Photochemical rearrangement of 4,4‐diphenylcyclohexa‐2,5‐dien‐1‐one I can see how the option A and B are correct as it is just a standard dienone–phenol type reaction which involves a phenyl shift. However, the given answer is A, B, C and D.

I do not see how C and D are formed. Does the presence of light affect the reaction? If so, how?

The formation of structure 6 (your D) occurs as follows. Photolysis of dienone 1 affords excited state 2 that forms cyclopropane 3. In turn, 3 is capable of breaking a different cyclopropane bond to form 5-membered ring 4. Structure 5 is a resonance structure of 4. Structure 5 closes to the [3.1.0] product 6 (your D). I don't know how you formed A (12 here) but further photolysis of 6 can form 12 via a bridged phenyl migration as shown in structure 9. A similar migration in structure 2 leads to your B. Of course, structure C has only five carbons and was added to the choices as a trap. See reference 1 for an example of a dienone phenol rearrangement.

1) A. G. Schultz and S. A. Hardinger, J. Org. Chem., 1991, 56, 1105.


If losing electrons means getting oxidized, then does it mean that ALL metals are reducing agents, since they themselves get oxidized? I was just going through my last year's school textbooks and while I was reading this chapter on Redox, this above question crossed my mind.
Shouldn't it be true that since ALL metals lose electrons to get stable, they should also get oxidized. And since they getting oxidized themselves, they should act as reducing agents... Always?
If the answer is no, please provide an example of the exception.

Your reasoning is sound. All (pure) metals can act as reducing agents, but they do not have to all the time. There are a few examples of metals in negative oxidation states, where they would act as oxidising agent.

Probably the most prominent examples are Zintl phases, where alkaline or earth alkaline metals reduce post transition metals (or metalloids like antimony). One of the first reactions observed was the sodium lead reactions in liquid ammonia in 1907. Back in the day they proposed the following reaction equation:
$$\ce{Na + excess~Pb ~->[\ce{NH3}][\mathrm{-50^\circ C}]~ Na+ + $“$~Pb2^{-}~$”$}$$
It was later found that the actual composition is $\ce{Na4Pb9}$. With crown ethers and/or cryptants it is possible to isolate discrete anions, too.

There are more examples like this, and a nice overview is given in John E. Ellis, Inorg. Chem., 2006, 45 (8), 3167–3186.


Why are all the orbitals that have the same principal number in Hydrogen degenerate? In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number'l' (0,1...n-1 or s,p,d..). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state?

The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.

Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?

In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.

In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.


Why are the atomic radii of iron, cobalt, nickel, and copper almost the same? We have to select the set of elements which will have almost the same atomic radii:

$\ce{O, S, Se, Te}$
$\ce{Li, Be, B, C}$
$\ce{Fe, Co, Ni, Cu}$

I googled the atomic radii of the third set and found them to be extremely close to one another: all of them are nearly $\pu{125 pm}$ (according to empirical data from Wikipedia). Is there a specific reason for this, or is it just a coincidence?

I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer.

As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong.

Effective nuclear charge increases across a period because the nuclear charge increases but the shielding stays roughly the same (at least until you get to transition metals). Therefore atomic radius will decrease across the group and so option 2 cannot be correct and the answer must be option 3.


Electronic configuration of the actinides I was reading about actinides and I need some help in understanding this paragraph from Concise Inorganic Chemistry by JD Lee -

  The electronic structures of the actinides do not follow the simple pattern found in the lanthanides.
  Immediately after La the 4f orbitals become appreciably lower in energy than the 5d orbitals. Thus in the lanthanides the electrons fill the 4f orbitals in a regular way (apart from minor differences where it is possible to attain a half filled shell). It might have been expected that after Ac the 5f orbitals would become lower in energy than the 6d orbitals. However, for the first
  four actinide elements Th, Pa, U and Np the difference in energy between 5f and 6d orbitals is small. Thus in these elements (and their ions) electrons may occupy the 5f or the 6d levels, or sometimes both. Later in the actinide series the 5f orbitals do become appreciably lower in energy. Thus from Pt
  onwards the 5f shell fills in a regular way, and the elements become very similar.

How do we know when the energy of the orbitals of a subshell become lower or higher than the orbitals of another subshell? Can this just be found experimentally or is there some theoretical explanation too?

It is difficult to calculate the energy levels of the orbitals beyond hydrogen. "The first experimental evidence for the filling of the 5f shell in actinides was obtained by McMillan and Abelson in 1940," from Wikipedia

So the answer, as you suspected, is that experimental evidence from spectroscopy helped determine the electron configuration of the lanthanides and actinides, in particular.


What happens to a molecule while it is reacting? I recently learned about the graph of activation energy that looks like this (source):

I was wondering, what actually happens to the reactants as time passes on the graph? If we were to look at the molecules when the graph was at its peak, what would we see? And lastly, can the reaction be stopped somewhere in between, say at the peak, or does it always have to stop either at one end as a reactant, or at the other as a product?

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products.

(source)

As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and with some effort you might be able to stop your reaction right there, that is, to isolate the activated complex. The discovery of its structure may shed some light on the reaction mechanism and earn you a lot of likes from your colleagues.


Reaction between RCH2=CH2 and CCl4 -- Why does it form RCHClCH2CH(R)CH2CCl3 along with RCHClCH2CCl3? I am reading Organic Chemistry from Morrison & Boyd and there is this problem in it:

  Problem 6.8 From the addition of $\ce{CCl4}$ to alkenes (in presence of peroxides), $\ce{RCH2=CH2}$, there is obtained not only $\ce{RCHClCH2CCl3}$ but also $\ce{RCHClCH2CH(R)CH2CCl3}.$ Using only the kind of reactions you have already encountered, suggest a mechanism for formation of this second product.

I know that the addition of CCl4 to alkenes in presence of peroxides is a free radical reaction. So what I have thought is, while the reaction is going on, there are the following species present: RCH=CH2 (reactant), RC^HCH2CCl3 (^ represents the electron on the free radical) and the product $\ce{RCHClCH2CCl3}$.

When the free radical reacts with $\ce{RCH=CH2}$ it gives the main product $\ce{RCHClCH2CCl3}$ and $\ce{.CCl3}$. So probably the second product is formed by reaction of $\ce{.CH(R)CH2CCl3}$ with $\ce{RCHClCH2CCl3}$?

$$\ce{RCHClCH2-CCl3 + .CH(R)CH2CCl3 ->[???] RCHClCH2CH(R)CH2CCl3 + .CCl3}$$

However this reaction I have never encountered while the question explicitly states that I have to use only the kind of reactions I have already encountered. I don't think anyway that this reaction is possible.

There must be something big that I am missing here ... Can someone give me a little hint?

The propagation series starts with the addition of the trichloromethyl radical to the alkene to give the more stable secondary radical. This radical can abstract a chloride from carbon tetrachloride to give the expected product. Alternatively, this radical can add to the alkene to give an a new carbon-centered radical with an extended chain. Abstraction of chloride gives the 'unexpected' product.

The key realization here, as permeakra suggested in the comments, is that any radical in solution can theoretically add to the alkene. This means that the penultimate radical intermediate could add to the alkene as well extending the chain. This is the basis of a radical polymerization.


Quadrupole moment of a molecule What is a quadrupole moment of a molecule and how does it arise? How is it measured for a particular molecule?

I've read the Wikipedia article on quadrupoles and understand that it has to do with the distribution of charge, but I'd like a more chemistry-oriented explanation.

What is a quadrupole moment of a molecule and how does it arise?

This explanation is geared at someone with one year of chemistry. It captures the gist but is not rigorous:

To test for monopoles, dipoles or quadrupoles in a molecule or ion, assign charges or partial charges to each atoms. For partial charges, δ+ and δ- is fine (if you want estimated values for partial charges of simple molecules, use a tool like MolCalc).

To test for monopoles (ions), add up all the charges. If they are non-zero, it's an ion.

If the net charge is zero, test for presence of a dipole: If you can draw a line through the molecule such that more negative partial charges are on one side of the line and more positive partial charges are on the other side of the line, it has a dipole moment. An example for water is below.

If it does not have a dipole, you can check for a quadrupole moment: If you can draw an "X" through the molecule such that more negative partial charges are in opposite quadrants and more positive partial charges are in the other two quadrants, it has a quadrupole moment.

The animated GIF shows the model of carbon dioxide rotated in the presence of such an "X" (or "+" in this orientation if you will). In one orientation (where it is diagonal), the negative partial charges of oxygen are in opposite quadrants, while the positive partial charge is in the center (neither quadrant), showing that it has a quadrupole moment.

For a definition of the quadrupole moment (i.e. how to calculate it from a given charge distribution), take a look at this section of the wikipedia article on multipole expansions, which I found when reading the answer to Can nonpolar molecules exhibit dipole-dipole forces?.

  How is it measured for a particular molecule?

The accepted answer for this question cites a paper describing one method.


What is the degree of unsaturation of oxalate anion? I was practicing a DPP for JEE Mains & Advanced exam. There is a question in organic chemistry, a reaction sequence was given whose answer was oxalate anion.
But the next question was what is the degree of unsaturation of the compound formed in the previous question .
I simply did it 2 because there are 2 C=O bond but the answer was given 3. Please help me in making out, why it is so.

The Degree of Unsaturation can be considered as the number of Pairs of Hydrogen Atoms that could be added to the structure. One can simplify the structure to consist of two carbon-oxygen double bonds and two carbon-oxygen single bonds in one resonance contributor. Note that the single bonded oxygens have a negative charge and thus could accept a hydrogen each. This accounts for 1 degree of unsaturation. There are also two double bonds, and a pair of hydrogens could be added across these also. This gives two degrees of unsaturation. Hence there are 3 degrees of unsaturation in total (as the book says).

In order to calculate the degree of unsaturation for any compound, one can use the formula:
((2* Number of Carbons) + 2 - (Number of Hydrogens and halides) + (number of nitrogens)) / 2.
In this case, the formula gives 3, which is the degree of unsaturation.


What causes the colour when conducting flame tests on solid salts? The normal technique for flame tests is to dip a clean nichrome or platinum wire into a solution of the relevant salt, and observe the resulting flame colour when inserting the wire into a non-luminous Bunsen flame.

But what is the species undergoing the electron transition that emits the light of characteristic frequency?

One explanation suggests vaporisation and then atomisation of the sample, with the light emission occurring from the excited atom, rather than ion. Another suggests that it’s the ions undergoing the electronic transitions. The Wikipedia article is somewhat vague on the question.

Now, I was doing some flame tests for a student, and in the absence of an available nichrome or platinum wire I simply took a pinch of the solid salts ($\ce{NaCl}$ and $\ce{KCl}$ respectively), and threw it into a non-luminous Bunsen flame.

The characteristic yellow and lilac flame colours were instantly and repeatably observed.

My question is this:

Given that the outer regions of a Bunsen flame are at less than $1000\ \mathrm{^\circ C}$, and that the boiling points of $\ce{NaCl}$ and $\ce{KCl}$ are $1413\ \mathrm{^\circ C}$ and $1420\ \mathrm{^\circ C}$ respectively (according to Wikipedia), it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.

So what is the species exhibiting the electron transitions? Is it the metal ions still within the crystal lattice?

Edit: I should emphasise that the flame colour was observed in the outer regions of the flame, not the super-hot tip of the inner cone. So $1000\ \mathrm{^\circ C}$, not $1500\ \mathrm{^\circ C}$ by any stretch of the imagination. And I don’t at all buy the notion that the colours were simply the glow of hot solid particles emitting black body radiation. The colours were the characteristic sodium yellow and potassium lilac, instantly recognisable by anyone that’s familiar with simple tests like these.

There is a very detailed discussion of what cause the emission lines in flames with $\ce{NaCl}$ in The Origin of the Spectra (ironically by P. Darwin Foote et al.) 1922, starting around page 165.
They are saying that for $\ce{NaCl}$ most of the transitions are from atomic $\ce{Na}$, the $\ce{NaCl}$ having vaporized, and Na+ then captured an electron, and a small fraction of the transitions are from the act itself of Na+ capturing the electron.
They calculate that a Na atom emits 2000 D-line quanta per second.
They also calculate the fraction of Na atoms with electrons in the $3p$ state is $\ce{10^{-11}}$ at 1000K, rising to $\ce{10^{-7}}$ at 1500K. (Note: the book says "$2p$", but $2p$ in Fowler notation is $3p$ in modern notation)
Also, they say the vapor pressure of $\ce{NaCl}$ in a Bunsen flame is on the order of $\ce{10^{-6} \ mmHg}$.
So the color is from vaporized $\ce{NaCl}$ (vapor phase $\ce{NaCl}$ being molecular), that has dissociated and formed atomic $\ce{Na}$.
Also in The Absorption of Light by Flames Containing Sodium Phys. Rev. 38, 699 (emphasis added):

The dissociation probably occurs in two or more steps. Possibly the
molecules break up into sodium and chlorine ions which then lose their
charges becoming uncharged atoms. There may also be one or more
intermediate steps in the recombination.  But certainly the sodium
giving the light which we are studying consists of uncharged atoms, as
is evident from its spectrum. Evidence that there are comparatively
few charged sodium atoms in the flame is given by a study of the
conductivity of flames. Wilson, (p. 85) for example, estimates that
only 1.6 percent of the sodium in a fame is ionized even when the
concentration is small.
It also seems reasonable to assume that at the
edge of the flame there is more complete dissociation of the salt than
in the center, and that this is the cause of the difference between
the concentration ratios obtained with light from the center and from
the edges. A fairly definite proof that there is more dissociation at
the edges is given by the appearances of a fame into which copper
chloride is being sprayed. Such a fame gives almost no green color
except at the edges, where it is a bright green. This is the more
noticeable if there is a small amount of sodium impurity in the copper
chloride. The center of the flame is then yellow with a distinct border
of green.
The greater dissociation at the surface is no doubt in some way due to the greater amount of air there, but whether it is caused by a higher temperature
or by a difference in chemical action can not at present be stated.


How diphenylcarbazide is used to detect Mercury II ions? Diphenylcarbazide is used in test strips for heavy metals detection in drinking water but I can't find much info on this reagent, apart from it's MSDS and a preparation for detecting Chromium VI ions, in which diphenylcarbazide 250 mg is dissolved in 50ml acetone and few drops added to test sol which is then mixed with sulfuric acid. Also, "Vogel's qualitative inorganic analysis" references this reagent but it is very vague explanation.

If you can elaborate on how the reagent works to detect mercury, and possibly how it is stored and synthesized, that would be really appreciated.

Also, how are the test strips for heavy metals in drinking water made from this compound, I'm curious?

The Diphenylcarbazide Test for the presence of mercury was used as early as 1920s. However, since the test is very delicate (the test can detect $\pu{250 \mu g/L}$ mercury in solution), it has been used predominantly in neutral solutions, until it have been adapted to Quantitative Analysis Scheme by Scott in 1929 (Ref.1) to ditect in acidic medium:

  The mercury is precipitated in Group I is dissolved in aqua regia, that of Group II is dissoved in hydrochloric acid and sodium chlorate. In either case, the solution is evaporate to a volume of about $\pu{1 cc.}$, placed in a test tube, and diluted with $5$ to $\pu{6 cc.}$ of water. Four to eight drops of a saturated alcoholic solution of diphenylcarbazide is added and a large excess of solid sodium carbonate is gradually dropped into the solution. When the mercury is present, the form produced on neutralization assumes a blue tinge, and after the addition of an excess of carbonate the entire solution turns blue.

The author stated at the end that neither sodium acetate nor sodium bicarbonate can be successfully substituted for the sodium carbonate, indicating the sensitivity of the test for $\mathrm{pH}$. It is worth noting that the test outlined here (vide supra) has been successfully used in Quantitative Analysis Courses in University of Georgia, Athens, GA for two years before publication.

The major drawback of the test is the deterioration of stock solution of diphenylcarbazide in different solvents. Thus, it needs to be freshly prepared. To avoid this problem, a solution has been found in the publication by Urone by 1955 (Ref.2). Accordingly, non-aqueous ethyl acetate and acetone were the better solvents, diphenylcarbazide solutions of which are stable for months. Diphenylcarbazide solutions of Methyl ethyl ketone, methyl cellosolve (2-methoxyethanol), and 2-propanol are usable for 1-2 weeks. Aqueous solutions and solvents tending to be basic such as methanol and ethanol, and those containing traces of water and basic impurities, do not make good solvents for stock solutions of the colorimetric reagent (Ref.2).

Notes: (1) All solutions tested (diphenylcarbazide solutions in seven different solvents) in Ref.2 are $0.25\%$ in diphenylcarbazide. (2) Also, you don't have to synthesize 1,5-diphenylcarbazide; it is commercially available in Millipore-Sigma. (3) I don't think anybody can answer your question about "how are the test strips for heavy metals in drinking water made from this compound," because it should be a corporate secret, similar to $\ce{KI}$ papers for oxidant check.

References:

Alfred W. Scott, “Adaptation of the Diphenylcarbazide Test for Mercury to the Scheme of Qualitative Analysis,” J. Am. Chem. Soc. 1929, 51(11), 3351-3352 (https://doi.org/10.1021/ja01386a503).
P. F. Urone, “Stability of Colorimetric Reagent for Chromium, s-Diphenylcarbazide, in Various Solvents,” Anal. Chem. 1955, 27(8), 1354-1355 (https://doi.org/10.1021/ac60104a048).


False theory of how a battery works There is an explanation of how a battery works that says that inside the battery, the electrons do a complete loop and can return to their starting point. When the electrons arrive in the positive terminal and they lost all of their potential energy, the battery does work on the electrons to  put them back in the negative terminal so they are full of energy and reenter the circuit.

This explanation is very great because it answers questions : why does putting 2 batteries in series double the voltage? (because work has to be done two times on the electrons in the battery, so the potential energy doubles). Why connecting - of battery 1 to + of battery 2 without connecting + of battery 1 to - of battery 2 doesn't create electricity (because the electrons have to do a complete loop in the circuit and return to their starting point to make electricity for a long time. There is no external wire connecting - of battery 2 to + of battery 1 so there isn't a complete loop).

However, from what I understand of batteries, there is no electron that moves from the positive terminal to the negative terminal of the battery. The electrons of the circuit all come from the negative terminal and once they reach positive terminal there is nothing else happening except making the electrolyte ions neutral. They don't return to their starting point.

So could someone prove to me that inside a battery, electrons from the positive terminal are done work on and go to the negative terminal.

Or if this theory is false, could someone answer the two questions above that are easily answered by this theory in another way?

For a moment I would suggest you to forget about electrons,their flow ,batteries and stuff. Assuming that you have the basic intuitions regarding the principles of Newtonian mechanics and work done and all I would like to proceed in a different way. Suppose you are standing now on a ground and you have a ball in front of you lying on the ground. It doesn't lift itself up in the air, which is a common observation we all have made by now. But when you wish to lift it up to some height,be it h, then you expend your ATPs and chemical energies in your muscle to lift the ball slowly to that height h , and in doing so you did a work exactly equalling mgh ( neglecting air drag) . Now you take a tube of some arbitrary shape ( just to give a proper intuition) such that the ball is at its one end and the other end meets the ground. Then you leave the ball and it cleverly curves through the tube and gets back to ground level. This is exactly what we find in a battery system. It pumps electrons, against an opposing conservative field and then allows them to fall back. Here ,there's no man power but the chemicals do the trick by getting indulged into some redox reactions. Now ,if you think of those questions from this perspective hope you will be able to answer them efficiently.


Solubility of succinic acid in water vs ethyl acetate? Today I completed an organic lab practical test about discontinuous extraction. I had to extract succinic acid from a water solution using ethyl acetate as the solvent extractor.
After the extraction, I calculated the partition coefficientto be 0.3.
So, succinic acid is more soluble in water than ethyl acetate, but why?

Does it have to do with the ionization of the succinic in water that may not be possible in ethyl acetate or its related to the intermolecular interations?

Hydrogen bonding interactions are also much stronger between succinic acid and water as opposed to EA since EA doesn't have any H-atoms that can participate in the hydrogen bonding.

Ionization is also a significant point to consider, but without pH data we can't say for sure.


Acid catalysed bond hydrolysis in 3-bromobut-1-ene I came across this doubt while solving a problem. I can't understand to which path (1 or 2) I should give priority, as both sound equally good to me.

Without looking up data, I would firstly assume that the $\ce{-CHBrCH3}$ group acts like any other alkyl group, and that path I would be preferred (by Markovnikov's rule). Bromine is electron-withdrawing via the inductive effect, but only very weakly.

I did go and look up data for this particular substrate. The reference is J. Org. Chem. 1936, 1 (4), 393–404. On page 399 the authors describe the addition of $\ce{HBr}$ to the above substrate.

  In the absence of air and peroxides, however, 3-bromo-1-butene gives mostly (60 per cent.) 2,3-dibromobutane.

2,3-Dibromobutane is the product arising from path I, which confirms my assumption.


Why H2O is a liquid (water) and H2S is a gas? I always think about this question.
Is it due to attraction or repulsion or is there something else that makes one of them a gas and the other a liquid (water)?

The reason is hydrogen bonding.

The relatively large electronegativity difference between oxygen and hydrogen (about 2.3) is the reason why the partially positively charged (try to think why they are partially positively charged) hydrogens are attracted to the lone pairs on the oxygen atoms of other water molecules.

The relatively smaller electronegativity difference between sulphur and hydrogen (about 0.5) is too small to polarise the covalent bond to a required extent. Secondly, the lone pairs on the sulphur atom are delocalised in the sp3 orbitals and the d orbitals - an option unavailable to oxygen. So, the “negative charge” of the lone pairs is also distributed and the interaction is lowered.

So, the hydrogen bonding in water causes the water molecules to be associated with each other due to intermolecular forces of attraction, whereas in the case of hydrogen sulphide, these forces of attraction area much weaker.

So, lesser energy is required to overcome the forces of interaction between the hydrogen sulphide molecules than those between water molecules. This energy is available at room temperature and hence, hydrogen sulphide is a gas, while water is still a liquid.


How does HF dissolve glass? By what mechanism does HF proceed in dissolving glass? Why is it the only acid that has this capability? Is it because of the small size and high electronegativity of fluorine?

$\ce{HF}$ reacts with glass ($\ce{SiO2}$).[1]

$$\ce{SiO2 + 4HF -> SiF4 + 2H2O}$$

$\ce{SiF4}$ is not a solid that consists of vertex-connected tetrahedra like $\ce{SiO2}$ but is a gas at room temperature. Technically, $\ce{HF}$ is not a solvent since in this case it reacts with the glass vessel.

According to Spierings:[2]

  $\ce{HF2-}$ ions are adsorbed on surface silanol groups, the $\ce{HF}$ molecules on vicinal silanol groups and $\ce{H+}$ on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as $\ce{\bond{#}Si-F}$ and $\ce{\bond{#}Si-O-SiF3}$. The adsorption of $\ce{HF}$ and $\ce{HF2-}$ increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more $\ce{H+}$ ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of $\ce{H+}$ ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions.

Hollemann, A. F.; Wiberg, E. Hollemann–Wiberg: Lehrbuch der Anorganischen
Chemie, 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.
Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. J. Mater. Sci. 1993, 28 (23), 6261–6273. DOI: 10.1007/BF01352182.


Lewis model and modern physics Electron affinity of chlorine = -349 kJ per mole. And, first ionization energy of sodium = 496 kJ per mole while energy required for combining= 147 kJ per mole.

Isn't it a contradiction against genereal physics and G.N. Lewis' "electronic theory of valency"?

The Born-Haber Cycle might help you visualizing how those values are used. Notice the values used bellow are a bit different from the ones you got.

Electron affinity

  The amount of energy released when an electron is added to a neutral
  atom or molecule to form a negative ion.

Ionization energy

  The amount of energy required to remove an electron from the atom.

Enthalpy of formation (ΔH)

  The change of enthalpy that accompanies the formation of 1 mole of the
  compound from its elements.


What new material is D3O? I am a PhD student in materials engineering. What is D3O? It is not a chemical, but rather a "polyurethane energy-absorbing material" (from Wikipedia). I would wish to know and study the chemical composition of this impact shock absorbing materials.

The inventor, Richard M. Palmer, has filed at number of patents on "energy absorbing materials", one of which is US patent (7,381,460).

On the other hand, there does not seem to be any peer-reviewed publication by Mr Palmer.

There's always the option to ask the manufacturer D30 for further information and/or material samples.

The possible downside of such a cooperation is that you might not be allowed to report on the progress of your experimental work on conferences or in publications due to an NDA previously signed!


What products do 4-nitrobenzaldehyde and 3-nitrobenzaldehyde give in a crossed Cannizzaro reaction and why? What are the products of  reaction of 4-nitrobenzaldehyde with 3-nitrobenzaldehyde in the presence of $\ce{NaOH}$ when they show Cannizzaro reaction?

To my understanding, 4-nitrobenzaldehyde has more electrophilic carbonyl because the nitro group is an electron withdrawing group and at para position it exerts a very strong mesomeric effect, so it should form the carboxylic acid. To my surprise, I was told that it's the other way around.

Can someone explain me why this is true?

The first thing to point out is that the products of the Cannizzaro reaction do not necessarily correspond to the most stable products, i.e. the reaction is not necessarily thermodynamically controlled. Instead, one should consider the transition state—a six-membered ring as taught generally in organic chemistry—and note that after the reaction the compound which formed the hydrate will end up as the carboxylic acid and the one which did not form a hydrate is reduced to the alcohol. The transition state is shown in figure 1 below.

Figure 1: transition state of the Cannizzaro reaction

Thus, the underlying question is which compound is more likely to form an aldehyde hydrate. You argued with the nitro group of 4-nitrobenzaldehyde exhibiting a strong $-M$ (mesomeric) effect. However, this mesomeric effect cannot, in fact, cross the $\ce{C-C(=O)}$ bond and thus cannot influence the carbonyl group as much as the simple $-M$ label implies.

Figure 2: some mesomeric structures of 4-nitrobenzaldehyde, demonstrating that the mesomeric effect cannot ‘cross the $\ce{C-C}$ bond’.

Now it is known that aldehydes with a more electron deficient carbon are more likely to form hydrates, hence why we are doing a mesomeric/inductive discussion in the first place. But since we actually do not have any $-M$ arguments that can help our case, we need to turn to $-I$ arguments. When considering inductive effects, distance matters much more so 3-nitrobenzaldehyde experiences a stronger $-I$ effect than 4-nitrobenzaldehyde. Therefore, this compound is hydrated more favourably and form 3-nitrobenzoic acid while the reduced product should be 4-nitrobenzyl alcohol.


How dangerous are these soil organic carbon digestion methods? I have several ground up soil samples that I need to get fractions from. The fractions are based on the Century Soil Model. Briefly, the fractions relate to the approximate time frame it takes for an organic carbon pool to be metabolized in the soil. Fast pools ≈1 year, slow pools ≈10+ years, etc.

I have looked at several papers to try to get an idea of the chemical methods used for digesting the soil samples to get at the more stable carbon pools. For more easily digested organic carbon pools, hot water seems to work, but for the middling and more recalcitrant pools, other reagents are used.

Some methods mentioned include:

dissolve in $\pu{0.1 M}~\ce{NaOH} + \pu{0.1 M} ~\ce{Na4P2O7}$
dissolve in $\ce{HCl}$
reflux with $\pu{6 M} ~\ce{HCl}$
"stepwise chemical digestion with cold and hot $\ce{H2SO4}$
(95 - 98% concentration)"
oxidation with $\ce{KMnO4}$ ($33\text{ or }\pu{333 mM}$)  (this one I see mentioned most frequently)

I am not a chemist. I took the required chemistry classes for my degree and passed, but that's years ago now. When I did do chemistry labs, my results hardly ever came out as they were supposed to, so my skill in this field is minimal. Besides these methods not working, and thus wasting time and money, my real concern is how potentially dangerous these chemicals are. Would someone be so kind as to give me an indication of how risky these chemicals might be? We have a functioning fume hood, will buy PPE, and be sure there are safety procedures in place, but I'm still wondering how much I should worry if a flask breaks or something spills.

$\pu{0.1 M}~\ce{NaOH}$ is potentially corrosive when in contact with metals or human skin (see here), so I advise precautions such as wearing protective standard gloves and keep metals away from feasible damage.

$\ce{HCl}$ and $\ce{H2SO4}$, especially when they're hot and are in higher concentrations, are very corrosive acids that are dangerous in contact with many materials. They can also release noxious gases in the reactions, but their weakness is glass. Due to reasons irrelevant, these acids can't corrode glass. I recommend extreme precaution. You can work under a "glassy" protection and as Yomen mentions, under a fume hood. Gloves are very important here. Here's what Caltech thinks about precautions using $\ce{HCl}$ and here's what Berekley does about sulfuric acid.

$\ce{KMNO4}$ is a very strong oxidizer. It may cause fire in contact with different materials (we had a volcano experiment in elementary school where we mixed it with glycerine and the result was very exothermic), and it has biohazards such as kidney damage. Here it's explained more comprehensively.


Does magnesium carbonate exist? what makes magnesium bicarbonate different from calcium bicarbonate In my book, I found the following reactions:

$\ce{Ca(HCO3)2 ->[\Delta] CaCO3 + … }$

$\ce{Mg(HCO3)2 ->[\Delta] Mg(OH)2 +...}$

What makes these two reactants different? Why is one forming hydroxide while the other carbonate?

Both of these reactions are thermal decomposition reactions, but the difference basically occurs in the polarizing power of the cations $\ce{Ca^2+}$ and $\ce{Mg^2+}$ .

The thermal decomposition of $\ce{Ca(HCO3)2}$ is just the reverse reaction of the lime water test for the detection of $\ce{CO2}$ , where the carbonate combines with $\ce{CO2}$ and $\ce{H2O}$ to give the bicarbonate, so it stands to reason that the bicarbonate will break down to give the reactants we started with.

We can therefore, imagine that a similar decomposition will occur with $\ce{Mg(HCO3)2}$ to yield $\ce{CO2}$, $\ce{H2O}$ and $\ce{MgCO3}$. And in fact, a similar reaction does occur on drying the bicarbonate solution.

However, a small amount of $\ce{Mg(OH)2}$ is also obtained as a side product.
The reason is that, $\ce{CO3^2-}$ is relatively quite a large-sized anion.
Also, if you recall the diagonal relationship between $\ce{Li}$ and $\ce{Mg}$, then sizes of both $\ce{Li+}$ and $\ce{Mg^2+}$ will end up being similarly quite small relative to other members of their respective groups. Thereby,Fajan's rule will become important to consider here, and the carbonates of both these cations will experience similar internal polarization, and tendency to decompose relatively.

Hence, since we know that $\ce{Li2CO3}$ is the only Group 1 carbonate to decompose at laboratory temperatures to give $\ce{Li2O}$ and $\ce{CO2}$ ,(see this) we can expect $\ce{MgCO3}$ to also  break down into $\ce{MgO}$ and $\ce{CO2}$ on prolonged heating.

$\ce{CO2}$ being a gas will escape, and since Group 2 bicarbonates exist only in aqueous form, so the magnesium oxide produced will hydrolyse within the solution to produce the hydroxide, that is, $\ce{Mg(OH)2}$.

Note: Decomposition of $\ce{Li2CO3}$ is only "easy" relative to other members of Group 1. It still has a pretty high decomposition temperature of 1300 °C approximately

This value of 1300 °C might still not be accurate, see the green Note under 'Heating the carbonates' for more insight.


Combining Herbal Powder Extract If I mixed two powdered herbal extracts together and put them through distillation process, then would I end up with a high concentrated form of the two extracts combined together? I'm trying to make an all natural energy drink. I'm not sure if i should just mix the powdered herbs or use some chemical process to combined the herbs into a concentrated form. I figured I would need high doses of the herbs in order for the drink to have an affect. So, should a chemical process to combined my ingredient, or just use herb tictures to avoid a powdery taste?

If this is an energy drink for humans then the active are either soluble in water or ethanol. The standard factory method is to use supercritical CO2 extraction. Check the method for production of instant coffee. Briefly, they mix coffee beans with CO2 at 40C (above critical temperature and under very high pressure. Then they filter it and drop the pressure. Components precipitate out. You can drop it to certain value, filter to second chamber, drop pressure slightly, filter move to third chamber, drop pressure to 1 Atm. In this case second chamber can have a single component. In case of coffee caffeine precipitates in second chamber. Now mix stuff from first and third chamber to get decaf.

Ethanol extraction works ok and is used by pharmacists. It is hard to concentrate it however. And ethanol is an active ingredient on its own.

Water extraction is the dirtiest way mainly because it is difficult to concentrate your material.


Mechanism of oxidative dearomatisation with hypervalent iodine The following step was taken from the synthesis of Kinamycin C on SynArchive. It employs the use of a peculiar reagent, that is, bis(acetoxy)iodobenzene (BAIB), also known as phenyliodine(III) diacetate (PIDA). Being a high-valent iodine species, it is used as an oxidising agent in organic chemistry. So clearly, there is some sort of oxidation taken place in this step, along with nucleophilic aromatic substitution involving an $\ce {MeOH}$ molecule. However, I am quite uncertain as to what is the mechanism by which this step takes place.

The conventional mechanism is as follows:

The phenol displaces one acetate group on iodine – this makes the iodine itself act as what is essentially a fancy leaving group. Nucleophilic attack at either the ortho or para position allows the loss of iodobenzene and another molecule of acetic acid. What I can't explain right now is why it chooses to go para rather than ortho; it probably depends on the exact substitution pattern as I have seen examples of both. A good reference for this is Zhdankin, Hypervalent Iodine Chemistry (2013).


How to quickly diffuse a dye through still water I'm doing an engineering experiment soon and would like to be able to quickly dye water. The water is still (i.e. not moving) and I would prefer not to physically disturb the water as best I can.

What techniques could I employ to quickly diffuse a dye through water? I was thinking that electrophoresis could be a possibility. If so, what dyes (or what could I add to a dye) to help speed up the process.

Thanks in advance.

If you want a homogeneous distribution of dye, you will have to stir the solution.
If you want to use electrophoresis, you need a charged dye.

Azo dyes used for hair coloring have colour and charge, for instance Basic Red 51 and Basic Yellow 87:

Of course, heat will increase the diffusion rate but will also disturb the water.


If something contains large amounts of cyclopentane and cyclohexane, what is it called? If something contains large amounts of cyclopentane and cyclohexane, what is it called? I'd like to know what is the something as well. It's a chemistry olympiad question for this year. See here, Question 3-g. Just let google translate it for you. I find some questions quite weird as well, but what can you do. Otherwise its pretty good.

The most likely name would be naphthenes. It could also be called cycloalkanes.

Added:
Since the reference to the question is about the coal and oil production, the answer would be naphthenes. When the coal and oil were distilled, a layer (called naphthene) was removed which was found to contain mainly cyclopentane and cyclohexane.

I would complete the statement as:
If a fraction of distillate contains large amounts of cyclopentane and cyclohexane, it is called naphthene.


Does higher Oxidation/ reduction potential translate to higher energy storage in battery? Some articles refer to the fact that Lithium is preferred metal of choice for batteries because their half cell voltage is high (slightly above 3V)

If that is the case why can't we use other metals like Mn which can have higher oxidation potential (Mn can have +7 oxidation state which translates to higher oxidation potential)?

Energy storage depends on the electromotive potential (i.e. difference between species in the electromotive series) and on the number of electrons available.

Li, for example, has an oxidation potential of ~3.04 V relative to hydrogen, but Al has one of 1.66 V, so Li has the greater potential. On the other hand, Li has only one freely available outer electron, but Al has three, so Al stores more charge per mol, giving three times the capacity in coulombs on that basis. On the gripping hand, however, Li has a mass of ~7 amu, and Al is ~27 amu, so an aluminum cell carries around the dead weight of unused neutrons and inner electron/protons.

So, depending on the needs, Li batteries weigh relatively little and provide comparatively high potential, aluminum batteries would hold more charge at a bit lower voltage, and even Zn is useful, though deficient compared to those others in volts and coulombs (it also has a bunch of inert neutrons, but there is no charge for them), because it is relatively inexpensive.


Freely available software for generating 3D coordinates of molecules As part of some research I am doing, I want to do a head-to-head comparison of software for generating 3D atomic coordinates for molecules. Is there a list anywhere of chemical software that does this?

I was able to settle on the following constraints: freely available, or at least free for academic users; can be run on the command line (either Windows or *nix); and, can handle small to medium-sized molecules (I.e. I am not expecting the software to handle oligopeptides or other sizable biomolecules).

The input could be any of 1. a MOL file without 3D information; or 2. line notation like SMILES or InChI. I was envisioning doing tests for at least those formats, unless I am missing anything.

If there is no such list, would you be able to recommend any software for this task?

I am already using OpenBabel and RDKit to generate coordinates, so I got the idea to look for other software to compare with it.

(Survey papers that have previously done comparisons of this sort would also be welcome and appreciated, so that I can see what hasn't already been covered by previous researchers.)

Summary: RDKit ETKDG is currently the best open source tool

This is an interesting question, since a reviewer just asked us to do such a benchmark.

The trick is that there's a fine distinction between "generating 3D coordinates" (i.e., one conformation) and "generating 3D coordinates and doing some conformer sampling" presumably with the latter helping to find a low-energy conformer, thermally-accessible ensemble, or a pose that might match an experimental structure or drug docking pose.

I'm not sure there are many papers on the first (i.e., generate one 3D structure - how good is it)? For one, it would depend a lot on your definition of 'good.' I personally prefer 'low energy conformer' because it's easy to define. Most people usually mean 'close to the experimental crystal structure' which is not usually the same as 'lowest energy'.

I digress. There are a few benchmark papers - largely grabbing a few conformers, usually including a mix of free or open source tools and commercial tools.

Ebejer, J.-P.; Morris, G. M.; Deane, C. M. "Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.
Friedrich, N.-O.; Meyder, A.; de Bruyn Kops, C.; Sommer, K.; Flachsenberg, F.; Rarey, M.; Kirchmair, J. "High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57 (3), 529-539.
Nils-Ole Friedrich, Christina de Bruyn Kops, Florian Flachsenberg, Kai Sommer, Matthias Rarey, and Johannes Kirchmair. "Benchmarking Commercial Conformer Ensemble Generators." J Chem Inf Model 2017, 57 (11) , 2719-2728.
Naruki Yoshikawa, Geoffrey R. Hutchison. "Fast, efficient fragment-based coordinate generation for Open Babel" J. Cheminf. 2019 11, art. 49.

While the newest method in Open Babel (development for v3) is faster than RDKit, particularly without using force field minimization, RDKit's ETKDG is very, very good - competitive with most commercial packages.


Can a supersaturated solution get cooler during crystallization? Suppose we have a supersaturated solution that is isolated from the environment. Some of the solutes condense into a crystal in the solution. Is it possible for the solution to get cooler during this process?

If so, why doesn't it violate the second law?

The system sodium sulfate (Na2SO4) / sodium sulfate decahydrate (Na2SO4 * 10 H2O) / Water might be an example.

Dissolving Na2SO4 in water is exothermic due to hydration.
Dissolving Na2SO4 * 10 H2O is endothermic (entropy effect).

Below 32.384°C Na2SO4 * 10 H2O crystallizes from aqueous solution, above 32.384°C water free Na2SO4.

The solubility of sodium sulphate shows the following unusual shape with a maximum at 32.384°C.

If you would now prepare a saturated solution of sodium sulphate at 32.384°C, and would then be able increase the temperature without precipitation of Na2SO4, you would get a supersaturated solution of sodium sulphate.

Putting a seed crystal of Na2SO4 into it should result in crystallization with the solution getting cooler during this process.

Note that the crystallization frees water molecules that were previously bonded through hydration. This will lead to an increase of entropy.


How do Invert Sugars deter crystallization in Sucrose solutions? Invert Sugars, by themselves, are notorious for their reputation to crystallize fast (think honey), but their addition to solutions of Sucrose (think simple syrup) deters the crystallization process. Take Caramel, Sucrose that has already decomposed into Fructose and Glucose and subsequently into the numerous products of caramelization: It crystallizes or, at least, turns grainy without the addition of Invert Sugars or a weak acid (Citric, Tartaric) (although the crystallization process is retarded, taking days even at low temperatures). Why do Invert Sugars deter crystallization when added in small quantities in solutions such as Caramel, which, I think, do already contain Invert Sugars as they are already largely decomposed? I would like to know of how Invert Sugars physically interact with the other sugar molecules and prevent them from coming together, even though they're not really resistant to crystallization (as pointed out above).

An additional (and optional) question: Why does the rate of cooling of a sugar solution affect its crystallization? So, while a slowly cooled solution crystallizes more, a faster cooled solution is more likely to crystallize less spontaneously.

I would like to know of how Invert Sugars physically interact with the other sugar molecules and prevent them from coming together

Glucose and fructose molecules (being essentially parts of sucrose) can merge into sucrose crystals during their formation, slowing or even completely stopping their growth in certain directions (if their concentration is high enough).

  even though they're not really resistant to crystallization (as pointed out above)

Glucose and fructose crystallization process is much slower than that of sucrose. From confectionery process's point of view, once the fast growing sucrose crystal structure is disrupted and its growth hindered no crystallization occurs.

  Why does the rate of cooling of a sugar solution affect its crystallization?

Because crystals' growth takes some time. If cooled fast, neither existing crystals will have a chance to grow, nor new seeds will appear. If cooled really fast, saturated sucrose syrup can even turn into glass with no crystals at all.

[literature]: Sgualdino, G., Aquilano, D., Vaccari, G., Mantovani, G., & Salamone, A. (1998). Growth morphology of sucrose crystals. Journal of Crystal Growth, 192(1-2), 290–299.


Can mechanical agitation increase the reaction rate? The specific example I saw recently was cleansing of concrete (portland cement) film from brick using a diluted muriatic acid. It appeared that the more scrubbing was used, the quicker the cleaning process would go. That seems to match up with everyday experience.

However what is not clear is how that mechanical agitation was specifically improving the reaction rate.

More generally, are there a 'class' of reactions or reagents where mechanical agitation will alter the reaction rate? And is that affecting the actual reaction or more just the process efficiency?

If the reaction is heterogeneous (multiple phases involved), mechanical agitation helps. If the reaction is homogeneous (all reactants and products in the same phase), mechanical agitation usually does not help. An exception is when the process is auto-catalytic (i.e. starts in one corner and then propagates).

In the example mentioned, agitation will help to transport concrete away from the brick surface and to bring fresh acid to the surface. Scrubbing also helps if there is no chemical reaction and the removal is mechanical in nature.


How to identify the structure of an organic acid based on its reactivity, mass and NMR spectrum? Taken from OCR A Chemistry Unified Chemistry Sample Paper.

  Information about a monobasic organic acid D is shown below.

  D reacts by both electrophilic substitution and electrophilic addition.
  The molecular formula of D is $\ce{C_xH_yO2}$.
  The mass spectrum of D has a molecular ion peak at $m/z = 148$.
  The $\ce{^13C}$ NMR spectrum of D contains seven peaks.

  Determine and draw a possible structure for D. Explain your reasoning from the evidence provided.

I understand that D must contain a benzene ring (electrophilic substitution) and a double carbon bond (electrophilic addition). However, I'm at a loss about how to deduce the molecular formula. Taking away $M_r(\ce{O2})$ from $\pu{148 g mol^-1}$, gives $M_r(\ce{C_xH_y})= \pu{116 g mol^-1}$. The mark scheme gives the molecular formula as $\ce{C9H8O2}$. I can't see how the values of $x$ and $y$ were deduced. All you know is that D contains a $\ce{C=C}$, benzene ring and most likely a $\ce{-COOH}$ group (due to monobasic nature).

I'm at a loss about how to deduce the molecular formula.

Compounds containing only C, H, and O have a certain number of hydrogens (h) compared to the number of carbons (c).

If there are no double bonds and no rings, h = 2c + 2. This is true for linear chain alkanes. Adding oxygens to those (hydroxyl groups, ethers) does not change the number of hydrogens. Turning the linear chain to a branched structure does not change the number of hydrogens.

For every double bond you add, you lose two hydrogens (think of it as turning sigma bonds to hydrogen into pi bonds between the heavier atoms). For every ring you close, you lose two hydrogens (one additional C-C bond instead of each carbon bound to a hydrogen, for example).

So for molecules with double bonds (d) and rings (r), you get:

h = 2c + 2 - 2d - 2r

  All you know is that D contains a C=C, benzene ring and most likely a -COOH group

So the benzene ring accounts for three double bonds and a ring, the C=C for a double bond and the -COOH for a double bond (C=O). At this point, we don't know if there are even more double bonds or rings, so we have d >= 5 and r >= 1 and the following inequality

h = 2c + 2 - 2d - 2r < 2c - 10

So the molecule will not have the maximum possible number of hydrogen atoms (2c + 2) but much less.

  $M_r(\ce{C_xH_y})= \pu{116 g mol^{-1}}$

Just like porphyrin said in his comment, there is only one solution to x and y that makes chemical sense: 9 carbons and 8 hydrogens. You don't even have to think about the rings and double bonds like I discussed above, but it is a nice check. For a hydrocarbon with 9 carbon atoms we would expect 20 hydrogen atoms in the absence of rings and double bonds. We have 12 less than that, so the number of rings and double bonds in the molecules is 6. That fits with the expectation of a benzene ring (accounting for 4), another C=C double bond (1 more) and a carboxylate group (1 more).


Energy-efficiency aside, what are the chemical constraints on CO₂ capture and methanation? Synthesising $\ce {CH4}$ from air and water (in a non-biological process) has been proposed as one form of energy storage. What are the chemical constraints at play here? That is to say, what sort of catalysts might be used, and what are the performance / lifetime issues with them?

More specifically:

$\ce {CO2}$ capture and recycling (CCR) has been suggested as one form of round-trip electricity storage, with a round trip being:

electricity
$\to$ something that can hang around for hours to months
$\to$ electricity some time later

The proposal goes something like this: capture the $\ce {CO2}$ produced from the combustion of $\ce {CH4}$ in air in a gas-turbine electricity plant, and store it. Then, when there's surplus electricity, use it to synthesise $\ce {CH4}$ from the $\ce {CO2}$ plus water (I'm guessing electrolysis of water and a Fischer-Trop / Sabatier type reaction). And next time there's a deficit of electricity, burn the $\ce {CH4}$ to generate electricity, and capture the $\ce {CO2}$, and go around again. So, for the sake of this question, I'm specifically interested in the non-biological synthesis of methane.

And let's put aside for one moment the issue of energy efficiency, as that's effectively an economics question. (The question being "is the round-trip efficiency less than the ratio of off-peak to peak electricity prices?") Let's just agree that round-trip energy efficiency is one of the factors that influences the economics, and leave it at that for now.

So the question is: what sort of catalysts might be used, and what are the performance / lifetime issues with them?

I'm only directly attacking your direct question, "what sort of catalysts might be used, and what are the performance / lifetime issues with them?" because the rest is ill-defined. I'll mention some electrochemical conversion aspects. I hope to illustrate why your question is problematic.

There are several different products one could imagine reducing carbon dioxide to:

methane
methanol
formaldehyde
carbon monoxide
formic acid
ethylene

Each will require different materials and conditions, but they all require energy for this to occur. Not to mention, these are all difficult reactions. Thermodynamically speaking carbon dioxide is at the bottom of this energy well and to get out of it requires spending energy.

One approach involves directly electrolyzing on a metal. Cu in aqueous solutions can give methane as a major product, but it also evolves hydrogen and many of the others I listed above. There are a couple of problems with this but the major problem is that it often requires large overpotentials at rates of interest for industrial purposes and this means more cost in terms of electrical input. Another major problem surrounding nearly all of these studies is the selectivity of a catalyst: in water at pH 7 evolving dihydrogen is thermodynamically more favorable than reducing carbon dioxide to everything except methane and methanol. Unfortunately these two are the most complex, as far as chemical transformations are concerned, to make. So the other products are much more likely to be produced along the way. This is why people have focused on making simpler things such as carbon monoxide or formic acid.

Where does this energy come from? Right now coal. So we would end up creating more carbon dioxide than we reduced to other species. It reminds me of how diesel-electric transmission works in freight trains: a dirty engine turns an electric generator that generates electricity that is used on motors instead of the motor directly driving the wheels; I think we can call this diesel-electric transmission an indirect drive whereas the latter is a direct drive. It turns out that making the former is far easier in engineering rather than the latter. The same is true here. It's easier to generate electricity somewhere else then reduce CO2 rather than have energy directly fed to reduce carbon dioxide. You can get efficiencies better by going the latter, but engineering/technical complexities increase immensely.

As a means to address this electrical burden via the direct-drive analogy, people have looked at these artificial photosynthesis approaches, i.e. using light to  drive these reactions or at least help ease the energy burden. This requires photocatalysts and often with these come semiconductors. Some semiconductors are naturally good at reducing carbon dioxide, such as CdTe, but this better performance for generating methanol is in turn related to an actual corrosion process of the semiconductor. Semiconductors are very complicated and expensive to manufacture well. They tend to corrode easily when they work well with sunlight. They require catalysts and this introduces huge compatibility issues and manufacturing complexities. We'll just leave the indirect approach at the state  the field is, complicated and incomplete.

As for the direct approach:

it's complicated by the fact that carbon dioxide is fundamentally limited in solubility at room temperature. Hence why other people prefer non-water systems. These unfortunately add other complexities...

Actually while typing this I came to the conclusion that I could write a much more succinct response to you rather than continuing to babble on: The best candidate will involve a transition metal, there is no doubt about that. It may end up being a molecular species, but there are so many possible structures that it is hard to say if we won't find a simpler heterogenous catalyst that is more capable first. By Murphy's law, it will be the most expensive material imaginable. But right now, considering everything, you can always look at all extensive work that has been done on copper, but realize it isn't very efficient or selective.


Product of an intramolecular aldol condensation When nonane-2,8-dione is treated with base, the product 1-(2-methylcyclopent-1-en-1-yl)ethan-1-one forms according to the following scheme: The arrow denotes the hydrogen that is first attacked by the base. But why does the reaction not take place like this? —
Why is the arrow-marked hydrogen less acidic than the previous one?

The two protons are of similar pKa and you can get deprotonation at both sites with equal ease, however it is important to understand that most intramolecular aldol reactions are reversible and thus give rise to the thermodynamic product i.e. the most stable product. [6]-ring formation is always preferred over [8]-ring formation 1, 2. If you prepare the [8] product by other means, and subject it to the aldol conditions you will get the [6] product. The only exception to this is when you set up the reaction to be under kinetic control e.g. by using LDA, then you can get a mixture.


Bromination with NBS on cyclic structures I found this question and the answer as shown in my book. I didn't understand why the bromine particularly entered the position shown in the answer.

First of all, if a free radical was generated there it would be 2°, and also will have resonance. Is any other product possible or this is the only major product?

Also, is this allylic bromination? Every allylic position is also a vinylic position — I am not getting it.

As a follow-up to the excellent comments provided thus far, the following thoughts are added. Electrostatic Potential Map of azulene (1a) shows that there is high electron density in the 5-membered ring (red color signifies high electron density). The 5-membered ring is reminiscent of the aromaticity of cyclopentadienide anion and the 7-membered ring bears a positive charge and has aromaticity similar to cycloheptatrienyl cation. This phenomenon is represented in structure 1b. The reaction does not involve bromine atom (a radical) because a bromine atom would be expected to attack the electropositive ring, not the electron-rich ring. The reaction is ionic and likely involves bromine rather than N-bromosuccinimide (NBS, 4) directly. Bromine is liberated from NBS by trace amounts of HBr formed by the slow decomposition of NBS. In the event, bromination of azulene (1c) produces a resonance stabilized cation 2a. Its canonical structure is 2b that is identical to 2b'. Loss of a proton from 2b (or 2b') provides bromoazulene 3.


Variation in atomic sizes in the transition elements From left to right across a period, effective nuclear charge increases in the transition series, just like in the p block, and so atomic size decreases. Here the electrons are being added to the d orbital, where shielding is relatively poor compared to the p orbital. It should follow that the increase in the effective nuclear charge is more significant for the transition elements than the p block in a given period, leading to greater size variation among successive members - but the size variation is smaller in the transition elements instead. In fact atomic size increases from Ni to Zn. Is there any logical explanation for the fact?

From Ni to Zn there is increase in size because of the completely filled 3d and 4s orbitals. Here the nuclear charge is increased with atomic number but due to more electron -electron repulsion in 3d and 4s orbitals  as compared to force of  attraction between nucleus and outer orbital electrons there is little expansion in size as compared to other elements in 3d series.


What is θ in Bent's rule equation (%s - character)? I was learning about Bent's Rule. I came across a formula
$$\cos\theta = \frac{s}{s-1}$$
I am quite confused about $\theta$. I know that it represents bond angle. But for a compound like $\ce{PCl5}$, which bonds angle has to be considered? For the equatorial bonds, $\theta = 2 \pi/2$ was used and $\theta = \pi/2$ was used for axial bonds. My doubt is that why we are not using $\theta = \pi/2$ for equatorial bonds as they also have form $\pi/2$  angle with with axial bonds.
I asked my mentor about this. He said that $\ce{PCl5}$ has non-equivalent hybridization and $\mathrm{sp^3d}$ is actually $\mathrm{sp^2 + pd}$. The equatorial bonds are $\mathrm{sp^2}$ hybridized and have bond angle as $120^\circ$. The bond angle for axial bonds is the angle formed by the axial bond and the plane formed by chlorine atoms which connected to phosphorus through equatorial bonds.
After this, I became more confused. Can anyone please tell me the meaning of $\theta$ in the formula? Is it really the bond angle? If yes, then which angle has to be considered in $\ce{PCl5}$ and why?

TL;DR: $\theta$ is the angle between any two equivalent bonds and using  Coulson's theorem for anything beyond the second period is out of its scope.

First things first, Coulson's theorem (the generalization of said formula) cannot be applied in cases of spndm. Therefore using $\ce{PCl5}$ as a comparison is not the right idea. Also, the formula that you've stated is only when finding the bond angle between two equivalent bonds as would be proved below.
This equation is supposedly a simplification of Coulson's theorem that states,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Here, $\lambda_1$ stands for the hybridization index of one bonded orbital, $\lambda_2$ denotes the hybridization index of the second bonded orbital and $\theta$ signifies the bond angle between the two orbitals. However, this rule as said above can only be applied to spn hybridisation compounds where $\sqrt{n}$ is the hybridisation index. Its formal proof is given in Wikipedia if you are curious and want to find out.
The simplified idea behind Coulson's theorem (as I understand it), is as follows.
Each set of equivalent bonds is made up of a linear combination between s and p orbitals. This is what hybridization is. Given any pair of bonds, we can find the bond angle between the two through their hybridization indexes ($\lambda_1$ and $\lambda_2$) and vice versa. Let's take the example of $\ce{CH4}$,
$\ce{CH4}$ has four equivalent $\ce{C-H}$ bonds and we know that the bond angle is $109.5^\circ$and so according to Coulson's theorem, the hybridisation of the $\ce{C=H}$ bond can be found to be,
$$1+\lambda_\ce{C-H}\cdot\lambda_\ce{C-H}\cos(\theta_\ce{H-C-H}) = 0 $$
$$1 +\lambda_\ce{C-H}^2\cos(\theta_\ce{H-C-H})  = 0 $$
\begin{align}
\therefore \lambda_\ce{C-H}^2 &= \frac{-1}{\cos(109.5^\circ)} \\
&\approx 3
\end{align}
Therefore, the $\ce{C-H}$ bonds are sp3 in nature.

Proof of $\cos \theta = \frac{s}{s-1}$
From Coulson's theorem, we have,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,
$$1 + \lambda^2\cos(\theta) = 0$$
This implies,
$$\cos \theta = \frac{-1}{\lambda^2}$$
Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be
$$\text{%s character} = \frac{1}{1 + \lambda^2}$$
Solving for $\lambda^2$, we get:
$$\lambda^2 = \frac{1-s}{s}$$
Substituting this in the equation for $\cos \theta$, we get:
$$\cos \theta = \frac{s}{s-1}$$
Note: An assumption that I took here is that the bonds are equivalent. If this is in fact true, then this rule can only be used to find the bond angle between any two equivalent bonds and hence in the case of $\ce{PCl5}$, only the axial-axial ($\ce{P-Cl}$) bonds or equatorial-equatorial $\ce{(P-Cl)}$ bonds can be compared. θ then is the angle between any two equivalent bonds.


Cause of uniform glow in cathode ray tubes I got the following extract from a chemistry book (emphasis mine):

  It is observed that current does not flow through the gas at ordinary pressure even at high voltage of 5000 volts. When the pressure inside the tube is reduced and a high voltage of 5000–10000 volts is applied, then an electric discharge takes place through the gas producing a uniform glow inside the tube. When the pressure is reduced further to about 0.01 torr, the original glow disappears. Some rays are produced which create fluorescence on the glass wall opposite to the cathode.

This excerpt is about cathode rays discovery in a chapter on atomic structure. Here, the author is talking about a uniform glow appearing when the pressure was reduced. Then he says that the rays (which he then refers to as cathode rays) appear after the disappearance of the glow. I have never noticed this glow in any experiment. Does anybody know what this is?

The uniform glow is due to ionization and recombination of the residual gas; it's called a glow discharge. At higher pressure, a spark or arc discharge occurs at much higher current density. Fluorescent lamps and neon lamps operate in the glow-discharge region, and high-pressure xenon lamps use an arc discharge.

Elements of the residual gas can be identified by the color of the glow discharge. Nitrogen and argon in air, for example, glow purplish blue.

As pressure decreases further, density is too low (and path too short) to produce a visible glow discharge, but electrons, AKA "cathode rays", hurled off the electrodes impact the walls of the container, which may glow green in borosilicate glass, and in very low-pressure vacuum tubes, such as the 1G3GT high-voltage rectifier, the electrons impacting the anode produce X-rays (high energy photons), which may also create fluorescence in the glass shell.

The difference is easy to see in this video, which demonstrates the change in appearance from an arc discharge to glow discharge, and finally to the fluorescence of the glass envelope due to X-rays or electron bombardment. The arc starts at ~20 seconds, the glow discharge at ~26 seconds, striations form ~36 seconds, and by ~46 seconds, the glass envelope glows green from electron bombardment (though it might rather be the green of residual oxygen). This was one of my favorite demonstrations to show students the basis for spectroscopy!


A question about geometry of benzenetetrol In the 3d conformation of 1,2,4,5 - benzenetetrol, how could i determine the direction to which hydrogen atoms of hydroxyl groups would point? and if there were tertiary butyl groups in positions 3 & 6 would it be more stable for hydrogen atoms to point in the direction away from t-butyl groups as a matter of steric hindrance ?

!

Source

The melting point of 1,2,4,5-benzenetetrol is 210C (https://www.chemicalbook.com/ProductChemicalPropertiesCB41307665_EN.htm), so there is very probably significant intermolecular attraction between molecules, probably in three dimensions, which would involve the hydrogens pointing up and/or down, and not in-plane.

For a crystal structure, https://pubchem.ncbi.nlm.nih.gov/compound/1_2_4_5-Benzenetetrol suggests that 3 hydroxyls are planar while one is as out of plane as you would expect for cyclohexanol. Believe it or not. (I don't believe it.)

The (predicted) NMR spectrum shows two peaks, one very sharp for the aromatic protons, the other fairly sharp for the hydroxyls, but no experimental data seems to be available. The sharpness of the predicted spectrum suggests that the prediction is that free rotation is likely in solution.


Integrated rate law graphs Why does the graph of $\ln[\ce A]$ vs $t$ produce a straight line for the first order rate law?

Also why does the graph of $t$ vs $\frac{1}{B}$ produce a straight line for the second order rate law?

The short answer is that the equations you describe as graphs are the solutions of the differential equations that define the rate laws in question.  If you don't understand what a differential equation is, any sensible answer is going to be problematic.

Assuming you do know what a differential equation is, we proceed!

For a first order rate law, the differential equation is something like:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$

You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:

$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

Similarly, the differential equation for the second order rate law is:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$

And again integrate to get:

$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

This all comes from the following Wikipedia page, which has a lot more details: Rate equation


A question about net ionic equations I have a question regarding net ionic equations. In a solution, sodium fluoride and hydrochloric acid are mixed together. The "correct" net ionic equation is shown below. However, how can this actually BE a net ionic equation, since none of the ions change state (all aqueous)?

To answer your question it is important to know which compound dissociate completely in water when in aqueous solution. For example, consider following reaction:

$$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}} \tag1$$

When ionic compound dissolves in water, each molecule separates to corresponding ions, which influenced by water. All of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ are ionic compounds, which are soluble in water. $\ce{AgCl}$ is, onthe other hand, an ionic compound, but it is insoluble in water (refer to solubility rules).  Hence the reaction $(1)$ can be rewrite as follows:

$$\ce{Ag+_{(aq)} + NO3-_{(aq)} + Na+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} + Na+_{(aq)} + NO3-_{(aq)}} \tag2$$

Like in algebra, you can canceled out common ions from each side of the equation $(2)$, which leaves:

$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} } \tag3$$

We called the equation $(3)$ a net ionic equation.

Now, let's look at your equation:

$$\ce{HCl_{(aq)} + NaF_{(aq)} -> HF_{(aq)} + NaCl_{(aq)}} \tag4$$

Now, $\ce{NaF}$ and $\ce{NaCl}$ are ionic compounds, which are soluble in water, hence they separated into ions in aqueous medium. How about $\ce{HCl}$, and $\ce{HF}$? They are soluble in water, but are they ionic? Since both are acids, to answer this question, we should look into their respective $\mathrm{p}K_\mathrm{a}$ values (or $K_\mathrm{a}$ values). Ref.1 is listed followings:

$$
\begin{array}{c|ccc}
\text{acid/base (chemical formula)} & \mathrm{p}K_\mathrm{a} & K_\mathrm{a} & \\
\hline
\text{hydrogen iodide }(\ce{HI}) & -10.0 & 1.0 \times 10^{10} \\
\text{hydrogen bromide }(\ce{HBr}) & -9.0 & 1.0 \times 10^{9} \\
\text{hydrogen chloride }(\ce{HCl}) & -8.0 & 1.0 \times 10^{8} \\
\text{hydronium ion }(\ce{H3O+}) & -1.7 & 50.1 \\
\text{hydrogen fluoride }(\ce{HF}) & 3.2 & 6.3 \times 10^{-4} \\
\text{acetic acid }(\ce{CH3CO2H}) & 4.7 & 2.0 \times 10^{-5} \\
\hline
\end{array}
$$

Accordingly, $\ce{HCl}$ should be ionic in aqueous solutions and should dissolve in water according to following equation:

$$\ce{HCl_{(aq)} + H2O_{(l)} -> H3O+_{(aq)} + Cl-_{(aq)}} \tag5$$

I put forward arrow because it'd dissociate completely, since $K_\mathrm{a} = \frac{\ce{[H3O+][Cl-]}}{\ce{[HCl]}} = 1.0 \times 10^{8}$.

However, $\ce{HF}$ is a different story. Since its  $K_\mathrm{a} = 6.3 \times 10^{-4} \lt \lt 1$, its dissociation is not complete, and most of $\ce{HF_{(aq)}}$ molecules stay intact (keep its covalent nature):

$$\ce{HF_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + F-_{(aq)}} \tag6$$

Note that $K_\mathrm{a}^\ce{HF} = \frac{\ce{[H3O+][F-]}}{\ce{[HF]}} = 6.3 \times 10^{-4}$, and hence $\ce{[HF]} \approx 1.6 \times 10^{3}\ce{[H3O+][F-]}$. Like sugar molecules, which are soluble in water but did not dissociate, $\ce{HF}$ stays as $\ce{HF_{(aq)}}$ in water. Thus, the equation $(4)$ can be rewritten as:

$$\ce{H+_{(aq)} + Cl-_{(aq)} + Na+_{(aq)} + F-_{(aq)} -> HF_{(aq)} + Na+_{(aq)} + Cl-_{(aq)}} \tag7$$

When you made common ions cancelled out, the net ionic equation renains as:

$$\ce{H+_{(aq)} + F-_{(aq)} -> HF_{(aq)} }$$

References:

K. Peter C. Vollhardt, Neil E. Schore, In Organic Chemistry: Structure and Function; Fifth Edition; W. H. Freeman & Co.: New York, NY, 2007 (ISBN: 0-7167-9949-9).


What happens to the gas inside cold drinks as the cold drink is frozen? According to Henry's law the amount of gas dissolved is directly proportional to the pressure exerted by a gas on the surface of the liquid.

And we know if we increase the temperature, the amount of dissolved gas decreases as it is an exothermic process and due to increase in entropy of system too , and if we decrease the temperature the amount of gas dissolved increase.

So what will happen to the gas inside the bottle if we freeze the cold drink bottle

  Does all the carbon-dioxide dissolve??

Assuming the bottle doesn't burst

Well, you are certainly right in that the solubility of the gas will grow if you decrease the temperature, and so it will continue all the way down to the freezing point. That's when things get hairy.

Ice is quite different from liquid water. It does not dissolve any $\ce{CO2}$ to speak of.(*) So pretty much all $\ce{CO2}$ will leave the solution and go to the gas phase, adding its pressure to that of ice in an effort to tear the bottle open. (Some plastic bottles manage to withstand it, some don't.) The dissolved gas and whatever little salts are there will lower the freezing point of water and in fact make it an interval rather than a point, but that's another story.

After thawing, much of $\ce{CO2}$ will happily go back to the solution (though not instantly), and the drink will be more or less restored to its original state.

(*) For any physical quantity, there are layers upon layers of scientists from different fields, each confined to a certain range of the said quantity and calling everything above it "infinity" and everything below it "zero". The levels of dissolved $\ce{CO2}$ in the ice core samples from lake Vostok are quite a big deal for paleoclimatology. Not my field, though.


Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution? If we make a resonance structure for the anilinium ion, with positive charge at either the ortho or the para position, we get a pentavalent nitrogen, which is not possible. So, how is the $\ce{-NH3+}$ group meta-directing in electrophilic aromatic substitution reactions, e.g. in nitration of aniline under acidic conditions?

how is anilinium ion meta directing for electrophiles?

Actually, anilinium is not meta directing (I know it is often taught that way), but rather it inductively deactivates the entire aromatic ring.

To explain the electrophilic substitution pattern observed with any aromatic system we must consider both resonance and inductive effects.

Resonance Effects: You are correct that you can't delocalize the positive charge on the anilinium nitrogen into the benzene ring through resonance.  As you point out, resonance structures with 5 bonds to nitrogen aren't allowed.

Therefore resonance effects are not involved in explaining the reactivity of the anilinium ion towards electrophilic aromatic substitution.

Inductive Effects: Inductively we expect the anilinium nitrogen to strongly withdraw electron density from the ortho position since it is closest to the positively charged nitrogen, this makes the ortho position the most deactivated.  Inductive effects die off quickly with distance, so we would expect the meta position to also be deactivated, but less so than the ortho position.  Finally, we would still expect some deactivation at the distant para position, but slightly less than we saw at the meta position. Overall, in terms of reactivity we would expect para > meta > ortho, with the difference between para and meta being much smaller than the difference between meta and ortho due to the effect distance has on inductive electron withdrawal.

If we examine the nitration of aniline in strong acid ($\ce{HNO3}$, $\ce{H2SO4}$), under these conditions the strong acid will completely protonate the aniline and reaction will proceed through the anilinium ion.  We see that, as expected, there is very little ortho product and slightly more para product than meta product.  This is pretty much as we expected and is consistent with the electron withdrawing inductive effect of the positive anilinium nitrogen.


Why does nitrous oxide have 300 times the global warming potential of CO2? Both nitrous oxide and carbon dioxide have roughly the same atmospheric lifetime (nitrous oxide slightly longer), so I thought that it would have to do with the infrared absorbance associated with each of them. So I looked at their spectrum.

Perhaps this is misleading because it is my understanding that not much infrared light at around the 2300 $\pu{cm^{-1}}$ region exists. But even so, that means that the two peaks that matter are $\ce{CO2}$'s 700 peak, and $\ce{N2O}$'s 1300 peak. It is my understanding that the blackbody radiation of earth peaks at around 700, and though sizable, is much less at 1300. So, I don't see why nitrous oxide would have a greater global warming potential than carbon dioxide.

There are two really important properties of N2O that make it such an important greenhouse gas, both of which you have raised in your question but need some clarification.

Infrared absorption - it's good to think in terms of gaps here. There's a lot of overlap between CO2, CH4 and (a frequently ignored GHG, but very important) water vapour in the wavelengths which they absorb. At the 4-5 micron and 7-8 micron range, N2O is extremely efficient at absorbing infrared radiation. These are important regions, precisely because there is no overlap with other GHGs in these regions. N2O is absorbing infrared very efficiently, and is doing so without "competition".
Lifetime - you noted that the lifetime of N2O and CO2 are comparable, which is sort of true. N2O, as a stable, inert molecule that is well mixed in the atmosphere, has a very long lifetime of 120 years compared to some other GHGs (CH4, for example is 8 years, based on the fact that its removed by the hydroxyl OH radical). It's possible to determine this fairly accurately as the only sinks for N2O are its stratospheric photo-dissociation and reaction with O(1D) radical. CO2 by comparison has a lifetime that varies anywhere between 5 and 200 years. The comparable part comes from the fact that for calculations for global warming potential, CO2 is given an "averaged" lifetime.

With N2O, therefore, we have a very efficient infrared absorber at two "windows" at which there is no overlap with other GHGs, and a long lifetime. The Global Warming Potential is the time integrated radiative forcing for a 1 kg pulse emission of the compound, where the upper limit of the integration is time horizon (usually set at 100 yr), relative to the same quantity of the reference compound CO2. It becomes clear from this definition why a stable, inert molecule with a long lifetime and which is very efficient at absorbing at particular infrared wavelengths would have a greater global warming potential than CO2 (shorter lifetime, absorbing in "windows" shared by other compounds) on a kg to kg comparison.


Best Cheminformatics API for drawing SMARTS reactions? I have a need to automatically draw SMARTS reactions. Rdkit's drawings are suboptimal. OpenBabel's drawings don't work properly on macOS. Indigo's drawings are cut off half the page.
I would love to use ChemDraw, but I can't seem to find an API for it to draw reactions with SMILES/SMARTS. I would absolutely be willing to write a wrapper to do this if there is a nice API to actually programmatically do drawings headlessly.

OpenEye is probably the most powerful and fast depiction SDK, but it costs money. Also OpenEye is a low-level, cumbersome, complicated library - so I'd consider it only if you have very sophisticated needs. It's written in C++, but it has Python/Java wrappers.
CDK (Java) has an OS-agnostic depiction API, works great for simple needs, haven't tried anything super complicated.
I tried Indigo several years ago and it worked fine, maybe you misconfigured it? Though it did require an additional OS-level dependency for fonts (Cargo?) on Linux. Don't remember if there were problems with other OS's.


How is the energy of reactants in a chemical reaction found ? A pretty rudimentary question, but I'm reviewing the concept of energy diagrams and activation energy and I'm curious to know how chemists find the energy of reactants and of a transition state in a laboratory.

What kinds of instruments are used ? As well, aren't transition states only something we can imagine ? So how is it that chemistry are able to find their energy in a lab ?

Energy itself is usually not measured, but rather change in energy ($\Delta E$) is the quantity of practical importance. One may be interested in change in free energy as well, $\Delta G$, changes in enthaly $\Delta H$, and even change in entropy $\Delta S$.

A common way to evaluate the increased energy of transition state ($\Delta E^{\ddagger}$) above the reactant is via the Arhennius Plot.  Here, one evaluates the rates of reaction ($k$) at a set of different temperatures ($T$ in Kelvin).  It is fairly easy to show that a plot of the $\ln(k)$ vs. $1/T$ gives a line whose slope is related to $\Delta H^{\ddagger}$, and where the intercept is related to $\Delta S^{\ddagger}$. (The ideal gas constant is in here too, but it's easy to accomodate.)

(There is an analogous plot for energy change for a reaction, which is called the van't Hoff plot, but it generally requires both reactant and product to be in equilibrium such that the concentrations of both are observable.)

Especially the enthalpy of reaction can be found by methods such as bomb calorimetry.  This measures the heat that is liberated upon reaction.


Electron density in benzene ring What would be the order of electron density in the benzene ring in the following compounds?

Methyl Benzene

Ethyl Benzene

t-butyl benzene

I would say that t-butyl benzene would have the highest electron density as the alkyl group attached to it is bulkier than methyl and ethyl thus it will exert a greater +I effect pushing the electrons into the ring making it electron rich. Using same analogy ethyl benzene would have a greater electron density in the ring compared to methyl benzene.

However that is not the case. Something counter intuitive seems to be happening here. The order is methyl benzene >ethyl benzene >t-butyl benzene.

I think this phenomenon can somehow be explained using hyper conjugation, but i don't get it how?

Well, tetrahydrocannabinol, you said it correct only the correct reason/explanation is hyperconjugation only.
More the number of (alpha) hydrogen atoms on a C=C [isolated as in alkenes or in resonance as in benzene], more the possibility of hyperconjugation.

In benzene, this will shift the electron density from C-H bond of CH3 group towards the ring, making it electron rich.

Contrary to what could have been thought from +I effect of corresponding alkyl groups on the ring, hyperconjugation is more in CH3.

Since hyperconjugation involves complete shifting of electron from valence shell of atoms to the other, it is more "effective" that INDUCTIVE which just pushes electron little more towards more EN (more correctly, towards group with more -I) atom.


Cleaning with vinegar and sodium bicarbonate Mixing sodium bicarbonate with vinegar (and sometimes adding dishwasher detergent) is quite a common tip on the internet. But I have never understood the reason.
I saw there was a question about using sodium bicarbonate and vinegar here to unclog drains, where the answer referred to the production of $\ce{CO2}$ being able to push whatever gunk blocking the drain.
However, for cleaning, the reaction is done when you use the mixture. Is there something I am missing? I am not a chemist, but my intuition is that using either bicarbonate or vinegar + dishwasher detergent alone would be more effective? Can somebody explain if the cleaning tip actually makes sense?

Your intuition is on point. Pretty much all the "lifehacks" praising cleaning abilities of vinegar and sodium bicarbonate are ineffective. The reaction appears vigorous due to the formation of carbon dioxide gas, but what is left is a diluted solution of sodium acetate:
$$\ce{CH3COOH(aq) + NaHCO3(aq) -> NaCH3COO(aq) + H2O(l) + CO2(g)}$$
Sodium acetate doesn't have any pronounced cleaning abilities. It doesn't act like a surfactant, doesn't drastically increase solubility of the impurities, or participate in complexation helping to dissolve metal-containing compounds.
As for unclogging abilities—yes, this is something that my grandmother used to use, but these days there are much better alternatives.
The idea was to pour some hot water into the system and then put a couple of spoons of baking soda into the drain creating warm concentrated solution in the siphon.
This solution would also be mildly basic due to hydrolysis and help to dissolve some fat and thin hair:
$$\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH-(aq)}$$
After some time (approx. 30 min) it would make sense to add some vinegar hoping the reaction would cause mixing up the constituents of the clog and excessive pressure would help to push the waste further down the drain.


Is carbon dioxide soluble in petroleum ether, and why? I googled it, and I'm finding some forum posts that say no, but no authoritative reference.
CO2 is somewhat soluble in water (1.7 g/L at 20°C) and I understand that it's mostly NOT because it forms carbonic acid: it's just dissolved CO2 molecules. Why wouldn't the same thing happen in petroleum ether?
Has anyone tried putting petroleum ether in a SodaStream device?

Carbon dioxide is, in fact, more soluble in nonpolar solvents than in water.

From the Wikipedia page on carbon dioxide: Solubility in water
1.45 g/L at 25 °C (77 °F), 100 kPa
(Alternative source: CRC handbook. Solubility is about twice as much at 0°C.)

From IUPAC, Solubility Data Series: Carbon Dioxide in Non-Aqueous Solvents at Pressures Less Than 200 kPa, chart on page 3: mole fraction of CO2 in hexane at 101.3 kPa, 300 K = 0.012, which corresponds to 4 g/L.
(Solubility is only slightly (10-20%) higher at 0°C.)

So the forum posts I read were wrong.

As to why, I suppose it's sufficient to argue that CO2 is a nonpolar molecule, and like dissolves like.
(It's true because it's Latin: Similia similibus solvuntur! )


Is ammonium hydroxide safe to be used for sanitizing food? At least one big fast food chain uses ammonium hydroxide for making "pink slime" edible by killing off harmful micro organisms: https://en.wikipedia.org/wiki/Ammonium_hydroxide

Is this product from this process safe to consume?

In pure form, aqueous solutions of ammonia ($\ce{NH3}$) cannot be consumed. Solutions containing 5 % ammonia or more are considered “corrosive” (causing severe skin burns and eye damage).

However, in reasonable quantities or in dilute form, aqueous solutions of ammonia can be generally used safely in food. They are sometimes named “ammonium hydroxide”, “ammonia aqueous”, or “aqua ammonia”. The food additive code (the E number, which is commonly found on food labels) of aqueous solutions of ammonia is E 527.

This food additive is approved in the European Union as well as in the United States.

For example, aqueous solutions of ammonia are used in food processing as pH control agent (acidity regulator), leavening agent, surface-finishing agent, or additive to boiling water.


Chemical Formulae and the Periodic Table of Elements Reading about heavy water, I have found that the chemical formula for it can be written 2H2O or D2O.

  Deuterium is an isotope of hydrogen that is found in large quantities in water, more than one atom per ten thousand hydrogen atoms has a deuterium nucleus. The isotope is denoted “2H” or “D”, and is normally known as “heavy hydrogen”. Deuterium is used in a number of conventional nuclear reactors in the form of heavy water (D2O), and it will probably also be used as fuel in fusion reactors in the future (Source: ScienceDaily).

Encyclopædia Britannica says the formula can be 2H2O.  Is this correct?

2H (or 2H) = D is interesting to me as D is not in the periodic table of elements.

I thought that formulae should be formed using the symbols within the periodic table.  What are the rules surrounding these chemical formulae?

According to this entry in the IUPAC Gold book

https://goldbook.iupac.org/html/D/D01648.html

the symbol for Deuterium is H together with the isotopes notation.

However deuterium and tritium are routinely indicated by D and T respectively. Their use is dictated by the framework. Besides nuclear reactions, when only straightforward chemical properties* are considered, it is convenient to use a single letter rather than H and super/sub-scripts.

In organic nomenclature, while writing names for molecules, the presence of deuterium is signalled by the symbol (d).

*note that isotopic effect might however affect chemical properties at fine levels. The toxicity of heavy water is a notable striking example.


What is "Steel Reinforced Epoxy" No, it's not epoxy for reinforcing steel. It's "steel reinforced epoxy."

http://www.jbweld.com/products/j-b-weld-twin-tube

How in the world does steel reinforce epoxy?

I can understand how carbon nanotubes dispersed in epoxy can help strength it, quote from paper below:

  The mechanical behavior of multiwalled carbon nanotube/epoxy
  composites was studied in both tension and compression. It was found
  that the compression modulus is higher than the tensile modulus,
  indicating that load transfer to the nanotubes in the composite is
  much higher in compression. In addition, it was found that the Raman
  peak position, indicating the strain in the carbon bonds under
  loading, shifts significantly under compression but not in tension. It
  is proposed that during load transfer to multiwalled nanotubes, only
  the outer layers are stressed in tension whereas all the layers
  respond in compression.

But back to steel - as far as I know, steel is a crystalline material and it doesn't possess any property akin to CNTs. Can someone explain how this particular epoxy works?

There's no "steel" in that stuff. It is simply filled with inorganic particles, probably (partly?) iron dust. I'm using jbweld too. Quick test with a bar magnet showed the tube with the "steel" component as well as the cured epoxy are indeed slightly ferromagnetic.
Most importantly, the filler reduces macroscopic shrinking during curing. That's a general principle: The density of the filler particles stays constant, while that if the glue component goes down, remarkably. The filler material type is not really important, only its surface must be compatible with the epoxy.
Plus the filler particles impede the propagation of cracks, and reduce the permeability to water, oxygen, etc. Again, that's a general property of fillers, not specific to this one.
It's hard to tell what the specific advantage of using iron/steel powder is, instead of some inorganic oxide (vulgo "sand"). A metal might catch environmental oxygen that diffuses in, increasing the temperature stability of the epoxy. If that's the case I don't know, but this stuff is remarkably temperature-resistant.


Why doesn't HCl form when you dissolve NaCl in water? People say it dissolves, but shouldn't the acid form?

$$
\ce{NaCl + H2O ->[?] HCl + ???}
$$

If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed.

The reaction that you propose -

$\ce{Cl- + H2O -> HCl + HO-}$

is highly thermodynamically unfavorable.

We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a much (as in almost a trillion trillion times) stronger acid than water.

Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water.

So even if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH.

As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a lot bigger).

$\ce{Cl- + H2O  <<=> HCl + HO-}$