the guest molecule you want to insert. These open-cage endofullerenes, which are relatively stable under mild conditions, are introduced to the guest molecules. Elevated temperature and pressure often helps get the molecules into the cage. Finally the fullerene is closed using chemical methods. In this way, H$_2$O@C$_{60}$ and H$_2$@C$_{60}$ have been made. Just in the past year, HF@C$_{60}$ was made with similar techniques.


How would you add saccharide residue information to PDB files? I have multiple PDB files containing a single molecules of polysaccharides, with a single chain and a long list of all the atoms -they are not divided by residue.
The PDB files were created from ChemDraw files using OpenBabel and Corina applet form molecular-networks. The workflow was: .cdx <-> OpenBabel <-> smiles/sdf <-> Corina <-> pdb.
What i end up with is this:

How do i easily break up the list into individual residues(individual monosaccharides)? I trying to something similar to 1CAP

This is what the PDB file looks like after CORINA:

HEADER    UNK                                     15-05-08    1UNK
REMARK   1 corina 3.48 0000  08.02.2010
HETATM    1  C1  UNK     1       6.576  -2.524   4.463  1.00 20.00
HETATM    2  H2  UNK     1       6.573  -2.048   5.444  1.00 20.00
HETATM    3  C3  UNK     1       5.876  -1.614   3.449  1.00 20.00
HETATM    4  C4  UNK     1       4.411  -1.438   3.863  1.00 20.00
HETATM    5  C5  UNK     1       3.748  -2.815   3.964  1.00 20.00
HETATM    6  H6  UNK     1       3.760  -3.296   2.986  1.00 20.00
HETATM    7  O7  UNK     1       4.467  -3.621   4.900  1.00 20.00
HETATM    8  C8  UNK     1       5.829  -3.859   4.541  1.00 20.00
HETATM    9  C9  UNK     1       6.487  -4.751   5.595  1.00 20.00
HETATM   10  O10 UNK     1       5.864  -6.037   5.587  1.00 20.00
HETATM   11  O11 UNK     1       2.398  -2.662   4.404  1.00 20.00
HETATM   12  C12 UNK     1       1.655  -3.882   4.435  1.00 20.00
HETATM   13  C13 UNK     1       0.198  -3.584   4.795  1.00 20.00
HETATM   14  C14 UNK     1      -0.566  -4.899   4.964  1.00 20.00
HETATM   15  H15 UNK     1      -0.496  -5.481   4.045  1.00 20.00
HETATM   16  C16 UNK     1      -2.036  -4.592   5.265  1.00 20.00
HETATM   17  C17 UNK     1      -2.607  -3.729   4.136  1.00 20.00
HETATM   18  C18 UNK     1      -1.764  -2.458   4.000  1.00 20.00
...
...
HETATM  286  H   UNK     1      14.974  11.280   1.314  1.00 20.00
HETATM  287  H   UNK     1      15.417  11.215  -0.409  1.00 20.00
HETATM  288  H   UNK     1      15.300  13.559   0.392  1.00 20.00
HETATM  289  H   UNK     1      14.052  13.241  -0.836  1.00 20.00
HETATM  290  H   UNK     1      13.609  13.306   0.886  1.00 20.00
HETATM  291  H   UNK     1       9.253  -4.179   6.144  1.00 20.00
HETATM  292  H   UNK     1      11.359  -2.412   4.593  1.00 20.00
CONECT    1    2    8    3  115
CONECT    2    1
CONECT    3    1    4   86  120
CONECT    4    3    5   69  121
CONECT    5    4    6    7   11
CONECT    6    5
CONECT    7    5    8
CONECT    8    7    1    9  122
...
CONECT  289  114
CONECT  290  114
CONECT  291  118
CONECT  292  119
END

Short answer: your workflow doesn't include residue perception, so you would need to do this manually.

I don't use CORINA, but I suspect the answer is the same as if you generated coordinates using Open Babel.

You're using ChemDraw, which as far as I know, doesn't include any sort of residue information in the file, then creating SMILES and/or SD files, which definitely don't include residue information.

Then you create 3D coordinates with CORINA.

But nowhere along that chain is there a step to create / perceive / retain residue information. So I'm not surprised that you get one large unknown residue.

Open Babel does have code that does amino acid and nucleic acid residue perception when translating to PDB or Mol2 or other format that allows storing residue information. It does not have support for perceiving saccharides. Indeed, I'm not even sure what algorithms exist for automatically perceiving sugars.

My suggestion would be to either:

Number / label the saccharides by hand (yuck)
Write some sort of script in Python to label the residues - since presumably the atoms come in a particular order
Use a software builder that creates the residue information with the 3D coordinates.

For the latter case, a quick search suggested a few possibilities, although I haven't tried any of them:

Glycam
Yasara


Binary diagram and two phase region I know that to find the composition in a two phase region of a binary diagram we have to draw the isotherm and to look at the intersection with phase boundaries.

But why? Is there a demonstration?

I will assume that the y-axis in your graph is temperature, and that the pressure of the system is fixed.

Consider the physical system: a mixture of two species in two phases is at equilibrium. Gibbs' phase rule tells us that we have two degrees of freedom, which can be chosen from the set $\{P, T, X_B^{liq}, X_B^\text{solid}\}$. By fixing $P$ (first degree of freedom) and letting $T$ (second degree of freedom) vary over a prescribed range, we produce the functions $X_B^{liq} = X_B^{liq}(T)$ and $X_B^\text{solid} = X_B^\text{solid}(T)$, and this is exactly what is represented by a phase diagram.


Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene? According to my course of organic chemistry, para-nitro-aminobenzene has to break his aromaticity to delocalize the electrons of the amino-group in the nitro-group.
I don't really see this, when I draw the resonance structures.

Your teacher is correct, some of the resonance structures in your figure are not aromatic.  Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic.  The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a "quinone" rather than an aromatic.

o- and p-benzoquinone are examples of quinones; they are not aromatic compounds (ref, note the sentence "Quinones are conjugated but not aromatic").

Likewise, their carbon analogues, o- and p-xylylene are also non-aromatic quinoid-like compounds.

For a compound to be aromatic it must be

cyclic
contain 4n+2 pi electrons
have a continuous loop of p-orbitals
and you must be able to draw double bonds - internal to the ring - from each ring carbon atom.

It is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds.

Finally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, [6]-paracyclophane)


Which would be more acidic, benzene or cyclohexane? If benzene were to donate a hydrogen ion, then could the negative charge still be delocalised around the ring? Wouldn't that break aromaticity and thus destabilise the molecule?

See: Relative acidities of alkanes, alkenes, and alkynes. Essentially the same explanation holds here, as benzene carbons are sp2-hybridised.

The newly formed lone pair is in a σ-type orbital (blue) which does not overlap with the π system (red), hence no delocalisation is possible.


A 1.00 L gas at 100 °C and 500 Torr contains 70.0 % He and 30.0 % Ne by mass. What is the partial pressure of He? A $1.00~\mathrm{L}$ gas at $100~^\circ\mathrm{C}$ and $500~\mathrm{Torr}$ contains $70.0\ \%~\ce{He}$ and $30.0\ \%~\ce{Ne}$ by mass. What is the partial pressure of $\ce{He}$?

My answer is $350~\mathrm{Torr}$ whereas the correct answer is $421~\mathrm{Torr}$.

Here is my solution (source; commas , should be points .; I used to write it this way.)

$$\begin{aligned}
  n_\mathrm{total} &= \frac{pV}{RT} = \frac{(1\ \mathrm l)(0.652\ \mathrm{atm})}{(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}=0.0215~\mathrm{mol}\\
  70\ \%~\ce{He} &\implies (70\ \%)(0.0215\ \mathrm{mol}) = 0.0151~\mathrm{mol}~\ce{He}\\
  p_\ce{He} &= \frac{nRT}{V}\\
            &= \frac{(0.0151\ \mathrm{mol})(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}{1\ \mathrm l} = 0.460~\mathrm{atm}\end{aligned}$$

$$0.460~\mathrm{atm}\cdot\frac{101325~\mathrm{Pa}}{1~\mathrm{atm}}\cdot\frac{1~\mathrm{Torr}}{133.322~\mathrm{Pa}}\approx 350~\mathrm{Torr}$$

You're getting a different answer because partial pressure is proportional to
the fraction of each gas in moles, rather than by weight. Why is that?
Looking at $pV=nRT$, you can see that the ratio between partial
pressures is the same as the ratio between number of moles (since $V,
R, T$ are the same for both gases).

So, if you respect proportions,
your answer will come out the same whether you consider $1.00\ \mathrm{L}$ of the mix
or some unspecified number of liters. This lets you avoid a messy
calculation with $pV/RT$.
You may as well assume that there are 70 g of helium and 30 g of
neon. Find how many moles of each gas that represents. The partial
pressures of each gas are in the same proportion as the number of
moles. With this and the total pressure, you'll get the answer.


was the octet rule disproven? so according to my understanding orbitals can hold s can hold 2 electrons and p can hold 8 electron d hold 18 and f can hold 32. wouldn't this violate the octet rule? which say only the first shell can hold 2 electrons and the rest of the shells 8. im also confused why some websites says that each orbital has 2 electrons, s has 1 orbital  (2 electrons) p has 3 orbitals (6 electrons) d has 5 (10 electrons) and f has 7 orbits ( 14 electrons)

im so confused guys plz help me as simple as possible

The octet rule has no theoretical basis. It has been observed to work quite often. But that is all we could say. It cannot be proven. The octet rule works nearly perfectly in organic chemistry,

There are many examples where this rule is not verified. For example the Boron hydrides, the simplest being $B_2H_6$. This compound is commercial, but it does not follow the octet rule. Another example is the familiar molecule of Oxygen $O_2$, which has no double bond as often given in elementary chemistry courses. Other examples : $PCl_5, SF_6$. Another simple example is Nitrogen oxide $NO$. And a lot of compounds made with metals from the transition range.


What is the simplest explanation of the ASTM G71 procedures that can be done in a laboratory? In my corrosion testing, I am to use the galvanic corrosion testing by ASTM G71. But due to limited knowledge about this field, I can't understand fully the procedures needed to conduct the test in a laboratory scale. What is the simplest explanation of the ASTM G71 procedures that can be done in a laboratory?

ASTM maintains a variety of publications. To the practitioner, usually the most useful are the test Methods. I have also found the guides (such as G71) and glossaries to be useful background. G71 is not a test method, and it lacks any "procedures", the only answer to your question is: "There aren't any procedures in ASTM G71." This may seem confusing since the Guide contains a diagram which seems to imply that a laboratory test involves two equally sized sample rectangles partially submersed in an electrolyte and connected via an external wire.

Since you are unfamiliar with ASTM publications, allow me to use an analogy which may be more intuitive. Take for example a "Guide for baking bread" which recommends that only a circulating air oven, with an air flow of at least 4 cu. meters per minute and a volume of less than 0.1 cu. meter capable of controlling internal temperature in the range of 50° to 250°C to within ± 2° and that the baker should wash his/her hands prior to starting while keeping the tools, ingredients and preparation surfaces at 20°C. You'll note that my guide (I made it up, so don't take it too literally) completely avoids telling you:

what the ingredients are,
How to mix and process those ingredients, and
how to bake bread.

That is, it is not a "method"; it is a series of best practices (or preferred, consistent practices). When is something like G71 appropriate? It is appropriate when the test needs to be customized to fit the problem. The guide recommends the metal be "as large as possible" (meaning up to full commercial size) while being practical in either lab or field test conditions.

There is also the need for the exposed surfaces to be in the same area ratio, and to avoid edge effects, especially those which will not or do not appear on the "real" assembly. The guide mentions 4 techniques to evaluate corrosion: visual (and photographic), mass loss, current, and potential (voltage). Which of those to use (if any) is up to the individual. It is vague because it can't be specific. It can't be specific because it is intended to apply to a great variety of different galvanic corrosion scenarios.


To find the value of x in a metal chloride MClx A solution containing 0.1 mol of a metal chloride MClx requires 500 ml of 0.8 M AgNO3 solution for complete precipitation. The value of x is:

(a) 1
(b) 2
(c) 4
(d) 3

Please answer with an explanation.

$$\ce{MCl_x + xAgNO3 -> xAgCl + M(NO3)_x}$$

$500$ mL of $0.8$ M is $0.4$ mol, so the mole ratio between $\ce{MCl_x}$ and $\ce{AgNO3}$ used is $1:4$.

From the equation above, the mole ratio is $1:x$.

Therefore, $1:x=1:4 \implies x=4$.


Which is more acidic acetic acid or propanoic acid? I got the answer from $K_a$ value that acetic acid is more acidic, but I have also read that inductive effect decreases as the length of alkyl group increases so acetic acid having a methyl group should have greater +I effect than ethyl group in propanoic acid.

And this comparitive large +I effect of methyl group in acetic acid should make it less acidic (as polarity of $\ce{O-H}$ bond decreases) than propanoic acid with ethyl group with less +I effect.

My best guess is you are interpreting the statement "inductive effect decreases as the length of alkyl group increases" somewhat out of context.

So, let's compare formic acid to acetic acid.  Formic acid has hydrogen bonded to the carbonyl, while acetic acid has a methyl group.  The methyl group is inductively donating, and as a result acetic acid is less acidic than formic acid. $\mathrm{p}K_\mathrm{a}$ difference of $1$ full unit.

So now, let us compare methyl acetic acid (a precise, but incorrect, alternate name for propanoic acid) to acetic acid. Methyl acetic acid has an inductive donating group at the $\ce{CH2}$ carbon. As such, we expect some donation, and a lower acidity for methyl acetic acid as compared to acetic acid. True, the donating group is one group removed from the acidic site, which means the effect of the methyl group on methyl acetic acid's acidity is diminished as compared to an electron donating group directly attached to the carbonyl, so the acidity reduction will be smaller. $\mathrm{p}K_\mathrm{a}$ difference, $0.12$ units.

After you have this concept in your head, purge the abomination of nomenclature that is methyl acetic acid from your memory.

Ron has some views on hyperconjugation that may also be useful in this context, but strictly speaking, I don't find them necessary.


Theoretical p-Phenylenediamine synthesis I am thinking on an alternative synthesis of pPDA, in which I need counsel. I was thinking about these easy reactions, and I want to ask if it can be done, or maybe you can hint me some reference. I am trough the reading of nitration and reduction of aniline, or from 4-nitrochlorobenzene or the DuPont synthesis.

The first route smells a bit fishy, but I think with enough heat, and molar ratios it can be done.

The second reaction is more promising for me since it is made from economic reagents, I'm fairly sure the ascorbic acid reduces the imine to the amine form. The only problem that could occur is with the reaction of 1,4-benzoquinone with ammonia. Can you validate my thoughts? I see no reason why it should not work.

Please if you have any suggestions or help, I gladly welcome it.

EDIT 1:

The comment from Waylander in wich he/she states the step from 1,4-benzoquinone to the imine form is probably not achievable. Howered i found also an another route from benzoquinone, wich is to react it with hydroxil-amine forming the oxime. *1 Then the question left behind is the step from oxime to amine, maybe the ascorbic acid still can do the job, however i dont find any source for it, but my assumption is that it still works.

I'd say the suggested reaction of benzene-1,4-diol (hydroquinone) with $\ce{PCl3}$ to get 1,4-dichlorobenzene would not work (see  Waylander's comment). However, you can buy 1,4-dichlorobenzene very cheap (e.g., USD 21.50 for $\pu{100 g}$ bottle from Sigma-Aldrich) so why you need this trouble?

The suggested reaction of 1,4-dichlorobenzene with $\ce{NH3}$ to get benzene-1,4-diamine (p-phenylenediamine) would work, however, because there is a literature evidence for that (Ref.1). To do so, you need to use copper(I) chloride catalyst ($\ce{CuCl}$) and use 28% $\ce{NH4OH}$ solution as $\ce{NH3}$ source. Also need to heat the reaction mixture above $\pu{200 ^\circ C}$ in a pressure bomb (you may need to read this reference carefully before you start the reaction).

References:

Armand J. Quick, "The Preparation of p-Phenylenediamine and Aniline from their Corresponding Chlorobenzenes," J. Am. Chem. Soc. 1920, 42(5), 1033-1042 (https://doi.org/10.1021/ja01450a018).


Can energy savings be estimated from activation energy differences? Imagine two companies X and Y. Both companies use very similar technical equipment to carry out a biotechnological process where a chemical reaction is catalyzed by an enzyme. Company X uses an enzyme with a reaction barrier of, say, 15 kcal/mol, while company Y uses a slightly different enzyme to catalyze the same reaction but this enzyme has an activation energy of only 12 kcal/mol.
The reaction energies in the two reactors are, obviously, the same since both companies do $\ce{A} \rightarrow \ce{B}$. Still, as I see it, company X has to input 3 kcal per mole of product more into activating the enzyme.

Does company Y therefore require 3 kcal less energy to produce 1 mole of product?

The statement of this questions mixes a number of issues, so there is, in my opinion, no definite answer to this question.

First of all, one must take care that activation energies will have an influence on the kinetics (rate constant) of the reaction that is catalyzed. In a very general point of view, the overall energetic requirement for a chemical reaction does not directly depend on the activation energy (for instance the equilibrium constant is not changed).

Thus if two separate processes are considered, a different activation energy will mostly mean a reaction rate that will be slower, so overall a process that might take longer but not necessarily need a higher energy input.

Now if one would add to this problem a constraint that would be that the reaction rates of bith processes should be similar, the main method to achieve such a result would be to change the temperature at which the process takes place. This is related to the activation energy through the Arrhenius equation:

$$k = A e^{-\frac{E_{a}}{RT}}$$

If one assumes that the pre-exponential factor is the same for both processes (in effect meaning that the transition states in the catalyzed reaction have similar constraints) then one can reach some conclusions on both processes.

For instance, the reaction rates at constant temperature can be calculated. At 298 K, 3 kcal/mol difference would mean a ratio $k_{1}/k_{2}$ of about 150, so the process with lower activation energy would proceed 150 times faster than the other one. An other option would then be to increase the temperature of the slower process. In such a case, the temperature would depend on the ratio of the activation energies (and not the difference) as :

$$T_{2} = \frac{E_{a2}}{E_{a1}} T_{1}$$

If one takes the values of the qustion and considers a process that would take place at 298 K, one would need to heat at 373 K to reach the same rate constant. From this, one could consider the additional energy input required to heat the reaction medium up to this temperature.


Why is the Jacobsen's catalyst with tert-Butyl groups a better catalyst for epoxidation? In an experiment I synthesised the below Jacobsen's catalst:

I used it to create styrene oxide through epoxidation. Chiral GC analysis showed that the enantiometric excess was lower compared to using Jacobsen's catalyst with tert-Butyl groups (shown below). This leads me to believe that it is more enantioselective, but why is this?

This type of oxygen transfer reaction consists of a two step
catalytic cycle, wherein, first an intermediate (salen)$\ce{Mn^V}$ oxo complex is generated, and then the activated oxygen is added to the olefinic double bond.

Electron withdrawing and electron donating group influence the strength of the $\ce{Mn=O}$ bond, with electron withdrawing substituents leading to a weaker $\ce{Mn=O}$, and electron donating substituents leadings to a stronger $\ce{Mn=O}$ bond.

An argument based on Hammond Postulate can be applied where in the case of electron withdrawing substituents (weak $\ce{Mn=O}$ bond), one expects an early transition state, and in the case of electron donating substituents (stronger $\ce{Mn=O}$) a late transition state is seen.

Enantioselectivity is governed by interactions of the salen ligand, which in the transition state geometry adapts a folded geometry forming a chiral pocket, and the incoming olefin.

A late transition state demands a certain proximity between the
olefin and the manganese oxo complex. This leads to a less  spatial separation between substrate and catalyst and consequently allows for better differentiation of stereoisomeric transition structures.

Thus the attenuation of the reactivity of the oxo-species by electro donating groups, leading to a comparatively late transition state is the reason for higher enantioselectivity.

References

This article looks at the electronic effects of 5,5' substituents using DFT calculations.
Cavallo, Luigi, and Heiko Jacobsen. "Electronic Effects in (salen)Mn-Based Epoxidation Catalysts." - The Journal of Organic Chemistry, 2008, 68(16), pp 6202–6207


Deciding higher priority order between a triple bond and tertiary carbon I want to know which of the groups will be given higher priority $\ce{-C#CH}$ or $\ce{-C(CH3)3}$ and why. I tried resolving the triple bond to single bonds such that they are phantom atoms. But that gave me a structure in which there are no atoms after the first point of difference. How is this (see image) wrong and if this is correct then isn't it a contradiction? After first point of difference, there are two difference one is nothing v/s $\ce H$ and other is $\ce C$ vs $\ce H$ and both situation seem to contradict in giving their respective group higher priority order.

From the Wikipedia article on CIP rules.

  If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a "ghost atom" that is a duplicate of B (has the same atomic number) but is not attached to anything except A.

Ghost atoms are attached to resolve the triple/double bond.

So according to CIP rules the two compounds can be resolved as

 The pink atoms are ghost carbons

From here we can see that ethyne has a greater preference over isobutyl as after the first point of difference we move over to the next highest order carbon.In ethyne the second carbon is connected to $\ce{C2H}$ whereas the next highest order carbon in isobutyl is connected to $\ce{H3}$.


Axial Equatorial NMR graph difference I wanted to ask a question about NMR for Axial and Equatorial molecules.

I was asked to describe how to separate the following two product molecules from this reaction:

and the answer that my colleague mentioned was NMR.

He stated that for the product on the left, the $\ce{P}$ atoms were in the same environment (axial) so they would show only $1$ peak but on the right hand side product, the $\ce{P}$ atoms were in different environments (equatorial and axial) hence two different peaks would be shown.

I had a previous introductory course to NMR where coupling constants were discussed, and as the maximum bond coupling "length" was considered to be $3$, this made perfect sense.

My question is, how are axial and equatorial substituents (if they are the same substituent as in this case) in such different environments? Is it to do with the surrounding substituents deshielding by varying amounts? What is the case?

I failed to find a definitive answer on Google or StackExchange, but only statistical interpretations were given, no actual explanations.

When discussing about same environment and different environment, not only the different position should be noted (axial/equatorial), but also the different electronic effects implied.

In the first case:

In short: the molecule has a plane of symmetry, and the substituents are "mirrored" through the plane.
"Long" explanation: This means that they "see" the same electronic environment: the
equatorial substituents have the same electronic effects on the two
axial substituents. This means that their resonance frequency, is the same.

In the second case:

Short answer: no symmetry exists between the substituents
"Long" answer: every group neighboring the substituents is different. One of the two substituents feels the effects of $\ce{CO}$, the other one feels the effect of -$\ce{O}$-. Except in some unlikely and unlucky cases, this means two different signals.

Note to the reader: shielding, resonance frequencies and NMR behavior arise from (way more) complex interactions: I did not mention, above, any possible effect of non-neighbouring substituents, for instance, neither I talked about the shielding or deshielding effects of near substituents.

Nonetheless, a perception of the "symmetry" of the involved groups might give you a "dirt cheap" and operational insight on what could be the outcome of an NMR experiment, in cases as simple as this.


Why sponge can't absorb mercury Recently I have watched a video demonstration with a sponge and mercury and the sponge can't absorb mercury like it does with water.

Why is that?

I think the key feature is the surface tension of mercury-air which is extremely high (about 7 times water-air which is already high), combined with the chemical nature of the sponge, which is typically cellulose.

Since cellulose is a big polysaccharide it has a reasonably favorable interaction with both water (due to the OH groups) and alcohols and alkanes (due to the carbon chain). This causes the contact angle of water, alcohol and alkanes with the sponge to be low (at least lower than 90 degrees) thus allowing capillary action to suck the liquid into the sponges pores. The metallic nature of mercury has an unfavorable interaction with the sponge thus resulting in a contact angle above 90 degrees. This will not allow capillary action because the cosine of angles above 90 degrees is negative, therefore the sponge cannot soak up the mercury.

A sponge made out of metal (silver is common I believe) would be able to soak up the mercury because it has a favorable interaction in terms of surface energy.


What are the fundamental principles that are found throughout chemistry? In the same way that biology has the fundamental idea that all organisms want to reproduce and pass on their genetic information, and that evolution accounts for the diversity of life, what are some of the ideas that tie chemistry together? Apart from all the specifics, what is chemistry really about at its heart?

Apologies if this sounds like a really strange question.

This is an inherently subjective question. It's very difficult to come up with an answer that isn't just a fundamental principle of physics. My personal opinion is the following:

  Chemistry is the study of interacting electrons.

Now, obviously the electrons are only near enough to interact because the positive charges of nuclei don't let the electrons fly away. Fundamentally, however, all of chemistry can be described in terms of electron densities, which is the basis of density functional theory.

This is not just a theoretical construct either. In every field, people are talking about different types of bonds and different electronegativities. If you think about organic chemistry, reaction mechanisms are described in terms of electron-pushing, which is some highly abstract way of talking about the dynamics one might find if you could solve the time-dependent Schrodinger equation for arbitrary lengths of time.

Even things that seem to have nothing to do with electrons, such as nuclear magnetic resonance, are often useful because they key in information about the where electrons are localized via chemical shifts.

Also, this definition manages to distinguish chemists from physicists because very often physicists don't care about solving for the energy of something to the sixth decimal place. Whereas, as a computational chemist, I sometimes worry that I might actually need to get that electronic energy out to the ninth decimal place...

So, my opinion is that chemistry is the study of interacting electrons, and chemists care very deeply about the details of those interactions.


Does solubility of acetic acid depend on the pH? Does the solubility of acetic acid increase or decrease in water if I adjust the pH?

Conversely, a water-insoluble molecule (like retinol), can its solubility be increased by adjustment of the pH?

I would say no (in both cases) because the protonating or deprotonating these molecules does not change their overall hydrophilic or hydrophobic character.

You are correct in both cases, though your reasoning isn't quite on target.

Regardless of the pH, retinol will not be protonated or deprotonated to any significant degree (which would greatly affect it's solubility if it were). It's just a nonpolar molecule that will stay that way regardless of pH.

Acetic acid also would not be protonated to any significant degree at low pH, but that's a mute point as it's fully miscible in water anyway.

The one situation in which you would see deprotonation of the compound is that of acetic acid at high pH. For the most part, both acetic acid and the acetate ion produced by the basic conditions should remain highly soluble. It is possible however, that if you added enough strong base, say NaOH, to a concentrated solution of acetic acid, you could eventually precipitate sodium acetate.

Summary, TL;DR:
You are correct that the answer to both questions is no. Acetic acid is miscible with water (so it's solubility definitely can't be increased) and it's solubility simply won't be affected by pH. If retinol could be protonated or deprotonated, that would increase it's solubility, but it can't be protonated or deprotonated to any significant degree so it is insoluble in water, regardless of pH.


Making sense of J coupling data for Hydronium I found J coupling data for hydronium using DFT and GIAO but I need help making sense of the data. Can someone with help?

I read this as indicating 2JH-H = 9.6Hz and 1JO-H = -163.8Hz. That's a large O-H coupling.

For water, experimental values have been reported as 2JH-H = -7.3Hz and 1JO-H = -80.6Hz 1, and calculated values vary, but within an order of magnitude. Gas phase studies for 1JO-H also provide values around -80Hz. I can't explain the large discrepancy here; perhaps there is a significant change in bond angle for hydronium? Or perhaps a poorly refined calculation set?


Spin spin coupling in a proton NMR of an ester? I am learning about proton NMR and spin-spin coupling, and am confused about whether splitting occurs over an ester bond. Specifically, in the case of ethyl methanoate, HCOOCH2CH3, if I were to number the proton environments going from left to right as 1,2 and 3, would 1 and 2 cause splitting of each other?
If they would not cause splitting, then I would expect a singlet, a quartet and a triplet respectively. If they do cause splitting, then would you get a triplet, a quartet of doublets, and a triplet for 1,2 and 3 respectively? I know that with -OH and -NH the protons do not couple (although I am not sure why), so I was wondering whether the same principle may perhaps apply here...

In the most common case you can only observe 3J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds.

In this case the spins are too far away from each other to cause a visible splitting, you are correct that it would result in a singlet a quartet and a triplet.

In the case of OH and NH protons, chemical exchange can play a role. This is a bit more complicated, but the simple version is that protons that can exchange quickly with the solvent are often not observable in NMR (or just as a very broad signal) and due to that also don't couple. Though in the most common cases the OH proton is usually too far away from the other protons in the molecule to show a coupling anyway.


Naturally Occurring Sulfites or Sulphites form In the following link, it says that the product

  May contain trace amounts of naturally occurring sulfite residue

I asked the manufacturer about it and the explanation I got was:

  The trace amounts of naturally occurring sulfites are present in hydrolyzed collagen (as are most other proteins as well). The sulfur that holds the bonds between the molecules is broken up from these processes and changes them to sulfite

Does anybody knows the exact chemical form of the sulfite that this product contain?

Is it in the form of an anion or in the form of a compound?

When collagen is hydrolyzed, sulfite ions ($\ce{SO3^2-}$) are released. Usually manufacturers filter out the residues, but may some still remain behind, which is why they put the warning on the label.


Why coefficient doesn't change amount of substance $\ce{3Mg + 2As -> Mg3As2}$ ($10\ \mathrm g$ of $\ce{Mg}$ and $10\ \mathrm g$ of $\ce{As}$)

I don't understand why coefficient doesn't change the amount of substance, here's what I did:

$n_{\ce{Mg}}=3\cdot\frac{10\ \mathrm g}{24.31\ \mathrm{g/mol}}=1.233\ \mathrm{mol}$ of $\ce{Mg}$

But the correct solution is:

$n_{\ce{Mg}}=\frac{10\ \mathrm g}{24.31\ \mathrm{g/mol}}=0.411\ \mathrm{mol}$ of $\ce{Mg}$

When you say number of moles, it is a number to express the number of constituent species (atoms, in this case) in the given substance. It does not have any correlation to where that substance will be used and what will be its 'coefficient' in that reaction.

When you say 10g Mg, you mean 0.411 moles of Mg (and vice versa) irrespective of whatever you are going to do with those 10 grams.

Even if you were to use 10g Mg in a reaction such as $$\ce{ 2Mg + O2 -> 2MgO} $$, you would still have 0.411 moles of Mg as it is still equivalent to 10g. Nothing changes on the basis of the 'coefficient'.

You can further try to look up more information to get a better understanding about 'The Mole Concept' because that is where I think your problem lies.


Nomenclature with Complex Substituents For the above molecule, I know that the longest carbon chain has 7 carbons, so the backbone is heptane.

The two substituents are at the 4 position on the heptane.

One of the substituents is a methyl group. That much is obvious.

The other appears to be a complex substituent and is a 1-chloroethyl group.

Thus, would the name be:

4,4-(1-chloroethyl)methylheptane?

I think the issue is what happens when we have different substituents with the same loctant. I've looked in one book and at all its practice problems, and this scenario never occurs.

I know that 4-methylheptane is a real compound and that its name is correct. So I am leaning toward 4,4-(1-chloroethyl)methylheptane as being a correct name; the complex substituent is correctly named; both locatants are present; everything is alphabetized; my only issue is whether I am missing some convention (e.g. should there be a dash somewhere, a comma somewhere, or what?)

Close. But each (different type of) group added to the heptane backbone is numbered separately: 4-(1-chloroethyl)-4-methylheptane. Also, you have an asymmetric carbon, so the chirality needs to be specified (I will leave that to you).


What determines the boiling point of esters of the same number of carbon atoms but different structural formula? The boiling point of esters increases with molecular weight (and therefore, number of carbon atoms) because of more dispersion forces.
But if you had two different esters with the same number of carbon atoms but different structures (e.g. methyl-ethanoate and ethyl-methanoate), what determines which one will have a higher boiling point?
Does it have something to do with the number of carbon atoms in the alkyl group in comparison to the acyl group?

In this case you have consider the degree of branching of the isomer, this influences the strength of intermolecular forces. Think of how tree logs with lots of branches sticking out cannot stack together with each other as easily as a pile of logs with no branches. Branched chain isomers have less contact and therefore weaker intermolecular forces and lower boiling point and straight chained isomers have better contact and thus stronger molecular forces and higher boiling point.


overall energy and the boiling point i need help with the fractional distillation method of a couple of substances at least more than two substances , just like air actually what am i looking for is that why do substances in air with a low boiling point like nitrogen reach higher the fractionating column even if it looks like that substances with higher boiling point will absorb more energy and thus eventually result in greater kinetic energy (at least to me). please help . heres how i came to know that nitrogen rises higher the fractionating column

The gases in the distillation column boil off according to their boiling point.

   Gas          Boiling point
   Nitrogen     77.355 K (−195.795 °C)
   Argon        87.302 K (−185.848 °C)
   Oxygen       90.188 K (−182.962 °C)

The distillation column shown is setup for continuous distillation. But imagine that a pot distillation was being done. Nitrogen would boil off first, then argon, then oxygen. So the various trays in the distillation column represent different temperatures. Nitrogen boils off at the top since it has the lowest boiling point, and oxygen collects in the "pot" since it has the highest boiling point.


Increasing Covalent character increases Colour intensity I observed some trends as follows:-

$\ce{PbCl2}$ white and $\ce{PbI2}$ yellow
$\ce{SnCl2}$ white, $\ce{SnCl4}$ red and $\ce{SnI4}$ black
$\ce{AgCl}$ white, $\ce{AgI}$, $\ce{AgBr}$ and $\ce{Ag2CO3}$ yellow

In all these cases as the size of anion increases ( polarisability increases)  or charge on cation increases (polarising power increases) colour exhibited becomes darker.. How does increasing covalent character affect this?  Or is there  any other reason?

In molecular orbital theory, as the covalent character increases the homo-lumo gap decreases, and the excitation happens more easily. Thus the colour intensity increases.


Crystal structure vs melting point Calcium oxide has a melting point of $\ce{2700^\circ C}$ and sodium chloride has a melting point of $\pu{801^\circ C}$.

If they have the same crystal structure and ions are about the same distance in each crystal, why is there such a big difference?

Calcium oxide is double charged, meaning the force required to separate ion pairs is much larger than Sodium chloride.

Melting an ionic solid requires you give an ion sufficient energy to become free of the ionic bonding of its neighbors.  In a simplified view, each ion has 6 neighbors in bulk (both are fcc, I believe) or 5 at surface face, 3 on corner and 4 on edge.  You need to add at least 3-5 times as much thermal energy to liberate the ions in calcium oxide as sodium chloride due to the doubling of charge from each neighbor.


What do 'sigma' and 'pi' in front of ligands in the formula of a coordination compound mean? The following coordination compound was given in the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha) under the topic 'Effective Atomic Number (EAN)':

$\ce{[Ti(\sigma -C5H5)2 (\pi  -C5H5)2]^0}$

What is meant by the quantifiers $\sigma$ and $\pi$ in front of the ligand?  I have not seen this format with other ligands? If this is to signify $\sigma$ and $\pi$ bonds, why is it not used in the formulas of metal carbonyls which have synergic bonds (a combination of both $\sigma$ and $\pi$ bonds)?

In all the 67 occurrences of 'cyclopentadienyl' in the IUPAC Red Book, I was unable to see $\ce{C5H5-}$ having a $\sigma$ or $\pi$ in front of it. Instead I only saw $\eta$ being used in the name. I came to know that it is meant to denote 'hapticity'. Is this related to the presence of $\sigma$, $\pi$?

Concise Inorganic Chemistry1 has the following to say about this tetra(cyclopentadienyl) titanium complex:

  Tetra(Cyclopentadienyl) compounds such as $\ce{Ti(C5H5)4}$ may be made from $\ce{TiCl4}$ and $\ce{Na(C5H5)}$. The formula may be written as $\ce{[Ti(\eta_5-C5H5)2(\eta_1-C5H5)2]}$ where two cyclopentadienyl rings are attached by 5 C-atoms ($\pi$ - bonded) and two rings are attached by one C-atom ($\sigma$ - bonded).

  NMR studies on these tetra(cyclopentadienyl) compounds suggest that in the $\eta_1$ rings the C bonded to Ti continually changes, and the $\eta_1$ and $\eta_5$ rings change their roles.

The first paragraph explains quite well what the author intends by the quantifiers $\sigma$ and $\pi$. Two rings are bonded by $\pi$ donation with a hapticity of 5 and two rings are $\sigma$ bonded with a hapticity of 1. This picture should clear things up

References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers & Blackwell Science, pp. 696, 1996


Which is not true about Ideal gas Interaction between atom or molecule is neglected
The transformation to liquid or solid is observed under appropriate condition
The volume of constituent atom or molecule is neglected
The compressibility, PV/RT is independent of pressure

Why is 2 wrong? In formula there is P,V,T which means under certain condition?

Why is 3 true? Then what volume of gas do we calculate if it is not from atom or molecules?

Why are 1 and 4 true? Do atoms in gas interact? Why is the compressibility independent of pressure?

Interaction between atom or molecule is neglected

True. In an ideal gas the gas particles are unaware of each other's existence.

  The transformation to liquid or solid is observed under appropriate condition

False. As mentioned in the answer to (1), the particles don't sense each other. Since there are no attractive interactions there is nothing to prevent the gas particles from flying apart, there is nothing to encourage association with other particles. This can of course be a bad assumption in reality, particularly if the gas is highly compressed, but you should not be applying the ideal gas law under such conditions anyway!

  The volume of constituent atom or molecule is neglected

True. This is a central tenet in the definition of an ideal gas. A correction for a definite particle volume is included in other gas laws, such as van der Waals'.

  The compressibility, PV/RT is independent of pressure

True. This follows directly from the definition, $PV=nRT$


What are the biochemical ramifications of ingesting of distilled water? Are there any distinct biochemical reactions that are affected by the ingestion of distilled water?

The water is "distilled" in its literal sense.

As @DavePhD has noted, drinking small amounts of distilled water does not pose any significant adverse health consequences.  However, the World Health Organization (WHO) has studied the effects of long-term drinking of distilled water and issued a report.  They investigated the following five areas of potential adverse human health impact:

Direct effects on the intestinal mucous membrane, metabolism and mineral homeostasis or other body functions.
Little or no intake of calcium and magnesium from low-mineral water.
Low intake of other essential elements and microelements.
Loss of calcium, magnesium and other essential elements in prepared food.
Possible increased dietary intake of toxic metals.

In their conclusion, they state,

  "Drinking water should contain minimum levels of certain essential
  minerals (and other components such as carbonates). Unfortunately,
  over the two past decades, little research attention has been given to
  the beneficial or protective effects of drinking water substances."

and

  "Demineralised water that has not been remineralized, or low-mineral
  content water – in the light of the absence or substantial lack of
  essential minerals in it – is not considered ideal drinking water."

(emphasis mine)


Creation and Combustion of Nitrocellulose I am intending to demonstrate the formation and combustion of nitrocellulose with my high school students, however I am concerned about the potential products of the combustion equation. I have been able to satisfy the safety requirements for the creation which I believe follows this equation:

$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$

The only equation for the combustion I have found is this one:

$$\ce{2C6H7N3O11 + 9/2 O2→12CO2 + 3N2 + 7H2O}$$

I am concerned about the possibility of $\ce{NO2}$ gas being created instead of or in addition to $\ce{N2}$ gas, my understanding of chemistry if fairly basic so I was hoping someone could confirm whether or not the production of $\ce{NO2}$ is likely.

Flash paper is commonly used in classroom demonstrations, I think, and it is made of nitrocellulose.  A journal article from 1995 describes rapid creation of flash paper through nitration with nitric acid.

I have never done this procedure myself, but looking through it the biggest safety risk to my eyes is using and handling concentrated nitric and sulfuric acids.  $\ce{H2SO4}$ and $\ce{HNO3}$ are very dangerous especially when concentrated.  Full PPE when handling these is an absolute must.  I'm glad to hear in your question that you are already thinking about safety requirements at this step.

Another important risk is the inherent risk of lighting anything on fire.  Nitrocellulose is an energetic material.  Since it is heavily nitrated, it needs less oxygen gas to completely burn than things like wood, paper, or even gasoline: a lot of oxidizer is already built into the molecule.  This risk is best managed by only ever igniting small quantities of nitrocellulose, and of course by storing any larger quantities of the nitrocellulose far away from sources of ignition.  Full PPE is a good idea too.  Perform the combustion in a well-ventilated area.

I'm less worried about the particular risk incomplete combustion forming dangerous levels of $\ce{NO_{x}}$, especially if you limit the combustion to small sheets of nitrocellulose at a time.  High levels of $\ce{NO2}$ are unlikely to form via incomplete combustion; if large amounts did form, since $\ce{NO2}$ is a strong oxidizer, it would react with remaining nitrocellulose to less toxic, more reduced nitrogen oxides or more likely to nitrogen gas.

Small amounts of $\ce{NO2}$ do form from high-temperature combustion of anything in air, but if you use small amounts of flash paper, exposure of your students to $\ce{NO2}$  is likely to be background sources in the atmosphere will vastly exceed anything produced by your demonstration.

All that said, don't take too much safety advice from the internet.  Talking to experienced chemists in person who have done this demonstration is probably the single best thing you can do to stay safe.

Update: I should explicitly say that although I have never made flash paper, I have burned small pieces of it on numerous occasions, never with any problems or even noticeable amounts of smoke or fumes.


Interstitial defects in crystals: "high energy configurations"? In the Wikipedia article on crystallographic defects, a type of point defect known as an interstitial defect is described as follows:

  Interstitial defects are atoms that occupy a site in the crystal structure at which there is usually not an atom. They are generally high energy configurations. Small atoms (mostly impurities) in some crystals can occupy interstices without high energy, such as hydrogen in palladium.

What is meant here by "high energy configurations"? If these defects possess a "high energy configuration", then it sounds like they would be difficult to come about? If so, then how do these defects come about (where do they gain the energy from to occur)?

I would greatly appreciate it if people would please take the time to clarify this.

Yes, interstitial defects increase potential energy, as stated in Wikipedia: "In interstitial defects three or more atoms may share one lattice site, thereby increasing its total energy." See Physics on StackExchange for an explanation of potential energy in crystals.

How can they form? One example is during rapid crystallization, where there is insufficient time for atoms to segregate. Think of this as "freezing in place" a defect. If warmed and cooled again, the defect can be excluded and a more-perfect crystal forms. This is the basis for the zone refining, Czochralski process and float-zone purification, where extremely slow crystal growth allows time for impurities to be pushed out.

On the other hand, one can intentionally induce interstitial defects through repeated folding, heating, and cooling to make metals harder and tougher, such as Wootz or Damascus steel. It is easy to see how this laborious process requires energy to insert the defects.


How can potassium disappear after cooking? I read this at this webpage: http://www.whfoods.com/genpage.php?dbid=90&tname=nutrient#impactcookingstorageprocessing

  How do cooking, storage, or processing affect potassium?

  Potassium losses from cooking of high-potassium foods can be
  significant. In the case of spinach for example, potassium levels have
  been shown to drop from 6.9 to 3.0 grams in 3 and 1/2 ounces of
  spinach after blanching for several minutes (a loss of about 56%).

I don't understand this. How can potassium disappear after cooking ?

As Eric has mentioned in his answer, the potassium present in foods will leach out into water when you cook them.

Potassium is present in food in the form of potassium ions $(\ce{K+})$. These $\ce{K+}$ ions happen to be extremely soluble in water.

When food is boiled, the cells break down and release their contents more easily which makes the $\ce{K+}$ ions free to move into the water. The $\ce{K+}$ ions remain in the water since they are so highly soluble in it.

This happens to all water-soluble nutrients like the B vitamins, vitamin C, sodium, etc. Some ways to avoid the loss of all these nutrients would be to steam such foods instead of boiling them since the food wouldn’t come into contact with water.

Side note: The same problem occurs when you deep fry foods containing fat-soluble nutrients like vitamins A and E. These leach out easily into oil and so all those nutrients would also be lost in the process.


Sign of change in enthalpy and change in entropy Why is it wrong to assert that the change in entropy and the change in enthalpy must always have the same sign?
What makes me think that they must have the same sign is the fact that every reaction invariably comes to equilibrium under suitable conditions; and so we have the corresponding temperature equal to $ΔH/ΔS$ (setting $∆G = 0$ in the equation $ΔG = ΔH - TΔS).$

TL;DR
In general the entropy of reaction can be written as
$$  T\Delta _r S=  \Delta_r H  + RT\log \left(\frac{Q_e}{Q}\right)  $$
At equilibrium $Q_e=Q$ and
$$  T\Delta _r S_e=  \Delta_r H   $$

Consider a simple reaction that behaves ideally (occurs under ideal solution conditions).
If it is carried out at constant T and p we can write
$$\Delta_r G = \Delta_r G^\circ + RT \log Q \tag{1}$$
where Q is the reaction quotient.
But we can also write that
$$\Delta_r G = \Delta_r H - T\Delta_r S\tag{2a}$$
and
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ\tag{2b}$$
Equation (1) can then be written as
$$\Delta_r G = \Delta_r H^\circ - T(\Delta_r S^\circ-R \log Q) \tag{3}$$
Matching terms in equations (2a) and (3) we have that
$$\Delta_r H = \Delta_r H^\circ \tag{4a}$$
and
$$\Delta_r S = \Delta_r S^\circ - R\log Q \tag{4b}$$
When the reaction is at equilibrium $Q=Q_e$ (the reaction quotient is then equal to the equilibrium constant, here written $Q_e$) and $\Delta_r G = 0$ which means, combining equations (2a) and (4a) that
$$ T\Delta _r S_e =  \Delta _r H^ \circ \tag{5} $$
and
$$ T\Delta _r S^ \circ = T\Delta_r S_e + RT\log Q_e  \tag{6}$$
so that
$$  T\Delta _r S=  T\Delta_r S_e + RT\log Q_e  - RT\log Q \tag{7a} $$
or
$$  T\Delta _r S=  \Delta_r H ^\circ + RT\log\left(\frac{Q_e}{Q}\right)  \tag{7b} $$
Now compare equations (5) and (7b). Equation (5) holds at equilibrium and says, sure enough, that the reaction entropy and enthalpy are equal in sign at this point in the reaction coordinate. However, equation (7b) - which is the more general expression - says that $\Delta_r S$ can in fact differ in sign from $\Delta_r H^\circ$, depending on the magnitude of the reaction quotient Q. It turns out that while the enthalpy of a reaction in an ideal solution is a constant, the entropy of reaction can be tuned by modifying Q.


Why is there a difference in available energy in saturated vs. unsaturated fatty acids? On Wikipedia it states:

  In cellular metabolism, unsaturated fat molecules contain somewhat less energy (i.e., fewer calories) than an equivalent amount of saturated fat.

Can someone explain why that is true in simple words? (that is, if possible, in a way understandable by someone who is not conversant with biology nor chemistry)

The total energy stored in macromolecules is commonly quantified by the amount of $\ce{ATP}$ produced, as $\ce{ATP}$ is used throughout the body to couple reactions and make them thermodynamically favorable.

First, as the oxidation of a double bond in an fully saturated fatty acid requires a reducing agent, you here produce $\ce{FADH2}$ from $\ce{FAD+}$ which is equivalent to a $\ce{1.5 ATP}$ gain.[1]

Also, unsaturated fat has double bonds in cis positions that are unfavorable to beta oxidation.[2] An enzyme coupled reaction is needed for the double bond shift. Therefore, an enzyme coupled reaction that requires energy (in the reductant form $\ce{NADPH}$) is needed to make the molecule a trans fat with a shift in bond position in order to continue oxidizing the molecule completely. $\ce{NADPH}$, usually is defined as $\ce{2.5ATP}$ equivalents[1] and this extra energy expenditure in shifting the double bond is why less energy is ultimately produced.  Note, this $\ce{NADPH}$ is only used when unsaturated fats have an even numbered bond position.[2] This happens with polyunsaturated fats when the first bond is shifted.

For example:

18:2 linoleic acid will yield $1.5 + 1.5 + 2.5 = 5.5$ less $\ce{ATP}$ than 18:0 stearic acid.

References:

[1]: http://www.saylor.org/site/wp-content/uploads/2013/04/BIO101A-6.2.3-CellularRespiration.pdf

[2]: https://en.wikipedia.org/wiki/Beta_oxidation


Oil-water emulsion doesn't separate I've steam distilled a couple of different powdered seeds/herbs to obtain the essential oils from them.

Somehow some of the extracts won't separate and stay in an emulsion.

I've seen that some fixed oils that where extracted too are separating in cool conditions, but I can't see any essential oil layer forming. Even when letting the extract (emulsion) sit for 2 weeks.

Can anyone explain why this is happening (or not) and how I could break the emulsion? Acidifying or salting it out?

Since if I would try to boil the water of off the emulsion, its vapors would carry the oils with it, so is there any way to evaporate the water but leaving the oils?

I live in a country where it's hard to obtain some chemicals, so are there any easily obtainable chemicals/solvents to use for further or better processing?
I have access to $95~\%$ bio ethanol, $95~\%$ acetone, sodium hydroxide, sodium chloride and I've managed to get some naphtha.

So my possibilities are limited as you can imagine …

I've thought about using a Soxhlet extraction with ethanol or acetone where I could simply distill the extracted and concentrated solution to recover the solvent and be left with the extract (essential oil, fixed oil, etc.).

What can I then do to separate the fixed oils from the essential oils that I want?

I have encountered the same problem with plant extracts. I think the term is hydro/sol when the extract is suspended in aqueous solution. Anyway, I've had some success with heavy sodium chloride brine then chilling the solution to near freezing for an extended time (a week or so). Even then the return hasn't been that impressive. In most cases making a tincture (or gas)extraction then distilling the product (if you still need to) produces a much better yield.


What is the pH of this H2SO4 solution? I'm trying to solve a simple problem which is driving me crazy. It says:

  Estimate the pH of a solution that contains 1 gram of $\ce{H_2SO_4}$ dissolved in 1 liter of water.

When I solve the problem, I dissociate the $\ce{H_2SO_4}$ like this: $$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}$$

And I get a pH of around 1.99

However, following my book's solution, the dissociation occurs as follows: $$\ce{H2SO4 <=> 2H+ + {SO4}^2-}$$
And they say the pH is around 1.69

Now, my question is: why is my dissociation equation wrong?

Your problem is that you only accounted for the first dissociation of $\ce{H2SO4}$, a polyprotic acid -- your book needed the extra specificity from the second dissociation. I will walk through the entire process, including the parts that you know already.

Begin by finding the molar mass of $\ce{H2SO4}$ in order to find out to how many moles one gram of it is equivalent. Then, convert to molarity (concentration) using the given volume of water.

$$\ce{MM_{H_2SO_4} = 2*1.01 g + 1*32.06 g + 4*16.00 g = 98.08 g}$$

$$\ce{\frac{1 g H2SO4}{1}\times\frac{1 mol H2SO4}{98.08 g H2SO4} = 1.0\times10^{-2} mol H2SO4}$$

$$\ce{\frac{1.0\times10^{-2} mol H2SO4}{1 L H2O} = 1.0\times10^{-2} M H2SO4}$$

Although the ICE-box is a formality for such a strong acid, it can still be shown.

\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 0 & 0 \\ \hline
 & \ce{H2SO4} & \ce{H2O} & \ce{H3O+} & \ce{HSO4-}\\ \hline
\text{Change}: & -x &  & +x & +x \\ \hline
\text{Equilibrium}: & 0 &  & 1.0\times10^{-2} & 1.0\times10^{-2}\\ \hline
\end{array}

The second ICE-box is a good way of organizing the second dissociation. Transfer the equilibrium concentrations from the first table. All calculations up to the line are for finding the change (using $\ce{K_{a(2)} = 1.2\times10^{-2}}$). Note that after $y$ is found, it is used again in the second ICE-box to determine the equilibrium concentrations after the second dissociation. Also note that you cannot neglect the $y$ after the second equation because of the similar magnitudes of the molarity and the $K_a$ and must use the quadratic formula.

\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 1.0\times10^{-2} & 0 \\ \hline
 & \ce{HSO4-} & \ce{H2O} & \ce{H3O+} & \ce{SO4^{2-}}\\ \hline
\text{Change}: & -y &  & +y & +y \\ \hline
\text{Equilibrium}: & 0.5\times10^{-2} &  & 1.5\times10^{-2} & 4.8\times10^{-3}\\ \hline
\end{array}

$$\ce{K_a = \frac{[H3O+][SO4^{2-}]}{[HSO4-]}}$$

$$\ce{1.2\times10^{-2} = \frac{(1.0\times10^{-2} + y)(y)}{1.0\times10^{-2} - y}}$$

$$\ce{1.2\times10^{-4} - (1.0\times10^{-2})y = (1.0\times10^{-2})y + y^2}$$

$$\ce{1.2\times10^{-4} = (2.0\times10^{-2})y + y^2}$$

$$\ce{0 = y^2 + (2.0\times10^{-2})y - 1.2\times10^{-4}}$$

\begin{split}
\ce{y} & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& = \frac{-(2.0\times10^{-2})\pm\sqrt{(2.0\times10^{-2})^2-4(1)(-1.2\times10^{-4})}}{2(1)}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{4.0\times10^{-4}+4.8\times10^{-4}}}{2}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{8.8\times10^{-4}}}{2}\\
& \approx 4.8\times10^{-3}
\end{split}

Plug into the p function to determine the pH.

$$-\log(1.5\times10^{-2})=1.82$$

Note that $-\log(2\times10^{-2})=1.69$ so your book probably rounded to one significant figure (which would make sense given the way the problem is worded).


Can I use gasoline as a non-polar solvent? I was wondering if I can use gasoline for a non-polar extraction of organic substances instead of prescribed diethyl ether, which is a bit hard to get in my country. If not, are there any products that can be found at hardware stores for example and can be used as a non-polar solvent?

Gasoline is a mixture of several compounds, paraffins, cycloalkanes, and olefins (alkenes), the purification can be a challenge.


What is an algorithm for making a net neutral molecule from a set of guessed positive charges? Given a set of partial charges for the atoms in a molecule, which are educated guesses but do not add up to zero, we would like to zero them in an intelligent way.

For example, if the net charge is negative and we see a +0.1 atom bonded to a -0.9 atom, we should probably shift the -0.9 towards zero to make that section of the molecule more neutral.

There is also chemical information which can be used, eg, keep oxygen atoms positive and don't let the charge on hydrogen be greater than 1.0.

Is there a name for this class of problem? What algorithms can be applied?

Although you did not mentioned specifically, it is likely that you want to perform some Molecular mechanics force field calculations. The charges are important quantity, so you need more than educated guess to obtain usable results. Nonetheless, the problem of charge equilibration is solved by "Polarizable force fields".

It should be noted, that they are often more expensive, less reliable and less accurate than standard FFs.


How can sodium be reduced in the Castner–Kellner process? In the Castner–Kellner process, why is sodium amalgam formed instead of hydrogen gas at the cathode?

If we electrolyse brine solution, then it will liberate hydrogen at cathode, not sodium, because $\ce{Na+}$ has a negative reduction potential while that of $\ce{H+}$ is (by definition) 0. What factor here causes the sodium to get reduced to form an amalgam in spite of the reduction potentials?

I am not sure if you are familiar with the term "overpotential." With mercury electrode, as a cathode, it takes more negative potential to reduce hydrogen ions to hydrogen gas, rather it is easier for sodium ions to get reduced on a mercury surface and form an amalgam (an alloy of Hg and Na). This is a peculiarity of the mercury electrode.

Mercury cell in chlor-alkali industry


How to calculate the molar mass of a protein from its osmotic pressure? I am unsure how to go about solving this question. I know how to get the osmotic pressure, but where do I go from there?
I can't use the gas law since the protein is not a gas.

$\pu{2.38 g}$ of a protein is dissolved in $\pu{100 mL}$ of water at $\pu{25^\circ C}$. The solution rises to a height of $\pu{8.5 cm}$ in an osmometer (a device for measuring osmotic pressure). What is the molar mass, in grams per mole, of the protein?
Data: assuming the water ls pure, then the osmotic pressure $\pi = \rho gh$, where $h$ is the height of the water, $\rho$ is the density of water ($\pu{1 g/cm^3}$),and $g$ is the acceleration due to gravity ($\pu{9.81 m/s^2})$.

UPDATE: I've spoken with a masters student (my TA) about this and it turns out that the way to solve this is to use an osmotic pressure equation which is identical to the ideal gas law, but only because it is an approximation. If this question is unlocked I would be happy to share my answer.

Use osmotic pressure $Π=iMRT$
From this get your concentration and calculate molarity of solution.
By the way if you are looking more on this topic then you must consider the chapter on solutions and their colligative properties in your chemistry textbooks


Defining equilibrium position vs equilbrium constant My chemistry textbook says that $K_{p}$ indicates the equilibrium position. It then says that $kp$ will not change if pressure is increased, but it also says that the position of equilibrium does change if pressure is increased.

Let me restate that: $K_{p}$ tells you the equilibrium position. Increasing pressure does not change $K_p$ (indicating the equilibrium position does not change) but does change the equilibrium position. This is clearly a contradiction, so how should I properly define equilibrium position and equilibrium constants? I've never really had a clear understanding of what equilibrium position means.

You're using the word "pressure" too casually. There are partial pressures of species and total pressure (which is the sum of all partial pressures of all species, regardless of if they're participating in the equilibrium or not).

Partial pressures of reactants and products factor into the equilibrium constant $K_{\mathrm{p}}$. If you increase the partial pressure of a gas that is not participating in the equilibrium, none of the other partial pressures change, so the equilibrium is unaffected, but the total pressure will change. Thus, depending on the circumstances, changing the total pressure may or may not change the position of equilibrium.

The key is you should focus only on the partial pressures of interest. All other pressures are irrelevant, unless they somehow tell you about a change in a partial pressure of interest.


Why is the enthalpy change not zero in an adiabatic process? Which of the following must be true for adiabatic processes?

  $C_V = C_p$
  $\Delta H = 0$
  $\Delta U = 0$
  $\Delta S = 0$
  $q = 0$

(Source: Chemistry GRE)

The answer is $q = 0$. From what I can find, an adiabatic process is when there is no transfer of heat, but then why is the enthalpy change not zero?

The Enthalpy $H$ is defined as $H=U+PV$. Therefore,

$$\Delta H=\Delta U + P\Delta V +V\Delta P$$

For an adiabatic process, $q=0$. therefore from the first law of thermodynamics,
$$\Delta U = q +w =q-P\Delta V$$
$$\Delta U=w=-P\Delta V$$
Substituting this in the first equation you get,
$$\Delta H=V\Delta P$$
If $\Delta P$ is zero during the process (Isobaric) then the enthalpy change will indeed be zero, but it will be non-zero if the process is not isobaric.


Why do electron donating groups stabilize free radicals? Why do electron donating groups stabilize free radicals? I have read this fact in various websites and it is not mentioned why. Isn't a free radical like a highly reactive individual molecule? Why does it require electrons? And won't the requirement of electrons depend on the octet of the central atom of the free radical?

Throughout this answer, I'm assuming that you mean carbon free radicals, as you tagged this with organic chemistry.

A free radical is actually more than just a molecule. It is somewhere between an ion and a stable molecule, but has no formal charge. However, it is highly reactive, as you mentioned. You've also identified the right reason for the need of electrons, it is the octet rule, at least at an elementary level. They behave as electrophiles, in order to complete their octet.

A general free radical of carbon has 7 valence electrons, (like the methyl, ethyl free radical etc.) and while other free radicals of carbon do exist (like carbene, benzyne, and a few I don't know about), most of these are hypo octet, i.e. they have fewer electrons than necessary to complete their octet.

Hence, they get stabilised by electron releasing groups, which "quench" the requirement of electrons for the free radicals. This can happen via induction, or by resonance,or even by hyperconjugation (which is basically the same as resonance).

I hope this answers your question.


Why is bleaching with chlorine permanent but with sulphur dioxide temporary? I was reading the properties of chlorine when I came across this statement:

Chlorine is a powerful bleaching agent; bleaching action is due to oxidation. Chlorine bleaches vegetable or organic matter in the presence of moisture. Bleaching effect of chlorine is permanent. $$ \ce { Cl_2 +H_2O -> 2HCl + O} $$ $$ \ce { Colored Substance + O -> Colorless Substance } $$

I searched it up and found out that bleaching with $\ce{Cl2}$ is an oxidation and is made powerful with the help of the nascent oxygen, hence it is permanent, but bleaching with $\ce{SO2}$ is a reduction and hence it is temporary.
Could someone please help me understand why the oxidation is permanent and the reduction is not? Is it powerful to an extent of the reaction not being able to get reversed? If yes/no, is there any quantitative/qualitative measure for the same?

The bleaching action of Chlorine is permanent because it involves the process of oxidation. $\ce{Cl2}$ reacts with water to produce nascent oxygen which combines with the material in question (by oxidising it, as pointed out by @Mathew Mahindaratne
) and makes it colourless.
The bleaching action of $\ce{SO2}$ is temporary because it involves the process of reduction. $\ce{SO2}$ removes oxygen from the coloured substance to make it colourless. Atmospheric oxygen slowly takes place of the removed oxygen, because of which the material regains it's colour.


Sodium orthovanadate synthesis My father* wants to synthesize Sodium orthovanadate, and has found out it is prepared from Vanadium pentoxide and a strongly alkaline solution of NaOH, or by fusion with NaOH but cannot find a detailed method, e.g. about quantities, separation from excess alkali, drying, etc.

I** haven't been able to find out much by googling or searching Pubmed: most methods I've found deal with stabilizing the solution, not with the complete synthesis. Could you point us to a complete protocol?

*chemist, but his English is not so great.

**not a chemist anymore

From Supniewski, J. "Inorganic Preparative Chemistry", 1958 (in Polish):

  Melt $\pu{36.4 g}$ of vanadium pentoxide $\ce{(V2O5)}$ and $\pu{63.6 g}$ of sodium carbonate $\ce{(Na2CO3)}$ a nickel crucible until you get a uniform, clear melt with no bubbles of $\ce{CO2}$ coming out. Pour the melt on a cold plate and wait until it solidifies to a colorless mass (during the cooling process it should turn dark green, then yellow, then colorless). Pulverize the melt and dissolve it in a minimum amount of cold water, filter, and layer with several volumes of ethyl alcohol. After several hours, colorless crystals of $\ce{Na3VO4 * 16 H2O}$ should separate. Wash the product with ethanol and dry in a desiccator.

Some clarifications as requested in the comments:

"Minimum amount" means that you need to add only so much water to dissolve the salt completely, no more. Just add water in small portions and stir well - if the salt is still visible add some more water, etc. Suppose it took you about $\pu{50 mL}$ of water to dissolve your substance. Now you need to add twice or thrice (i.e., $\pu{100 .. 150 mL}$) the amount of ethanol ("several volumes"). Add the alcohol slowly so the solutions do not mix immediately but form two separate layers, and leave it undisturbed for several hours (I think it may take a day or two).


Does naphthalene in moth balls really sublimate or does it evaporate? The triple point of naphthalene is $\pu{-56 °C}, \pu{1.05 kPa}$ (which is much less than $\pu{1 atm}$). So how can a naphthalene ball sublimate at $\pu{1 atm}$, when considering that its melting point at $\pu{1atm}$ is $\pu{78.2 °C}$ (and the boiling point is $\pu{218 °C}$)?

Naphthalene, as well as any other solid, may sublimate at any temperature and pressure. The triple point parameters are irrelevant. The pressure of other components (that is, air) is irrelevant as well. What is relevant is the vapor pressure of naphthalene itself in the surrounding air, and that is pretty close to $0$. Until it reaches the saturated vapor pressure (or until your mothball is all gone, whatever happens sooner), naphthalene will sublimate.

The solids which we perceive as incapable of sublimation (like steel, glass, or rocks) have very low vapor pressure, so their sublimation would take forever. Other than that, everything above applies to them as well.

Then again, some solids melt upon heating and some do not, but that's another story. We aren't talking about any heating, are we?


Performing XPS on Pt–Ru/C nanoparticles From what I understand, to get clear peaks, $\ce{Pt}$ is analyzed at $\ce{Pt}$ 4f; and $\ce{Ru}$ at $\ce{Pt}$ 3d.
However, if i'm not mistaken, charge correction is needed.
But $\ce{Ru}$ 3d(280/284 eV) overlaps with $\ce{C}$ 1s(284 eV), therefore I cannot use $\ce{C}$ 1s for charge correction. Hence, I mix in some $\ce{Au}$ powder and use $\ce{Au}$ 4f(84 eV) for charge correction and then analyze $\ce{Ru}$ at $\ce{Ru}$ 3p(461eV).
But now, $\ce{Au}$ 5p(74 eV) overlaps with $\ce{Au}$ 4f(74 eV) so I cannot analyze at $\ce{Pt}$ 4f
and I try to analyze at $\ce{Pt}$ 4p(519 eV).
However, my problem is that I'm not getting clear enough peaks. The obvious thing I can think to do is increase the number of scans or lengthen the dwell time.
But I was wondering if there are other solutions other people usually do that I might be missing.
I do see some papers that try to deconvolute $\ce{Pt}$ and $\ce{C}$ in the same region, but my professor says that this is an incorrect method because it is impossible to consider the correct charge correction that way.
Does anyone have any suggestion on how I should proceed?

However, my problem is that I'm not getting clear enough peaks. The obvious thing I can think of to do is increase the number of scans or lengthen the dwell time.
But I was wondering if there are other solutions other people usually
do that I might be missing. I do see some papers that try to
deconvolute Ru and C in the same region but my professor says that
this is an incorrect method because it is impossible to consider the
correct charge correction that way.

Do you have an Atlas of XPS spectrum of elements? Check the instrument's manual. I do recall CASA used to have element by element spectrum.
Thermo Fisher has one periodic XPS spectrum here. If gold is causing problems, choose another metal whose peaks do not overlap with your sample.
The only solution for weak signals is that you collect the data for a longer time (ensemble averaging).
Don't try deconvolution in XPS especially when it is peak fitting. Your mentor is right. It is not reliable. You can fit as many peaks as you like. There must be a rationale for fitting peaks in an analytical signal (you exactly know peak shapes, peak model, baseline etc.)


How to dissolve steel without affecting aluminium alloy Disclaimer: I have very limited knowledge of chemistry

My problem is the following: I have motorcycle part made from some sort of Aluminium alloy - most probably 6005, 6061-T6 or 7005. And inside of this aluminium part there is steel part which is simply stuck there and it's really hard to get it out physically without damaging the aluminium part.

After a bit of googling, I've found out several ways of doing that:

By putting aluminium part with stuck steel part into water and start adding concentrated nitric acid until steel part starts to dissolve. But I've also read that dillute nitric acid (which will occur after mixing with water) will also dissolve aluminium or I'm wrong here?
By using some sort of Alum: either by using Potassium alum or Ammonium alluminium sulfate. Some guy on the forum suggested to dissolve above alum in hot tap water and then put aluminium part with steel part into resulting solution and let it stay there for a day. It should dissolve or at least damage steel part without affecting aluminium part which may ease physical removal of steel part.

Could you please advise which of above solutions is better? And in case of using alum - which of them should I buy - potassium or ammonium one?

Thanks in advance for any help!

The part of the steel insert that is stuck to the aluminum is not accessible to water.  Let's look at linear coefficients of thermal expansion,

http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html

Aluminum is 22.2 ppm/K. Steels are smaller, the largest being 304 austentic at 17.3 ppm.  Ambient temp is around 24 C, a 400 F kitchen oven is around 204 C.  That should not be hot enough to lose temper in the alloys.  If the aluminum hole is an inch in diameter, worst case increasing the temp 180 K times 4.9 ppm/K net gives you 882 ppm or 22 microns of slack.  The steel insert pushes out.


Which compound is more reactive towards acetylation, 4-aminophenol or 4-aminocyclohexanol? Does the delocalisation of the lone pair of electrons in the NH2 group mean that they are less available to attack the carbon making the 4-aminocyclohexanol more reactive? Or does the polarisation of the neighbouring molecule due to the higher electron density of the benzene ring make the 4-aminophenol more reactive?

4-aminocyclohexanol should be more reactive towards acetylation because the reactivity depends on the availability of the lone pair on N to be able to attack the C atom during acetylation. Since in 4-aminophenol the lone pair resonates in the benzene ring, its availability to attack will get reduced making it lesser reactive towards acetylation.


Comparing acidity between groups I tried to do this using the stability of conjugate base. I can see (2) will be most stable due to resonance however my confusion lies in (1) and (3).
While in (3), the negative charge is unstable but in (1) I can't tell if it will take part in resonance because in pyridine the lone pair doesn't take part in resonance so is the same thing happening here?

The answer should be A.

Since 3 is vinylic hydrogen, therefore it is least acidic.
Both 1 and 2 are allylic and are at the alpha position with respect to the carbonyl group.
Nitrogen is quite electronegative, therefore hydrogen 1 is more acidic than hydrogen 2. Hence the result.


How can a methyl cation exist without 8 valence electrons So far I have been taught that everything wants to achieve noble gas configuration - 8 electrons in the outer shell.

I came across a methyl cation. Which contains a positive carbon, no lone pairs and 3 bonding pairs of electrons so 6 electrons in total. 6 electrons are obviously less than 8 electrons so how can it exist as it defies apparently the rules I have been taught.

Thanks,

Daniel

The way you put it that everything wants to achieve noble gas configuration is a good rule of thumb for predicting molecules that are stable and long-lived. However, not everything is in equilibrium. A methyl cation certainly exists, and it certainly wants to achieve the noble gas configuration, but it still takes time to do so. So we can conclude that the methyl cation is an unstable, short-lived, highly reactive molecule, which is true. And you are totally right to explain this with the tendency of having a closed electronic shell.


Hydrogenation stereochemistry-Pd,Pt, Ni catalysts When my textbook talks about hydrogenation using Pt, Pd or Ni heterogeneous catalysts, it never mentions if it is anti or syn addition.
It simply jumps on to say that for alkynes, NiB2 (P-2) catalysts and Lindlar's catalysts are syn hydrogenation.
Can someone tell me if using Pt gives syn or anti for alkynes?

Using any transition metal catalyst (Pt, Pd, Ni) for hydrogenation of alkynes results in syn hydrogenation to not alkenes but alkanes. Using Lindlar's catalyst results in syn hydrogenation of an alkyne to an alkene.

In the case of sodium metal (not a transition metal) in ammonia, hydrogenation of alkynes results in a trans alkene.


Is there a term that denotes 'molecules and molecular constituents'? In my notes, I've used phrases such as 'atom, particle, or molecule' more often than I'd like. For example, Wikipedia defines atomic mass as

  the mass of an atomic particle, sub-atomic particle, or molecule.

I remember concise definitions more easily, so a single word for 'molecules and molecular constituents' would be useful.

Is there such a term?

This is pretty much link-only, but anyway...

When searching for official definitions, nomenclature, or recommended procedures, the organisation providing all these is the IUPAC (International Union of Pure and Applied Chemistry)!

An IUPAC book series, known as the Colour Books, provides the normative definitions and guidelines on various topics. These are the results of what chemists in the different committees in the IUPAC divisions have eventually agreed upon.

To my knowledge, all these Colour Books are available online.

As kindly mentioned by Loong in a comment, the online versions are not necessary the latest ones! The online version of the IUPAC Nomenclature of Organic Chemistry is from 1993. In addition, a 2004 draft copy is legally available.

Organic chemists will often refer to the Blue Book (Nomenclature of Organic Chemistry), inorganic chemists to the Red Book (Nomenclature of Inorganic Chemistry), etc.

As far as your question is concerned, the Gold Book (IUPAC Compendium of Chemical Terminology) provides the definitions for molecule and molecular entity.


How is silicic acid formed? My book gives the following reaction:
$$\ce{SiCl4 + H2O -> HCl + Si(OH)4}$$
Since equilibrium favours the formation of weaker acids. How is this reaction able to proceed if a strong acid like HCl is being formed? Am I missing some point here? Or is my interpretation or equilibrium in this case is faulty?

As this video clearly confirms, the hydrolysis of SiCl4 is highly exothermic from breaking of chlorine bonds. The solution boils and solid insoluble Si(OH)4 is produced.
Also, per Wikipedia on the hydrolysis, to quote:

The reaction can be noticed on exposure of the liquid to air, the vapour produces fumes as it reacts with moisture to give a cloud-like aerosol of hydrochloric acid.[5]

As such, the balanced reaction is likely best expressed as:
$\ce{ SiCl4 (l) + 4 H2O (l) ⟶ 4 HCl (g) + Si(OH)4 (s)}$
where the gaseous and solid precipitates formed can push the reaction to the right.
There is an interesting comment in Atomistry, to quote:

Consider, for example, the two reactions:

$\ce{Ca(OH)2 + 2HCl → CaCl2 + 2H2O}$

$\ce{Si(OH)4 + 4HCl ← SiCl4 +4H2O}$

The difference between the metal and the non-metal is signalised by the direction of the reaction. This difference, however, is seldom absolute, and with elements of intermediate character it disappears. In the case of stannic tin, for example, the reaction is reversible:

$\ce{Sn(OH)4 + 4HCl ⇔ SnCl4 + 4H2O;}$

and stannic chloride can be prepared either by dissolving stannic hydroxide in hydrochloric acid, or by passing chlorine over heated tin or a mixture of stannic oxide and carbon.

So, part of the answer also relates to the non-metal characteristics of Silicon as well.


Are p‐xylylene and fulvene aromatic? Are the following structures aromatic?

I have not seen this type of question anywhere so I do not know the answer.
This is just a problem that has been roaming around in my head that I may have seen somewhere.
My approach: since aromaticity leads to stability the carbon-carbon bond in both structure should be polar leading to aromaticity like this:

But this is just a thought and I do not know if it is right.
Is any of the two structures aromatic?

To decide if a molecule is aromatic or not you must know some rules-

The Molecule Must Be Cyclic and the system should be conjugated.
The molecule must follow Huckle's rule i.e. it should have (4n+2) Pi Electrons participating in resonance.

3.The molecule must be planar.

For details see-https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_15%3A_Benzene_and_Aromaticity/15.03_Aromaticity_and_the_Huckel_4n___2_Rule

Now if you try to draw the resonance structure for your 1st molecule then you will see that 4 electrons are participating in resonance at a time. So the system is not fully conjugated. It will be a cross conjugated system.So both the structures you have mentioned are non aromatic.

Stability is a result of aromaticity. You can not comment that if a compound is stable then it is due to aromaticity.

Here is an image showing resonance in para xylylene.

It can be seen from the resonating structures that some of the resonance structures are not aromatic making the molecule non-aromatic in total.

Any recommendations on making the image better are welcomed.


Final temperature of a adiabatic process One molar of a perfect gas of $C_v = \pu{20.18 J/K}$ at $\pu{3.25atm}$ and $\pu{310 K}$ undergoes adiabatic expansion to reach a final state of $\pu{2.50atm}$. Calculate final volume and temperature.

Since $PV^{\gamma} = const$ and $\displaystyle T^{C_v/R}V = const$,

$$V_f = V_i\left(P_i\over P_f \right)^{1\over \gamma}$$

Which I got as $\pu{0.0113m^3}$ and $V_i = \pu{0.00782 m^3}$ .

To get final temperature I used
$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0113 \right)^{8.314\over 20.8} = \pu{267K}$$

But when I use equation of state to get final temperature I get $\pu{344K}$ which is the correct answer.

Why did not equation of state and $\displaystyle T^{C_v/R}V= const$ match in this case ?

Your error stems from a miscalculation of $\mathrm{V_f}$ in the first calculation. As mentioned in the comments, it appears that you actually exponentiated by $\gamma$ rather than $\frac 1 \gamma$. When you differentiate by $\frac 1 \gamma$ , you should obtain $\mathrm{V_f}=0.0094 \mathrm{m^3}$.

Plugging this into your expression for $\mathrm{T_f}$, you should obtain:$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0094 \right)^{8.314\over 20.8} = 288\mathrm{K}$$

Similarly, inserting this into the ideal gas equation, you should obtain: $$\mathrm{T_f= {P_fV_f \over R}}={(2.5)(0.0094) \over(8.206\times10^{-5})}=288\mathrm{K}$$ (The ideal equation will be slightly off as written due to rounding).

In regards to the textbook solution, the final temperature could not possibly be higher than the initial. An adiabatic expansion always reduces the temperature. Intuitively, we can understand this by noting that an expansion with no heat exchange should decrease the internal energy and thus decrease the temperature of the system.