A thorough benchmark of DFT functionals without dispersion corrections So, I'm looking for a benchmark of DFT functionals not including dispersion corrections. Something similar to Grimme's and coworkers GMTKN30 database, but for benchmark calculations done without dispersion corrections.

As in the case of GMTKN30 database, I'm interested in the same properties: general main group thermochemistry & kinetics. But

I already have calculations done many years ago without dispersion corrections since this feature simply was not implemented in most of the packages that time.
My workhorse is G3/G4 method which relies on geometry obtained without dispersion corrections anyway.
I quickly run few calculations including dispersion corrections and found no noticeable difference. Don't have time to redo all the work anyway.

I basically need to justify the choice of just a few DFT functional for my own benchmarks based on thorough literature reviews.

There is a benchmark of the GMTKN30 database with and without  the D3 dispersion correction. See Figure 4 here: "Effect of the damping function in dispersion corrected density functional theory"

However, your statement

  I quickly run few calculations including dispersion corrections and found no noticeable difference.

is not supported by this.

Another example is this paper: "Improving intermolecular interactions in DFTB3 using extended polarization from chemical-potential equalization" where we compare PBE to PBE-D3 for 11 different datasets (see the SI). For many dispersion-dominated complexes, DFT (no D3) is qualitatively wrong, and in many other cases, the error is comparable to the mean interaction energy.

In general, I would strongly recommend against DFT without an empirical dispersion correction unless you are only focusing on electronic properties, but then you wouldn't be very interested in GMTKN30 anyway.

Since you say you are using G3/G4 you might also be interested in these recent benchmark papers:

"Benchmarking Compound Methods (CBS-QB3, CBS-APNO, G3, G4, W1BD) against the Active Thermochemical Tables: Formation Enthalpies of Radicals"
"Accurate reaction barrier heights of pericyclic reactions: Surprisingly large deviations for the CBS-QB3 composite method and their consequences in DFT benchmark studies"


Parts of a manometer There is a poor quality photo in the diagram of a manometer below:

The picture is too poor for me to understand what they mean by "glass pointer marks calibrated volume".  I would be interested if anyone (a) knows what this is - preferably with a manufacturer's number so I can inquire further or (b) has a higher quality image of the part.

I can't tell you a manufacturer, but I can tell you the purpose of the pointer.

When the height of the column of mercury changes, it changes the height of the mercury exposed to the atmosphere. Since the ruler is (usually) fixed in place, the level of the mercury needs to be adjusted to the zero on the ruler. The pointer shows you the zero point. You adjust the level of mercury so that it barely touches the point.


How many coordinates should be used for the potential energy surface of a non-rotating linear molecule? (3N-6) or (3N-5)? Take, for example, $\ce{CO2}$.  It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.

So if I wanted to build a potential energy surface for this molecule that was valid near and far from equilibrium, it should depend on these coordinates.  But how do I rectify this with the fact that $\ce{CO2}$ has four vibrational modes? As we learn in undergraduate chemistry, it has two stretches and two degenerate bends.

If I build a Hamiltonian in the three valence coordinates, could it possibly describe the ground state energy correctly?  That is, would it contain the zero-point energy for all four modes?  Would it be able to find the right low-lying excited states of the bending mode(s)?

EDIT:
Wildcat has brought up the point of symmetry and point groups below, and perhaps that can help get to the heart of my question.  When we say that $\ce{CO2}$ is a linear molecule, that means that the lowest energy configuration has a bond angle of $\pi$.  But in no way do I want to restrict the symmetry of the molecule.  I want to work with a potential surface that is valid for symmetric and asymmetric geometries.

But here is where I get confused.  Let's say I did restrict it to be linear.  I would then have a two-dimensional potential, a function of the two bond lengths.  Of the seven normal modes, I only keep two: the symmetric stretch and anti-symmetric stretch.  The ground state energy would be the sum of the zero-point energy of these two modes.   So therefore the number of modes is equal to the number of coordinates I used to write my Hamiltonian.

If I further constrict the symmetry, and impose the $D_{\infty h}$ point group, then the potential is now a function of one parameter.  I will now only have one vibrational mode, the symmetric stretch, and the ground state energy is the zero-point energy of this mode.  So, again, the number of modes is equal to the number of coordinates I use to write my Hamiltonian.

But what if the symmetry were not restricted in any way?  I can write the potential as a function three internal coordinates, be they bond lengths and angle, othogonormal rectilinear coordinates, jacobi coordinates, etc.  This potential, and the Hamiltonian I construct with it, should find the same modes described above, as well as the bending modes.  But will it find both bending modes?  I don't see why it should.  Based on the two reduced-dimensional examples above, I would expect the number of modes to equal the number of coordinates.  I would expect therefore, not to find the right ground-state energy, etc.

Specifically, I am concerned about the accuracy of dynamical simulations performed using this three-dimensional potential.

Likewise, what about a tetratomic liner molecule like $\ce{C2H2}$? I would use three bond lengths, two bond angles, and one dihedral angle, for a total of 6 coordinates - but it has 7 vibrational modes.  Would I model the dynamics properly?

The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation
$$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$
where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.

In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.

So the Hessian generated in the three internal coordinates has all the information you need.


What is the priority for numbering alkyl side chains? I'm wondering why the above image is named 5-ethyl-3,3-dimethylheptane rather than 3-ethyl-5,5-dimethylheptane. I thought an ethyl substituent has a higher priority (I know this is true when arranging the final name, but what about numbering the parent chain)?

The process for naming this molecule correctly follows the steps listed here:

Identify the parent hydrocarbon, which will be the one with the longest chain. It is correctly identified as the 7-carbon chain, and as such is a derivative of heptane.
Identify the side-chain functional groups. We have 2 methyl groups and an ethyl group.  When listing them, they will be written alphabetically.  So, we have:

X-ethyl-Y,Y-dimethylheptane and not Y,Y-dimethyl-X-ethylheptane
Numbering (which is what you are asking about) is determined by numbering the carbon atoms consecutively from the end of the chain nearer the functional group of the highest priority. In terms of molecule numbering, alkane groups have the same functional group priority. We would only need to consider relative priorities of the ethyl and methyl groups if there was only one methyl substituent, in which case you would have 3-ethyl-5-methylheptane.

In our case, the process is to count in from the ends of the primary chain. Here, the first substituents occur at an equal distance from each end of the chain, both the ethyl and the methyl occur at position 3. We now need to invoke one of the IUPAC sub-rules for nomenclature. If the first substituents occur at an equal distance from each end of the chain, we need to number from the end nearer a second substituent. The second substituent is the extra methyl group, and so this is the end that gains numbering priority.

There are many nice examples of nomenclature on the internet. One very good example, which highlights your question is this page. Have a look at Rule 2 and sub-rule 2


Is it true that all of the physical laws and mathematics needed to describe chemistry are known? This question comes from a rather famous quote by Paul Dirac which goes like this,

  The underlying physical laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble. It therefore becomes desirable that approximate practical methods of applying quantum mechanics should be developed, which can lead to an explanation of the main features of complex atomic systems without too much computation.
  -Paul M. Dirac

I have seen this quote a number of times and it never struck me as being peculiar because to my knowledge the actual physical laws describing everything we care about in chemistry are known. However, I was recently at a talk (it was a presentation to one of my classes, but the speaker gave a talk about his research) about density functional theory, and this quote was shown as a sort of introduction for why density functional theory was developed and is applied as a form of approximation. After showing this quote, the speaker said "well that's not strictly speaking true..." and then moved on and didn't say anything else.

I meant to ask what he meant by this after the talk, but I forgot because I was asking about something else.

So, to expand on this:

Are there any physical laws which are unknown (or believed to be unknown) which directly relate to chemistry?

Are there any mathematics which have not been developed which are necessary to fully describe some aspect of known chemistry?

First, it is one thing to know the basis for a "law", and quite another to mathematically calculate the effects of that law. Consider just carbon, for example... chains may be made of thousands of atoms, with various functional groups attached to each. Though, as Dirac stated, it helps to have "shortcuts" to computation such as Fast Fourier transforms, there are still problems that cannot be solved in a "reasonable" time.

Second, if there are unknown laws, how would we know about it (not to quote Rumsfeld on unknown unknowns)?

And finally, even if all physical laws were known and understood, it would still be impossible to predict everything: Kurt Gödel's incompleteness theorems show that in a complex system (it does not have to be very complex; basic grammar-school mathematics qualifies), questions can be asked that cannot be proved true or false. This extends to chemistry and physics.


What is the mole fraction in mixture before combustion? What is the mole fraction of $\ce{C2H6} / \ce{C3H8}$ in mixture, if both compounds reacted with  $\ce{O2}$ in stoichiometric ratio? Mass of produced $\ce{CO2}$ is $1.819$ times larger than the mass of $\ce{H2O}$.

$$\ce{C2H6_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$

$$\ce{C3H8_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$

I balanced the equations and tried the following:

$\frac{X4M(CO_2)+(X-1)4M(CO_2)}{X8M(H_2O)+(X-1)6M(H_2O)}=1.819$

Which led me to the wrong result.

Could anyone kick me in the right direction?

It's always helpful to show as much work as you can so that we can see where you went off track,  In any case it appears you made an effort, so let's start at the beginning.

First we need to write out the balanced equations
\begin{aligned}
\ce{&C2H6(g) + $3.5$~ O2(g) -> 2 CO2(g) + 3 H2O(g)} \\
\ce{&C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)}
\end{aligned}

Second we're given that

mass ($\ce{CO2}$) = 1.819 mass ($\ce{H2O}$)

we can converting this to a molar ratio

$$\ce{\frac{[mass ~(CO2)]}{[$44.009$]}~=~ $1.819$ * \frac{[mass~ (H2O)]}{[$18.015$]}}$$

and we find that the molar ratio of
$$\ce{\frac{[moles~(CO2)]}{[moles~(H2O)]}~=~ $0.744604$}$$

Now, just to get a feel for things, let's look at the ethane case and we see that we have a mole ratio of $\ce{CO2~/~ H2O}$ of 2/3, whereas in the propane case the ratio is 3/4.  The observed molar ratio of 0.74 indicates that the propane reaction is the major contributor. We can write a general equation
$$\ce{\frac{[2/3 + (r * 3/4)]}{[1+r]}~=~$0.744604$}$$
where "r" is the molar ratio of propane to ethane being burned. Solving for "r" yields
$$\ce{r~=~$13.59$}$$

In other words $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$


Dominant contribution of v = 0 → v = 1 transition in vibrational spectroscopy Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.

After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($\Delta v = \pm 1$), my professor's notes state that:

  Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 \to v = 1$ transition and is called fundamental transition.

I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 \to v = 1$ will take place?

The reference should really be to the Boltzmann distribution where the  population is proportional to $\exp(-E_n/(k_\mathrm BT))$  where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_\mathrm BT$ at room temperature $\approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $n\gt 0$.


Why does the crystal sublime more quickly at the corners where there is a higher vapour gradient? The crystal responds to the larger vapor gradients near the edges by
  sublimating faster there and thus rounding the crystal.

  J. Nelson, J. Atmos. Sci., 1998, 55, 910-119.

The change in shape according to the gradient makes intuitive sense to me but I can't for the life of me figure out why it does this. Maybe I'm being really foolish here, but surely a higher concentration of vapour at the corners would decrease the rate of sublimation there.

The vapor gradient reflects the fact there is much more space at the corner than in the middle of a face. The vapor builds up near the surface but because there is much more space near a corner, the vapor pressure drops off, where as the vapor pressure near a point on the middle of a face is reinforced by sublimation at points near it.


Molality and dissociation For $\ce{HCl(aq) -> H^{+}(aq) + Cl^{-}(aq)}$, if $\ce{HCl}$ has a concentration of 1 molar then $\ce{H+}$ and $\ce{Cl-}$ also have 1 molar. But for $\ce{HF(aq) <=> H^{+}(aq) + F^{-}(aq)}$, if $\ce{HF}$ has a concentration of 1.5 molar, then why doesn't $\ce{H+}$ and $\ce{F-}$ have 1.5 molar as well?

The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$.  Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$.

$\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution.
$\ce{pK_a}$ does measure the relative strength of an acid, look at the following equilibrium

$$\ce{H2O + HX<=>H3O^+ + X^-}$$
$$\ce{so~~~K_{eq}=\frac{[H3O^+][X^{-}]}{[H2O][HX]}}$$ but in dilute solutions the concentration of water stays constant so we can rewrite this equation as
$$\ce{K_{a}=\frac{[H3O^+][X^{-}]}{[HX]}}$$ and finally, $$\ce{pK_{a}~=~-Log_{10}K_{a}}$$ Now we can see that the more highly dissociated our acid $\ce{HX}$, the higher the concentrations of $\ce{H3O^+}$ and $\ce{X^{-}}$, and in turn the larger the value for $\ce{K_{a}}$, and the more negative the value of $\ce{pK_{a}}$.

Bottom line: highly dissociated acids have large negative $\ce{pK_{a}}$ values

Now, here is a Table from SparkNotes.  It compares the $\ce{pK_{a}s}$ (degree of ionization or strength) of the various haloacids.  You can see that as we move down the Table from $\ce{HF}$ to $\ce{HI}$ the $\ce{pK_{a}}$ becomes much more negative; our acids become much more dissociated in water.  So we see that $\ce{HF}$ is not very dissociated ($\ce{K_{a}=10^{-3.1}}$, equilibrium lies to the left). On the other hand, 1 mole of $\ce{HCl}$ will produce about 1 mole of $\ce{H^+}$ ($\ce{K_{a}=10^{+7}}$, equilibrium lies far to the right).  Note too that the $\ce{pK_{a}s}$ of these acids correlate with their bond strength which makes sense.  As the $\ce{H-X}$ bond gets weaker, it becomes easier for the acid to ionize (dissociate).


How to obtain pure aniline from a mixture of phenol and aniline? There is a mixture of aniline and phenol. I need to obtain (just the procedure) pure aniline.

So I think the mixture should heat with Zn powder(to convert phenol to benzene) and then I thought to synthesize aniline from benzene.

I am not sure about this method?

Please help me

First dissolve phenol and aniline in diethyl ether and add excess HCl solution .
Then aniline will react with HCl and dissolve in water layer.
Easily remove organic layer.

Take the water layer and add excess NaOH to it and aniline will be precipitated.


Reaction of H2o2 on Ozone in jar washing Tank in packaged drinking Water Company What is reaction of H2o2 when added in Jar Washing Tank  where ozone is also being dosed at 0.5ppm in a Packaged Drinking Water Industry ?

Both ozone and H2O2 are reactive oxygen species. They don't necessarely react with each other, but oxidize organics and act as bacteriocides. Majority of bacteria are killed by H2O2 (which is cheaper and safer). In case some spots were not cleaned ozone is added. Producing enough ozone to clean it all would be difficult. Guaranteeing that H2O2 reached every spot is also hard. this is why they use two chemicals.

0.5ppm (1/2 parts per million) is a relatively low dose, but working with ozone at higher concentrations is dangerous. Ozone is more toxic than chlorine.

So, this is much like mixing dishwashing liquid with laundry detergent. There is no reaction between them.

Note that in a lab you can have O3+H2O2 -> H2O3 + H2O, but you have to run it at low temperature, the yield is low and the product decomposes fast.


Why is sulphuric acid added in the ripper titration? Ripper titration can be used to determine free $\ce{SO2}$ in wine by titration with iodine.

The method requires sulphuric acid to be added to wine before titration. Why is this the case?

There are 3 forms:

sulfur dioxide (molecular) $$\ce{SO2}$$  ,
Bisulphite $$\ce{HSO3-}$$
Sulphite $$\ce{SO3^2-}$$
(Additionally, bisulphite can become oxidized with water to form sulphuric acid)

These species exist in equilibrium, and shifts in pH with shift the concentration of each species.

The addition of sulphuric acid lowers pH and favors the molecular form. The molecular form can react with iodine and water. When there is no longer any of the molecular form to react with the iodine, the end point its reached and the color changes.

$$\ce{SO2 + I2 + 2H2O → H2SO4 + 2HI}$$


Glucose spatial arrangement It was written in my textbook that the spatial arrangement of hydroxy groups was determined by Fischer by studying certain properties. But I wanted to know to how the relative configurations of hydroxy groups were determined.
Glucose is dextrorotatary but that could be possible with many arrangements.

It's not certain properties, but a very careful observation if and with what products glucose reacts to other chemicals. Took many years. And even then Fischer did not know the absolute configuration, because at his time nobody knew the absolute configuration of anything.

Only when the first absolute configurations were known from X-ray diffraction experiments, one could start comparing.

Today we can calculate the absolute optical activity via quantum mechanics, but Fischer solved the Glucose structure in 1891, that's a quarter century before quantum mechanics.


Bond length and bond strength correlation Is bond strength inversely proportional to bond length? As the bond length decreases, does the bond strength increase?

Decrease in bond length and increase in strength is only a trend, that is observed only between the same pairs of atoms, so its approximately true for C C bonds, single, double triple and similarly for N N bonds. But in general this is not true as can be seen in the following list of bonds with  approximately the same length e.g.
$\ce{H-Br,141 pm 366} \pu{ kJ mol^{-1}}$
$\ce{F-F, ~142 , ~~~~158}$
$\ce{C-O, 143 ,  ~~~~360}$
$\ce{O-O, 148 ,  ~~~145}$
$\ce{N-N, 145 ,  ~~~170}$


Amber solvatebox: How do I remove water that has traveled away I used Amber's solvatebox function with a small polypeptide and tip3p water with a distance of 15 angstroms between the molecule and the edge of the box.

After I ran dynamics, I tried to view the output by removing water from view in Jmol, but I had a really hard time rotating the polypeptide. When I viewed it with water, I found that the difficulty in rotating was because the radius of rotation (are those the right words?) was extremely large because some waters floated off in the distance.

I want to remove these stray waters. How would I go about doing that? Would I just crack open the output and remove those waters by deleting lines that contain information for those particular waters?

This has nothing to do with box walls (there are no walls!) or restraining the waters. You do not want to do that!

Do not confuse periodic boundary conditions with wrapping to the central cell. Look at the amber tutorials, it is all explained there.

You either need to set iwrap=1 in your amber minor put your trajectory trough cpptraj and use autoimage.

Again, all in the tutorial and manuals.


Extracting sodium from salt At my school, my teacher tell us to extract sodium from normal salt ($\ce{NaCl}$) but without using electrolysis of molten sodium chloride. So is there a way to do that?

I don't think it is safe or realistic to do that without extensive training and equipment, but yes, historically sodium metal has been made by chemical reduction of sodium salts at high temperature and condensing sodium vapor produced.  For example, sodium carbonate was heated with charcoal.  Even though equillibrium does not favor sodium metal, the fact that the small fraction of sodium produced is removed takes advantage of Le Chatelier's principle to keep making more sodium.


What criteria have to be met to term a carboxylic acid a 'fatty acid'? So we were doing the chapter Biomolecules in class the other day, and my teacher told us that every carboxylic acid that consists of an aliphatic chain and a total carbon atom count that's four or more, is a fatty acid.

Now owing to my teacher's (rather annoying) habit of grossly over-simplifying things, I decided to look up 'fatty acids' in my trusty Oxford Science Dictionary (printed in 2003). According to it,

  Chain length ranges from one hydrogen atom (in methanoic acid) to nearly 30 carbon atoms.

But a quick Wikipedia search, gives me this:

  Most naturally occurring fatty acids have an unbranched chain of an even number of carbon atoms, from 4 to 28.

So now I'm completely lost. Fine, chuck what my teacher said, but the other two sources seem to contradict each other. So I'd really appreciate it, if someone can help me get the following straight (I'd prefer a reliable source, if you're citing anything):

Q1: What is the minimum number of carbons in a carboxylic acid as a whole (includes the carboxyl group)? So is methanoic acid or butanoic acid that's the smallest fatty acid?

Q2: Is there an 'upper limit' on the total carbon atom count for fatty acids? As in; can I synthesize, say, an 80 carbon aliphatic carboxylic acid and still correctly call it a fatty acid?

Q3: Is there any other criterion that's necessary to classify something as a fatty acid that hasn't been mentioned by my teacher or the other sources? If so, what is it?

I suggest to consult the IUPAC Compendium of Chemical Terminology, aka the Gold Book for a definition of fatty acids!

In a narrow sense, fatty acids are defined as

  aliphatic monocarboxylic acids derived from or contained in esterified form in an animal or vegetable fat, oil or wax.

The definition continues to mention that

  natural fatty acids commonly have a chain of 4 to 28 carbons […]

The paragraph closes with a wider definition, namely:

  By extension, the term is sometimes used to embrace all acyclic aliphatic carboxylic acids.

Personally, I have used the term fatty acids only for natural fatty acids.


Effect of pressure on the freezing temperature of water A few lines are written in a book by B.H. Mahan regarding the change in freezing point of water on changing the pressure.

  The vapor pressure of water at triple point of water is 4.579 mm. What would happen if the external pressure applied to the system by the piston were to be increased above 4.579 mm ? First of all the va pour present would be completely converted into liquid and solid, to restore the equilibrium pressure.

I'm in total agreement of this. But the next lines appear to be odd.

  It is found experimentally that as the pressure on the system is further increased the temperature of the system must be decreased, in order for both  ice and water to remain in equilibrium.

I don't understand  the second statement. If the equilibrium is further disturbed why the system should not  behave the same as before? Why do we have to change the equilibrium constant  to restore the equilibrium?
Thanks in advance.

From the triple point (273.16K, 611.657 Pa), if pressure is increased and temperature maintained at 273.16K, all the water being liquid is the equilibrium condition.

To maintain the solid and liquid phases in equilibrium, as pressure is increased above the triple point pressure, temperature must decrease slightly (to 273.15K if the pressure has increased to atmospheric pressure, for example).


How to compare the bond lengths of an peroxide complex before and after oxidation? The complex $\ce{K6[(CN)5Co-O-O-Co(CN)5]}$ is oxidised by bromine into $\ce{K5[(CN)5Co-O-O-Co(CN)5]}$.

In this we have to comment on the change on length of $\ce{O–O}$.

But for that we have to find the oxidation number of $\ce{O2}$, so that we can know bond order.

But I am confused, how can I find the oxidation number?

The short and unpleasant answer is that we cannot derive the (spectroscopic) oxidation states based on theory alone. We need experiments (preferably crystal structures and some way to determine the oxidation states of cobalt which are accessible by experiment).

Oxidation states are unfortunately only easy for inorganic molecular compounds or simple salts. In your case, you have a bidentate bridging dioxygen ligand. This could be any of the species $\ce{O2, O2^.-, O2^2-}$ and potentially even more (with an even weaker oxygen–oxygen bond). This is what is called a noninnocent ligand: it can undergo redox chemistry with the central atom. The difficulties of finding out what the actual oxidation state is, is outlined in this question concerning the nature of the $\ce{Fe-O2}$ bond in haemoglobin.

For these cases, IUPAC has defined a formal oxidation state which can be derived according to a set of rules so that one may at least name the complex in oxidation nomenclature. In these rules, both the cyanido and the peroxido ligand would be seen as anions, peroxido being $\ce{O2^2-}$ and cyanido obviously $\ce{CN-}$. By considering the total charge of the $\ce{[Co2(CN)10\mu{-}(O2)]^6-}$ we can allocate cobalt an oxidation state of $\mathrm{+III}$ in the reactant where the maths adds up nicely.

Upon oxidation two things can happen:

one of the two cobalt ions supplies the electron, resulting in a decacyanido-μ-peroxidodicobaltate(II,III)
the peroxido bridge supplies the electron resulting in a superoxido complex with again two cobaltate(III) nuclei.

The resulting crystal structures would be indistinguishable except if one is able to synthesise both products and compare the bond lengths. Therefore, one must resort to other methods of determining the oxidation states. The most practicable way in my humble opinion would be to examine the UV/Vis spectra of the cobalt centres. These should remain low-spin due to the ten cyanido ligands. If we happen to have two $\mathrm{d^6}$, then cobalt was not oxidised. If one of the cobalt centres turns out to be $\mathrm{d^7}$ after the reaction, this should result in a noticable change from the original reactant spectrum.

In the absence of experimental evidence we need to resort to hidden clues in the question. You were asked to consider the oxygen–oxygen bond length. If the metal centre was oxidised, this should not change significantly. However, if the peroxido ligand was supplying the electron for oxidation, the bond order would increase when the species transforms from peroxido to superoxido and this should result in a reduced bond length.

Only with this meta-knowledge we can a posteriori deduce that oxygen’s oxidation state must have gone from $\mathrm{-I}$ to $-\frac12$.


Hydrolysis of β-1,2 glycosidic bond Are there any easy, not enzymatic, ways to hydrolyse $\beta$-1,2 glycosidic bonds like they occur in steviosids?

I thought of something like vinegar, which is commonly used to cleave the $\alpha$-1,2 glycosidic bond of saccharose when producing inverted sugar $-$ could that also work for the $\beta$-type?

Thank you in advance.

There are non-enzymatic methods for removing the three glucose moieties of stevioside 1. Unfortunately, dilute mineral acid hydrolysis not only removes the glucose entities but also effects Wagner-Meerwein rearrangement of the aglycone steviol 2 to isosteviol 3.1 While vinegar (5% aqueous acetic acid) is a weaker acid than hydrochloric acid, I am certain that it would have been tried to avoid the rearrangement of steviol 2. Exposure of stevioside 1 with aqueous KOH effects saponification of the ester group leading to steviobioside 4.---continued---

A more fruitful procedure that avoids rearrangement involves periodic acid cleavage of the five vicinal diols of the glucose units (marked in red in structure 1 below).2 This procedure liberates hexa-aldehyde 5 in addition to two equivalents of formic acid.3  Vigorous base treatment is required to saponify the hindered ester while base also liberates tetra-aldehyde entity by β-elimination forming steviol 2 in 75% yield. And NO, you can't do this at home.

1) Avent, a. G., Hanson, J. R., de Oliveira, Bras H., Phytochemistry, 1990, 29, 2712.
2) Ogawa, T., Nozaki, M., Matsui, M., Tetrahedron, 1980, 36, 2641.
3) On periodic acid cleavage of vicinal diols, see: Periodic cleavage of carbohydrates


Finding out the final product The question is to find out the final product $T$ in the following series of reactions.

I am provided the following options to choose from

With alcoholic $\ce{KOH}$ elimination will occur and with $\ce{KOH}$ syn diol will be formed in place of double bonds.I am unable to see what will $\ce{CaO}$ do to the syn diol.I had an idea that it is used for decarboxylation hence the final product should have 1 carbon less but I am unable to get the right answer.Any ideas?Thanks.

Lets see this step-by-step.

Bromocyclohexane is reacted with alcoholic KOH. This, as you would know, results in the formation of an alkene by elimination reaction. Product P would be Cyclohexene.

Cyclohexene is heated with KMnO4. You might think a cis-1,2-cyclohexanediol would form, and indeed it does. But vigorous oxidation provided by the heat cleaves the C-C bond between the two hydroxy groups and essentially converts into 1,6-hexanedioic acid (product Q, a.k.a adipic acid).

Calcium Oxide, being a pretty strong base,  reacts with the dicarboxylic acid and forms a saltcalcium adipate (product R)

Heating a calcium salt of any carboxylic acid results in the elimination of a $\ce{CaCO3}$ molecule, and forms a ketone. If you take calcium adipate, upon heating would give cyclopentanone(product S).

Now, cyclopentanone, just like any other carbonyl compound with α-hydrogens, undergoes an aldol reaction, heating which would result in the formation of the compound given in option (B).

Here is the complete reaction mechanisms:


Fe 2+ and MnO4- titration? As a lab experiment, I am preparing a mixture of iron tablets dissolved in sulfuric acid so they can be titrated against $\ce{MnO4-}$. The tablet contain insoluble filler material.

Why is it important this 'filler' be removed when preparing the iron tablets solution?

Insoluble doesn't mean unreactive. The filler must be something which would be oxidized by permanganate in acidic solution.


Chloroform-acid wash If I wash chloroform (stabilized with amylene) with sulphuric acid in a separating funnel, where will the chloroform be found (top or bottom)? I think the chloroform will be at the bottom, because it is heavier than sulphuric acid.

Usually, you shake chloroform with concentrated sulfuric acid, wash with water, dry over $\ce{CaCl2}$, and distill.

The density of chloroform is approximately $\rho = 1.48\ \mathrm{g\ cm^{-3}}$, whereas the density of concentrated sulfuric acid is approximately $\rho = 1.84\ \mathrm{g\ cm^{-3}}$.
Therefore, after you allow the layers to separate, the concentrated sulfuric acid is in the lower layer and the chloroform is in the upper layer.

After that, however, when you wash with water ($\rho = 1.00\ \mathrm{g\ cm^{-3}}$), the chloroform is in the lower layer and the water is in the upper layer.

If there is any doubt about the identity of a layer, add a drop of the layer in question to a small amount of water in a test tube. If the drop does not dissolve, the layer in question is the organic layer.

Caution: The mixture may become hot when the remaining sulfuric acid is removed with water, and chloroform boils at $61.2\ \mathrm{^\circ C}$.


Copper powder amount dissoved in water How much copper powder dispersed into 2 liters of water will result in a 10 ppm concentration? I am not a chemist.

If you want 10 ppm copper in 2 liters of water, the best way to make it would be to start with some soluble copper salt, like $\ce{CuSO4.5H2O}$, make up a convenient concentration (like 1%), then dilute it to 10 ppm.

As was mentioned in the comments, copper metal is not noticeably soluble in water. When rainwater washing thru copper downspouts leaves a blue stain on concrete, the copper metal is dissolving because of tiny solubility of copper oxides or carbonates. It might be 10 ppm, or 1 ppm in a lot more water.


How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups? It is my understanding that transition states in an SN2 reaction have some filled and some empty orbital character, as we have the backside of the C-X σ* beginning to fill with electrons from the HOMO of the nucleophile, forming a C-Nu σ bond in the process.

If we have an adjacent carbonyl group to the α carbon, I have learnt that the TS is then only stabilised by the  C-O π*, i.e. by electron withdrawal and not by the C-O π.

However, with adjacent double bonds, I have learnt that the TS is stabilised by both the C-C π and the C-C π*.

I was wondering why this would be the case, as looking at the shape of the orbitals, there seems no reason that the C-O π could not align itself so that the partly-empty p orbital in the TS is parallel to the lobes of this MO, therefore stabilising the TS as new lower energy MOs form.

Also, with regard to the best representation of these structures, I always see partial negative charges on the Leaving group and the Nucleophile in the transition state, but never the positive charge located on the α carbon. Surely if with adjacent carbonyl groups we only have stabilisation by electron withdrawal, it is valid to put the partial positive charge there?

Unsaturated groups conjugated to the site of attack (aromatic rings, alkenes, carbonyl groups) all tend to have the effect of increasing the rate of an SN2 reaction at that centre.

As an example, displacement by nucleophilic iodide in acetone as a solvent is ca. 30x faster for allyl chloride than ethyl chloride, with benzyl chloride being faster still.[1]

The effect was initially thought to be steric, given that unsaturated groups are sp2 hybridised (and therefore planar) and allow the nucleophile a less hindered approach, but many empirical and computational studies now dispute that, with a stereoelectronic effect[*] widely accepted (though sterics may play a small role).

Empirical Evidence for a Stereoelectronic Effect

One of the earliest indications that the effect was stereoelectronic was via a series of experiments conducted by Payne (shown below) [2].The open (acyclic) compound reacts ca. 8000x faster than the closed (cyclic) compound. To understand why, we have to consider the orbitals involved...

The Orbital Explanation

If we accept a stereoelectronic effect (coming back to Payne’s findings later),  we then have to decide whether the effect is a ground state effect (weakening of the C-X bond, raising the energy) or a transition state effect (stabilisation of the TS, lowering the energy). To quote Fleming (emphasis my own):[3]

  [This conjugation] may partly operate by raising the energy of the starting material, but most probably [the conjugating groups] are primarily operating to lower the energy of the transition structures.

In practice, computational studies [4] prove this to be the case, with the pi bond interacting with the principal axis of the SN2 transition state (sigma-pi conjugation):

Calculation of bond lengths also agrees with this mode of conjugation, with the C-C bond length decreasing upon stabilisation of the TS:

Hopefully, by this point, Payne's experimental results are obvious. In the cyclic structure, there is no way of allowing the pi-orbitals on benzene to correctly align with the SN2 transition state (given that we know the TS is linear, not bent).

One final thing to notice is that in both the alkene and carbonyl case (as in your question) the dominant effect is thought to be as above (sigma-pi conjugation).

—

[1]: March, J.; Smith, M. B. Advanced Organic Chemistry Wiley: Hoboken, New Jersey, 2007

[2]: King, J. F.; Tsang, G. T. Y.; Abdel-Malik, M. M.; Payne, N.C. J. Am. Chem. Soc. 1985, 107, 3224

[3]: Fleming, I. Molecular Orbitals & Organic Chemical Reactions:Reference Edition Wiley: Hoboken, New Jersey, 2010

[4]: Bach, R. D.; Soddens, B. A.; Wolber, G. J. J. Org. Chem. 1986, 51, 1030

[*]: Sorry Jan, couldn’t help myself with the bold, I’m not shouting, I promise…


Why is Ka constant when volume is increased? The following graph shows $\ce{Ka}$  vs ${V}$  (Volume) at constant temperature, as ethanoic acid is diluted:

However, I thought it would instead produce the following graph (see below).

$\ce{Ka=[H3O+]*[A-]/[HA]}$

As ${V}$  increases, I would think all of $\ce{[H3O+]}$,$\ce{[A-]}$ and $\ce{[HA]}$ would decrease similarly. As there are two concentrations decreasing in the numerator, and only one decreasing in the denominator, this leads me to believe I can write this as $\ce{Ka ∝ [X]}$, where $\ce{[X]}$ decreases as ${V}$ increases (i.e. that I could cancel out one of the concentration in both the numerator and denominator, leaving only $1$ concentration in the numerator; noting that during ionisation $1$ mole of $\ce{[HA]}$ turns into $1$ mole of $\ce{[H3O+]}$and $1$ mole of $\ce{[-A]}$)

Giving the following graph:

Why is the top and not the bottom graph produced?

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}}$$

The above equation defines the $K_\mathrm{a}$, or acid dissociation constant, of an acid. The reason it is even a value worth measuring is because it is an intrinsic property of the acid, i.e. it's value does not change from sample to sample (temperature aside).

While it is true that as a solution of acid becomes more dilute, $\ce{[A-]}$, $\ce{[HA]}$ and $\ce{[H3O+]}$ all decrease, they do not all decrease by the same amount. Consider the effect of adding water to the following equilibrium:

$$\ce{HA + H2O <=>A- +H3O+}$$

The number of $\ce{HA}$ molecules will decrease, and the number of $\ce{A-}$ and $\ce{H3O+}$ molecules will increase—a conclusion we arrive at via Le Châtelier's principle. Separating the volume of water from the $K_\mathrm{a}$ equation, we get:

$$K_\mathrm{a}=\ce{\frac{mol(A-)\cdot mol(H3O+)}{mol(HA)}}\cdot\frac{1}{V}$$

What we are left with is two larger numbers over a smaller and a larger number. While it's not obvious that these numbers should change to give precisely the same quotient, I can assure you they do.


What is a hydrogen bond acceptor and donor? I recently read about hydrogen bonding. One thing about which I am really confused is how to distinguish between hydrogen bond acceptors and donors.

I looked it up on SE and found this post, which said that water can form four hydrogen bonds as it has 2 hydrogen bond acceptors (the H atoms) and two H bond donors (the two lone pair of electrons on O atom).

When I looked it up on Wikipedia, it said:

  In a hydrogen bond, the electronegative atom not covalently attached to the hydrogen is named proton acceptor, whereas the one covalently bound to the hydrogen is named the proton donor.

And both of them seem to be contradictory. Which is correct?

Dissenter's answer which you linked appears to be wrong (as of now). I have left a comment.

The conventional way in which we talk about hydrogen bond acceptors/donors is: in a hydrogen bond $\ce{X-H...Y}$, the hydrogen bond donor is $\ce{X-H}$ and the hydrogen bond acceptor is $\ce{Y}$. (See for example Chem LibreTexts, or this other question on hydrogen bonds.)

Of course, water is capable of acting as both an acceptor and as a donor:

I wouldn't call it a "proton donor" like Wikipedia does, though, as that suggests some kind of acid-base reaction occurring, like

$$\ce{\color{red}{H2O} + \color{blue}{H2O} <=> \color{red}{OH-} + \color{blue}{H3O+}}$$

In this reaction the red water molecule is acting as a proton donor (Brønsted acid) and the blue water molecule is acting as a proton acceptor (Brønsted base). But these terms should only be used in the context of a proton transfer, and not hydrogen bonding.


Why is cysteine more acidic than serine? How is that possible that the -I effect of $\ce{-SH}$ is greater than $\ce{-OH}$ group?

In case of cysteine $\mathrm{p}K_\mathrm{a}$ value of $\ce{-COOH}$ is $1.96$, whereas in case of serine it is $2.1$; a similar trend can be observed in case of $\mathrm{p}K_\mathrm{a}$'s of $\ce{-NH3+}$ group.

$$
\begin{array}{lccc} \hline
\text{Amino acid} & \mathrm pK_{\mathrm{a}1} & \mathrm pK_{\mathrm{a}2} & \mathrm pK_{\mathrm{a}3} & \mathrm{pI} \\
\hline
\text{Serine} & \color{red}{2.21} & \color{red}{9.15} & - & 5.68 \\
\text{Cysteine} & \color{red}{1.96} & \color{red}{8.18} & - & 5.07 \\ \hline
\end{array}
$$

Good question. First let's recognize that thiols (-SH) are more acidic functional groups than the corresponding alcohols (-OH) (some rationale here). So if we were talking about pKr we would see around 8 for cysteine's thiol side chain and around 15 for serine's alcohol side chain.

Now, to address your question about the carboxyl acidity, pK1, of these compounds. Serine's OH side chain has the more electronegative atom than cysteine's SH group, so we would indeed expect a stronger electron-withdrawing inductive effect in serine, possibly increasing the carboxyl acidity. However, the O atom is two carbons away from the carboxyl group, so this effect may be minimal to negligible... Not to mention of course the empirical data you provided implies that cysteine has the more acidic hydroxyl, not serine.

So I'm just speculating without looking at the actual experiments used to establish the pK1s, but it might be possible that when attempting to measure the pK1 of cysteine's carboxyl, the acidity of its thiol may have slightly skewed the measurement to make the carboxyl appear more acidic. I'm not sure if this was controlled for in the pK1 determination, but either chemically or computationally removing the contribution of the thiol to cysteine's acidity might result in a slight discrepancy compared to what it "really" is.

Additionally, one might imagine that the acidic thiol of could intramolecularly protonate cysteine's amino group, creating a proton-labile positive charge (-NH3^+) quite close to the carboxyl, and this might serve to stabilize the negative charge of the carboxylate conjugate base (-COO^-) thus increasing the carboxyl group's acidity. This effect would be more prominent in cysteine (due to the acidity of the thiol) than in serine (whose hydroxyl would be less likely to protonate the amine to stabilize the carboxylate). So this is a possible (but perhaps oversimplified) mechanistic rationale of why cysteine's carboxyl is more acidic than serine's.

As for the amino group, being more often protonated by the side chain in cysteine, it may also appear to be more acidic in serine.

This is just conjecture, for a complete understanding I would recommend looking at some experiments in the literature to see how they handle determining pK1 and pK2 when an acidic (or basic) side chain is present. It's possible that common practice is to eliminate the side chain's effect by computational or chemical methods, so without examining the primary sources one cannot be certain exactly what the reported pKa values actually represent.


How does refrigerator actually work? In refrigerator, heat energy got extracted from cold sink and dumped into hot source. This is not spontaneous so work must be done. I came across the following while reading a textbook:

  For refrigeration, if no work is done,
   ΔS= qh/Th+qc/Tc;

  -qh=qc, implying ΔS= qh/Th-qh/Tc =qh(1/Th-1/Tc)

  qh>0, 1/Th-1/Tc<0, so ΔS<0 so not spontaneous. Therefore, we need to add heat energy to the stream of heat energy flowing from the cold sink to hot source, so that the entropy increase at hot source 'covers up' the decrease in entropy at cold sink, and the total entropy increases.

It does indeed make sense if we look at the value of entropy, but how can this be feasible physically? You want to move heat energy from cold sink to hot source. In terms of molecular motions, the molecules at hot source always vibrate more vigorously and give energy to molecules at cold sink spontaneously. To achieve your purpose, you literally make the molecules at hot source vibrate faster?

If you still do not get what I am asking, imagine you have a flask of water and you want it to flow from A to B. If you punch a hole at B to remove some water at B, the water will flow to B, but adding water to B definitely will not cause more water to flow to B. Now replace water with heat energy, A with cold sink and B with hot source, I think adding heat energy to hot source should not cause cooling of cold sink.

What is my misconception here?

Refrigerators work by exploiting some properties of compressible gases.

To simplify greatly, gases tend to heat up when they are compressed and liquefied. And, when the liquid revapourises this process absorbs energy. |In a refrigerator those two processes are physically separated and this enables the transfer of heat from one place (the inside, which gets colder) to the outside (which gets warmer).

The way this works in a typical refrigerator is to have a compressor on the outside of the cold area that takes gas and compresses it while letting the extra heat dissipate into the outside environment. This combined process of compression and cooling (or, more accurately shedding the extra heat generated by compression) liquefy the gas. This liquid is then allowed to flow into a heat exchanger on the inside of the refrigerator where it is allowed to evaporate. This evaporation absorbs energy from inside the fridge making that environment colder. This gas is then allowed to flow into the compressor, completing the cycle.

This allows the fridge to convert mechanical energy in the compressor into a clever way of transferring energy from the inside of the fridge to the outside. This amounts to exploiting a phase change in the refrigerant into a pump that moves energy from cold to hot. Though this only works because we can separate the inside from the outside thermally enabling the two side of the process to behave normally (heat flows from hot to cold on both sides) but the net flow ends up taking energy from cold to hot by exploiting the energy used to cause a phase change in the refrigerant.


What happens when an electron in a metal is excited? For special glasses, crystals, or gases absorb energy from an electrical current or another laser they become "excited." The excited electrons move from a lower-energy orbit to a higher-energy orbit around the atom's nucleus. When they return to their normal or "ground" state, the electrons emit photons.

I think the excited electron is always the valence electron. However In metals the valence electrons are not contained to orbitals, they exist in a sea of electrons that flow freely through the object. Does this mean that valence electrons in metals can not be excited or can not move up orbitals? or perhaps the next electron in line moves up an orbital. If so does it join the sea of electrons? (I'm just throwing out ideas here).

Maybe I'm misunderstanding what it means to be 'excited' because, as it says, energy from current can excite electrons.

Any insight is appreciated.

You seem to be misunderstanding what is a "sea of electrons". In fact, this is a metaphor upon a metaphor upon an abstraction. There is no sea. There is a huge bunch of orbitals. (Sure, the solid state people prefer to call them "states", but that's not really important.) The whole piece of metal is a giant molecule. It is not all that different from ordinary small molecules, except that it is very big, and many orbitals span the entire molecule (but then again, that's what they often do in normal molecules).

All these orbitals tend to have different energies. They are everywhere on the energy scale, very close to each other. You point your finger at any given energy, and you find an orbital with that energy. We can't really tell them apart. They kinda blend into a continuous spectrum. And that's what we metaphorically call the sea of electrons.

Electrons are not free, they are confined to some states. When a photon hits, any electron can get excited all right. It will move up to one of the empty states. In fact, it has an immense freedom compared to an electron in a small molecule. Instead of a few discrete states, it has an entire sea of them above. It can get excited to any of them, which in particular means that metal as a whole can absorb a photon of any energy. This, BTW, is also an explanation of metallic luster.

The excited electron is not always the valence electron (you may fire an X-ray photon at a substance and excite an electron from core orbitals deep down below), but that's irrelevant.


How do you separate microcrystalline cellulose from amino acids in a powder mixture? I am looking to separate out the microcrystalline cellulose that has been mixed in with amino acids in a powder mixture.
I have tried suspending the powder mixture in water and then filtering out the microcrystalline cellulose afterwards. The amino acids in the mixture dissolve cleanly without a trace in water after stirring. Although microcrystalline cellulose is supposed to be insoluble in water, which mostly does form a layer in the solution once the fine particles settle after a while, somehow I find that a slight amount of it seems to gets dissolved as well. Either that or it requires an even finer filter, although I have already tried using a filter of 0.1 micron pore size, which should have been enough to filter out any microcrystalline cellulose particles that are still suspended in the solution, but it seems that some traces still remain after filtering.
I would like to cleanly separate out the microcrystalline cellulose from the amino acids in the mixture. Any suggestions as to how I would be able to do this, such as using water or other solvents, or through other means to separate out the microcrystalline cellulose from the amino acids in the powder mixture?

I would try to use a 5-10 cm layer of fine SiO2 particles above a slow filter in a Buchner funnel. I'd be surprised if it didn't do the trick.
Consider that the cellulose particles you see after filtration could come from the paper filter as well (I sometimes see such tiny tiny particles floating around after filtration), in which case it may help to submerge the filter in a beaker of hot water before use, and in the worst case, carry out the SiO2 filtration AFTER the filter paper (in a column with a fritted disk for instance, above larger sized SiO2 not to waste the fritted glass with tiny particles).


How can I mask the Chromium (III) and Chromium (VI) ions in my solution? I want to conduct a quantitative analysis on my acid etching solution for my Aluminum pre-treatment process. The easiest way for me to conduct my quantitative analysis of Al is through complexometric titration with EDTA and back-titration with standardized Zinc solution.
My problem is that there's also Chromium ions in my solution and it can bind with the EDTA which reduces the volume of Zn solution I utilize in my back-titration. [I have confirmed my dilemma because the solution I am using was analyzed through Atomic Absorption Spectroscopy (AAS).]
How can I mask the chromium ions in my solution so I can proceed with the complexometric titration?

If I summarize, you want to make an EDTA titration of $\ce{Al^{3+}}$ in a solution containing also $\ce{Cr^{3+}}$ and $\ce{Cr^{6+}}$. The $\ce{Cr^{6+}}$ ions are usually in the form $\ce{CrO4^{2-}}$ and they don't interfere with $\ce{Al^{3+}}$ in EDTA titrations. But they can be eliminated by precipitation. The best thing to do is to oxidize $\ce{Cr^{3+}}$ into $\ce{CrO4^{2-}}$. In the old Treadwell's book, they propose to take $5$ mL $\ce{Cr^{3+}}$ solution, add $0.5$ mL of a saturated solution of $\ce{Br_2}$ + $0.5$ mL $\ce{KOH}$ $4$ mol/L. The reaction is finished at room temperature in $5$ minutes, and $\ce{Al^{3+}}$ ions do not interfere. $$\ce{2 Cr^{3+} + 3 Br2 + 16 OH^- -> 2 CrO4^{2-} + 6 Br^- + 8 H2O}$$ At the end of the reaction, a solution of barium chloride $\ce{BaCl2}$ or baryum hydroxide $\ce{Ba(OH)2}$ is slowly added to get rid of the chromate ions, by the reaction $$\ce{Ba^{2+} + 2 CrO4^{2-} -> BaCrO4(s)}$$ After filtration of the insoluble $\ce{BaCrO4}$, the obtained solution does not contain any more chromium, it can be acidified and the $\ce{Al^{3+}}$ ions can be titrated by EDTA.


Why can't a nucleophile attack the intermediate of an electrophilic aromatic substitution? During an electrophilic aromatic substitution, the aromatic pi cloud interacts with an electrophile to form an arenium ion (AKA sigma-complex). The following image (from my organic chemistry class) suggests that aromaticity is always restored and that a nucleophile can't interact with the resonanced carbocation of the arenium ion.

If I were to replace the base in the above image with a nucleophile that is an extremely poor base (such as iodine), could the nonaromatic cyclodiene be formed?
Or would the reaction just halt instead?

Follow up question: If electrophilic aromatic subsitution doesn't work, how would one go about making the cyclodiene (with Nu and E substituents) from an arene? Is it possible?

The intermediate is non-aromatic arenium cation which is stabilized by resonance. Deprotonation of sigma complex is a fast process as it leads to much thermodynamically stable i.e., aromatic ring. Aromatization or deprotonation step is fast as Kinetic Isotopic effect is not observed. Hence even a weak base will extract the proton from the substituted site which is sp3 hybridized.

As for another nucleophile to attack the $\sigma$ complex, the positive charge is delocalized and deprotonation step is much faster. I think that it would be easier for the base to extract proton than to attack the Delocalized positive charge. The proton already has a $\delta$+ charge due to Hyperconjugation effect. I think that even if the base if a strong acid, it can dissociate to give H+ but the non-aromatic intermediate will not remain the solution as it is.


Charge density is the amount of charge per unit volume but what is charge? I was studying about x ray crystallography and my book says that charge density is measured in electrons per nano-meter cubed and this means that charge is the amount of electrons. But i am not sure if this definition (charge density = the number of electrons in a unit volume) applies when we are studying crystallography only. So to make sure, is charge simply the amount of electrons ?

An electron has a charge (lets say $e [C]$). If you have a certain amount of electrons in a certain volume ($V [m^3]$), you have a certain charge per volume $\rho = e/V [\frac{C}{m^3}]$. Which is a density.


Reaction of potassium cyanide with 2-(chloromethyl)furan What would the mechanism for the reaction between 2-(chloromethyl)furan and potassium cyanide be, as detailed below?

I know that for 6-membered rings like pyridine, a $\mathrm{S_NAr}$ reaction can occur. However, I'm not sure as to how the mechanism differs between the two in a way that the cyanide group is added at the 5-position instead of the methyl group.

Although the answer was already appropriately provided, for anyone interested in further reading, the reaction is covered in detail by Divald and coworkers:1

Essentially, the proposed mechanism starts from the resonance structures of 2-chloromethyl furan, where the chlorine bond is treated as ionic and the electron-donating property of oxygen stabilizes the shared positive charge similar to a benzyl cation. The 5' position is therefore electrophilic (and is presumably further activated by the inductive effects of a neighboring oxygen), resulting in a 60/40 yield of 5/2-methyl substitution.

It is however mentioned early in the paper that 5' substitution is only seen in protic solvents. This could be explained in the proposed mechanism, where after addition to the 5' position, elimination of the geminal hydrogen and tautomerization proceed through exchange with a protic solvent.

Reference

Divald, S.; Chun, M. C.; Joullié, M. M. Chemistry of 2-(chloromethyl)furans. Reaction of 2-(chloromethyl)furans with aqueous potassium cyanide and other nucleophiles. J. Org. Chem. 1976, 41 (17), 2835–2846. DOI: 10.1021/jo00879a010.


Mixture of water, methanol and dichloromethane If I have a mixture of 25 vol% methanol in water and I mix it with an equal volume of dichloromethane (which is miscible with methanol but not water) can I roughly say that both phases (i.e. water and dichloromethane) will have the same methanol concentration upon reaching equilibrium? Or will it vary significantly based on the activity of each mixture?

This is a partial phase diagram. That seems to indicate that systems with roughly 50% methanol are miscible.

Source: http://shodhganga.inflibnet.ac.in/bitstream/10603/9995/11/11_chapter%206.pdf


Acid Accident Treatment What are the general ways to treat an accident involving highly concentrated or similarly dangerous acids?

I have a lab instructor who recently debated with our group about treating an acid spill on the skin. According to him, one should immediately rinse off the acid with running water for about 15-20 minutes. He says that one should never apply bases to the site because an Acid-Base is highly exothermic and would only increase the damage.

However, I clearly remember learning once that one should apply baking soda (sodium bicarbonate) to a bee sting or ant bite in order to neutralize the acid and relieve the burning.
I also know that the dilution of an acid in water is done by adding the acid slowly to the water containing vessel because of the heat released in the process. And especially when preparing a dilute solution of sulfuric acid, one wears safety goggles and gently stirs the solution to prevent splashing.

So which is it? And does it depend on the severity of the burn or type acid?

My Lab First Aid book$^{*1}$ tells the following:

  The local therapy mainly consists of the instant, intense rinsing (at least 10 minutes) with tap water (ideally 15 to 20 °C). Don't try to neutralize the chemical burn, because the heat generation could lead to further tissue damages.

This means, that your lab instructor is absolutely right.

The rinsing water should wash the cold water away faster than the heat from the acid base reaction could harm you. It's not for nothing that there are emergency showers and eye douches in every lab.

I have a friend who is a chemical lab assistant and once had to carry concentrated sulfuric acid in a big glass beaker. Suddenly the bottom of the beaker broke off and all the acid poured over her etc. She not even thought of anything else than getting out of her clothes and under the next emergency shower.

$^{*1}$ The book is: R. Rossi, Erste hilfe bei akuten Notfällen - Begleitheft zu Jander - Blasius, S. Hirzel Verlag, 2005, p. 11f.. The text is in german and I translated it.


How do I calculate the proportion of respective conjugates of one base mixed with several acids (and/or conversely) at different temperatures? How do I calculate the proportion of respective conjugates of one base mixed with several acids (and/or conversely) at different temperatures, and is there a free software tool that would allow me to interact with the computations therefor (e.g. search the temperature or portion that maximises one conjugate over another)?

If you want a general solution for complex mixtures, where simplified formulas do not apply, you may need to involve numerical mathematics.

As solving the equation set may lead to polynomial equation of higher order you may like.

Enumerate equations of the known.

First, charge balance.
Then, mass balances.
Finally, dissociation equilibriums.

Involve a function of [H+] with a minimum at the value being searched for.

About temperature, there is problem the  constants are rarely provided for other but the standard temperature $\pu{25 ^{\circ}C}$


How can I predict the product of the reaction of 2-methyltetrahydrofuran with HI? I am supposed to explain why treatment of 2-methyltetrahydrofuran with hydrogen iodide leads to the formation of 5-iodopentan-2-ol. However, I am getting stuck as it seems that the less stable primary carbocation is being preferentially formed, leading to the product. Is my mechanism incorrect?

Initially, the oxygen in the ring gets protonated. However, there is no formation of a carbocation. The ring opening proceeds by an SN2 pathway, where iodide ion attacks at the less hindered carbon:


Work done when P,V, and T are changing simultaneously I am given the following P-V graph for an ideal gas:

I have been asked to find the work done in the process from C to A, for which $P$ is directly proportional to $V$. This may be calculated in two ways:

Work done= Area under the line CA, which is the area of the trapezium, given by

$$w = -\frac{1}{2}(P_0+2P_0)V_0 = -\frac{3P_0V_0}{2}$$

(where the negative sign has been introduced since the work done is negative due to decrease in volume.)

Work done = $\int -P\,\mathrm{d}v$. However, in this case, since $P$ is directly proportional to $V$, how to integrate this?

How to calculate the work done in a process where all 3 variables, $P$, $V$, and $T$ are changing simultaneously and $P$ is directly proportional to $V$?

As far as I am aware, there is no generic formula for when all three variables are changing at once.

In this case, the easiest way is to simply use the area of a trapezium.

If you insist on integrating, then you need to know what $P$ is as a function of $V$. This derives from basic principles of integration - if you have a function $f(x)$, then $\int_a^b f(x)\,\mathrm{d}x$ gives the area under the curve of $f(x)$ from $x = a$ to $x = b$.

The line $\mathrm{C \rightarrow A}$ can be easily found since the beginning and end points are given. The gradient of the line is

$$m = \frac{2P_0 - P_0}{2V_0 - V_0} = \frac{P_0}{V_0}$$

and the the intercept is clearly 0 from the diagram (you can work it out anyway). In general the equation of a line $y = y(x)$ is given by $y - y_i = m(x - x_i)$ where $(x_i,y_i)$ is any point on the line.

That was perhaps an unnecessarily long explanation but the point is that you now know the relation

$$P = mV$$

holds (where $m$ was determined earlier). Therefore

$$\begin{align}
w &= -\int_{2V_0}^{V_0} P\,\mathrm{d}V \\
&= -\int_{2V_0}^{V_0} -mV\,\mathrm{d}V \\
&= m\left[\frac{1}{2}V^2\right]_{2V_0}^{V_0} \\
&= m\left(\frac{1}{2}V_0^2 - 2V_0^2\right) \\
&= \frac{P_0}{V_0}\left(-\frac{3}{2}V_0^2\right) \\
&= -\frac{3}{2}P_0V_0
\end{align}$$

which is the same as you found via the first method (you also forgot to put the negative sign in the right-hand side of the equation, but I assume that is just a typo).


How does water defy gravity in paper towels? How can water be absorbed into a paper towel such that it continues to be absorbed up into the material against the force of gravity? What property of paper towels/water causes this? I can understand that cohesive forces cause water molecules to follow other water molecules, but why do the molecules begin this climb in the first place?

This is simply an example of capillary action expanded to a sheet of many tiny capillaries. In a narrow glass tube immersed in water, water will climb the inside of the tube to a level higher than that of the water outside because it is energetically favourable for the water to wet the surface due to adhesive forces and the cohesive forces of the water pull the bulk of the water up with the contact line of the water and the glass. Glass and water have strong adhesive forces as glass has many silicate groups which can interact with water through hydrogen bonding and other polar interactions. As the liquid rises, the volume of water above the surrounding liquid increases, thus increasing the weight of the column. When the weight of the column equals the adhesive and cohesive forces drawing the water upwards, the system is at equilibrium and the water will rise no further.

This can be calculated from: $$h = \frac{\gamma\cos\theta}{\rho gr}$$ where h is the height of the column, $\gamma$ is the liquid-air surface tension, $\theta$ is the equilibrium contact angle between the glass and the water, $\rho$ is the density of water, g is the acceleration due to gravity, and r is the radius of the tube. As you can see, increasing the surface tension (which comes from water's cohesive forces) or decreasing the contact angle (the lower the contact angle, the stronger the adhesive forces between liquid and surface) will increase the height the column will reach, and conversely a denser liquid or larger tube will decrease the height (greater increase in weight for a given rise).

I won't do the whole calculation but for a 1 mm radius glass tube with water ($\theta \approx 50°$, $\gamma = 72\ \mathrm{mN/m}$), the equilibrium height is 4.7 mm. Now, if we shrink the tube much further to approximate something like a pore in a piece of paper towel, we can make the liquid go much further. (it's obviously not perfectly cylindrical, but close enough for illustration and the cellulose-water contact angle seems to be close enough to glass' as cellulose's abundance of hydroxyl groups is similarly adhesive to water) At a diameter of 10 µm, we get a height of 47 cm.

When we expand this concept to a large sheet of interconnected pores, we find that capillary action can draw a relatively large volume of water across considerable distances, which in addition to being great for cleaning up household spills, opens the possibility of using paper as a substrate for analytical devices that don't need pumps, including the lateral flow assay (most commonly seen as the pregnancy test stick) and paper microfluidic devices.


Chain with double bond or triple bond In this hydrocarbon there can be two main chains: one with two double bonds and one with a double and a triple bond

Both main chains contain seven carbon atoms.
Which will be considered correct and why?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the first relevant criterion to be considered in choosing a principal chain is the length of the chain.
When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)

Since both possible chains in this case have two multiple bonds, Rule (a) is not enough to make a choice.
According to Rule (b), the principal chain is the hepta-1,6-diene because it has the greater number of double bonds.
Therefore, the preferred IUPAC name (PIN) is 4-(prop-2-yn-1-yl)hepta-1,6-diene.


Can ionic and covalent bond be explained in terms of orbitals? Covalent bonding is stated as overlap of atomic orbitals .In ionic bonding the composite ions are held together by their electrostatic charges that result from the transfer of electrons from one atom to the other. What is exactly meant by "transfer"
in terms of orbitals?

The "transfered" electron will be in an orbital that is mostly on the side of only one atom, i.e. the anion. That orbital will be slightly deformed by the presence of the cations around it, but no significant overlap with any of their outer orbitals (those which are as good as empty, because the electron has been lost to the other atom).

(Im writing "mostly" and "as good as" because there can always be a covalent contribution to the bond.)


why can't I get my absorbance values to come out right on my oxybenzone uv-vis assay test? I am attempting to assay a raw material sample, oxybenzone, against oxybenzone, USP using UV-vis. The requirement for acceptance is NLT 99% mg C14H12O3/mg raw material sample. The method instructs a standard preparation of 10ug/mL in CH3OH, and then a sample preparation by 100 mg sample oxybenzone in 100 mL, pipette one mL into another 100 mL flask, dilute to volume with CH3OH. I used these same steps to prepare my standard solution as well. Scan and use A value to calculate mg C14H12O3 in the sample using 10(C)(At/As) with At being A of the sample and As being A of the standard, C being concentration of the standard solution in ug/mL

assuming I know my sample should pass, my results make no sense.

As=.8341
At=.6304
C= 10

I used methanol as the blank per the method.

I am sure I must have made a mistake somewhere, but I am stuck. The only thing I can think of is if I made a mistake preparing the standard solution and it was too concentrated perhaps A would go up, but I don't think I did. I weighed 100 mg oxybenzone USP standard, diluted to volume, removed 1 mL, diluted to 100 mL again, same as I did for the sample solution prep.

The method, and your result, both make sense to me: it indicates that the test sample is 75.57% pure, although as you point out the result is not consistent with the assumption that the sample should pass the requirement for acceptance. However, why would you run the test if you knew ahead of time that the material would pass?

I would recommend that you repeat your test 2 more times from scratch - that way you would know if there was an error in your technique. If you get the same result (within experimental error), then you would have to admit that it is hard to argue with the data.


Is there an explanation to the atomisation energy trend of transition metals? I am familiar with the double hump trend for atomisation of transition metals, and I can explain the unusually low atomisation of $\ce{Mn}$ with the exchange principle, specifically that in the solid state the spins are more localised whereas for the gas/vapour the spins are able to match and hence there is a gain in exchange energy (stabilising).

However, what I am struggling to understand is (ignoring the middle drop due to exchange):

  Why there is a hump in the first place?

I have come across a few ideas, some suggesting that it is a bonding/antibonding situation, whereby, first you fill a d-band gap up to $\ce{Mn}$ and then an antibonding contribution comes in.

Is there a more eloquent way of explaining this trend?

The hump is apparent because of the dip. To be more clear, it is good to know that all these attributes follow the same trends:

Metallic character
Strength of metallic bonding
Enthalpy of atomisation
Melting point
Density

If you plot a graph with any of these attributes on the $y$ axis, you'll notice that more or less in each of these cases, the graph rises first reaches a maxima just before the middle of the series, undergoes a dip in the middle again rises and then finally decreases till the end of the series.

It can be somewhat explained (these observations are made on an experimental basis) by the strength of metallic bonding. As the number of unpaired electrons increase, the metallic bonding becomes stronger. Note that $\ce{Cr}$ has the highest number of unpaired $e^-$s: $6$. This explains the hump.

$\ce{Mn}$ on the other hand has a half-filled electronic configuration. Thus, it is more stable. Stability order on the grounds of electronic configuration follows the order:

  Fully filled > Half filled > Partially filled

A lesser strength of metallic bonding, plus the stability collectively account for the dip.

Also, as a quick roundup of densities, density of $\ce{Os}$ is the highest: $\pu{22.59 gm cm^{-3}}$ and $\ce{Ir}$ stands quite close $\pu{22.56 gm cm^{-3}}$. Another fun fact, gold and mercury are also quite dense: $\pu{\rho_{Au} = 19.32 gm cm^{-3}}$, $\pu{\rho_{Hg} = 13.56 gm cm^{-3}}$. As a comparison (as to how dense I actually mean): $\pu{\rho_{Pb} = 11.34 gm cm^{-3}}$.

The low melting points of $\ce{Mn}$ and $\ce{Tc}$ are also mapped to half-filled stable configurations.


Do hydroxymethyl-substituted cyclopropylmethyl cations undergo ring expansion? In the image, what would happen after step 2?

Will a double bond be formed between carbon 3 and 4? If this is the case, then we would get benzene in the end.
But, if the positive charge in shifted to carbon 1, then won't it be a more stable carbocation?
Also, if a double bond were to be formed between carbon 1 and 4 now, it would be more substituted than a double bond between carbon 3 and 4.

In spite of all these factors, the double bond is formed between carbons 3 and 4, and benzene is formed.

Kindly explain how this process takes place.And, please tell me how to name this compound.

The following takes place-

And then H3O+ can be assumed to give H+ and H2O.
Try it out and you'll get to the product.

If the positive charge is shifted to carbon 1, it would form a more stable alkene since it is more substituted. But after formation of C1-C4 double bond, no further reaction will occur since after expansion + charge will be on a double bonded carbon which would be highly unstable. Since Benzene is more stable than the product which will be formed after shifting of + charge to Carbon 1, and since heat is also applied, there will be enough energy to form more stable benzene and thus the reaction takes place in such a manner.


Converting pounds to grams I wanted to figure out how many grams were to a lb and instead of looking it up I used my textbook and found 1 kg = 1000 g and 1 kg = 2.205 lb. I set up a cross multiplication equation to solve for x. The number I got was close but off by a few hundredths. I got 453.5147 g = 1 lb. I looked to see if I was right and found that the actual number is 453.59 g to lb. What did I do wrong?


Does melting HDPE plastic produce toxic fumes? (melting, not burning) ● The question isn't about burning HDPE but melting it at the proper temperature. (At 120 to 180°C depending on it's density, it becomes gooey. According to the source below the "extrusion" temperature range of HDPE is 176 to 260°C ― at 260°C, I guess it becomes a very viscous liquid).

Is it perfectly safe (eg: indoor melting, in a kitchen oven). Or does it produce toxic fumes ― before 180°C? Before 260°C? If so which one?

● And what about melting it canola oil? (This technic seems more energy efficient, and easier to control the temperature).
This experimenter tried with Canola oil, which has a smoke point of over 204 °C. He melted the HDPE in the oil at 176°C (the output needs to form a block and expresses the oil).

Is it safer to melt HDPE in oil ?

○ Bonus: does the Canola oil degrade the HDE? (The experimenter found out that the oil "gets really thick when cool" and he thinks it's because some HDPE dissolves on it. Someone comment "The canola oil probably acts like that because some of the HDPE is converted into paraffin wax which is soluble in canola oil. HDPE isn't lipid soluble".

Any information on the topic is welcome

It's not a duplicated:
This question has been asked about specifically melting empty silicone tubes (made of HDPE). It's off topic since not about HDPE only (it's a complex issue since there are many different types of silicone on the market)

This table indicates that polyethylene (HD/LD not specified) "melts at" 135°C, decomposes in the range 335-450°C, and produces vapors that will ignite between 341-357°C.

I'm sure melting/softening temperature is more complex than that, but I'm not sure that you should expect much decomposition into toxic gases at temperatures below 260°C.

However, if you're melting it in an oven, it's possible to overshoot the setpoint temperature. Your oven's thermostat might be slow to kick on and off. Direct thermal radiation from the heating element can heat an object in the oven well beyond the air temperature. Your plastic can even spill/drip onto the heating element.

I wouldn't like the idea of disposing of canola oil contaminated with unknown HDPE breakdown products. I'd be happier using some sort of double-boiler arrangement, where the HDPE is in a container that's heated by a bath of some other fluid. I'd suggest paraffin as a heat-transfer fluid, but its flash point is a bit low (200-240°C?)


What is exactly meant by position of equilibrium? I'm confused with the terms "position of equilibrium" and "equilibrium shifting towards left/right". I do have a vague picture that if a reaction shifts to the right, for example, it basically means that the reaction is now biased towards the products. But I'm not thorough with it. Does it mean that the rate of forward reaction is now faster than the rate of backward reaction, which would mean that the reaction isn't at equilibrium anymore?

The term "equilibrium shifting" is mostly used for a reaction that was already at equilibrium, but we changed something (in the simplest case adding one reactant or product to the equilibrium mixture so the reaction is no longer at equilibrium). The "equilibrium is shifting" is a shortcut for saying there is a net reaction to re-establish equilibrium.

The term is used both in cases where the equilibrium constant does not change (change in one of the concentrations) and in cases where the equilibrium constant does change (change in temperature, for example).

  Does it mean that the rate of forward reaction is now faster than the rate of backward reaction, which would mean that the reaction isn't at equilibrium anymore?

Yes, for some reason the reaction is out of equilibrium, otherwise there would be no net reaction (no "shift").

  I'm confused with the terms "position of equilibrium" and "equilibrium shifting towards left/right".

"Position of equilibrium" could be roughly equated with the reaction quotient Q or more generally with the set of all concentrations at equilibrium.

Taking a change in temperature as an example, you start with one set of equilibrium concentrations ("old position of equilibrium" corresponding to some reaction quotient $Q_1$). As you change the temperature, the reaction is no longer at equilibrium because the equilibrium constant changes (lets say from $K_{T_1}$ to $K_{T_2}$), so $Q_1$ is equal to the former equilibrium constant $K_{T_1}$, but not to the current equilibrium constant $K_{T_2}$). There is a net reaction until you reach the new equilibrium ("new position of equilibrium" corresponding to some other reaction quotient $Q_2$). At the first temperature $T_1$ after equilibrium is established, $Q_1 = K_{T_1}$. At the second temperature, $Q_2 = K_{T_2}$ once the equilibrium is re-established.

For adding or removing one of the species, you start with the old position of equilibrium, then the addition or removal of a species changes Q, which no longer equals K. As a consequence, there is a net reaction that changes Q back to K. Now, you have a new set of concentrations, but the reaction quotient is back to what it was at equilibrium. I'm not sure whether to call this a new position of equilibrium, or to say the reaction went back to equilibrium (this might be confusing because the equilibrium corresponds to a third set of concentrations - first set: original equilibrium concentrations; second set: concentrations right after the manipulation; third set: concentrations after the net reaction re-establishes equilibrium).


Doesn't methyl amine give carbylamine test? A compound $\ce X$ of formula $\ce{C2H5NO2}$ on treatment with $\ce {HNO2}$
   $(\ce{NaNO2}+\ce{HCl})$ gives off an inactive gas $\ce Y$ and forms $\ce Z$. When heated with soda lime, $\ce X$ gives $\ce{A(CH5N)}$ soluble in acid. Find the true statements :

  $\ce A$ answers carbylamine test.
  $\ce X$ is an amino acid.
  $\ce Y$ is $\ce{CO2}$.
  $\ce Z$ is a hydroxyl acid.

My attempt :

$\ce X$ is glycine.

$\ce A$ is methyl amine.

$\ce Y$ is dinitrogen.

$\ce Z$ is 2-hydroxy ethanoic acid.

Hence, correct options should be : $1,2,4$. But the answer given is $2,4$.

Can anyone help me to understand if and where I went wrong?

You are indeed correct.

$\ce{HOOC-CH2-NH2(X) ->[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$

$\ce{HOOC-CH2-NH2(X) ->[soda lime] CH2-NH2(A)}$

Also, $ \ce{(A) + CHCl3 + 3KOH ->[Heat] CH3-NC + 3KCl + 3H2O}$.

Methylamine does give the carbylamine test.

So clearly, the answer given must be wrong.

The correct options are 1,2 and 4.


Understanding the solubility of Ca(HCO3)2 According to wikipedia $\ce{Ca(HCO3)2}$ has the following solubility values:

$\pu{16.1 g/100 mL} \,(\pu{0^\circ C)}$
$\pu{16.6 g/100 mL} \,(\pu{20^\circ C)}$
$\pu{18.4 g/100 mL} \,(\pu{100^\circ C)}$

So I assume $\ce{Ca(HCO3)2}$ would precipitate beyond this level. However,  the wikipedia page for calcium bicarbonate states that attempts to prepare compounds such as solid calcium bicarbonate by evaporating its solution to dryness invariably yield instead the solid $\ce{CaCO3}$:

$\ce{Ca(HCO3)2(aq) → CO2(g) + H2O(l) + CaCO3(s)}$

which leaves me wondering if, when the solution is saturated beyond ~$\pu{16-18 g/100 mL}$, a precipitate of $\ce{Ca(HCO3)2}$ will exist  in solution? or if it also dissociating beyond that point while still in solution?

Calcium bicarbonate $\ce{Ca(HCO3)2}$ simply does not exist. It is impossible to fill up any container with pure $\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing $\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$, by bubbling $\ce{CO2}$ into a solution of calcium hydroxyde $\ce{Ca(OH)2}$. The following reaction will happen:

$$\ce{Ca(OH)2 + 2CO2 -> Ca^{2+} + 2HCO3^-}$$

The result is exactly what could be expected if some $\ce{Ca(HCO3)2}$ had been dissolved in water. And of course, it makes sense to admit that by evaporating the solution, the calcium bicarbonate could be obtained dry. Unfortunately it is not the case. When trying to evaporate this solution, $\ce{Ca(HCO3)2}$ is always and totally decomposed according to the following equation:

$$\ce{Ca^{2+} + 2HCO3^- -> CaCO3 + H2O + CO2}$$

And all you obtain by trying to evaporate a solution containing he ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$ is a deposit of calcium carbonate $\ce{CaCO3}$, whatever the method used for carrying out this evaporation.

Of course you can calculate the maximum amount of Calcium and of $\ce{HCO3^-}$ ions that may exist simultaneously in such a solution. You may even calculate the solubility of $\ce{Ca(HCO3)2}$. But this does not prove that $\ce{Ca(HCO3)2}$ exists as a pure compound. $\ce{Ca(HCO3)2}$ is a compound that only exist in solutions.


Order of filling electrons in Jahn-Teller energy diagram When filling electrons in atomic or molecular orbitals we fill first the lower energy orbitals and then the higher, according to Pauli and Hund's rule. But this seem not to be the case when considering metal complexes. According to Wikipedia

For example why the $d_{xy}$ orbital in the $d^4$ high spin is not fully occupied? Is the pairing energy that changes the order of filling the orbitals?

This was just an illustrative example. The tendency of a complex to favor low-spin or high-spin configurations is dictated by the relative magnitudes of the pairing energy, P, and the splitting energy, $\Delta$, between $e_g$ and $t_{2g}$.*

If $\Delta$ is greater than the energy required to pair two electrons, then electrons pair in $t_{2g}$ before occupying $e_g$. In the example you've shown of compression, this would be the $d_{xy}$ orbital.

*For octahedral complexes. Jahn-Teller can also happen for tetrahedral complexes.


Enthalpy of formation of ammonium bicarbonate I'm helping with correcting lab reports, and one of the experiments the students did was heating of ammonium bicarbonate, $\ce{NH_3HCO_3}$ in powder form, over Bunsen burners, which turns the compound into water, carbon dioxide and ammonia (all gasses). They are supposed to say, in their report, whether this is an endothermic or exothermic reaction.

Now, I would be very surprised if this reaction turned out to be exothermic. An hour of googling seems to agree with me, but I can't find a definite answer. In the course we've talked about enthalpy of formation and Gibbs free energy, so it would fit into the curriculum if we could find the relevant data.

Unfortunately, it's apparently very difficult to find the enthalpy of formation of sodium bicarbonate through google. Many small tables do not list the chemical at all, and the other sources that I've found does not list enthalpy of formation. I've found one number, which is -32 kcal, without specifying whether it's per mole, g, kg, ounce or pound.

Does anyone have a source on the enthalpy of formation of solid $\ce{NH3HCO3}$ at room temperature?

−2580 cal/g.  Here is a reference that lists heats of formation:

https://engineering.purdue.edu/~propulsi/propulsion/comb/propellants.html


Why Zn has highest ionisation enthalpy in 3d series? Zn which has the highest ionisation enthalpy in 3d series.The reason given in my textbook is:

  The value of zinc is higher because it represent ionisation from 4s level.

This is not correct because for every 3d element ionisation is from 4s orbital. The explanation I found for this is from this site is:

https://www.chemguide.co.uk/atoms/properties/ies.html

  Increase in nuclear charge outweigh the electron electron repulsion of 4s electrons rendering Zn an higher ionisation enthalpy than Cu.

This seems like a probable explanation but if Zn has higher ionisation enthalpy than Cu due to its increased nuclear charge.Why can't Zn have smaller atomic radius than Cu due to its increased nuclear charge which can decrease the size of the atom. But experimentally it has been found that Zn is bigger in size than Cu.

So why does Zn has the highest ionisation enthalpy in 3d series?

TEXT SOURCE:NCERT textbook (chapter:d and f block)

There is band structure formed in the metallic lattice  in the crystals of d and f block elements due to partially filled orbitals. But as you move along the periodic table to 'Zn' , you will find that the orbitals are completely filled hence there is no possibility for inter-orbital mixing. Another thing worth noting is that due to the filled d-orbital, zn and copper behaves more like s-block rather than d-block elements.

"The filled d-shells in these elements contribute little to interatomic interactions, which are dominated by the s-electrons through metallic bonds. Unlike metals with incomplete d-shells, metallic bonds in copper are lacking a covalent character and are relatively weak. "

An alternate explanation can be on the metallic lattice:
Copper has a face centered cubic structure and zinc has a hexagonal close packed structure

reference : wikipedia


Why can't all electrophiles act as Lewis acids? We know that all Lewis acids are electrophiles but I have a confusion whether the reverse is also true. Electrophiles are molecules that are in search of electrons while Lewis acids are short of electrons. So can't all electrophiles act as Lewis acids?

The definition of Lewis acids states:

  A molecular entity (and the corresponding chemical species ) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:

  (source: iupac.org)

In principle, since all electrophiles can be electron acceptors, they can usually act as Lewis bases, too. However, sometimes sterically demanding moieties stabilise the electrophile, so that it is impossible to form a Lewis adduct. These compound are then one part of a frustrated Lewis pair.


How easy is it to make animal horns transparent? I never knew that HORN in HORNbook refers to animal horns, or animal horns can be made transparent? If Medievalists can make it transparent, the transparency process must've been cheap easy and quick, right?
The Horn Book | Why is it called "The Horn Book"?

Back in the sixteenth century, English monks began to make hornbooks to help their pupils learn to read. Usually a wooden paddle with an alphabet and a verse glued to the surface, hornbooks derived their name from the piece of transparent horn protecting the verse. The picture to the right shows a modern replica of a hornbook.

Hornbook - Wikipedia

A hornbook is a book that serves as primer for study. The hornbook originated in England as long ago as 1450,[1] or earlier.[2] The term has been applied to a few different study materials in different fields. In children's education, in the years before modern educational materials were used, it referred to a leaf or page displaying the alphabet, religious materials, etc., covered with a transparent sheet of horn (or mica) and attached to a frame provided with a handle.[3]

I found this description of how the horn is prepared in a post by Tammy L. Austin, at the University of Notre Dame's website (https://www3.nd.edu/~rbarger/www7/hornbook.html):

The horn of oxen and sheep were used to make the laminating structure. The horn was left in cold water for several weeks, which separated the usable part from the bone. It was then heated, first in boiling water then by fire, and pressed by plates and machines to make it smooth and transparent.


What's the most efficient way to find crystal structures of drugs? I'd like to be able to find the crystal structures of drugs, for example at the moment I want to find the crystal structure of (R)-fluoxetine so that I can use Jmol to create an accurate 3D model of it. This is so I can then upload it to Wikimedia Commons so it may be used alongside Benjah-bmm27's structure of (S)-fluoxetine on the Wikipedia page for fluoxetine. I have tried this database [1] but when I searched for the SMILES of (R)-fluoxetine, CNCCC@HOC2=CC=C(C=C2)C(F)(F)F (from ChemSpider [2]), I obtained no search results.

I think my first question is why you need the crystal structure?

For organic molecules, the definitive resource is the proprietary (and fairly expensive) Cambridge Structure Database.

The Crystallographic Open Database which you linked is fairly new and expanding, but does not have the many, many molecular crystals in the CSD. For example, American Chemical Society journals required submission to the CSD for publication.

My suggestion would be to start with the NIH Chemical Resolver or PubChem. The latter has a 3D computed structure of (R)-fluoxetine, which I'm sure is pretty decent. Jmol should be able to load these directly, or you could save SDF files and open them in Jmol.

I'll point out that Avogadro can also do these tasks, but since I'm the creator, I'm biased.


Why are elements on the right side of the periodic table nonmetallic and gas at room temperature? Elements on the left side of the periodic table tend to be solid and metallic, elements on the right side of the periodic table are nonmetal and tend to be gases at room temperature, and the semi-metals are in between.  Why is that?  Why are the elements on the right side nonmetallic and in a gas phase?

The elements on the left-side of the periodic table are relatively electron deficient (i.e., they have few valence electrons), and due to their comparatively low effective nuclear charges (the net positive charge of the protons minus the shielding core electrons below the valence level), their electrostatic hold on these electrons are weak. Consequently, the electrons of metal atoms tend to be highly delocalized, not forming localized covalent bonds or possessing tightly held lone pairs. In metals, the electrons are effectively shared between many atoms, loosely bound in what is often described as an electron "sea," and the atoms of metals arrange to achieve maximally efficient packing in a crystal lattice, through which the electrons (particularly the most extensively delocalized electrons in the s and p orbitals at the highest main energy level) move freely. The tight packing of the crystal lattice accounts for the density and hardness of metals, while the extensive delocalization of electrons partially explains their malleability and conductivity.

Elements further right on the period table, on the other hand, have higher effective nuclear charges and stabilize electrons more effectively, leading to localized covalent bonding and the formation of molecules (vs. the ionic bonding and crystal lattices of metals). Electrostatic intermolecular forces of attraction are generally weaker than ionic bonding or delocalized metallic bonding, so non-metals and molecular compounds are more likely to exist in the liquid or gaseous phase at comparatively low temperatures than metals or ionic compounds.


Lewis structure of ions In class, we learned about the Lewis structure of some elements such as carbon, but we didn't learn how to do the Lewis structure of cations and anions.

Do we just draw the number of electrons in the ion, or do we do something different?

So for example, what would be the Lewis structure of $\ce{Li+}$ and $\ce{Li-}$? How would you draw it out?

The charge on a centre in Lewis structures is not of special relevance, except that it helps to provide the information which is useful: the number of electrons on that centre.

A positive charge of $X$ means $X$ electrons were lost, while a negative charge, $-Y$ means $Y$ electrons gained.

From this you can either simply remove/add the corresponding number of dots from/to the diagram, or calculate how many valence electrons are now present, and then draw the diagram ab initio from this information.

A "normal" lithium centre has 1 valence electron, so the positive ion just loses it (leaving no dots) while the negative ion gains one more (and thus has two dots on the diagram).


What is Cβ (C-beta) deviation? Molprobity and some other protein structure validation tools report a Cβ deviation statistic and offer plots for it (example below).  Apparently if the Cβ is greater than 0.25 Å, some attention should be given to the residue.

The statistic is given in units of angstroms, but I'm not sure what it's actually referring to.  With the Rama plot, it's the peptide dihedral angles, but what is this?  And why are the graphs 2-D despite the one dimension given?

Just browsing through this "Unanswered Question" now (in late 2014). I hope it's not too late?

In protein structure validation, Ramachandran outliers are a strong, but not absolute, indicator that the backbone confirmation of your protein structure deviates from all the high resolution structures known to date, and hence need your attention during the progress of refinement.

If all the phi-psi angles (backbone conformation) all fall into the "allowed/favoured" regions on the Ramachandran plot, your structure model is probably, but not necessarily, esp in the case of low resolution, correct. Hence there are other weaker criteria in protein structure validation, and Cbeta deviation is just one of them. In the light of high resolution structures out there, it turns out that Cbeta also has a favoured range of dihedral angles, suggesting that any misfit between your structure and the "probable" side chain-backbone conformation also deserves your attention during structure refinement.

In MolProbity, the justification and validation procedure are described here.