What is the process of turning ionic silver solution into colloidal silver solution? I've looked at a number of different colloidal/ionic "silver solution generators."  After my research I'm under a few impressions and if they're wrong please let me know.

I also am wondering what the product is that one generator sells to add to the ionic silver to create colloidal silver and is it safe to ingest?  I've already seen enough information to believe that ionic silver is safe. I'm wondering if the additive that turns ionic silver into colloidal silver is safe.

Impressions:

When you set up a silver generator, either homemade or store bought, you send a current through two pieces of silver in distilled water.  The particles released into the water are ONLY ionic silver.

Will colloidal silver form over time?  Will ionic silver precipitate or remain suspended?

I could easily build a machine to do this rather than buy one from a dealer that seems to withhold info to either prevent comparisons or to prevent laymen from building their own.  I'm looking for any and all comments related to this topic.  Thanks

To reply to GM's answer: I hesitate to put up pictures or links to the commercial generators since I'm not sure if that would violate the TOU and I have no interest in advertising for anyone.  I imagine if the commercial generators, which work on same principal as the home made ones, didn't generate at least ionic silver they'd have some explaining (lawsuits) to do to someone and they've been on the market for years.

If you provide a scheme of this mysterious  "colloidal silver generator" would be very interesting, from what I see from googling I see only things like this.

Up to my knowledge this "colloidal silver generator" has no sense. From instructions notes it seems there are two silver bar in a solution of pure water, as you say, this in fact should create a very slow (because there are no electrolytes) electrolysis of water into $O_2$ and $H_2$ but I don't see any reasons for the producing of colloidal silver.
In chemistry laboratory to produce colloidal silver usually you start from a solution of ionic silver $Ag^{+}$ and reduce it to metallic silver $Ag$.
$$\ce{Ag^{+} + e^{-} -> Ag} $$
Caution the following is a reaction not a recipe do not ingest the ultimate product of this reaction!!!
Once time I've made a true colloidal silver with Lee-Meisel method (J.Phys. CHem 1982,86,3391-3395) if you look at the original paper there are many methods however the following (method c) is used most.

(c) $AgNO_3$ ($90 mg$) was dissolved in $500 mL$ of $H_2O$ and brought to boiling. A solution of $1 %$ sodium citrate ($10 mL$) was added. The solution was kept on boiling for cs. 1h.

The solution should be greenish yellow. If you examine the reaction the $AgNO_3$ is the source of $Ag^{+}$, sodium citrate is the reducing agent that reduce $Ag^{+}$ to $Ag$ and stabilize it. I think colloidal silver generator would be for anyone who makes colloidal solution a dream I assure you is a very complex and tedious process.
Moreover both $Ag^{+}$ and colloidal silver are very difficult to store, the first may  undergo photo-reduction to the metal the second is unstable as many colloidal solution because colloidal solution have grater energy due their larger surface (More energy many times means more instability). Without any stabilizer (such as t-BuONa or citric acid) I don't think the colloid will last long.
However I don't see why you should drink something like this. In fact $Ag^{+}$ has antiseptic properties such as  Mercurochrome,HydrogenPeroxide or Lysol(R) but this don't mean drinking Lysol, Hydrogenperoxide or mercurochorme every morning will make you feel better. I strongly suggest you to do not drink such thing ever.


Do really the electrons could only orbit in specific orbits with a fixed radius? According to Bohr's model of atom

  The electrons could only orbit the nucleus in specific orbits or shells with a fixed radius. Only shells with a radius given by the equation below would be allowed, and the electron could not exist in between these shells.
  r(n)=n2
   ⋅r(1)

But when I learnt about different orbitals inside the atom in my textbook, it was said that you cannot precisely tell the location of an electron, as you know the velocity, according to Heisenberg's uncertainty principle. Besides in a video of khanacademy.org Sal said that orbitals means that if we took a number of snapshots of atoms, we would see that the electrons are denser in those orbitals but that does not necessarily mean that we wouldn't find it outside of the orbitals. The probability of finding electrons is higher in the orbitals but not 100%. But Bohr said electrons are quantized. It must be in a fixed orbital with a fixed radius. Isn't that contradictory? Which one is true? Or what is wrong in my understanding?

Bohr said electrons are quantized. It must be in a fixed Orbital with a fixed radius. Isn't that contradictory?

No it is not contradictory.

"Quantized" means that a continuous range of a variable, such as energy, is not allowed.  Only discrete values of the variable are allowed.

In Bohr's model, the radii and energies were quantized.

  Which one is true?

It is not true that electrons orbit in fixed circular orbits.  Subsequent to the Bohr theory, there was Sommerfeld theory, which involved elliptical orbits and was a relativistic theory that correctly predicted the fine structure of the hydrogen atom (which Bohr's theory did not). Then there was Schrodinger's theory, which did away with the fixed orbits, but was non-relativistic and gave the same energy levels as Bohr's theory.  This was followed by Dirac's theory which was a relativistic theory and gave the same energy levels as Sommerfeld's theory.  However, the hyperfine structure of the hydrogen atom, such as the Lamb shift, was still not explained, until quantum electrodynamics was developed. Further developments included the recognition that the proton is not a point particle, but is composite.  So there has been no finding the truth, just more accurate, but more complex, approximations of the truth.

  Sal said that orbitals means that if we took a number of snapshots of atoms, we would see that the electrons are denser in those orbitals but that does not necessarily mean that we wouldn't find it outside of the orbitals. The probability of finding electrons is higher in the orbitals but not 100%

Orbitals have an infinite spatial extent.  There are nodes where the wavefunction is zero for the Schrodinger equation solutions of the hydrogen atom, but nodes are specific planes, conical surfaces or spherical surfaces, not 3-dimensional regions.  People depict the orbitals as having finite volumes, but these depictions are the high probability regions not the entire orbital.  So if Sal said that, he should have said that the electrons can be outside of the high-probability regions of orbitals that people draw, rather than saying outside the orbitals.


Why can't terminal alkynes take part in Birch reduction? According to my book terminal alkynes are not observed in Birch reduction.Why is this so? The only property I know about them is that they are slightly acidic. Does this have something to do with it? Please help me understand this.

The alkyne proton is acidic enough to react with the dissolving metal to give the anion, giving the metal alkynide as it shown by these procedures sodium acetylide prep,  sodium phenylacetylide prep; this does not reduce.


Why is the atomic mass unit less than the mass of both a neutron and a proton? The atomic mass unit is $1.6605 \times 10^{-27}$ kg.
This is less than the mean of the masses of 6 protons and 6 neutrons.
How do we account for the lower mass ?
My understanding is that some of the mass is in the form of energy somewhere, in the bonding maybe or in the kinetic energy.
Can someone clarify where this mass is please ?

The source of the discrepancy is the mass defect, which is described in the article below as :
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation 4, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The equation four referenced is just E=mc^2.  When these nucleons form carbon-12, which is the reference isotope for the definition of the amu, the mass defect appears, as you mentioned, by conversion to binding energy.

https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/2%3A_Atomic_Structure/2.07_Mass_Defect_-_The_Source_of_Nuclear_Energy


Is sodium acetate strong or a weak electrolyte? Since it is formed of a strong base and a weak acid it should be a strong electrolyte but a chemistry teacher said that it is a weak electrolyte.....

Sodium acetate is a strong electrolyte in water.  The status of sodium hydroxide as a strong base triumphs; it makes even a weak acid such as acetic acid dissociate essesitally 100% into ions (by stealing the protons).  And when something is essentially 100% dissociated into ions, by definition it's a strong electrolyte.


Water molecule bent shape and electron configuration The hydrogen atoms in a water molecule are negatively charged, so they should repulse each other completely due to the electromagnetic force, forcing the hydrogen atoms to be on opposite sides of the oxygen atom. However, this is not the case, as the H-O-H bond angle is $104.5 ^\circ$.

Why does the lone electron pairs repulse the hydrogen atoms to an angle of $104.5 ^\circ$ in $\ce{H2O}$ instead of $180 ^\circ$?

According to VSEPR theory (there are other models, but they give the same qualitative results), since there are four electron pairs around oxygen they will be in a tetrahedral configuration. Assume the hydrogen atoms are at A and B as you've described them in your question; C and D (the lone pairs) are in a plane perpendicular to the one containing A and B. Below is a ball-and-stick model of water, showing the lone-pair electrons as yellow "atoms".
 (Public domain image from Wikimedia Commons)

It sounds as if you're thinking in two dimensions with the hydrogen atoms and lone pairs in the same plane. If water existed in only two dimensions it could conceivably be linear, but it really is in three dimensions. (The molecule itself is flat because the lone pairs affect the shape but are not counted as part of it.)

All of the above applies to water under normal conditions. I'm not aware of any excited or exotic states of water in which the VSEPR prediction doesn't apply, but I can't guarantee that there aren't any.


Sketch the Resonance structure What are the resonance structure of thioformate ion and  selenocyanat ion?
Give the structure of each resonance?
With formal charge?

As can be seen from the periodic table below, oxygen, sulphur, and selenium belong to the same group, and they are all non-metals, so one should expect their chemical properties to be similar.

Having established their chemical similarity, one can transform the two ions in question into more familiar forms.

In thioformate anion ($\ce{HCSO-}$), we are not too familiar with sulphur $\ce{S}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes formate ion ($\ce{HCOO-}$), whose resonance structure is well-known:

Just transform one oxygen back to sulphur and we are done:

In selenocyanate anion ($\ce{[SeCN]-}$), we are not too familiar with selenium $\ce{Se}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes cyanate ion ($\ce{[OCN]-}$), whose resonance structure is well-known:

Just transform the oxygen back to selenium and we are done:

Copyright issues:

The first image is copyrighted by Michael Dayah.
The second image is from Wikipedia.
The third image is a derivative work of the second image.
The fourth image is taken by HGTCChem.
The fifth image is a derivative work of the fourth image.


Why are DCM and chloroform so resistant towards nucleophilic substitution? In the book Organic Chemistry by  J. Clayden, N. Greeves, S. Warren, and P. Wothers I found the following reasoning:

  You may have wondered why it is that, while methyl chloride (chloromethane) is a reactive electrophile that takes part readily in substitution reactions, dichloromethane (DCM) is so unreactive that it can be used as a solvent in which substitution reactions of,other alkyl halides take place. You may think that this is a steric effect: Indeed, $\ce{Cl}$ is bigger than $\ce{H}$.
  But $\ce{CH2Cl2}$ is much less reactive as an electrophile than ethyl chloride or propyl chloride: there must be more to its unreactivity. And there is: Dichloromethane benefits from a sort of 'permanent anomeric effect.' One lone pair of each chlorine is always anti-periplanar to the other $\ce{C–Cl}$ bond so
  that there is always stabilization from this effect.

So, in MO-terms the situation would look something like this.

The reasoning looks plausible to me. The interaction between the free electron pair on $\ce{Cl}$ and the $\sigma^{*}$ orbital of the neighboring $\ce{C-Cl}$ bond, which would be the LUMO of DCM, lowers the energy of the free-electron-pair-orbital, thus stabilizing the compound and it raises the energy of the LUMO, thus making DCM less reactive towards nucleophiles.

But how important is this anomeric effect actually for explaining the unreactiveness of DCM toward nucleophiles, especially compared to the steric effect? And how important is the steric effect actually: Is the steric hindrance exerted by the second $\ce{Cl}$ in $\ce{CH2Cl2}$ really that much larger than the steric hindrance exerted by the methyl group in $\ce{CH3CH2Cl}$ (which is a good electrophile for $\mathrm{S_N2}$ reactions)?

I'm a little skeptical because if this anomeric effect was very important I would have expected that the $\ce{C-Cl}$ bond length in DCM would be slightly higher than in methyl chloride because there should be some transfer of electron density from the free electron pair into the antibonding $\sigma^{*}$ orbital, which should weaken the $\ce{C-Cl}$ bond. But the actual bond lengths don't show this. They show rather the contrary:

$$\begin{array}{c|c}
\hline
\text{Species} & \text{Average }\ce{C-Cl}\text{ bond length / Å} \\
\hline
\ce{CH3Cl} & 1.783 \\
\ce{CH2Cl2} & 1.772 \\
\ce{CHCl3} & 1.767 \\
\ce{CCl4} & 1.766 \\
\hline
\end{array}
$$
(source: Wikipedia)

Now, I know that the stronger polarization of the $\ce{C}$ atomic orbitals in di-, tri-, and tetrachlorinated methane as compared to methyl chloride (due to the electronegativity of $\ce{Cl}$) should lead to an overall strengthening of the $\ce{C-Cl}$ bonds in those compounds. But I would have expected a trend that would show only a slight decrease (or maybe even an increase) of bond length when going from methyl chloride to DCM and then a more pronounced decrease when going from DCM to chloroform followed by a similar decrease when going from chloroform to tetrachloromethane. But instead the polarization effect seems to only slightly increase on adding more chlorine atoms.

Introduction (and Abstract, TLDR)

In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise.

Analysis of Molecular Orbitals

I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane.

First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article.

\begin{array}{lr}\hline
\text{Compound} & \mathbf{d}(\ce{C-Cl})\\\hline
\ce{ClCH3}      &  1.797\\
\ce{Cl2CH2}     &  1.786\\
\ce{Cl3CH}      &  1.783\\\hline
\ce{ClCH2CH3}   &  1.797\\\hline
\end{array}

In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals.

In the  mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack.

One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted.
The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond.
\begin{array}{lrr}\hline
\text{Compound} &\sigma-\ce{C-Cl} & \sigma^*-\ce{C-Cl}\\\hline
\ce{ClCH3}      &  45\ce{C}(21s79p) 55\ce{Cl}(14s85p)
                &  55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\
\ce{Cl2CH2}     &  46\ce{C}(22s77p) 54\ce{Cl}(14s85p)
                &  54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\
\ce{Cl3CH}      &  48\ce{C}(24s76p) 52\ce{Cl}(14s86p)
                &  52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline
\ce{ClCH2CH3}   &  44\ce{C}(19s81p) 56\ce{Cl}(14s85p)
                &  56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline
\end{array}
As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own.

The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well.
In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the "backside" of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon.
In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange.

(Note that in chloroform two lone pair orbitals are shown.)

Conclusion

Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of Bent's Rule (compare also: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon.


How do we know that we have isolated a substance? In The Historical Background of Chemistry by Henry M. Leicester page 131, the author says:

  Carl Wilhelm Scheele (1742-1786) [...] isolated many organic compounds, including hydrogen cyanide, lactic, citric, and malic acids, and glycerol.

But how did Scheele know that those substances were isolated?

According to me, we can’t be sure of that without doing some very high-tech analyses.

While modern tools make this easier, there are plenty of physical characteristics that help identify that a substance is pure

Modern tools like NMR, X-ray structure determination and others make identifying pure substances and their structures relatively easy, there are plenty of other characteristics than can support the identification of a pure substance.

Pure compounds tend to have sharply defined boiling points or melting points, for example. So if you can generate crystals, that is a good start. Crystallisation from solution, for example, tends to yield relatively pure compounds and this can be checked by measuring the melting point. Repeated fractional crystallisation can confirm this if the melting point doesn't change. For liquids you can distill them and observe the boiling point. Fractional distillation can separate mixtures and you can check that the results boil at a single temperature. Gases also have physical properties that can be checked after fractional distillation or other separation techniques.

Here is an example where attention to detail yielded the discovery of a new element. Rayleigh and Ramsay were experimenting with the separation of gases and noticed that nitrogen isolated from air (by removing all the other known gases) seemed to have different properties to nitrogen formed from chemical reactions (it was about 0.5% denser, for example). By careful fractional distillation, Ramsay was able to isolate a much denser gas from the nitrogen which we now call Argon. This was verified to be an element by observing its discharge spectrum which had new lines in it not present for other elements. So careful separation techniques and carful observation of physical properties could yield a new substance with different properties to nitrogen.

Before we had modern instrumentation this careful application of known chemical and physical techniques was the basis for the development of the laws of chemistry and the identification of new pure substances. It takes more time, but it works.


Benzylic Oxidation of Cyclic Molecules Using KMnO₄ What is the product of the permanganate oxidation of 1-methylindane (1-methyl-2,3-dihydro-1H-indene)?

There are two choices of benzylic protons (one proton on the carbon with the methyl group at the top of the cyclopentane, and two protons on the carbon in the cyclopentane at the bottom of the molecule).

I figure that the ketone is added to the carbon that has two benzylic protons, and that the other carbon with a benzylic proton is untouched. Am I correct?

For the homogenous permanganate oxidation of aromatic side chains, as long as there is at least one benzylic hydrogen, then the side chain will be oxidized to the corresponding benzoic acid.  Here are some examples.

(image source)

Look at the third example, the permanganate oxidation of tetralin to phthalic acid.  It is very analogous to your molecule.  Your molecule would also be oxidized to phthalic acid.


Hazard and precautionary statements Is there a scientific database where one can find all the H and P statements for all chemicals? I would like to avoid using wikipedia but I can't find a database from a trusted source.

The most comprehensive one (so not "all" chemicals) that I've seen is the Aldrich website. By law they have to supply MSDS for everything they sell. However, they may ask for a lot number or the function may not work unless you're logged in as a customer.


Dieckmann condensation's final protonation step This is the Wikipedia image for the reaction mechanism:

In the final protonation step, why does the enol tautomerize to the keto form?

In the enol form the carbon-carbon double bond is in resonance with the ester's carbonyl double bond. Also the enol's $\ce{-OH}$ group would form a H-bond.

Shouldn't the enol form be more stable/predominant?

why does the enol tautomerizes to the keto form?

In 1,3-dicarbonyl systems, like the one pictured above, we would generally expect to have a measurable amount of the enol form present.  The keto-enol interconversion is an equilibrium and ratio of keto and enol will be guided by thermodynamic factors.  Let's take a look.

If we compare the structural elements in the keto and enol forms, we see that the differences between the two tautomers are as follows:

The keto form has an extra carbonyl (178), a $\ce{C-C}$ single bond (83) and a $\ce{C-H}$ bond (99)
The enol form has a $\ce{C-O}$ bond (86), a $\ce{O-H}$ bond (111) and a $\ce{C=C}$ double bond (146)

The numbers in parenthesis are standard bond energies in kcal/mol.  If we add up the keto values we get 360 kcal/mol; while for the enol it totals to 343 kcal/mol.  The numbers strongly favor the keto form (by 17 kcal/mol).  That is why is most simple systems the keto form strongly predominates.

However as you point out, both resonance and hydrogen bonding can further stabilize the enol form in the current example.

The hydrogen bond strength in $\ce{O..H-O}$ systems is typically 5-6 kcal/mol. In this system, with an intramolecular bond and a 6-membered transition state the hydrogen bond strength is likely closer to 9-10 kcal/mol. Resonance stabilization in 1,3-pentadiene is approximately 6.2 kcal/mol. Due to increased polarization in the enol from the carbonyl replacing a carbon-carbon double bond, the resonance stabilization probably increases to something around 8 kcal/mol.

A 9 kcal/mol effect due to hydrogen bonding and a 8 kcal/mol push due to resonance, shifts the 17 kcal/mol stabilization of the keto form to something very close to where the keto and enol forms have comparable energy. Therefore, in  1,3-dicarbonyl compounds we would expect a delicate balance between keto and enol forms, where both forms may be detectable.


Thermal reduction of other metal oxides beside Fe2O3 In the firing of ceramic glazes you can make an effect called oil spot.
This is done by making two layers of glaze. A bottom layer rich in iron oxide (+15%) and a top white layer that's fairly permeable. Once your reach 1225° C, the iron loses it's bond with the oxygen and goes into reduction, despite being in an oxygen rich electric kiln. So the oxygen bubbles up to the surface, taking along a bit of the iron. It then turns into FeO, and it also has something to do with magnetite which I don't understand.

Anyway, the result is that you get a white glaze with dark spots surrounded with grey/green parts, or some yellow.This effect is very nice but it only works with iron oxide as far as I know.

So my question is: are there other oxides that can have the same thermal reduction effect, and if so at what temperature. I know somebody found a way to make it in blue (it took him 4 years) so maybe he used cobalt oxide, though that looks to me to be to obvious to take 4 years. Anyway, he's not a chemist, he's a potter.

For a potter this is hard to find out but maybe for a chemist this is not that difficult to resolve, so I thought why not go to the people who know this stuff much better than us potters.

One candidate is manganese dioxide, $\ce{MnO2}$.  This decomposes at 535°C, releasing oxygen.  As the phase diagram from Ref. [1] shows below, the reduced manganese fuses with excess silica between 1250 and 1300°C, forming rhodonite ($\ce{MnSiO3}$).  You would see this as a pink color.  But, of course, I'm a chemist rather than a potter, so I make no guarantees on whether or how it works in pottery!

Reference:

1.
Y.-B. Kang, I.-H. Jung, H.-G. Lee, "Critical thermodynamic evaluation and optimization of the MnO–SiO2–TiO2–Ti2O3" system, Calphad
Volume 30, Issue 3, September 2006, Pages 226-234.  https://doi.org/10.1016/j.calphad.2006.05.002


Difference between 18 electron rule and effective atomic number rule Is there any difference between 18-electron rule and effective atomic number rule? Both rules are based on attainment of a noble gas configuraion as a criteria for stability of complexes. Besides, both rules were proposed by Sidgwick. I was wondering if the mere fact that 18-electron rule applies mainly to organometallic complexes is a good enough difference.

Both of these rules are equivalent and are related to transition metals attaining a noble gas configuration of electrons in coordination complexes. The difference between them is the number of electrons given to the central metal.

The effective atomic number uses all of the metals electrons. For example, in an $Fe^{2+}$ complex, the central iron atom has 24 electrons, and you would want the ligands coordinated to the metal to contribute another 12 electrons to reach 36 electrons, the next noble gas configuration.

The 18-electron rule is a simplification of the EAN. There is always a noble gas configuration of electrons for the metals core orbitals (e.g. 1s2s2p3s3p for $Fe^{2+}$), so you can ignore these electrons in the counting scheme. For transition metals, this means that the only important orbitals are the valence nd(n+1)s(n+1)p orbitals, which can hold a total of 18 electrons. For $Fe^{2+}$, there are 6 electrons in these orbitals. Again, the ligands would need to donate 12 electrons to reach the 18 electron limit.


Dissolving Metal Reductions of Organic Compounds I'm having trouble understanding why dissolving metal reductions, like (for example) sodium with alkenes in liquid ammonia, sodium with aldehydes/ketones in ethanol, or the Bouveault–Blanc reduction, reduce the substrate instead of the solvent. If you put sodium in ethanol (or any protic solvent), the ethanol is reduced and hydrogen gas is released. I can't see how the presence of a ketone (for example) will create a more favorable reaction.

For once, it doesn't seem thermodynamically favorable - reduction of the ethanol will release hydrogen gas, largely increasing the entropy, while reduction of the ketone produces no gas. Even if it is thermodynamically favorable, however, reduction of the ethanol seems much more favorable kinetically, since there are much more solvent molecules than substrate molecules. What is going on here?

There are plenty of examples where ethanol is present at over 10ml/g for this reaction. The key thing is that this is transfer of a single electron from Na initially. Where would this go onto ethanol? Clearly, it is faster to transfer an electron to an unsaturated compound i. e. The carbonyl compound. Once this occurs, protons are transferred. The movement of electrons and protons is fast compared to getting the two requisite H's together to form hydrogen and drive the alternative reaction.


Are all naturally occuring reactions exothermic? If a reaction is endothermic, then the reaction must take energy from the environment to occur (as far as I know), and if this is the case, then to me the reaction seems energetically unfavorable, as there is only so much energy you can get from the environment. Does this means all naturally occurring reactions, (that don't just happen once in a blue moon but at a steady rate) are exothermic? This seems pretty silly, so if not, why not?

It's not necessary that all naturally occurring reactions are exothermic. The definition of Gibbs free energy is:

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

A spontaneous reaction needs to have $\Delta G<0$, and therefore for spontaneous endothermic reactions to occur we only need that $\Delta H < T\Delta S$.


Can formaldehyde be stored at a higher temperature than room temperature? I have studied that formaldehyde stored at temperature lower than room temperature convert into 1,3,5-oxane. So, it's not stored at colder temperatures.

But my question is - Can a similar degrading reaction happen if methanal is stored at temperature higher than room temperature?

Trimerization of formaldehyde can happen at high temperatures, but appear to require catalysts. There is a patent for synthesizing 1,3,5-trioxide at high temperature. This patent uses phosphoric acid catalysts.
According to the patent:

  The process of the invention is advantageously carried out at reaction temperatures of from 80 to 160° C., preferably from 90 to 145° C., at an inlet partial pressure of the formaldehyde of from 0.5 to 5 bar absolute, preferably from 0.5 to 2 bar, and at an inlet concentration of formaldehyde of from 1 to 100% by volume, preferably from 20 to 100% by volume

There is also an early study which explored the kinetics of the reaction but for a different catalyst. In "Vapour-phase trimerization of formaldehyde to trioxane catalysed by 1-vanado-11-molybdophosphoric acid" by Kern and Emig, they use a molybdophosphoric acid catalyst. They used a tube reactor and varied the temperature from
102.5 to 110° C. According to their study:

  The
  conversion
  observed
  increases
  extremely
  with
  decreasing
  temperature.

However, they comment that

  the
  temperature
  dependence
  cannot
  be
  attributed
  to
  thermodynamic
  effects
  only.
  It
  has
  to
  be
  related
  to
  the
  properties
  of
  the
  heteropoly
  compound.

Another study is "Synthesis of Trioxane Using Heteropolyacids as
Catalysts" by Mazamoto et al. Their process uses heteropolyacids as catalysts and they were able to obtain a conversion % of 22.9 at 110° C. Where
$$\text{Conversion %}=\frac{3[\text{trioxane}]+2[\text{methyl formate}]+[\text{methylal}]}{[\text{feed formaldehyde}]}\cdot 100$$


cyclopropane's reaction with lithium aluminium hydride what is cyclopropane's reaction with lithium aluminium hydride? I think it must lead to ring-opening but I didn't find any article favoring this in my texts.

Cyclopropanes do not react with LiAlH4. Substituents of cyclopropanes rings may be reduced with std LiAlH4 conditions with the cyclopropane ring being untouched

example in this patent


Can acidic conditions break disulfide bonds I am denaturing a protein using organic solvent and acid (49:49:2% water:methanol:acetic acid), but I want to maintain the disulfide bonds.

My chemistry knowledge isn't good but disulfides are broken under reducing conditions and the reaction also uses free protons:

So I wanted to check if acids could break disulfide bonds without reducing conditions.

No, acidic conditions (i.e. low pH) shouldn't be enough by itself to break a disulfide bond. The main reaction that breaks a S–S bond is cleavage by reduction. This is in a way similar to peroxides, which are stable even at very low pH (see e.g. mixtures of sulfuric acid and hydrogen peroxide).


Effect of chloromethyl substituent vs methyl substituent on yields of nitration The effects of $\ce {-CH3}$ and $\ce {-CCl3}$ substituents on the product distribution of an electrophilic aromatic substitution reaction are unambiguous. The $\ce {-CH3}$ group exhibits a clear electron-donating effect and directs electrophiles to the 2- and 4- positions. Similarly, the $\ce {-CCl3}$ group exhibits a clear electron-withdrawing effect and directs to the 3- position. Additionally, due to the steric bulk of the three chlorine atoms, there is significant steric hindrance that electrophiles would face when they attack the 2- position. Hence, the proportion of ortho nitration products would be very low.

Some data on the production distributions for nitrations of particular substituted benzenes, taken from Carey & Sundberg (2007), p. 786, is shown below. Observe that the ratios given are consistent with what we have discussed above.

\begin{array}{|c|c|c|c|} \hline
\text{Substituent } & ortho & meta & para \\ \hline
\ce{$\ce {-CH3}$} & 56-63 & 2-4 & 34-41 \\ \hline
\ce{$\ce {-CCl3}$} & 7 & 64 & 29\\ \hline
\end{array}

However, we are interested in the $\ce {-CH2Cl}$ substituent, not the $\ce {-CCl3}$ substituent. We would, of course, expect the steric effect to be lesser than that of the $\ce {-CCl3}$ substituent and we would also expect different electronic effects exerted on the ring. If we were to look at more data from p. 786, as shown below, we would observe that for a substituent like $\ce {-CHCl2}$, it still remains as an electron-donating group as the proportion of meta nitration is still lower than that of ortho and para nitration. It seems that even with one $\ce {C-H}$ bond, there is still sufficient hyperconjugation with the benzene ring to outweigh the inductive pull of the two electronegative chlorine atoms.

I have no data for the $\ce {-CH2Cl}$ substituent but I believe that based on the above analysis, we predict that $\ce {-CH2Cl}$ would also be an electron-donating group that is more electron-donating than $\ce {-CHCl2}$ but less so than $\ce {-CH3}$. Additionally, we would expect to also display directing effects to the 2- and 4- positions However, I have great uncertainty regarding the steric effect of this substituent. How would the ratio of ortho to para nitration be like?

Reference

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.

The following is the abstract of K.M. Zhang "Selective Nitration Of Benzyl Chloride And Its Application In The Synthesis Of Nitrobenzaldehyde"(link). The full pdf is behind paywall.

  The selective nitration of benzyl chloride and its application in the synthesis of 4-nitrobenzaldehyde were studied. Nitration of benzyl chloride with $\ce{HNO3}$ in the presence of $\ce{Nb2O5}$(6% weight percent) at 0-5℃ for 1.5 hours gave mononitro products in 94% yield with 77-80% para selectivty. $\ce{Nb2O5}$ was recovered by filtration and reused in the nitration. The yield of mononitro products and para selectivity were above 90% and 76% respectively for 5 times repeated use of $\ce{Nb2O5}$. Nitration of benzyl chloride with urea nitrate(UN) in $\ce{H2SO4}$ was also studied. The nitration was carried out at 25℃ for 24 hours and gave mononitro products in 98% yield with 90% para selectivity. Base catalyzed hydrolysis of 4-nitrobenzyl chloride at 75-80℃ for 22 hours gave 4-nitrobenzyl alcohol in 77% yield with 98% purity. Oxidation of 4-nitrobenzyl alcohol in dilute nitric acid at 55℃ for 3 hours gave 86% 4-nitrobenzaldehyde, compared with industrial oxidation of 4-nitrobenzyl alcohol,whichwas carried out at 70℃for 4 hours and gave 70% 4-nitrobenzaldehyde. 4-nitrobenzaldehyde was also obtained in 86% yield with 92% purity via ionic liquid immobilized $\ce{TEMPO/CuCl}$ catalyzed oxidation of 4-nitrobenzyl alcohol with molecular oxygen in ionic liquid $\ce{[bmim]PF6}$. The reaction condition is mild, the selectivity is high, and no 4-nitrobenzoic acid is detected in any case.


Is it possible to have an airborn fluorescent particle that reacts to lightning? I want to measure the expansion of the atmosphere when it is heated and impacted by lightning, to have an idea of the propagation of sound waves made by lightning.

To do this, the ideal would be to have a box of static air which changes color or causes a durable fluorescence indicating the reaction of the electricity with the air and showing the patterns of expansion of air that has been heated by electricity.

Please can y ou suggest some chemicals which can be used to make the effects of lightning on air visible?

Maybe something that fluoresces in the presence of ozone? A quick search gave this article. Lightning is known to convert oxygen into ozone, so I think this should work or at least give a good starting point for future development.


How to partially hydrolyze triglycerides? How to partially hydrolyze triglycerides for food application?

I want to make a mixture of mono- and diglycerides starting from triglycerides.

Over the last decades, partial hydrolysis of triglycerides has been examined in various contexts.

Note that this isn't my field of work and that there might hundreds of better and/or more instructive recipes, but a cursory search yielded the following freely available references:

THE DIGESTION IN VITRO OF TRIGLYCERIDES BY PANCREATIC LIPASE,
F. H. Mattson, and L. W. Beck,
J. Biol. Chem., 1955, 214, 115-125 (PDF)

A stereospecific analysis of triglycerides,
H. Brockerhoff, Journal of Lipid Research, 1965, 6, 10-15
(PDF).

The products from lipase-catalysed hydrolysis of bovine milkfat kill
Helicobacter pylori in vitro,
Cynthia Q. Sun, Charmian J. O’Connor, Alastair K. H. MacGibbon, and Anthony M. Roberton, FEMS Immunology & Medical Microbiology, 2007, 49, 235-242
(DOI)

It seems that the digestion of triglycerides by lipases in buffered solutions is a viable approach.


Li2O + 2KCl =2 LiCl +K2O: Is it a feasible Reaction? $$\ce{Li2O + 2KCl <=> 2LiCl + K2O}$$

I got this reaction on a test question paper. With three other options, it was asked whether this reaction is feasible or not?

I googled it and found the reverse reaction of at every webpage I looked. So, it seems to me it isn't feasible.

I think that the reverse reaction is feasible because $\ce{LiCl}$ is quite covalent in nature(Fajan's rule) So, for becoming more ionic in nature $\ce{Li+}$ makes bond with oxygen ion. (I now think it's a bogus reasoning)

What could be the reason behind it? Is it feasible?

Although every reaction is reversible in nature, let's try to understand where the equilibrium might be biased, i.e. whether the reactant side favoured or the product side.

A rule of thumb while predicting feasibility is to remember, a reaction where you can get a weak acid from a strong acid is feasible.

$$\ce{Strong Acid -> Weak acid}$$

In your case we're dealing with $\ce{KCl}$ and $\ce{LiCl}$. The acidic or basic nature or salts is determined by the extent of hydrolysis of its ions. Ions from a weaker species undergo hydrolysis while ions from a stronger species don't. Another way to look at this is to compare the acidic and basic strengths of the acids and bases which formed them in the first place.
$$\ce{LiOH + HCl -> LiCl + H2O}$$
$$\ce{KOH + HCl -> KCl + H2O}$$
Notice that $\ce{HCl}$ is common to both of these species, so we have to just compare the basic strengths of $\ce{LiOH}$ and $\ce{KOH}$. Down the group (in alkali metal hydroxides) basic strength increases. So $\ce{KOH}$ is a better base. And thus $\ce{KCl}$ is a weaker acid. Since we're making a stronger acid from a weaker acid, the reaction can be termed as unfeasible.


Reaction with tin(IV) iodide and hydrogen peroxide Today I reacted tin(IV) iodide $\ce{SnI4}$ (dissolved in water) with hydrogen peroxide. This produced a yellow/brown colour (I'm guessing due to formation of $\ce{I2}?).$

I then added some petroleum spirit which gave a bi-phasic system which was a clear/pink/ light purple colour on top and an orange-brown yellow underneath.

I'm struggling to find out what reactions have occurred here, can anyone please help?

Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk).

Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid:

$$\ce{SnI4 + 6 H2O <=> H2[Sn(OH)6] + 4 HI}$$

Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(IV) is used as a stabiliser for hydrogen peroxide), and is likely going to oxidize hydroiodic acid to iodine (which attributes to the brown color) instead:

$$\ce{H2O2 + 2 HI -> I2 + 2 H2O}$$

Hexahydroxostannic acid only exists in solution and tends to form cloudy/milky gel over time [2, p. 1021]; this can be simplified as formation of colloidal particles of tin(IV) oxide [hydrates]:

$$\ce{H2[Sn(OH)6] <=> SnO2 · n H2O + $(4-n)$ H2O}$$

When petroleum ether is added, iodine is extracted to organic phase (on top due to lower density of the ether). Tin(IV) oxide sol remains in aqueous phase and attributes to its slightly pink color due to Mie scattering of colloidal solution.*

* Phase order doesn't match OP's description which suggests opposite coloring of the phases (additional clarification on the type of petroleum spirit or a photo of the binary phase mixture would help).

References

Inorganic Syntheses; Bailar, J. C., Ed.; Inorganic Syntheses; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1953; Vol. 4. https://doi.org/10.1002/9780470132357.
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin New York, 2007. ISBN 978-3-11-017770-1. (in German)


Does mechanotransduction of cells play a role in the biocompatibility of titanium? What exactly is mechanotransduction? The definition seems to vary from source to source. Some indicate that mechanotransduction is the underlying principle where cells pull on the surface they grow on to test their stiffness, so they can stretch and facilitate cell growth and differentiation. Others indicate that mechanotransduction is how cells detect mechanical stimuli and turn them into chemical signals like our four senses (touch, hearing, smell, etc).

Also, is this the main reason titanium is biocompatible to replace bones because it has a low surface resistance and cells can stretch on it?

Mechanotransduction (mechano-relating to mechanics + transduction-to convert into another form) is indeed the second definition you listed. Mechanoreceptors in somatosensory nerve cells the ability to detect mechanical pressure or distortion. These facilitate our touch, hearing, and proprioception, or the sense of how our body is positioned. Olfaction, however, is actually facilitated through chemoreceptors, and depends on how molecules bind to them.

Cellular response to a foreign object is the result of largely chemoreception. Interaction at the implant-cell surfaces determines how an implant will be perceived by the body. Titanium's biocompatability is a result of the protective oxide film that forms on the surface of the metal in the presence of oxygen. This oxide film is strongly adhered, insoluble in water, and chemically impermeable, preventing reactions between the metal and the surrounding environment. Titanium's high dielectric constant allows it to be easily integrated into cells and bone tissue$^{[1]}$.

From Wikipedia:

  Titanium naturally passivates, forming an oxide film that becomes heterogeneous and polarized as a function of exposure time to bodily environments. This leads to the increased adsorption of hydroxyl groups, lipoproteins, and glycolipids over time. The adsorption of these compounds changes how the material interacts with the body and can improve biocompatibility.

[1] Mishra, B. Review Of Extraction, Processing, Properties, and Applications of Reactive Metals 1999 TMS Annual Meeting, San Diego, CA, February 28 - March 15, 1999; Wiley: Hoboken, 2013.


What chemicals would make lime sulfur dip to produce a green colored smoke? I work at an animal shelter that uses lime sulfur dip to treat ringworm. We had an accident where someone mixed bleach instead of water with the lime sulfur dip which caused us to go through decontamination with Hazmat. Are there any other chemicals or acids that could cause this to happen?

Yes, mixing lime sulfur (calcium polysulfide) with an acid will react via the following reaction:

$$\ce{S^{2-}\:(aq) +2H+\:(aq) ->H2S\:(g)}$$

Which is both extremely toxic and extremely flammable. You should always be extremely cautious and mindful when mixing any chemicals and always be sure to label everything!


Oxidation of alkenes with KMnO4 This is a weird question but it's finals week and it's impossible to get in contact with my prof. I have no idea what he means by the last line where he says a C-O bond is formed for every C-C and C-H bond. Can someone just clarify this for me?

Your professor obviously wants you to consider the $\ce{C-C}$ bonds of the $\ce{C=C}$ double bond plus any $\ce{C-H}$ bond to either of the alkene carbons.

The first step in double bond oxidation by permanganate will always be replacing a $\ce{C=C}$ by $\ce{C=O + O=C}$. If you consider the double bond two bonds, then for each of those bonds each carbon atom gets one $\ce{C-O}$ bond to again give a double bond.

After we have generated the inital carbonyl, we can turn our attention to the other bonds the former alkene carbon has. Remember that aldehydes are easily oxidised to carboxylic acids while ketones are rather resistant to oxidation. The difference is the presence of a $\ce{C-H}$ bond in the aldehyde; permanganate can oxidise this to another $\ce{C-O(-H)}$ bond. This is the second part of your rule.

If accidentally the carboxylic acid species generated is formic acid, which has another $\ce{C-H}$ bond, this can again be oxidised to give carbonic acid. Since carbon’s oxidation state is maximal in $\ce{H2CO3}$, this is truly the end of the oxidation.


How does a carbocation have a positive charge? When an alkene bonds with an H+ ion, the electron pair from the pi bond goes towards a new dative covalent bond with the hydrogen ion, leaving, on one side of the old double bond, a carbon bonded to an extra hydrogen, and on the other side, a carbocation with a positive charge. If this gave, for example, a tertiary carbocation, I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge? I understand that it has lost electrons that were previously in the pi bond, but I don't see how it has a positive charge given that it still has more electrons than protons.

I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?

You need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?


What are the preferred IUPAC names of these two compounds? In a recent homework assignment we are asked to give IUPAC name to the following two compounds:

I am not asking this question for homework help -- for that purpose I got both names correct, to the extent of the standard given in my orgo textbook. What intrigues me is that compound naming softwares seem unable to agree upon the names of these compounds.

For structure I: (MarvinSketch is consistent with homework answer, because bromo comes alphabetically before bromobutyl)

ACD/ChemSketch (freeware) 2017.1.2: 4-(2-bromobutyl)-6-methyloctan-3-ol
MarvinSketch 18.4: 6-bromo-4-(2-methylbutyl)octan-3-ol

For structure II: (ACD/ChemSketch is consistent with homework answer, because this name recognizes 3 substituents, while the other recognizes only 2)

ACD/ChemSketch (freeware) 2017.1.2: 4-(2-bromobutyl)-6,6-dimethyloctan-3-ol
MarvinSketch 18.4: 6-bromo-4-(2,2-dimethylbutyl)octan-3-ol

Is it because IUPAC rules are so complicated they can be only followed by hand, or because the tie-breaking rules presented in textbooks are too simplistic to give preferred names in every case?

We can only speculate about the various automatic algorithms that are available to generate names for chemical structures. Anyway, all known programs have limitations or shortcomings. In particular, the algorithms may not include all rules or they may use obsolete rules. Note that the IUPAC recommendations have changed several times and that the current (2013) recommendations have more than 1500 pages.

In my experience and also in this case, ACD/ChemSketch is more reliable than MarvinSketch. However, such programs are typically updated frequently, and some issues might be solved in a new version.

The compounds that are given in the question are not trivial cases. In order to find the preferred IUPAC names, we have to dig a bit deeper and consider the rules given in Section P-45 of the current recommendations. These rules are applied if two or more names may result according to the basic nomenclature rules, because of different substitution patterns or multiple occurrences of the same senior parent structure. Apparently, these rules are taken into account in the present version of ACD/ChemSketch but not in the present version of MarvinSketch.

Thus, ACD/ChemSketch correctly gives the preferred IUPAC name for the first compound as 4-(2-bromobutyl)-6-methyloctan-3-ol.

The name ‘6-bromo-4-(2-methylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not wrong, it is just not the preferred name. Remarkably, this name was in accordance with obsolete IUPAC recommendations from 1979 (retained in 1993). Therefore, some homework exercises might still expect this answer. However, this does not necessarily mean that MarvinSketch is using old IUPAC recommendations from 1979 or 1993; it could be a mere coincidence as the second example below shows.

ACD/ChemSketch also correctly gives the preferred IUPAC name for the second compound as 4-(2-bromobutyl)-6,6-dimethyloctan-3-ol.

The name ‘6-bromo-4-(2,2-dimethylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not in accordance with IUPAC recommendations (neither 1979 nor 1993 nor 2013). Apparently, the relevant (past or present) tie-breaking rules for such cases are not included at all in the present version of MarvinSketch.


What is trans-addition? A question from JEE-Advanced(2017) :

Bromination of both Z-but-2ene and E-but-2-ene involve syn and anti addition. So, there would be 4 products as suggested in the question. But, I don't understand the terminology of 'trans-addition', mentioned in the question.

What is actually trans addition?

Just like @QuantumAMERICCINO has pointed out, it's probably anti addition.

Furthermore, you seem to wonder why a syn compound doesn't form. It all lies in the mechanism of bromination:

The double bond in cis-2-butene polarises the $\ce{Br-Br}$ bond in a nearby $\ce{Br2}$ molecule. A $\ce{Br+}$ ion attacks the double bond and forms a bromonium ion.

The bromonium ion makes the adjacent carbon atoms electrophilic. The bromide may attack the most electrophilic out of them. Here, however, they are equivalent.

You should now be able to understand how the mechanism works.


Will a film of oil on water retard evaporation (of water) and prevent air and atmospheric impurities from diffusing into it? I vaguely recall hearing that a layer of oil on water will slow down the evaporation of water, as well as serving as a barrier that would prevent (or retard) the diffusion of air and atmospheric impurities into it. How accurate is this claim?

Some Background:

I want to make zinc/air batteries. $\ce{KOH}$ is commonly used in the electrolyte, but $\ce{CO2}$ in the air can react with the $\ce{KOH}$ and form carbonates which hinders the battery's performance. It seems like oil on top will shield the electrolyte from atmospheric $\ce{CO2}$. The $\ce{H2O}$ also must be kept from evaporating, which would otherwise result in changes to  concentration. $\ce{KOH}$ is also very hygroscopic and tends to attract water from air which would also cause variations in concentration.

When the batteries get discharged, $\ce{ZnO}$ is a final product. This should almost certainly precipitate and collect on the bottom of the tank since its considered insoluble in water.

Solar panels harness a vast amount of energy from the sun during the Summer. Storing surplus enrgy for use in Winter is a daunting task. A method that's been doing rounds these days: One could collect and reduce the $\ce{ZnO}$ back to $\ce{Zn}$, deposit it on plates/electrodes, and store these plates for Winter use. The plates could be put back in the batteries and the electrolyte (a fresh solution of $\ce{KOH}$) would have to be replenished.

I don't know about oil (fun fact, putting oil on water was a traditional approach to calming waves, thus the phrase "pour oil on troubled waters").  But cetyl alcohol has been used to decrease evaporation from reservoirs, which might be what you heard about.  For instance, here is a report on this, with lots of references.  They used a mix of steryl and cetyl alcohols (long-chain alcohols that will form a monolayer on the water).


Major product in dehydration of alcohol Consider the dehydration of the following molecule with conc. $\ce{H2SO4}$ at $443 \text K$:

The possible products are:

The first one is the Zaitsev product and the second one Hofmann  product.
Which compound is the major product?
There are 2 factors to consider here:

Zaitsev product is more stable than Hofmann product due to
hyperconjugation
Anti elimination is more favorable than syn elimination due to steric requirements

Obviously both the factors clash with each other, so I couldn't come to a definite conclusion. Can you help me out?

The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When the OH is protonated and it leaves, we form a secondary carbocation, but that can be further more made stabilized via an hydride shift from the adjacent carbon producing a tertiary carbocation. Now either of the base $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ can capture the hydrogen from the adjacent carbon leading to the desired Zaitsev product. (Also the stereochemistry isn't necessary to worry about here, as the reaction proceeds through E1.)


Phospors and Phosphorus I'm not a chemist.

I understand that phosphorus comes in multiple forms, one of which is white, volatile, and highly toxic.  I know that phosphors come in multiple chemical forms, two of which are zinc sulfide and strontium aluminate for glow-in-the-dark products like light sticks, etc, many of which kids use.

In doing some research for a magazine article I'm writing, I read a sentence that said "Chemists have been able to mimic phosphors …" and the word "mimic" leaves me wondering if this was done because original phosphors were (and still can be) from phosphorus?

Phosphers have never been made with phosphorous just like lead pencils have never been made with lead. The words have a common origin but there is very little chemical relationship.

Phosphorus is a Latin name for the planet Venus and, figuratively, for any glowing object. Early chemists in the 17th century used the word to refer to any substance which glowed spontaneously or when heated. They characterised and named several different types of phosphoruses. When the element phosphorus was isolated from urine in the late 17th century it was called “English Phosphorus” or “Phosphorus of Urine” and later simply “Phosphorus”.

The older meaning of “phosphorus” and “phosphor” continued into the 19th century and, in the early 20th century, “phosphor” began to be used with more scientific rigor to refer to a material exhibiting phosphorescence.

Phosphorus (the element) is not phosphorescent; it emits light only insomuch as it spontaneously ignites in air.

Graphite (a.k.a. plumbago) was used as a substitute for black lead hence it was the "lead" in the pencil.


Nomenclature of acid anhydride I have been struggling hard for writing IUPAC naming of this compound.

Anhydride with benzene having five membered ring can be named as benzene dicarboxylic anhydride and five membered ring without benzene can be named as pentane -1,5-dioic anhydride. But what can I do for this compound?

It looks like a isochroman-1,3-dione to me.

For the reference there is a similar compound (published in French): Kakou-Yao, R.; Saba, A.; Ebby, N.; Pierrot, M.; Aycard, J. P. Tautomérie de la 4-(hydroxyphénylméthylène) isochroman-1,3 dione à l'état solide, Acta Cryst C 1999, 55 (9), 1591–1593. DOI: 10.1107/S0108270199006782


Constructing potential energy surface of a periodic system, how to deal with atoms moving out of the supercell? I'm using NVE ensemble for doing molecular dynamics (MD) simulations. The periodic system I'm working with contains $50$ atoms and the "calculator" for calculating potential energies and forces is a trained neural network. The platform I'm using for MD simulation is ase (atomic simulation environment).

After I ran the MD simulation for about $3$ps under $380$K, the atoms are moving out of the supercell. The ase module for MD simulation just faithfully updates the positions of atoms in the supercell (the supercell contains 50 atoms) based on the velocity. From its source code, it seems the module does not take periodicity of the system into account.

EDIT

@LonelyProf 's helpful comment reminds me periodic boundary conditions (PBC) should be implemented in the neural network calculator. The neural network calculator consists of two parts:

Generate Behler-Parrinello fingerprint given the Cartesian coordinates of atoms in the structure model. The fingerprint is actually a vector of symmetry functions that describes two-body and three-body interactions. Two-body symmetry functions are functions of atomic spacing, three-body symmetry functions are functions of both atomic spacing and angle.

In my original implementation (decorate the neural network potential to be a readable calculator by ase), there is not a module that adjusts the Cartesian coordinates of atoms "run wild" so as to move them back to the supercell.
Calculate the energy of a structure given its fingerprints. The calculated fingerprint of a structure in step one is treated as the neural network input to calculate the energy of the structure.

When constructing the potential energy surface of a periodic system whose structure model has a fixed supercell, should I always keep the atoms within the supercell?

I think that it is best to consider the calculation of the potential energy (respecting the periodic boundary conditions) as a separate matter from keeping the atoms within the supercell. For simplicity I'm going to assume a cuboidal box with sides $(L_x,L_y,L_z)$, with periodicity in all three Cartesian directions.

Some people like to put the atoms back into the primary cell whenever they move out. This is done by replacing the atomic position vector $(x,y,z)$ by $(x,y,z)+(n_xL_x,n_yL_y,n_zL_z)$, where $(n_x,n_y,n_z)$ are integers, chosen to make the new coordinate values lie in the desired range, often $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. If you do this every MD step, the integers will usually just need to take the values $0,\pm1$. Other people prefer not to apply this correction, and just keep track of the copy of an atom that diffuses into the periodic-image supercells.

Even if you do this correction to the atom positions, you still need to apply a similar correction every time you calculate a vector between atoms, for the purposes of computing two-body, three-body, potentials and forces. So, a properly written calculator for these quantities will not rely on the atoms lying within the same supercell. If the atomic coordinates are allowed to evolve freely, without periodic boundary corrections, then the integers $(n_x,n_y,n_z)$ will take whatever values are needed to bring the corrected interatomic vector into "minimum image" form, with components in the range $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. There are standard prescriptions for doing this.

Often, the energy/force calculation is speeded up by a neighbour list calculation, which involves assigning each atom to a subcell of the original supercell. On the face of it, this seems to require that we put the atoms back into the supercell, before doing the assignment. But actually, a properly written function to identify nearest neighbours will have this mapping built in to it, and may just work with copies of the original coordinates. So, again, there should be no need to put the atoms back into the original supercell "by hand", if the neighbour identification is done this way, and if the vectors between neighbours are corrected as just described.

My impression, from a cursory look at ase, is that this mapping is indeed built in to the functions used to identify neighbours "within range". There is a simple Lennard-Jones calculator, for instance, which seems to rely on the neighbour list function for this purpose. The relevant lines are

        neighbors, offsets = self.nl.get_neighbors(a1)
        cells = np.dot(offsets, cell)
        d = positions[neighbors] + cells - positions[a1]

where I think that offsets contains the integers $(n_x,n_y,n_z)$ and I think that cell contains the cell lengths $(L_x,L_y,L_z)$ in the form of a $3\times3$ matrix (just the diagonal elements would be nonzero in this case).

If you are not using such a function to locate neighbours "within range" and pre-calculate the necessary minimum image corrections to the interatomic vectors, you will need to put it into your calculation of pair and three-body energies and forces, somewhere, yourself. As I mentioned, this is not the same as putting the atoms back into the supercell, when they move out.

Caveat! I am not an expert in ase, and if you want specific advice on that package, you should look at the package documentation, and at simple examples of calculators provided within it. Generally speaking, I believe that advice on debugging programs, and on specific packages, is considered off-topic for this site, so I think that's as far as I should go in any case.


How to neutralize the stink of decomposed fish? Recently I discovered for myself an interesting way to easily produce home made plant fertilizer that is quite effective (tested myself on the plants). It's fish emulsion (so can be used also in a hydroponic system) - 1 part fish wastes(skin/bowels) to 2 parts water placed in an airtight container in a sunny spot for ca. 1-2 weeks. The problem - it stinks unbearably. Both - the gas that is produced during the decomposition and the liquid(/fats).

Is it possible to neutralize the stench somehow without loosing the fertilizing qualities of the emulsion? Especially of phosphorus.
1.1 Ideally of both - the gas and the liquid. But at least of the liquid - as that is what is used actually and then continues to spread the smell after being added to the plants.
1.2 Ideally it should be done using some usual household substances/wastes - maybe adding of lemon/orange peels along with the fish wastes? Or maybe ash/charcoals?

Ideally the resulting emulsion should be acidic as many plants like the pH range of 5 - 6.5 .

Plants like fully decomposed (composted) organic materials just like humans prefer cooked food. The solution which you are preparing, is going to hurt the plants rather badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in the ground or soil, mix it with old leaves and let it compost for 1-2 months? No smell should arise because soil bacteria and perhaps worms/insects will take care of it. Above all everything will be well covered with soil.
Do not add vinegar or lemon juice to your liquid (which should go to waste anyway). Vinegar has salt and plants don't grow well with salty water.


Hydrogenation of alkenes How many structural isomeric alkenes ($\ce{C_6H_{10}}$) can be hydrogenated to give methylcyclopentane? Would there be three? I drew out what I thought was correct but I just want to confirm that there are three.

There are 4 structural isomeric alkenes ($\ce{C_6H_{10}}$) that can be hydrogenated to give methylcyclopentane.

1-Methycyclopentene

3-Methylcyclopentene

4-Methylcyclopentene

Methylenecyclopentane


effect of overpressure on lowering viscosity of polysaccharide solution it has been reported that agarose dissolved at 110C under slight over-pressure has a viscosity at least 10X smaller than dissolved at atmospheric pressure at 90C (ref:https://doi.org/10.1016/j.ceramint.2004.06.007).
I have not independently verified this result (but I plan to).

The context of this result is ceramic but the solution viscosity result is independent of this context.
I see this result as quite interesting with many other applications BUT
I cannot find ANY other reference of similar results with agarose gel or any other polysaccharide for that matter, which surprise me.

Do I miss something?

I am looking for references to better understand this result and the physico-chemistry behind it.
If somebody could provide some pointers it will be greatly appreciated

Thanks

JF

The pressure is irrelevant, it is only applied so the solution can reach 110°C. Water boils at 100°C at ambient pressure.

Agarose is a polysugar, and at 110°C in water, I expect it breaks down into smaller molecules, disugars, monosugars etc.

Solutions of smaller molecules have lower viscosity. https://en.wikipedia.org/wiki/Mark%E2%80%93Houwink_equation

For typical polymers, the scaling exponent $a$ of Mark and Houwink is 0.5-0.8, but that's for infinitely low concentrations. It goes up drastically for higher concentrations, reaching a typical 3.4 in the melt.

(That's a bit sloppy, Mark-Houwink predicts the intrinsic viscosity, while in an entangled melt the actual viscosity scales with $M^{\approx3.4}$.)

So, if you have a saturated solution, and the average agarose polymer molecule breaks just twice, a factor ten in viscosity decrease is very reasonable.


How does radiocarbon dating work? I'd like to better understand radiocarbon dating, however, I do not know what resources I must use in this regard. Could someone walk me through the process of radiocarbon dating with a few examples? (I'm already well-versed with the basics of nuclear chemistry).

The radiocarbon method is in principle fairly simple. $^{14}\ce{C}$ is an isotope of Carbon that has a relative abundance of around $10^{-10} \%$ on earth.

Addition by @MaxW:

  $^{14}\ce{C}$ is formed in the upper atmosphere from $^{14}\ce{N}$ so
  that the relative amount of $^{14}\ce{C}$ in the atmospheric
  $\ce{CO_2}$ is reasonably constant. Since plants get most of their
  carbon from atmospheric $\ce{CO_2}$, the relative amount of
  $^{14}\ce{C}$ to $\ce{^{12}C}$ is thus constant.

It is also not stable and decays $^{14}\ce{C ->}^{14}\ce{N} + e + v$.

Now a living organism based on carbon (like us) always has a stable amount of $^{14}\ce{C}$ in his body, because he has to eat and so there is a stable balance between $^{12}\ce{C}$, $^{13}\ce{C}$ and $^{14}\ce{C}$.

Now if that organism dies, it does not eat anymore, so there is no new income of $^{14}\ce{C}$. Since that decays and the other $\ce{C}$-isotopes don't, over the years, the percentage of $^{14}\ce{C}$ is getting lower and lower at a constant rate (you can calculate that rate with the half-time of $^{14}\ce{C}$).

Now when you measure the relative amount of $^{14}\ce{C}$ in a skeleton, you know since when it hasn't eaten anymore, so you know how old it is. The method can be used around 600-50k years to the past.

There are different methods of measuring isotope amounts. You can measure how fast a radioactive decay still is and calculate from that the amount of $^{14}\ce{C}$, which is the method from Libby.

You can also use Mass-Spectroscopy, because $^{14}\ce{C}$ is heavier than  $^{12}\ce{C}$ and $^{13}\ce{C}$ you can "easily" measure the amount of $^{14}\ce{C}$.

There are other methods, you can look this up in Wikipedia.


What is a laboratory set-up for the synthesis of ethanol from ethene? I am doing a chemistry investigation and one component of that requires me to synthesize ethanol from ethene. Ethene itself is gaseous at room temperature and pressure so I was wondering if someone could suggest how I can carry out this synthesis with the use of acids to obtain ethanol.

To give you an idea of what I am actually doing for this investigation, I am trying to find the yield of ethanol when ethene is hydrated using different acid catalysts. However, because ethene is gaseous, I don't know how I am going to actually do this experiment.

To further clarify, I know what reagents I need to use and I know the mechanism of the reaction, I just don't know how to do it practically within a laboratory setting.

You should try bubbling the ethene gas through dilute sulfuric acid. It should form ethyl sulfate and get hydrated with water forming ethanol. As the gas will bubble through the solution, you should have an ample amount of liquid so that most of the gas can maximally react with the sulfuric acid before bubbling out as gas. You should do the reaction in a fume, hood too. Then I think that you could separate the ethanol from the water by distillation of the ethanol from the mixture, or you could separate it by exploiting the great difference in the freezing points of ethanol and water.

If you could connect tubes in such a way with special chemistry glassware, then maybe you could make some kind of reflux apparatus like an acid reflux apparatus where the gas funnels back into the bottom of the liquid and cycles through several times then you could increase your yield. Just if you use some cyclic mechanism, don't release the gas too quickly or you could break your apparatus.

I'm not exactly sure what the experimental design is. But due to the concentration of gas in the reaction chamber, it's important to have an escape valve for the gas for safety purposes of equalizing pressure. If you know what you want, an experienced glass-blower can make your apparatus.


Why does attraction always cause release of energy making the system more stable? Suppose $\ce{Na^+}$ and $\ce{Cl^-}$ form an ionic bond. In doing so they will come closer by attraction. Now as they come closer their speed will increase and hence the kinetic energy increases. So basically the electrostatic potential energy will be controverted into kinetic energy.

  My question is why do these processes that involve attraction cause a release of energy from the system? And after this release why is the system said to be more stable?

My understanding is there might be two concepts involved.

First, as the ions approach each other their speed increase, but as soon as they touch each other they collide and reflect back. So the ions keep on oscillating. These oscillations are what we know as heat and the temperature of $\ce{NaCl}$ will be higher than its surroundings so it will release its energy in the form of heat.

Second, it might be because any accelerating charge releases electromagnetic waves, and so as long as the ions oscillate they'll keep releasing energy in the form of EM waves which may be trapped by other ions or may escape.

After the energy will be released from the system(NaCl ion here) our system must become more stable. Stable in the sense that NaCl crystal will not convert back into ionic form separated from other ions because for coming back to that state the system must be provided the lost energy back by some external agency. This is the way how I understand the term stabilty; I don't what does this term mean in chemistry. I also don't know why " every system in nature wants to be more stable" -- This is the phrasing that I heard in school days from chemistry teacher. This fact must be related to some thermodynamic fact of kind that every system wants to release energy, but why?

Attraction is a shorthand way of saying "potential energy gets lower as the distance gets lower".  The opposite is true of repulsion.  The potential energy of two like-charged particles, e.g. two $\ce{Na+}$ atoms, will increase as they are brought closer together.

Attraction (and its opposite, repulsion), are forces with units (in SI) of Newtons, or $\frac{\rm kg\cdot m}{\rm s^2}= \rm N$.

Energy has units of Joules, or $\frac{\rm kg\cdot m^2}{\rm s^2} = {\rm N\cdot m} $.

The relationship between a force $F$ between two particles separated by a distance $r$ and the potential energy $E$ of those particles is $F = -\frac{dE}{dr}$.

If the energy gets lower ($dE$ is negative) when two particles get closer together ($dr$ is negative), then the force $F$ is attractive. If the energy gets higher ($dE$ is positive) as the distance gets lower ($dr$ is negative), then the force is repulsive.

Force and energy are two different mathematical ways of looking at the same thing.  That's why they are always related.$%edit$


Is there a way to oxidize salicyl alcohol to salicylic acid without using chromium trioxide? I'm doing a project for school and I need to convert salicyl alcohol to salicylic acid, but I'm not allowed to use $\ce{CrO3}$ because of its toxicity. I know it is also possible to oxidize a primary alcohol to a carboxylic acid with potassium permanganate, but I read it might be a too strong oxidizer. Is there any good alternative?

@Waylander: A general Procedure given in J. Org. Chem. 1999, 64(7), 2564–2566 as follows:

  Typical Procedure. A mixture of alcohol 1 ($\pu{40 mmol}$), TEMPO ($\pu{436 mg}$, $\pu{2.8 mmol}$), $\ce{MeCN}$ ($\pu{200 mL}$), and sodium phosphate buffer ($\pu{150 mL}$, $\pu{0.67 M}$, $\pu{pH = 6.7}$) is heated to $\pu{35 °C}$. Then mixtures of $\pu{9.14 g}$ of 80% sodium chlorite ($\ce{NaClO2}$, $\pu{80.0 mmol}$) in $\pu{40 mL}$ of water and 2.0 mol% dilute bleach ($\pu{1.06 mL}$ of 5.25% $\ce{NaOCl}$ diluted into $\pu{20 mL}$ with water) are added simultaneously over $\pu{2 h}$ (Caution! Do not mix bleach and $\ce{NaClO2}$ before being added to the reaction mixture). The mixture is stirred at $\pu{35 °C}$ until the reaction is complete (<2% starting material, $\pu{2−5 h}$ reaction time), then cooled to room temperature. Water ($\pu{300 mL}$) is added, and the $\mathrm{pH}$ is adjusted to $8.0$ with $\pu{2.0 N}$ $\ce{NaOH}$ (about $\pu{48 mL}$). The reaction is quenched by pouring into cold ($\pu{0 °C}$) $\ce{Na2SO3}$ solution ($\pu{12.2 g}$ in $\pu{200 mL}$ of water) maintained at $\pu{<20 °C}$. The $\mathrm{pH}$ of the aqueous layer should be $8.5$−$9.0$. After stirring for $\pu{0.5 h}$ at room temperature, methyl tert-butyl ether (MTBE, $\pu{200 mL}$) is added. The organic layer is separated and discarded. More MTBE ($\pu{300 mL}$) is added, and the aqueous layer is acidified with $\pu{2.0 N}$ $\ce{HCl}$ (~$\pu{100 mL}$) to $\pu{pH = 3−4}$. The organic layer is separated, washed with water ($2 \times \pu{100 mL}$) and brine ($\pu{150 mL}$), and then concentrated to give the crude carboxylic acid 2 in 85−100% yield. The products may be purified by silica gel column chromatography or crystallization.


Reaction rate coefficient constant multiplier for equation with multiple atoms consumed? For the elementary reaction $\ce{2O -> O2}$, when calculating the rate of concentration change of atomic oxygen as $\frac{d[O]}{dt}$, would I use the standard rate coefficient $k_f$ or would I have to multiply by 2 since 2 oxygen molecules are being consumed?

$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -2 \cdot k_f \cdot [O] \cdot [O]$

or

$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -k_f \cdot [O] \cdot [O]$

Or would the rate constant define the dependency on atomic oxygen?

In general if the reaction is $\ce{aA + bB \to cC +dD}$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$$.

If the reaction is indeed second order in oxygen atoms as stated by the OP, then the first equation given by the OP is correct, but unusual (you could just leave the factor two in the denominator on the left hand side instead of multiplying by two on the right hand side).


Is there a way to turn cellulose into sugar at home? I know that once you have sugars, you can ferment, distill, and dehydrate to create 95+% ethanol. I've also read about cellulosic ethanol, with the new step there being the breakdown of the cellulose into simpler sugars.  Is there a relatively cheap way to do that yet, or it still only in expensive research labs?

A method used to break down cellulose into sugars is hydrolysis :

You could use enzymes to hydrolyze the cellulose to produce glucose, but enzymes cost a lot and as you are looking for a cheap way, using enzymes is not an option. A second method is to use an acid (e.g. sulfuric acid) to produce the glucose (just google "cellulose acid hydrolysis"  and you'll find some useful links)


Reason for prescence of oil bubbles while frying When frying with oil, it's generally assumed that heavy bubbling indicates frying, and little to no bubbling indicates a lack thereof. What's the connection between bubbles and frying? If you create bubbles in some other way, baking soda and vinegar in oil perhaps, will you increase frying?

I am extremely skeptical that a typical cooking oil and water form a significant azeotrope. Such a claim demands an authoritative reference. Most known azeotropes have boiling points only fractions of a degree below the lower boiler, making little difference in the boiling point. Likewise, most known azeotropes contain over 99 wt% of the lower boiler again making most azeotropes insignificant.

It is simply my opinion, but I believe the oil transported from the bubbling liquid is in the form of an aerosol or perhaps even larger droplets.

Those issues aside I agree the bubbling is boiling of lower bp chemicals, almost certainly mostly water. As cooking occurs, decomposition into lower bp compounds is possible (and if significant amounts form, they may form azeotropes with the oil (but see comments above)).

I am also not aware of the solubility of the various atmospheric gasses (N2, O2, CO2) in oil. Gasses are in general less soluble as temperature increases (eg carbonated beverages), so they also would boil out if present in significant amounts (I'd guess they aren't very soluble at 1 atm pressure).

I should also add that while most azeotropes aren't "significant", most liquids do form them.


Axial Chirality in Carbodiimides Do carbodiimides exhibit axial chirality or do the nitrogens invert rapidly enough, as in trisubstituted amines with different alkyl groups?

Yes, carbodiimides exhibit axial chirality similar to allenes. While there are significant differences in the allene - carbodiimide systems (see below) there are also some similarities. For example, just like in amines and imines the barrier to racemization in generally carbodiimidesis low, typically in the 6-10 kcal/m range.

Anet did a study on 1,3-diisopropylcarbodimide ("Energy barrier of racemization in diisopropylcarbodiimide", Frank A. L. Anet, J. C. Jochims, C. H. Bradley
J. Am. Chem. Soc., 1970, 92 (8), pp 2557–2558, DOI: 10.1021/ja00711a064).

In the chiral compound each isopropyl group would contain two diastereotopic methyl groups split by the isopropyl proton so we would expect to see 2 doublets. On the other hand, if racemization is rapid all methyl groups become equivalent and we would then expect to see only one doublet. Experimentally it turns out that around -150 C the doublet broadens into 2 doublets giving rise to a barrier height for racemization around 6.7 kcal/m.

Important Notes:

1) Just like with amines and imines, placing very electronegative groups on the carbodiimide nitrogens (fluorine, for example) raises the barrier to racemization.  Incorporating the carbodiimide structure into a small ring (see here for an example) can also raise the barrier to enatiomerization.  Both techniques have been used to permit isolation of chiral carbodiimides.

2) Carbodiimides are not linear, but rather the $\ce{N-C-N}$ angle is around 170°.  This is attributed to the terminal nitrogens being more electronegative than the central carbon and consequently affecting the hybridization of the central carbon (e.g. changing it from $\ce{sp^1}$ to $\ce{sp^{x}}$ where 1            < x < 2)

3) Racemization in carbodiimides can be achieved by two different mechanisms

by traditional nitrogen inversion involving a planar nitrogen atom, and/or
by traditional cis-trans isomerization about the $\ce{C=N}$ double bond

both mechanisms are thought to play a role in carbodiimide racemization.


Directional emission from molecules This might be a silly question (probably is).

When a molecule absorbs it has a transition dipole moment, a measure of how strongly a certain transition will interact with light will depend on how closely a the light is to resonance with this dipole moment. The same in reverse applies for emission.

Typically in a bulk material, these dipoles are randomly spread, so luminescence is in all directions. Some molecules can be aligned, like in LCDs.

Is there a way I could align some luminescent materials such that they might emit preferentially in a certain direction? Has anyone done any work on this, what would this be called?

To get directional emission from a molecular substance, just make it into a LASER. Actually, the predecessor to the LASER, the first MASER, was based on microwave emission from ammonia. A $\ce{CO2}$ LASER is quite efficient and fairly easy to make, and the atmosphere of Mars is a natural, solar-pumped laser. So if you want directional emission from molecules, it's not even necessary to align them -- just provide suitable end-reflectors, or even a linear medium with enough gain.

That said, your question is about aligned materials... and there is a strong correlation between alignment and emission. An interesting example is found in topological insulators, which conduct only in certain directions and locations. It is possible to make a LASER from topological insulators. Some of these are as simple as $\ce{Bi2Se3}$, and others are more complex such as bis(methylene) adamantyl carbocation.


HF and water miscible? According to Wikipedia, HF is miscible with water. But isn't hydrogen fluoride a gas at RT and a sensible pressure? What would happen if you just kept adding more HF gas to some water?

Aqueous hydrofluoric acid of commerce is 48 wt-% HF, which averages tight to hydronium fluoride, $\ce{H3O^+ F^-}$.  You can snug 100% HF as $\ce{3 HF}$ being in equilibrium with $\ce{H2F^+ + {FHF}^{−}}$.  However, there is no concentration of HF that will separate into two liquid phases or a solid phase with solution at ambient temperature and pressure.

Hydrofluoric acid is exceptionally aggressive and lethal toward tissue by contact.  Effects can be delayed but not reversed.  Topical or infused calcium gluconate can ameliorate further damage.


Is there any connection between "static correlation" and the Born-Oppenheimer approximation? Beware: High probability for that I am mixing two concepts that should not be mixed, just because the same term is used when explaining the two concepts!

For some reason I get really confused when I think about static, or nondynamical, electron correlation. I somehow want to make a connection to the Born-Oppenheimer (BO) approximation.

As we know, the BO approximation is no longer accurate when two electronic states come close in energy, due to the increase in the nonadiabatic coupling terms. These terms couple the electronic motion to the nuclear motion, and arise because the electronic and nuclear motion is not fully separable in the Schrödinger equation.

Static correlation, sometimes called a "near-degeneracy effect", describes the correlated movement of electrons belonging in different electronic configurations. Certain systems are not described properly by just a single-determinant wave function. One example is the H-H molecule at long bond distances, for which the bonding and anti-bonding $\sigma$-orbitals become degenerate. (this was probably a bad explanation, as I struggle with this concept!)

So, as we have just seen, the BO approximation breaks down when two electronic states come close in energy (near-degeneracy), and the static correlation becomes extremely important in molecules with near-degenerate electronic states. Is there a connection here?

Imagine we are in the ground state at a point on some potential energy surface, and the first excited state is very close in energy. At this point, there should be a high coupling between the electronic and nuclear positions (which means what, exactly??). Also, at this nuclear geometry, the electronic wave function should be described as a linear combination of these two near-degenerate electronic states.

It is the BO approximation that introduces the idea of a potential energy surface (PES). The PES itself can be thought of as an approximation, as the nuclei should ideally be fully described by quantum mechanics. The nuclear positions should be described probabilistically by a wave function, and the very idea of a molecular geometry then becomes fuzzy.

What is the connection, if there is any?

Credit to Rudi_Birnbaum for coming up in a comment to the question with what I think is the key here.

Static correlation is a difficult to grasp concept of which even a clear definition is beyond me. However, it seems clear that a full Configuration Interaction (FCI) calculation yields all correlation energy, thus, must correctly handle static correlation. From the discussion of the Born-Oppenheimer approximation (BO) by Jensen,$^{[1]}$ I gather that the BO approximation is essentially a modification of the Hamilton operator. This shortcoming cannot be healed by a FCI calculation that only considers the electronic problem. So, in this sense, the two concepts are unrelated.

(And now for the speculative part of my answer.)

However, I cannot speak as to whether they frequently show up together in chemical systems. Though I cannot find a reference for it, I recall that the beryllium atom requires a multi-reference treatment. Being a single atom that is rather heavy, the BO approximation could be acceptable for it. Thus, this would be a case where the two are not related.


Reaction of sodium hydroxide with ethanol I am carrying out a titration to determine the solubility of a carboxylic acid in mixtures of water and ethanol. I will be titrating a diluted (previously) saturated solution of water, ethanol and the acid against sodium hydroxide.

Can the sodium hydroxide interact with the ethanol so as to give an incorrect result?

The hydroxyl group in ethanol has a pKa of about 16 (see wiki), which means you will not be able to deprotonate it using sodium hydroxide. Thus, the result of your titration should not be altered by the amount of alcohol in the solution.

Depending on the amount of ethanol I could envision that you are not determing solubility in pure water but in the sepcific mixture with ethanol, but that seems to be assumed in this kind of experiment, right ?


Resonance and mesomeric effect Can you please explain to me why are A and E not relevant resonance structures of the molecule F?
And is B aromatic?

For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.


How exactly does copper react with hydrogen peroxide, salt and citric acid? Context
I have no knowledge in chemistry. I am looking for information about etching circuit boards with citric acid instead of hydrochloric acid because it seems it could be a safer, easier way as citric acid is way safer to handle and not that toxic and harmful.
I am struggling to find much information online about the process since it is not as popular as other etchants.
The process
It seems to work really well if I mix 100ml of 3% hydrogen peroxide solution in water with quite a lot of citric acid powder(4-6 teaspoons) and a bit of salt(around 1 teaspoon), heat everything in a microwave oven to a warmer temperature(not boiling, around 70-80° C) so everything dissolves and then while it's still warm I put the copper clad in it and this seems to work just as good as hydrochloric acid and hydrogen peroxide in room temperature - etches 10x15cm PCB in 15-20 minutes.
Concerns/questions
I have done it a few times in well ventilated room, but I have no knowledge about what exactly happens in the reaction and if it is dangerous in any way, so I would like some insight into what exactly might be happening, what materials are produced and how safe/toxic they are.

The acids, hydrochloric or citric, have a common property : they produce $\ce{H+}$ in water. These ions are necessary to react with copper and hydrogen peroxide $\ce{H2O2}$. Indeed metallic copper reacts with hydrogen peroxide $\ce{H2O2}$ and an acid in the following way : $$\ce{Cu + H2O2 + 2 H+ -> Cu^{2+} + 2 H2O}$$ So whatever the acid chosen, hydrochloric or citric, the copper will be dissolved the same way. With citric acid, the dissolution should be slower. But it leads to the same final result : metallic copper will be destroyed. The only difference is the final composition of the liquid. With HCl, the final liquid is made of $\ce{Cu^{2+}}$ and $\ce{Cl-}$ ions, which may be partially recombined. With citric acid, the final liquid is made of citrate ions, which will make a complex with $\ce{Cu^{2+}}$. Citral ions from citric acid will make a small change of color, but has no other effect. Both solutions are equally dangerous, or equally non-dangerous if you prefer, because copper solutions are always toxic. You should take care of the obtained copper solutions in the same way, whatever their compositions.
Personally I would recommend using citric acid instead of hydrochloric acid. It is less corrosive, and has nearly the same effect on copper + $\ce{H2O2}$.


Order of second iozination energies of lithium,berillium,neon,carbon and boron? The correct arrange of them is like this:

Be < B < Li ... But why?

Li will have the greatest IE2 (second ionization energy) because that will involve removing a core electron, but what I am confused about is about Be and B.

I think Be will have greater IE2 than B, because B has 2 electrons in the same orbital, so the repulsive force between them will make it easier to remove one of them, but in Be there is only one electron in the orbital and there is no repulsive force in the same orbital, therefore
$$\ce{ B < Be}$$

What do you think?

To reach the second ionization potential we remove a second electron from an already ionized element
$$\ce{M^{+1} -> M^{+2}}$$
Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol)
\begin{aligned}
\ce{& Li^{+1} & 1s^2~~ &~~ 7297}\\
\ce{& Be^{+1} & 1s^2~ 2s^1~~&~~ 1757}\\
\ce{& B^{+1} & 1s^2~ 2s^2~~&~~ 2426}\\
\end{aligned}
We see that both $\ce{Li^{+1}}$ and $\ce{B^{+1}}$ have completely filled shells, the stable "inert gas" configuration.  We would expect it to be difficult to remove one electron from these two ions and destroy this stable configuration.  Further, we would expect removal of one electron from $\ce{Li^{+1}}$ to be more difficult than from $\ce{B^{+1}}$ because in the former case we are removing a 1s electron which is closer to the nucleus than the 2s electron in the case of $\ce{B^{+1}}$.  On the other hand, if $\ce{Be^{+1}}$ loses one electron it will have a $\ce{1s^2}$ electron configuration and achieve a stable "inert gas" configuration.  This should make it much easier to remove one electron from $\ce{Be^{+1}}$ than from the other two ions.


How to name tricyclo carbon compounds? Recently, I read about the nomenclature of bicyclic compounds, by mentioning the number of carbons between two bridgeheads. But, then I came across the structure of a weird type of compound, whose IUPAC name is tricyclo-[2.2.1.02,6]-heptane. As the name suggests it must be a polycyclic compound, having three alicycles. 2.2.1 suggests that there are 2, 2 and 1 carbons between the bridgeheads. But what about the next part? What is the actual rule for naming tricyclo compounds?

The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible. The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring. Bridges other than the main bridge are called secondary bridges. Thus, tricyclic hydrocarbons consist of the main ring, the main bridge, and one secondary bridge.
The nomenclature of tricyclic hydrocarbons is based on the nomenclature of bicyclic systems. The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons
P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.
The name of the tricyclic system is then constructed by citing:
(a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;
(b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,
(c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

For example tricyclo[2.2.1.02,6]heptane

(The main ring is shown in red, the main bridge is shown in black, and the secondary bridge is shown in blue.)